751.24 Retaining Walls: Difference between revisions

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|[[media:751.24 LFD Retaining Walls Sept 2011.pdf|'''Printable Version of September 2011 LFD Retaining Walls Info''']]
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|align="left"|EPG 751.24 LFD Retaining Walls presents the very latest information, but this pdf file may be helpful for those wanting to easily print the LFD seismic information as it was in September 2011.
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==751.24.1 General==
==751.24.1 General==
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|'''Additional Information'''
| '''Additional Information'''
|-
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|AASHTO 5.1
| LRFD 11
|}
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Retaining wall shall be designed to withstand lateral earth and water pressures, including any live and dead load surcharge, the self weight of the wall, temperature and shrinkage effect, live load and collision forces, and earthquake loads in accordance with the general principles of AASHTO Section 5 and the general principles specified in this article.
For understanding the equivalency of seismic design category (SDC) and seismic zone for LRFD, see [[751.9_Bridge_Seismic_Design#751.9.1.1_Applicability_of_Guidelines_and_Seismic_Design_Philosophy|EPG 751.9.1.1]]  and [https://epg.modot.org/forms/general_files/BR/Bridge_Seismic_Design_Flowchart.pdf Bridge Seismic Design Flowchart].
 
Retaining wall shall be designed to withstand lateral earth and water pressures, including any live and dead load surcharge, the self weight of the wall, temperature and shrinkage effect, live load and collision forces, and earthquake loads in accordance with the general principles of LRFD Section 11 and the general principles specified in this article.
 
Seismic analysis provisions shall not be ignored for walls that support another structure (i.e. support abutment fill or building) in SDC B or C (seismic zone 2 or 3). No-seismic-analysis provisions may be considered for walls that do not support another structure (i.e. most of District walls) in SDC B or C (seismic zone 2 or 3) in accordance with LRFD 11.5.4.2 and Geotech report. Seismic analysis provisions shall not be ignored for walls in SDC D (seismic zone 4).


===751.24.1.1 Wall Type Selection===
===751.24.1.1 Wall Type Selection===
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|'''Additional Information'''
| '''Additional Information'''
|-
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|AASHTO 5.2.1
| LRFD 11
|}
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Selection of wall type shall be based on an assessment of the magnitude and direction of loading, depth to suitable foundation support, potential for earthquake loading, presence of deleterious environmental factors, wall site cross-sectional geometry, proximity of physical constraints, tolerable and differential settlement, facing appearance and ease and cost of construction.
Selection of wall type shall be based on an assessment of the magnitude and direction of loading, depth to suitable foundation support, potential for earthquake loading, presence of deleterious environmental factors, wall site cross-sectional geometry, proximity of physical constraints, tolerable and differential settlement, facing appearance and ease and cost of construction.
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|'''Additional Information'''
| '''Additional Information'''
|-
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|AASHTO 5.2.1.4 & 5.8
|LRFD 11.10,</br>FHWA-NHI-10-024 and 025
|}
|}
MSE retaining walls use precast block or panel like facing elements combined with either metallic or geosynthetic tensile reinforcements in the soil mass. MSE walls are preferred over cast-in-place walls because they are usually more economical. Other advantages include a wide variety of design styles, ease and speed of installation, and their ability to accommodate total and differential settlements. Wall design heights upwards of 80 ft. are technically feasible (FHFW-SA-96-071). MSE walls may be used to retain fill for end bents of bridge structures.
MSE retaining walls use precast block or panel like facing elements combined with either metallic or geosynthetic tensile reinforcements in the soil mass. MSE walls are preferred over cast-in-place walls because they are usually more economical. Other advantages include a wide variety of design styles, ease and speed of installation, and their ability to accommodate total and differential settlements. Wall design heights upwards of 80 ft. are technically feasible (FHFW-SA-96-071). MSE walls may be used to retain fill for end bents of bridge structures.


Situations exist where the use of MSE walls is either limited or not recommended. Some obstacles such as drop inlets, sign truss pedestals or footings, and fence posts may be placed within the soil reinforcement area, however, these obstacles increase the difficulty and expense of providing sufficient soil reinforcement for stability. Box culverts and highway drainage pipes may run through MSE walls, but it is preferable not to run the pipes close to or parallel to the walls. Utilities other than highway drainage should not be constructed within the soil reinforcement area. Be cautious when using MSE walls in a flood plain. A flood could cause scouring around the reinforcement and seepage of the backfill material. Soil reinforcements should not be used where exposure to ground water contaminated by acid mine drainage or other industrial pollutants as indicated by a low pH and high chlorides and sulfates exist. Galvanized metallic reinforcements shall not be used where stray electrical ground currents could occur as would be present near an electrical substation.  
Situations exist where the use of MSE walls is either limited or not recommended. Some obstacles such as drop inlets, sign truss pedestals or footings, and fence posts may be placed within the soil reinforcement area, however, these obstacles increase the difficulty and expense of providing sufficient soil reinforcement for stability. Box culverts and highway drainage pipes may run through MSE walls, but it is preferable not to run the pipes close to or parallel to the walls. Utilities other than highway drainage should not be constructed within the soil reinforcement area. Be cautious when using MSE walls in a floodplain. A flood could cause scouring around the reinforcement and seepage of the backfill material. Soil reinforcements should not be used where exposure to ground water contaminated by acid mine drainage or other industrial pollutants as indicated by a low pH and high chlorides and sulfates exist. Galvanized metallic reinforcements shall not be used where stray electrical ground currents could occur as would be present near an electrical substation.  


Sufficient right of way is required to install the soil reinforcement which extends into the backfill area at least 8 ft., 70 % of the wall height or as per design requirements, whichever is greater. For more information regarding soil reinforcement length and excavation limits see [[751.6 General Quantities#751.6.2.17 Excavation|EPG 751.6.2.17 Excavation]]. Finally, barrier curbs constructed over or in line with the front face of the wall shall have adequate room provided laterally between the back of the wall facing and the curb or slab so that load is not directly transmitted to the top wall facing units.  
Sufficient right of way is required to install the soil reinforcement which extends into the backfill area at least 8 feet, 70 percent of the wall height or as per design requirements set forth in [[751.6 General Quantities#751.6.2.17 Excavation|EPG 751.6.2.17 Excavation]], whichever is greater. For more information regarding soil reinforcement length, excavation limits and Minimum Embedment Depth of MSEW, see [[751.6 General Quantities#751.6.2.17 Excavation|EPG 751.6.2.17 Excavation]].
 
Finally, barrier curbs constructed over or in line with the front face of the wall shall have adequate room provided laterally between the back of the wall facing and the curb or slab so that load is not directly transmitted to the top of MSE wall or facing units.
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<div style="text-align: center; margin-left: auto; margin-right: auto;">
<gallery mode=packed widths=200px heights=200px>
File:751.24.1.1_barrier_top_MSE_wall-01.jpg| '''Barrier at Top of MSE Wall'''
File:blank.jpg|
File:751.24.1.1_barrier_front_MSE_wall-01.jpg| '''Barrier in Front of MSE Wall'''
</gallery>
</div>
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'''Concrete Cantilever Wall on Spread Footing'''
'''Concrete Cantilever Wall on Spread Footing'''
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Concrete cantilever walls derive their capacity through combinations of dead weight and structural resistance. These walls are constructed of reinforced concrete.
Concrete cantilever walls derive their capacity through combinations of dead weight and structural resistance. These walls are constructed of reinforced concrete.


Concrete cantilever walls are used when MSE walls are not a viable option. Cantilever walls can reduce the rock cut required and can also provide solutions when there are right of way restrictions. Concrete walls also provide better structural capacity when barrier curbs on top of the walls are required.
Concrete cantilever walls are used when MSE walls are not a viable option. Cantilever walls can reduce the rock cut required and can also provide solutions when there are right of way restrictions. Concrete walls also provide better structural capacity when barrier or railing on top of the walls are required.


Counterforts are used on rare occasions. Sign-board type retaining walls are a special case of counterfort retaining walls. They are used where the soil conditions are such that the footings must be placed well below the finished ground line. For these situations the wall is discontinued 12 in. below the ground line or below the frost line. Counterforts may also be a cost-savings option when the wall height approaches 20 ft. (''Foundation Analysis and Design'' by Joseph E. Bowles, 4th ed., 1988). However, other factors such as poor soil conditions, slope of the retained soil, wall length and uniformity in wall height should also be considered before using counterforts.
Counterforts are used on rare occasions. Sign-board type retaining walls are a special case of counterfort retaining walls. They are used where the soil conditions are such that the footings must be placed well below the finished ground line. For these situations the wall is discontinued 12 in. below the ground line or below the frost line. Counterforts may also be a cost-savings option when the wall height approaches 20 ft. (''Foundation Analysis and Design'' by Joseph E. Bowles, 4th ed., 1988). However, other factors such as poor soil conditions, slope of the retained soil, wall length and uniformity in wall height should also be considered before using counterforts.
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'''Concrete Cantilever Wall on Pile Footing'''
'''Concrete Cantilever Wall on Pile Footing'''


Concrete cantilever walls on pile footings are used when the soil conditions do not permit the use of spread footings. These walls are also used when an end bent requires wings longer than 22 feet. In these cases a stub wing is left attached to the end bent and the rest of the wing is detached to become a retaining wall.
Concrete cantilever walls on pile footings are used when the soil conditions do not permit the use of spread footings. These walls are also used when an end bent requires wings longer than 22 feet for seismic category A or 17 ft. for seismic category B, C or D. In these cases a stub wing is left attached to the end bent and the rest of the wing is detached to become a retaining wall as shown in [[751.35_Concrete_Pile_Cap_Integral_End_Bents#751.35.3.5_Wing_and_Detached_Wing_Walls|751.35.3.5 Wing and Detached Wing Walls]].


'''Concrete L-Shaped Retaining Wall on Spread Footings'''
'''Concrete L-Shaped Retaining Wall on Spread Footings'''
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===751.24.1.2 Loads===
===751.24.1.2 Loads===
Conventional retaining walls: Loads shall be determined in accordance with LRFD 3 and 11.6.
MSE retaining walls: Loads shall be determined in accordance with LRFD 3 and 11.10.
Note: For guidance, follow the [[751.40_LFD_Widening_and_Repair#751.40.8.15_Cast-In-Place_Concrete_Retaining_Walls|751.40.8.15 Cast -In-Place Concrete Retaining Walls]] and modify guidance of ASD as necessary to meet LRFD requirements until this section is modified for LRFD.


'''Dead Loads'''
'''Dead Loads'''
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Dead loads shall be determined from the unit weights in [[751.2 Loads#751.2.1.1 Dead Load |EPG 751.2.1.1 Dead Load]].
Dead loads shall be determined from the unit weights in [[751.2 Loads#751.2.1.1 Dead Load |EPG 751.2.1.1 Dead Load]].


'''Equivalent Fluid Pressure (Earth Pressures)'''
==751.24.2 Mechanically Stabilized Earth (MSE) Walls==
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===751.24.2.1 Design===
|'''Additional Information'''
 
|-
Designs of Mechanically Stabilized Earth (MSE) walls shall be completed by consultants or contractors in accordance with Section 11.10 of LRFD specifications, FHWA-NHI-10-024 and FHWA-NHI-10-025 for LRFD. [https://www.modot.org/bridge-pre-qualified-products-list Bridge Pre-qualified Products List (BPPL)] provided on MoDOT's web page and in Sharepoint contains a listing of facing unit manufacturers, soil reinforcement suppliers, and wall system suppliers which have been approved for use. See [http://www.modot.org/business/standards_and_specs/SpecbookEPG.pdf#page=11 Sec 720] and [http://www.modot.org/business/standards_and_specs/SpecbookEPG.pdf#page=14 Sec 1010] for additional information. The Geotechnical Section is responsible for checking global stability of permanent MSE wall systems, which should be reported in the Foundation Investigation Geotechnical Report. For MSE wall preliminary information, see [[751.1_Preliminary_Design#751.1.4.3_MSE_Walls|EPG 751.1.4.3 MSE Walls]]. For design requirements of MSE wall systems and temporary shoring (including temporary MSE walls), see [[:Category:720_Mechanically_Stabilized_Earth_Wall_Systems#720.2_Design_Requirements|EPG 720 Mechanically Stabilized Earth Wall Systems]]. For staged bridge construction, see [[751.1_Preliminary_Design#751.1.2.11_Staged_Construction|EPG 751.1.2.11 Staged Construction]].
|AASHTO 3.20.1
|}


For determining equivalent earth pressures for Group Loadings I through VI the Rankine Formula for Active Earth Pressure shall be used.
For seismic design requirements, see [https://epg.modot.org/forms/general_files/BR/Bridge_Seismic_Design_Flowchart.pdf Bridge Seismic Design Flowchart]. References for consultants and contractors include Section 11.10 of LRFD, FHWA-NHI-10-024 and FHWA-NHI-10-025.


Rankine Formula: <math>P_a = \frac{1}{2}C_a\gamma_sH^2</math> where:
'''Design Life'''  
:''C<sub>a</sub>'' = <math>\cos\delta\Bigg[\frac{\cos\delta - \sqrt{\cos^2\delta - \cos^2\phi}}{\cos\delta + \sqrt{\cos^2\delta - \cos^2\phi}}\Bigg]</math> = coefficient of active earth pressure


:''P<sub>a</sub>'' = equivalent active earth pressure
* 75 year minimum for permanent walls (if retained foundation require 100 year than consider 100 year minimum design life for wall).


:''H'' = height of the soil face at the vertical plane of interest
'''Global stability:'''


:<math>\boldsymbol{\gamma_s}</math> = unit weight of soil
Global stability will be performed by Geotechnical Section or their agent.


:<math>\boldsymbol{\delta}</math>= slope of fill in degrees
'''MSE wall contractor/designer responsibility:'''


:<math>\boldsymbol{\phi}</math> = angle of internal friction of soil in degrees
MSE wall contractor/designer shall perform following analysis in their design for all applicable limit states.
[[image:751.24.1.2.jpg|center|485px]]


'''Example'''
:* External Stability
::* Limiting Eccentricity
::* Sliding
::* Factored Bearing Pressure/Stress ≤ Factored Bearing Resistance
:* Internal Stability
::* Tensile Resistance of Reinforcement
::* Pullout Resistance of Reinforcement
::* Structural Resistance of Face Elements
::* Structural Resistance of Face Element Connections
:* Compound Stability
:: Capacity/Demand ratio (CDR) for bearing capacity shall be ≥ 1.0
:: <math>Bearing\ Capacity\ (CDR)  = \frac{Factored\ Bearing\ Resistance}{Maximum\ Factored\ Bearing\ Stress} \ge 1.0</math>
:: Strength Limit States:
:: Factored bearing resistance = Nominal bearing resistance from Geotech report X Minimum Resistance factor (0.65, Geotech report)  LRFD Table 11.5.7-1 


Given:
:: Extreme Event I Limit State:
:: Factored bearing resistance = Nominal bearing resistance from Geotech report X Resistance factor
:: Resistance factor = 0.9  LRFD 11.8.6.1


:''δ'' = 3:1 (H:V) slope
:: Factored bearing stress shall be computed using a uniform base pressure distribution over an effective width of footing determined in accordance with the provisions of LRFD 10.6.3.1 and 10.6.3.2, 11.10.5.4  and Figure 11.6.3.2-1 for foundation supported on soil or rock.


:''ϕ'' = 25°
:: B’ = L – 2e


:''γ<sub>s</sub>'' = 0.120 kcf
:: Where,
::: L = Soil reinforcement length (For modular block use B in lieu of L as per LRFD 11.10.2-1)
::: B’ = effective width of footing
::: e = eccentricity
::: Note: When the value of eccentricity e is negative then B´ = L.  


:''H'' = 10 ft
::Capacity/Demand ratio (CDR) for overturning shall be ≥ 1.0
::<math>Overtuning\ (CDR)  = \frac{Total\ Factored\ Resisting\ Moment}{Total\ Factored\ Driving\ Moment} \ge 1.0</math>


''δ'' = arctan<math>\Big[\frac{1}{3}\Big]</math> = 18.4°
::Capacity/Demand ratio (CDR) for eccentricity shall be ≥ 1.0
::<math>Eccentricity\ (CDR)  = \frac{e_{Limit}}{e_{design}} \ge 1.0</math>


''C<sub>a</sub>'' = <math>\cos (18.4^\circ)\Bigg[\frac{\cos(18.4^\circ) - \sqrt{\cos^2(18.4^\circ) - \cos^2(25^\circ)}}{\cos(18.4^\circ) + \sqrt{\cos^2(18.4^\circ) - \cos^2(25^\circ)}}\Bigg]</math> = 0.515
::Capacity/Demand ratio (CDR) for sliding shall be ≥ 1.0 &nbsp;&nbsp;&nbsp;&nbsp; LRFD 11.10.5.3 & 10.6.3.4
::<math>Sliding\ (CDR) = \frac{Total\ Factored\ Sliding\ Resistance}{Total\ Factored\ Active\ Force} \ge 1.0</math>


''P<sub>a</sub>'' = (1/2)(0.515)(0.120 kips/ft<sup>3</sup>)(10 ft)<sup>2</sup> = 3.090 kips per foot of wall length
::Capacity/Demand ratio (CDR) for internal stability shall be ≥ 1.0


The ''ϕ'' angle shall be determined by the Materials Division from soil tests. If the ''ϕ'' angle cannot be provided by the [http://wwwi/intranet/cm/default.htm Construction and Materials Division] a ''ϕ'' angle of 27 degrees shall be used.
::Eccentricity, (e) Limit for Strength Limit State: &nbsp;&nbsp;&nbsp;&nbsp; LRFD 11.6.3.3 & C11.10.5.4
::: For foundations supported on soil or rock, the location of the resultant of the reaction forces shall be within the middle two-thirds of the base width, L or (e ≤ 0.33L).


Drainage shall be provided to relieve water pressure from behind all cast-in-place concrete retaining walls. If adequate drainage can not be provided then walls shall be designed to resist the maximum anticipated water pressure.
::Eccentricity, (e) Limit for Extreme Event I (Seismic): &nbsp;&nbsp;&nbsp;&nbsp; LRFD 11.6.5.1
:::For foundations supported on soil or rock, the location of the resultant of the reaction forces shall be within the middle two-thirds of the base width, L or (e ≤ 0.33L) for  γ<sub>EQ</sub> = 0.0 and middle eight-tenths of the base width, L or (e ≤ 0.40L) for  γ<sub>EQ</sub> = 1.0.  For γ<sub>EQ</sub>  between 0.0 and 1.0, interpolate e value linearly between 0.33L and 0.40L. For γ<sub>EQ</sub>  refer to LRFD 3.4.


'''Surcharge Due to Point, Line and Strip Loads'''
:::Note: Seismic design shall be performed for γ<sub>EQ</sub> = 0.5


Surcharge due to point and line loads on the soil being retained shall be included as dead load surcharge. The effect of these loads on the wall may be calculated using Figure 5.5.2B from AASHTO.
::Eccentricity, (e) Limit for Extreme Event II:
:::For foundations supported on soil or rock, the location of the resultant of the reaction forces shall be within the middle eight-tenths of the base width, L or (e ≤ 0.40L).


Surcharge due to strip loads on the soil being retained shall be included as a dead load surcharge load. The following procedure as described in ''Principles of Foundation Engineering'' by Braja M. Das (1995) shall be applied to calculate these loads when strip loads are applicable. An example of this application is when a retaining wall is used in front of an abutment so that the wall is retaining the soil from behind the abutment as a strip load on the soil being retained by the wall.
'''General Guidelines'''


[[image:751.24.1.2 retaining.jpg|center|255px|thumb|<center>'''Retaining Wall in front of an Abutment'''</center>]]
* Drycast modular block wall (DMBW-MSE) systems are limited to a 10 ft. height in one lift.


The portion of soil that is in the active wedge must be determined because the surcharge pressure only affects the wall if it acts on the active wedge. The actual failure surface in the backfill for the active state can be represented by ABC shown in the figure below. An approximation to the failure surface based on Rankine's active state is shown by dashed line AD. This approximation is slightly unconservative because it neglects friction at the pseudo-wall to soil interface.
* Wetcast modular block wall (WMBW-MSE) systems are limited to a 15 ft. height in one lift.


The following variables are shown in the figure below:
* For Drycast modular block wall (DMBW-MSE) systems and Wetcast modular block wall (WMBW-MSE) systems, top cap units shall be used and shall be permanently attached by means of a resin anchor system.


:''β'' = slope of the active failure plane in degrees
* For precast modular panel wall (PMPW-MSE) systems, capstone may be substituted for coping and either shall be permanently attached to wall by panel dowels.
:''δ'' = slope of fill in degrees
:''H'' = height of the pseudo-wall (fom the bottom of the footing).
:''L<sub>1</sub>'' = distance from back of stem to back of footing heel
:''L<sub>2</sub>'' = distance from footing heel to intersection of failure plane with ground surface


[[image:751.24.1.2 wedges.jpg|center|575px|thumb|<center>'''Determination of Active Wedges'''</center>]]
* For precast modular panel wall (PMPW-MSE) systems, form liners are required to produce all panels. Using form liner to produce panel facing is more cost effective than producing flat panels. Standard form liners are specified on the [https://www.modot.org/mse-wall-msew MSE Wall Standard Drawings]. Be specific regarding names, types and colors of staining, and names and types of form liner.


In order to determine ''β'', the following equation which has been derived from Rankine's active earth pressure theory must be solved by iteration:
* MSE walls shall not be used where exposure to acid water may occur such as in areas of coal mining.


:<math>\tan (-\beta) + \frac{1}{\tan (\beta - \phi)} - \frac{1}{\tan (\beta - \delta)} + \frac{1}{\tan (90^\circ + \phi + \delta - \beta)} = 0</math>
* MSE walls shall not be used where scour is a problem.


:''ϕ'' = angle of internal friction of soil in degrees
* MSE walls with metallic soil reinforcement shall not be used where stray electrical ground currents may occur as would be present near electrical substations.


A good estimate for the first iteration is to let ''β'' = 45° + (ϕ/2). In lieu of iterating the above equation a conservative estimate for ''β'' is 45°. Once β has been established, an estimate of L<sub>1</sub> is needed to determine L<sub>2</sub>. From the geometry of the variables shown in the above figure:
* No utilities shall be allowed in the reinforced earth if future access to the utilities would require that the reinforcement layers be cut, or if there is a potential for material, which can cause degradation of the soil reinforcement, to leak out of the utilities into the wall backfill, with the exception of storm water drainage.


:<math>L_2 = H\frac{\cos\delta\cos\beta}{\sin(\beta - \delta)}</math>
* All vertical objects shall have at least 4’-6” clear space between back of the wall facing and object for select granular backfill compaction and soil reinforcement skew limit requirements. For piles, see pipe pile spacers guidance.


The resultant pressure due to the strip load surcharge and its location are then determined. The following variables are shown in the figure below:
* The interior angle between two MSE walls should be greater than 70°. However, if unavoidable, then place [[751.50_Standard_Detailing_Notes#J._MSE_Wall_Notes_.28Notes_for_Bridge_Standard_Drawings.29|EPG 751.50 J1.41 note]] on the design plans.


:''q'' = load per unit area
* Drycast modular block wall (DMBW-MSE) systems and Wetcast modular block wall (WMBW-MSE) systems may be battered up to 1.5 in. per foot. Modular blocks are also known as “segmental blocks”.
:''P<sub>s</sub>'' = resultant pressure on wall due only to surcharge earth pressure
:<math>\bar{z}</math> = location of P<sub>s</sub> measured from the bottom of the footing
:''L<sub>3</sub>'' = distance from back of stem to where surcharge pressure begins


[[image:751.24.1.2 surcharge.jpg|center|625px|thumb|<center>'''Surcharge Pressure on Retaining Wall'''</center>]]
* The friction angle used for the computation of horizontal forces within the reinforced soil shall be greater than or equal to 34°.


From the figure:
* For epoxy coated reinforcement requirements, see [[751.5 Structural Detailing Guidelines#751.5.9.2.2 Epoxy Coated Reinforcement Requirements|EPG 751.5.9.2.2 Epoxy Coated Reinforcement Requirements]].


:P<sub>s</sub> = <math>\frac{q}{90}\big[H(\theta_2 - \theta_1)\big]</math> where
* All concrete except facing panels or units shall be CLASS B or B-1.


:<math>\theta_1 = \arctan\Big[\frac{L_3}{H}\Big] \ and \ \theta_2 = \arctan\Big[\frac{L_2}{H}\Big]</math>
* The friction angle of the soil to be retained by the reinforced earth shall be listed on the plans as well as the friction angle for the foundation material the wall is to rest on.


:<math>\bar{z} = \frac{H^2(\theta_2 - \theta_1) - (R - Q) + 57.03L_4H}{2H(\theta_2 - \theta_1)}</math> where
* The following requirement shall be considered (from 2009_FHWA-NHI-10-024 MSE wall 132042.pdf, page 200-201) when seismic design is required:  
:* For seismic design category, SDC C or D (Zones 3 or 4), facing connections in modular block faced walls (MBW) shall use shear resisting devices (shear keys, pin, etc.) between the MBW units and soil reinforcement, and shall not be fully dependent on frictional resistance between the soil reinforcement and facing blocks. For connections partially dependent on friction between the facing blocks and the soil reinforcement, the nominal long-term connection strength T<sub>ac</sub>, should be reduced to 80 percent of its static value.


:<math>R = (L_2)^2(90^\circ - \theta_2) \ and \ Q = (L_3)^2(90^\circ - \theta_1)</math>
* Seismic design category and acceleration coefficients shall be listed on the plans for categories B, C and D. If a seismic analysis is required that shall also be noted on the plans. See [[751.50_Standard_Detailing_Notes#A._General_Notes|EPG 751.50 A1.1 note]].


When applicable, P<sub>s</sub> is applied to the wall in addition to other earth pressures. The wall is then designed as usual.
* Plans note ([[751.50_Standard_Detailing_Notes#J._MSE_Wall_Notes_.28Notes_for_Bridge_Standard_Drawings.29|EPG 751.50 J1.1]]) is required to clearly identify the responsibilities of the wall designer.


'''Live Load Surcharge'''
* Do not use Drycast modular block wall (DMBW-MSE) systems in the following locations:
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|'''Additional Information'''
|-
|AASHTO 3.20.3 & 5.5.2
|}
Live load surcharge pressure of not less than two feet of earth shall be applied to the structure when highway traffic can come within a horizontal distance equal to one-half of the wall height, measured from the plane where earth pressure is applied.


[[image:751.24.1.2 live load1.jpg|center|475px]]
::* Within the splash zone from snow removal operations (assumed to be 15 feet from the edge of the shoulder).


[[image:751.24.1.2 live load surcharge.jpg|center|575px|thumb|<Center>'''Live Load Surcharge'''</center>]]
::* Where the blocks will be continuously wetted, such as around sources of water.


:''P<sub>LLS</sub>'' = (2 ft.) ''γ<sub>s</sub> C<sub>a</sub> H'' = pressure due to live load surcharge only
::* Where blocks will be located behind barrier or other obstacles that will trap salt-laden snow from removal operations.


:''γ<sub>s</sub>'' = unit weight of soil (Note: AASHTO 5.5.2 specifies a minimum of 125 pcf for live load surcharge, MoDOT policy allows 120 pcf as given from the unit weights in [[751.2 Loads#751.2.1.1 Dead Load |EPG 751.2.1.1 Dead Load]].)
* Do not use Drycast modular block wall (DMBW-MSE) systems or Wetcast modular block wall (WMBW-MSE) systems in the following locations:
:''C<sub>a</sub>'' = coefficient of active earth pressure


:''H'' = height of the soil face at the vertical plane of interest
::* For structurally critical applications, such as containing necessary fill around structures.


The vertical live load surcharge pressure should only be considered when checking footing bearing pressures, when designing footing reinforcement, and when collision loads are present.
::* In tiered wall systems.
 
* For locations where Drycast modular block wall (DMBW-MSE) systems and Wetcast modular block wall (WMBW-MSE) systems are not desirable, consider coloring agents and/or architectural forms using precast modular panel wall (PMPW-MSE) systems for aesthetic installations.


'''Live Load Wheel Lines'''
* For slab drain location near MSE Wall, see [[751.10 General Superstructure#General Requirements for Location and Spacing of Slab Drains|EPG 751.10.3.1 Drain Type, Alignment and Spacing]] and [[751.10 General Superstructure#751.10.3.3 General Requirements for Location of Slab Drains|EPG 751.10.3.3 General Requirements for Location of Slab Drains]].


Live load wheel lines shall be applied to the footing when the footing is used as a riding or parking surface.
* Roadway runoff should be directed away from running along face of MSE walls used as wing walls on bridge structures.
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
|-
|'''Additional Information'''
|-
|AASHTO 3.24.5.1.1 & 5.5.6.1
|}


Distribute a LL<sub>WL</sub> equal to 16 kips as a strip load on the footing in the following manner.
* Drainage:


:*Gutter type should be selected at the core team meeting.


:P = LL<sub>WL</sub>/E
:* When gutter is required without fencing, use Type A or Type B gutter (for detail, see [https://www.modot.org/media/16880 Std. Plan 609.00]).


::where E = 0.8X + 3.75
:* When gutter is required with fencing, use Modified Type A or Modified Type B gutter (for detail, see [https://www.modot.org/media/16871 Std. Plan 607.11]).


::::X = distance in ft. from the load to the front face of the wall
:* When fencing is required without gutter, place in tube and grout behind the MSE wall (for detail, see [https://www.modot.org/bridge-standard-drawings MSE Wall Standard Drawings - MSEW], Fence Post Connection Behind MSE Wall (without gutter).
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
|-
|'''Additional Information'''
|-
|AASHTO 3.24.2 & 3.30
|}


The wheel lines shall move 1 ft. from the barrier curb or wall to 1 ft. from the toe of the footing.
:* Lower backfill longitudinal drainage pipes behind all MSE walls shall be two-6” (Min.) diameter perforated PVC or PE pipe (See Sec 1013) unless larger sizes are required by design which shall be the responsibility of the District Design Division. Show drainage pipe size on plans. Outlet screens and cleanouts should be detailed for any drain pipe (shown on MoDOT MSE wall plans or roadway plans). Lateral non-perforated drain pipes (below leveling pad) are permitted by Standard Specifications and shall be sized by the District Design Division if necessary. Lateral outlet drain pipe sloped at 2% minimum.


[[image:751.24.1.2 wheel.jpg|center|350px]]
::* Identify on MSE wall plans or roadway plans drainage pipe point of entry, point of outlet (daylighting), 2% min. drainage slopes in between points to ensure positive flow and additional longitudinal drainage pipes if required to accommodate ground slope changes and lateral drainage pipes if required by design.


'''Collision Forces'''
::* Adjustment in the vertical alignment of the longitudinal drainage pipes from that depicted on the MSE wall standard drawings may be necessary to ensure positive flow out of the drainage system.
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
|-
::* Identify on MSE wall plans or roadway plans the outlet ends of pipes which shall be located to prevent clogging or backflow into the drainage system. Outlet screens and cleanouts should be detailed for any drain pipe.
|'''Additional Information'''
|-
|AASHTO Figure 2.7.4B
|}


Collision forces shall be applied to a wall that can be hit by traffic. Apply a point load of 10 kips to the wall at a point 3 ft. above the finished ground line.
:* For more information on drainage, see [[#Drainage at MSE Walls|Drainage at MSE Walls]].


[[image:751.24.1.2 collision section.jpg|center|450px|thumb|<center>'''Section'''</center>]]
'''MSE Wall Construction: Pipe Pile Spacers Guidance'''


Distribute the force to the wall in the following manner:
For bridges not longer than 200 feet, pipe pile spacers or pile jackets shall be used at pile locations behind mechanically stabilized earth walls at end bents. Corrugated pipe pile spacers are required when the wall is built prior to driving the piles to protect the wall reinforcement when driving pile for the bridge substructure at end bents(s). Pile spacers or pile jackets may be used when the piles are driven before the wall is built. Pipe pile spacers shall have an inside diameter greater than that of the pile and large enough to avoid damage to the pipe when driving the pile. Use [[751.50 Standard Detailing Notes#E1. Excavation and Fill|EPG 751.50 Standard Detailing Note E1.2a]] on bridge plans.


:Force per ft of wall = (10 kips)/2L
For bridges longer than 200 feet, pipe pile spacers are required and the pile spacer shall be oversized to mitigate the effects of bridge thermal movements on the MSE wall. For HP12, HP14, CIP 14” and CIP 16” piles provide 24-inch inside diameter of pile spacer for bridge movement. Minimum pile spacing shall be 5 feet to allow room for compaction of the soil layers. Use [[751.50 Standard Detailing Notes#E1. Excavation and Fill|EPG 751.50 Standard Detailing Note E1.2b]] on bridge plans.


[[image:751.24.1.2 collision profile.jpg|center|350px|thumb|<center>'''Profile'''</center>]]
The bottom of the pipe pile spacers shall be placed 5 ft. min. below the bottom of the MSE wall leveling pad. The pipe shall be filled with sand or other approved material after the pile is placed and before driving. Pipe pile spacers shall be accurately located and capped for future pile construction.  


When considering collision loads, a 25% overstress is allowed for bearing pressures and a factor of safety of 1.2 shall be used for sliding and overturning.
Alternatively, for bridges shorter than or equal to 200 feet, the contractor shall be given the option of driving the piles before construction of the mechanically stabilized earth wall and placing the soil reinforcement and backfill material around the piling. In lieu of pipe pile spacers contractor may place pile jackets on the portion of the piles that will be in the MSE soil reinforced zone prior to placing the select granular backfill material and soil reinforcement. The contractor shall adequately support the piling to ensure that proper pile alignment is maintained during the wall construction. The contractor’s plan for bracing the pile shall be submitted to the engineer for review.  


'''Wind and Temperature Forces'''
Piling shall be designed for downdrag (DD) loads due to either method. Oversized pipe pile spacers with sand placed after driving or pile jacket may be considered to mitigate some of the effects of downdrag (DD) loads. Sizing of pipe pile spacers shall account for pile size, thermal movements of the bridge, pile placement plan, and vertical and horizontal placement tolerances.


These forces shall be disregarded except for special cases, consult the Structural Project Manager.
When rock is anticipated within the 5 feet zone below the MSE wall leveling pad, prebore into rock and prebore holes shall be sufficiently wide to allow for a minimum 10 feet embedment of pile and pipe pile spacer. When top of rock is anticipated within the 5 to 10 feet zone below the MSE wall leveling pad, prebore into rock to achieve a minimum embedment (pile only) of 10 feet below the bottom of leveling pad. Otherwise, the pipe pile spacer requires a minimum 5 feet embedment below the levelling pad. Consideration shall also be given to oversizing the prebore holes in rock to allow for temperature movements at integral end bents.  


When walls are longer than 84 ft., an expansion joint shall be provided.  
For bridges not longer than 200 feet, the minimum clearance from the back face of MSE walls to the front face of the end bent beam, also referred to as setback, shall be 4 ft. 6 in. (typ.) unless larger than 18-inch pipe pile spacer required. The 4 ft. 6 in. dimension is based on the use of 18-inch inside diameter pipe pile spacers & FHWA-NHI-10-24, Figure 5-17C, which will help ensure that soil reinforcement is not skewed more than 15° for nut and bolt reinforcement connections. Similarly, the minimum setback shall be determined when larger diameter pile spacers are required. For bridges longer than 200 feet, the minimum setback shall  be 5 ft. 6 in. based on the use of 24-inch inside diameter of pipe pile spacers. Other types of connections may require different methods for splaying. In the event that the minimum setback cannot be used, the following guidance for pipe pile spacers clearance shall be used: pipe pile spacers shall be placed 18 in. clear min. from the back face of MSE wall panels; 12 in. minimum clearance is required between pipe pile spacers and leveling pad and 18 in. minimum clearance is required between leveling pad and pile.


Contraction joint spacing shall not exceed 28 feet.
'''MSE Wall Plan and Geometrics'''


'''Seismic Loads'''
* A plan view shall be drawn showing a baseline or centerline, roadway stations and wall offsets. The plan shall contain enough information to properly locate the wall. The ultimate right of way shall also be shown, unless it is of a significant distance from the wall and will have no effect on the wall design or construction.


Retaining walls in Seismic Performance Category A (SPC A) and SPC B that are located adjacent to roadways may be designed in accordance with AASHTO specifications for SPC A. Retaining walls in SPC B which are located under a bridge abutment or in a location where failure of the wall may affect the structural integrity of a bridge shall be designed to AASHTO specifications for SPC B. All retaining walls located in SPC C and SPC D shall be designed in accordance to
* Stations and offsets shall be established between one construction baseline or roadway centerline and a wall control line (baseline). Some wall designs may contain a slight batter, while others are vertical. A wall control line shall be set at the front face of the wall, either along the top or at the base of the wall, whichever is critical to the proposed improvements. For battered walls, in order to allow for batter adjustments of the stepped level pad or variation of the top of the wall, the wall control line (baseline) is to be shown at a fixed elevation. For battered walls, the offset location and elevation of control line shall be indicated. All horizontal breaks in the wall shall be given station-offset points, and walls with curvature shall indicate the station-offsets to the PC and PT of the wall, and the radius, on the plans.  
AASHTO specifications for the corresponding SPC.


In seismic category B, C and D determine equivalent fluid pressure from Mononobe-Okabe static method.
* Any obstacles which could possibly interfere with the soil reinforcement shall be shown. Drainage structures, lighting, or truss pedestals and footings, etc. are to be shown, with station offset to centerline of the obstacle, with obstacle size. Skew angles are shown to indicate the angle between a wall and a pipe or box which runs through the wall.  
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
|-
|'''Additional Information'''
|-
|1992 AASHTO Div. IA Eqns. C6-3 and C6-4
|}
''P<sub>AE</sub>'' = equivalent active earth pressure during an earthquake


''P<sub>AE</sub>'' = 0.5 γ<sub>s</sub>H<sup>2</sup>(1 - k<sub>v</sub>)K<sub>AE</sub> where
* Elevations at the top and bottom of the wall shall be shown at 25 ft. intervals and at any break points in the wall.


''K<sub>AE</sub>'' = seismic active pressure coefficient
* Curve data and/or offsets shall be shown at all changes in horizontal alignment. If battered wall systems are used on curved structures, show offsets at 10 ft. (max.) intervals from the baseline.


:<math>K_{AE} = \frac{\cos^2(\phi - \theta - \beta)}{\cos\theta\cos^2\beta\cos(\delta + \beta + \theta)\Big\{1 + \sqrt{\frac{\sin(\phi + \delta)\sin(\phi - \theta - i)}{\cos(\delta + \beta + \theta)\cos(i - \beta)}}\Big\}^2}</math>
* Details of any architectural finishes (formliners, concrete coloring, etc.).


* Details of threaded rod connecting the top cap block.


''γ<sub>s</sub>'' = unit weight of soil
* Estimated quantities, total sq. ft. of mechanically stabilized earth systems.


{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
* Proposed grade and theoretical top of leveling pad elevation shall be shown in constant slope. Slope line shall be adjusted per project. Top of wall or coping elevation and stationing shall be shown in the developed elevation per project. If leveling pad is anticipated to encounter rock, then contact the Geotechnical Section for leveling pad minimum embedment requirements.
|-
|'''Additional Information'''
|-
|AASHTO 5.2.2.3 & Div. IA 6.4.3
|}
''k<sub>v</sub>'' = vertical acceleration coefficient


''k<sub>h</sub>'' = horizontal acceleration coefficient which is equal to 0.5A for all walls,
'''MSE Wall Cross Sections'''
:::but 1.5A for walls with battered piles where
:::A = seismic acceleration coefficient


The following variables are shown in the figure below:
* A typical wall section for general information is shown.


''ϕ'' = angle of internal friction of soil
* Additional sections are drawn for any special criteria. The front face of the wall is drawn vertical, regardless of the wall type.


''θ'' = <math>\arctan\ \Big(\frac{k_h}{1 - k_v}\Big)</math>
* Any fencing and barrier or railing are shown.


''β'' = slope of soil face
* Barrier if needed are shown on the cross section. Barriers are attached to the roadway or shoulder pavement, not to the MSE wall. Standard barriers are placed along wall faces when traffic has access to the front face of the wall over shoulders of paved areas.


''δ'' = angle of friction between soil and wall in degrees
<div id="Drainage at MSE Walls"></div>
'''Drainage at MSE Walls'''


''i'' = backfill slope angle in degrees
*'''Drainage Before MSE Wall'''


''H'' = distance from the bottom of the part of the wall to which the pressure is applied to the top of the fill at the location where the earth pressure is to be found.
:Drainage is not allowed to be discharged within 10 ft. from front of MSE wall in order to protect wall embedment, prevent erosion and foundation undermining, and maintain soil strength and stability.


[[image:751.24.1.2 active soil.jpg|center|450px|thumb|<center>'''Active Soil Wedge'''</center>]]
*'''Drainage Behind MSE Wall'''


<div id="Group Loads"></div>
::'''Internal (Subsurface) Drainage'''
'''Group Loads'''


For SPC A and B (if wall does not support an abutment), apply AASHTO Group I Loads only. Bearing capacity, stability and sliding shall be calculated using working stress loads. Reinforced concrete design shall be calculated using load factor design loads.
::Groundwater and infiltrating surface waters are drained from behind the MSE wall through joints between the face panels or blocks (i.e. wall joints) and two-6 in. (min.) diameter pipes located at the base of the wall and at the basal interface between the reinforced backfill and the retained backfill.
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
|-
|'''Additional Information'''
|-
|AASHTO Table 3.22.1A
|}


AASHTO Group I Load Factors for Load Factor Design of concrete:
::Excessive subsurface draining can lead to increased risk of backfill erosion/washout through the wall joints and erosion at the bottom of walls and at wall terminal ends. Excessive water build-up caused by inadequate drainage at the bottom of the wall can lead to decreased soil strength and wall instability. Bridge underdrainage (vertical drains at end bents and at approach slabs) can exacerbate the problem.
''γ'' = 1.3


''β<sub>D</sub>'' = 1.0 for concrete weight
::Subsurface drainage pipes should be designed and sized appropriately to carry anticipated groundwater, incidental surface run-off that is not collected otherwise including possible effects of drainage created by an unexpected rupture of any roadway drainage conveyance or storage as an example.


''β<sub>D</sub>'' = 1.0 for flexural member
::'''External (Surface) Drainage'''


''β<sub>E</sub>'' = 1.3 for lateral earth pressure for retaining walls
::External drainage considerations deal with collecting water that could flow externally over and/or around the wall surface taxing the internal drainage and/or creating external erosion issues. It can also infiltrate the reinforced and retained backfill areas behind the MSE wall.  


''β<sub>E</sub>'' = 1.0 for vertical earth pressure
::Diverting water flow away from the reinforced soil structure is important. Roadway drainage should be collected in accordance with roadway drainage guidelines and bridge deck drainage should be collected similarly.


''β<sub>LL</sub>'' = 1.67 for live load wheel lines
*'''Guidance'''


''β<sub>LL</sub>'' = 1.67 for collision forces
:ALL MSE WALLS


{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
:1. Appropriate measures to prevent surface water infiltration into MSE wall backfill should be included in the design and detail layout for all MSE walls and shown on the roadway plans.  
|-
|'''Additional Information'''
|-
|AASHTO 5.14.2
|}


''β<sub>E</sub>'' = 1.67 for vertical earth pressure resulting from live load surcharge
:2. Gutters behind MSE walls are required for flat or positive sloping backfills to prevent concentrated infiltration behind the wall facing regardless of when top of backfill is paved or unpaved. This avoids pocket erosion behind facing and protection of nearest-surface wall connections which are vulnerable to corrosion and deterioration. Drainage swales lined with concrete, paved or precast gutter can be used to collect and discharge surface water to an eventual point away from the wall. If rock is used, use impermeable geotextile under rock and align top of gutter to bottom of rock to drain. (For negative sloping backfills away from top of wall, use of gutters is not required.)


''β<sub>E</sub>'' = 1.3 for horizontal earth pressure resulting from live load surcharge
:District Design Division shall verify the size of the two-6 in. (min.) diameter lower perforated MSE wall drain pipes and where piping will daylight at ends of MSE wall or increase the diameters accordingly.  This should be part of the preliminary design of the MSE wall. (This shall include when lateral pipes are required and where lateral drain pipes will daylight/discharge).
:BRIDGE ABUTMENTS WITH MSE WALLS


For SPC B (if wall supports an abutment), C, and D apply AASHTO Group I Loads and seismic loads in accordance with AASHTO Division IA - Seismic Design Specifications.
:Areas of concern: bridge deck drainage, approach slab drainage, approach roadway drainage, bridge underdrainage:  vertical drains at end bents and approach slab underdrainage, showing drainage details on the roadway and MSE wall plans


{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
:3. Bridge slab drain design shall be in accordance with [[751.10 General Superstructure#751.10.3 Bridge Deck Drainage - Slab Drains |EPG 751.10.3 Bridge Deck Drainage – Slab Drains]] unless as modified below.
|-
|'''Additional Information'''
|-
|AASHTO Div. IA 4.7.3
|}


When seismic loads are considered, load factor for all loads = 1.0.
:4. Coordination is required between the Bridge Division and District Design Division on drainage design and details to be shown on the MSE wall and roadway plans.  


==751.24.2 Mechanically Stabilized Earth (MSE) Walls==
:5. Bridge deck, approach slab and roadway drainage shall not be allowed to be discharged to MSE wall backfill area or within 10 feet from front of MSE wall.  
::*(Recommended) Use of a major bridge approach slab and approach pavement is ideal because bridge deck, approach slab and roadway drainage are directed using curbs and collected in drain basins for discharge that protect MSE wall backfill. For bridges not on a major roadway, consideration should be given to requiring a concrete bridge approach slab and pavement incorporating these same design elements (asphalt is permeable).


===751.24.2.1 Design===
::*(Less Recommended) Use of conduit and gutters:


Designs of Mechanically Stabilized Earth (MSE) walls are completed by consultants or contractors in accordance with Section 5 of the AASHTO Specifications. MoDOT Internet site contains a listing of facing unit manufacturers, soil reinforcement suppliers, and wall system suppliers which have been approved for use. See [http://www.modot.org/business/standards_and_specs/SpecbookEPG.pdf#page=11 Sec 720] and [http://www.modot.org/business/standards_and_specs/SpecbookEPG.pdf#page=14 Sec 1010] for additional information. The [http://sharepoint/systemdelivery/CM/geotechnical/default.aspx Geotechnical Section] is responsible for checking global stability, which should be reported on the Foundation Investigation Geotechnical Report. For MSE wall preliminary information, see [http://epg.modot.mo.gov/index.php?title=751.1_Preliminary_Design#751.1.4.3_MSE_Walls EPG 751.1.4.3 MSE Walls].
:::* Conduit: Drain away from bridge and bury conduit daylighting to natural ground or roadway drainage ditch at an eventual point beyond the limits of the wall. Use expansion fittings to allow for bridge movement and consider placing conduit to front of MSE wall and discharging more than 10 feet from front of wall or using lower drain pipes to intercept slab drainage conduit running through backfill.


'''General Guidelines'''
:::* Conduit and Gutters: Drain away from bridge using conduit and 90° elbow (or 45° bend) for smoothly directing drainage flow into gutters and that may be attached to inside of gutters to continue along downward sloping gutters along back of MSE wall to discharge to sewer or to natural drainage system, or to eventual point beyond the limits of the wall.  Allow for independent bridge and wall movements by using expansion fittings where needed. See [[751.10 General Superstructure#751.10.3.1 Type, Alignment and Spacing|EPG 751.10.3.1 Type, Alignment and Spacing]] and [[751.10 General Superstructure#751.10.3.3 General Requirements for Location of Slab Drains|EPG 751.10.3.3 General Requirements for Location of Slab Drains]].


* Small block walls are limited to a 10 ft. height in one lift.
:6. Vertical drains at end bents and approach slab underdrainage should be intercepted to drain away from bridge end and MSE wall.


* For small block walls, top cap units shall be used and shall be permanently attached by means of a resin anchor system.
:7. Discharging deck drainage using many slab drains would seem to reduce the volume of bridge end drainage over MSE walls.


* For large block walls, capstone may be substituted for coping and either shall be permanently attached to wall by panel dowels.
:8. Drain flumes at bridge abutments with MSE walls do not reduce infiltration at MSE wall backfill areas and are not recommended.


* For large block walls, form liners are required to produce all panels. Using form liner to produce panel facing is more cost effective than producing flat panels. Standard form liners are specified on the [http://www.modot.org/business/consultant_resources/bridgestandards.htm Bridge Standard Drawings - MSE Wall MSEW]. Be specific regarding names, types and colors of staining, and names and types of form liner.
:DISTRICT DESIGN DIVISION MSE WALLS


* MSE walls shall not be used where exposure to acid water may occur such as in areas of coal mining.
:Areas of concern: roadway or pavement drainage, MSE wall drainage, showing drainage details on the roadway and MSE wall plans.


* MSE walls shall not be used where scour is a problem.
:9. For long MSE walls, where lower perforated drain pipe slope become excessive, non-perforated lateral drain pipes, permitted by Standard Specifications, shall be designed to intercept them and go underneath the concrete leveling pad with a 2% minimum slope. Lateral drain pipes shall daylight/discharge at least 10 ft. from front of MSE wall. Screens should be installed and maintained on drain pipe outlets.


* MSE walls with metallic soil reinforcement shall not be used where stray electrical ground currents may occur as would be present near electrical substations.
:10. Roadway and pavement drainage shall not be allowed to be discharged to MSE wall backfill area or within 10 feet from front of MSE wall.  


* No utilities shall be allowed in the reinforced earth if future access to the utilities would require that the reinforcement layers be cut, or if there is a potential for material, which can cause degradation of the soil reinforcement, to leak out of the utilities into the wall backfill, with the exception of storm water drainage.
:11. For district design MSE walls, use roadway or pavement drainage collection pipes to transport and discharge to an eventual point outside the limits of the wall.


* The interior angle between two walls should be greater than 70°. However, if unavoidable, then place [[751.50 Standard Detailing Notes#J1.31|EPG 751.50 J1.31 note]] on the plans.
:Example: Showing drain pipe details on the MSE wall plans.


* Small block walls may be battered up to 1.5 in. per foot.
<gallery mode=packed widths=300px heights=300px>
File:751.24.2.1_elev_drain_pipe-01.png| <big>'''ELEVATION SHOWING DRAIN PIPE'''</big>
File:751.24.2.1_elev_drain_pipe_alt-01.png| <big>'''Alternate option'''</big>
</gallery>
<gallery mode=packed widths=400px heights=400px>
File:751.24.2.1_sec_A-A-02.png| <big>'''Section A-A'''</big>
</gallery>
{| style="text-align: left; margin-left: auto; margin-right: auto;"
|
Notes:</br>
(1) To be designed by District Design Division.</br>
(2) To be designed by District Design Division if needed. Provide non-perforated lateral drain pipe under leveling pad at 2% minimum slope. (Show on plans).</br>
(3) Discharge to drainage system or daylight screened outlet at least 10 feet away from end of wall (typ.). (Skew in the direction of flow as appropriate).</br>
(4) Discharge to drainage system or daylight screened outlet at least 10 feet away from front face of wall (typ.). (Skew in the direction of flow as appropriate).</br>
(5) Minimum backfill cover = Max(15”, 1.5 x diameter of drain pipe).</br>
|}


* The friction angle used for the computation of horizontal forces within the reinforced soil shall be greater than or equal to 34°.
===751.24.2.2 Excavation===


* For epoxy coated reinforcement requirements, see [[751.5 Structural Detailing Guidelines#751.5.9.2.2 Epoxy Coated Reinforcement Requirements|EPG 751.5.9.2.2 Epoxy Coated Reinforcement Requirements]].
For estimating excavation and minimum soil reinforcement length, see [[751.6 General Quantities#751.6.2.17 Excavation|EPG 751.6.2.17 Excavation]].


* All concrete except facing panels or units shall be CLASS B or B-1.  
For division responsibilities for preparing MSE wall plans, computing excavation class, quantities and locations, see [[:Category:747 Bridge Reports and Layouts#747.2.6.2 Mechanically Stabilized Earth (MSE) Wall Systems|EPG 747.2.6.2 Mechanically Stabilized Earth (MSE) Wall Systems]].


* The friction angle of the soil to be retained by the reinforced earth shall be listed on the plans as well as the friction angle for the foundation material the wall is to rest on.
===751.24.2.3 Details===
<center>
{| border="1" class="wikitable" style="margin: 1em auto 1em auto" style="text-align:center"
|+
| style="background:#BEBEBE" width="300" |'''[https://www.modot.org/bridge-standard-drawings Bridge Standard Drawings]'''
|-
|align="center"| MSE Wall - MSEW
|}
</center>


* Seismic performance category and acceleration coefficient shall be listed on the plans.
<gallery mode=packed widths=625px heights=625px>
File:751.24.2.3_mse_wall-01.png | <big>'''Fig. 751.24.2.3.1 MSE Wall Developed Elevation and Plan'''</big>
</gallery>
{| style="text-align: left; margin-left: auto; margin-right: auto;"
|
(1) Minimum embedment = maximum (2 feet; or embedment based on Geotechnical Report and global stability requirements;</br>or FHWA-NH1-10-0124, Table 2-2); or as per Geotechnical Report if rock is known to exist from Geotechnical Report.
|}


* Factors of Safety for MSE walls shall be 2.0 for overturning, 1.5 for sliding, 2.0 for ultimate bearing capacity and 1.5 for pullout resistance.
'''Drycast Modular Block Wall Systems and Wetcast Modular Block Wall Systems'''


* Factors of Safety for seismic design shall be 1.5 for overturning and 1.1 for sliding.
Battered mechanically stabilized earth wall systems may be used unless the design layout specifically calls for a vertical wall (precast modular panel wall systems shall not be battered and drycast modular block wall systems or wetcast modular block wall systems may be built vertical). If a battered MSE wall system is allowed, then [[751.50_Standard_Detailing_Notes#J1._General|EPG 751.50 J1.19]] note shall be placed on the design plans:


* Do not use small block walls in the following locations:
For battered walls, note on the plans whether the horizontal offset from the baseline is fixed at the top or bottom of the wall. Horizontal offset and corresponding vertical elevation shall be noted on plans.


::* Within the splash zone from snow removal operations (assumed to be 15 ft. from the edge of the shoulder).
<gallery mode=packed widths=400px heights=400px>
File:751.24.2.3_typ_dmbw-mse-01.png | <big>'''Fig. 751.24.2.3.2 Typical Section Through Generic Drycast Modular Block Wall (DMBW-MSE) System or Wetcast Modular Block Wall (WMBW-MSE) System'''</big>
</gallery>
{| style="text-align: left; margin-left: auto; margin-right: auto;"
|
<nowiki>*</nowiki> The maximum vertical spacing of reinforcement should be limited to two times the block depth or 32 in., whichever is less.</br>For large modular block (block height > 16 in.), maximum vertical spacing of reinforcement equal to the block height.  
|}


::* Where the blocks will be continuously wetted, such as around sources of water.
'''Fencing (See [https://www.modot.org/bridge-standard-drawings Bridge Standard Drawing] for details)'''


::* Where blocks will be located behind barrier curbs or other obstacles that will trap salt-laden snow from removal operations.
Fencing may be installed on the Modified Type A or Modified Type B Gutter or behind the MSE Wall.


::* For structurally critical applications, such as containing necessary fill around structures.
For Modified Type A and Modified Type B Gutter and Fence Post Connection details, see [https://www.modot.org/media/16871 Standard Plan 607.11].


::* In tiered wall systems.
==751.24.3 Cast-In-Place Concrete Retaining Walls==


* For locations where small block walls are not desirable, consider coloring agents and/or architectural forms using large block walls for aesthetic installations.
===751.24.3.1 Unit Stresses===
<div id="For slab drain location near MSE Wall,"></div>


* For slab drain location near MSE Wall, see [[751.10 General Superstructure#General Requirements for Location and Spacing of Slab Drains|EPG 751.10.3.1 Drain Type, Alignment and Spacing]] and [[751.10 General Superstructure#751.10.3.3 General Requirements for Location of Slab Drains|EPG 751.10.3.3 General Requirements for Location of Slab Drains]].
'''Concrete'''
Concrete for retaining walls shall be Class B Concrete (f'c = 3000 psi) unless the footing is used as a riding surface in which case Class B-1 Concrete (f'c = 4000 psi) shall be used.


* Roadway runoff should be directed away from running along face of MSE walls used as wing walls on bridge structures.
'''Reinforcing Steel'''


* Drainage:
Reinforcing Steel shall be Grade 60 (fy = 60,000 psi).


:*Gutter type should be selected at the core team meeting.
'''Pile Footing'''


:* When gutter is required without fencing, use Type A or Type B gutter (for detail, see [http://www.modot.mo.gov/business/standards_and_specs/documents/60900.pdf Std. Plan 609.00]).
For steel piling material requirements, see the unit stresses in [[751.50 Standard Detailing Notes#A1. Design Specifications, Loadings & Unit Stresses and Standard Plans|EPG 751.50 A1.3 note]].


:* When gutter is required with fencing, use Modified Type A or Modified Type B gutter (for detail, see [http://www.modot.mo.gov/business/standards_and_specs/documents/60711.pdf Std. Plan 607.11]).
'''Spread Footing'''


:* When fencing is required without gutter, place in tube and grout behind the MSE wall (for detail, see Fig. 751.24.2.1.7, Fence Post Connection Behind MSE Wall (without gutter).
For foundation material capacity, see Foundation Investigation Geotechnical Report.


:* Lower backfill longitudinal drainage pipes behind all MSE walls shall be  two-6” (Min.) diameter perforated PVC or PE pipe (See Sec 1013) unless larger sizes are required by design which shall be the responsibility of the District Design Division. Show drainage pipe size on plans. Outlet screens and cleanouts should be detailed for any drain pipe (shown on MoDOT MSE wall plans or roadway plans). Lateral non-perforated drain pipes (below leveling pad) are permitted by Standard Specifications and shall be sized by the District Design Division if necessary. Lateral outlet drain pipe sloped at 2% minimum.
===751.24.3.2 Design===


::* Identify on MSE wall plans or roadway plans drainage pipe point of entry, point of outlet (daylighting), 2% min. drainage slopes in between points to ensure positive flow and additional longitudinal drainage pipes if required to accommodate ground slope changes and lateral drainage pipes if required by design.
Note: For design concepts and guidance, follow the design process ([[751.40_LFD_Widening_and_Repair#751.40.8.15_Cast-In-Place_Concrete_Retaining_Walls|EPG 751.40.8.15]]) and modify design/details of ASD as necessary to meet LRFD requirements until [https://epgtest.modot.org/index.php/751.24_Retaining_Walls EPG 751.24] is updated for LRFD.


::* Adjustment in the vertical alignment of the longitudinal drainage pipes from that depicted on the MSE wall standard drawings may be necessary to ensure positive flow out of the drainage system.
Capacity/Demand ratio (CDR) for bearing capacity shall be ≥ 1.0
: <math>Bearing\ Capacity\ (CDR) = \frac{Factored\ Bearing\ Resistance}{Maximum\ Factored\ Bearing\ Stress} \ge 1.0</math>
::* Identify on MSE wall plans or roadway plans the outlet ends of pipes which shall be located to prevent clogging or backflow into the drainage system. Outlet screens and cleanouts should be detailed for any drain pipe.
: Strength Limit States:
: Factored bearing resistance = Nominal bearing resistance from Geotech report X
: Minimum Resistance factor (0.55, Geotech report) &nbsp;&nbsp;&nbsp;&nbsp; LRFD Table 11.5.7


:* For more information on drainage, see [[#Drainage at MSE Walls|Drainage at MSE Walls]].
: Extreme Event I and II Limit State:
: Factored bearing resistance = Nominal bearing resistance from Geotech report X Resistance factor
: Resistance factor = 0.8 &nbsp;&nbsp;&nbsp;&nbsp; LRFD 11.5.8


<div id="MSE Wall Construction:"></div>
: When wall is supported by soil:
'''MSE Wall Construction: Pipe Pile Spacers Guidance'''
: Factored bearing stress per LRFD eq. 11.6.3.2-1


Pipe pile spacers shall be used at pile locations behind mechanically stabilized earth walls to protect the wall reinforcement when driving pile for the bridge substructure at end bents(s). Pipe pile spacers shall have an inside diameter greater than that of the pile and large enough to avoid damage to the pipe when driving the pile. The bottom of the  pipe pile spacers shall be placed 5 ft. min. below the bottom of the MSE wall leveling pad. The pipe shall be filled with sand or other approved material after the pile is placed and before driving. Pipe pile spacers shall be accurately located and capped for future pile construction.  
: When wall is supported by a rock foundation:
: Factored bearing stress per LRFD eq. 11.6.3.2-2 and 11.6.3.2-3


Alternatively, the contractor shall be given the option of driving the piles before construction of the mechanically stabilized earth wall and placing the soil reinforcement and backfill material around the piling. The contractor shall adequately support the piling to ensure that proper pile alignment is maintained during the wall construction. The contractor’s plan for bracing the pile shall be submitted to the engineer for review.  
: Note: When the value of eccentricity e is negative then ''use e = 0''.  


Piling shall be designed for downdrag (DD) loads due to either method. Oversized  pipe pile spacers with sand placed after driving may be considered to mitigate some of the effects of downdrag (DD) loads. Sizing of  pipe pile spacers shall account for pile size, thermal movements of the bridge, pile placement plan, and vertical and horizontal placement tolerances.  
Capacity/Demand ratio (CDR) for overturning shall be ≥ 1.0
: <math>Overtuning\ (CDR) = \frac{Total\ Factored\ Resisting\ Moment}{Total\ Factored\ Driving\ Moment} \ge 1.0</math>


The minimum clearance from the back face of MSE walls to the front face of the end bent beam shall be 3 ft. 9 in. (typ.). The 3 ft. 9 in. dimension is based on the use of 18 in.  Pipe pile spacers& FHWA-NHI-10-24, Figure 5-17C, which will help ensure that soil reinforcement is not skewed more than 15° for nut and bolt reinforcement connections. Other types of connections may require different methods for splaying. In the event that the 3 ft. 9 in. dimension or setback cannot be used, the following guidance for  pipe pile spacers clearance shall be used: pipe pile spacers shall be placed 18 in. clear min. from the back face of MSE wall panels; 12 in. minimum clearance is required between  pipe pile spacers and leveling pad and 18 in. minimum clearance is required between leveling pad and pile.  
Capacity/Demand ratio (CDR) for eccentricity shall be ≥ 1.0
: <math>Eccentricity\ (CDR) = \frac{e_{Limit}}{e_{design}} \ge 1.0</math>


'''MSE Wall Plan and Geometrics'''
Capacity/Demand ratio (CDR) for sliding shall be ≥ 1.0
: <math>Sliding\ (CDR) = \frac{Total\ Factored\ Sliding\ Resistance}{Total\ Factored\ Active\ Force} \ge 1.0</math>


* A plan view shall be drawn showing a baseline or centerline, roadway stations and wall offsets. The plan shall contain enough information to properly locate the wall. The ultimate right of way shall also be shown, unless it is of a significant distance from the wall and will have no effect on the wall design or construction.
: Sliding shall be checked in accordance with LRFD 11.6.3.6 and 10.6.3.4


* Stations and offsets shall be established between one construction baseline or roadway centerline and a wall control line (baseline). Some wall designs may contain a slight batter, while others are vertical. A wall control line shall be set at the front face of the wall, either along the top or at the base of the wall, whichever is critical to the proposed improvements. For battered walls, in order to allow for batter adjustments of the stepped level pad or variation of the top of the wall, the wall control line (baseline) is to be shown at a fixed elevation. For battered walls, the offset location and elevation of control line shall be indicated. All horizontal breaks in the wall shall be given station-offset points, and walls with curvature shall indicate the station-offsets to the PC and PT of the wall, and the radius, on the plans.  
Eccentricity, (e) Limit for Strength Limit State: &nbsp;&nbsp;&nbsp;&nbsp; LRFD 11.6.3.3
:* For foundations supported on soil, the location of the resultant of the reaction forces shall be within the middle two-thirds of the base width, B or (e ≤ 0.33B).
:* For foundations supported on rock, the location of the resultant of the reaction forces shall be within the middle nine-tenths of the base width, B or (e ≤ 0.45B).


* Any obstacles which could possibly interfere with the soil reinforcement shall be shown. Drainage structures, lighting, or truss pedestals and footings, etc. are to be shown, with station offset to centerline of the obstacle, with obstacle size. Skew angles are shown to indicate the angle between a wall and a pipe or box which runs through the wall.  
Eccentricity, (e) Limit for Extreme Event I (Seismic): &nbsp;&nbsp;&nbsp;&nbsp; LRFD 11.6.5.1
:* For foundations supported on soil or rock, the location of the resultant of the reaction forces shall be within the middle two-thirds of the base width, B or (e ≤ 0.33B) for  γ<sub>EQ</sub> = 0.0 and middle eight-tenths of the base width, B or (e ≤ 0.40B) for  γ<sub>EQ</sub> = 1.0.  For γ<sub>EQ</sub> between 0.0 and 1.0, interpolate e value linearly between 0.33B and 0.40B. For γ<sub>EQ</sub> refer to LRFD 3.4.


* Elevations at the top and bottom of the wall shall be shown at 25 ft. intervals and at any break points in the wall.
:Note: Seismic design shall be performed for γ<sub>EQ</sub> = 0.5


* Curve data and/or offsets shall be shown at all changes in horizontal alignment. If battered wall systems are used on curved structures, show offsets at 10 ft. (max.) intervals from the baseline.
Eccentricity, (e) Limit for Extreme Event II:
:* For foundations supported on soil or/and rock, the location of the resultant of the reaction forces shall be within the middle eight-tenths of the base width, B or (e ≤ 0.40B).


* Details of any architectural finishes (formliners, concrete coloring, etc.).
For epoxy coated reinforcement requirements, see [[751.5 Structural Detailing Guidelines#751.5.9.2.2 Epoxy Coated Reinforcement Requirements|EPG 751.5.9.2.2 Epoxy Coated Reinforcement Requirements]].


* Details of threaded rod connecting the top cap block.
If the height of the wall or fill is a variable dimension, then base the structural design of the wall, toe, and heel on the high quarter point between expansion joints.


* Estimated quantities, total sq. ft. of mechanically stabilized earth systems.
[[image:751.24.3.2.jpg|center|600px|thumb|<center>'''Fig. 751.24.3.2'''</center>]]


* Proposed grade and theoretical top of leveling pad elevation shall be shown in constant slope. Slope line shall be adjusted per project. Top of wall or coping elevation and stationing shall be shown in the developed elevation per project. If leveling pad is anticipated to encounter rock, then contact the Geotechnical Section for leveling pad minimum embedment requirements.


'''MSE Wall Cross Sections'''


* A typical wall section for general information is shown.


* Additional sections are drawn for any special criteria. The front face of the wall is drawn vertical, regardless of the wall type.


* Any fencing and barrier curb are shown.


* Barriers if needed are shown on the cross section. Concrete barriers are attached to the roadway or shoulder pavement, not to the MSE wall. Standard Type B barrier curbs are placed along wall faces when traffic has access to the front face of the wall over shoulders of paved areas.
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|'''Additional Information'''
|-
|AASHTO 5.5.5
|}


<div id="Drainage at MSE Walls"></div>
'''Drainage at MSE Walls'''


*'''Drainage Before MSE Wall'''
====751.24.3.2.1 Spread Footings====


:Drainage is not allowed to be discharged within 10 ft. from front of MSE wall in order to protect wall embedment, prevent erosion and foundation undermining, and maintain soil strength and stability.
'''Location of Resultant'''


*'''Drainage Behind MSE Wall'''
The resultant of the footing pressure must be within the section of the footing specified in the following table.


::'''Internal (Subsurface) Drainage'''
{| border="1" class="wikitable" style="margin: 1em auto 1em auto"
|+
! style="background:#BEBEBE" |When Retaining Wall is Built on: !! style="background:#BEBEBE"|AASHTO Group Loads I-VI !! style="background:#BEBEBE"|For Seismic Loads
|-
|  align="center" |Soil<sup>a</sup> || align="center"|Middle 1/3||  align="center"|Middle 1/2 <sup>b</sup>
|-
|  align="center"|Rock<sup>c</sup> || align="center"|Middle 1/2||align="center"|Middle 2/3
|-
|colspan="3"|<sup>'''a'''</sup> Soil is defined as clay, clay and boulders, cemented gravel, soft shale, etc. with allowable bearing values less than 6 tons/sq. ft.
|-
|colspan="3"|<sup>'''b'''</sup> MoDOT is more conservative than AASHTO in this requirement.
|-
|colspan="3"|<sup>'''c'''</sup> Rock is defined as rock or hard shale with allowable bearing values of 6 tons/sq. ft. or more.
|}


::Groundwater and infiltrating surface waters are drained from behind the MSE wall through joints between the face panels or blocks (i.e. wall joints) and two-6 in. (min.) diameter pipes located at the base of the wall and at the basal interface between the reinforced backfill and the retained backfill.
Note: The location of the resultant is not critical when considering collision loads.


::Excessive subsurface draining can lead to increased risk of backfill erosion/washout through the wall joints and erosion at the bottom of walls and at wall terminal ends. Excessive water build-up caused by inadequate drainage at the bottom of the wall can lead to decreased soil strength and wall instability. Bridge underdrainage (vertical drains at end bents and at approach slabs) can exacerbate the problem.
'''Factor of Safety Against Overturning'''
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|'''Additional Information'''
|-
|AASHTO 5.5.5
|}


::Subsurface drainage pipes should be designed and sized appropriately to carry anticipated groundwater, incidental surface run-off that is not collected otherwise including possible effects of drainage created by an unexpected rupture of any roadway drainage conveyance or storage as an example.
AASHTO Group Loads I - VI:
* F.S. for overturning ≥ 2.0 for footings on soil.
* F.S. for overturning ≥ 1.5 for footings on rock.


::'''External (Surface) Drainage'''
For seismic loading, F.S. for overturning may be reduced to 75% of the value for AASHTO Group Loads I - VI. For seismic loading:
* F.S. for overturning ≥ (0.75)(2.0) = 1.5 for footings on soil.
* F.S. for overturning ≥ (0.75)(1.5) = 1.125 for footings on rock.


::External drainage considerations deal with collecting water that could flow externally over and/or around the wall surface taxing the internal drainage and/or creating external erosion issues. It can also infiltrate the reinforced and retained backfill areas behind the MSE wall.  
For collision forces:
* F.S. for overturning ≥ 1.2.


::Diverting water flow away from the reinforced soil structure is important. Roadway drainage should be collected in accordance with roadway drainage guidelines and bridge deck drainage should be collected similarly.
'''Factor of Safety Against Sliding'''
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|'''Additional Information'''
|-
|AASHTO 5.5.5
|}


*'''Guidance'''
Only spread footings on soil need be checked for sliding because spread footings on rock or shale are embedded into the rock.
* F.S. for sliding ≥ 1.5 for AASHTO Group Loads I - VI.
* F.S. for sliding ≥ (0.75)(1.5) = 1.125 for seismic loads.
* F.S. for sliding ≥ 1.2 for collision forces.


:ALL MSE WALLS
The resistance to sliding may be increased by:
* adding a shear key that projects into the soil below the footing.
* widening the footing to increase the weight and therefore increase the frictional resistance to sliding.


:1. Appropriate measures to prevent surface water infiltration into MSE wall backfill should be included in the design and detail layout for all MSE walls and shown on the roadway plans.
'''Passive Resistance of Soil to Lateral Load'''


:2. Gutters behind MSE walls are required for flat or positive sloping backfills to prevent concentrated infiltration behind the wall facing regardless of when top of backfill is paved or unpaved. This avoids pocket erosion behind facing and protection of nearest-surface wall connections which are vulnerable to corrosion and deterioration. Drainage swales lined with concrete, paved or precast gutter can be used to collect and discharge surface water to an eventual point away from the wall. If rock is used, use impermeable geotextile under rock and align top of gutter to bottom of rock to drain. (For negative sloping backfills away from top of wall, use of gutters is not required.)
The Rankine formula for passive pressure can be used to determine the passive resistance of soil to the lateral force on the wall. This passive pressure is developed at shear keys in retaining walls and at end abutments.


:District Design Division shall verify the size of the two-6 in. (min.) diameter lower perforated MSE wall drain pipes and where piping will daylight at ends of MSE wall or increase the diameters accordingly.  This should be part of the preliminary design of the MSE wall. (This shall include when lateral pipes are required and where lateral drain pipes will daylight/discharge).
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|-
:BRIDGE ABUTMENTS WITH MSE WALLS
|'''Additional Information'''
|-
|AASHTO 5.5.5A
|}


:Areas of concern: bridge deck drainage, approach slab drainage, approach roadway drainage, bridge underdrainage:  vertical drains at end bents and approach slab underdrainage, showing drainage details on the roadway and MSE wall plans
The passive pressure against the front face of the wall and the footing of a retaining wall is loosely compacted and should be neglected when considering sliding.


:3. Bridge slab drain design shall be in accordance with [[751.10 General Superstructure#751.10.3 Bridge Deck Drainage - Slab Drains |EPG 751.10.3 Bridge Deck Drainage – Slab Drains]] unless as modified below.
Rankine Formula: <math>P_p = \frac{1}{2}C_p\gamma_s[H^2-H_1^2]</math> where thefollowing variables are defined in the figure below
:
:''C<sub>p</sub>'' = <math>\tan \big( 45^\circ + \frac{\phi}{2}\big)</math>


:4. Coordination is required between the Bridge Division and District Design Division on drainage design and details to be shown on the MSE wall and roadway plans.
:''y<sub>1</sub> = <math>\frac{H_1y_2^2 + \frac{2}{3}y_2^3}{H^2 - H_1^2}</math>


:5. Bridge deck, approach slab and roadway drainage shall not be allowed to be discharged to MSE wall backfill area or within 10 feet from front of MSE wall.
:''P<sub>p</sub>'' = passive force at shear key in pounds per foot of wall length
::*(Recommended) Use of a major bridge approach slab and approach pavement is ideal because bridge deck, approach slab and roadway drainage are directed using curbs and collected in drain basins for discharge that protect MSE wall backfill. For bridges not on a major roadway, consideration should be given to requiring a concrete bridge approach slab and pavement incorporating these same design elements (asphalt is permeable).


::*(Less Recommended) Use of conduit and gutters:
:''C<sub>p</sub>'' = coefficient of passive earth pressure


:::* Conduit: Drain away from bridge and bury conduit daylighting to natural ground or roadway drainage ditch at an eventual point beyond the limits of the wall.  Use expansion fittings to allow for bridge movement. (can consider placing conduit to front of MSE wall and discharging more than 10 feet from front of wall or using lower drain pipes to intercept slab drainage conduit running through backfill.
:<math>\boldsymbol{\gamma_s}</math> = unit weight of soil


:::* Conduit and Gutters: Drain away from bridge using conduit and 90° elbow (or 45° bend) for smoothly directing drainage flow into gutters and that may be attached to inside of gutters to continue along downward sloping gutters along back of MSE wall to discharge to sewer or to natural drainage system, or to eventual point beyond the limits of the wall.  Allow for independent bridge and wall movements by using expansion fittings where needed. See [[751.10 General Superstructure#751.10.3.1 Type, Alignment and Spacing|EPG 751.10.3.1 Type, Alignment and Spacing]] and [[751.10 General Superstructure#751.10.3.3 General Requirements for Location of Slab Drains|EPG 751.10.3.3 General Requirements for Location of Slab Drains]].
:''H'' = height of the front face fill less than 1 ft. min. for erosion


:6. Vertical drains at end bents and approach slab underdrainage should be intercepted to drain away from bridge end and MSE wall.
:''H<sub>1</sub>'' = H minus depth of shear key


:7. Discharging deck drainage using many slab drains would seem to reduce the volume of bridge end drainage over MSE walls.
:''y<sub>1</sub>'' = location of ''P<sub>p</sub>'' from bottom of footing


:8. Drain flumes at bridge abutments with MSE walls do not reduce infiltration at MSE wall backfill areas and are not recommended.
:<math>\boldsymbol{\phi}</math> = angle of internal friction of soil


:DISTRICT DESIGN DIVISION MSE WALLS
[[image:751.24.3.2.1 passive.jpg|center|500px]]
 
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:Areas of concern: roadway or pavement drainage, MSE wall drainage, showing drainage details on the roadway and MSE wall plans.
|-
 
|'''Additional Information'''
:9. For long MSE walls, where lower perforated drain pipe slope become excessive, non-perforated lateral drain pipes, permitted by Standard Specifications, shall be designed to intercept them and go underneath the concrete leveling pad with a 2% minimum slope. Lateral drain pipes shall daylight/discharge at least 10 ft. from front of MSE wall. Screens should be installed and maintained on drain pipe outlets.
|-
|AASHTO 5.5.2
|}
The resistance due to passive pressure in front of the shear key shall be neglected unless the key extends below the depth of frost penetration.
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|-
|'''Additional Information'''
|-
|[http://sp/sites/cm/Pages/default.aspx MoDOT Materials Division]
|}


:10. Roadway and pavement drainage shall not be allowed to be discharged to MSE wall backfill area or within 10 feet from front of MSE wall.  
Frost line is set at 36 in. at the north border of Missouri and at 18 in. at the south border.


:11. For district design MSE walls, use roadway or pavement drainage collection pipes to transport and discharge to an eventual point outside the limits of the wall.
'''Passive Pressure During Seismic Loading'''


:Example: Showing drain pipe details on the MSE wall plans.
During an earthquake, the passive resistance of soil to lateral loads is slightly decreased. The Mononobe-Okabe static method is used to determine the equivalent fluid pressure.


[[image:751.24.2.1.jpg|center|825px]]
:''P<sub>PE</sub>'' = equivalent passive earth pressure during an earthquake
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|-
|'''Additional Information'''
|-
|1992 AASHTO Div. IA Eqns. C6-5 and C6-6
|}
:<math>P_{PE} = \frac{1}{2}\gamma_sH^2(1 - k_v)K_{PE}</math> where:


:''K<sub>PE</sub>'' = seismic passive pressure coefficient


[[image:751.24.2.1 alternate.jpg|center|600px]]
:<math>K_{PE} = \frac{\cos^2(\phi - \theta - \beta)}{\cos\theta\cos^2\beta\cos(\delta + \beta + \theta)\Bigg[1 + \sqrt{\frac{\sin(\phi + \delta)\sin(\phi - \theta - i)}{\cos(\delta + \beta + \theta)\cos(i - \beta)}}\Bigg]^2}</math>


::<math>\boldsymbol{\gamma}_s</math> = unit weight of soil


[[image:751.24.2.1 section AA.jpg|center|700px]]
:''H'' = height of soil at the location where the earth pressure is to be found


:''k<sub>V</sub>'' = vertical acceleration coefficient


::Notes:
:<math>\boldsymbol{\phi}</math> = angle of internal friction of soil
::'''*''' To be designed by District Design Division.


::'''**''' To be designed by District Design Division if needed. Provide non-perforated lateral drain pipe under leveling pad at 2% minimum slope. (Show on plans).
:<math>\boldsymbol{\theta} =  arctan \big[\frac{k_h}{1 - k_V}\big]</math>


::'''***''' Discharge to drainage system or daylight screened outlet at least 10 feet away from end of wall (typ.). (Skew in the direction of flow as appropriate).
:''k<sub>H</sub>'' = horizontal acceleration coefficient


::'''****''' Discharge to drainage system or daylight screened outlet at least 10 feet away from front face of wall (typ.). (Skew in the direction of flow as appropriate).
:<math>\boldsymbol{\beta}</math> = slope of soil face in degrees


===751.24.2.2 Excavation===
:''i'' = backfill slope angle in degrees


For estimating excavation, see [[751.6 General Quantities#751.6.2.17 Excavation|EPG 751.6.2.17 Excavation]].
:<math>\boldsymbol{\delta}</math> = angle of friction between soil and wall


For division responsibilities for preparing MSE wall plans, computing excavation class, quantities and locations, see [[:Category:747 Bridge Reports and Layouts#747.2.6.2 Mechanically Stabilized Earth (MSE) Wall Systems|EPG 747.2.6.2 Mechanically Stabilized Earth (MSE) Wall Systems]].
'''Special Soil Conditions'''


===751.24.2.3 Details===
Due to creep, some soft clay soils have no passive resistance under a continuing load. Removal of undesirable material and replacement with suitable material such as sand or crushed stone is necessary in such cases. Generally, this condition is indicated by a void ratio above 0.9, an angle of internal friction (<math>\boldsymbol{\phi}</math>) less than 22°, or a soil shear less than 0.8 ksf. Soil shear is determined from a standard penetration test.


<center>
:Soil Shear <math>\Big(\frac{k}{ft^2}\Big) = \frac{blows \ per\ 12\ in.}{10}</math>
{| border="1" class="wikitable" style="margin: 1em auto 1em auto" style="text-align:center"
|+
| style="background:#BEBEBE" width="300" |'''[http://www.modot.org/business/consultant_resources/bridgestandards.htm Bridge Standard Drawings]'''
|-
|align="center"|[http://www.modot.org/business/standard_drawings2/mse_wall_new_title_block.htm MSE Wall]
|}


</center>
'''Friction'''


[[image:751.24.2.2.jpg|center|825px|thumb|<center>'''Fig. 751.24.2.2.1 MSE Wall Developed Elevation and Plan'''</center>]]
In the absence of tests, the total shearing resistance to lateral loads between the footing and a soil that derives most of its strength from internal friction may be taken as the normal force times a coefficient of friction. If the plane at
which frictional resistance is evaluated is not below the frost line then this resistance must be neglected.


[[image:751.24.3.2.1 friction 2016.jpg|center|450px|thumb|<center>'''When A Shear Key Is Not Used'''</center>]]
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
|-
|'''Additional Information'''
|-
|AASHTO 5.5.2B
|}


'''Battered Small Block Walls'''
Sliding is resisted by the friction force developed at the interface between the soil and the concrete footing along the failure plane. The coefficient of friction for soil against concrete can be taken from the table below. If soil data
is not readily available or is inconsistent, the friction factor (f) can be taken as


Battered mechanically stabilized earth wall systems may be used unless the design layout specifically calls for a vertical wall (large block walls shall not be battered and small block walls may be built vertical). If a battered MSE wall system is allowed, then the following note shall be placed on the design plans:
: ''f'' =<math>tan \Big(\frac{2\phi}{3}\Big)</math> where <math>\boldsymbol{\phi}</math> is the angle of internal friction of the soil (''Civil Engineering Reference Manual'' by Michael R. Lindeburg, 6th ed., 1992).


:"The top and bottom of wall elevations are given for a vertical wall. If a battered small block wall system is used, the height of the wall shall be adjusted as necessary to fit the ground slope. If fence is built on an extended gutter, then the height of the wall shall be adjusted further."
{| border="1" class="wikitable" style="margin: 1em auto 1em auto"
 
|+
For battered walls, note on the plans whether the horizontal offset from the baseline is fixed at the top or bottom of the wall. Horizontal offset and corresponding vertical elevation shall be noted on plans.
!style="background:#BEBEBE" colspan="2"|Coefficient of Friction Values for Soil Against Concrete
 
|-
[[image:751.24.2.2 battered.jpg|center|700px|thumb|<center>'''Fig. 751.24.2.2.2 Typical Section Through Generic Small Block Wall'''</center>]]
! style="background:#BEBEBE" |Soil Type<sup>a</sup> !! style="background:#BEBEBE"|Coefficient of Friction
 
|-
'''Fencing (See [http://www.modot.org/business/standard_drawings2/mse_wall_new_title_block.htm Standard Drawing] for details)'''
|  align="center" |coarse-grained soil without silt || align="center"|0.55
|-
|  align="center"|coarse-grained soil with silt  || align="center"|0.45
|-
|align="center"|silt (only)||  align="center"|0.35
|-
|align="center"|clay|| align="center"|0.30<sup>b</sup>
|-
|colspan="2"|<sup>'''a'''</sup> It is not necessary to check rock or shale for sliding due to embedment.
|-
|colspan="2"|<sup>'''b'''</sup> Caution should be used with soils with <math>\boldsymbol{\phi}</math> < 22° or soil shear < 0.8 k/sq.ft. (soft clay soils). Removal and replacement of such soil with suitable material should be considered.
|}


Fencing may be installed on the Modified Type A or Modified Type B Gutter or behind the MSE Wall.
[[image:751.24.3.2.1 soil and soil.jpg|center|450px|thumb|<center>'''When A Shear Key Is Used'''</center>]]


For Modified Type A and Modified Type B Gutter and Fence Post Connection details, see [http://www.modot.mo.gov/business/standards_and_specs/documents/60711.pdf Standard Plan 607.11].
When a shear key is used, the failure plane is located at the bottom of the shear key in the front half of the footing. The friction force resisting sliding in front of the shear key is provided at the interface between the stationary layer of soil and the moving layer of soil, thus the friction angle is the internal angle of friction of the soil (soil against soil). The friction force resisting sliding on the rest of the footing is of that between the concrete and soil. Theoretically
the bearing pressure distribution should be used to determine how much normal load exists on each surface, however it is reasonable to assume a constant distribution. Thus the normal load to each surface can be divided out between the two surfaces based on the fractional length of each and the total frictional force will be the sum of the normal load on each surface
multiplied by the corresponding friction factor.


==751.24.3 Cast-In-Place Concrete Retaining Walls==
'''Bearing Pressure'''
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
|-
|'''Additional Information'''
|-
|AASHTO 4.4.7.1.2 & 4.4.8.1.3
|}


===751.24.3.1 Unit Stresses===
:'''Group Loads I - VI'''


'''Concrete'''
:The bearing capacity failure factor of safety for Group Loads I - VI must be greater than or equal to 3.0. This factor of safety is figured into the allowable bearing pressure given on the "Design Layout Sheet".
Concrete for retaining walls shall be Class B Concrete (f'c = 3000 psi) unless the footing is used as a riding surface in which case Class B-1 Concrete (f'c = 4000 psi) shall be used.


'''Reinforcing Steel'''
:The bearing pressure on the supporting soil shall not be greater than the allowable bearing pressure given on the "Design Layout Sheet".


Reinforcing Steel shall be Grade 60 (fy = 60,000 psi).  
:'''Seismic Loads'''
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
|-
|'''Additional Information'''
|-
|AASHTO Div. IA 6.3.1(B) and AASHTO 5.5.6.2
|}


'''Pile Footing'''
:When seismic loads are considered, AASHTO allows the ultimate bearing capacity to be used. The ultimate capacity of the foundation soil can be conservatively estimated as 2.0 times the allowable bearing pressure given on the "Design Layout".


For piling capacities, see the unit stresses in [[751.50 Standard Detailing Notes#A1. Design Specifications, Loadings & Unit Stresses |EPG 751.50 Standard Detailing Notes]].
:'''Stem Design'''
:The vertical stem (the wall portion) of a cantilever retaining wall shall be designed as a cantilever supported at the base.


'''Spread Footing'''
:'''Footing Design'''
 
For foundation material capacity, see the Unit Stresses Section of the Bridge Manual and the Design Layout Sheet.
 
===751.24.3.2 Design===
For epoxy coated reinforcement requirements, see [[751.5 Structural Detailing Guidelines#751.5.9.2.2 Epoxy Coated Reinforcement Requirements|EPG 751.5.9.2.2 Epoxy Coated Reinforcement Requirements]].
 
If the height of the wall or fill is a variable dimension, then base the structural design of the wall, toe, and heel on the high quarter point between expansion joints.
 
[[image:751.24.3.2.jpg|center|600px|thumb|<center>'''Fig. 751.24.3.2'''</center>]]
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"  
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"  
|-
|-
|'''Additional Information'''
|'''Additional Information'''
|-
|-
|AASHTO 5.5.5
|AASHTO 5.5.6.1
|}
|}


====751.24.3.2.1 Spread Footings====
::'''Toe'''
 
::The toe of the base slab of a cantilever wall shall be designed as a cantilever supported by the wall. The critical section for bending moments shall be taken at the front face of the stem. The critical section for shear shall be taken at a distance d (d = effective depth) from the front face of the stem.


'''Location of Resultant'''
::'''Heel'''


The resultant of the footing pressure must be within the section of the footing specified in the following table.
::The rear projection (heel) of the base slab shall be designed to support the entire weight of the superimposed materials, unless a more exact method is used. The heel shall be designed as a cantilever supported by the wall. The critical section for bending moments and shear shall be taken at the back face of the stem.


{| border="1" class="wikitable" style="margin: 1em auto 1em auto"
:'''Shear Key Design'''
|+
! style="background:#BEBEBE" |When Retaining Wall is Built on: !! style="background:#BEBEBE"|AASHTO Group Loads I-VI !! style="background:#BEBEBE"|For Seismic Loads
|-
|  align="center" |Soil<sup>a</sup> || align="center"|Middle 1/3||  align="center"|Middle 1/2 <sup>b</sup>
|-
|  align="center"|Rock<sup>c</sup> || align="center"|Middle 1/2||align="center"|Middle 2/3
|-
|colspan="3"|<sup>'''a'''</sup> Soil is defined as clay, clay and boulders, cemented gravel, soft shale, etc. with allowable bearing values less than 6 tons/sq. ft.
|-
|colspan="3"|<sup>'''b'''</sup> MoDOT is more conservative than AASHTO in this requirement.
|-
|colspan="3"|<sup>'''c'''</sup> Rock is defined as rock or hard shale with allowable bearing values of 6 tons/sq. ft. or more.
|}


Note: The location of the resultant is not critical when considering collision loads.
:The shear key shall be designed as a cantilever supported at the bottom of the footing.


'''Factor of Safety Against Overturning'''
====751.24.3.2.2 Pile Footings====
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
|-
|'''Additional Information'''
|-
|AASHTO 5.5.5
|}


AASHTO Group Loads I - VI:
Footings shall be cast on piles when specified on the "Design Layout Sheet". If the horizontal force against the retaining wall cannot otherwise be resisted, some of the piles shall be driven on a batter.
* F.S. for overturning ≥ 2.0 for footings on soil.
* F.S. for overturning ≥ 1.5 for footings on rock.


For seismic loading, F.S. for overturning may be reduced to 75% of the value for AASHTO Group Loads I - VI. For seismic loading:
:'''Pile Arrangement'''
* F.S. for overturning ≥ (0.75)(2.0) = 1.5 for footings on soil.
* F.S. for overturning ≥ (0.75)(1.5) = 1.125 for footings on rock.


For collision forces:
:For retaining walls subject to moderate horizontal loads (walls 15 to 20 ft. tall), the following layout is suggested.
* F.S. for overturning ≥ 1.2.


'''Factor of Safety Against Sliding'''
[[image:751.24.3.2.2 batter piles.jpg|center|300px|thumb|<center>'''Section'''</center>]]
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
|-
|'''Additional Information'''
|-
|AASHTO 5.5.5
|}


Only spread footings on soil need be checked for sliding because spread footings on rock or shale are embedded into the rock.
[[image:751.24.3.2.2 plan 2016.jpg|center|450px|thumb|<center>'''Plan'''</center>]]
* F.S. for sliding ≥ 1.5 for AASHTO Group Loads I - VI.
* F.S. for sliding ≥ (0.75)(1.5) = 1.125 for seismic loads.
* F.S. for sliding ≥ 1.2 for collision forces.


The resistance to sliding may be increased by:
:For higher walls and more extreme conditions of loading, it may be necessary to:
* adding a shear key that projects into the soil below the footing.
* widening the footing to increase the weight and therefore increase the frictional resistance to sliding.


'''Passive Resistance of Soil to Lateral Load'''
:* use the same number of piles along all rows


The Rankine formula for passive pressure can be used to determine the passive resistance of soil to the lateral force on the wall. This passive pressure is developed at shear keys in retaining walls and at end abutments.
:* use three rows of piles


{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
:* provide batter piles in more than one row
|-
|'''Additional Information'''
|-
|AASHTO 5.5.5A
|}


The passive pressure against the front face of the wall and the footing of a retaining wall is loosely compacted and should be neglected when considering sliding.
::'''Loading Combinations for Stability and Bearing'''


Rankine Formula: <math>P_p = \frac{1}{2}C_p\gamma_s[H^2-H_1^2]</math> where thefollowing variables are defined in the figure below
::The following table gives the loading combinations to be checked for stability and pile loads. These abbreviations are used in the table:
:
:''C<sub>p</sub>'' = <math>\tan \big( 45^\circ + \frac{\phi}{2}\big)</math>


:''y<sub>1</sub> = <math>\frac{H_1y_2^2 + \frac{2}{3}y_2^3}{H^2 - H_1^2}</math>
:::DL = dead load weight of the wall elements


:''P<sub>p</sub>'' = passive force at shear key in pounds per foot of wall length
:::SUR = two feet of live load surcharge


:''C<sub>p</sub>'' = coefficient of passive earth pressure
:::E = earth weight


:<math>\boldsymbol{\gamma_s}</math> = unit weight of soil
:::EP = equivalent fluid earth pressure


:''H'' = height of the front face fill less than 1 ft. min. for erosion
:::COL = collision force


:''H<sub>1</sub>'' = H minus depth of shear key
:::EQ = earthquake inertial force of failure wedge


:''y<sub>1</sub>'' = location of ''P<sub>p</sub>'' from bottom of footing
{| border="1" class="wikitable" style="margin: 1em auto 1em auto"
 
|+
:<math>\boldsymbol{\phi}</math> = angle of internal friction of soil
!style="background:#BEBEBE" rowspan="2"|Loading Case !!style="background:#BEBEBE" rowspan="2"|Vertical Loads !!style="background:#BEBEBE" rowspan="2"|Horizontal Loads !!style="background:#BEBEBE" rowspan="2"|Overturning Factor of Safety !!style="background:#BEBEBE" colspan="2"|Sliding Factor of Safety
 
|-
[[image:751.24.3.2.1 passive.jpg|center|500px]]
!style="background:#BEBEBE" |Battered Toe Piles !!style="background:#BEBEBE" |Vertical Toe Piles
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"  
|-
|-
|'''Additional Information'''
|align="center"|I<sup>a</sup>||align="center"| DL+SUR+E ||align="center"|EP+SUR||align="center"| 1.5||align="center"| 1.5||align="center"|2.0
|-
|-
|AASHTO 5.5.2
|align="center"|II||align="center"| DL+SUR+E ||align="center"|EP+SUR+COL||align="center"| 1.2|| align="center"|1.2||align="center"| 1.2
|}
The resistance due to passive pressure in front of the shear key shall be neglected unless the key extends below the depth of frost penetration.
{|style="padding: 0.3em; margin-right:7px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="left"  
|-
|-
|'''Additional Information'''
|align="center"|III||align="center"| DL+E||align="center"| EP||align="center"| 1.5||align="center"| 1.5||align="center"| 2.0
|-
|-
|[http://wwwi/intranet/cm/ MoDOT Materials Division]
|align="center"|IV<sup>b</sup>||align="center"| DL+E ||align="center"|None||align="center"| -||align="center"| -||align="center"| -
|}
|-
 
|align="center"|V<sup>c</sup>||align="center"| DL+E||align="center"| EP+EQ||align="center"| 1.125||align="center"| 1.125||align="center"| 1.5
Frost line is set at 36 in. at the north border of Missouri and at 18 in. at the south border.
|-
 
|colspan="6"|<sup>'''a'''</sup> Load Case I should be checked with and without the vertical surcharge.
'''Passive Pressure During Seismic Loading'''
 
During an earthquake, the passive resistance of soil to lateral loads is slightly decreased. The Mononobe-Okabe static method is used to determine the equivalent fluid pressure.
 
:''P<sub>PE</sub>'' = equivalent passive earth pressure during an earthquake
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"  
|-
|-
|'''Additional Information'''
|colspan="6"|<sup>'''b'''</sup> A 25% overstress is allowed on the heel pile in Load Case IV.
|-
|-
|1992 AASHTO Div. IA Eqns. C6-5 and C6-6
|colspan="6"|<sup>'''c'''</sup> The factors of safety for earthquake loading are 75% of that used in Load Case III. Battered piles are not recommended for use in seismic performance categories B, C, and D. Seismic design of retaining walls is not required in SPC A and B. Retaining walls in SPC B located under a bridge abutment shall be designed to AASHTO Specifications for SPC B.
|}
|}
:<math>P_{PE} = \frac{1}{2}\gamma_sH^2(1 - k_v)K_{PE}</math> where:


:''K<sub>PE</sub>'' = seismic passive pressure coefficient
::'''Pile Properties and Capacities'''


:<math>K_{PE} = \frac{\cos^2(\phi - \theta - \beta)}{\cos\theta\cos^2\beta\cos(\delta + \beta + \theta)\Bigg[1 + \sqrt{\frac{\sin(\phi + \delta)\sin(\phi - \theta - i)}{\cos(\delta + \beta + \theta)\cos(i - \beta)}}\Bigg]^2}</math>
::For Load Cases I-IV in the table above, the allowable compressive pile force may be taken from the pile capacity table in the Piling Section of the Bridge Manual which is based in part on AASHTO 4.5.7.3. Alternatively, the allowable compressive pile capacity of a friction pile may be determined from the ultimate frictional and bearing capacity between the soil and pile divided by a safety factor of 3.5 (AASHTO Table 4.5.6.2.A). The maximum amount of tension allowed on a heel pile is 3 tons.


::<math>\boldsymbol{\gamma}_s</math> = unit weight of soil
::For Load Case V in the table above, the allowable compressive pile force may be taken from the pile capacity table in the Piling Section of the Bridge Manual multiplied by the appropriate factor (2.0 for steel bearing piles, 1.5 for friction piles). Alternatively, the allowable compressive pile capacity of a friction pile may be determined from the ultimate frictional and bearing capacity between the soil and pile divided by a safety factor of 2.0. The allowable tension force on a bearing or friction pile will be equal to the ultimate friction capacity between the soil and pile divided by a safety factor of 2.0.


:''H'' = height of soil at the location where the earth pressure is to be found
::To calculate the ultimate compressive or tensile capacity between the soil and pile requires the boring data which includes the SPT blow counts, the friction angle, the water level, and the soil layer descriptions.


:''k<sub>V</sub>'' = vertical acceleration coefficient
::Assume the vertical load carried by battered piles is the same as it would be if the pile were vertical. The properties of piles may be found in the Piling Section of the Bridge Manual.


:<math>\boldsymbol{\phi}</math> = angle of internal friction of soil
:::'''Neutral Axis of Pile Group'''


:<math>\boldsymbol{\theta} =  arctan \big[\frac{k_h}{1 - k_V}\big]</math>
:::Locate the neutral axis of the pile group in the repetitive strip from the toe of the footing at the bottom of the footing.


:''k<sub>H</sub>'' = horizontal acceleration coefficient
:::'''Moment of Inertia of Pile Group'''


:<math>\boldsymbol{\beta}</math> = slope of soil face in degrees
:::The moment of inertia of the pile group in the repetitive strip about the neutral axis of the section may be determined using the parallel axis theorem:
 
::::I = Σ(I<sub>A</sub>) + Σ(Ad<sup>2</sup>) where :


:''i'' = backfill slope angle in degrees
::::''I<sub>A</sub>'' = moment of inertia of a pile about its neutral axis


:<math>\boldsymbol{\delta}</math> = angle of friction between soil and wall
::::''A'' = area of a pile


'''Special Soil Conditions'''
::::''d'' = distance from a pile's neutral axis to pile group's neutral axis


Due to creep, some soft clay soils have no passive resistance under a continuing load. Removal of undesirable material and replacement with suitable material such as sand or crushed stone is necessary in such cases. Generally, this condition is indicated by a void ratio above 0.9, an angle of internal friction (<math>\boldsymbol{\phi}</math>) less than 22°, or a soil shear less than 0.8 ksf. Soil shear is determined from a standard penetration test.
:::''I<sub>A</sub>'' may be neglected so the equation reduces to:


:Soil Shear <math>\Big(\frac{k}{ft^2}\Big) = \frac{blows \ per\ 12\ in.}{10}</math>
::::''I'' =  Σ(Ad<sup>2</sup>)


'''Friction'''
::'''Resistance To Sliding'''


In the absence of tests, the total shearing resistance to lateral loads between the footing and a soil that derives most of its strength from internal friction may be taken as the normal force times a coefficient of friction. If the plane at
::Any frictional resistance to sliding shall be ignored, such as would occur between the bottom of the footing and the soil on a spread footing.
which frictional resistance is evaluated is not below the frost line then this resistance must be neglected.


[[image:751.24.3.2.1 friction 2016.jpg|center|450px|thumb|<center>'''When A Shear Key Is Not Used'''</center>]]
::'''Friction or Bearing Piles With Batter (Case 1)'''
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
|-
|'''Additional Information'''
|-
|AASHTO 5.5.2B
|}


Sliding is resisted by the friction force developed at the interface between the soil and the concrete footing along the failure plane. The coefficient of friction for soil against concrete can be taken from the table below. If soil data
::Retaining walls using friction or bearing piles with batter should develop lateral strength (resistance to sliding) first from the batter component of the pile and second from the passive pressure against the shear key and the piles.
is not readily available or is inconsistent, the friction factor (f) can be taken as


: ''f'' =<math>tan \Big(\frac{2\phi}{3}\Big)</math> where <math>\boldsymbol{\phi}</math> is the angle of internal friction of the soil (''Civil Engineering Reference Manual'' by Michael R. Lindeburg, 6th ed., 1992).
::'''Friction or Bearing Piles Without Batter (Case 2)'''


{| border="1" class="wikitable" style="margin: 1em auto 1em auto"
::Retaining walls using friction or bearing piles without batter due to site constrictions should develop lateral strength first from the passive pressure against the shear key and second from the passive pressure against the pile below the bottom of footing. In this case, the shear key shall be placed at the front face of the footing.
|+
!style="background:#BEBEBE" colspan="2"|Coefficient of Friction Values for Soil Against Concrete
|-
! style="background:#BEBEBE" |Soil Type<sup>a</sup> !! style="background:#BEBEBE"|Coefficient of Friction
|-
|  align="center" |coarse-grained soil without silt || align="center"|0.55
|-
|  align="center"|coarse-grained soil with silt  || align="center"|0.45
|-
|align="center"|silt (only)||  align="center"|0.35
|-
|align="center"|clay||  align="center"|0.30<sup>b</sup>
|-
|colspan="2"|<sup>'''a'''</sup> It is not necessary to check rock or shale for sliding due to embedment.
|-
|colspan="2"|<sup>'''b'''</sup> Caution should be used with soils with <math>\boldsymbol{\phi}</math> < 22° or soil shear < 0.8 k/sq.ft. (soft clay soils). Removal and replacement of such soil with suitable material should be considered.
|}


[[image:751.24.3.2.1 soil and soil.jpg|center|450px|thumb|<center>'''When A Shear Key Is Used'''</center>]]
::'''Concrete Pedestal Piles or Drilled Shafts (Case 3)'''


When a shear key is used, the failure plane is located at the bottom of the shear key in the front half of the footing. The friction force resisting sliding in front of the shear key is provided at the interface between the stationary layer of soil and the moving layer of soil, thus the friction angle is the internal angle of friction of the soil (soil against soil). The friction force resisting sliding on the rest of the footing is of that between the concrete and soil. Theoretically
::Retaining walls using concrete pedestal piles should develop lateral strength first from passive pressure against the shear key and second from passive pressure against the pile below the bottom of the footing. In this case, the shear key shall be placed at the front of the footing. Do not batter concrete pedestal piles.
the bearing pressure distribution should be used to determine how much normal load exists on each surface, however it is reasonable to assume a constant distribution. Thus the normal load to each surface can be divided out between the two surfaces based on the fractional length of each and the total frictional force will be the sum of the normal load on each surface
multiplied by the corresponding friction factor.


'''Bearing Pressure'''
[[image:751.24.3.2.2 cases.jpg|center|450px]]
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
|-
|'''Additional Information'''
|-
|AASHTO 4.4.7.1.2 & 4.4.8.1.3
|}


:'''Group Loads I - VI'''
::'''Resistance Due to Passive Pressure Against Pile'''


:The bearing capacity failure factor of safety for Group Loads I - VI must be greater than or equal to 3.0. This factor of safety is figured into the allowable bearing pressure given on the "Design Layout Sheet".
::The procedure below may be used to determine the passive pressure resistance developed in the soil against the piles. The procedure assumes that the piles develop a local failure plane.


:The bearing pressure on the supporting soil shall not be greater than the allowable bearing pressure given on the "Design Layout Sheet".
:::''F'' = the lateral force due to passive pressure on pile


:'''Seismic Loads'''
:::<math>F = \frac{1}{2}\gamma_s C_P H^2 B </math> , where: <math> C_P = tan^2\Big[45 + \frac{\phi}{2}\Big]</math>
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
|-
|'''Additional Information'''
|-
|AASHTO Div. IA 6.3.1(B) and AASHTO 5.5.6.2
|}


:When seismic loads are considered, AASHTO allows the ultimate bearing capacity to be used. The ultimate capacity of the foundation soil can be conservatively estimated as 2.0 times the allowable bearing pressure given on the "Design Layout".
:::<math>\boldsymbol{\gamma_s}</math> = unit weight of soil


:'''Stem Design'''
:::''H'' = depth of pile considered for lateral resistance (H<sub>max</sub>= 6B)
:The vertical stem (the wall portion) of a cantilever retaining wall shall be designed as a cantilever supported at the base.


:'''Footing Design'''
:::''C<sub>P</sub>'' = coefficient of active earth pressure
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
|-
|'''Additional Information'''
|-
|AASHTO 5.5.6.1
|}


::'''Toe'''
:::''B'' = width of pile


::The toe of the base slab of a cantilever wall shall be designed as a cantilever supported by the wall. The critical section for bending moments shall be taken at the front face of the stem. The critical section for shear shall be taken at a distance d (d = effective depth) from the front face of the stem.
:::<math>\boldsymbol{\phi}</math> = angle of internal friction of soil


::'''Heel'''
[[image:751.24.3.2.2 resistance passive.jpg|center|450px]]


::The rear projection (heel) of the base slab shall be designed to support the entire weight of the superimposed materials, unless a more exact method is used. The heel shall be designed as a cantilever supported by the wall. The critical section for bending moments and shear shall be taken at the back face of the stem.
::'''Resistance Due to Pile Batter'''


:'''Shear Key Design'''
::Use the horizontal component (due to pile batter) of the allowable pile load as the lateral resistance of the battered pile. (This presupposes that sufficient lateral movement of the wall can take place before failure to develop the ultimate strength of both elements.)


:The shear key shall be designed as a cantilever supported at the bottom of the footing.
[[image:751.24.3.2.2 12.jpg|center|125px]]


====751.24.3.2.2 Pile Footings====
:::''b'' = the amount of batter per 12 inches.


Footings shall be cast on piles when specified on the "Design Layout Sheet". If the horizontal force against the retaining wall cannot otherwise be resisted, some of the piles shall be driven on a batter.
:::<math> c = \sqrt{(12 in.)^2 + b^2}</math>


:'''Pile Arrangement'''
:::<math>P_{HBatter} = P_T \Big(\frac{b}{c}\Big)</math> (# of battered piles) where:


:For retaining walls subject to moderate horizontal loads (walls 15 to 20 ft. tall), the following layout is suggested.
:::''P<sub>HBatter</sub>'' = the horizontal force due to the battered piles


[[image:751.24.3.2.2 batter piles.jpg|center|300px|thumb|<center>'''Section'''</center>]]
:::''P<sub>T</sub>'' = the allowable pile load


[[image:751.24.3.2.2 plan 2016.jpg|center|450px|thumb|<center>'''Plan'''</center>]]
::Maximum batter is 4" per 12".


:For higher walls and more extreme conditions of loading, it may be necessary to:
::'''Resistance Due to Shear Keys'''


:* use the same number of piles along all rows
::A shear key may be needed if the passive pressure against the piles and the horizontal force due to batter is not sufficient to attain the factor of safety against sliding. The passive pressure against the shear key on a pile footing is found in the same manner as for spread footings.


:* use three rows of piles
::'''Resistance to Overturning'''


:* provide batter piles in more than one row
::The resisting and overturning moments shall be computed at the centerline of the toe pile at a distance of 6B (where B is the width of the pile) below the bottom of the footing. A maximum of 3 tons of tension on each heel pile may be assumed to resist overturning. Any effects of passive pressure, either on the shear key or on the piles, which resist overturning, shall be ignored.


::'''Loading Combinations for Stability and Bearing'''
[[image:751.24.3.2.2 resistance overturning.jpg|center|450px]]


::The following table gives the loading combinations to be checked for stability and pile loads. These abbreviations are used in the table:
::'''Pile Properties'''


:::DL = dead load weight of the wall elements
:::'''Location of Resultant'''


:::SUR = two feet of live load surcharge
:::The location of the resultant shall be evaluated at the bottom of the footing and can be determined by the equation below:


:::E = earth weight
::::<math>e = \frac{\Sigma M}{\Sigma V}</math>  where:


:::EP = equivalent fluid earth pressure
::::e = the distance between the resultant and the neutral axis of the pile group


:::COL = collision force
::::''ΣM'' = the sum of the moments taken about the neutral axis of the pile group at the bottom of the footing


:::EQ = earthquake inertial force of failure wedge
::::''ΣV'' = the sum of the vertical loads used in calculating the moment


{| border="1" class="wikitable" style="margin: 1em auto 1em auto"
:::'''Pile Loads'''
|+
!style="background:#BEBEBE" rowspan="2"|Loading Case !!style="background:#BEBEBE" rowspan="2"|Vertical Loads !!style="background:#BEBEBE" rowspan="2"|Horizontal Loads !!style="background:#BEBEBE" rowspan="2"|Overturning Factor of Safety !!style="background:#BEBEBE" colspan="2"|Sliding Factor of Safety
|-
!style="background:#BEBEBE" |Battered Toe Piles !!style="background:#BEBEBE" |Vertical Toe Piles
|-
|align="center"|I<sup>a</sup>||align="center"| DL+SUR+E ||align="center"|EP+SUR||align="center"| 1.5||align="center"| 1.5||align="center"|2.0
|-
|align="center"|II||align="center"| DL+SUR+E ||align="center"|EP+SUR+COL||align="center"| 1.2|| align="center"|1.2||align="center"| 1.2
|-
|align="center"|III||align="center"| DL+E||align="center"| EP||align="center"| 1.5||align="center"| 1.5||align="center"| 2.0
|-
|align="center"|IV<sup>b</sup>||align="center"| DL+E ||align="center"|None||align="center"| -||align="center"| -||align="center"| -
|-
|align="center"|V<sup>c</sup>||align="center"| DL+E||align="center"| EP+EQ||align="center"| 1.125||align="center"| 1.125||align="center"| 1.5
|-
|colspan="6"|<sup>'''a'''</sup> Load Case I should be checked with and without the vertical surcharge.
|-
|colspan="6"|<sup>'''b'''</sup> A 25% overstress is allowed on the heel pile in Load Case IV.
|-
|colspan="6"|<sup>'''c'''</sup> The factors of safety for earthquake loading are 75% of that used in Load Case III. Battered piles are not recommended for use in seismic performance categories B, C, and D. Seismic design of retaining walls is not required in SPC A and B. Retaining walls in SPC B located under a bridge abutment shall be designed to AASHTO Specifications for SPC B.
|}


::'''Pile Properties and Capacities'''
:::The loads on the pile can be determined as follows:


::For Load Cases I-IV in the table above, the allowable compressive pile force may be taken from the pile capacity table in the Piling Section of the Bridge Manual which is based in part on AASHTO 4.5.7.3. Alternatively, the allowable compressive pile capacity of a friction pile may be determined from the ultimate frictional and bearing capacity between the soil and pile divided by a safety factor of 3.5 (AASHTO Table 4.5.6.2.A). The maximum amount of tension allowed on a heel pile is 3 tons.
::::<math>P = \frac{\Sigma V}{A} \pm \frac{Mc}{I}</math> where:


::For Load Case V in the table above, the allowable compressive pile force may be taken from the pile capacity table in the Piling Section of the Bridge Manual multiplied by the appropriate factor (2.0 for steel bearing piles, 1.5 for friction piles). Alternatively, the allowable compressive pile capacity of a friction pile may be determined from the ultimate frictional and bearing capacity between the soil and pile divided by a safety factor of 2.0. The allowable tension force on a bearing or friction pile will be equal to the ultimate friction capacity between the soil and pile divided by a safety factor of 2.0.
:::::''P'' = the force on the pile


::To calculate the ultimate compressive or tensile capacity between the soil and pile requires the boring data which includes the SPT blow counts, the friction angle, the water level, and the soil layer descriptions.
:::::''A'' = the areas of all the piles being considered


::Assume the vertical load carried by battered piles is the same as it would be if the pile were vertical. The properties of piles may be found in the Piling Section of the Bridge Manual.
:::::''M'' = the moment of the resultant about the neutral axis


:::'''Neutral Axis of Pile Group'''
:::::''c'' = distance from the neutral axis to the centerline of the pile being investigated


:::Locate the neutral axis of the pile group in the repetitive strip from the toe of the footing at the bottom of the footing.
:::::''I'' = the moment of inertia of the pile group
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
|-
|'''Additional Information'''
|-
|AASHTO 5.5.6.2
|}


:::'''Moment of Inertia of Pile Group'''
:::'''Stem Design'''


:::The moment of inertia of the pile group in the repetitive strip about the neutral axis of the section may be determined using the parallel axis theorem:
:::The vertical stem (the wall portion) of a cantilever retaining wall shall be designed as a cantilever supported at the base.


::::I = Σ(I<sub>A</sub>) + Σ(Ad<sup>2</sup>) where :
:::'''Footing Design'''


::::''I<sub>A</sub>'' = moment of inertia of a pile about its neutral axis
::::'''Toe'''
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
|-
|'''Additional Information'''
|-
|AASHTO 5.5.6.1
|}


::::''A'' = area of a pile
::::The toe of the base slab of a cantilever wall shall be designed as a cantilever supported by the wall. The critical section for bending moments shall be taken at the front face of the stem. The critical section for shear shall be taken at a distance d (d = effective depth) from the front face of the stem.


::::''d'' = distance from a pile's neutral axis to pile group's neutral axis
::::'''Heel'''
::::The top reinforcement in the rear projection (heel) of the base slab shall be designed to support the entire weight of the superimposed materials plus any tension load in the heel piles (neglect compression loads in the pile), unless a more exact method is used. The bottom reinforcement in the heel of the base slab shall be designed to support the maximum compression load in the pile neglecting the weight of the superimposed materials. The heel shall be designed as a cantilever supported by the wall. The critical sections for bending moments and shear shall be taken at the back face of the stem.


:::''I<sub>A</sub>'' may be neglected so the equation reduces to:
:::'''Shear Key Design'''
:::The shear key shall be designed as a cantilever supported at the bottom of the footing.


::::''I'' = Σ(Ad<sup>2</sup>)
====751.24.3.2.3 Counterfort Walls====


::'''Resistance To Sliding'''
'''Assumptions:'''


::Any frictional resistance to sliding shall be ignored, such as would occur between the bottom of the footing and the soil on a spread footing.
(1) Stability
The external stability of a counterfort retaining wall shall be determined in the same manner as described for cantilever retaining walls. Therefore refer to previous pages for the criteria for location of resultant, factor of safety for sliding and bearing pressures.
(2) Stem
[[image:751.24.3.2.3 counterfort.jpg|center|800px]]


::'''Friction or Bearing Piles With Batter (Case 1)'''
:<math>P = C_a \boldsymbol \gamma_s</math>


::Retaining walls using friction or bearing piles with batter should develop lateral strength (resistance to sliding) first from the batter component of the pile and second from the passive pressure against the shear key and the piles.
:where:
::''C<sub>a</sub>'' = coefficient of active earth pressure


::'''Friction or Bearing Piles Without Batter (Case 2)'''
::<math>\boldsymbol \gamma_s</math> = unit weigt of soil


::Retaining walls using friction or bearing piles without batter due to site constrictions should develop lateral strength first from the passive pressure against the shear key and second from the passive pressure against the pile below the bottom of footing. In this case, the shear key shall be placed at the front face of the footing.
Design the wall to support horizontal load from the earth pressure and the liveload surcharge (if applicable) as outlined on the previous pages and as designated in AASHTD Section 3.20, except that maximum horizontal loads shall be the calculated equivalent fluid pressure at 3/4  height of wall [(0.75 H)P] which shall be considered applied uniformly from the lower quarter point to the bottom of wall.


::'''Concrete Pedestal Piles or Drilled Shafts (Case 3)'''
In addition, vertical steel In the fill face of the bottom quarter of the wall shall be that required by the vertical cantilever wall with the equivalent fluid pressure of that (0.25 H) height.


::Retaining walls using concrete pedestal piles should develop lateral strength first from passive pressure against the shear key and second from passive pressure against the pile below the bottom of the footing. In this case, the shear key shall be placed at the front of the footing. Do not batter concrete pedestal piles.
Maximum concrete stress shall be assumed as the greater of the two thus obtained.
The application of these horizontal pressures shall be as follows:
[[image:751.24.3.2.3 counterfort wall.jpg|center|800px|thumb|<center>'''Counterfort Wall Section'''</center> <center>Moments are to be determined by analysis as a continuous beam. The counterforts are to be spaced so as to produce approximately equal positive and negative moments.</center>]]


[[image:751.24.3.2.2 cases.jpg|center|450px]]
(3)  Counterfort
Counterforts shall be designed as T-beams, of which the wall is the flange and the counterfort is the stem. For this reason the concrete stresses ane normally low and will not control.


::'''Resistance Due to Passive Pressure Against Pile'''
For the design of reinforcing steel in the back of the counterfort, the effective d shall be the perpendicular distance from the front face of the wall (at point that moment is considered), to center of reinforcing steel.


::The procedure below may be used to determine the passive pressure resistance developed in the soil against the piles. The procedure assumes that the piles develop a local failure plane.
[[image:751.24.3.2.3 moment.jpg|center|500px]]


:::''F'' = the lateral force due to passive pressure on pile
(4) Footing


:::<math>F = \frac{1}{2}\gamma_s C_P H^2 B </math> , where: <math> C_P = tan^2\Big[45 + \frac{\phi}{2}\Big]</math>
The footing of the counterfort walls shall be designed as a continuous beam of spans equal to the distance between the counterforts.


:::<math>\boldsymbol{\gamma_s}</math> = unit weight of soil
The rear projection or heel shall be designed to support the entire weight of the superimposed materials, unless a more exact method is used. Refer to AASHTD Section 5.5.6.


:::''H'' = depth of pile considered for lateral resistance (H<sub>max</sub>= 6B)
Divide footing (transversely) into four (4) equal sections for design footing pressures.


:::''C<sub>P</sub>'' = coefficient of active earth pressure
Counterfort walls on pile are very rare and are to be treated as special cases.  See Structural Project Manager.


:::''B'' = width of pile
(5)  Sign-Board type walls


:::<math>\boldsymbol{\phi}</math> = angle of internal friction of soil
The Sign-Board type of retaining walls are a special case of the counterfort retaining walls.  This type of wall is used where the soiI conditions are such that the footings must be placed a great distance below the finished ground line.  For this situation, the wall is discontinued approximately 12 in. below the finished ground line or below the frost line.


[[image:751.24.3.2.2 resistance passive.jpg|center|450px]]
Due to the large depth of the counterforts, it may be more economical to use a smaller number of counterforts than would otherwise be used.
All design assumptions that apply to counterfort walls will apply to sign-board walls with the exception of the application of horizontal forces for the stem (or wall design), and the footing design which shall be as follows:


::'''Resistance Due to Pile Batter'''
:'''Wall'''


::Use the horizontal component (due to pile batter) of the allowable pile load as the lateral resistance of the battered pile. (This presupposes that sufficient lateral movement of the wall can take place before failure to develop the ultimate strength of both elements.)
[[image:751.24.3.2.3 load.jpg|center|550px]]


[[image:751.24.3.2.2 12.jpg|center|125px]]
:'''Footing'''


:::''b'' = the amount of batter per 12 inches.
:The individual footings shall be designed transversely as cantilevers supported by the wall.  Refer to AASHTO Section 5.


:::<math> c = \sqrt{(12 in.)^2 + b^2}</math>
===751.24.3.3 Example 1: Spread Footing Cantilever Wall===


:::<math>P_{HBatter} = P_T \Big(\frac{b}{c}\Big)</math> (# of battered piles) where:
[[image:751.24.3.3.jpg|center|750px|thumb|<Center>'''Typical Section thru Wall</center><center>(Spread Footing)</center>''']]


:::''P<sub>HBatter</sub>'' = the horizontal force due to the battered piles
:f'<sub>c</sub> = 3,000 psi
:f<sub>y</sub> = 60,000 psi
:''φ'' = 24 in.
:''γ<sub>s</sub>'' = 120 pcf (unit wgt of soil)
:Allowable soil pressure = 2 tsf
:''γ<sub>c</sub>'' = 150 pcf (unit wgt of concr.)
:Retaining wall is located in Seismic Performance Category (SPC) B.
:A = 0.1 (A = seismic acceleration coefficient)


:::''P<sub>T</sub>'' = the allowable pile load
{| style="margin: 1em auto 1em auto"
|-
|<math>P_a = \frac{1}{2}\gamma_s C_a H^2</math>||width=50| ||<math>P_p = \frac{1}{2}\gamma_s C_p H_2^2 - H_1^2</math>
|}


::Maximum batter is 4" per 12".
'''Assumptions'''


::'''Resistance Due to Shear Keys'''
* Retaining wall is under an abutment or in a location where failure of the wall may affect the structural integrity of a bridge. Therefore, it must be designed for SPC B.


::A shear key may be needed if the passive pressure against the piles and the horizontal force due to batter is not sufficient to attain the factor of safety against sliding. The passive pressure against the shear key on a pile footing is found in the same manner as for spread footings.
* Design is for a unit length (1 ft.) of wall.


::'''Resistance to Overturning'''
* Sum moments about the toe at the bottom of the footing for overturning.


::The resisting and overturning moments shall be computed at the centerline of the toe pile at a distance of 6B (where B is the width of the pile) below the bottom of the footing. A maximum of 3 tons of tension on each heel pile may be assumed to resist overturning. Any effects of passive pressure, either on the shear key or on the piles, which resist overturning, shall be ignored.
*For Group Loads I-VI loading:
:* F.S. for overturning ≥ 2.0 for footings on soil.
:* F.S. for sliding ≥ 1.5.
* Resultant to be within middle 1/3 of footing.


[[image:751.24.3.2.2 resistance overturning.jpg|center|450px]]
* For earthquake loading:
:* F.S. for overturning ≥ 0.75(2.0) = 1.5.
:* F.S. for sliding ≥ 0.75(1.5) = 1.125.
:* Resultant to be within middle 1/2 of footing.


::'''Pile Properties'''
* Base of footing is below the frost line.


:::'''Location of Resultant'''
* Neglect top one foot of fill over toe when determining passive pressure and soil weight.
 
* Use of a shear key shifts the failure plane to "B" where resistance to sliding is provided by passive pressure against the shear key, friction of soil along failure plane "B" in front of the key, and friction between soil and concrete along the footing behind the key.


:::The location of the resultant shall be evaluated at the bottom of the footing and can be determined by the equation below:
* Soil cohesion along failure plane is neglected.


::::<math>e = \frac{\Sigma M}{\Sigma V}</math>  where:
* Footings are designed as cantilevers supported by the wall.
:* Critical sections for bending are at the front and back faces of the wall.
:* Critical sections for shear are at the back face of the wall for the heel and at a distance d (effective depth) from the front face for the toe.


::::e = the distance between the resultant and the neutral axis of the pile group
* Neglect soil weight above toe of footing in design of the toe.


::::''ΣM'' = the sum of the moments taken about the neutral axis of the pile group at the bottom of the footing
* The wall is designed as a cantilever supported by the footing.


::::''ΣV'' = the sum of the vertical loads used in calculating the moment
* Load factors for AASHTO Groups I - VI for design of concrete:
:* ''γ'' = 1.3.
:* ''β<sub>E</sub>'' = 1.3 for horizontal earth pressure on retaining walls.
:* ''β<sub>E</sub>'' = 1.0 for vertical earth pressure.


:::'''Pile Loads'''
* Load factor for earthquake loads = 1.0.


:::The loads on the pile can be determined as follows:
'''Lateral Pressures Without Earthquake'''


::::<math>P = \frac{\Sigma V}{A} \pm \frac{Mc}{I}</math> where:
:''C<sub>a</sub>'' = <math>\cos\delta\Bigg[\frac{\cos\delta - \sqrt{\cos^2\delta - \cos^2\phi}}{\cos\delta + \sqrt{\cos^2\delta - \cos^2\phi}}\Bigg]</math>


:::::''P'' = the force on the pile
:''C<sub>a</sub>'' = <math>\cos 18.435^\circ \Bigg[\frac{\cos\ 18.435^\circ - \sqrt{\cos^2\ 18.435^\circ - \cos^2\ 24^\circ }}{\cos\ 18.435^\circ  + \sqrt{\cos^2\ 18.435^\circ  - \cos^2\ 24^\circ }}\Bigg]</math> = 0.546


:::::''A'' = the areas of all the piles being considered
:<math>C_p = tan^2 \big( 45^\circ + \frac{\phi}{2}\big)  = tan^2 \big( 45^\circ + \frac{24^\circ}{2}\big) = 2.371</math>


:::::''M'' = the moment of the resultant about the neutral axis
:<math>P_A = \frac{1}{2}\big[0.120\frac{k}{ft^3}\big](1 ft)(0.546)(10.667 ft)^2 = 3.726k</math>


:::::''c'' = distance from the neutral axis to the centerline of the pile being investigated
:<math>P_P = \frac{1}{2}\big[0.120\frac{k}{ft^3}\big](1 ft)(2.371)\big[(5.0)^2 - (2.5)^2\big] = 2.668k</math>


:::::''I'' = the moment of inertia of the pile group
:<math>P_{AV} = P_A (sin \delta) = 3.726k (sin 18.435^\circ ) = 1.178k</math>
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"  
 
:<math>P_{AH} = P_A (cos \delta) = 3.726k (cos 18.435^\circ ) = 3.534k</math>
 
{| border="1" class="wikitable" style="margin: 1em auto 1em auto"
|+
!style="background:#BEBEBE" |Load !!style="background:#BEBEBE" |Area (ft<sup>2</sup>) !!style="background:#BEBEBE" |Force (k) = (Unit Wgt.)(Area) !!style="background:#BEBEBE" |Arm (ft.) !!style="background:#BEBEBE"|Moment (ft-k)
|-
|align="center"|(1)||align="center"| (0.5)(6.667ft)(2.222ft) = 7.407||align="center"| 0.889||align="center"| 7.278 ||align="center"|6.469
|-
|align="center"|(2)||align="center"| (6.667ft)(6.944ft) = 46.296||align="center"| 5.556||align="center"| 6.167||align="center"| 34.259
|-
|align="center"|(3) ||align="center"|(0.833ft)(8.000ft) + (0.5)(0.083ft)(8.000ft) = 7.000||align="center"|1.050||align="center"| 2.396||align="center"| 2.515
|-
|align="center"|(4) ||align="center"|(1.500ft)(9.500ft) = 14.250||align="center"| 2.138 ||align="center"|4.750 ||align="center"|10.153
|-
|align="center"|(5) ||align="center"|(2.500ft)(1.000ft) = 2.500||align="center"| 0.375||align="center"| 2.500||align="center"| 0.938
|-
|align="center"|(6) ||align="center"|(1.000ft)(1.917ft)+(0.5)(0.010ft)(1.000ft) = 1.922||align="center"|<u>0.231</u>||align="center"| 0.961||align="center"|<u>0.222</u>
|-
|-
|'''Additional Information'''
|align="center"|Σ ||align="center"| -  ||align="center"|ΣV = 10.239 ||align="center"| - ||align="center"|ΣM<sub>R</sub> = 54.556
|-
|align="center"|P<sub>AV</sub>||align="center"| -  ||align="center"|<u>1.178</u>||align="center"| 9.500 ||align="center"|<u>11.192</u>
|-
|align="center"|Σ resisting ||align="center"| - ||align="center"|ΣV = 11.417||align="center"| - ||align="center"| ΣM<sub>R</sub> = 65.748
|-
|align="center"|P<sub>AH</sub> ||align="center"| - ||align="center"|3.534 ||align="center"|3.556 ||align="center"|12.567
|-
|align="center"|P<sub>P</sub>||align="center"| -  ||align="center"|2.668 ||align="center"|1.389<sup>1</sup>||align="center"| -
|-
|-
|AASHTO 5.5.6.2
|colspan="5"|'''<sup>1</sup>''' The passive capacity at the shear key is ignored in overturning checks,since this capacity is considered in the factor of safety against sliding. It is assumed that a sliding and overturning failure will not occur simultaneously. The passive capacity at the shear key is developed only if the wall does slide.
|}
|}


:::'''Stem Design'''
[[image:751.24.3.3 passive.jpg|right|150px]]
<math>\bar{y} = \frac{H_1y^2 + \frac{2}{3}y^3}{H_2^2 - H_1^2} = \frac{(2.5 ft)(2.5 ft)^2 + \frac{2}{3}(2.5 ft)^3}{(5.0 ft)^2 - (2.5 ft)^2}</math> = 1.389 ft.


:::The vertical stem (the wall portion) of a cantilever retaining wall shall be designed as a cantilever supported at the base.
:'''Overturning'''


:::'''Footing Design'''
:F.S. = <math>\frac{M_R}{M_{OT}} = \frac{65.748(ft-k)}{12.567(ft-k)} = 5.232 \ge 2.0 </math> <u>o.k.</u>


::::'''Toe'''
:where: M<sub>OT</sub> = overturning moment; M<sub>R</sub> = resisting moment
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
|-
|'''Additional Information'''
|-
|AASHTO 5.5.6.1
|}


::::The toe of the base slab of a cantilever wall shall be designed as a cantilever supported by the wall. The critical section for bending moments shall be taken at the front face of the stem. The critical section for shear shall be taken at a distance d (d = effective depth) from the front face of the stem.
:'''Resultant Eccentricity'''


::::'''Heel'''
:<math>\bar{x} = \frac{(65.748 - 12.567)(ft-k)}{11.417k}</math> = 4.658 ft.
::::The top reinforcement in the rear projection (heel) of the base slab shall be designed to support the entire weight of the superimposed materials plus any tension load in the heel piles (neglect compression loads in the pile), unless a more exact method is used. The bottom reinforcement in the heel of the base slab shall be designed to support the maximum compression load in the pile neglecting the weight of the superimposed materials. The heel shall be designed as a cantilever supported by the wall. The critical sections for bending moments and shear shall be taken at the back face of the stem.


:::'''Shear Key Design'''
:<math>e = \frac{9.500 ft}{2} - 4.658 ft. = 0.092 ft.</math>
:::The shear key shall be designed as a cantilever supported at the bottom of the footing.
:<math>\frac{L}{6} =\frac{9.500 ft}{6} = 1.583 ft > e</math> <u>o.k.</u>


====751.24.3.2.3 Counterfort Walls====
:'''Sliding'''


'''Assumptions:'''
:Check if shear key is required for Group Loads I-VI:


(1) Stability
:F.S. = <math>\frac{\Sigma V(tan\phi_{s-c})}{P_{AH}} = \frac{11.042k(tan \frac{2}{3}(24^\circ)}{3.534k} </math>= 0.896 <u>no good - shear key req'd</u>
The external stability of a counterfort retaining wall shall be determined in the same manner as described for cantilever retaining walls. Therefore refer to previous pages for the criteria for location of resultant, factor of safety for sliding and bearing pressures.
(2) Stem
[[image:751.24.3.2.3 counterfort.jpg|center|800px]]


:<math>P = C_a \boldsymbol \gamma_s</math>
:where: ''φ<sub>s-c</sub>'' = angle of friction between soil and concrete = (2/3)''φ<sub>s-s</sub>''


:where:
:F.S. = <math>\frac{P_P + (\Sigma V) \Big(\frac{L_2}{L_1} tan \phi_{s-s}+\frac{L_3}{L_1} tan \phi_{s-c}\Big)}{P_{AH}}</math>
::''C<sub>a</sub>'' = coefficient of active earth pressure


::<math>\boldsymbol \gamma_s</math> = unit weigt of soil
:where: ''φ<sub>s-s</sub>''  = angle of internal friction of soil


Design the wall to support horizontal load from the earth pressure and the liveload surcharge (if applicable) as outlined on the previous pages and as designated in AASHTD Section 3.20, except that maximum horizontal loads shall be the calculated equivalent fluid pressure at 3/4 height of wall [(0.75 H)P] which shall be considered applied uniformly from the lower quarter point to the bottom of wall.
:F.S. = <math>\frac{2.668k + (11.417k) \Big[\Big(\frac{2 ft}{9.50 ft}\Big) tan 24^\circ + \Big(\frac{7.50 ft}{9.50 ft} tan \Big(\frac{2}{3}(24^\circ)\Big)\Big]}{3.534 k}</math> = 1.789 ≥ 1.5 <u>o.k.</u>


In addition, vertical steel In the fill face of the bottom quarter of the wall shall be that required by the vertical cantilever wall with the equivalent fluid pressure of that (0.25 H) height.
:'''Footing Pressure'''


Maximum concrete stress shall be assumed as the greater of the two thus obtained.
:<math>P = \frac{\Sigma V}{bL} \Big[1 \pm \frac{6e}{L}\Big]</math>
The application of these horizontal pressures shall be as follows:
[[image:751.24.3.2.3 counterfort wall.jpg|center|800px|thumb|<center>'''Counterfort Wall Section'''</center> <center>Moments are to be determined by analysis as a continuous beam.  The counterforts are to be spaced so as to produce approximately equal positive and negative moments.</center>]]


(3) Counterfort
:P<sub>H</sub> = pressure at heel <math>P_H = \frac{11.417 k}{(1 ft)9.50 ft} \Big[1 - \frac{6 (0.092 ft)}{9.50 ft}\Big]</math> = 1.132 k/ft<sup>2</sup>
Counterforts shall be designed as T-beams, of which the wall is the flange and the counterfort is the stem. For this reason the concrete stresses ane normally low and will not control.


For the design of reinforcing steel in the back of the counterfort, the effective d shall be the perpendicular distance from the front face of the wall (at point that moment is considered), to center of reinforcing steel.  
:P<sub>T</sub> = pressure at toe <math>P_T = \frac{11.417 k}{(1 ft)9.50 ft} \Big[1 + \frac{6 (0.092 ft)}{9.50 ft}\Big]</math> = 1.272 k/ft<sup>2</sup>


[[image:751.24.3.2.3 moment.jpg|center|500px]]
:Allowable pressure = 2 tons/ft<sup>2</sup> = 4 k/ft<sup>2</sup> ≥ 1.272 k/ft<sup>2</sup> <u>o.k.</u>


(4) Footing
'''Lateral Pressures With Earthquake'''


The footing of the counterfort walls shall be designed as a continuous beam of spans equal to the distance between the counterforts.
k<sub>h</sub> = 0.5A = 0.5 (0.1) = 0.05


The rear projection or heel shall be designed to support the entire weight of the superimposed materials, unless a more exact method is used. Refer to AASHTD Section 5.5.6.
k<sub>v</sub> = 0


Divide footing (transversely) into four (4) equal sections for design footing pressures.
<math>\theta = arctan \Big[\frac{k_h}{1 - k_v}\Big] = arctan \Big[\frac{0.05}{1 - 0}\Big] = 2.862^\circ</math>


Counterfort walls on pile are very rare and are to be treated as special cases.  See Structural Project Manager.
:'''Active Pressure on Psuedo-Wall'''


(5) Sign-Board type walls
:''δ'' = ''φ'' = 24° (''δ'' is the angle of friction between the soil and the wall. In this case, ''δ'' = ''φ'' = because the soil wedge considered is next to the soil above the footing.)


The Sign-Board type of retaining walls are a special case of the counterfort retaining walls.  This type of wall is used where the soiI conditions are such that the footings must be placed a great distance below the finished ground line.  For this situation, the wall is discontinued approximately 12 in. below the finished ground line or below the frost line.
:''i'' = 18.435°


Due to the large depth of the counterforts, it may be more economical to use a smaller number of counterforts than would otherwise be used.
:''β'' = 0°
All design assumptions that apply to counterfort walls will apply to sign-board walls with the exception of the application of horizontal forces for the stem (or wall design), and the footing design which shall be as follows:


:'''Wall'''
:<math>K_{AE} = \frac{cos^2(\phi - \theta - \beta)}{cos \theta cos^2 \beta cos(\delta + \beta + \theta)\Big(1 + \sqrt\frac{sin(\phi + \delta) sin (\phi - \theta - i)}{cos (\delta + \beta + \theta) cos(I - \beta)}\Big)^2}</math>


[[image:751.24.3.2.3 load.jpg|center|550px]]
:<math>K_{AE} = \frac{cos^2(24^\circ - 2.862^\circ - 0^\circ)}{cos (2.862^\circ) cos^2 (0^\circ) cos(24^\circ + 0^\circ + 2.862^\circ)\Big(1 + \sqrt\frac{sin(24^\circ + 24^\circ) sin (24^\circ - 2.862^\circ - 18.435^\circ)}{cos (24^\circ + 0^\circ + 2.862^\circ) cos(18.435^\circ - 0^\circ)}\Big)^2}</math>


:'''Footing'''
:K<sub>AE</sub> = 0.674


:The individual footings shall be designed transversely as cantilevers supported by the wall.  Refer to AASHTO Section 5.
:P<sub>AE</sub> = ½''γ<sub>s</sub>H<sup>2</sup>''(1 − ''k<sub>v</sub>'')''K<sub>AE</sub>''


===751.24.3.3 Example 1:  Spread Footing Cantilever Wall===
:P<sub>AE</sub> = ½[0.120 k/ft<sup>3</sup>](10.667 ft)<sup>2</sup>(1 ft.)(1 - 0)(0.674) = 4.602k


[[image:751.24.3.3.jpg|center|750px|thumb|<Center>'''Typical Section thru Wall</center><center>(Spread Footing)</center>''']]
:P<sub>AEV</sub> = P<sub>AE</sub>(sin''δ'') = 4.602k(sin24°) = 1.872k


:f'<sub>c</sub> = 3,000 psi
:P<sub>AEH</sub> = P<sub>AE</sub>(cos''δ'') = 4.602k(cos 24°) = 4.204k
:f<sub>y</sub> = 60,000 psi
:''φ'' = 24 in.
:''γ<sub>s</sub>'' = 120 pcf (unit wgt of soil)
:Allowable soil pressure = 2 tsf
:''γ<sub>c</sub>'' = 150 pcf (unit wgt of concr.)
:Retaining wall is located in Seismic Performance Category (SPC) B.
:A = 0.1 (A = seismic acceleration coefficient)


{| style="margin: 1em auto 1em auto"
:P'<sub>AH</sub> = P<sub>AEH</sub> − P<sub>AH</sub> = 4.204k − 3.534k = 0.670k
|-
|<math>P_a = \frac{1}{2}\gamma_s C_a H^2</math>||width=50| ||<math>P_p = \frac{1}{2}\gamma_s C_p H_2^2 - H_1^2</math>
|}


'''Assumptions'''
:P'<sub>AV</sub> = P<sub>AEV</sub> − P<sub>AV</sub> = 1.872k − 1.178k = 0.694k


* Retaining wall is under an abutment or in a location where failure of the wall may affect the structural integrity of a bridge. Therefore, it must be designed for SPC B.
:where: P'<sub>AH</sub> and P'<sub>AV</sub> are the seismic components of the active force.


* Design is for a unit length (1 ft.) of wall.
:'''Passive Pressure on Shear Key'''


* Sum moments about the toe at the bottom of the footing for overturning.
:''δ'' = ''φ'' = 24° (''δ'' = ''φ'' because the soil wedge considered is assumed to form in front of the footing.)


*For Group Loads I-VI loading:
:''i'' = 0
:* F.S. for overturning ≥ 2.0 for footings on soil.
:* F.S. for sliding ≥ 1.5.
* Resultant to be within middle 1/3 of footing.


* For earthquake loading:
:''β'' = 0
:* F.S. for overturning ≥ 0.75(2.0) = 1.5.
:* F.S. for sliding ≥ 0.75(1.5) = 1.125.
:* Resultant to be within middle 1/2 of footing.


* Base of footing is below the frost line.
:<math>K_{PE} = \frac{cos^2(\phi - \theta + \beta)}{cos \theta cos^2 \beta cos(\delta - \beta + \theta)\Big(1 - \sqrt\frac{sin(\phi - \delta) sin (\phi - \theta + i)}{cos (\delta - \beta + \theta) cos(I - \beta)}\Big)^2}</math>


* Neglect top one foot of fill over toe when determining passive pressure and soil weight.
:<math>K_{PE} = \frac{cos^2(24^\circ - 2.862^\circ + 0^\circ)}{cos (2.862^\circ) cos^2 (0^\circ) cos(24^\circ - 0^\circ + 2.862^\circ)\Big(1 - \sqrt\frac{sin(24^\circ - 24^\circ) sin (24^\circ - 2.862^\circ + 0^\circ)}{cos (24^\circ - 0^\circ + 2.862^\circ) cos(0^\circ - 0^\circ)}\Big)^2}</math>


* Use of a shear key shifts the failure plane to "B" where resistance to sliding is provided by passive pressure against the shear key, friction of soil along failure plane "B" in front of the key, and friction between soil and concrete along the footing behind the key.
:K<sub>PE</sub> = 0.976


* Soil cohesion along failure plane is neglected.
:P<sub>PE</sub> = ½''γ<sub>s</sub>H<sup>2</sup>''(1 − ''k<sub>v</sub>'')''K<sub>PE</sub>''


* Footings are designed as cantilevers supported by the wall.
:P<sub>PE</sub> =  ½[0.120 k/ft<sup>3</sup>][(5.0 ft)<sup>2</sup> - (2.5 ft<sup>2</sup>)](1 ft.)(1 - 0)(0.976) = 1.098k
:* Critical sections for bending are at the front and back faces of the wall.
:* Critical sections for shear are at the back face of the wall for the heel and at a distance d (effective depth) from the front face for the toe.


* Neglect soil weight above toe of footing in design of the toe.
{| border="1" class="wikitable" style="margin: 1em auto 1em auto"
 
|+  
* The wall is designed as a cantilever supported by the footing.
!style="background:#BEBEBE" |Load !!style="background:#BEBEBE" |Force (k) !!style="background:#BEBEBE" |Arm (ft) !!style="background:#BEBEBE" |Moment (ft-k)
 
* Load factors for AASHTO Groups I - VI for design of concrete:
:* ''γ'' = 1.3.
:* ''β<sub>E</sub>'' = 1.3 for horizontal earth pressure on retaining walls.
:* ''β<sub>E</sub>'' = 1.0 for vertical earth pressure.
 
* Load factor for earthquake loads = 1.0.
 
'''Lateral Pressures Without Earthquake'''
 
:''C<sub>a</sub>'' = <math>\cos\delta\Bigg[\frac{\cos\delta - \sqrt{\cos^2\delta - \cos^2\phi}}{\cos\delta + \sqrt{\cos^2\delta - \cos^2\phi}}\Bigg]</math>
 
:''C<sub>a</sub>'' = <math>\cos 18.435^\circ \Bigg[\frac{\cos\ 18.435^\circ - \sqrt{\cos^2\ 18.435^\circ - \cos^2\ 24^\circ }}{\cos\ 18.435^\circ  + \sqrt{\cos^2\ 18.435^\circ  - \cos^2\ 24^\circ }}\Bigg]</math> = 0.546
 
:<math>C_p = tan^2 \big( 45^\circ + \frac{\phi}{2}\big)  = tan^2 \big( 45^\circ + \frac{24^\circ}{2}\big) = 2.371</math>
 
:<math>P_A = \frac{1}{2}\big[0.120\frac{k}{ft^3}\big](1 ft)(0.546)(10.667 ft)^2 = 3.726k</math>
 
:<math>P_P = \frac{1}{2}\big[0.120\frac{k}{ft^3}\big](1 ft)(2.371)\big[(5.0)^2 - (2.5)^2\big] = 2.668k</math>
 
:<math>P_{AV} = P_A (sin \delta) = 3.726k (sin 18.435^\circ ) = 1.178k</math>
 
:<math>P_{AH} = P_A (cos \delta) = 3.726k (cos 18.435^\circ ) = 3.534k</math>
 
{| border="1" class="wikitable" style="margin: 1em auto 1em auto"
|+  
!style="background:#BEBEBE" |Load !!style="background:#BEBEBE" |Area (ft<sup>2</sup>) !!style="background:#BEBEBE" |Force (k) = (Unit Wgt.)(Area) !!style="background:#BEBEBE" |Arm (ft.) !!style="background:#BEBEBE"|Moment (ft-k)
|-
|-
|align="center"|(1)||align="center"| (0.5)(6.667ft)(2.222ft) = 7.407||align="center"| 0.889||align="center"| 7.278 ||align="center"|6.469
|align="center"|Σ (1) thru (6) ||align="center"| 10.239||align="center"| - ||align="center"| 54.556
|-
|-
|align="center"|(2)||align="center"| (6.667ft)(6.944ft) = 46.296||align="center"| 5.556||align="center"| 6.167||align="center"| 34.259
|align="center"|P<sub>AV</sub>||align="center"| 1.178 ||align="center"|9.500||align="center"| 11.192
|-
|-
|align="center"|(3) ||align="center"|(0.833ft)(8.000ft) + (0.5)(0.083ft)(8.000ft) = 7.000||align="center"|1.050||align="center"| 2.396||align="center"| 2.515
|align="center"|P'<sub>AV</sub> ||align="center"|0.694 ||align="center"|9.500||align="center"| 6.593
|-
|-
|align="center"|(4) ||align="center"|(1.500ft)(9.500ft) = 14.250||align="center"| 2.138 ||align="center"|4.750 ||align="center"|10.153
|align="center"|Σ<sub>resisting</sub> ||align="center"|ΣV = 12.111 ||align="center"| - ||align="center"|ΣM<sub>R</sub> = 72.341
|-
|-
|align="center"|(5) ||align="center"|(2.500ft)(1.000ft) = 2.500||align="center"| 0.375||align="center"| 2.500||align="center"| 0.938
|align="center"|P<sub>AH</sub> ||align="center"|3.534 ||align="center"|3.556 ||align="center"|12.567
|-
|-
|align="center"|(6) ||align="center"|(1.000ft)(1.917ft)+(0.5)(0.010ft)(1.000ft) = 1.922||align="center"|<u>0.231</u>||align="center"| 0.961||align="center"|<u>0.222</u>
|align="center"|P'<sub>AH</sub> ||align="center"|0.670||align="center"| 6.400<sup>a</sup>||align="center"| 4.288
|-
|-
|align="center"|Σ ||align="center"| ||align="center"|ΣV = 10.239 ||align="center"| - ||align="center"|ΣM<sub>R</sub> = 54.556
|align="center"|P<sub>PEV</sub> ||align="center"|0.447<sup>b</sup>||align="center"| 0.000||align="center"| 0.000
|-
|-
|align="center"|P<sub>AV</sub>||align="center"| -  ||align="center"|<u>1.178</u>||align="center"| 9.500 ||align="center"|<u>11.192</u>
|align="center"|P<sub>PEH</sub> ||align="center"|1.003<sup>b</sup> ||align="center"|1.389<sup>c</sup>||align="center"| <u>0.000</u>
|-
|-
|align="center"|Σ resisting ||align="center"| - ||align="center"|ΣV = 11.417||align="center"| - ||align="center"| ΣM<sub>R</sub> = 65.748
|align="center"| - ||align="center"| - ||align="center"| - ||align="center"|ΣM<sub>OT</sub> = 16.855
|-
|-
|align="center"|P<sub>AH</sub> ||align="center"| - ||align="center"|3.534 ||align="center"|3.556 ||align="center"|12.567
|colspan="4"|<sup>'''a'''</sup> P'<sub>AH</sub> acts at 0.6H of the wedge face (1992 AASHTO Div. IA Commentary).
|-
|-
|align="center"|P<sub>P</sub>||align="center"| -  ||align="center"|2.668 ||align="center"|1.389<sup>1</sup>||align="center"| -
|colspan="4"|<sup>'''b'''</sup> P<sub>PEH</sub> and P<sub>PEH</sub> are the components of P<sub>PE</sub> with respect to ''δ'' (the friction angle). P<sub>PE</sub> does not contribute to overturning.
|-
|-
|colspan="5"|'''<sup>1</sup>''' The passive capacity at the shear key is ignored in overturning checks,since this capacity is considered in the factor of safety against sliding. It is assumed that a sliding and overturning failure will not occur simultaneously. The passive capacity at the shear key is developed only if the wall does slide.
|colspan="4"|<sup>'''c'''</sup> The line of action of P<sub>PEH</sub> can be located as was done for P<sub>P</sub>.
|}
|}


[[image:751.24.3.3 passive.jpg|right|150px]]
:'''Overturning'''
<math>\bar{y} = \frac{H_1y^2 + \frac{2}{3}y^3}{H_2^2 - H_1^2} = \frac{(2.5 ft)(2.5 ft)^2 + \frac{2}{3}(2.5 ft)^3}{(5.0 ft)^2 - (2.5 ft)^2}</math> = 1.389 ft.


:'''Overturning'''
:<math>F.S._{OT} = \frac{72.341ft-k}{16.855ft-k} = 4.292 > 1.5</math> <u>o.k.</u>


:F.S. = <math>\frac{M_R}{M_{OT}} = \frac{65.748(ft-k)}{12.567(ft-k)} = 5.232 \ge 2.0 </math> <u>o.k.</u>
:'''Resultant Eccentricity'''


:where: M<sub>OT</sub> = overturning moment; M<sub>R</sub> = resisting moment
:<math>\bar{x} = \frac{72.341ft-k - 16.855ft-k}{12.111k} = 4.581 ft.</math>


:'''Resultant Eccentricity'''
:<math>e = \frac{9.5 ft.}{2}\ - 4.581 ft. = 0.169 ft.</math>


:<math>\bar{x} = \frac{(65.748 - 12.567)(ft-k)}{11.417k}</math> = 4.658 ft.
:<math>\frac{L}{4} = \frac{9.5 ft.}{4} = 2.375 ft. > e</math> <u>o.k.</u>


:<math>e = \frac{9.500 ft}{2} - 4.658 ft. = 0.092 ft.</math>
:<math>\frac{L}{6} =\frac{9.500 ft}{6} = 1.583 ft > e</math> <u>o.k.</u>


:'''Sliding'''
:'''Sliding'''


:Check if shear key is required for Group Loads I-VI:
:<math>F.S. = \frac{1.003k + 12.111k \Big[(\frac{2}{9.5})tan 24^\circ + (\frac{7.5}{9.5}) tan \Big( \frac{2}{3}(24^\circ) \Big)\Big]}{4.204 k} = 1.161 > 1.125</math> <u>o.k.</u>


:F.S. = <math>\frac{\Sigma V(tan\phi_{s-c})}{P_{AH}} = \frac{11.042k(tan \frac{2}{3}(24^\circ)}{3.534k} </math>= 0.896 <u>no good - shear key req'd</u>


:where: ''φ<sub>s-c</sub>'' = angle of friction between soil and concrete = (2/3)''φ<sub>s-s</sub>''  
:'''Footing Pressure'''


:F.S. = <math>\frac{P_P + (\Sigma V) \Big(\frac{L_2}{L_1} tan \phi_{s-s}+\frac{L_3}{L_1} tan \phi_{s-c}\Big)}{P_{AH}}</math>
:for e ≤ L/6:


:where: ''φ<sub>s-s</sub>''  = angle of internal friction of soil
:<math>P = \frac{\Sigma V}{bL} \Big[ 1 \pm \frac{6e}{L}\Big] </math>


:F.S. = <math>\frac{2.668k + (11.417k) \Big[\Big(\frac{2 ft}{9.50 ft}\Big) tan 24^\circ + \Big(\frac{7.50 ft}{9.50 ft} tan \Big(\frac{2}{3}(24^\circ)\Big)\Big]}{3.534 k}</math> = 1.789 ≥ 1.5  <u>o.k.</u>
:<math>P_H = pressure\ at\ heel\ P_H = \frac{12.111 k}{(1 ft.)9.50 ft.} \Big[1 - \frac{6(0.169 ft.)}{9.50 ft}\Big]</math> = 1.139 k/ft<sup>2</sup>


:'''Footing Pressure'''
:<math>P_TH = pressure\ at\ toe\ P_T = \frac{12.111 k}{(1 ft.)9.50 ft.} \Big[1 + \frac{6(0.169 ft.)}{9.50 ft}\Big]</math> = 1.411 k/ft<sup>2</sup>


:<math>P = \frac{\Sigma V}{bL} \Big[1 \pm \frac{6e}{L}\Big]</math>
:Allowable soil pressure for earthquake = 2 (allowable soil pressure)


:P<sub>H</sub> = pressure at heel <math>P_H = \frac{11.417 k}{(1 ft)9.50 ft} \Big[1 - \frac{6 (0.092 ft)}{9.50 ft}\Big]</math> = 1.132 k/ft<sup>2</sup>
:(2)[4 k/ft<sup>2</sup>] = 8 k/ft<sup>2</sup> > 1.411 k/ft<sup>2</sup> <u>o.k.</u>


:P<sub>T</sub> = pressure at toe <math>P_T = \frac{11.417 k}{(1 ft)9.50 ft} \Big[1 + \frac{6 (0.092 ft)}{9.50 ft}\Big]</math> = 1.272 k/ft<sup>2</sup>
'''Reinforcement-Stem'''


:Allowable pressure = 2 tons/ft<sup>2</sup> = 4 k/ft<sup>2</sup> ≥ 1.272 k/ft<sup>2</sup> <u>o.k.</u>
[[image:751.24.3.3 reinforcement stem.jpg|center|200px]]


'''Lateral Pressures With Earthquake'''
d = 11" - 2" - (1/2)(0.5") = 8.75"


k<sub>h</sub> = 0.5A = 0.5 (0.1) = 0.05
b = 12"


k<sub>v</sub> = 0
f'<sub>c</sub> = 3,000 psi


<math>\theta = arctan \Big[\frac{k_h}{1 - k_v}\Big] = arctan \Big[\frac{0.05}{1 - 0}\Big] = 2.862^\circ</math>
:'''Without Earthquake'''


:'''Active Pressure on Psuedo-Wall'''
:P<sub>AH</sub> = ½ [0.120 k/ft<sup>3</sup>](0.546)(6.944 ft.)<sup>2</sup>(1 ft.)(cos 18.435°) = 1.499k


:''δ'' = ''φ'' = 24° (''δ'' is the angle of friction between the soil and the wall. In this case, ''δ'' = ''φ'' = because the soil wedge considered is next to the soil above the footing.)
:''γ'' = 1.3


:''i'' = 18.435°
:''β<sub>E</sub>'' = 1.3 (active lateral earth pressure)


:''β'' =
:M<sub>u</sub> = (1.3)(1.3)(1.499k)(2.315ft) = 5.865 (ft-k)


:<math>K_{AE} = \frac{cos^2(\phi - \theta - \beta)}{cos \theta cos^2 \beta cos(\delta + \beta + \theta)\Big(1 + \sqrt\frac{sin(\phi + \delta) sin (\phi - \theta - i)}{cos (\delta + \beta + \theta) cos(I - \beta)}\Big)^2}</math>
:'''With Earthquake'''


:<math>K_{AE} = \frac{cos^2(24^\circ - 2.862^\circ - 0^\circ)}{cos (2.862^\circ) cos^2 (0^\circ) cos(24^\circ + 0^\circ + 2.862^\circ)\Big(1 + \sqrt\frac{sin(24^\circ + 24^\circ) sin (24^\circ - 2.862^\circ - 18.435^\circ)}{cos (24^\circ + 0^\circ + 2.862^\circ) cos(18.435^\circ - 0^\circ)}\Big)^2}</math>
:k<sub>h</sub> = 0.05


:K<sub>AE</sub> = 0.674
:k<sub>v</sub> = 0
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
|-
|'''Additional Information'''
|-
|1992 AASHTO Div. IA Commentary
|}


:P<sub>AE</sub> = ½''γ<sub>s</sub>H<sup>2</sup>''(1 − ''k<sub>v</sub>'')''K<sub>AE</sub>''
:''θ'' = 2.862°


:P<sub>AE</sub> = ½[0.120 k/ft<sup>3</sup>](10.667 ft)<sup>2</sup>(1 ft.)(1 - 0)(0.674) = 4.602k
:''δ'' = ''φ''/2 = 24°/2 = 12° for angle of friction between soil and wall. This criteria is used only for seismic loading if the angle of friction is not known.


:P<sub>AEV</sub> = P<sub>AE</sub>(sin''δ'') = 4.602k(sin24°) = 1.872k
:''φ'' = 24°


:P<sub>AEH</sub> = P<sub>AE</sub>(cos''δ'') = 4.602k(cos 24°) = 4.204k
:''i'' = 18.435°


:P'<sub>AH</sub> = P<sub>AEH</sub> − P<sub>AH</sub> = 4.204k − 3.534k = 0.670k
:''β'' =


:P'<sub>AV</sub> = P<sub>AEV</sub> − P<sub>AV</sub> = 1.872k − 1.178k = 0.694k
:K<sub>AE</sub> = 0.654


:where: P'<sub>AH</sub> and P'<sub>AV</sub> are the seismic components of the active force.
:P<sub>AEH</sub> = 1/2 '<sub>s''</sub>K<sub>AE</sub>H<sup>2</sup>cos''δ''


:'''Passive Pressure on Shear Key'''
:P<sub>AEH</sub> = 1/2 [0.120k/ft](0.654)(6.944 ft.)<sup>2</sup>(1 ft.) cos(12°) = 1.851k


:''δ'' = ''φ'' = 24° (''δ'' = ''φ'' because the soil wedge considered is assumed to form in front of the footing.)
:M<sub>u</sub> = (1.499k)(2.315 ft.) + (1.851k − 1.499k)(0.6(6.944 ft.)) = 4.936(ft−k)


:''i'' = 0
:The moment without earthquake controls:


:''β'' = 0
:<math>R_n = \frac{M_u}{\phi bd^2} = \frac{5.865(ft-k)}{0.9(1 ft.)(8.75 in.)^2}\Big(1000 \frac{lb}{k}\Big)</math> = 85.116 psi
 
:''ρ'' = <math>\frac{0.85f'_c}{f_y} \Big[1 - \sqrt{1 - \frac{2R_n}{0.85f'_c}}\Big]</math>
 
:''ρ'' = <math>\frac{0.85 (3.000 psi}{60,000 psi} \Bigg[1 - \sqrt{1 - \frac{2 (85.116 psi}{0.85 (3000 psi)}}\Bigg]</math> = 0.00144
 
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
|-
|'''Additional Information'''
|-
|AASHTO 8.17.1.1 & 8.15.2.1.1
|}
 
:''ρ<sub>min</sub>'' = <math> 1.7 \Bigg[\frac{h}{d}\Bigg]^2 \frac{\sqrt{f'_c}}{f_y} = 1.7 \Bigg[\frac{11 in.}{8.75 in.}^2 \frac{\sqrt{3000 psi}}{60,000 psi}\Bigg]</math> = 0.00245


:<math>K_{PE} = \frac{cos^2(\phi - \theta + \beta)}{cos \theta cos^2 \beta cos(\delta - \beta + \theta)\Big(1 - \sqrt\frac{sin(\phi - \delta) sin (\phi - \theta + i)}{cos (\delta - \beta + \theta) cos(I - \beta)}\Big)^2}</math>
:Use ''ρ'' = 4/3 ''ρ'' = 4/3 (0.00144) = 0.00192


:<math>K_{PE} = \frac{cos^2(24^\circ - 2.862^\circ + 0^\circ)}{cos (2.862^\circ) cos^2 (0^\circ) cos(24^\circ - 0^\circ + 2.862^\circ)\Big(1 - \sqrt\frac{sin(24^\circ - 24^\circ) sin (24^\circ - 2.862^\circ + 0^\circ)}{cos (24^\circ - 0^\circ + 2.862^\circ) cos(0^\circ - 0^\circ)}\Big)^2}</math>
:''A<sub>S<sub>Req</sub></sub>'' = ''ρbd'' = 0.00192 (12 in.)(8.75 in.) = 0.202 in.<sup>2</sup>/ft


:K<sub>PE</sub> = 0.976
:One #4 bar has A<sub>S</sub> = 0.196 in<sup>2</sup>


:P<sub>PE</sub> = ½''γ<sub>s</sub>H<sup>2</sup>''(1 − ''k<sub>v</sub>'')''K<sub>PE</sub>''
:<math>\frac{s}{0.196 in.^2} = \frac{12 in.}{0.202 in.^2}</math>


:P<sub>PE</sub> = ½[0.120 k/ft<sup>3</sup>][(5.0 ft)<sup>2</sup> - (2.5 ft<sup>2</sup>)](1 ft.)(1 - 0)(0.976) = 1.098k
:''s'' = 11.64 in.


{| border="1" class="wikitable" style="margin: 1em auto 1em auto"
:<u>Use #4's @ 10" cts.</u>
|+
!style="background:#BEBEBE" |Load !!style="background:#BEBEBE" |Force (k) !!style="background:#BEBEBE" |Arm (ft) !!style="background:#BEBEBE" |Moment (ft-k)
|-
|align="center"|Σ (1) thru (6) ||align="center"| 10.239||align="center"| - ||align="center"| 54.556
|-
|align="center"|P<sub>AV</sub>||align="center"| 1.178 ||align="center"|9.500||align="center"| 11.192
|-
|align="center"|P'<sub>AV</sub> ||align="center"|0.694 ||align="center"|9.500||align="center"| 6.593
|-
|align="center"|Σ<sub>resisting</sub> ||align="center"|ΣV = 12.111 ||align="center"| - ||align="center"|ΣM<sub>R</sub> = 72.341
|-
|align="center"|P<sub>AH</sub> ||align="center"|3.534 ||align="center"|3.556 ||align="center"|12.567
|-
|align="center"|P'<sub>AH</sub> ||align="center"|0.670||align="center"| 6.400<sup>a</sup>||align="center"| 4.288
|-
|align="center"|P<sub>PEV</sub> ||align="center"|0.447<sup>b</sup>||align="center"| 0.000||align="center"| 0.000
|-
|align="center"|P<sub>PEH</sub> ||align="center"|1.003<sup>b</sup> ||align="center"|1.389<sup>c</sup>||align="center"| <u>0.000</u>
|-
|align="center"| - ||align="center"| - ||align="center"| - ||align="center"|ΣM<sub>OT</sub> = 16.855
|-
|colspan="4"|<sup>'''a'''</sup> P'<sub>AH</sub> acts at 0.6H of the wedge face (1992 AASHTO Div. IA Commentary).
|-
|colspan="4"|<sup>'''b'''</sup> P<sub>PEH</sub> and P<sub>PEH</sub> are the components of P<sub>PE</sub> with respect to ''δ'' (the friction angle). P<sub>PE</sub> does not contribute to overturning.
|-
|colspan="4"|<sup>'''c'''</sup> The line of action of P<sub>PEH</sub> can be located as was done for P<sub>P</sub>.
|}


:'''Overturning'''
:'''Check Shear'''


:<math>F.S._{OT} = \frac{72.341ft-k}{16.855ft-k} = 4.292 > 1.5</math> <u>o.k.</u>
:V<sub>u</sub> ≥ ''φ'' V<sub>n</sub>


:'''Resultant Eccentricity'''
::'''Without Earthquake'''


:<math>\bar{x} = \frac{72.341ft-k - 16.855ft-k}{12.111k} = 4.581 ft.</math>
::V<sub>u,</sub> = (1.3)(1.3)(1.499k) = 2.533k


:<math>e = \frac{9.5 ft.}{2}\ - 4.581 ft. = 0.169 ft.</math>
::'''With Earthquake'''


:<math>\frac{L}{4} = \frac{9.5 ft.}{4} = 2.375 ft. > e</math> <u>o.k.</u>
::V<sub>u</sub> = 1.851k


:The shear force without earthquake controls.


:'''Sliding'''
:<math>\frac{\nu_u}{\phi} = \frac{2.533k}{0.85(12 in.)(8.75 in.)} (1000 lb/k)</math> = 28.4 psi


:<math>F.S. = \frac{1.003k + 12.111k \Big[(\frac{2}{9.5})tan 24^\circ + (\frac{7.5}{9.5}) tan \Big( \frac{2}{3}(24^\circ) \Big)\Big]}{4.204 k} = 1.161 > 1.125</math> <u>o.k.</u>
:<math>\nu_c = 2 \sqrt{3,000 psi}</math> = 109.5 psi > 28.4 psi <u>o.k.</u>


'''Reinforcement-Footing-Heel'''


:'''Footing Pressure'''
[[image:751.24.3.3 heel.jpg|center|250px]]


:for e ≤ L/6:
Note: Earthquake will not control and will not be checked.


:<math>P = \frac{\Sigma V}{bL} \Big[ 1 \pm \frac{6e}{L}\Big] </math>
''β<sub>E</sub>'' = 1.0 (vertical earth pressure)


:<math>P_H = pressure\ at\ heel\ P_H = \frac{12.111 k}{(1 ft.)9.50 ft.} \Big[1 - \frac{6(0.169 ft.)}{9.50 ft}\Big]</math> = 1.139 k/ft<sup>2</sup>
d = 18" - 3" - (1/2)(0.750") = 14.625"


:<math>P_TH = pressure\ at\ toe\ P_T = \frac{12.111 k}{(1 ft.)9.50 ft.} \Big[1 + \frac{6(0.169 ft.)}{9.50 ft}\Big]</math> = 1.411 k/ft<sup>2</sup>
b = 12"


:Allowable soil pressure for earthquake = 2 (allowable soil pressure)
''f'<sub>c</sub>'' = 3,000 psi


:(2)[4 k/ft<sup>2</sup>] = 8 k/ft<sup>2</sup> > 1.411 k/ft<sup>2</sup> <u>o.k.</u>
''M<sub>u</sub>'' = 1.3 [(5.556k + 1.500k)(3.333ft) + 0.889k(4.444ft) + 1.178k(6.667ft)]


'''Reinforcement-Stem'''
''M<sub>u</sub>'' = 45.919(ft−k)


[[image:751.24.3.3 reinforcement stem.jpg|center|200px]]
<math>R_n = \frac{45.919(ft-k)}{0.9(1 ft.)(14.625 in.)^2}(1000\frac{lb}{k})</math> = 238.5 psi


d = 11" - 2" - (1/2)(0.5") = 8.75"
''ρ'' = <math>\frac{0.85(3000)psi}{60,000 psi} \Bigg[ 1 - \sqrt{1 - \frac{2(238.5 psi)}{0.85(3000psi)}}\Bigg]</math> = 0.00418


b = 12"
''ρ<sub>min</sub>'' = <math> 1.7 \Big[\frac{18 in.}{14.625 in.}\Big]^2 \frac{\sqrt{3000 psi}}{60,000 psi}</math> = 0.00235


f'<sub>c</sub> = 3,000 psi
''A<sub>S<sub>Req</sub></sub>'' = 0.00418 (12 in.) (14.625 in.) = 0.734 in<sup>2</sup>/ft.


:'''Without Earthquake'''


:P<sub>AH</sub> = ½ [0.120 k/ft<sup>3</sup>](0.546)(6.944 ft.)<sup>2</sup>(1 ft.)(cos 18.435°) = 1.499k
<u>Use #6's @ 7" cts.</u>


:''γ'' = 1.3
:'''Check Shear'''


:''β<sub>E</sub>'' = 1.3 (active lateral earth pressure)
:Shear shall be checked at back face of stem.


:M<sub>u</sub> = (1.3)(1.3)(1.499k)(2.315ft) = 5.865 (ft-k)
:''V<sub>u</sub>'' = 1.3 (5.556k + 1.500k + 0.889k + 1.178k) = 11.860k


:'''With Earthquake'''
:<math>\frac{\nu_u}{\phi} = \frac{11.860k}{0.85(12 in.)(14.625 in.)}(1000 \frac{lb}{k} ) = 79.5 psi < 2 \sqrt{3,000 psi}</math> = 109.5 psi  o.k.


:k<sub>h</sub> = 0.05
'''Reinforcement-Footing-Toe'''


:k<sub>v</sub> = 0
[[image:751.24.3.3. toe.jpg|center|350px]]
{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
|-
|'''Additional Information'''
|-
|1992 AASHTO Div. IA Commentary
|}


:''θ'' = 2.862°
d = 18" - 4" = 14"


:''δ'' = ''φ''/2 = 24°/2 = 12° for angle of friction between soil and wall. This criteria is used only for seismic loading if the angle of friction is not known.
b = 12"


:''φ'' = 24°
:'''Without Earthquake'''


:''i'' = 18.435°
::'''Apply Load Factors'''


:''β'' =
::load 4 (weight) = 0.431k(1.3)(1.0) = 0.560k


:K<sub>AE</sub> = 0.654
::''β<sub>E</sub>'' = 1.3 for lateral earth pressure for retaining walls.


:P<sub>AEH</sub> = 1/2 ''γ<sub>s''</sub>K<sub>AE</sub>H<sup>2</sup>cos''δ''
::''β<sub>E</sub>'' = 1.0 for vertical earth pressure.


:P<sub>AEH</sub> = 1/2 [0.120k/ft](0.654)(6.944 ft.)<sup>2</sup>(1 ft.) cos(12°) = 1.851k
::''ΣM<sub>OT</sub>'' = 12.567(ft−k)(1.3)(1.3) = 21.238(ft−k)


:M<sub>u</sub> = (1.499k)(2.315 ft.) + (1.851k − 1.499k)(0.6(6.944 ft.)) = 4.936(ft−k)
::''ΣM<sub>R</sub>'' = [54.556(ft−k) + 11.192(ft−k)](1.3)(1.0) = 85.472(ft−k)


:The moment without earthquake controls:
::''ΣV'' = 11.417k(1.3)(1.0) = 14.842k


:<math>R_n = \frac{M_u}{\phi bd^2} = \frac{5.865(ft-k)}{0.9(1 ft.)(8.75 in.)^2}\Big(1000 \frac{lb}{k}\Big)</math> = 85.116 psi
:<math>\bar{x} = \frac{85.472(ft-k) - 21.238(ft-k)}{14.842k}</math> = 4.328 ft.


:''ρ'' = <math>\frac{0.85f'_c}{f_y} \Big[1 - \sqrt{1 - \frac{2R_n}{0.85f'_c}}\Big]</math>
:''e'' = (9.5 ft./2) − 4.328 ft. = 0.422 ft.


:''ρ'' = <math>\frac{0.85 (3.000 psi}{60,000 psi} \Bigg[1 - \sqrt{1 - \frac{2 (85.116 psi}{0.85 (3000 psi)}}\Bigg]</math> = 0.00144
:<math>P_H = \frac{14.842k}{(1 ft.)(9.5 ft.)} \Big[1 - \frac{6(0.422 ft.)}{9.5 ft.}\Big]</math> = 1.146k/ft<sup>2</sup>


{|style="padding: 0.3em; margin-left:5px; border:2px solid #a9a9a9; text-align:center; font-size: 95%; background:#f5f5f5" width="160px" align="right"
:<math>P_T = \frac{14.842k}{(1 ft.)(9.5 ft.)} \Big[1 + \frac{6(0.422 ft.)}{9.5 ft.}\Big]</math> = 1.979k/ft<sup>2</sup>
|-
|'''Additional Information'''
|-
|AASHTO 8.17.1.1 & 8.15.2.1.1
|}


:''ρ<sub>min</sub>'' = <math> 1.7 \Bigg[\frac{h}{d}\Bigg]^2 \frac{\sqrt{f'_c}}{f_y} = 1.7 \Bigg[\frac{11 in.}{8.75 in.}^2 \frac{\sqrt{3000 psi}}{60,000 psi}\Bigg]</math> = 0.00245
:<math>P =\Bigg[\frac{1.979 \frac{k}{ft.} - 1.146 \frac{k}{ft.}}{9.5 ft.}\Bigg](7.583 ft.) + 1.146\frac{k}{ft.}</math> = 1.811k/ft.


:Use ''ρ'' = 4/3 ''ρ'' = 4/3 (0.00144) = 0.00192
:<math>M_u = 1.811\frac{k}{ft.}\frac{(1.917 ft.)^2}{2} + \frac{1}{2}(1.917 ft.)^2\Big[1.979\frac{k}{ft.} - 1.811\frac{k}{ft.}\Big]\frac{2}{3} - 0.560k(0.958 ft.)</math>


:''A<sub>S<sub>Req</sub></sub>'' = ''ρbd'' = 0.00192 (12 in.)(8.75 in.) = 0.202 in.<sup>2</sup>/ft
:''M<sub>u</sub>'' = 2.997(ft−k)


:One #4 bar has A<sub>S</sub> = 0.196 in<sup>2</sup>
:'''With Earthquake'''


:<math>\frac{s}{0.196 in.^2} = \frac{12 in.}{0.202 in.^2}</math>
:''P<sub>H</sub>'' = 1.139 k/ft


:''s'' = 11.64 in.
:''P<sub>T</sub>'' = 1.411 k/ft


:<u>Use #4's @ 10" cts.</u>
:<math>P = \Bigg[\frac{1.411\frac{k}{ft.} - 1.139\frac{k}{ft.}}{9.5 ft.}\Bigg](7.583 ft.) + 1.139\frac{k}{ft.}</math> = 1.356 k/ft


:'''Check Shear'''
:<math>M_u = 1.356\frac{k}{ft.}\frac{(1.917 ft.)^2}{2} + \frac{1}{2}(1.917 ft.)^2 \Bigg[1.411\frac{k}{ft.} - 1.356\frac{k}{ft.}\Bigg]\frac{2}{3} - 0.431k (0.958 ft.)</math>


:V<sub>u</sub> ≥ ''φ'' V<sub>n</sub>
:''M<sub>u</sub>'' = 2.146(ft−k)


::'''Without Earthquake'''
:The moment without earthquake controls.


::V<sub>u,</sub> = (1.3)(1.3)(1.499k) = 2.533k
:<math>R_n = \frac{2.997(ft-k)}{0.9(1 ft.)(14.0 in.)^2}(1000\frac{lb}{k})</math> = 16.990 psi


::'''With Earthquake'''
:''ρ'' = <math>\frac{0.85(3000 psi)}{60,000 psi}\Bigg[1 - \sqrt{1 - \frac{2(16.990 psi)}{0.85(3000psi)}}\Bigg]</math> = 0.000284


::V<sub>u</sub> = 1.851k
:''ρ<sub>min</sub>'' = <math>1.7\Big[\frac{18 in.}{14.0 in.}\Big]^2 \frac{\sqrt{3,000 psi}}{60,000 psi}</math> = 0.00257


:The shear force without earthquake controls.
:Use ''ρ'' = 4/3 ''ρ'' = <math>\frac{4}{3}(0.000284)</math> = 0.000379


:<math>\frac{\nu_u}{\phi} = \frac{2.533k}{0.85(12 in.)(8.75 in.)} (1000 lb/k)</math> = 28.4 psi
:''A<sub>S<sub>Req</sub></sub>'' = 0.000379 (12 in.)(14.0 in.) = 0.064 in.<sup>2</sup>/ft.


:<math>\nu_c = 2 \sqrt{3,000 psi}</math> = 109.5 psi > 28.4 psi <u>o.k.</u>


'''Reinforcement-Footing-Heel'''
:<math>\frac{12 in.}{0.064 in.^2} = \frac{s}{0.196 in.^2}</math>


[[image:751.24.3.3 heel.jpg|center|250px]]
:''s'' = 36.8 in.


Note: Earthquake will not control and will not be checked.
:Minimum is # 4 bars at 12 inches. These will be the same bars that are in the back of the stem. Use the smaller of the two spacings.


''β<sub>E</sub>'' = 1.0 (vertical earth pressure)
:<u>Use # 4's @ 10" cts.</u>


d = 18" - 3" - (1/2)(0.750") = 14.625"
:'''Check Shear'''


b = 12"
:Shear shall be checked at a distance "d" from the face of the stem.


''f'<sub>c</sub>'' = 3,000 psi
::'''Without Earthquake'''


''M<sub>u</sub>'' = 1.3 [(5.556k + 1.500k)(3.333ft) + 0.889k(4.444ft) + 1.178k(6.667ft)]
::<math>P_d =\Bigg[\frac{1.979\frac{k}{ft.} - 1.146\frac{k}{ft.}}{9.5 ft.}\Bigg](8.750 ft.) + 1.146\frac{k}{ft.}</math> = 1.913k/ft.


''M<sub>u</sub>'' = 45.919(ft−k)
::<math>V_u =\frac{1.979\frac{k}{ft.} + 1.913\frac{k}{ft.}}{2}(0.750 ft.) - 1.3\Big[0.225\frac{k}{ft.}\Big](0.750 ft.)</math> = 1.240k


<math>R_n = \frac{45.919(ft-k)}{0.9(1 ft.)(14.625 in.)^2}(1000\frac{lb}{k})</math> = 238.5 psi
::'''With Earthquake'''


''ρ'' = <math>\frac{0.85(3000)psi}{60,000 psi} \Bigg[ 1 - \sqrt{1 - \frac{2(238.5 psi)}{0.85(3000psi)}}\Bigg]</math> = 0.00418
::<math>P_d =\Bigg[\frac{1.411\frac{k}{ft.} - 1.139\frac{k}{ft.}}{9.5 ft.}\Bigg](8.750 ft.) + 1.139\frac{k}{ft.}</math> = 1390k/ft.


''ρ<sub>min</sub>'' = <math> 1.7 \Big[\frac{18 in.}{14.625 in.}\Big]^2 \frac{\sqrt{3000 psi}}{60,000 psi}</math> = 0.00235
::<math>V_u =\frac{1.411\frac{k}{ft.} + 1.139\frac{k}{ft.}}{2}(0.750 ft.) - \Big[0.225\frac{k}{ft.}\Big](0.750 ft.)</math> = 0.788k


''A<sub>S<sub>Req</sub></sub>'' = 0.00418 (12 in.) (14.625 in.) = 0.734 in<sup>2</sup>/ft.
:Shear without earthquake controls.


:<math>\frac{\nu_u}{\phi} = \frac{1.240k}{0.85(12 in.)(14.0 in.)}(1000\frac{lb}{k} ) = 8.7 psi < 2\sqrt{3000 psi}</math> = 109.5 psi <u>o.k.</u>


<u>Use #6's @ 7" cts.</u>
'''Reinforcement-Shear Key'''


:'''Check Shear'''
[[image:751.24.3.3 shear key.jpg|center|250px]]


:Shear shall be checked at back face of stem.
The passive pressure is higher without earthquake loads.


:''V<sub>u</sub>'' = 1.3 (5.556k + 1.500k + 0.889k + 1.178k) = 11.860k
''γ'' = 1.3


:<math>\frac{\nu_u}{\phi} = \frac{11.860k}{0.85(12 in.)(14.625 in.)}(1000 \frac{lb}{k} ) = 79.5 psi < 2 \sqrt{3,000 psi}</math> = 109.5 psi  o.k.
''β<sub>E</sub>'' = 1.3 (lateral earth pressure)


'''Reinforcement-Footing-Toe'''
d = 12"-3"-(1/2)(0.5") = 8.75"


[[image:751.24.3.3. toe.jpg|center|350px]]
b = 12"


d = 18" - 4" = 14"
''M<sub>u</sub> = (3.379k)(1.360 ft.)(1.3)(1.3) = 7.764(ft−k)


b = 12"
<math>R_n = \frac{7.764(ft-k)}{0.9(1 ft.)(8.75 in.)^2} (1000\frac{lb}{k})</math> = 112.677 psi


:'''Without Earthquake'''
''ρ'' = <math>\frac{0.85(3000 psi)}{60,000 psi}\Bigg[1 - \sqrt{1 - \frac{2(112.677 psi)}{0.85(3000psi)}}\Bigg]</math> = 0.00192


::'''Apply Load Factors'''
''ρ<sub>min</sub> = <math>1.7\Big[\frac{12 in.}{8.75 in.}\Big]^2 \frac{\sqrt{3000 psi}}{60,000 psi}</math> = 0.00292


::load 4 (weight) = 0.431k(1.3)(1.0) = 0.560k
Use ''ρ'' = 4/3 ''ρ'' = 4/3 (0.00192) = 0.00256


::''β<sub>E</sub>'' = 1.3 for lateral earth pressure for retaining walls.
A<sub>S<sub>Req</sub></sub> = 0.00256(12 in.)(8.75 in.) = 0.269 in.<sup>2</sup>/ft.


::''β<sub>E</sub>'' = 1.0 for vertical earth pressure.


::''ΣM<sub>OT</sub>'' = 12.567(ft−k)(1.3)(1.3) = 21.238(ft−k)
<u>Use # 4 @ 8.5 in cts.</u>


::''ΣM<sub>R</sub>'' = [54.556(ft−k) + 11.192(ft−k)](1.3)(1.0) = 85.472(ft−k)
Check Shear


::''ΣV'' = 11.417k(1.3)(1.0) = 14.842k
:<math>\frac{\nu_u}{\phi} = \frac{1.3(3.379k)(1.3)}{0.85(12 in.)(8.75.)}(1000\frac{lb}{k} ) = 64.0 psi < 2\sqrt{3000 psi}</math> = 109.5 psi <u>o.k.</u>


:<math>\bar{x} = \frac{85.472(ft-k) - 21.238(ft-k)}{14.842k}</math> = 4.328 ft.
'''Reinforcement Summary'''


:''e'' = (9.5 ft./2) − 4.328 ft. = 0.422 ft.
[[Image:751.24.3.3 summary.jpg|500px|center]]


:<math>P_H = \frac{14.842k}{(1 ft.)(9.5 ft.)} \Big[1 - \frac{6(0.422 ft.)}{9.5 ft.}\Big]</math> = 1.146k/ft<sup>2</sup>
===751.24.3.4 Example 2: L-Shaped Cantilever Wall===


:<math>P_T = \frac{14.842k}{(1 ft.)(9.5 ft.)} \Big[1 + \frac{6(0.422 ft.)}{9.5 ft.}\Big]</math> = 1.979k/ft<sup>2</sup>
[[image:751.24.3.4.jpg|center|650px|thumb|<center>'''Typical Section thru Wall</center><center>(Spread Footing)</center>''']]


:<math>P =\Bigg[\frac{1.979 \frac{k}{ft.} - 1.146 \frac{k}{ft.}}{9.5 ft.}\Bigg](7.583 ft.) + 1.146\frac{k}{ft.}</math> = 1.811k/ft.
''f'<sub>c</sub>'' = 4000 psi


:<math>M_u = 1.811\frac{k}{ft.}\frac{(1.917 ft.)^2}{2} + \frac{1}{2}(1.917 ft.)^2\Big[1.979\frac{k}{ft.} - 1.811\frac{k}{ft.}\Big]\frac{2}{3} - 0.560k(0.958 ft.)</math>
''f<sub>y</sub>'' = 60,000 psi


:''M<sub>u</sub>'' = 2.997(ft−k)
''φ'' = 29°


:'''With Earthquake'''
''γ<sub>s</sub> = 120 pcf


:''P<sub>H</sub>'' = 1.139 k/ft
Allowable soil pressure = 1.5 tsf = 3.0 ksf


:''P<sub>T</sub>'' = 1.411 k/ft
Retaining wall is located in Seismic Performance Category (SPC) A.


:<math>P = \Bigg[\frac{1.411\frac{k}{ft.} - 1.139\frac{k}{ft.}}{9.5 ft.}\Bigg](7.583 ft.) + 1.139\frac{k}{ft.}</math> = 1.356 k/ft
<math>\delta = tan^{-1}\frac{1}{2.5}</math> = 21.801°


:<math>M_u = 1.356\frac{k}{ft.}\frac{(1.917 ft.)^2}{2} + \frac{1}{2}(1.917 ft.)^2 \Bigg[1.411\frac{k}{ft.} - 1.356\frac{k}{ft.}\Bigg]\frac{2}{3} - 0.431k (0.958 ft.)</math>
<math>C_a = cos \delta\Bigg[\frac{cos \delta - \sqrt{cos^2\delta - cos^2\phi}}{cos \delta + \sqrt{cos^2\delta - cos^2\phi}}\Bigg]</math> = 0.462


:''M<sub>u</sub>'' = 2.146(ft−k)
<math>C_p = tan^2\Big[45 + \frac{\phi}{2}\Big]</math> = 2.882


:The moment without earthquake controls.
''P<sub>A</sub>'' = 1/2 ''γ<sub>s</sub>'' C<sub>a</sub>H<sup>2</sup> = 1/2 (0.120 k/ft<sup>3</sup>)(0.462)(4.958 ft.)<sup>2</sup> = 0.681k


:<math>R_n = \frac{2.997(ft-k)}{0.9(1 ft.)(14.0 in.)^2}(1000\frac{lb}{k})</math> = 16.990 psi
For sliding, P<sub>P</sub> is assumed to act only on the portion of key below the frost line that is set at an 18 in. depth on the southern border.


:''ρ'' = <math>\frac{0.85(3000 psi)}{60,000 psi}\Bigg[1 - \sqrt{1 - \frac{2(16.990 psi)}{0.85(3000psi)}}\Bigg]</math> = 0.000284
''P<sub>P</sub>'' = 1/2 (0.120 k/ft<sup>3</sup>)(2.882)[(2.458 ft.)<sup>2</sup> − (1.500 ft.)<sup>2</sup>] = 0.656k


:''ρ<sub>min</sub>'' = <math>1.7\Big[\frac{18 in.}{14.0 in.}\Big]^2 \frac{\sqrt{3,000 psi}}{60,000 psi}</math> = 0.00257
'''Assumptions'''


:Use ''ρ'' = 4/3 ''ρ'' = <math>\frac{4}{3}(0.000284)</math> = 0.000379
* Design is for a unit length (1 ft.) of wall.


:''A<sub>S<sub>Req</sub></sub>'' = 0.000379 (12 in.)(14.0 in.) = 0.064 in.<sup>2</sup>/ft.
* Sum moments about the toe at the bottom of the footing for overturning.


* F.S. for overturning ≥ 2.0 for footings on soil.


:<math>\frac{12 in.}{0.064 in.^2} = \frac{s}{0.196 in.^2}</math>
* F.S. for sliding ≥ 1.5 for footings on soil.


:''s'' = 36.8 in.
* Resultant of dead load and earth pressure to be in back half of the middle third of the footing if subjected to frost heave.


:Minimum is # 4 bars at 12 inches. These will be the same bars that are in the back of the stem. Use the smaller of the two spacings.
* For all loading combinations the resultant must be in the middle third of the footing except for collision loads.


:<u>Use # 4's @ 10" cts.</u>
* The top 12 in. of the soil is not neglected in determining the passive pressure because the soil there will be maintained.


:'''Check Shear'''
* Frost line is set at 18 in. at the south border for Missouri.


:Shear shall be checked at a distance "d" from the face of the stem.
* Portions of shear key which are above the frost line are assumed not to resist sliding by passive pressure.


::'''Without Earthquake'''
* Use of a shear key shifts the failure plane to "B" where resistance to sliding is also provided by friction of soil along the failure plane in front of the shear key. Friction between the soil and concrete behind the shear key will be neglected.


::<math>P_d =\Bigg[\frac{1.979\frac{k}{ft.} - 1.146\frac{k}{ft.}}{9.5 ft.}\Bigg](8.750 ft.) + 1.146\frac{k}{ft.}</math> = 1.913k/ft.
* Soil cohesion along the failure plane is neglected.


::<math>V_u =\frac{1.979\frac{k}{ft.} + 1.913\frac{k}{ft.}}{2}(0.750 ft.) - 1.3\Big[0.225\frac{k}{ft.}\Big](0.750 ft.)</math> = 1.240k
* Live loads can move to within 1 ft. of the stem face and 1 ft. from the toe.


::'''With Earthquake'''
* The wall is designed as a cantilever supported by the footing.


::<math>P_d =\Bigg[\frac{1.411\frac{k}{ft.} - 1.139\frac{k}{ft.}}{9.5 ft.}\Bigg](8.750 ft.) + 1.139\frac{k}{ft.}</math> = 1390k/ft.
* Footing is designed as a cantilever supported by the wall. Critical sections for bending and shear will be taken at the face of the wall.


::<math>V_u =\frac{1.411\frac{k}{ft.} + 1.139\frac{k}{ft.}}{2}(0.750 ft.) - \Big[0.225\frac{k}{ft.}\Big](0.750 ft.)</math> = 0.788k
* Load factors for AASHTO Groups I-VI for design of concrete are:


:Shear without earthquake controls.
::*''γ'' = 1.3.


:<math>\frac{\nu_u}{\phi} = \frac{1.240k}{0.85(12 in.)(14.0 in.)}(1000\frac{lb}{k} ) = 8.7 psi < 2\sqrt{3000 psi}</math> = 109.5 psi <u>o.k.</u>
::*''β<sub>E</sub>'' = 1.3 for horizontal earth pressure on retaining walls.


'''Reinforcement-Shear Key'''
::*''β<sub>E</sub>'' = 1.0 for vertical earth pressure.


[[image:751.24.3.3 shear key.jpg|center|250px]]
::*''β<sub>LL</sub>'' = 1.67 for live loads and collision loads.


The passive pressure is higher without earthquake loads.
'''Dead Load and Earth Pressure - Stabilty and Pressure Checks'''


''γ'' = 1.3
{| border="1" class="wikitable" style="margin: 1em auto 1em auto"
 
|+
''β<sub>E</sub>'' = 1.3 (lateral earth pressure)
|-
 
!colspan="4" style="background:#BEBEBE" |Dead Load and Earth Pressure - Stabilty and Pressure Checks
d = 12"-3"-(1/2)(0.5") = 8.75"
|-
 
!style="background:#BEBEBE" |Load !!style="background:#BEBEBE" |Force (k) !!style="background:#BEBEBE" |Arm (in.) !!style="background:#BEBEBE"|Moment (ft-k)
b = 12"
|-
 
|align="center"|(1)||align="center"| (0.833 ft.)(5.167 ft.)(0.150k/ft<sup>3</sup>) = 0.646||align="center"| 5.333||align="center"| 3.444
''M<sub>u</sub> = (3.379k)(1.360 ft.)(1.3)(1.3) = 7.764(ft−k)
|-
 
|align="center"|(2)||align="center"| (0.958ft)(5.750ft)(0.150k/ft3) = 0.827||align="center"| 2.875||align="center"| 2.376
<math>R_n = \frac{7.764(ft-k)}{0.9(1 ft.)(8.75 in.)^2} (1000\frac{lb}{k})</math> = 112.677 psi
|-
 
|align="center"| (3)||align="center"|  (1.000ft)(1.500ft)(0.150k/ft3) = 0.22534.259||align="center"| 4.250 ||align="center"| 0.956
''ρ'' = <math>\frac{0.85(3000 psi)}{60,000 psi}\Bigg[1 - \sqrt{1 - \frac{2(112.677 psi)}{0.85(3000psi)}}\Bigg]</math> = 0.00192
|-
|align="center" colspan="3"|ΣV = 1.698 ||align="center"| ΣM<sub>R</sub> = 6.776
|-
|align="center"| P<sub>AV</sub>||align="center"|  0.253 ||align="center"| 5.750 ||align="center"| 1.455
|-
|align="center" colspan="3"| ΣV = 1.951||align="center"|  ΣM<sub>R</sub> = 8.231
|-
|align="center"| P<sub>AH</sub> ||align="center"| 0.633 ||align="center"| 1.653 ||align="center"| 1.045
|-
|align="center"| P<sub>P</sub>||align="center"|  0.656 ||align="center"| 1.06<sup>1</sup>||align="center"| -
|-
|colspan="4" align="right"|ΣM<sub>OT</sub> = 1.045
|-
|colspan="4"|<sup>'''1'''</sup> The passive pressure at the shear key is ignored in overturning checks.
|}


''ρ<sub>min</sub> = <math>1.7\Big[\frac{12 in.}{8.75 in.}\Big]^2 \frac{\sqrt{3000 psi}}{60,000 psi}</math> = 0.00292
:'''Overturning'''


Use ''ρ'' = 4/3 ''ρ'' = 4/3 (0.00192) = 0.00256
:<math>F.S. = \frac{\Sigma M_R}{\Sigma M_{OT}} = \frac{8.231(ft-k)}{1.045(ft-k)}</math> = 7.877 ≥ 2.0 <u>o.k.</u>


A<sub>S<sub>Req</sub></sub> = 0.00256(12 in.)(8.75 in.) = 0.269 in.<sup>2</sup>/ft.
:'''Location of Resultant'''


:MoDOT policy is that the resultant must be in the back half of the middle third of the footing when considering dead and earth loads:


<u>Use # 4 @ 8.5 in cts.</u>
:<math>\Bigg[\frac{5.750 ft.}{2} = 2.875 ft.\Bigg] \le \bar{x} \le \Bigg[\Bigg(\frac{5.750 ft.}{2} + \frac{5.750 ft.}{6}\Bigg) = 3.833 ft.\Bigg] </math>


Check Shear
:<math>\bar{x} = \frac{M_{NET}}{\Sigma V} = \frac{8.231(ft-k) - 1.045(ft-k)}{1.951k}</math> = 3.683 ft. <u>o.k.</u>


:<math>\frac{\nu_u}{\phi} = \frac{1.3(3.379k)(1.3)}{0.85(12 in.)(8.75.)}(1000\frac{lb}{k} ) = 64.0 psi < 2\sqrt{3000 psi}</math> = 109.5 psi <u>o.k.</u>
:'''Sliding'''


'''Reinforcement Summary'''
:<math>F.S. = \frac{P_P + \Sigma V \Bigg[\Big(\frac{L_2}{L_1}\Big)tan\phi_{s-s} + \Big(\frac{L_3}{L_1}\Big)tan\phi_{s-c}\Bigg]}{P_{AH}}</math>


[[Image:751.24.3.3 summary.jpg|500px|center]]
:where:
::''φ<sub>s-s</sub>'' = angle of internal friction of soil


===751.24.3.4 Example 2: L-Shaped Cantilever Wall===
::''φ<sub>s-c</sub>'' = angle of friction between soil and concrete = (2/3)''φ<sub>s-s</sub>''


[[image:751.24.3.4.jpg|center|650px|thumb|<center>'''Typical Section thru Wall</center><center>(Spread Footing)</center>''']]
:<math>F.S. = \frac{0.656k +(1.951k)\Big[\Big(\frac{3.75 ft.}{5.75 ft.}\Big)tan 29^\circ + \Big(\frac{1 ft.}{5.75 ft.}\Big) tan\Big(\frac{2}{3}(29^\circ)\Big)\Big]}{0.633 k}</math> = 2.339 ≥ 1.5 <u>o.k.</u>


''f'<sub>c</sub>'' = 4000 psi
:'''Footing Pressure'''


''f<sub>y</sub>'' = 60,000 psi
:<math>P = \frac{\Sigma V}{bL} \Big[1 \pm \frac{6e}{L}\Big]</math>


''φ'' = 29°
:<math>e = \bar{x} - \frac{L}{2} = 3.683 ft. - \frac{5.75 ft.}{2}</math> = 0.808 ft.


''γ<sub>s</sub> = 120 pcf
:Heel: <math>P_H = \frac{1.951k}{(1 ft.)(5.75 ft.)}\Big[1 + \frac{6(0.808 ft.)}{5.75 ft.}\Big]</math> = 0.625 ksf < 3.0 ksf <u>o.k.</u>


Allowable soil pressure = 1.5 tsf = 3.0 ksf
:Toe: <math>P_T = \frac{1.951k}{(1 ft.)(5.75 ft.)}\Big[1 - \frac{6(0.808 ft.)}{5.75 ft.}\Big]</math> = 0.053 ksf < 3.0 ksf <u>o.k.</u>


Retaining wall is located in Seismic Performance Category (SPC) A.
'''Dead Load, Earth Pressure, and Live Load - Stability and Pressure Checks'''


<math>\delta = tan^{-1}\frac{1}{2.5}</math> = 21.801°
Stability is not an issue because the live load resists overturning and increases the sliding friction force.


<math>C_a = cos \delta\Bigg[\frac{cos \delta - \sqrt{cos^2\delta - cos^2\phi}}{cos \delta + \sqrt{cos^2\delta - cos^2\phi}}\Bigg]</math> = 0.462
[[image:751.24.3.4 checks.jpg|center|250px]]


<math>C_p = tan^2\Big[45 + \frac{\phi}{2}\Big]</math> = 2.882
The live load will be distributed as:


''P<sub>A</sub>'' = 1/2 ''γ<sub>s</sub>'' C<sub>a</sub>H<sup>2</sup> = 1/2 (0.120 k/ft<sup>3</sup>)(0.462)(4.958 ft.)<sup>2</sup> = 0.681k
<math> F_{LL} = \frac{LL_{WL}}{E}</math>  


For sliding, P<sub>P</sub> is assumed to act only on the portion of key below the frost line that is set at an 18 in. depth on the southern border.
:where E = 0.8X + 3.75


''P<sub>P</sub>'' = 1/2 (0.120 k/ft<sup>3</sup>)(2.882)[(2.458 ft.)<sup>2</sup> − (1.500 ft.)<sup>2</sup>] = 0.656k
::X = distance in feet from the load to the front face of wall


'''Assumptions'''
The live load will be positioned as shown by the dashed lines above. The bearing pressure and resultant location will be determined for these two positions.


* Design is for a unit length (1 ft.) of wall.
:'''Live Load 1 ft From Stem Face'''


* Sum moments about the toe at the bottom of the footing for overturning.
::'''Resultant Eccentricity'''


* F.S. for overturning ≥ 2.0 for footings on soil.
::X = 1 ft.


* F.S. for sliding ≥ 1.5 for footings on soil.
::E = 0.8(1 ft.) + 3.75 = 4.55 ft.


* Resultant of dead load and earth pressure to be in back half of the middle third of the footing if subjected to frost heave.
::<math>F_{LL} = \frac{16k}{4.55 ft.} (1 ft.)</math> = 3.516k


* For all loading combinations the resultant must be in the middle third of the footing except for collision loads.
::<math>\bar{x} = \frac{M_{NET}}{\Sigma V} = \frac{8.231(ft-k) + (3.516k)(3.917 ft.) - 1.045(ft-k)}{1.951k + 3.516k}</math> = 3.834 ft.


* The top 12 in. of the soil is not neglected in determining the passive pressure because the soil there will be maintained.
::<math>e = \bar{x} - \frac{L}{2} = 3.834 ft. - \frac{5.75 ft.}{2} = 0.959 ft. \le \frac{L}{6}</math> = 5.75 ft. <u>o.k.</u>


* Frost line is set at 18 in. at the south border for Missouri.
::'''Footing Pressure'''


* Portions of shear key which are above the frost line are assumed not to resist sliding by passive pressure.
::<math>P = \frac{\Sigma V}{bL} \Big[1 \pm \frac{6e}{L}\Big]</math>


* Use of a shear key shifts the failure plane to "B" where resistance to sliding is also provided by friction of soil along the failure plane in front of the shear key. Friction between the soil and concrete behind the shear key will be neglected.
::Allowable Pressure = 3.0 ksf


* Soil cohesion along the failure plane is neglected.
::Heel: <math>P_H = \frac{5.467k}{(1 ft.)(5.75 ft.)}\Big[1 + \frac{6(0.959 ft.)}{5.75 ft.}\Big]</math> = 1.902 ksf


* Live loads can move to within 1 ft. of the stem face and 1 ft. from the toe.
::Toe: <math>P_T = \frac{5.467k}{(1 ft.)(5.75 ft.)}\Big[1 - \frac{6(0.959 ft.)}{5.75 ft.}\Big]</math> = 0.000ksf


* The wall is designed as a cantilever supported by the footing.
:'''Live Load 1 ft From Toe'''


* Footing is designed as a cantilever supported by the wall. Critical sections for bending and shear will be taken at the face of the wall.
::'''Resultant Eccentricity'''


* Load factors for AASHTO Groups I-VI for design of concrete are:
::X = 3.917 ft.


::*''γ'' = 1.3.
::E = 0.8(3.917 ft.) + 3.75 = 6.883 ft.


::*''β<sub>E</sub>'' = 1.3 for horizontal earth pressure on retaining walls.
::<math>F_{LL} = \frac{16k}{6.883 ft} (1 ft.)</math> = 2.324k


::*''β<sub>E</sub>'' = 1.0 for vertical earth pressure.
::<math>x = \frac{8.231(ft-k) + (2.324k)(1ft.) - 1.045(ft-k)}{1.951k + 2.324k}</math> = 2.225 ft.  


::*''β<sub>LL</sub>'' = 1.67 for live loads and collision loads.
::<math>e = \frac{L}{2} - \bar{x} = \frac{5.75 ft.}{2} - 2.225 ft. = 0.650 ft. \le \frac{L}{6} = \frac{5.75 ft.}{6}</math> = 0.958 ft. <u>o.k.</u>


'''Dead Load and Earth Pressure - Stabilty and Pressure Checks'''
::'''Footing Pressure'''


{| border="1" class="wikitable" style="margin: 1em auto 1em auto"
::Allowable Pressure = 3.0ksf
|+
|-
!colspan="4" style="background:#BEBEBE" |Dead Load and Earth Pressure - Stabilty and Pressure Checks
|-
!style="background:#BEBEBE" |Load !!style="background:#BEBEBE" |Force (k) !!style="background:#BEBEBE" |Arm (in.) !!style="background:#BEBEBE"|Moment (ft-k)
|-
|align="center"|(1)||align="center"| (0.833 ft.)(5.167 ft.)(0.150k/ft<sup>3</sup>) = 0.646||align="center"| 5.333||align="center"| 3.444
|-
|align="center"|(2)||align="center"| (0.958ft)(5.750ft)(0.150k/ft3) = 0.827||align="center"| 2.875||align="center"| 2.376
|-
|align="center"| (3)||align="center"|  (1.000ft)(1.500ft)(0.150k/ft3) = 0.22534.259||align="center"| 4.250 ||align="center"| 0.956
|-
|align="center" colspan="3"|ΣV = 1.698 ||align="center"| ΣM<sub>R</sub> = 6.776
|-
|align="center"| P<sub>AV</sub>||align="center"|  0.253 ||align="center"| 5.750 ||align="center"| 1.455
|-
|align="center" colspan="3"| ΣV = 1.951||align="center"|  ΣM<sub>R</sub> = 8.231
|-
|align="center"| P<sub>AH</sub> ||align="center"| 0.633 ||align="center"| 1.653 ||align="center"| 1.045
|-
|align="center"| P<sub>P</sub>||align="center"|  0.656 ||align="center"| 1.06<sup>1</sup>||align="center"| -
|-
|colspan="4" align="right"|ΣM<sub>OT</sub> = 1.045
|-
|colspan="4"|<sup>'''1'''</sup> The passive pressure at the shear key is ignored in overturning checks.
|}


:'''Overturning'''
::Heel: <math>P_H = \frac{4.275k}{(1 ft.)(5.75 ft.}\Big[1 - \frac{6 (0.650 ft.)}{5.75 ft.}\Big]</math> = 0.239ksf <u>o.k.</u>


:<math>F.S. = \frac{\Sigma M_R}{\Sigma M_{OT}} = \frac{8.231(ft-k)}{1.045(ft-k)}</math> = 7.877 ≥ 2.0 <u>o.k.</u>
::Toe: <math>P_T = \frac{4.275k}{(1 ft.)(5.75 ft.}\Big[1 + \frac{6 (0.650 ft.)}{5.75 ft.}\Big]</math> = 1.248ksf <u>o.k.</u>


:'''Location of Resultant'''
'''Dead Load, Earth Pressure, Collision Load, and Live Load - Stability and Pressure Checks'''


:MoDOT policy is that the resultant must be in the back half of the middle third of the footing when considering dead and earth loads:
During a collision, the live load will be close to the wall so check this combination when the live load is one foot from the face of the stem. Sliding (in either direction) will not be an issue. Stability about the heel should be checked although it is unlikely to be a problem. There are no criteria for the location of the resultant, so long as the footing pressure does not exceed 125% of the allowable. It is assumed that the distributed collision force will develop an equal and opposite force on the fillface of the back wall unless it exceeds the passive pressure that can be developed by soil behind the wall.
 
''F<sub>LL</sub>'' = 3.516k
 
[[image:751.24.3.4 collision.jpg|center|250px]]


:<math>\Bigg[\frac{5.750 ft.}{2} = 2.875 ft.\Bigg] \le \bar{x} \le \Bigg[\Bigg(\frac{5.750 ft.}{2} + \frac{5.750 ft.}{6}\Bigg) = 3.833 ft.\Bigg] </math>
''F<sub>COLL</sub>'' = <math>\frac{10k}{2(3 ft.)}(1 ft.)</math> = 1.667k


:<math>\bar{x} = \frac{M_{NET}}{\Sigma V} = \frac{8.231(ft-k) - 1.045(ft-k)}{1.951k}</math> = 3.683 ft. <u>o.k.</u>
<math>C_P = cos \delta \Bigg[\frac{cos \delta + \sqrt{cos^2 \delta - cos^2 \phi}}{cos \delta - \sqrt{cos^2 \delta - cos^2 \phi}}\Bigg]</math> = 1.867


:'''Sliding'''
<math>P_{PH} = \frac{1}{2}\gamma_s C_P H^2 cos\delta = \frac{1}{2}(0.120kcf)(1.867)(4.958ft)^2 cos(21.801^\circ)</math>


:<math>F.S. = \frac{P_P + \Sigma V \Bigg[\Big(\frac{L_2}{L_1}\Big)tan\phi_{s-s} + \Big(\frac{L_3}{L_1}\Big)tan\phi_{s-c}\Bigg]}{P_{AH}}</math>
''P<sub>PH</sub>'' = 2.556k > ''F<sub>COLL</sub>''  Thus the soil will develop an equal but opp. force.


:where:
:'''Overturning About the Heel'''
::''φ<sub>s-s</sub>'' = angle of internal friction of soil


::''φ<sub>s-c</sub>'' = angle of friction between soil and concrete = (2/3)''φ<sub>s-s</sub>''
:F.S. = <math>\frac{(0.646k)(0.417 ft.) + (0.827k)(2.875 ft.) + (0.225k)(1.500 ft.) + (3.516k)(1.833 ft.) + (1.667k)\big(\frac{4.958 ft.}{3}\big)}{(1.667k)(3.958 ft.)}</math>


:<math>F.S. = \frac{0.656k +(1.951k)\Big[\Big(\frac{3.75 ft.}{5.75 ft.}\Big)tan 29^\circ + \Big(\frac{1 ft.}{5.75 ft.}\Big) tan\Big(\frac{2}{3}(29^\circ)\Big)\Big]}{0.633 k}</math> = 2.339 ≥ 1.5 <u>o.k.</u>
:F.S. = <math>\frac{12.184(ft-k)}{6.598(ft-k)}</math> = 1.847 ≥ 1.2 <u>o.k.</u>


:'''Footing Pressure'''
:'''Footing Pressure'''


:<math>P = \frac{\Sigma V}{bL} \Big[1 \pm \frac{6e}{L}\Big]</math>
:<math>\bar{x} = \frac{12.184(ft-k) - 6.598(ft-k)}{1.951k + 3.516k}</math> = 1.022 ft. from heel


:<math>e = \bar{x} - \frac{L}{2} = 3.683 ft. - \frac{5.75 ft.}{2}</math> = 0.808 ft.
:''e'' = <math>\frac{5.75 ft.}{2} - 1.022 ft.</math> = 1.853 ft.


:Heel: <math>P_H = \frac{1.951k}{(1 ft.)(5.75 ft.)}\Big[1 + \frac{6(0.808 ft.)}{5.75 ft.}\Big]</math> = 0.625 ksf < 3.0 ksf <u>o.k.</u>
:Allowable Pressure = (1.25)(3.0ksf) = 3.75ksf


:Toe: <math>P_T = \frac{1.951k}{(1 ft.)(5.75 ft.)}\Big[1 - \frac{6(0.808 ft.)}{5.75 ft.}\Big]</math> = 0.053 ksf < 3.0 ksf <u>o.k.</u>
:Heel: <math> P_H =\frac {2(\Sigma V)}{3b[\frac{L}{2} - e]} = \frac {2(5.467k)}{3(1 ft.)\big[\frac{5.75 ft.}{2} - 1.853 ft.\big]}</math> = 3.566ksf <u>o.k.</u>


'''Dead Load, Earth Pressure, and Live Load - Stability and Pressure Checks'''
'''Stem Design-Steel in Rear Face'''


Stability is not an issue because the live load resists overturning and increases the sliding friction force.
[[image:751.24.3.4 steel in rear face.jpg|center|250px]]


[[image:751.24.3.4 checks.jpg|center|250px]]
''γ'' = 1.3


The live load will be distributed as:
''β<sub>E</sub>'' = 1.3 (active lateral earth pressure)


<math> F_{LL} = \frac{LL_{WL}}{E}</math>
d = 10 in. − 2 in. − (0.5 in./2) = 7.75 in.


:where E = 0.8X + 3.75
<math>P_{AH} = \frac{1}{2}\gamma_s C_a H^2 cos\delta = \frac{1}{2}\Bigg[0.120 \frac{k}{ft^3}\Bigg](0.462)(4 ft.)^2(1 ft.) cos 21.801^\circ</math>


::X = distance in feet from the load to the front face of wall
''P<sub>AH</sub>'' = 0.412k


The live load will be positioned as shown by the dashed lines above. The bearing pressure and resultant location will be determined for these two positions.
''M<sub>u</sub>'' = (1.333 ft.)(0.412k)(1.3)(1.3) = 0.928(ft−k)


:'''Live Load 1 ft From Stem Face'''
<math>R_n = \frac{M_u}{\phi b d^2} = \frac{0.928(ft-k)}{(0.9)(1 ft.)(7.75 in.)^2}\Big(1000\frac{lb}{k}\Big)</math> = 17.160psi


::'''Resultant Eccentricity'''
<math>\rho = \frac{0.85f_c}{f_y}\Bigg[1 - \sqrt{1 - \frac{2R_n}{0.85 f_c}}\Bigg]</math>


::X = 1 ft.
<math>\rho = \frac{4000 psi}{60,000 psi}\Bigg[1 - \sqrt{1 - \frac{2(17.160 psi)}{0.85 (4000psi)}}\Bigg]</math> = 0.000287


::E = 0.8(1 ft.) + 3.75 = 4.55 ft.
<math>\rho_{min} = 1.7 \Bigg[\frac{h}{d}\Bigg]^2 \frac{\sqrt{f_c}}{f_y}</math>


::<math>F_{LL} = \frac{16k}{4.55 ft.} (1 ft.)</math> = 3.516k
<math>\rho_{min} = 1.7 \Bigg[\frac{10 in.}{7.75 in.}\Bigg]^2 \frac{\sqrt{4000 psi}}{60000 psi}</math> = 0.00298


::<math>\bar{x} = \frac{M_{NET}}{\Sigma V} = \frac{8.231(ft-k) + (3.516k)(3.917 ft.) - 1.045(ft-k)}{1.951k + 3.516k}</math> = 3.834 ft.
Use ''ρ'' = (4/3)ρ = (4/3)(0.000287) = 0.000382


::<math>e = \bar{x} - \frac{L}{2} = 3.834 ft. - \frac{5.75 ft.}{2} = 0.959 ft. \le \frac{L}{6}</math> = 5.75 ft. <u>o.k.</u>
<math>A_{S_{Req}} = \rho bd = 0.000382(12 in.)(7.75 in.) = 0.036 \frac{in^2}{ft.}</math>


::'''Footing Pressure'''
One #4 bar has A<sub>S</sub> = 0.196 in<sup>2</sup>, so the required minimum of one #4 bar every 12 in. controls.


::<math>P = \frac{\Sigma V}{bL} \Big[1 \pm \frac{6e}{L}\Big]</math>
<u>Use #4's @ 12 in. (min)</u>


::Allowable Pressure = 3.0 ksf
(These bars are also the bars in the bottom of the footing so the smaller of the two required spacings will be used.)


::Heel: <math>P_H = \frac{5.467k}{(1 ft.)(5.75 ft.)}\Big[1 + \frac{6(0.959 ft.)}{5.75 ft.}\Big]</math> = 1.902 ksf
:'''Check Shear'''


::Toe: <math>P_T = \frac{5.467k}{(1 ft.)(5.75 ft.)}\Big[1 - \frac{6(0.959 ft.)}{5.75 ft.}\Big]</math> = 0.000ksf
:<math>\frac{\nu_u}{\phi} \le V_n</math>


:'''Live Load 1 ft From Toe'''
:<math>\frac{\nu_u}{\phi} = \frac{(1.3)(1.3)(0.412k)}{0.85(12 in.)(7.75 in.)}(1000\frac{lb}{k})</math>  = 8.8 psi


::'''Resultant Eccentricity'''
:<math>\nu_c = 2 \sqrt{f'_c}</math>


::X = 3.917 ft.
:<math>\nu_c = 2 \sqrt{4, 000 psi}</math> = 126.5 psi > 8.8 psi <u>o.k.</u>


::E = 0.8(3.917 ft.) + 3.75 = 6.883 ft.
'''Stem Design-Steel in Front Face (Collision Loads)'''


::<math>F_{LL} = \frac{16k}{6.883 ft} (1 ft.)</math> = 2.324k
[[image:751.24.3.4 steel in front face.jpg|center|300px]]


::<math>x = \frac{8.231(ft-k) + (2.324k)(1ft.) - 1.045(ft-k)}{1.951k + 2.324k}</math> = 2.225 ft.


::<math>e = \frac{L}{2} - \bar{x} = \frac{5.75 ft.}{2} - 2.225 ft. = 0.650 ft. \le \frac{L}{6} = \frac{5.75 ft.}{6}</math> = 0.958 ft. <u>o.k.</u>
The soil pressure on the back of the stem becomes passive soil pressure during a collision, however this pressure is ignored for reinforcement design.


::'''Footing Pressure'''
''γ'' = 1.3


::Allowable Pressure = 3.0ksf
''β<sub>LL</sub>'' = 1.67


::Heel: <math>P_H = \frac{4.275k}{(1 ft.)(5.75 ft.}\Big[1 - \frac{6 (0.650 ft.)}{5.75 ft.}\Big]</math> = 0.239ksf <u>o.k.</u>
<math>d = 10 in. - 1.5 in. - 0.5 in. - \frac{0.5 in.}{2}</math> = 7.75 in.


::Toe: <math>P_T = \frac{4.275k}{(1 ft.)(5.75 ft.}\Big[1 + \frac{6 (0.650 ft.)}{5.75 ft.}\Big]</math> = 1.248ksf <u>o.k.</u>
<math>F_{COLL} = \frac{10k}{2L} = \frac{10k}{(2)(3 ft.)}</math> = 1.667 k/ft.


'''Dead Load, Earth Pressure, Collision Load, and Live Load - Stability and Pressure Checks'''
''M<sub>u</sub>'' = 1.667k/ft. (1 ft.)(3 ft.)(1.3)(1.67) = 10.855(ft−k)


During a collision, the live load will be close to the wall so check this combination when the live load is one foot from the face of the stem. Sliding (in either direction) will not be an issue. Stability about the heel should be checked although it is unlikely to be a problem. There are no criteria for the location of the resultant, so long as the footing pressure does not exceed 125% of the allowable. It is assumed that the distributed collision force will develop an equal and opposite force on the fillface of the back wall unless it exceeds the passive pressure that can be developed by soil behind the wall.
<math>R_n = \frac{10.855(ft-k)}{0.9(1 ft.)(7.75 in.)^2} (1000\frac{lb}{k})</math> = 200.809 psi


''F<sub>LL</sub>'' = 3.516k
<math>\rho = \frac{0.85(4000 psi)}{60,000 psi}\Bigg[1 - \sqrt{1 - \frac{2(200.809 psi)}{0.85(4000psi)}}\Bigg]</math> = 0.00345


[[image:751.24.3.4 collision.jpg|center|250px]]
<math>\rho_{min} = 1.7\Bigg[\frac{10 in.}{7.75 in.}\Bigg]^2 \frac{\sqrt{4000 psi}}{60,000 psi}</math> = 0.00298


''F<sub>COLL</sub>'' = <math>\frac{10k}{2(3 ft.)}(1 ft.)</math> = 1.667k
<math>A_{S_{Req}} = 0.00345 (12 in.)(7.75 in.) = 0.321 \frac{in.^2}{ft.}</math>


<math>C_P = cos \delta \Bigg[\frac{cos \delta + \sqrt{cos^2 \delta - cos^2 \phi}}{cos \delta - \sqrt{cos^2 \delta - cos^2 \phi}}\Bigg]</math> = 1.867
One #4 bar has A<sub>S</sub> = 0.196 in<sup>2</sup>.


<math>P_{PH} = \frac{1}{2}\gamma_s C_P H^2 cos\delta = \frac{1}{2}(0.120kcf)(1.867)(4.958ft)^2 cos(21.801^\circ)</math>
<math>\frac{s}{0.196 in.^2} = \frac{12 in.}{0.321 in.^2}</math>


''P<sub>PH</sub>'' = 2.556k > ''F<sub>COLL</sub>''  Thus the soil will develop an equal but opp. force.
''s'' = 7.3 in.


:'''Overturning About the Heel'''
<u>Use #4's @ 7 in.</u>


:F.S. = <math>\frac{(0.646k)(0.417 ft.) + (0.827k)(2.875 ft.) + (0.225k)(1.500 ft.) + (3.516k)(1.833 ft.) + (1.667k)\big(\frac{4.958 ft.}{3}\big)}{(1.667k)(3.958 ft.)}</math>
:'''Check Shear'''


:F.S. = <math>\frac{12.184(ft-k)}{6.598(ft-k)}</math> = 1.847 ≥ 1.2 <u>o.k.</u>
:<math>\frac{\nu_u}{\phi} = \frac{(1.3)(1.67)(1.667k)}{(0.85)(12 in.)(7.75 in.)} (1000\frac{lb}{k})</math> = 45.8 psi < 126.5 psi <u>o.k.</u>


:'''Footing Pressure'''
'''Footing Design - Bottom Steel'''


:<math>\bar{x} = \frac{12.184(ft-k) - 6.598(ft-k)}{1.951k + 3.516k}</math> = 1.022 ft. from heel
It is not considered necessary to design footing reinforcement based upon a load case which includes collision loads.


:''e'' = <math>\frac{5.75 ft.}{2} - 1.022 ft.</math> = 1.853 ft.
:'''Dead Load and Earth Pressure Only'''


:Allowable Pressure = (1.25)(3.0ksf) = 3.75ksf
[[image:751.24.3.4 dead load.jpg|center|250px]]


:Heel: <math> P_H =\frac {2(\Sigma V)}{3b[\frac{L}{2} - e]} = \frac {2(5.467k)}{3(1 ft.)\big[\frac{5.75 ft.}{2} - 1.853 ft.\big]}</math> = 3.566ksf <u>o.k.</u>
:''Footing wt.'' = <math>\Big[\frac{11.5}{12}ft.\Big](4.917 ft.)\Big[0.150 \frac{k}{ft.^3}\Big](1 ft.)</math> = 0.707k


'''Stem Design-Steel in Rear Face'''
:''β<sub>E</sub>'' = 1.3 (lateral earth pressure)


[[image:751.24.3.4 steel in rear face.jpg|center|250px]]
:''γ'' = 1.3


''γ'' = 1.3
:Apply Load Factors:


''β<sub>E</sub>'' = 1.3 (active lateral earth pressure)
:''ΣV'' = 1.951k (1.3) = 2.536k


d = 10 in. − 2 in. − (0.5 in./2) = 7.75 in.
:''ΣM<sub>R</sub>'' = 8.231(ft−k)(1.3) = 10.700(ft−k)


<math>P_{AH} = \frac{1}{2}\gamma_s C_a H^2 cos\delta = \frac{1}{2}\Bigg[0.120 \frac{k}{ft^3}\Bigg](0.462)(4 ft.)^2(1 ft.) cos 21.801^\circ</math>
:''ΣM<sub>OT</sub>'' = 1.045(ft−k)(1.3)(1.3) = 1.766(ft−k)


''P<sub>AH</sub>'' = 0.412k
:''Footing wt.'' = 0.707k (1.3) = 0.919k


''M<sub>u</sub>'' = (1.333 ft.)(0.412k)(1.3)(1.3) = 0.928(ft−k)
:<math>\bar{x} = \frac{10.700(ft-k) - 1.766(ft-k)}{2.536k}</math> = 3.523 ft.


<math>R_n = \frac{M_u}{\phi b d^2} = \frac{0.928(ft-k)}{(0.9)(1 ft.)(7.75 in.)^2}\Big(1000\frac{lb}{k}\Big)</math> = 17.160psi
:<math>e = 3.523 ft. - \frac{5.75ft}{2}</math> = 0.648 ft.


<math>\rho = \frac{0.85f_c}{f_y}\Bigg[1 - \sqrt{1 - \frac{2R_n}{0.85 f_c}}\Bigg]</math>
:<math>P_H = \frac{2.536k}{(1 ft.)(5.75 ft.)}\Bigg[1 + \frac{6(0.648 ft.)}{5.75 ft.}\Bigg]</math> = 0.739 ksf


<math>\rho = \frac{4000 psi}{60,000 psi}\Bigg[1 - \sqrt{1 - \frac{2(17.160 psi)}{0.85 (4000psi)}}\Bigg]</math> = 0.000287
:<math>P_T = \frac{2.536k}{(1 ft.)(5.75 ft.)}\Bigg[1 - \frac{6(0.648 ft.)}{5.75 ft.}\Bigg]</math> = 0.143ksf


<math>\rho_{min} = 1.7 \Bigg[\frac{h}{d}\Bigg]^2 \frac{\sqrt{f_c}}{f_y}</math>
:<math>P_W = 0.143 ksf + [0.739 ksf - 0.143 ksf]\Bigg[\frac{4.917 ft.}{5.75 ft.}\Bigg]</math> = 0.653 ksf


<math>\rho_{min} = 1.7 \Bigg[\frac{10 in.}{7.75 in.}\Bigg]^2 \frac{\sqrt{4000 psi}}{60000 psi}</math> = 0.00298
:Moment at Wall Face:


Use ''ρ'' = (4/3)ρ = (4/3)(0.000287) = 0.000382
:<math>M_W = \Big[0.143\frac{k}{ft.}\Big]\Bigg[\frac{(4.917 ft.)^2}{2}\Bigg] + \frac{1}{3}(4.917 ft.)^2 \Bigg[0.653\frac{k}{ft.} - 0.143\frac{k}{ft.}\Bigg]\frac{1}{2} -  0.919k \Bigg[\frac{4.917 ft.}{2}\Bigg]</math> = 1.524(ft−k)


<math>A_{S_{Req}} = \rho bd = 0.000382(12 in.)(7.75 in.) = 0.036 \frac{in^2}{ft.}</math>
:'''Dead Load, Earth Pressure, and Live Load'''


One #4 bar has A<sub>S</sub> = 0.196 in<sup>2</sup>, so the required minimum of one #4 bar every 12 in. controls.
::'''Live Load 1 ft. From Stem Face'''


<u>Use #4's @ 12 in. (min)</u>
[[image:751.24.3.4 live load.jpg|center|300px]]


(These bars are also the bars in the bottom of the footing so the smaller of the two required spacings will be used.)
::''β<sub>E</sub>'' = 1.3 (lateral earth pressure)


:'''Check Shear'''
::''β<sub>LL</sub>'' = 1.67


:<math>\frac{\nu_u}{\phi} \le V_n</math>
::''γ'' = 1.3


:<math>\frac{\nu_u}{\phi} = \frac{(1.3)(1.3)(0.412k)}{0.85(12 in.)(7.75 in.)}(1000\frac{lb}{k})</math>  = 8.8 psi
::Apply Load Factors:


:<math>\nu_c = 2 \sqrt{f'_c}</math>
::''F<sub>LL</sub>'' = 3.516k(1.3)(1.67) = 7.633k


:<math>\nu_c = 2 \sqrt{4, 000 psi}</math> = 126.5 psi > 8.8 psi <u>o.k.</u>
::''ΣV'' = 7.633k + 1.951k(1.3) = 10.169k


'''Stem Design-Steel in Front Face (Collision Loads)'''
::''ΣM<sub>OT</sub>'' = 1.045(ft−k)(1.3)(1.3) = 1.766(ft−k)


[[image:751.24.3.4 steel in front face.jpg|center|300px]]
::''ΣM<sub>R</sub>'' = 8.231(ft−k)(1.3) + 3.917 ft.(7.633k) = 40.599(ft−k)


::<math>\bar{x} = \frac{40.599(ft-k) - 1.766(ft-k)}{10.169k}</math> = 3.819 ft.


The soil pressure on the back of the stem becomes passive soil pressure during a collision, however this pressure is ignored for reinforcement design.
::''e'' = 3.819 ft. − (5.75 ft./2) = 0.944 ft.


''γ'' = 1.3
::<math>P_T = \Bigg[\frac{10.169k}{(1 ft.)(5.75 ft.)}\Bigg]\Bigg[{1 - \frac{ 6(0.944 ft.)}{5.75 ft.}}\Bigg]</math> = 0.026 ksf


''β<sub>LL</sub>'' = 1.67
::<math>P_H = \Bigg[\frac{10.169k}{(1 ft.)(5.75 ft.)}\Bigg]\Bigg[{1 + \frac{ 6(0.944 ft.)}{5.75 ft.}}\Bigg]</math> = 3.511 ksf


<math>d = 10 in. - 1.5 in. - 0.5 in. - \frac{0.5 in.}{2}</math> = 7.75 in.
::<math>P_W = 0.026 ksf + [3.511 ksf - 0.026 ksf]\Big[\frac{4.917 ft.}{5.75 ft.}\Big]</math> = 3.006 ksf


<math>F_{COLL} = \frac{10k}{2L} = \frac{10k}{(2)(3 ft.)}</math> = 1.667 k/ft.
::<math>P_{LL} = 0.026 ksf + [3.511 ksf - 0.026 ksf]\Bigg[\frac{3.917 ft.}{5.75 ft.}\Bigg] </math> = 2.400 ksf


''M<sub>u</sub>'' = 1.667k/ft. (1 ft.)(3 ft.)(1.3)(1.67) = 10.855(ft−k)
::Footing wt. from face of wall to toe:


<math>R_n = \frac{10.855(ft-k)}{0.9(1 ft.)(7.75 in.)^2} (1000\frac{lb}{k})</math> = 200.809 psi
::''Footing wt.'' = <math>1.3\Bigg[\frac{11.5}{12} ft.\Bigg](4.917 ft.)\Bigg[0.150 \frac{k}{ft^3}\Bigg](1 ft.)</math> = 0.919k


<math>\rho = \frac{0.85(4000 psi)}{60,000 psi}\Bigg[1 - \sqrt{1 - \frac{2(200.809 psi)}{0.85(4000psi)}}\Bigg]</math> = 0.00345
::Footing wt. from LL<sub>WL</sub> to toe:


<math>\rho_{min} = 1.7\Bigg[\frac{10 in.}{7.75 in.}\Bigg]^2 \frac{\sqrt{4000 psi}}{60,000 psi}</math> = 0.00298
::''Footing wt.'' = <math>1.3\Bigg[\frac{11.5}{12} ft.\Bigg](3.917 ft.)\Bigg[0.150 \frac{k}{ft^3}\Bigg](1 ft.)</math> = 0.732k


<math>A_{S_{Req}} = 0.00345 (12 in.)(7.75 in.) = 0.321 \frac{in.^2}{ft.}</math>
::Moment at Wall Face:


One #4 bar has A<sub>S</sub> = 0.196 in<sup>2</sup>.
::''M<sub>W</sub> = <math>0.026\frac{k}{ft} \frac{(4.917 ft.)^2}{2} - 7.633k (1 ft.) + \frac{1}{2}\Bigg[3.006\frac{k}{ft} - 0.026\frac{k}{ft}\Bigg](4.917 ft.)^2\Big[\frac{1}{3}\Big] - 0.919k\frac{(4.917 ft.)}{2}</math>


<math>\frac{s}{0.196 in.^2} = \frac{12 in.}{0.321 in.^2}</math>
::M<sub>W</sub> = 2.430(ft−k)


''s'' = 7.3 in.
::Moment at LL<sub>WL</sub>:


<u>Use #4's @ 7 in.</u>
::''M<sub>LL</sub>'' = <math>0.026\frac{k}{ft} \frac{(3.917 ft.)^2}{2} - 0.732k \frac{(3.917 ft.)}{2} + \frac{1}{2}\Bigg[2.400\frac{k}{ft} - 0.026\frac{k}{ft}\Bigg](3.917 ft.)^2\Big[\frac{1}{3}\Big] </math> = 4.837(ft−k)


:'''Check Shear'''
::'''Live Load 1 ft. From Toe'''


:<math>\frac{\nu_u}{\phi} = \frac{(1.3)(1.67)(1.667k)}{(0.85)(12 in.)(7.75 in.)} (1000\frac{lb}{k})</math> = 45.8 psi < 126.5 psi <u>o.k.</u>
[[image:751.24.3.4 toe.jpg|center|250px]]


'''Footing Design - Bottom Steel'''
::Apply Load Factors:


It is not considered necessary to design footing reinforcement based upon a load case which includes collision loads.
::''F<sub>LL</sub>'' = 2.324k(1.3)(1.67) = 5.045k


:'''Dead Load and Earth Pressure Only'''
::''ΣV'' = 5.045k + 1.951k(1.3) = 7.581k


[[image:751.24.3.4 dead load.jpg|center|250px]]
::''ΣM<sub>OT</sub>'' = 1.045(ft−k)(1.3)(1.3) = 1.766(ft−k)


:''Footing wt.'' = <math>\Big[\frac{11.5}{12}ft.\Big](4.917 ft.)\Big[0.150 \frac{k}{ft.^3}\Big](1 ft.)</math> = 0.707k
::''ΣM<sub>R</sub>'' = 8.231(ft−k)(1.3) + 5.045k(1ft.) = 15.745(ft−k)


:''β<sub>E</sub>'' = 1.3 (lateral earth pressure)
::<math>\bar{x} = \frac{15.745(ft-k)- 1.766(ft-k)}{7.581k}</math> = 1.844 ft.


:''γ'' = 1.3
::<math>e = \frac{5.75 ft.}{2} - 1.844 ft.</math> = 1.031 ft.


:Apply Load Factors:
::''P<sub>H</sub>'' = 0 ksf


:''ΣV'' = 1.951k (1.3) = 2.536k
::<math>P_T = \frac{2(7.581k)}{3(1 ft.)\big[\frac{5.75 ft.}{2} - 1.031 ft.\big]}</math> = 2.741 ksf


:''ΣM<sub>R</sub>'' = 8.231(ft−k)(1.3) = 10.700(ft−k)
::''L<sub>1</sub>'' = 3[(L/2)− e]


:''ΣM<sub>OT</sub>'' = 1.045(ft−k)(1.3)(1.3) = 1.766(ft−k)
::''L<sub>1</sub>'' = 3[(5.75 ft./2)1.031 ft.] = 5.532 ft.


:''Footing wt.'' = 0.707k (1.3) = 0.919k
::<math>P_W = 2.741 ksf \Big[\frac{0.615 ft.}{5.532 ft.}\Big]</math> = 0.305 ksf


:<math>\bar{x} = \frac{10.700(ft-k) - 1.766(ft-k)}{2.536k}</math> = 3.523 ft.
::<math>P_{LL} = 2.741 ksf \Big[\frac{4.432 ft.}{5.532 ft.}\Big]</math> = 2.196 ksf


:<math>e = 3.523 ft. - \frac{5.75ft}{2}</math> = 0.648 ft.
::Moment at Wall Face:


:<math>P_H = \frac{2.536k}{(1 ft.)(5.75 ft.)}\Bigg[1 + \frac{6(0.648 ft.)}{5.75 ft.}\Bigg]</math> = 0.739 ksf
::''M<sub>W</sub>'' = <math> -5.045k (3.917 ft.) - 0.919k\Bigg[\frac{4.917 ft.}{2}\Bigg] + \frac{1}{2}(0.305\frac{k}{ft.})(4.917 ft.)^2 + \frac{1}{2}(4.917 ft.)^2 \Bigg[2.741\frac{k}{ft.} - 0.305\frac{k}{ft.}\Bigg]\Bigg[\frac{2}{3}\Bigg]</math> = 1.298(ft−k)


:<math>P_T = \frac{2.536k}{(1 ft.)(5.75 ft.)}\Bigg[1 - \frac{6(0.648 ft.)}{5.75 ft.}\Bigg]</math> = 0.143ksf
::Moment at LL<sub>WL</sub>:


:<math>P_W = 0.143 ksf + [0.739 ksf - 0.143 ksf]\Bigg[\frac{4.917 ft.}{5.75 ft.}\Bigg]</math> = 0.653 ksf
::''M<sub>LL</sub>'' = <math>-0.187k(0.5 ft.) + 2.196\frac{k}{ft.}\frac{(1 ft.)^2}{2} +\frac{1}{2}(1 ft.)\Bigg[2.741\frac{k}{ft.}  - 2.196\frac{k}{ft.}\Bigg]\Bigg[\frac{2}{3}\Bigg](1 ft.)</math> = 1.186(ft−k)


:Moment at Wall Face:
:'''Design Flexural Steel in Bottom of Footing'''


:<math>M_W = \Big[0.143\frac{k}{ft.}\Big]\Bigg[\frac{(4.917 ft.)^2}{2}\Bigg] + \frac{1}{3}(4.917 ft.)^2 \Bigg[0.653\frac{k}{ft.} - 0.143\frac{k}{ft.}\Bigg]\frac{1}{2} -  0.919k \Bigg[\frac{4.917 ft.}{2}\Bigg]</math> = 1.524(ft−k)
:''d'' = 11.5 in. 4 in. = 7.500 in.


:'''Dead Load, Earth Pressure, and Live Load'''
:''M<sub>u</sub>'' = 4.837(ft−k) (controlling moment)


::'''Live Load 1 ft. From Stem Face'''
:<math>R_n = \frac{4.837(ft-k)}{0.9(1 ft.)(7.5 in.)^2}</math> = 0.096 ksi


[[image:751.24.3.4 live load.jpg|center|300px]]
:<math>\rho = \frac{0.85(4000 psi)}{60,000 psi}\Bigg[1 - \sqrt{1 - \frac{2(0.096 ksi)}{0.85(4 ksi)}}\Bigg] </math> = 0.00162


::''β<sub>E</sub>'' = 1.3 (lateral earth pressure)
:<math>\rho_{min} = 1.7\Bigg[\frac{11.5 in.}{7.5 in.}\Bigg]^2\frac{\sqrt{4000 psi}}{60,000 psi}</math> = 0.00421


::''β<sub>LL</sub>'' = 1.67
:Use ''ρ'' = (4/3)''ρ'' = (4/3)(0.00162) = 0.00216


::''γ'' = 1.3
:''A<sub>S<sub>Req</sub></sub>'' = 0.00216(12 in.)(7.5 in.) = 0.194 in<sup>2</sup>/ft.


::Apply Load Factors:


::''F<sub>LL</sub>'' = 3.516k(1.3)(1.67) = 7.633k
:<math>\frac{s}{0.196 in^2} = \frac{12 in.}{0.194 in^2}</math>


::''ΣV'' = 7.633k + 1.951k(1.3) = 10.169k
:''s'' = 12.1 in.


::''ΣM<sub>OT</sub>'' = 1.045(ft−k)(1.3)(1.3) = 1.766(ft−k)
:<u>Use #4's @ 12 in. cts.</u> (Also use this spacing in the back of the stem.)


::''ΣM<sub>R</sub>'' = 8.231(ft−k)(1.3) + 3.917 ft.(7.633k) = 40.599(ft−k)
:'''Check Shear'''


::<math>\bar{x} = \frac{40.599(ft-k) - 1.766(ft-k)}{10.169k}</math> = 3.819 ft.
::'''Dead Load and Earth Pressure Only'''


::''e'' = 3.819 ft. (5.75 ft./2) = 0.944 ft.
::<math>V_W = 0.143\frac{k}{ft.}(4.917 ft.) + \frac{1}{2}(4.917 ft.)\Big[0.653\frac{k}{ft.} - 0.143\frac{k}{ft.}\Big] - 0.919k</math>


::<math>P_T = \Bigg[\frac{10.169k}{(1 ft.)(5.75 ft.)}\Bigg]\Bigg[{1 - \frac{ 6(0.944 ft.)}{5.75 ft.}}\Bigg]</math> = 0.026 ksf
::''V<sub>W</sub>'' = 1.038k


::<math>P_H = \Bigg[\frac{10.169k}{(1 ft.)(5.75 ft.)}\Bigg]\Bigg[{1 + \frac{ 6(0.944 ft.)}{5.75 ft.}}\Bigg]</math> = 3.511 ksf
::'''Live Load 1 ft. From Stem Face'''


::<math>P_W = 0.026 ksf + [3.511 ksf - 0.026 ksf]\Big[\frac{4.917 ft.}{5.75 ft.}\Big]</math> = 3.006 ksf
::Shear at the wall can be neglected for this loading case.


::<math>P_{LL} = 0.026 ksf + [3.511 ksf - 0.026 ksf]\Bigg[\frac{3.917 ft.}{5.75 ft.}\Bigg] </math> = 2.400 ksf
::<math>V_{LL} = 0.026\frac{k}{ft.}(3.917 ft.) + \frac{1}{2}(3.917 ft.)\Big[2.400\frac{k}{ft.} - 0.026\frac{k}{ft.}\Big] - 0.732k</math>


::Footing wt. from face of wall to toe:
::''V<sub>LL</sub>'' = 4.019k


::''Footing wt.'' = <math>1.3\Bigg[\frac{11.5}{12} ft.\Bigg](4.917 ft.)\Bigg[0.150 \frac{k}{ft^3}\Bigg](1 ft.)</math> = 0.919k
::'''Live Load 1 ft. From Toe'''


::Footing wt. from LL<sub>WL</sub> to toe:
::<math>V_W = 0.305\frac{k}{ft.}(4.917 ft.) + \frac{1}{2}(4.917 ft.)\Big[2.741\frac{k}{ft.} - 0.305\frac{k}{ft.}\Big] - 0.919k - 5.045k</math>


::''Footing wt.'' = <math>1.3\Bigg[\frac{11.5}{12} ft.\Bigg](3.917 ft.)\Bigg[0.150 \frac{k}{ft^3}\Bigg](1 ft.)</math> = 0.732k
::''V<sub>W</sub>'' = 1.525k


::Moment at Wall Face:
::<math>V_{LL} = 2.196\frac{k}{ft.}(1ft) + \frac{1}{2}(1ft)\Big[2.741\frac{k}{ft.} - 2.196\frac{k}{ft.}\Big] - 0.187k</math>


::''M<sub>W</sub> = <math>0.026\frac{k}{ft} \frac{(4.917 ft.)^2}{2} - 7.633k (1 ft.) + \frac{1}{2}\Bigg[3.006\frac{k}{ft} - 0.026\frac{k}{ft}\Bigg](4.917 ft.)^2\Big[\frac{1}{3}\Big] - 0.919k\frac{(4.917 ft.)}{2}</math>
::''V<sub>LL</sub>'' = 2.282k


::M<sub>W</sub> = 2.430(ft−k)
:Use ''V<sub>U</sub>'' = 4.019k


::Moment at LL<sub>WL</sub>:
:<math>\frac{\nu_u}{\phi} = \frac{4019(lbs)}{0.85(12 in.)(7.5 in.)} = 52.5 psi < 2\sqrt{4000 psi}</math> = 126.5 psi


::''M<sub>LL</sub>'' = <math>0.026\frac{k}{ft} \frac{(3.917 ft.)^2}{2} - 0.732k \frac{(3.917 ft.)}{2} + \frac{1}{2}\Bigg[2.400\frac{k}{ft} - 0.026\frac{k}{ft}\Bigg](3.917 ft.)^2\Big[\frac{1}{3}\Big] </math> = 4.837(ft−k)
'''Shear Key Design'''


::'''Live Load 1 ft. From Toe'''
[[image:751.24.3.4 shear key.jpg|center|300px]]


[[image:751.24.3.4 toe.jpg|center|250px]]
For concrete cast against and permanently exposed to earth, minimum cover for reinforcement is 3 inches.


::Apply Load Factors:
<math>d = 12 in. - 3 in. - \frac{1}{2}\Big[\frac{1}{2}in.\Big]</math> = 8.75 in.


::''F<sub>LL</sub>'' = 2.324k(1.3)(1.67) = 5.045k
<math>P_1 = 0.120\frac{k}{ft^3}(1 ft.)(2.882)\Big[\frac{11.5}{12}ft.\Big]</math> = 0.331 k/ft.


::''ΣV'' = 5.045k + 1.951k(1.3) = 7.581k
<math>P_2 = 0.120\frac{k}{ft^3}(1 ft.)(2.882)\Big[\frac{29.5}{12}ft.\Big]</math> = 0.850 k/ft.


::''ΣM<sub>OT</sub>'' = 1.045(ft−k)(1.3)(1.3) = 1.766(ft−k)
<math>M_u = (1.3)(1.3)\Bigg\{0.331\frac{k}{ft.}\frac{(1.5 ft.)^2}{2} + \frac{1}{2}(1.5 ft.)\Big[0.850\frac{k}{ft.} - 0.331\frac{k}{ft}\Big]\Big[\frac{2}{3}\Big](1.5 ft.)\Bigg\}</math>


::''ΣM<sub>R</sub>'' = 8.231(ft−k)(1.3) + 5.045k(1ft.) = 15.745(ft−k)
''M<sub>u</sub>'' = 1.287(ft−k)


::<math>\bar{x} = \frac{15.745(ft-k)- 1.766(ft-k)}{7.581k}</math> = 1.844 ft.
<math>R_n = \frac{1.287(ft-k)}{0.9(1ft.)(8.75in.)^2}</math> = 0.0187 ksi


::<math>e = \frac{5.75 ft.}{2} - 1.844 ft.</math> = 1.031 ft.
<math>\rho = \frac{0.85(4000psi)}{60,000psi}\Bigg[1 - \sqrt{1 - \frac{2(0.0187ksi)}{0.85(4ksi)}}\Bigg]</math> = 0.000312


::''P<sub>H</sub>'' = 0 ksf
<math>\rho_{min} = 1.7\Big[\frac{12in.}{8.75in.}\Big]^2\frac{\sqrt{4000psi}}{60,000psi}</math> = 0.00337


::<math>P_T = \frac{2(7.581k)}{3(1 ft.)\big[\frac{5.75 ft.}{2} - 1.031 ft.\big]}</math> = 2.741 ksf
Use ''ρ'' = (4/3)''ρ'' = (4/3)(0.000312) = 0.000416


::''L<sub>1</sub>'' = 3[(L/2)− e]
''A<sub>S<sub>Req</sub></sub>'' = 0.000416 (12 in.)(8.75 in.) = 0.0437 in<sup>2</sup>/ft.


::''L<sub>1</sub>'' = 3[(5.75 ft./2)− 1.031 ft.] = 5.532 ft.


::<math>P_W = 2.741 ksf \Big[\frac{0.615 ft.}{5.532 ft.}\Big]</math> = 0.305 ksf
<math>\frac{s}{0.196 in.^2} = \frac{12in.}{0.0437in.^2}</math>


::<math>P_{LL} = 2.741 ksf \Big[\frac{4.432 ft.}{5.532 ft.}\Big]</math> = 2.196 ksf
''s'' = 53.8 in.


::Moment at Wall Face:
<u>Use #4's @ 18 in. cts. (min)</u>


::''M<sub>W</sub>'' = <math> -5.045k (3.917 ft.) - 0.919k\Bigg[\frac{4.917 ft.}{2}\Bigg] + \frac{1}{2}(0.305\frac{k}{ft.})(4.917 ft.)^2 + \frac{1}{2}(4.917 ft.)^2 \Bigg[2.741\frac{k}{ft.} - 0.305\frac{k}{ft.}\Bigg]\Bigg[\frac{2}{3}\Bigg]</math> = 1.298(ft−k)
:'''Check Shear'''


::Moment at LL<sub>WL</sub>:
:''V'' = 0.886k


::''M<sub>LL</sub>'' = <math>-0.187k(0.5 ft.) + 2.196\frac{k}{ft.}\frac{(1 ft.)^2}{2} +\frac{1}{2}(1 ft.)\Bigg[2.741\frac{k}{ft.}  - 2.196\frac{k}{ft.}\Bigg]\Bigg[\frac{2}{3}\Bigg](1 ft.)</math> = 1.186(ft−k)
:<math>\frac{\nu_u}{\phi} = \frac{(1.3)(1.3)(886 lbs)}{0.85(12 in.)(8.75 in.)}</math> = 16.8 psi < 126.5 psi <u>o.k.</u>


:'''Design Flexural Steel in Bottom of Footing'''
'''Reinforcement Summary'''


:''d'' = 11.5 in. 4 in. = 7.500 in.
[[image:751.24.3.4 summary.jpg|center|400px]]


:''M<sub>u</sub>'' = 4.837(ft−k) (controlling moment)
===751.24.3.5 Example 3: Pile Footing Cantilever Wall===


:<math>R_n = \frac{4.837(ft-k)}{0.9(1 ft.)(7.5 in.)^2}</math> = 0.096 ksi
[[image:751.24.3.5.jpg|center|850px]]


:<math>\rho = \frac{0.85(4000 psi)}{60,000 psi}\Bigg[1 - \sqrt{1 - \frac{2(0.096 ksi)}{0.85(4 ksi)}}\Bigg] </math> = 0.00162
''f’<sub>c</sub>'' = 3,000 psi


:<math>\rho_{min} = 1.7\Bigg[\frac{11.5 in.}{7.5 in.}\Bigg]^2\frac{\sqrt{4000 psi}}{60,000 psi}</math> = 0.00421
''f<sub>y</sub>'' = 60,000 psi


:Use ''ρ'' = (4/3)''ρ'' = (4/3)(0.00162) = 0.00216
''φ'' = 27°


:''A<sub>S<sub>Req</sub></sub>'' = 0.00216(12 in.)(7.5 in.) = 0.194 in<sup>2</sup>/ft.
''γ<sub>s</sub>'' = 120 pcf


Pile type: HP 10 x 42


:<math>\frac{s}{0.196 in^2} = \frac{12 in.}{0.194 in^2}</math>
Allowable pile bearing = 56 tons


:''s'' = 12.1 in.
Pile width = 10 inches


:<u>Use #4's @ 12 in. cts.</u> (Also use this spacing in the back of the stem.)
Toe pile batter = 1:3


:'''Check Shear'''
See [[751.12 Barriers, Railings, Curbs and Fences|EPG 751.12 Barriers, Railings, Curbs and Fences]] for weight and centroid of barrier.


::'''Dead Load and Earth Pressure Only'''
'''Assumptions'''


::<math>V_W = 0.143\frac{k}{ft.}(4.917 ft.) + \frac{1}{2}(4.917 ft.)\Big[0.653\frac{k}{ft.} - 0.143\frac{k}{ft.}\Big] - 0.919k</math>
:* Retaining wall is located such that traffic can come within half of the wall height to the plane where earth pressure is applied.


::''V<sub>W</sub>'' = 1.038k
:* Reinforcement design is for one foot of wall length.


::'''Live Load 1 ft. From Stem Face'''
:* Sum moments about the centerline of the toe pile at a distance of 6B (where B is the pile width) below the bottom of the footing for overturning.


::Shear at the wall can be neglected for this loading case.
:* Neglect top one foot of fill over toe in determining soil weight and passive pressure on shear key.


::<math>V_{LL} = 0.026\frac{k}{ft.}(3.917 ft.) + \frac{1}{2}(3.917 ft.)\Big[2.400\frac{k}{ft.} - 0.026\frac{k}{ft.}\Big] - 0.732k</math>
:* Neglect all fill over toe in designing stem reinforcement.


::''V<sub>LL</sub>'' = 4.019k
:* The wall is designed as a cantilever supported by the footing.


::'''Live Load 1 ft. From Toe'''
:* Footing is designed as a cantilever supported by the wall.


::<math>V_W = 0.305\frac{k}{ft.}(4.917 ft.) + \frac{1}{2}(4.917 ft.)\Big[2.741\frac{k}{ft.} - 0.305\frac{k}{ft.}\Big] - 0.919k - 5.045k</math>
:* Critical sections for bending are at the front and back faces of the wall.


::''V<sub>W</sub>'' = 1.525k
:* Critical sections for shear are at the back face of the wall for the heel and at a distance d (effective depth) from the front face for the toe.


::<math>V_{LL} = 2.196\frac{k}{ft.}(1ft) + \frac{1}{2}(1ft)\Big[2.741\frac{k}{ft.} - 2.196\frac{k}{ft.}\Big] - 0.187k</math>
:* For load factors for design of concrete, see [[#Group Loads|EPG 751.24.1.2 Group Loads]].


::''V<sub>LL</sub>'' = 2.282k
<math>C_A = cos\delta\Bigg[\frac{cos\delta - \sqrt{cos^2\delta - cos^2\phi}}{cos\delta + \sqrt{cos^2\delta - cos^2\phi}}\Bigg]</math>


:Use ''V<sub>U</sub>'' = 4.019k
''δ'' = 0, ''ϕ'' = 27° so ''C<sub>A</sub>'' reduces to:


:<math>\frac{\nu_u}{\phi} = \frac{4019(lbs)}{0.85(12 in.)(7.5 in.)} = 52.5 psi < 2\sqrt{4000 psi}</math> = 126.5 psi
<math>C_A = \frac{1 - sin\phi}{1 + sin\phi} = \frac{1 - sin 27^\circ}{1 + sin 27^\circ}</math> = 0.376
<math>C_P = tan^2\Bigg[45^\circ + \frac{\phi}{2}\Bigg] = tan^2\Bigg[ 45^\circ + \frac{27^\circ}{2}\Bigg]</math> = 2.663


'''Shear Key Design'''
Table 751.24.3.5.1 is for stability check (moments taken about C.L. of toe pile at a depth of 6B below the bottom of the footing).


[[image:751.24.3.4 shear key.jpg|center|300px]]
{| border="1" class="wikitable" style="margin: 1em auto 1em auto" style="text-align:center"
 
|+ '''''Table 751.24.3.5.1'''''
For concrete cast against and permanently exposed to earth, minimum cover for reinforcement is 3 inches.
! style="background:#BEBEBE" colspan="2"|Load !! style="background:#BEBEBE"|Force (kips/ft) !! style="background:#BEBEBE"|Arm about C.L. of toe pile at 6B below footing (ft.) !! style="background:#BEBEBE"|Moment (ft-kips) per foot of wall length
 
|-
<math>d = 12 in. - 3 in. - \frac{1}{2}\Big[\frac{1}{2}in.\Big]</math> = 8.75 in.
|rowspan="5"|'''Dead Load'''||(1)|| 0.340|| 2.542|| 0.864
 
|-
<math>P_1 = 0.120\frac{k}{ft^3}(1 ft.)(2.882)\Big[\frac{11.5}{12}ft.\Big]</math> = 0.331 k/ft.
|(2)|| (1.333 ft.)(7.000 ft.)(0.150k/ft<sup>3</sup>) = 1.400 ||2.833|| 3.966
 
|-
<math>P_2 = 0.120\frac{k}{ft^3}(1 ft.)(2.882)\Big[\frac{29.5}{12}ft.\Big]</math> = 0.850 k/ft.
|(3)|| (3.000 ft.)(8.500 ft.)(0.150k/ft<sup>3</sup>) = 3.825|| 4.417|| 16.895
 
|-
<math>M_u = (1.3)(1.3)\Bigg\{0.331\frac{k}{ft.}\frac{(1.5 ft.)^2}{2} + \frac{1}{2}(1.5 ft.)\Big[0.850\frac{k}{ft.} - 0.331\frac{k}{ft}\Big]\Big[\frac{2}{3}\Big](1.5 ft.)\Bigg\}</math>
|(4)|| (1.000 ft.)(1.750 ft.)(0.150k/ft<sup>3</sup>) = <u>0.263</u>|| 4.417|| <u>1.162</u>
|-
|Σ||ΣV = 5.828 || - ||ΣM<sub>R</sub> = 22.887
|-
|rowspan="3"|'''Earth Load'''||(5)|| (7.000 ft.)(5.167 ft.)(0.120k/ft<sup>3</sup>) = 4.340|| 6.083|| 26.400
|-
|(6)|| (2.000 ft.)(2.000 ft.)(0.120k/ft<sup>3</sup>) = <u>0.480</u>|| 1.167|| <u>0.560</u>
|-
|Σ ||ΣV = 4.820|| - ||ΣM<sub>R</sub> = 26.960
|-
|rowspan="2"|'''Live Load Surcharge'''||P<sub>SV</sub>|| (2.000 ft.)(5.167 ft.)(0.120k/ft<sub>3</sub>) = 1.240|| 6.083|| M<sub>R</sub> = 7.543
|-
|P<sub>SH</sub>||(2.000 ft.)(0.376)(10.000 ft.)(0.120k/ft<sup>3</sup>) = 0.902||10.000|| M<sub>OT</sub> = 9.020
|-
|rowspan="2"|'''Earth Pressure'''||P<sub>A</sub>||2.256<sup>'''1'''</sup>|| 8.333|| M<sub>OT</sub> = 18.799
|-
|P<sub>P</sub>|| 3.285<sup>'''2'''</sup> || - || -
|-
|colspan="2"|'''Collision Force''' (F<sub>COL</sub>)||(10.000k)/[2(7.000 ft.)] = 0.714|| 18.000 ||M<sub>OT</sub> = 12.852
|-
|colspan="2"|'''Heel Pile Tension''' (P<sub>HV</sub>)||(3.000 tons)(2 k/ton)(1 pile)/(12.000 ft.) = 0.500|| 7.167|| M<sub>R</sub> = 3.584
|-
|colspan="2"|'''Toe Pile Batter''' (P<sub>BH</sub>)|| 5.903<sup>'''3'''</sup>|| - || -
|-
|colspan="2"|'''Passive Pile Pressure''' (P<sub>pp</sub>)|| 0.832<sup>'''4'''</sup>|| - || -
|-
|colspan="5" align="left"|<sup>'''1'''</sup> <math>P_A = \frac{1}{2}\gamma_S C_A H^2 = \frac{1}{2}\Bigg[0.120\frac{k}{ft^3}\Bigg](0.376)(10 ft.)^3 = 2.256\frac{k}{ft}</math>
|-
|colspan="5" align="left"|<sup>'''2'''</sup> <math>P_P = \frac{1}{2}\gamma_S C_A\Big[H_2^2 - H_1^2\Big] = \frac{1}{2}\Bigg[0.120\frac{k}{ft^3}\Bigg](2.663)[(6.75 ft.)^2 - (5 ft.)^2] = 3.285\frac{k}{ft}</math>
|-
|colspan="5" align="left"|<sup>'''3'''</sup> <math>P_{BH} = \Big(56 \frac{tons}{pile}\Big)\Big( 2 \frac{k}{ton}\Big)(2 piles)\Bigg(\frac{4 in.}{\sqrt{(12 in.)^2 + (4 in.)^2}}\Bigg)\Big(\frac{1}{12 ft.}\Big) = 5.903 \frac{k}{ft}</math>
|-
|colspan="5" align="left"|<sup>'''4'''</sup> <math>P_{PP} = \frac{1}{2}(2.663)(5 ft.)^2\Big(0.120 \frac{k}{ft^3}\Big)(0.833 ft.)(3 piles)\Big(\frac{1}{12 ft.}\Big) = 0.832\frac{k}{ft}</math>
|}


''M<sub>u</sub>'' = 1.287(ft−k)


<math>R_n = \frac{1.287(ft-k)}{0.9(1ft.)(8.75in.)^2}</math> = 0.0187 ksi
Table 751.24.3.5.2 is for bearing pressure checks (moments taken about C.L of toe pile at the bottom of the footing).


<math>\rho = \frac{0.85(4000psi)}{60,000psi}\Bigg[1 - \sqrt{1 - \frac{2(0.0187ksi)}{0.85(4ksi)}}\Bigg]</math> = 0.000312
{| border="1" class="wikitable" style="margin: 1em auto 1em auto" style="text-align:center"
 
|+ '''''Table 751.24.3.5.2'''''
<math>\rho_{min} = 1.7\Big[\frac{12in.}{8.75in.}\Big]^2\frac{\sqrt{4000psi}}{60,000psi}</math> = 0.00337
! style="background:#BEBEBE" colspan="2"|Load !! style="background:#BEBEBE"|Force (kips/ft) !! style="background:#BEBEBE"|Arm about C.L. of toe pile at 6B below footing (ft.) !! style="background:#BEBEBE"|Moment (ft-kips) per foot of wall length
 
|-
Use ''ρ'' = (4/3)''ρ'' = (4/3)(0.000312) = 0.000416
|rowspan="5"|'''Dead Load'''||(1)|| 0.340|| 0.875|| 0.298
 
|-
''A<sub>S<sub>Req</sub></sub>'' = 0.000416 (12 in.)(8.75 in.) = 0.0437 in<sup>2</sup>/ft.
|(2)|| (1.333 ft.)(7.000 ft.)(0.150k/ft<sup>3</sup>) = 1.400 ||1.167|| 1.634
 
|-
 
|(3)|| (3.000 ft.)(8.500 ft.)(0.150k/ft<sup>3</sup>) = 3.825|| 2.750|| 10.519
<math>\frac{s}{0.196 in.^2} = \frac{12in.}{0.0437in.^2}</math>
|-
 
|(4)|| (1.000 ft.)(1.750 ft.)(0.150k/ft<sup>3</sup>) = <u>0.263</u>|| 2.750|| <u>0.723</u>
''s'' = 53.8 in.
|-
 
|Σ||ΣV = 5.828 || - ||ΣM<sub>R</sub> = 13.174
<u>Use #4's @ 18 in. cts. (min)</u>
|-
 
|rowspan="3"|'''Earth Load'''||(5)|| (7.000 ft.)(5.167 ft.)(0.120k/ft<sup>3</sup>) = 4.340|| 4.417|| 19.170
:'''Check Shear'''
|-
|(6)|| (2.000 ft.)(2.000 ft.)(0.120k/ft<sup>3</sup>) = <u>0.480</u>|| -0.500|| <u>-0.240</u>
|-
|Σ ||ΣV = 4.820|| - ||ΣM<sub>R</sub> = 18.930
|-
|rowspan="2"|'''Live Load Surcharge'''||P<sub>SV</sub>|| (2.000 ft.)(5.167 ft.)(0.120k/ft<sub>3</sub>) = 1.240|| 4.417|| M<sub>R</sub> = 5.477
|-
|P<sub>SH</sub>||(2.000 ft.)(0.376)(10.000 ft.)(0.120k/ft<sup>3</sup>) = 0.902||5.000|| M<sub>OT</sub> = 4.510
|-
|rowspan="2"|'''Earth Pressure'''||P<sub>A</sub>||2.256|| 3.333|| M<sub>OT</sub> = 7.519
|-
|P<sub>P</sub>|| 3.285 || - || -
|-
|colspan="2"|'''Collision Force''' (F<sub>COL</sub>)||(10.000k)/[2(7.000 ft.)] = 0.714|| 13.000 ||M<sub>OT</sub> = 9.282
|-
|colspan="2"|'''Heel Pile Tension''' (P<sub>HV</sub>)||(3.000 tons)(2 k/ton)(1 pile)/(12.000 ft.) = 0.500|| 5.500|| M<sub>R</sub> = 2.750
|-
|colspan="2"|'''Toe Pile Batter''' (P<sub>BH</sub>)|| 5.903|| - || -
|-
|colspan="2"|'''Passive Pile Pressure''' (P<sub>pp</sub>)|| 0.832|| - || -
|}


:''V'' = 0.886k
Investigate a representative 12 ft. strip. This will include one heel pile and two toe piles. The assumption is made that the stiffness of a batter pile in the vertical direction is the same as that of a vertical pile.


:<math>\frac{\nu_u}{\phi} = \frac{(1.3)(1.3)(886 lbs)}{0.85(12 in.)(8.75 in.)}</math> = 16.8 psi < 126.5 psi <u>o.k.</u>
Neutral Axis Location = [2piles(1.5 ft.) + 1pile(7 ft.)] / (3 piles) = 3.333 ft. from the toe.


'''Reinforcement Summary'''
[[image:751.24.3.5 neutral axis.jpg|center|350px]]


[[image:751.24.3.4 summary.jpg|center|400px]]
''I ''= Ad<sup>2</sup>


===751.24.3.5 Example 3: Pile Footing Cantilever Wall===
For repetitive 12 ft. strip:


[[image:751.24.3.5.jpg|center|850px]]
:Total pile area = 3A


''f’<sub>c</sub>'' = 3,000 psi
:''I ''= 2A(1.833 ft.)<sup>2</sup> + A(3.667 ft.)<sup>2</sup> = 20.167(A)ft.<sup>2</sup>


''f<sub>y</sub>'' = 60,000 psi
For a 1 ft. unit strip:


''φ'' = 27°
:<math>I = \frac{20.167(A)ft.^2}{12 ft.} = 1.681(A)ft.^2</math>


''γ<sub>s</sub>'' = 120 pcf
:Total pile area = (3A/12 ft.) = 0.250A


Pile type: HP 10 x 42
:'''Case I'''


Allowable pile bearing = 56 tons
:F.S. for overturning ≥ 1.5


Pile width = 10 inches
:F.S. for sliding ≥ 1.5


Toe pile batter = 1:3
::'''Check Overturning'''


Barrier curb weight = 340 lbs/ft. of length
::Neglect resisting moment due to P<sub>SV</sub> for this check.


Barrier curb resultant = 0.375 ft. from its flat back
::''ΣM<sub>R</sub>'' = 22.887(ft−k) + 26.960(ft−k) + 3.584(ft−k)


'''Assumptions'''
::''ΣM<sub>R</sub>'' = 53.431(ft−k)


:* Retaining wall is located such that traffic can come within half of the wall height to the plane where earth pressure is applied.
::''ΣM<sub>OT</sub>'' = 9.020(ft−k) + 18.799(ft−k) = 27.819(ft−k)


:* Reinforcement design is for one foot of wall length.
::''F.S.<sub>OT</sub>'' = <math>\frac{\Sigma M_R}{\Sigma M_{OT}} = \frac{53.431(ft-k)}{27.819(ft-k)}</math> = 1.921 > 1.5 <u>o.k.</u>


:* Sum moments about the centerline of the toe pile at a distance of 6B (where B is the pile width) below the bottom of the footing for overturning.
::'''Check Pile Bearing'''


:* Neglect top one foot of fill over toe in determining soil weight and passive pressure on shear key.
::Without P<sub>SV</sub> :


:* Neglect all fill over toe in designing stem reinforcement.
::''ΣV'' = 5.828k + 4.820k = 10.648k


:* The wall is designed as a cantilever supported by the footing.
::''e'' = <math>\frac{\Sigma M}{\Sigma V} = \frac{(13.174 + 18.930)(ft-k) - (4.510 + 7.519)(ft-k)}{10.648k}</math> = 1.885 ft.


:* Footing is designed as a cantilever supported by the wall.
::Moment arm = 1.885 ft. - 1.833 ft. = 0.052 ft.


:* Critical sections for bending are at the front and back faces of the wall.
::<math>P_T = \frac{\Sigma V}{A} - \frac{M_c}{I} = \frac{10.648k}{0.250A} - \frac{10.648k(0.052 ft.)(1.833 ft.)}{1.681(A)ft^2}</math>


:* Critical sections for shear are at the back face of the wall for the heel and at a distance d (effective depth) from the front face for the toe.
::<math>P_T = \frac{41.988}{A} k</math>


:* For load factors for design of concrete, see [[#Group Loads|EPG 751.24.1.2 Group Loads]].
::<math>P_H = \frac{10.648k}{0.250A} + \frac{10.648k(0.052 ft.)(3.667 ft.)}{1.681(A)ft^2}</math>
 
::<math>P_H = \frac{43.800}{A} k</math>
 
::Allowable pile load = 56 tons/pile. Each pile has area A, so:
 
::<math>P_T = 41.988\frac{k}{pile} = 20.944\frac{tons}{pile} </math> <u> o.k.</u>
 
::<math>P_H = 43.800\frac{k}{pile} = 21.900\frac{tons}{pile} </math> <u> o.k.</u>
 
::With P<sub>SV</sub>:
 
::''ΣV'' = 5.828k + 4.820k + 1.240k = 11.888k
 
::<math>e = \frac{(13.174 + 18.930 + 5.477)(ft-k) - (4.510 + 7.519)(ft-k)}{11.888k}</math> = 2.149 ft.


<math>C_A = cos\delta\Bigg[\frac{cos\delta - \sqrt{cos^2\delta - cos^2\phi}}{cos\delta + \sqrt{cos^2\delta - cos^2\phi}}\Bigg]</math>
::Moment arm = 2.149 ft. - 1.833 ft. = 0.316 ft.


''δ'' = 0, ''ϕ'' = 27° so ''C<sub>A</sub>'' reduces to:
::<math>P_T = \frac{11.888k}{0.250A} - \frac{11.888k(0.316 ft.)(1.833 ft.)}{1.681(A)ft^2} = 43.456k = 21.728\frac{tons}{pile}</math> <u> o.k.</u>


<math>C_A = \frac{1 - sin\phi}{1 + sin\phi} = \frac{1 - sin 27^\circ}{1 + sin 27^\circ}</math> = 0.376
::<math>P_H = \frac{11.888k}{0.250A} + \frac{11.888k(0.316 ft.)(3.667 ft.)}{1.681(A)ft^2} = 55.747k = 27.874\frac{tons}{pile}</math> <u> o.k.</u>
<math>C_P = tan^2\Bigg[45^\circ + \frac{\phi}{2}\Bigg] = tan^2\Bigg[ 45^\circ + \frac{27^\circ}{2}\Bigg]</math> = 2.663


Table 751.24.3.5.1 is for stability check (moments taken about C.L. of toe pile at a depth of 6B below the bottom of the footing).
::'''Check Sliding'''


{| border="1" class="wikitable" style="margin: 1em auto 1em auto" style="text-align:center"
::<math>F.S._{Sliding} = \frac{3.285k + 5.903k + 0.832k}{0.902 k + 2.256k}</math> = 3.173 ≥ 1.5 <u> o.k.</u>
|+ '''''Table 751.24.3.5.1'''''
! style="background:#BEBEBE" colspan="2"|Load !! style="background:#BEBEBE"|Force (kips/ft) !! style="background:#BEBEBE"|Arm about C.L. of toe pile at 6B below footing (ft.) !! style="background:#BEBEBE"|Moment (ft-kips) per foot of wall length
|-
|rowspan="5"|'''Dead Load'''||(1)|| 0.340|| 2.542|| 0.864
|-
|(2)|| (1.333 ft.)(7.000 ft.)(0.150k/ft<sup>3</sup>) = 1.400 ||2.833|| 3.966
|-
|(3)|| (3.000 ft.)(8.500 ft.)(0.150k/ft<sup>3</sup>) = 3.825|| 4.417|| 16.895
|-
|(4)|| (1.000 ft.)(1.750 ft.)(0.150k/ft<sup>3</sup>) = <u>0.263</u>|| 4.417|| <u>1.162</u>
|-
|Σ||ΣV = 5.828 || - ||ΣM<sub>R</sub> = 22.887
|-
|rowspan="3"|'''Earth Load'''||(5)|| (7.000 ft.)(5.167 ft.)(0.120k/ft<sup>3</sup>) = 4.340|| 6.083|| 26.400
|-
|(6)|| (2.000 ft.)(2.000 ft.)(0.120k/ft<sup>3</sup>) = <u>0.480</u>|| 1.167|| <u>0.560</u>
|-
|Σ ||ΣV = 4.820|| - ||ΣM<sub>R</sub> = 26.960
|-
|rowspan="2"|'''Live Load Surcharge'''||P<sub>SV</sub>|| (2.000 ft.)(5.167 ft.)(0.120k/ft<sub>3</sub>) = 1.240|| 6.083|| M<sub>R</sub> = 7.543
|-
|P<sub>SH</sub>||(2.000 ft.)(0.376)(10.000 ft.)(0.120k/ft<sup>3</sup>) = 0.902||10.000|| M<sub>OT</sub> = 9.020
|-
|rowspan="2"|'''Earth Pressure'''||P<sub>A</sub>||2.256<sup>'''1'''</sup>|| 8.333|| M<sub>OT</sub> = 18.799
|-
|P<sub>P</sub>|| 3.285<sup>'''2'''</sup> || - || -
|-
|colspan="2"|'''Collision Force''' (F<sub>COL</sub>)||(10.000k)/[2(7.000 ft.)] = 0.714|| 18.000 ||M<sub>OT</sub> = 12.852
|-
|colspan="2"|'''Heel Pile Tension''' (P<sub>HV</sub>)||(3.000 tons)(2 k/ton)(1 pile)/(12.000 ft.) = 0.500|| 7.167|| M<sub>R</sub> = 3.584
|-
|colspan="2"|'''Toe Pile Batter''' (P<sub>BH</sub>)|| 5.903<sup>'''3'''</sup>|| - || -
|-
|colspan="2"|'''Passive Pile Pressure''' (P<sub>pp</sub>)|| 0.832<sup>'''4'''</sup>|| - || -
|-
|colspan="5" align="left"|<sup>'''1'''</sup> <math>P_A = \frac{1}{2}\gamma_S C_A H^2 = \frac{1}{2}\Bigg[0.120\frac{k}{ft^3}\Bigg](0.376)(10 ft.)^3 = 2.256\frac{k}{ft}</math>
|-
|colspan="5" align="left"|<sup>'''2'''</sup> <math>P_P = \frac{1}{2}\gamma_S C_A\Big[H_2^2 - H_1^2\Big] = \frac{1}{2}\Bigg[0.120\frac{k}{ft^3}\Bigg](2.663)[(6.75 ft.)^2 - (5 ft.)^2] = 3.285\frac{k}{ft}</math>
|-
|colspan="5" align="left"|<sup>'''3'''</sup> <math>P_{BH} = \Big(56 \frac{tons}{pile}\Big)\Big( 2 \frac{k}{ton}\Big)(2 piles)\Bigg(\frac{4 in.}{\sqrt{(12 in.)^2 + (4 in.)^2}}\Bigg)\Big(\frac{1}{12 ft.}\Big) = 5.903 \frac{k}{ft}</math>
|-
|colspan="5" align="left"|<sup>'''4'''</sup> <math>P_{PP} = \frac{1}{2}(2.663)(5 ft.)^2\Big(0.120 \frac{k}{ft^3}\Big)(0.833 ft.)(3 piles)\Big(\frac{1}{12 ft.}\Big) = 0.832\frac{k}{ft}</math>
|}


:'''Case II'''
:F.S. for overturning ≥ 1.2
:F.S. for sliding ≥ 1.2
::'''Check Overturning'''
::''ΣM<sub>R</sub> ''= (22.887 + 26.960 + 7.543 + 3.584)(ft−k) = 60.974(ft−k)
::''ΣM<sub>OT</sub>'' = (9.020 + 18.799 + 12.852)(ft−k) = 40.671(ft−k)
::<math>F.S._{OT} = \frac{\Sigma M_R}{\Sigma M_{OT}} = \frac{60.974(ft-k)}{40.671(ft-k)}</math> = 1.499 ≥ 1.2  <u> o.k.</u>
::'''Check Pile Bearing'''
::<math>e = \frac{\Sigma M}{\Sigma V} = \frac{(13.174 + 18.930 + 5.477)(ft-k) - (4.510 + 7.519 + 9.282)(ft-k)}{(5.828 + 4.820 + 1.240)k}</math> = 1.369 ft.
::Moment arm = 1.833 ft. - 1.369 ft. = 0.464 ft.


Table 751.24.3.5.2 is for bearing pressure checks (moments taken about C.L of toe pile at the bottom of the footing).
::<math>P_T = \frac{\Sigma V}{A} + \frac{M_c}{I} = \frac{11.888k}{0.250A} + \frac{11.888k(0.464 ft.)(1.833 ft.)}{1.681(A)ft^2}</math>


{| border="1" class="wikitable" style="margin: 1em auto 1em auto" style="text-align:center"
::<math>P_T = 53.567\frac{k}{pile} = 26.783\frac{tons}{pile} \le 56\frac{tons}{pile}</math> <u> o.k.</u>
|+ '''''Table 751.24.3.5.2'''''
! style="background:#BEBEBE" colspan="2"|Load !! style="background:#BEBEBE"|Force (kips/ft) !! style="background:#BEBEBE"|Arm about C.L. of toe pile at 6B below footing (ft.) !! style="background:#BEBEBE"|Moment (ft-kips) per foot of wall length
|-
|rowspan="5"|'''Dead Load'''||(1)|| 0.340|| 0.875|| 0.298
|-
|(2)|| (1.333 ft.)(7.000 ft.)(0.150k/ft<sup>3</sup>) = 1.400 ||1.167|| 1.634
|-
|(3)|| (3.000 ft.)(8.500 ft.)(0.150k/ft<sup>3</sup>) = 3.825|| 2.750|| 10.519
|-
|(4)|| (1.000 ft.)(1.750 ft.)(0.150k/ft<sup>3</sup>) = <u>0.263</u>|| 2.750|| <u>0.723</u>
|-
|Σ||ΣV = 5.828 || - ||ΣM<sub>R</sub> = 13.174
|-
|rowspan="3"|'''Earth Load'''||(5)|| (7.000 ft.)(5.167 ft.)(0.120k/ft<sup>3</sup>) = 4.340|| 4.417|| 19.170
|-
|(6)|| (2.000 ft.)(2.000 ft.)(0.120k/ft<sup>3</sup>) = <u>0.480</u>|| -0.500|| <u>-0.240</u>
|-
|Σ ||ΣV = 4.820|| - ||ΣM<sub>R</sub> = 18.930
|-
|rowspan="2"|'''Live Load Surcharge'''||P<sub>SV</sub>|| (2.000 ft.)(5.167 ft.)(0.120k/ft<sub>3</sub>) = 1.240|| 4.417|| M<sub>R</sub> = 5.477
|-
|P<sub>SH</sub>||(2.000 ft.)(0.376)(10.000 ft.)(0.120k/ft<sup>3</sup>) = 0.902||5.000|| M<sub>OT</sub> = 4.510
|-
|rowspan="2"|'''Earth Pressure'''||P<sub>A</sub>||2.256|| 3.333|| M<sub>OT</sub> = 7.519
|-
|P<sub>P</sub>|| 3.285 || - || -
|-
|colspan="2"|'''Collision Force''' (F<sub>COL</sub>)||(10.000k)/[2(7.000 ft.)] = 0.714|| 13.000 ||M<sub>OT</sub> = 9.282
|-
|colspan="2"|'''Heel Pile Tension''' (P<sub>HV</sub>)||(3.000 tons)(2 k/ton)(1 pile)/(12.000 ft.) = 0.500|| 5.500|| M<sub>R</sub> = 2.750
|-
|colspan="2"|'''Toe Pile Batter''' (P<sub>BH</sub>)|| 5.903|| - || -
|-
|colspan="2"|'''Passive Pile Pressure''' (P<sub>pp</sub>)|| 0.832|| - || -
|}


Investigate a representative 12 ft. strip. This will include one heel pile and two toe piles. The assumption is made that the stiffness of a batter pile in the vertical direction is the same as that of a vertical pile.
::<math>P_H = \frac{11.888k}{0.250A} - \frac{11.888k(0.464 ft.)(3.667 ft.)}{1.681(A)ft^2}</math> = 35.519k


Neutral Axis Location = [2piles(1.5 ft.) + 1pile(7 ft.)] / (3 piles) = 3.333 ft. from the toe.
::<math>P_H = 17.760\frac{tons}{pile} \le 56\frac{tons}{pile} </math> <u> o.k.</u>


[[image:751.24.3.5 neutral axis.jpg|center|350px]]
::'''Check Sliding'''


''I ''= Ad<sup>2</sup>
::<math>F.S._{Sliding} = \frac{3.285k + 5.903k + 0.832k}{0.902k + 2.256k + 0.714k}</math> = 2.588 ≥ 1.2 <u> o.k.</u>


For repetitive 12 ft. strip:
:'''Case III'''


:Total pile area = 3A
:F.S. for overturning ≥ 1.5


:''I ''= 2A(1.833 ft.)<sup>2</sup> + A(3.667 ft.)<sup>2</sup> = 20.167(A)ft.<sup>2</sup>
:F.S. for sliding ≥ 1.5
 
::'''Check Overturning'''
 
::''ΣM<sub>R</sub>'' = (22.887 + 26.960 + 3.584)(ft−k) = 53.431(ft−k)


For a 1 ft. unit strip:
::''ΣM<sub>OT</sub>'' = 18.799(ft−k)


:<math>I = \frac{20.167(A)ft.^2}{12 ft.} = 1.681(A)ft.^2</math>
::<math>F.S._{OT} = \frac{\Sigma M_R}{\Sigma M_{OT}} = \frac{53.431(ft-k)}{18.799(ft-k)}</math> = 2.842 ≥ 1.5 <u> o.k.</u>


:Total pile area = (3A/12 ft.) = 0.250A
::'''Check Pile Bearing'''


:'''Case I'''
::<math>e = \frac{\Sigma M}{\Sigma V} = \frac{(13.174 + 18.930)(ft-k) - 7.519(ft-k)}{(5.828 + 4.820)k}</math> = 2.309 ft.


:F.S. for overturning ≥ 1.5
::Moment arm = 2.309 ft. - 1.833 ft. = 0.476 ft.


:F.S. for sliding ≥ 1.5
::<math>P_T = \frac{10.648k}{0.250A} - \frac{10.648k(0.476 ft.)(1.833 ft.)}{1.681(A)ft^2}</math> = 37.065k


::'''Check Overturning'''
::<math>P_T = 18.532\frac{tons}{pile} \le 56\frac{tons}{pile}</math> <u> o.k.</u>


::Neglect resisting moment due to P<sub>SV</sub> for this check.
::<math>P_H = \frac{10.648k}{0.250A} + \frac{10.648k(0.476 ft.)(3.667 ft.)}{1.681(A)ft^2}</math> = 53.649k


::''ΣM<sub>R</sub>'' = 22.887(ft−k) + 26.960(ft−k) + 3.584(ft−k)
::<math>P_H = 26.825\frac{tons}{pile} \le 56\frac{tons}{pile} </math> <u> o.k.</u>


::''ΣM<sub>R</sub>'' = 53.431(ft−k)
::'''Check Sliding'''


::''ΣM<sub>OT</sub>'' = 9.020(ft−k) + 18.799(ft−k) = 27.819(ft−k)
::<math>F.S._{Sliding} = \frac{3.285k+5.903k+0.832k}{2.256k}</math> = 4.441 ≥ 1.5 <u> o.k.</u>


::''F.S.<sub>OT</sub>'' = <math>\frac{\Sigma M_R}{\Sigma M_{OT}} = \frac{53.431(ft−-k)}{27.819(ft-−k)}</math> = 1.921 > 1.5 <u>o.k.</u>
:'''Case IV'''


::'''Check Pile Bearing'''
::'''Check Pile Bearing'''


::Without P<sub>SV</sub> :
::<math>e = \frac{\Sigma M}{\Sigma V} = \frac{(13.174 + 18.930)(ft-k)}{5.828k + 4.820k}</math> = 3.015 ft.


::''ΣV'' = 5.828k + 4.820k = 10.648k
::Moment arm = 3.015 ft. - 1.833 ft. = 1.182 ft.


::''e'' = <math>\frac{\Sigma M}{\Sigma V} = \frac{(13.174 + 18.930)(ft−-k) − (4.510 + 7.519)(ft-−k)}{10.648k}</math> = 1.885 ft.
::<math>P_H = \frac{\Sigma V}{A} + \frac{M_c}{I} = \frac{10.648k}{0.250A} + \frac{10.648k(1.182 ft.)(3.667 ft.)}{1.681(A)ft^2}</math>


::Moment arm = 1.885 ft. - 1.833 ft. = 0.052 ft.
::<math>P_H = 70.047k = 35.024 \frac{tons}{pile}</math>


::<math>P_T = \frac{\Sigma V}{A} -− \frac{M_c}{I} = \frac{10.648k}{0.250A} -− \frac{10.648k(0.052 ft.)(1.833 ft.)}{1.681(A)ft^2}</math>
::25% overstress is allowed on the heel pile:


::<math>P_T = \frac{41.988}{A} k</math>
::<math>P_H = 35.024\frac{tons}{pile} \le 1.25 (56\frac{tons}{pile}) = 70 \frac{tons}{pile}</math> <u> o.k.</u>


::<math>P_H = \frac{10.648k}{0.250A} + \frac{10.648k(0.052 ft.)(3.667 ft.)}{1.681(A)ft^2}</math>
::<math>P_T = \frac{10.648k}{0.250A} - \frac{10.648k(1.182 ft.)(1.833 ft.)}{1.681(A)ft^2}</math> = 28.868k


::<math>P_H = \frac{43.800}{A} k</math>
::<math>P_T = 14.434\frac{tons}{pile} \le 56\frac{tons}{pile} </math> <u> o.k.</u>


::Allowable pile load = 56 tons/pile. Each pile has area A, so:
:'''Reinforcement - Stem'''


::<math>P_T = 41.988\frac{k}{pile} = 20.944\frac{tons}{pile} </math> <u> o.k.</u>
[[image:751.24.3.5 reinforcement stem.jpg|300px|center]]


::<math>P_H = 43.800\frac{k}{pile} = 21.900\frac{tons}{pile} </math> <u> o.k.</u>
:b = 12 in.


::With P<sub>SV</sub>:
:cover = 2 in.


::''ΣV'' = 5.828k + 4.820k + 1.240k = 11.888k
:h = 16 in.


::<math>e = \frac{(13.174 + 18.930 + 5.477)(ft-−k) -− (4.510 + 7.519)(ft-−k)}{11.888k}</math> = 2.149 ft.
:d = 16 in. - 2 in. - 0.5(0.625 in.) = 13.688 in.


::Moment arm = 2.149 ft. - 1.833 ft. = 0.316 ft.
:''F<sub>Collision</sub>'' = 0.714k/ft


::<math>P_T = \frac{11.888k}{0.250A} -− \frac{11.888k(0.316 ft.)(1.833 ft.)}{1.681(A)ft^2} = 43.456k = 21.728\frac{tons}{pile}</math> <u> o.k.</u>
::<math>P_{LL} = \gamma_s C_A H(2.000 ft.) = (2.000 ft.)(0.376)(7.000 ft.)(0.120 \frac{k}{ft^3}) = 0.632\frac{k}{ft}</math>


::<math>P_H = \frac{11.888k}{0.250A} + \frac{11.888k(0.316 ft.)(3.667 ft.)}{1.681(A)ft^2} = 55.747k = 27.874\frac{tons}{pile}</math> <u> o.k.</u>
::<math>P_{A_{Stem}} = \frac{1}{2} \gamma_s C_A H^2 = \frac{1}{2}\Big[0.120 \frac{k}{ft^3}\Big](0.376)(7.000 ft.)^2 = 1.105\frac{k}{ft} </math>


::'''Check Sliding'''
:::'''Apply Load Factors'''


::<math>F.S._{Sliding} = \frac{3.285k + 5.903k + 0.832k}{0.902 k + 2.256k}</math> = 3.173 ≥ 1.5 <u> o.k.</u>
:::''F<sub>Col.</sub>'' = ''γβ<sub>LL</sub>''(0.714k) = (1.3)(1.67)(0.714k) = 1.550k


:'''Case II'''
:::''P<sub>LL</sub>'' = ''γβ<sub>E</sub>'' (0.632k) = (1.3)(1.67)(0.632k) = 1.372k


:F.S. for overturning ≥ 1.2
:::''P<sub>A<sub>Stem</sub></sub>'' = ''γβ<sub>E</sub>'' (1.105k) = (1.3)(1.3)(1.105k) = 1.867k


:F.S. for sliding ≥ 1.2
::''M<sub>u</sub>'' = (10.00 ft.)(1.550k) + (3.500 ft.)(1.372k) + (2.333 ft.)(1.867k)


::'''Check Overturning'''
::''M<sub>u</sub>'' = 24.658(ft−k)


::''ΣM<sub>R</sub> ''= (22.887 + 26.960 + 7.543 + 3.584)(ft−k) = 60.974(ft−k)
::<math>R_n = \frac{M_u}{\phi b d^2} = \frac{24.658(ft-k)}{(0.9)(1 ft.)(13.688 in.)^2}</math> = 0.146ksi


::''ΣM<sub>OT</sub>'' = (9.020 + 18.799 + 12.852)(ft−k) = 40.671(ft−k)
::<math>\rho = \frac{0.85f'_c}{f_y}\Bigg[1 - \sqrt{1 - \frac{2R_n}{0.85f'_c}}\Bigg] =  
\frac{0.85(3 ksi)}{60 ksi}\Bigg[1 - \sqrt{1 - \frac{2(0.146 ksi)}{0.85(3 ksi)}}\Bigg]</math> = 0.00251


::<math>F.S._{OT} = \frac{\Sigma M_R}{\Sigma M_{OT}} = \frac{60.974(ft-−k)}{40.671(ft-−k)}</math> = 1.499 ≥ 1.2  <u> o.k.</u>
::<math>\rho_{min} = 1.7\Big[\frac{h}{d}\Big]^2 \frac{\sqrt{f'_c}}{f_y} = 1.7\Big[\frac{16 in.}{13.688 in.}\Big]^2 \frac{\sqrt{3000 psi}}{60,000 psi}</math> = 0.00212


::'''Check Pile Bearing'''
::''ρ'' = 0.00251


::<math>e = \frac{\Sigma M}{\Sigma V} = \frac{(13.174 + 18.930 + 5.477)(ft-−k) -− (4.510 + 7.519 + 9.282)(ft-−k)}{(5.828 + 4.820 + 1.240)k}</math> = 1.369 ft.
::<math>A_{S_{Req.}} = \rho bd = (0.00251)(12 in.)(13.688 in.) = 0.412 \frac{in^2}{ft.}</math>


::Moment arm = 1.833 ft. - 1.369 ft. = 0.464 ft.
::One #5 bar has A<sub>S</sub> = 0.307 in<sup>2</sup>


::<math>P_T = \frac{\Sigma V}{A} + \frac{M_c}{I} = \frac{11.888k}{0.250A} + \frac{11.888k(0.464 ft.)(1.833 ft.)}{1.681(A)ft^2}</math>
::<math>\frac{s}{0.307 in^2} = \frac{12 in.}{0.412 in^2}</math>


::<math>P_T = 53.567\frac{k}{pile} = 26.783\frac{tons}{pile} \le 56\frac{tons}{pile}</math> <u> o.k.</u>
::''s'' = 8.9 in.


::<math>P_H = \frac{11.888k}{0.250A} -− \frac{11.888k(0.464 ft.)(3.667 ft.)}{1.681(A)ft^2}</math> = 35.519k
::<u>Use # 5 bars @ 8.5 in. cts.</u>


::<math>P_H = 17.760\frac{tons}{pile} \le 56\frac{tons}{pile} </math> <u> o.k.</u>
:::'''Check Shear'''


::'''Check Sliding'''
:::''V<sub>u</sub>'' ≤ ''φV<sub>n</sub>''


::<math>F.S._{Sliding} = \frac{3.285k + 5.903k + 0.832k}{0.902k + 2.256k + 0.714k}</math> = 2.588 ≥ 1.2 <u> o.k.</u>
:::''V<sub>u</sub>'' = ''F<sub>Collision</sub>'' + ''P<sub>LL</sub>'' + ''P<sub>A<sub>Stem</sub></sub>'' = 1.550k + 1.372k + 1.867k = 4.789k


:'''Case III'''


:F.S. for overturning ≥ 1.5
:::<math>\frac{\nu_u}{\phi} = \frac{v_u}{\phi bd} = \frac{4789 lbs}{0.85(12 in.)(13.688 in.)}</math> = 34.301 psi


:F.S. for sliding ≥ 1.5
:::<math> \nu_n = \nu_c = 2\sqrt{f'_c} = 2\sqrt{3000psi}</math> = 109.5 psi > 34.3 psi <u>o.k.</u>


::'''Check Overturning'''
::'''Reinforcement - Footing - Top Steel'''


::''ΣM<sub>R</sub>'' = (22.887 + 26.960 + 3.584)(ft−k) = 53.431(ft−k)
[[image:751.24.3.5 footing.jpg|300px|center]]


::''ΣM<sub>OT</sub>'' = 18.799(ft−k)
::b = 12 in.


::<math>F.S._{OT} = \frac{\Sigma M_R}{\Sigma M_{OT}} = \frac{53.431(ft−-k)}{18.799(ft−-k)}</math> = 2.842 ≥ 1.5 <u> o.k.</u>
::cover = 3 in.


::'''Check Pile Bearing'''
::h = 36 in.


::<math>e = \frac{\Sigma M}{\Sigma V} = \frac{(13.174 + 18.930)(ft-−k) -− 7.519(ft-−k)}{(5.828 + 4.820)k}</math> = 2.309 ft.
::d = 36 in. - 3 in. - 0.5(0.5 in.) = 32.750 in.


::Moment arm = 2.309 ft. - 1.833 ft. = 0.476 ft.
::Design the heel to support the entire weight of the superimposed materials.


::<math>P_T = \frac{10.648k}{0.250A} -− \frac{10.648k(0.476 ft.)(1.833 ft.)}{1.681(A)ft^2}</math> = 37.065k
::Soil(1) = 4.340k/ft.


::<math>P_T = 18.532\frac{tons}{pile} \le 56\frac{tons}{pile}</math> <u> o.k.</u>
::LL<sub>s</sub> = 1.240k/ft.


::<math>P_H = \frac{10.648k}{0.250A} + \frac{10.648k(0.476 ft.)(3.667 ft.)}{1.681(A)ft^2}</math> = 53.649k
::<math>Slab \ wt. = (3.000 ft.)\Big[0.150 \frac{k}{ft^3}\Big](5.167 ft.)</math> = 2.325k/ft.


::<math>P_H = 26.825\frac{tons}{pile} \le 56\frac{tons}{pile} </math> <u> o.k.</u>
:::'''Apply Load Factors'''


::'''Check Sliding'''
:::Soil(1) = ''γβ<sub>E</sub>''(4.340k) = (1.3)(1.0)(4.340k) = 5.642k


::<math>F.S._{Sliding} = \frac{3.285k+5.903k+0.832k}{2.256k}</math> = 4.441 ≥ 1.5 <u> o.k.</u>
:::''LL<sub>s</sub>'' = ''γβ<sub>E</sub>''(1.240k) = (1.3)(1.67)(1.240k) = 2.692k


:'''Case IV'''
:::Slab wt. = ''γβ<sub>D</sub>''(2.325k) = (1.3)(1.0)(2.325k) = 3.023k


::'''Check Pile Bearing'''
::''M<sub>u</sub>'' = (2.583 ft.)(5.642k + 2.692k + 3.023k) = 29.335(ft−k)


::<math>e = \frac{\Sigma M}{\Sigma V} = \frac{(13.174 + 18.930)(ft-−k)}{5.828k + 4.820k}</math> = 3.015 ft.
::<math>R_n = \frac{M_u}{\phi bd^2} = \frac{29.335(ft-k)}{(0.9)(1 ft.)(32.750 in.)^2}</math> = 0.0304 ksi
::<math>\rho = \frac{0.85(3ksi)}{60ksi}\Bigg[1 - \sqrt{1 - \frac{2(0.0304ksi)}{0.85(3ksi)}}\Bigg]</math> = 0.000510


::Moment arm = 3.015 ft. - 1.833 ft. = 1.182 ft.
::<math>\rho_{min} = 1.7\Big[\frac{36 in.}{32.750 in.}\Big]^2 \frac{\sqrt{3000 psi}}{60,000psi}</math> = 0.00188


::<math>P_H = \frac{\Sigma V}{A} + \frac{M_c}{I} = \frac{10.648k}{0.250A} + \frac{10.648k(1.182 ft.)(3.667 ft.)}{1.681(A)ft^2}</math>
::Use ''ρ'' = 4/3 ''ρ'' = 4/3 (0.000510) = 0.000680


::<math>P_H = 70.047k = 35.024 \frac{tons}{pile}</math>
::<math>A_{S_{Req}} = \rho bd = (0.000680)(12 in.)(32.750 in.) = 0.267\frac{in^2}{ft.}</math>


::25% overstress is allowed on the heel pile:
::One #4 bar has A<sub>s</sub> = 0.196 in.<sup>2</sup>


::<math>P_H = 35.024\frac{tons}{pile} \le 1.25 (56\frac{tons}{pile}) = 70 \frac{tons}{pile}</math> <u> o.k.</u>
::<math>\frac{s}{0.196 in^2} = \frac{12 in}{0.267 in.^2}</math>


::<math>P_T = \frac{10.648k}{0.250A} -− \frac{10.648k(1.182 ft.)(1.833 ft.)}{1.681(A)ft^2}</math> = 28.868k
::''s'' = 8.8 in.


::<math>P_T = 14.434\frac{tons}{pile} \le 56\frac{tons}{pile} </math> <u> o.k.</u>
::<u>Use #4 bars @ 8.5 in. cts.</u>


:'''Reinforcement - Stem'''
:::'''Check Shear'''


[[image:751.24.3.5 reinforcement stem.jpg|300px|center]]
:::<math>V_u = Soil(1) + LL_s + Slab \ wt. = 5.642k + 2.692k + 3.023k = 11.357k</math>


:b = 12 in.
:::<math>\frac{\nu_u}{\phi} = \frac{V_u}{\phi bd} = \frac{11357 lbs}{(0.85)(12 in.)(32.750 in.)}</math> = 33.998 psi ≤ 109.5 psi = ''ν<sub>c</sub>''  <u>o.k.</u>


:cover = 2 in.
::'''Reinforcement - Footing - Bottom Steel'''


:h = 16 in.
::Design the flexural steel in the bottom of the footing to resist the largest moment that the heel pile could exert on the footing. The largest heel pile bearing force was in Case IV. The heel pile will cause a larger moment about the stem face than the toe pile (even though there are two toe piles for every one heel pile) because it has a much longer moment arm about the stem face.


:d = 16 in. - 2 in. - 0.5(0.625 in.) = 13.688 in.
[[image: 751.24.3.5 heel pile.jpg|center|300px]]


:''F<sub>Collision</sub>'' = 0.714k/ft
::Pile is embedded into footing 12 inches.


::<math>P_{LL} = \gamma_s C_A H(2.000 ft.) = (2.000 ft.)(0.376)(7.000 ft.)(0.120 \frac{k}{ft^3}) = 0.632\frac{k}{ft}</math>
::''b'' = 12 in.


::<math>P_{A_{Stem}} = \frac{1}{2} \gamma_s C_A H^2 = \frac{1}{2}\Big[0.120 \frac{k}{ft^3}\Big](0.376)(7.000 ft.)^2 = 1.105\frac{k}{ft} </math>
::''h'' = 36 in.


:::'''Apply Load Factors'''
::''d'' = 36 in. - 4 in. = 32 in.


:::''F<sub>Col.</sub>'' = ''γβ<sub>LL</sub>''(0.714k) = (1.3)(1.67)(0.714k) = 1.550k
:::'''Apply Load Factors to Case IV Loads'''


:::''P<sub>LL</sub>'' = ''γβ<sub>E</sub>'' (0.632k) = (1.3)(1.67)(0.632k) = 1.372k
:::<math>\Sigma V = \gamma \beta_D\Big[5.828 \frac{k}{ft.}\Big] + \gamma \beta_E \Big[4.820 \frac{k}{ft.}\Big]</math>


:::''P<sub>A<sub>Stem</sub></sub>'' = ''γβ<sub>E</sub>'' (1.105k) = (1.3)(1.3)(1.105k) = 1.867k
:::<math>\Sigma V = 1.3(1.0)\Big[5.828\frac{k}{ft.}\Big] + 1.3(1.0)\Big[4.820\frac{k}{ft.}\Big]</math>


::''M<sub>u</sub>'' = (10.00 ft.)(1.550k) + (3.500 ft.)(1.372k) + (2.333 ft.)(1.867k)
:::''ΣV'' = 13.842 k/ft.


::''M<sub>u</sub>''  = 24.658(ft−k)
:::<math>\Sigma M = \gamma \beta_D\Big[13.174\frac{(ft-k)}{ft.}\Big] + \gamma \beta_E\Big[18.930\frac{(ft-k)}{ft.}\Big]</math>


::<math>R_n = \frac{M_u}{\phi b d^2} = \frac{24.658(ft−-k)}{(0.9)(1 ft.)(13.688 in.)^2}</math> = 0.146ksi
:::<math>\Sigma M = (1.3)(1.0)\Big[13.174\frac{(ft-k)}{ft.}\Big] + (1.3)(1.0)\Big[18.930\frac{(ft-k)}{ft.}\Big]</math>


::<math>\rho = \frac{0.85f'_c}{f_y}\Bigg[1 -− \sqrt{1 -− \frac{2R_n}{0.85f'_c}}\Bigg] =  
:::''ΣM'' = 41.735 (ft−k)/ft.
\frac{0.85(3 ksi)}{60 ksi}\Bigg[1 -− \sqrt{1 −- \frac{2(0.146 ksi)}{0.85(3 ksi)}}\Bigg]</math> = 0.00251


::<math>\rho_{min} = 1.7\Big[\frac{h}{d}\Big]^2 \frac{\sqrt{f'_c}}{f_y} = 1.7\Big[\frac{16 in.}{13.688 in.}\Big]^2 \frac{\sqrt{3000 psi}}{60,000 psi}</math> = 0.00212
::e = <math>\frac{\Sigma M}{\Sigma V} = \frac{41.735 (ft-k)}{13.842k}</math> = 3.015 ft.


::''ρ'' = 0.00251
::Moment arm = 3.015 ft. - 1.833 ft. = 1.182 ft.


::<math>A_{S_{Req.}} = \rho bd = (0.00251)(12 in.)(13.688 in.) = 0.412 \frac{in^2}{ft.}</math>
::<math>P_H = \frac{\Sigma V}{A} + \frac{M_c}{I} = \frac{13.842k}{0.250A} + \frac{13.842k (1.182 ft.)(3.667 ft.)}{1.681(A)ft^2}</math>


::One #5 bar has A<sub>S</sub> = 0.307 in<sup>2</sup>
::<math>P_H = 91.059 \frac{k}{pile}\Big(\frac{1}{12 ft.}\Big)</math> = 7.588 k/ft.


::<math>\frac{s}{0.307 in^2} = \frac{12 in.}{0.412 in^2}</math>
::<math>M_u = \Big(7.588\frac{k}{ft.}\Big)(3.667 ft.)</math> = 27.825(ft−k)/ft.


::''s'' = 8.9 in.
::<math>R_n = \frac{M_u}{\phi bd^2} = \frac{27.825(ft-k)}{(0.9)(1 ft.)(32 in.)^2}</math> = 0.0301 ksi


::<u>Use # 5 bars @ 8.5 in. cts.</u>
::<math>\rho = \frac{0.85(3 ksi)}{60ksi}\Bigg[1 - \sqrt{1 - \frac{2(0.0301 ksi)}{0.85(3 ksi)}}\Bigg]</math> = 0.000505
 
::<math>\rho_{min} = 1.7\Big[\frac{36 in.}{32 in.}\Big]^2 \frac{\sqrt{3000 psi}}{60,000 psi}</math> = 0.00196


:::'''Check Shear'''
::Use ''ρ'' = 4/3 ''ρ'' = 4/3 (0.000505) = 0.000673


:::''V<sub>u</sub>'' ≤ ''φV<sub>n</sub>''
::''A<sub>S<sub>Req</sub></sub> = ρbd'' = (0.000673)(12 in.)(32 in.) = 0.258 in<sup>2</sup>/ft.''


:::''V<sub>u</sub>'' = ''F<sub>Collision</sub>'' + ''P<sub>LL</sub>'' + ''P<sub>A<sub>Stem</sub></sub>'' = 1.550k + 1.372k + 1.867k = 4.789k


::One #4 bar has A<sub>s</sub> = 0.196 in<sup>2</sup>.


:::<math>\frac{\nu_u}{\phi} = \frac{v_u}{\phi bd} = \frac{4789 lbs}{0.85(12 in.)(13.688 in.)}</math> = 34.301 psi
::<math>\frac{s}{0.196 in.^2} = \frac{12 in.}{0.258 in.^2}</math>


:::<math> \nu_n = \nu_c = 2\sqrt{f'_c} = 2\sqrt{3000psi}</math> = 109.5 psi > 34.3 psi <u>o.k.</u>
::''s'' = 9.1 in.


::'''Reinforcement - Footing - Top Steel'''
::<u>Use #4 bars @ 9 in. cts.</u>


[[image:751.24.3.5 footing.jpg|300px|center]]
:::'''Check Shear'''


::b = 12 in.
:::The critical section for shear for the toe is at a distance d = 21.75 inches from the face of the stem. The toe pile is 6 inches from the stem face so the toe pile shear does not affect the shear at the critical section. The critical section for shear is at the stem face for the heel so all of the force of the heel pile affects the shear at the critical section. The worst case for shear is Case IV.


::cover = 3 in.
:::''V<sub>u</sub>'' = 7.588k


::h = 36 in.
:::<math>\frac{\nu_u}{\phi} = \frac{V_u}{\phi bd} = {7588 lbs}{0.85(12 in.)(32 in.)}</math> = 23.248 psi ≤ 109.5 psi = ''ν<sub>c</sub>'' <u>o.k.</u>


::d = 36 in. - 3 in. - 0.5(0.5 in.) = 32.750 in.
::'''Reinforcement - Shear Key'''


::Design the heel to support the entire weight of the superimposed materials.
::''b'' = 12 in.


::Soil(1) = 4.340k/ft.
::''h'' = 12 in.


::LL<sub>s</sub> = 1.240k/ft.
::cover = 3 in.


::<math>Slab \ wt. = (3.000 ft.)\Big[0.150 \frac{k}{ft^3}\Big](5.167 ft.)</math> = 2.325k/ft.
::''d'' = 12 in. - 3 in. - 0.5(0.5 in.) = 8.75 in.


:::'''Apply Load Factors'''
:::'''Apply Load Factors'''


:::Soil(1) = ''γβ<sub>E</sub>''(4.340k) = (1.3)(1.0)(4.340k) = 5.642k
:::''P<sub>P</sub> = γβ<sub>E</sub>'' (3.845k) = (1.3)(1.3)(3.845k) = 6.498k
 
::''M<sub>u</sub>'' = (0.912 ft.)(6.498k) = 5.926(ft−k)


:::''LL<sub>s</sub>'' = ''γβ<sub>E</sub>''(1.240k) = (1.3)(1.67)(1.240k) = 2.692k
::<math>R_n = \frac{M_u}{\phi bd^2} = \frac{5.926(ft-k)}{(0.9)(1 ft.)(8.75 in.)^2}</math> = 0.0860 ksi


:::Slab wt. = ''γβ<sub>D</sub>''(2.325k) = (1.3)(1.0)(2.325k) = 3.023k
::<math>\rho = \frac{0.85(3ksi)}{60ksi}\Bigg[1 - \sqrt{1 - \frac{2(0.0860ksi)}{0.85(3ksi)}}\Bigg]</math> = 0.00146


::''M<sub>u</sub>'' = (2.583 ft.)(5.642k + 2.692k + 3.023k) = 29.335(ft−k)
::<math>\rho_{min} = 1.7\Big[\frac{12 in.}{8.75 in}\Big]^2\frac{\sqrt{3000 psi}}{60,000 psi}</math> = 0.00292


::<math>R_n = \frac{M_u}{\phi bd^2} = \frac{29.335(ft-−k)}{(0.9)(1 ft.)(32.750 in.)^2}</math> = 0.0304 ksi
::Use ''ρ'' = 4/3 ''ρ'' = 4/3(0.00146) = 0.00195
::<math>\rho = \frac{0.85(3ksi)}{60ksi}\Bigg[1 -− \sqrt{1 -− \frac{2(0.0304ksi)}{0.85(3ksi)}}\Bigg]</math> = 0.000510


::<math>\rho_{min} = 1.7\Big[\frac{36 in.}{32.750 in.}\Big]^2 \frac{\sqrt{3000 psi}}{60,000psi}</math> = 0.00188
::''A<sub>S<sub>Req</sub></sub> = ρbd'' = (0.00195)(12 in.)(8.75 in.) = 0.205 in.<sup>2</sup>/ft.


::Use ''ρ'' = 4/3 ''ρ'' = 4/3 (0.000510) = 0.000680


::<math>A_{S_{Req}} = \rho bd = (0.000680)(12 in.)(32.750 in.) = 0.267\frac{in^2}{ft.}</math>
::One #4 bar has A<sub>s</sub> = 0.196 in<sup>2</sup>


::One #4 bar has A<sub>s</sub> = 0.196 in.<sup>2</sup>


::<math>\frac{s}{0.196 in^2} = \frac{12 in}{0.267 in.^2}</math>
::<math>\frac{s}{0.196 in.^2} = \frac{12 in.}{0.205 in.^2}</math>


::''s'' = 8.8 in.
::''s'' = 11.5 in.


::<u>Use #4 bars @ 8.5 in. cts.</u>
::<u>Use #4 bars @ 11 in. cts.</u>


:::'''Check Shear'''
:::'''Check Shear'''


:::<math>V_u = Soil(1) + LL_s + Slab \ wt. = 5.642k + 2.692k + 3.023k = 11.357k</math>
:::<math>\frac{\nu_u}{\phi} = \frac{V_u}{\phi bd} = \frac{6498 lbs}{0.85(12 in.)(8.75 in.)}</math> = 72.807 psi < 109.5 psi = ''ν<Sub>c</sub>''


:::<math>\frac{\nu_u}{\phi} = \frac{V_u}{\phi bd} = \frac{11357 lbs}{(0.85)(12 in.)(32.750 in.)}</math> = 33.998 psi ≤ 109.5 psi = ''ν<sub>c</sub>'' <u>o.k.</u>
::'''Reinforcement Summary'''


::'''Reinforcement - Footing - Bottom Steel'''
[[image:751.24.3.5 summary.jpg|center|350px]]


::Design the flexural steel in the bottom of the footing to resist the largest moment that the heel pile could exert on the footing. The largest heel pile bearing force was in Case IV. The heel pile will cause a larger moment about the stem face than the toe pile (even though there are two toe piles for every one heel pile) because it has a much longer moment arm about the stem face.
===751.24.3.6 Dimensions===
'''Cantilever Walls'''
Each section of wall shall be in increments of 4 ft. with a maximum length of 28'-0".
[[image:751.24.3.6 friction or bearing piles.jpg|center|800px]]


[[image: 751.24.3.5 heel pile.jpg|center|300px]]


::Pile is embedded into footing 12 inches.
Each section of wall shall be in increments of 4 ft. with a maximum length of 28'-0".
[[image:751.24.3.6 pile footing.jpg|center|800px]]


::''b'' = 12 in.


::''h'' = 36 in.
'''Cantilever Walls - L-Shaped'''


::''d'' = 36 in. - 4 in. = 32 in.
Each section of wall shall be in increments of 4 ft. with a maximum length of 28'-0".
[[image:751.24.3.6 L shaped.jpg|center|800px]]


:::'''Apply Load Factors to Case IV Loads'''


:::<math>\Sigma V = \gamma \beta_D\Big[5.828 \frac{k}{ft.}\Big] + \gamma \beta_E \Big[4.820 \frac{k}{ft.}\Big]</math>
'''Counterfort Walls'''
[[image:751.24.3.6 counterfort part elev.jpg|center|800px|thumb|
{| style="margin: 1em auto 1em auto style="text-align:left""
|-
|'''Notes:'''
|-
|'''Dimension "A"'''
|-
|• Maximum length = 28'-0".
|-
|• Each section to be in 4'-0" increments.
|-
|• (See [[#Rustication Recess|rustication recess details]].)
|-
|'''Dimensions "B" & "C"'''
|-
|• As required by the design to balance the negative and positive moments. (See the design assumptions).
|}]]


:::<math>\Sigma V = 1.3(1.0)\Big[5.828\frac{k}{ft.}\Big] + 1.3(1.0)\Big[4.820\frac{k}{ft.}\Big]</math>
[[image:751.24.3.6 counterfort typ section.jpg|center|800px|thumb|
 
{| style="margin: 1em auto 1em auto style="text-align:left""
:::''ΣV'' = 13.842 k/ft.
|-
 
|'''Notes:'''
:::<math>\Sigma M = \gamma \beta_D\Big[13.174\frac{(ft-−k)}{ft.}\Big] + \gamma \beta_E\Big[18.930\frac{(ft-−k)}{ft.}\Big]</math>
|-
|'''Batter  "D":'''
|-
|* As required to maintain 9" minimum at the top of the counterfort and 12" minimum edge distance at the top of the footing, between counterfort and footing edge.
|-
|* Batter to be given an eighth of an inch per foot of counterfort height.
|-
|'''Dimension "L":'''
|-
|* As required for stability.
|-
|* As an estimate, use "L" equal to 1/2 the height of "H".
|}]]


:::<math>\Sigma M = (1.3)(1.0)\Big[13.174\frac{(ft-−k)}{ft.}\Big] + (1.3)(1.0)\Big[18.930\frac{(ft-−k)}{ft.}\Big]</math>
'''Sign-Board Type Counterfort Walls'''
 
[[image:751.24.3.6 sign board part elev.jpg|center|800px|thumb|
:::''ΣM'' = 41.735 (ft−k)/ft.
{| style="margin: 1em auto 1em auto style="text-align:left""
 
|-
::e = <math>\frac{\Sigma M}{\Sigma V} = \frac{41.735 (ft-−k)}{13.842k}</math> = 3.015 ft.
|'''Notes:'''
 
|-
::Moment arm = 3.015 ft. - 1.833 ft. = 1.182 ft.
|'''Dimension "A"'''
 
|-
::<math>P_H = \frac{\Sigma V}{A} + \frac{M_c}{I} = \frac{13.842k}{0.250A} + \frac{13.842k (1.182 ft.)(3.667 ft.)}{1.681(A)ft^2}</math>
|* Maximum length = 28'-0".
 
|-
::<math>P_H = 91.059 \frac{k}{pile}\Big(\frac{1}{12 ft.}\Big)</math> = 7.588 k/ft.
|* Each section to be in 4'-0" increments.
 
|-
::<math>M_u = \Big(7.588\frac{k}{ft.}\Big)(3.667 ft.)</math> = 27.825(ft−k)/ft.
|* (See [[#Rustication Recess|rustication recess details]].)
|-
|'''Dimensions "B" & "C"'''
|-
|* As required by the design to balance the negative and positive moments. (See the design assumptions).
|-
|'''Dimension "E"'''
|-
|* (Sign-board type only)
|-
|* As required to maintain footing pressure within the allowable for existing foundation material. 12" minimum.


::<math>R_n = \frac{M_u}{\phi bd^2} = \frac{27.825(ft−-k)}{(0.9)(1 ft.)(32 in.)^2}</math> = 0.0301 ksi
|}]]


::<math>\rho = \frac{0.85(3 ksi)}{60ksi}\Bigg[1 -− \sqrt{1 -− \frac{2(0.0301 ksi)}{0.85(3 ksi)}}\Bigg]</math> = 0.000505
[[image:751.24.3.6 sign board typ section.jpg|center|800px|thumb|
{| style="margin: 1em auto 1em auto style="text-align:left""
|-
|'''Notes:'''
|-
|'''Batter  "D":'''
|-
|* As required to maintain 9" minimum at the top of the counterfort and 12" minimum edge distance at the top of the footing, between counterfort and footing edge.
|-
|* Batter to be given an eighth of an inch per foot of counterfort height.
|-
|'''Dimension "L":'''
|-
|* As required for stability.
|-
|* As an estimate, use "L" equal to 1/2 the height of "H".
|}]]


::<math>\rho_{min} = 1.7\Big[\frac{36 in.}{32 in.}\Big]^2 \frac{\sqrt{3000 psi}}{60,000 psi}</math> = 0.00196
===751.24.3.7 Reinforcement===
 
'''Cantilever Walls'''
::Use ''ρ'' = 4/3 ''ρ'' = 4/3 (0.000505) = 0.000673
[[image:751.24.3.7 friction.jpg|center|800px|thumb|
{| style="margin: 1em auto 1em auto"
|-
|'''(*)''' Alternate long and short bars at equal spaces.
|-
|'''(**)''' If collision forces are assumed, use #4 @ 12" cts. min. and extend at least development length into footing.  (See [[751.5 Structural Detailing Guidelines#751.5.9.2.8.1 Development and Lap Splice General|EPG 751.5.9.2.8.1 Development and Lap Splice General]].)
|-
|'''(***)''' Theo. cut-off for bending + development length.  (Wall height over 10' only.)
|}
]]


::''A<sub>S<sub>Req</sub></sub> = ρbd'' = (0.000673)(12 in.)(32 in.) = 0.258 in<sup>2</sup>/ft.''
[[image:751.24.3.7 pile footing.jpg|center|800px|thumb|
{| style="margin: 1em auto 1em auto"
|-
|'''(*)''' Alternate long and short bars at equal spaces.
|-
|'''(**)''' If collision forces are assumed, use #4 @ 12" cts. min. and extend at least development length into footing. (See [[751.5 Structural Detailing Guidelines#751.5.9.2.8.1 Development and Lap Splice General|EPG 751.5.9.2.8.1 Development and Lap Splice General]].)
|-
|'''(***)''' Theo. cut-off for bending + development length. (Wall height over 10' only.)
|-
|'''(****)''' Due to site constriction.
|}
]]


'''Cantilever Walls - L-Shaped'''


::One #4 bar has A<sub>s</sub> = 0.196 in<sup>2</sup>.
[[image:751.24.3.7 L shaped.jpg|center|800px|thumb|
{| style="margin: 1em auto 1em auto"
|-
|'''(*)''' Do not splice stress bars in the fill face at top of footing.
|-
|'''(**)''' If collision forces are assumed, use #4 @ 12" cts. min. and extend at least development length into footing.  (See [[751.5 Structural Detailing Guidelines#751.5.9.2.8.1 Development and Lap Splice General|EPG 751.5.9.2.8.1 Development and Lap Splice General]].)
|}
]]


::<math>\frac{s}{0.196 in.^2} = \frac{12 in.}{0.258 in.^2}</math>
'''Counterfort Walls'''
:'''Wall and Stem'''
[[image:751.24.3.7 counterfort.jpg|center|800px|thumb|
{| style="margin: 1em auto 1em auto"
|-
|<center>(For footing reinforcement, see the "Footing" diagram, below)</center>
|-
|'''(*)''' Use development length or standard hook in accordance with [[751.5 Structural Detailing Guidelines#751.5.9.2.8.1 Development and Lap Splice General|EPG 751.5.9.2.8.1 Development and Lap Splice General]].
|-
|'''(**)''' See lap splices Class B.  (See [[751.5 Structural Detailing Guidelines#751.5.9.2.8.1 Development and Lap Splice General|EPG 751.5.9.2.8.1 Development and Lap Splice General]].)
|}
]]


::''s'' = 9.1 in.
:'''Footing'''
[[image:751.24.3.7 footing.jpg|center|800px|thumb|
{| style="margin: 1em auto 1em auto"
|-
|'''(*)''' By design for loads and footing pressures on section under consideration.  (#5 @ 12" cts. is the minimum.)
|}
]]


::<u>Use #4 bars @ 9 in. cts.</u>
'''Counterfort Walls - Sign-Board Type'''
:'''Wall and Stem'''
:Refer to "Counterfort Walls, Wall and Stem", above.


:::'''Check Shear'''
:'''Spread Footing'''
[[image:751.24.3.7 sign board.jpg|center|800px]]


:::The critical section for shear for the toe is at a distance d = 21.75 inches from the face of the stem. The toe pile is 6 inches from the stem face so the toe pile shear does not affect the shear at the critical section. The critical section for shear is at the stem face for the heel so all of the force of the heel pile affects the shear at the critical section. The worst case for shear is Case IV.
:If the shear line is within the counterfort projected (longitudinally or transversely), the footing may be considered satisfactory for all conditions. If outside of the counterfort projected, the footing must be analyzed and reinforced for bending and checked for bond stress and for diagonal tension stress.


:::''V<sub>u</sub>'' = 7.588k
[[image:751.24.3.7 sign board footing.jpg|center|800px]]


:::<math>\frac{\nu_u}{\phi} = \frac{V_u}{\phi bd} = {7588 lbs}{0.85(12 in.)(32 in.)}</math> = 23.248 psi ≤ 109.5 psi = ''ν<sub>c</sub>'' <u>o.k.</u>
===751.24.3.8 Details===
'''Non-Keyed Joints'''


::'''Reinforcement - Shear Key'''
Each section of wall shall be in increments of 4 ft. with a maximum length of 28'-0".
[[image:751.24.3.8 nonkeyed.jpg|center|800px]]
<center>See [[751.50 Standard Detailing Notes|EPG 751.50 Standard Detailing Notes]] for appropriate notes.</center>


::''b'' = 12 in.
'''Keyed Joints'''
[[image:751.24.3.8 keyed.jpg|center|800px]]
<center>See [[751.50 Standard Detailing Notes|EPG 751.50 Standard Detailing Notes]] for appropriate notes.</center>


::''h'' = 12 in.


::cover = 3 in.
<div id="Rustication Recess"></div>
'''Rustication Recess'''
[[image:751.24.3.8 rustication.jpg|center|800px]]


::''d'' = 12 in. - 3 in. - 0.5(0.5 in.) = 8.75 in.


:::'''Apply Load Factors'''
'''Drains'''
[[image:751.24.3.8 drains.jpg|center|800px]]
<center>Note: French drains shall be used on all retaining walls, unless otherwise specified on the Design Layout.</center>


:::''P<sub>P</sub> = γβ<sub>E</sub>'' (3.845k) = (1.3)(1.3)(3.845k) = 6.498k
[[image:751.24.3.8 drop inlet.jpg|center|800px]]


::''M<sub>u</sub>'' = (0.912 ft.)(6.498k) = 5.926(ft−k)


::<math>R_n = \frac{M_u}{\phi bd^2} = \frac{5.926(ft-−k)}{(0.9)(1 ft.)(8.75 in.)^2}</math> = 0.0860 ksi
'''Construction Joint Keys:
:'''Cantilever Walls'''
[[image:751.24.3.8 cantilever.jpg|center|800px]]


::<math>\rho = \frac{0.85(3ksi)}{60ksi}\Bigg[1 -− \sqrt{1 -− \frac{2(0.0860ksi)}{0.85(3ksi)}}\Bigg]</math> = 0.00146


::<math>\rho_{min} = 1.7\Big[\frac{12 in.}{8.75 in}\Big]^2\frac{\sqrt{3000 psi}}{60,000 psi}</math> = 0.00292
:'''Counterfort Walls'''
[[image:751.24.3.8 counterfort.jpg|center|800px]]


::Use ''ρ'' = 4/3 ''ρ'' = 4/3(0.00146) = 0.00195


::''A<sub>S<sub>Req</sub></sub> = ρbd'' = (0.00195)(12 in.)(8.75 in.) = 0.205 in.<sup>2</sup>/ft.
::Key length:  Divide the length "A" into an odd number of spaces of equal lengths. Each space shall not exceed a length of 24 inches. Use as few spaces as possible with the minimum number of spaces equal to three (or one key).


::Key width = Counterfort width/3 (to the nearest inch)


::One #4 bar has A<sub>s</sub> = 0.196 in<sup>2</sup>
::Key depth = 2" (nominal)


:'''Sign-Board Walls'''
[[image:751.24.3.8 sign board.jpg|center|800px]]


::<math>\frac{s}{0.196 in.^2} = \frac{12 in.}{0.205 in.^2}</math>
::Key length = divide length "A" or "B" into an odd number of spaces of equal lengths. Each space length shall not exceed 24 inches. Use as few spaces as possible with the minimum number of spaces equal to three (or one key).
 
::''s'' = 11.5 in.


::<u>Use #4 bars @ 11 in. cts.</u>
:::'''Check Shear'''
:::<math>\frac{\nu_u}{\phi} = \frac{V_u}{\phi bd} = \frac{6498 lbs}{0.85(12 in.)(8.75 in.)}</math> = 72.807 psi < 109.5 psi = ''ν<Sub>c</sub>''
::'''Reinforcement Summary'''
[[image:751.24.3.5 summary.jpg|center|350px]]
===751.24.3.6 Dimensions===
'''Cantilever Walls'''
Each section of wall shall be in increments of 4 ft. with a maximum length of 28'-0".
[[image:751.24.3.6 friction or bearing piles.jpg|center|800px]]
Each section of wall shall be in increments of 4 ft. with a maximum length of 28'-0".
[[image:751.24.3.6 pile footing.jpg|center|800px]]
'''Cantilever Walls - L-Shaped'''
Each section of wall shall be in increments of 4 ft. with a maximum length of 28'-0".
[[image:751.24.3.6 L shaped.jpg|center|800px]]
'''Counterfort Walls'''
[[image:751.24.3.6 counterfort part elev.jpg|center|800px|thumb|
{| style="margin: 1em auto 1em auto style="text-align:left""
|-
|'''Notes:'''
|-
|'''Dimension "A"'''
|-
|* Maximum length = 28'-0".
|-
|* Each section to be in 4'-0" increments.
|-
|* (See [[#Rustication Recess|rustication recess details]].)
|-
|'''Dimensions "B" & "C"'''
|-
|* As required by the design to balance the negative and positive moments. (See the design assumptions).
|}]]
[[image:751.24.3.6 counterfort typ section.jpg|center|800px|thumb|
{| style="margin: 1em auto 1em auto style="text-align:left""
|-
|'''Notes:'''
|-
|'''Batter  "D":'''
|-
|* As required to maintain 9" minimum at the top of the counterfort and 12" minimum edge distance at the top of the footing, between counterfort and footing edge.
|-
|* Batter to be given an eighth of an inch per foot of counterfort height.
|-
|'''Dimension "L":'''
|-
|* As required for stability.
|-
|* As an estimate, use "L" equal to 1/2 the height of "H".
|}]]
'''Sign-Board Type Counterfort Walls'''
[[image:751.24.3.6 sign board part elev.jpg|center|800px|thumb|
{| style="margin: 1em auto 1em auto style="text-align:left""
|-
|'''Notes:'''
|-
|'''Dimension "A"'''
|-
|* Maximum length = 28'-0".
|-
|* Each section to be in 4'-0" increments.
|-
|* (See [[#Rustication Recess|rustication recess details]].)
|-
|'''Dimensions "B" & "C"'''
|-
|* As required by the design to balance the negative and positive moments. (See the design assumptions).
|-
|'''Dimension "E"'''
|-
|* (Sign-board type only)
|-
|* As required to maintain footing pressure within the allowable for existing foundation material.  12" minimum.
|}]]
[[image:751.24.3.6 sign board typ section.jpg|center|800px|thumb|
{| style="margin: 1em auto 1em auto style="text-align:left""
|-
|'''Notes:'''
|-
|'''Batter  "D":'''
|-
|* As required to maintain 9" minimum at the top of the counterfort and 12" minimum edge distance at the top of the footing, between counterfort and footing edge.
|-
|* Batter to be given an eighth of an inch per foot of counterfort height.
|-
|'''Dimension "L":'''
|-
|* As required for stability.
|-
|* As an estimate, use "L" equal to 1/2 the height of "H".
|}]]
===751.24.3.7 Reinforcement===
'''Cantilever Walls'''
[[image:751.24.3.7 friction.jpg|center|800px|thumb|
{| style="margin: 1em auto 1em auto"
|-
|'''(*)''' Alternate long and short bars at equal spaces.
|-
|'''(**)''' If collision forces are assumed, use #4 @ 12" cts. min. and extend at least development length into footing.  (See [[751.5 Structural Detailing Guidelines#751.5.9.2.7.1 Development and Lap Splice General|EPG 751.5.9.2.7.1 Development and Lap Splice General]].)
|-
|'''(***)''' Theo. cut-off for bending + development length.  (Wall height over 10' only.)
|}
]]
[[image:751.24.3.7 pile footing.jpg|center|800px|thumb|
{| style="margin: 1em auto 1em auto"
|-
|'''(*)''' Alternate long and short bars at equal spaces.
|-
|'''(**)''' If collision forces are assumed, use #4 @ 12" cts. min. and extend at least development length into footing.  (See [[751.5 Structural Detailing Guidelines#751.5.9.2.7.1 Development and Lap Splice General|EPG 751.5.9.2.7.1 Development and Lap Splice General]].)
|-
|'''(***)''' Theo. cut-off for bending + development length.  (Wall height over 10' only.)
|-
|'''(****)''' Due to site constriction.
|}
]]
'''Cantilever Walls - L-Shaped'''
[[image:751.24.3.7 L shaped.jpg|center|800px|thumb|
{| style="margin: 1em auto 1em auto"
|-
|'''(*)''' Do not splice stress bars in the fill face at top of footing.
|-
|'''(**)''' If collision forces are assumed, use #4 @ 12" cts. min. and extend at least development length into footing.  (See [[751.5 Structural Detailing Guidelines#751.5.9.2.7.1 Development and Lap Splice General|EPG 751.5.9.2.7.1 Development and Lap Splice General]].)
|}
]]
'''Counterfort Walls'''
:'''Wall and Stem'''
[[image:751.24.3.7 counterfort.jpg|center|800px|thumb|
{| style="margin: 1em auto 1em auto"
|-
|<center>(For footing reinforcement, see the "Footing" diagram, below)</center>
|-
|'''(*)''' Use development length or standard hook in accordance with [[751.5 Structural Detailing Guidelines#751.5.9.2.7.1 Development and Lap Splice General|EPG 751.5.9.2.7.1 Development and Lap Splice General]].
|-
|'''(**)''' See lap splices Class B.  (See [[751.5 Structural Detailing Guidelines#751.5.9.2.7.1 Development and Lap Splice General|EPG 751.5.9.2.7.1 Development and Lap Splice General]].)
|}
]]
:'''Footing'''
[[image:751.24.3.7 footing.jpg|center|800px|thumb|
{| style="margin: 1em auto 1em auto"
|-
|'''(*)''' By design for loads and footing pressures on section under consideration.  (#5 @ 12" cts. is the minimum.)
|}
]]
'''Counterfort Walls - Sign-Board Type'''
:'''Wall and Stem'''
:Refer to "Counterfort Walls, Wall and Stem", above.
:'''Spread Footing'''
[[image:751.24.3.7 sign board.jpg|center|800px]]
:If the shear line is within the counterfort projected (longitudinally or transversely), the footing may be considered satisfactory for all conditions.  If outside of the counterfort projected, the footing must be analyzed and reinforced for bending and checked for bond stress and for diagonal tension stress.
[[image:751.24.3.7 sign board footing.jpg|center|800px]]
===751.24.3.8 Details===
'''Non-Keyed Joints'''
Each section of wall shall be in increments of 4 ft. with a maximum length of 28'-0".
[[image:751.24.3.8 nonkeyed.jpg|center|800px]]
<center>See [[751.50 Standard Detailing Notes|EPG 751.50 Standard Detailing Notes]] for appropriate notes.</center>
'''Keyed Joints'''
[[image:751.24.3.8 keyed.jpg|center|800px]]
<center>See [[751.50 Standard Detailing Notes|EPG 751.50 Standard Detailing Notes]] for appropriate notes.</center>
<div id="level of service (LOS)"></div>
'''Rustication Recess'''
[[image:751.24.3.8 rustication.jpg|center|800px]]
'''Drains'''
[[image:751.24.3.8 drains.jpg|center|800px]]
<center>Note: French drains shall be used on all retaining walls, unless otherwise specified on the Design Layout.</center>
[[image:751.24.3.8 drop inlet.jpg|center|800px]]
'''Construction Joint Keys:
:'''Cantilever Walls'''
[[image:751.24.3.8 cantilever.jpg|center|800px]]
:'''Counterfort Walls'''
[[image:751.24.3.8 counterfort.jpg|center|800px]]
::Key length:  Divide the length "A" into an odd number of spaces of equal lengths.  Each space shall not exceed a length of 24 inches.  Use as few spaces as possible with the minimum number of spaces equal to three (or one key).
::Key width = Counterfort width/3 (to the nearest inch)
::Key depth = 2" (nominal)
:'''Sign-Board Walls'''
[[image:751.24.3.8 sign board.jpg|center|800px]]
::Key length = divide length "A" or "B" into an odd number of spaces of equal lengths.  Each space length shall not exceed 24 inches.  Use as few spaces as possible with the minimum number of spaces equal to three (or one key).




-->
[[Category:751 LRFD Bridge Design Guidelines]]
[[Category:751 LRFD Bridge Design Guidelines]]

Latest revision as of 15:13, 27 November 2024

751.24.1 General

Additional Information
LRFD 11

For understanding the equivalency of seismic design category (SDC) and seismic zone for LRFD, see EPG 751.9.1.1 and Bridge Seismic Design Flowchart.

Retaining wall shall be designed to withstand lateral earth and water pressures, including any live and dead load surcharge, the self weight of the wall, temperature and shrinkage effect, live load and collision forces, and earthquake loads in accordance with the general principles of LRFD Section 11 and the general principles specified in this article.

Seismic analysis provisions shall not be ignored for walls that support another structure (i.e. support abutment fill or building) in SDC B or C (seismic zone 2 or 3). No-seismic-analysis provisions may be considered for walls that do not support another structure (i.e. most of District walls) in SDC B or C (seismic zone 2 or 3) in accordance with LRFD 11.5.4.2 and Geotech report. Seismic analysis provisions shall not be ignored for walls in SDC D (seismic zone 4).

751.24.1.1 Wall Type Selection

Additional Information
LRFD 11

Selection of wall type shall be based on an assessment of the magnitude and direction of loading, depth to suitable foundation support, potential for earthquake loading, presence of deleterious environmental factors, wall site cross-sectional geometry, proximity of physical constraints, tolerable and differential settlement, facing appearance and ease and cost of construction.

The following wall types are the most commonly used in MoDOT projects

  • Mechanically Stabilized Earth Retaining Walls
  • Cast-In-Place Concrete Cantilever Retaining Walls
▪ Cantilever Walls on Spread Footings
▪ Cantilever Wall on Pile Footings
▪ L-Shaped Walls on Spread Footings

Mechanically Stabilized Earth (MSE) Retaining Walls

Additional Information
LRFD 11.10,
FHWA-NHI-10-024 and 025

MSE retaining walls use precast block or panel like facing elements combined with either metallic or geosynthetic tensile reinforcements in the soil mass. MSE walls are preferred over cast-in-place walls because they are usually more economical. Other advantages include a wide variety of design styles, ease and speed of installation, and their ability to accommodate total and differential settlements. Wall design heights upwards of 80 ft. are technically feasible (FHFW-SA-96-071). MSE walls may be used to retain fill for end bents of bridge structures.

Situations exist where the use of MSE walls is either limited or not recommended. Some obstacles such as drop inlets, sign truss pedestals or footings, and fence posts may be placed within the soil reinforcement area, however, these obstacles increase the difficulty and expense of providing sufficient soil reinforcement for stability. Box culverts and highway drainage pipes may run through MSE walls, but it is preferable not to run the pipes close to or parallel to the walls. Utilities other than highway drainage should not be constructed within the soil reinforcement area. Be cautious when using MSE walls in a floodplain. A flood could cause scouring around the reinforcement and seepage of the backfill material. Soil reinforcements should not be used where exposure to ground water contaminated by acid mine drainage or other industrial pollutants as indicated by a low pH and high chlorides and sulfates exist. Galvanized metallic reinforcements shall not be used where stray electrical ground currents could occur as would be present near an electrical substation.

Sufficient right of way is required to install the soil reinforcement which extends into the backfill area at least 8 feet, 70 percent of the wall height or as per design requirements set forth in EPG 751.6.2.17 Excavation, whichever is greater. For more information regarding soil reinforcement length, excavation limits and Minimum Embedment Depth of MSEW, see EPG 751.6.2.17 Excavation.

Finally, barrier curbs constructed over or in line with the front face of the wall shall have adequate room provided laterally between the back of the wall facing and the curb or slab so that load is not directly transmitted to the top of MSE wall or facing units.



Concrete Cantilever Wall on Spread Footing

Concrete cantilever walls derive their capacity through combinations of dead weight and structural resistance. These walls are constructed of reinforced concrete.

Concrete cantilever walls are used when MSE walls are not a viable option. Cantilever walls can reduce the rock cut required and can also provide solutions when there are right of way restrictions. Concrete walls also provide better structural capacity when barrier or railing on top of the walls are required.

Counterforts are used on rare occasions. Sign-board type retaining walls are a special case of counterfort retaining walls. They are used where the soil conditions are such that the footings must be placed well below the finished ground line. For these situations the wall is discontinued 12 in. below the ground line or below the frost line. Counterforts may also be a cost-savings option when the wall height approaches 20 ft. (Foundation Analysis and Design by Joseph E. Bowles, 4th ed., 1988). However, other factors such as poor soil conditions, slope of the retained soil, wall length and uniformity in wall height should also be considered before using counterforts.

Concrete Cantilever Wall on Pile Footing

Concrete cantilever walls on pile footings are used when the soil conditions do not permit the use of spread footings. These walls are also used when an end bent requires wings longer than 22 feet for seismic category A or 17 ft. for seismic category B, C or D. In these cases a stub wing is left attached to the end bent and the rest of the wing is detached to become a retaining wall as shown in 751.35.3.5 Wing and Detached Wing Walls.

Concrete L-Shaped Retaining Wall on Spread Footings

Concrete L-Shaped walls are cantilever walls without heels. These walls are used when there are space limitations for cantilever walls. Since there is no heel the height of these walls is limited to about 7 ft. depending on the soil conditions and the slope of the retained soil.

L-Shaped Walls are often used next to roadways where the footings are frequently used as shoulders and where the wall will require structural capacity for collision forces.

751.24.1.2 Loads

Conventional retaining walls: Loads shall be determined in accordance with LRFD 3 and 11.6.

MSE retaining walls: Loads shall be determined in accordance with LRFD 3 and 11.10.

Note: For guidance, follow the 751.40.8.15 Cast -In-Place Concrete Retaining Walls and modify guidance of ASD as necessary to meet LRFD requirements until this section is modified for LRFD.

Dead Loads

Dead loads shall be determined from the unit weights in EPG 751.2.1.1 Dead Load.

751.24.2 Mechanically Stabilized Earth (MSE) Walls

751.24.2.1 Design

Designs of Mechanically Stabilized Earth (MSE) walls shall be completed by consultants or contractors in accordance with Section 11.10 of LRFD specifications, FHWA-NHI-10-024 and FHWA-NHI-10-025 for LRFD. Bridge Pre-qualified Products List (BPPL) provided on MoDOT's web page and in Sharepoint contains a listing of facing unit manufacturers, soil reinforcement suppliers, and wall system suppliers which have been approved for use. See Sec 720 and Sec 1010 for additional information. The Geotechnical Section is responsible for checking global stability of permanent MSE wall systems, which should be reported in the Foundation Investigation Geotechnical Report. For MSE wall preliminary information, see EPG 751.1.4.3 MSE Walls. For design requirements of MSE wall systems and temporary shoring (including temporary MSE walls), see EPG 720 Mechanically Stabilized Earth Wall Systems. For staged bridge construction, see EPG 751.1.2.11 Staged Construction.

For seismic design requirements, see Bridge Seismic Design Flowchart. References for consultants and contractors include Section 11.10 of LRFD, FHWA-NHI-10-024 and FHWA-NHI-10-025.

Design Life

  • 75 year minimum for permanent walls (if retained foundation require 100 year than consider 100 year minimum design life for wall).

Global stability:

Global stability will be performed by Geotechnical Section or their agent.

MSE wall contractor/designer responsibility:

MSE wall contractor/designer shall perform following analysis in their design for all applicable limit states.

  • External Stability
  • Limiting Eccentricity
  • Sliding
  • Factored Bearing Pressure/Stress ≤ Factored Bearing Resistance
  • Internal Stability
  • Tensile Resistance of Reinforcement
  • Pullout Resistance of Reinforcement
  • Structural Resistance of Face Elements
  • Structural Resistance of Face Element Connections
  • Compound Stability
Capacity/Demand ratio (CDR) for bearing capacity shall be ≥ 1.0
Strength Limit States:
Factored bearing resistance = Nominal bearing resistance from Geotech report X Minimum Resistance factor (0.65, Geotech report) LRFD Table 11.5.7-1
Extreme Event I Limit State:
Factored bearing resistance = Nominal bearing resistance from Geotech report X Resistance factor
Resistance factor = 0.9 LRFD 11.8.6.1
Factored bearing stress shall be computed using a uniform base pressure distribution over an effective width of footing determined in accordance with the provisions of LRFD 10.6.3.1 and 10.6.3.2, 11.10.5.4 and Figure 11.6.3.2-1 for foundation supported on soil or rock.
B’ = L – 2e
Where,
L = Soil reinforcement length (For modular block use B in lieu of L as per LRFD 11.10.2-1)
B’ = effective width of footing
e = eccentricity
Note: When the value of eccentricity e is negative then B´ = L.
Capacity/Demand ratio (CDR) for overturning shall be ≥ 1.0
Capacity/Demand ratio (CDR) for eccentricity shall be ≥ 1.0
Capacity/Demand ratio (CDR) for sliding shall be ≥ 1.0      LRFD 11.10.5.3 & 10.6.3.4
Capacity/Demand ratio (CDR) for internal stability shall be ≥ 1.0
Eccentricity, (e) Limit for Strength Limit State:      LRFD 11.6.3.3 & C11.10.5.4
For foundations supported on soil or rock, the location of the resultant of the reaction forces shall be within the middle two-thirds of the base width, L or (e ≤ 0.33L).
Eccentricity, (e) Limit for Extreme Event I (Seismic):      LRFD 11.6.5.1
For foundations supported on soil or rock, the location of the resultant of the reaction forces shall be within the middle two-thirds of the base width, L or (e ≤ 0.33L) for γEQ = 0.0 and middle eight-tenths of the base width, L or (e ≤ 0.40L) for γEQ = 1.0. For γEQ between 0.0 and 1.0, interpolate e value linearly between 0.33L and 0.40L. For γEQ refer to LRFD 3.4.
Note: Seismic design shall be performed for γEQ = 0.5
Eccentricity, (e) Limit for Extreme Event II:
For foundations supported on soil or rock, the location of the resultant of the reaction forces shall be within the middle eight-tenths of the base width, L or (e ≤ 0.40L).

General Guidelines

  • Drycast modular block wall (DMBW-MSE) systems are limited to a 10 ft. height in one lift.
  • Wetcast modular block wall (WMBW-MSE) systems are limited to a 15 ft. height in one lift.
  • For Drycast modular block wall (DMBW-MSE) systems and Wetcast modular block wall (WMBW-MSE) systems, top cap units shall be used and shall be permanently attached by means of a resin anchor system.
  • For precast modular panel wall (PMPW-MSE) systems, capstone may be substituted for coping and either shall be permanently attached to wall by panel dowels.
  • For precast modular panel wall (PMPW-MSE) systems, form liners are required to produce all panels. Using form liner to produce panel facing is more cost effective than producing flat panels. Standard form liners are specified on the MSE Wall Standard Drawings. Be specific regarding names, types and colors of staining, and names and types of form liner.
  • MSE walls shall not be used where exposure to acid water may occur such as in areas of coal mining.
  • MSE walls shall not be used where scour is a problem.
  • MSE walls with metallic soil reinforcement shall not be used where stray electrical ground currents may occur as would be present near electrical substations.
  • No utilities shall be allowed in the reinforced earth if future access to the utilities would require that the reinforcement layers be cut, or if there is a potential for material, which can cause degradation of the soil reinforcement, to leak out of the utilities into the wall backfill, with the exception of storm water drainage.
  • All vertical objects shall have at least 4’-6” clear space between back of the wall facing and object for select granular backfill compaction and soil reinforcement skew limit requirements. For piles, see pipe pile spacers guidance.
  • The interior angle between two MSE walls should be greater than 70°. However, if unavoidable, then place EPG 751.50 J1.41 note on the design plans.
  • Drycast modular block wall (DMBW-MSE) systems and Wetcast modular block wall (WMBW-MSE) systems may be battered up to 1.5 in. per foot. Modular blocks are also known as “segmental blocks”.
  • The friction angle used for the computation of horizontal forces within the reinforced soil shall be greater than or equal to 34°.
  • All concrete except facing panels or units shall be CLASS B or B-1.
  • The friction angle of the soil to be retained by the reinforced earth shall be listed on the plans as well as the friction angle for the foundation material the wall is to rest on.
  • The following requirement shall be considered (from 2009_FHWA-NHI-10-024 MSE wall 132042.pdf, page 200-201) when seismic design is required:
  • For seismic design category, SDC C or D (Zones 3 or 4), facing connections in modular block faced walls (MBW) shall use shear resisting devices (shear keys, pin, etc.) between the MBW units and soil reinforcement, and shall not be fully dependent on frictional resistance between the soil reinforcement and facing blocks. For connections partially dependent on friction between the facing blocks and the soil reinforcement, the nominal long-term connection strength Tac, should be reduced to 80 percent of its static value.
  • Seismic design category and acceleration coefficients shall be listed on the plans for categories B, C and D. If a seismic analysis is required that shall also be noted on the plans. See EPG 751.50 A1.1 note.
  • Plans note (EPG 751.50 J1.1) is required to clearly identify the responsibilities of the wall designer.
  • Do not use Drycast modular block wall (DMBW-MSE) systems in the following locations:
  • Within the splash zone from snow removal operations (assumed to be 15 feet from the edge of the shoulder).
  • Where the blocks will be continuously wetted, such as around sources of water.
  • Where blocks will be located behind barrier or other obstacles that will trap salt-laden snow from removal operations.
  • Do not use Drycast modular block wall (DMBW-MSE) systems or Wetcast modular block wall (WMBW-MSE) systems in the following locations:
  • For structurally critical applications, such as containing necessary fill around structures.
  • In tiered wall systems.
  • For locations where Drycast modular block wall (DMBW-MSE) systems and Wetcast modular block wall (WMBW-MSE) systems are not desirable, consider coloring agents and/or architectural forms using precast modular panel wall (PMPW-MSE) systems for aesthetic installations.
  • Roadway runoff should be directed away from running along face of MSE walls used as wing walls on bridge structures.
  • Drainage:
  • Gutter type should be selected at the core team meeting.
  • When gutter is required without fencing, use Type A or Type B gutter (for detail, see Std. Plan 609.00).
  • When gutter is required with fencing, use Modified Type A or Modified Type B gutter (for detail, see Std. Plan 607.11).
  • When fencing is required without gutter, place in tube and grout behind the MSE wall (for detail, see MSE Wall Standard Drawings - MSEW, Fence Post Connection Behind MSE Wall (without gutter).
  • Lower backfill longitudinal drainage pipes behind all MSE walls shall be two-6” (Min.) diameter perforated PVC or PE pipe (See Sec 1013) unless larger sizes are required by design which shall be the responsibility of the District Design Division. Show drainage pipe size on plans. Outlet screens and cleanouts should be detailed for any drain pipe (shown on MoDOT MSE wall plans or roadway plans). Lateral non-perforated drain pipes (below leveling pad) are permitted by Standard Specifications and shall be sized by the District Design Division if necessary. Lateral outlet drain pipe sloped at 2% minimum.
  • Identify on MSE wall plans or roadway plans drainage pipe point of entry, point of outlet (daylighting), 2% min. drainage slopes in between points to ensure positive flow and additional longitudinal drainage pipes if required to accommodate ground slope changes and lateral drainage pipes if required by design.
  • Adjustment in the vertical alignment of the longitudinal drainage pipes from that depicted on the MSE wall standard drawings may be necessary to ensure positive flow out of the drainage system.
  • Identify on MSE wall plans or roadway plans the outlet ends of pipes which shall be located to prevent clogging or backflow into the drainage system. Outlet screens and cleanouts should be detailed for any drain pipe.

MSE Wall Construction: Pipe Pile Spacers Guidance

For bridges not longer than 200 feet, pipe pile spacers or pile jackets shall be used at pile locations behind mechanically stabilized earth walls at end bents. Corrugated pipe pile spacers are required when the wall is built prior to driving the piles to protect the wall reinforcement when driving pile for the bridge substructure at end bents(s). Pile spacers or pile jackets may be used when the piles are driven before the wall is built. Pipe pile spacers shall have an inside diameter greater than that of the pile and large enough to avoid damage to the pipe when driving the pile. Use EPG 751.50 Standard Detailing Note E1.2a on bridge plans.

For bridges longer than 200 feet, pipe pile spacers are required and the pile spacer shall be oversized to mitigate the effects of bridge thermal movements on the MSE wall. For HP12, HP14, CIP 14” and CIP 16” piles provide 24-inch inside diameter of pile spacer for bridge movement. Minimum pile spacing shall be 5 feet to allow room for compaction of the soil layers. Use EPG 751.50 Standard Detailing Note E1.2b on bridge plans.

The bottom of the pipe pile spacers shall be placed 5 ft. min. below the bottom of the MSE wall leveling pad. The pipe shall be filled with sand or other approved material after the pile is placed and before driving. Pipe pile spacers shall be accurately located and capped for future pile construction.

Alternatively, for bridges shorter than or equal to 200 feet, the contractor shall be given the option of driving the piles before construction of the mechanically stabilized earth wall and placing the soil reinforcement and backfill material around the piling. In lieu of pipe pile spacers contractor may place pile jackets on the portion of the piles that will be in the MSE soil reinforced zone prior to placing the select granular backfill material and soil reinforcement. The contractor shall adequately support the piling to ensure that proper pile alignment is maintained during the wall construction. The contractor’s plan for bracing the pile shall be submitted to the engineer for review.

Piling shall be designed for downdrag (DD) loads due to either method. Oversized pipe pile spacers with sand placed after driving or pile jacket may be considered to mitigate some of the effects of downdrag (DD) loads. Sizing of pipe pile spacers shall account for pile size, thermal movements of the bridge, pile placement plan, and vertical and horizontal placement tolerances.

When rock is anticipated within the 5 feet zone below the MSE wall leveling pad, prebore into rock and prebore holes shall be sufficiently wide to allow for a minimum 10 feet embedment of pile and pipe pile spacer. When top of rock is anticipated within the 5 to 10 feet zone below the MSE wall leveling pad, prebore into rock to achieve a minimum embedment (pile only) of 10 feet below the bottom of leveling pad. Otherwise, the pipe pile spacer requires a minimum 5 feet embedment below the levelling pad. Consideration shall also be given to oversizing the prebore holes in rock to allow for temperature movements at integral end bents.

For bridges not longer than 200 feet, the minimum clearance from the back face of MSE walls to the front face of the end bent beam, also referred to as setback, shall be 4 ft. 6 in. (typ.) unless larger than 18-inch pipe pile spacer required. The 4 ft. 6 in. dimension is based on the use of 18-inch inside diameter pipe pile spacers & FHWA-NHI-10-24, Figure 5-17C, which will help ensure that soil reinforcement is not skewed more than 15° for nut and bolt reinforcement connections. Similarly, the minimum setback shall be determined when larger diameter pile spacers are required. For bridges longer than 200 feet, the minimum setback shall be 5 ft. 6 in. based on the use of 24-inch inside diameter of pipe pile spacers. Other types of connections may require different methods for splaying. In the event that the minimum setback cannot be used, the following guidance for pipe pile spacers clearance shall be used: pipe pile spacers shall be placed 18 in. clear min. from the back face of MSE wall panels; 12 in. minimum clearance is required between pipe pile spacers and leveling pad and 18 in. minimum clearance is required between leveling pad and pile.

MSE Wall Plan and Geometrics

  • A plan view shall be drawn showing a baseline or centerline, roadway stations and wall offsets. The plan shall contain enough information to properly locate the wall. The ultimate right of way shall also be shown, unless it is of a significant distance from the wall and will have no effect on the wall design or construction.
  • Stations and offsets shall be established between one construction baseline or roadway centerline and a wall control line (baseline). Some wall designs may contain a slight batter, while others are vertical. A wall control line shall be set at the front face of the wall, either along the top or at the base of the wall, whichever is critical to the proposed improvements. For battered walls, in order to allow for batter adjustments of the stepped level pad or variation of the top of the wall, the wall control line (baseline) is to be shown at a fixed elevation. For battered walls, the offset location and elevation of control line shall be indicated. All horizontal breaks in the wall shall be given station-offset points, and walls with curvature shall indicate the station-offsets to the PC and PT of the wall, and the radius, on the plans.
  • Any obstacles which could possibly interfere with the soil reinforcement shall be shown. Drainage structures, lighting, or truss pedestals and footings, etc. are to be shown, with station offset to centerline of the obstacle, with obstacle size. Skew angles are shown to indicate the angle between a wall and a pipe or box which runs through the wall.
  • Elevations at the top and bottom of the wall shall be shown at 25 ft. intervals and at any break points in the wall.
  • Curve data and/or offsets shall be shown at all changes in horizontal alignment. If battered wall systems are used on curved structures, show offsets at 10 ft. (max.) intervals from the baseline.
  • Details of any architectural finishes (formliners, concrete coloring, etc.).
  • Details of threaded rod connecting the top cap block.
  • Estimated quantities, total sq. ft. of mechanically stabilized earth systems.
  • Proposed grade and theoretical top of leveling pad elevation shall be shown in constant slope. Slope line shall be adjusted per project. Top of wall or coping elevation and stationing shall be shown in the developed elevation per project. If leveling pad is anticipated to encounter rock, then contact the Geotechnical Section for leveling pad minimum embedment requirements.

MSE Wall Cross Sections

  • A typical wall section for general information is shown.
  • Additional sections are drawn for any special criteria. The front face of the wall is drawn vertical, regardless of the wall type.
  • Any fencing and barrier or railing are shown.
  • Barrier if needed are shown on the cross section. Barriers are attached to the roadway or shoulder pavement, not to the MSE wall. Standard barriers are placed along wall faces when traffic has access to the front face of the wall over shoulders of paved areas.

Drainage at MSE Walls

  • Drainage Before MSE Wall
Drainage is not allowed to be discharged within 10 ft. from front of MSE wall in order to protect wall embedment, prevent erosion and foundation undermining, and maintain soil strength and stability.
  • Drainage Behind MSE Wall
Internal (Subsurface) Drainage
Groundwater and infiltrating surface waters are drained from behind the MSE wall through joints between the face panels or blocks (i.e. wall joints) and two-6 in. (min.) diameter pipes located at the base of the wall and at the basal interface between the reinforced backfill and the retained backfill.
Excessive subsurface draining can lead to increased risk of backfill erosion/washout through the wall joints and erosion at the bottom of walls and at wall terminal ends. Excessive water build-up caused by inadequate drainage at the bottom of the wall can lead to decreased soil strength and wall instability. Bridge underdrainage (vertical drains at end bents and at approach slabs) can exacerbate the problem.
Subsurface drainage pipes should be designed and sized appropriately to carry anticipated groundwater, incidental surface run-off that is not collected otherwise including possible effects of drainage created by an unexpected rupture of any roadway drainage conveyance or storage as an example.
External (Surface) Drainage
External drainage considerations deal with collecting water that could flow externally over and/or around the wall surface taxing the internal drainage and/or creating external erosion issues. It can also infiltrate the reinforced and retained backfill areas behind the MSE wall.
Diverting water flow away from the reinforced soil structure is important. Roadway drainage should be collected in accordance with roadway drainage guidelines and bridge deck drainage should be collected similarly.
  • Guidance
ALL MSE WALLS
1. Appropriate measures to prevent surface water infiltration into MSE wall backfill should be included in the design and detail layout for all MSE walls and shown on the roadway plans.
2. Gutters behind MSE walls are required for flat or positive sloping backfills to prevent concentrated infiltration behind the wall facing regardless of when top of backfill is paved or unpaved. This avoids pocket erosion behind facing and protection of nearest-surface wall connections which are vulnerable to corrosion and deterioration. Drainage swales lined with concrete, paved or precast gutter can be used to collect and discharge surface water to an eventual point away from the wall. If rock is used, use impermeable geotextile under rock and align top of gutter to bottom of rock to drain. (For negative sloping backfills away from top of wall, use of gutters is not required.)
District Design Division shall verify the size of the two-6 in. (min.) diameter lower perforated MSE wall drain pipes and where piping will daylight at ends of MSE wall or increase the diameters accordingly. This should be part of the preliminary design of the MSE wall. (This shall include when lateral pipes are required and where lateral drain pipes will daylight/discharge).
BRIDGE ABUTMENTS WITH MSE WALLS
Areas of concern: bridge deck drainage, approach slab drainage, approach roadway drainage, bridge underdrainage: vertical drains at end bents and approach slab underdrainage, showing drainage details on the roadway and MSE wall plans
3. Bridge slab drain design shall be in accordance with EPG 751.10.3 Bridge Deck Drainage – Slab Drains unless as modified below.
4. Coordination is required between the Bridge Division and District Design Division on drainage design and details to be shown on the MSE wall and roadway plans.
5. Bridge deck, approach slab and roadway drainage shall not be allowed to be discharged to MSE wall backfill area or within 10 feet from front of MSE wall.
  • (Recommended) Use of a major bridge approach slab and approach pavement is ideal because bridge deck, approach slab and roadway drainage are directed using curbs and collected in drain basins for discharge that protect MSE wall backfill. For bridges not on a major roadway, consideration should be given to requiring a concrete bridge approach slab and pavement incorporating these same design elements (asphalt is permeable).
  • (Less Recommended) Use of conduit and gutters:
  • Conduit: Drain away from bridge and bury conduit daylighting to natural ground or roadway drainage ditch at an eventual point beyond the limits of the wall. Use expansion fittings to allow for bridge movement and consider placing conduit to front of MSE wall and discharging more than 10 feet from front of wall or using lower drain pipes to intercept slab drainage conduit running through backfill.
  • Conduit and Gutters: Drain away from bridge using conduit and 90° elbow (or 45° bend) for smoothly directing drainage flow into gutters and that may be attached to inside of gutters to continue along downward sloping gutters along back of MSE wall to discharge to sewer or to natural drainage system, or to eventual point beyond the limits of the wall. Allow for independent bridge and wall movements by using expansion fittings where needed. See EPG 751.10.3.1 Type, Alignment and Spacing and EPG 751.10.3.3 General Requirements for Location of Slab Drains.
6. Vertical drains at end bents and approach slab underdrainage should be intercepted to drain away from bridge end and MSE wall.
7. Discharging deck drainage using many slab drains would seem to reduce the volume of bridge end drainage over MSE walls.
8. Drain flumes at bridge abutments with MSE walls do not reduce infiltration at MSE wall backfill areas and are not recommended.
DISTRICT DESIGN DIVISION MSE WALLS
Areas of concern: roadway or pavement drainage, MSE wall drainage, showing drainage details on the roadway and MSE wall plans.
9. For long MSE walls, where lower perforated drain pipe slope become excessive, non-perforated lateral drain pipes, permitted by Standard Specifications, shall be designed to intercept them and go underneath the concrete leveling pad with a 2% minimum slope. Lateral drain pipes shall daylight/discharge at least 10 ft. from front of MSE wall. Screens should be installed and maintained on drain pipe outlets.
10. Roadway and pavement drainage shall not be allowed to be discharged to MSE wall backfill area or within 10 feet from front of MSE wall.
11. For district design MSE walls, use roadway or pavement drainage collection pipes to transport and discharge to an eventual point outside the limits of the wall.
Example: Showing drain pipe details on the MSE wall plans.

Notes:
(1) To be designed by District Design Division.
(2) To be designed by District Design Division if needed. Provide non-perforated lateral drain pipe under leveling pad at 2% minimum slope. (Show on plans).
(3) Discharge to drainage system or daylight screened outlet at least 10 feet away from end of wall (typ.). (Skew in the direction of flow as appropriate).
(4) Discharge to drainage system or daylight screened outlet at least 10 feet away from front face of wall (typ.). (Skew in the direction of flow as appropriate).
(5) Minimum backfill cover = Max(15”, 1.5 x diameter of drain pipe).

751.24.2.2 Excavation

For estimating excavation and minimum soil reinforcement length, see EPG 751.6.2.17 Excavation.

For division responsibilities for preparing MSE wall plans, computing excavation class, quantities and locations, see EPG 747.2.6.2 Mechanically Stabilized Earth (MSE) Wall Systems.

751.24.2.3 Details

Bridge Standard Drawings
MSE Wall - MSEW

(1) Minimum embedment = maximum (2 feet; or embedment based on Geotechnical Report and global stability requirements;
or FHWA-NH1-10-0124, Table 2-2); or as per Geotechnical Report if rock is known to exist from Geotechnical Report.

Drycast Modular Block Wall Systems and Wetcast Modular Block Wall Systems

Battered mechanically stabilized earth wall systems may be used unless the design layout specifically calls for a vertical wall (precast modular panel wall systems shall not be battered and drycast modular block wall systems or wetcast modular block wall systems may be built vertical). If a battered MSE wall system is allowed, then EPG 751.50 J1.19 note shall be placed on the design plans:

For battered walls, note on the plans whether the horizontal offset from the baseline is fixed at the top or bottom of the wall. Horizontal offset and corresponding vertical elevation shall be noted on plans.

* The maximum vertical spacing of reinforcement should be limited to two times the block depth or 32 in., whichever is less.
For large modular block (block height > 16 in.), maximum vertical spacing of reinforcement equal to the block height.

Fencing (See Bridge Standard Drawing for details)

Fencing may be installed on the Modified Type A or Modified Type B Gutter or behind the MSE Wall.

For Modified Type A and Modified Type B Gutter and Fence Post Connection details, see Standard Plan 607.11.

751.24.3 Cast-In-Place Concrete Retaining Walls

751.24.3.1 Unit Stresses

Concrete Concrete for retaining walls shall be Class B Concrete (f'c = 3000 psi) unless the footing is used as a riding surface in which case Class B-1 Concrete (f'c = 4000 psi) shall be used.

Reinforcing Steel

Reinforcing Steel shall be Grade 60 (fy = 60,000 psi).

Pile Footing

For steel piling material requirements, see the unit stresses in EPG 751.50 A1.3 note.

Spread Footing

For foundation material capacity, see Foundation Investigation Geotechnical Report.

751.24.3.2 Design

Note: For design concepts and guidance, follow the design process (EPG 751.40.8.15) and modify design/details of ASD as necessary to meet LRFD requirements until EPG 751.24 is updated for LRFD.

Capacity/Demand ratio (CDR) for bearing capacity shall be ≥ 1.0

Strength Limit States:
Factored bearing resistance = Nominal bearing resistance from Geotech report X
Minimum Resistance factor (0.55, Geotech report)      LRFD Table 11.5.7
Extreme Event I and II Limit State:
Factored bearing resistance = Nominal bearing resistance from Geotech report X Resistance factor
Resistance factor = 0.8      LRFD 11.5.8
When wall is supported by soil:
Factored bearing stress per LRFD eq. 11.6.3.2-1
When wall is supported by a rock foundation:
Factored bearing stress per LRFD eq. 11.6.3.2-2 and 11.6.3.2-3
Note: When the value of eccentricity e is negative then use e = 0.

Capacity/Demand ratio (CDR) for overturning shall be ≥ 1.0

Capacity/Demand ratio (CDR) for eccentricity shall be ≥ 1.0

Capacity/Demand ratio (CDR) for sliding shall be ≥ 1.0

Sliding shall be checked in accordance with LRFD 11.6.3.6 and 10.6.3.4

Eccentricity, (e) Limit for Strength Limit State:      LRFD 11.6.3.3

  • For foundations supported on soil, the location of the resultant of the reaction forces shall be within the middle two-thirds of the base width, B or (e ≤ 0.33B).
  • For foundations supported on rock, the location of the resultant of the reaction forces shall be within the middle nine-tenths of the base width, B or (e ≤ 0.45B).

Eccentricity, (e) Limit for Extreme Event I (Seismic):      LRFD 11.6.5.1

  • For foundations supported on soil or rock, the location of the resultant of the reaction forces shall be within the middle two-thirds of the base width, B or (e ≤ 0.33B) for γEQ = 0.0 and middle eight-tenths of the base width, B or (e ≤ 0.40B) for γEQ = 1.0. For γEQ between 0.0 and 1.0, interpolate e value linearly between 0.33B and 0.40B. For γEQ refer to LRFD 3.4.
Note: Seismic design shall be performed for γEQ = 0.5

Eccentricity, (e) Limit for Extreme Event II:

  • For foundations supported on soil or/and rock, the location of the resultant of the reaction forces shall be within the middle eight-tenths of the base width, B or (e ≤ 0.40B).

For epoxy coated reinforcement requirements, see EPG 751.5.9.2.2 Epoxy Coated Reinforcement Requirements.

If the height of the wall or fill is a variable dimension, then base the structural design of the wall, toe, and heel on the high quarter point between expansion joints.

Fig. 751.24.3.2