751.40 LFD Widening and Repair (CONTINUED)

751.40.8.12 Concrete Pile Cap Intermediate Bents

751.40.8.12.1 Design

751.40.8.12.1.1 Unit Stresses

(1)	Reinforced Concrete
	Class B Concrete (Substructure)	$f_c$ = 1,200 psi	$f{\text{'}}_c$ = 3,000 psi
	Reinforcing Steel (Grade 60	$f_s$ = 24,000 psi	$f_y$ = 60,000 psi
	$n$ = 10
	$E_c=3,122ksi(Ec=W^{1.5}\times 33\sqrt{f{\text{'}}_c},Ec=57,000\sqrt{f^{\prime }c}$
(2)	Structural Steel
	Structural Carbon Steel (ASTM A709 Grade 36)	$f_s$ = 20,000 psi	$f_y$ = 36,000 psi
(3)	Piling
(4)	Overstress
	The allowable overstresses as specified in AASHTO Article 3.22 shall be used where applicable for service loads.

751.40.8.12.1.2 Loads

(1)	Dead Loads
(2)	Live Load
	As specified on Bridge Memorandum.
	Impact of 30% is to be used for design of the beam. No impact is to be used for design of any other portion of bent including the piles.
(3)	Temperature, Wind and Frictional Loads

751.40.8.12.1.3 Distribution of Loads

(1)	Dead Loads
	Loads from stringers, girders, etc. shall be concentrated loads applied at the intersection of centerline of stringer and centerline of bearing. Loads from concrete slab spans shall be applied as uniformly, distributed loads along the centerline of bearing.
(2)	Live Load
	Loads from stringers, girders, etc. shall be applied as concentrated loads at the intersection of centerline of stringer and centerline of bearing. For concrete slab spans distribute two wheel lines over 10'-0" (normal to centerline of roadway) of substructure beam. This distribution shall be positioned on the beam on the same basis as used for wheel lines in Traffic Lanes for Substructure Design.
(3)	Temperature, Wind and Frictional Loads

751.40.8.12.1.4 Design Assumptions

LOADINGS

(1)	Beam
	The beam shall be assumed continuous over supports at centerline of piles.
	Intermediate bent beam caps shall be designed so that service dead load moments do not exceed the cracking moment of the beam cap (AASHTO Article 8.13.3, Eq. 8-2).
(2)	Piles
	(a)	Bending
		Stresses in the piles due to bending need not be considered in design calculations for Seismic Performance Category A.
	(b)	Dead Loads, etc.
		Dead load of superstructure and substructure will be distributed equally to all piles which are under the main portion of the bent.

751.40.8.12.2 Reinforcement

751.40.8.12.2.1 General

PRESTRESS DOUBLE-TEE STRUCTURES

BENTS WITH 3" THRU 6" CROWN

SECTION A-A	SECTION B-B

(*)	Channel shear connectors are to be used in Seismic Performance Categories B, C & D. For details not shown, see EPG 751.9 Bridge Seismic Design.
(**)	2'-6" Min. for Seismic Performance Category A. 2'-9" Min. for Seismic Performance Categories, B, C & D.
Note: Use square ends on Prestress Double-Tee Structures.

BENTS WITH CROWN OVER 6"

SECTION A-A	SECTION B-B

(*)	Channel shear connectors are to be used in Seismic Performance Categories B, C & D.
(**)	2'-6" Min. for Seismic Performance Category A. 2'-9" Min. for Seismic Performance Categories, B, C & D.
Note: Use square ends on Prestress Double-Tee Structures.

751.40.8.12.2.2 Anchorage of Piles for Seismic Performance Categories B, C & D

STEEL PILE

PART ELEVATION	SECTION THRU BEAM
PART PLAN

CAST-IN-PLACE PILE

PART ELEVATION	SECTION THRU BEAM
PART PLAN

751.40.8.12.2.3 Beam Reinforcement Special Cases

SPECIAL CASE I

If centerline bearing is 12" or less on either side of centerline piles, for all piles (as shown above), use 4-#6 top and bottom and #4 at 12" cts. (stirrups), regardless of pile size.

SPECIAL CASE II

When beam reinforcement is to be designed assuming piles to take equal force, design for negative moment in the beam over the interior piles.

(*) Dimensions shown are for illustration purposes only.

751.40.8.12.3 Details

751.40.8.12.3.1 Sway Bracing

Refer to EPG 751.32.3.2.1 Sway Bracing.

751.40.8.12.3.2 Miscellaneous Details for Prestressed Girder

PRESTRESSED GIRDERS (INTEGRAL INT. BENT)

DETAIL OF JOINT FILLER AT INT. BENTS
(Continuous Spans - No Longitudinal Beam Steps)

(*)	¼ Joint Filler for a P/S Double Tee Structure
	½ Joint Filler for a P/S I-Girder Structure

PRESTRESSED GIRDERS (NON-INTEGRAL INT. BENT)

DETAIL OF JOINT FILLER AT INT. BENTS
Longitudinal Beam Step and Shear Blocks shown)

DETAILS OF CONST. JOINT KEY

PART ELEVATION	PART SECTION THRU KEYS (P/S I-GIRDERS)	PART SECTION THRU KEYS (P/S DOUBLE TEE GIRDERS)

751.40.8.13 Concrete Pile Cap Non-Integral End Bents

751.40.8.13.1 Design

751.40.8.13.1.1 Unit Stresses

(1)	Reinforced Concrete
	Class B Concrete (Substructure)	$f_c$ = 1,200 psi	$f{\text{'}}_c$ = 3,000 psi
	Reinforcing Steel (Grade 60)	$f_s$ = 24,000 psi	$f_y$ = 60,000 psi
	$n$ = 10
	$E_c=W^{1.5}\times 33\sqrt{f{\text{'}}_c}$ AASHTO Article 8.7.1) (*)
(2)	Structural Steel
	Structural Carbon Steel (ASTM A709 Grade 36)	$f_s$ = 20,000 psi	$f_y$ = 36,000 psi
(3)	Piling
(4)	Overstress
	The allowable overstresses as specified in AASHTO Article 3.22 shall be used where applicable for Service Loads design method.
(*)	$E_c=57,000\sqrt{f^{\prime }c}W$ = 145 pcf., $Ec=60,625\sqrt{f^{\prime }c}$ for $W$ = 150 pcf.

751.40.8.13.1.2 Loads

(1)	Dead Loads
(2)	Live Load
	As specified on the Bridge Memorandum
	Impact of 30% is to be used for design of the beam. No impact is to be used for design of any other portion of bent including the piles.
(3)	Temperature, Wind and Frictional Loads
	Wind and temperature forces can be calculated based on longitudinal force distribution.

751.40.8.13.1.3 Distribution of Loads

(1)	Dead Loads
	Loads from stringers, girders, etc. shall be concentrated loads applied at the intersection of centerline of stringer and centerline of bearing.
(2)	Live Load
	Loads from stringers, girders, etc. shall be applied as concentrated loads at the intersection of centerline of stringer and centerline of earing.
(3)	Temperature
	The force due to expansion or contraction applied at bearing pads are not used for stability or pile bearing computations. However, the movement due to temperature should be considered in the bearing pad design and expansion device design.
(4)	Wing with Detached Wing Wall

SECTION A-A

DETAIL B

(*)	Detached wing wall shown is for illustration purpose only. Design detached wing wall as a retaining wall.
(**)	See retaining wall design.

751.40.8.13.1.4 Design Assumptions - Loadings

1)	Piles
	a.	Stresses in the piles due to bending need not be considered in design calculations except for seismic design in categories B, C and D.
	b.	The following four loading cases should be considered.
		Case Vertical Loads Horizontal Loads Special Consideration I DL + E + SUR EP + SUR - II DL + LL + E + SUR EP + SUR - III DL + LL + E EP - IV DL + LL + E None Allow 25% Overstress
		Where,
		LL	= live load
		DL	= dead load of superstructure, substructure and one half of the apporach slab
		SUR	= two feet of live load surcharge
		E	= dead load of earth fill
		EP	= equivalent fluid pressure of earth
		Maximum pile pressure = pile capacity
		Minimum pile pressure = 0 (tension on a pile will not be allowed for any combination of forces exept as noted)
2)	Analysis Procedure
	a.	Find the lateral stiffness of a pile, $K_{\delta }$:
		With fixed pile-head (i.e., only translation movement is allowed but no rotation allowed): The lateral stiffness of a pile can be estimated using Figures 1 and 3 or 2 and 3 for pile in cohesionless or cohesive soil, respectively. The method of using Figures 1, 2, and 3 to find lateral stiffness is called Linear Subgrade Modulus Method. Usually the significant soil-pile interaction zone for pile subjected to lateral movement is confined to a depth at the upper 5 to 10 pile diameters. Therefore, simplified single layer stiffness chart shown in Figure 3 is appropriate for lateral loading. The coefficient $f$ in Figures 1 and 2 is used to define the subgrade modulus $E_s$ at depth “z” representing the soil stiffness per unit pile length. For the purpose of selecting an appropriate $f$ value, the soil condition at the upper 5 pile diameters should be used. Since soil property, friction angle $\varphi$, or cohesion c, is needed when Figure 1 or 2 is used, determine soil properties based on available soil boring data. If soil boring data is not available, one can conservatively use $f$ value of 0.1 in Figure 3. Designer may also use soil properties to convert SPT N value to friction angle $\varphi$, or cohesion c, for granular or cohesive soil, respectively. Figures 1 and 2 were based on test data for smaller-diameter (12 inches) piles, but can be used for piles up to about 24 inches in diameter. In Figure 2, the solid line (by Lam et al. 1991) shall be used in design.
	b.	Find the axial stiffness of a pile, $K_a$:
		For friction pile, $K_a$ may be determined based on a secant stiffness approach as described in EPG 751.9 Bridge Seismic Design or by the in-house computer program “SPREAD” where $K_a$ is calculated as:
		$\frac{1}{K_a}=\frac{1}{AE/L^{\prime }}+\frac{1}{K_{Q_f}}+\frac{1}{K_{Q_b}}$   Equation (1)
		Where:
		$A$	= cross sectional area of pile
		$E$	= elastic modulus of pile
		$L^{\prime }$	= total length of pile
		$K_{Q_f}$	= secant stiffness due to ultimate friction capacity of the pile as described in EPG 751.9.2.6.3 Pile Axial Stiffness
		$K_{Q_f}$	= secant stiffness due to ultimate bearing capacity of the pile as described in EPG 751.9.2.6.3 Pile Axial Stiffness
	For HP bearing pile on rock $K_a$ shall be calculated as:
		$\frac{1}{K_a}=\frac{1}{AE/L^{\prime }}+\frac{1}{K_{Q_f}}$   Equation (2)
		Or Conservatively, $K_a$ may be determined as:
		$K_a=\frac{AE}{L^{\prime }}$   Equation (3)
Recommended Coefficient $f$ of Variation in Subgrade Modulus with Depth for Sand
Recommended Coefficient $f$ of Variation in Subgrade Modulus with Depth for Clay
PILE HEAD AT GRADE LEVEL   EMBEDDED PILE HEAD
Lateral Embedded Pile-Head Stiffness
	c.	Find the equivalent cantilever pile length, $L$
		For the structural model used in the structural analyses of loading cases I through IV. As shown in figure below, length L can be calculated as:
		$L=(\frac{12EI}{K_{\delta }}{)}^{1/3}$   Equation (4)
Structural Model
	d.	Find the equivalent pile area, $A_e$ :
		Once the equivalent cantilever pile length has been determined from step (c) above, the equivalent axial rigidity of the pile, $A_e\times E_e$ , can be calculated as $A_e\times E=K_{a}L$. Then, the equivalent pile area, $A_e$ , is equal to
		$A_e=\frac{K_{a}L}{E}$   Equation (5)
	e.	Perform structural analyses for loading cases I through IV.
		Use computer programs STRUCT3D, SAP2000 or any other program capable of running static analysis.
	f.	Check abutment movement at the top of backwall and at the bottom of beam cap
		Maximum movement away from the backfill shall not be greater than 1/8". Maximum movement toward the backfill shall not be greater than 1/4".
	g.	Check pile axial loads from the analysis with the allowable pile axial load capacity.
	h.	Check overturning of bent
		Conservatively, use the same equivalent cantilever pile length, $L$. Check overturning of bent at the bottom of toe pile for loading cases I and II(Figure of Structural model).
		Case I Point of Investigation Vertical Loads Horizontal Loads Factor of Safety (**) I Toe Pile DL + E EP + SUR 1.2 II Toe Pile DL + LL + E EP + SUR 1.5
5)	Deadman Anchorage System
	Deadman anchorage can be used when the abutment movement exceeds the allowable movement.
	The size and location of deadman anchorage shall be designed appropriately to maintain the stability of the abutment.
	The deadman forces may be used to resist overturning with the approval of the Structural Project Manager.
6)	Passive Pressure Shear Key (if applicable)
	Passive pressure shear key may be used when the abutment movement exceeds the allowable movement.
	The passive resistance of soil to the lateral force at shear keys may be used with the approval of structural project manager.

751.40.8.13.1.5 Deadman Anchors

Design Assumptions

	Length of Deadman = $(F_E+F_S/(P_P-P_A)$
	Number of tie rods required = $(F_E+F_S)/F_R$
	$P_A$ = Active earth pressure on deadman, in lb./ft. = (120 pcf) $K_{A}hT$
(**)	$P_P$ = Passive earth pressure on deadman, in lb./ft. = (120 pcf) $K_{P}hT$
	$F_E$ = Earth pressure on end bent, in lb. = 0.5(120 pcf)$K_A{H}^2$ (length of beam)
	$F_S$ = Surcharge on end bent, in lb. = $(120pcf)(2^{\prime })K_{A}H(lengthofbeam)$
	$K_A=Ta{n}^2(45^{\circ }-\varphi /2)$
	$K_P=Ta{n}^2(45^{\circ }-\varphi /2)$
(***)	$F_R$ = 8.0 kips for 7/8" Ø tie rod and 10.50 kips for 1" Ø tie rods (Capacity of the tie rods based on a maximum skew of 30°.)
*	If the number of 7/8" Ø tie rods causes too long of a deadman, then try 1" Ø tie rods.
**	For seismic loads only, use $P_P$ = 4 kips/sq.ft. as the ultimate capacity of compacted fill.
***	For seismic loads only, the allowable stress in the tie rod may be taken as the yield stress of the rod.

Notes:

No more than 20% of deadman may fall outside of the roadway shoulders. To prevent more than 20% limit, using a deeper deadman to reduce its length. If this is not possible, the total passive pressure resistance should be calculated by summing the resistance from the different fill depths.

When deadman anchors are to be used, design the piles for a factor of safety of 1.0 for sliding and design deadman anchors to resist all horizontal earth forces with a factor of safety of 1.0. This will result in a factor of safety for sliding of 2.0. For special cases, see the Structural Project Manager.

Design Example

Assume:
	Roadway width = 36', Out-Out slab width = 36' + 2 x 16" = 38.67'
	Skew = $15^{\circ }$, Length of Beam = $(38.67^{\prime })/(Cos15^{\circ })=40.03^{\prime }$
	Beam depth = $3^{\prime }-0^{\prime \prime }$, $\varphi =27^{\circ }$, $H=8.20^{\prime }$
	$\frac{H}{3}=\frac{8.20^{\prime }}{3}=2.73^{\prime }$
	$3^{\prime }-2.73^{\prime }=0.27^{\prime }<9^{\prime \prime }$, use $9^{\prime \prime }$
	$h=H-(Beamdepth)+9^{\prime \prime }=8.20^{\prime }-3^{\prime }+0.75=5.95^{\prime }$
	Assume $T=2^{\prime }-0^{\prime \prime }$ (Deadman anchor depth)

Determine Earth and Surcharge Forces
	$K_A$	=	$Ta{n}^2(45^{\circ }-\varnothing /2)=Ta{n}^2(45^{\circ }-27^{\circ }/2)=0.3755$
	$K_P$	=	$Ta{n}^2(45^{\circ }-\varnothing /2)=Ta{n}^2(45^{\circ }-27^{\circ }/2)=2.6629$
	$F_e$	=	$\frac{1}{2}(120K_A{H}^2)(LengthofBeam)$
		=	$(60lb./cu.ft.)(0.3755)(8.20^{\prime }{)}^2(40.03^{\prime })$
		=	$60,842lbs.$
	$F_s$	=	$(2^{\prime })(120K_{A}H)$$(LengthofBeam)$
		=	$(240lb./cu.ft.)(0.3755)(8.20^{\prime })(40.03^{\prime })$
		=	$297,582lbs.$
	$P_A$	=	$120K_{A}hT$
		=	$(120lb./cu.ft.)(0.3755)(5.95^{\prime })(2.0^{\prime })$
		=	$536lbs.perfootofDeadman$
	$P_P$	=	$120K_{P}hT$
		=	$(120lb./cu.ft.)(2.6629)(5.95^{\prime })(2.0^{\prime })$
		=	$3,803lbs.perfootofDeadman$

Determine number of Tie Rods required
	Try 7/8"Ø Rods: $F_R=8.0$ kips
	Number of Rods required = $(F_E+F_S)/F_R=(60,642+29,582)/8,000=11.29$
	Use 12-7/8"Ø Rie Rods.

Determine length of Deadman
	Length of Deadman required = $(F_E+F_S)/(P_P-P_A=(60,642+29,582)lbs./(3,803-536)lb/ft.=27.62^{\prime }$
	Tie Rod spacing = $(27.62^{\prime }-2.0^{\prime })/11=2.33^{\prime }say{2}^{\prime }-4^{\prime \prime }>12^{\prime \prime }$ minimum, ok.
	Length of Deadman provided = $(2^{\prime }-4^{\prime \prime })(11)+2.0^{\prime }=27^{\prime }-8^{\prime \prime }$

$\varphi =27^{\circ }$

$45^{\circ }-\frac{\varphi }{2}=31.5^{\circ }$

1)	Check tie rod skew angle at Fill Face of End Bent
	$(5.5spacing)(30.5^{\prime \prime }-28^{\prime \prime })=13.75^{\prime \prime },tan$	$\varphi =13.75^{\prime \prime }/(24.33\times 12^{\prime \prime })=0.471$
		$\varphi =2.70^{\circ }<30^{\circ }$, tie capacity ok.
2)	Check criteria for Deadman Anchors extending into Fill Slope

A)	Extension of Deadman into Fill Slope
	Length of Deadman extending into Fill Slope = $1.08^{\prime }tan15^{\circ }+$
		$(13.83^{\prime }-((15.04^{\prime }+3.87^{\prime })-24.33^{\prime }tan15^{\circ }))=1.73^{\prime }$
		0.2 (Length of Deadman) = $0.2(27.67^{\prime })=5.53^{\prime }$
			$1.73^{\prime }<5.53^{\prime }$
	Length of Deadman extending into Fill Slope $<0.2$ (Length of Deadman), ok
Note: See below for Section A-A details.
B)	Cover of Deadman in Fill Slope
	$1.44^{\prime }\times (cos15^{\circ })=1.39^{\prime }$

SECTION A-A
DETAIL AT FILL SLOPE

Note:

(*) Fill slope shown is for illustration purpose only, see roadway plans.

751.40.8.13.2 Reinforcement

751.40.8.13.2.1 Wide Flange Beams, Plate Girders and Prestressed Girders

END BENT WITH EXPANSION DEVICE

SECTION A-A
	PART ELEVATION

Notes:

(1) See details for reinforcement of end bent backwall.

(2) #6-H bars and #4-H bars in backwall of skewed bridges shall be bent in field if required.

(3) Center #5 bars in backwall.

Epoxy coat all reinforcing in end bents with expansion devices. See ______ for details of protective coating and sloping top of beam to drain.

	DETAIL OF #5 BARS SHAPE 19
PART PLAN B-B

END BENT WITHOUT EXPANSION DEVICE

SECTION A-A
	PART ELEVATION

(1) #5 Dowel bars are 2'-6" long and placed parallel to centerline roadway. (2) #6-H bars and #4-H bars in backwall of skewed bridges shall be bent in field. (3) For skewed bridges with no expansion device place a #4 bar along skew. (4) See details of end bent backwall for reinforcement. (5) Seal joint with joint sealant. See special provisions. Note: See Structural Project Manager before using this detail.

PART PLAN B-B

END BENT WING

|
SECTION A-A
	TYPICAL ELEVATION OF WING

Note: (1) Development length

h (2) (3) 2' or less #4 @ 12" #6 @ 6" Over 2' to 4' #5 @ 6" #7 @ 6" Over 4' to 6' #7 @ 5" #8 @ 5"
	SECTION B-B
PART SECTION THRU BENTS WITH PASSIVE PRESSURE
	HORIZONTAL SECTION THRU WING (K bars not shown for clarity)

END BENT BEAM HEEL

ELEVATION A-A (TYP.)	PART PLAN OF BEAM (SQUARE)
PART PLAN OF BEAM - SKEWS THRU 15° - LEFT ADVANCE SHOWN

	SECTION B-B
PART PLAN OF BEAM - SKEWS OVER 15° - LEFT ADVANCE SHOWN

Note:

Vertical spacing for #7 bars shown in Elevation A-A is typical for all types of end bent beams.

For a long distance between heel pile and bearing beam investigate for use of larger bars; e.g. larger skews where the shear line does not fall within the bearing beam.

Pile Load Not Greater	(1)$*$ Hair-Pin Stirrups	(2) Horizontal Rebar around Heel Pile
		Skew thru 30°	Skew 31° thru 45°	Skew 46° thru 60°	Skew over 60°
140 kips	#6 @ 9"	5-#7	5-#7	5-#8	By Design
194 kips	#6 @ 6"	5-#7	5-#8	By Design	By Design

$*$ Use 21" horizontal leg.

END BENT BACKWALL

PART SECTION THRU BACKWALL AND BEAM

V-BAR SIZE AND SPACING
h (feet)	t (inch)	Fill Face Reinforcement	Front Face Reinforcement
1-6	12	#5 @ 12"	#5 @ 12"
7	12	#5 @ 12"	#5 @ 12"
8	12	#5 @ 12"	#5 @ 12"
9	12	#6 @ 12"	#5 @ 12"
10	12	#6 @ 10"	#5 @ 12"
11	15	#6 @ 10"	#5 @ 12"
12	15	#6 @ 8"	#5 @ 12"
13	18	#6 @ 8"	#5 @ 12"
14	18	#6 @ 6"	#5 @ 12"

Note:

All reinforcement is grade 60.

Design is based on 45 lbs. per cu. ft. equivalent fluid pressure and 90 lbs. per sq. ft. live load surcharge.

Epoxy coat all reinforcing steel in beam and backwall on non-integral end bents with expansion devices.

751.40.8.14 Concrete Pile Cap Integral End Bents

751.40.8.14.1 Design

751.40.8.14.1.1 Design Unit Stresses

1. Reinforced Concrete
   • Class B Concrete (Substructure)   $f_c$   = 1,200 psi,   $f{\text{'}}_c$   = 3,000 psi
   • Reinforcing Steel (Grade 60)         $f_s$   = 24,000 psi   $f_y$   = 60,000 psi
   • $n$   = 10
   • $E_c$   =$w^{1.5}\times 33\sqrt{f{\text{'}}_c}$   (AASHTO Article 8.7.1)(*)
2. Structural Steel
   • Structural Carbon Steel (ASTM A709 Grade 36)   $f_s$   = 20,000 psi   $f_y$   = 36,000 psi
3. Piling
   • See the Bridge Memorandum if pile capacity is indicated.
4. Overstress
   • The allowable overstresses as specified in AASHTO Article 3.22 shall be used where applicable for Service Loads design method.

(*)   $E_c=57,000\sqrt{f{\text{'}}_c}forW=145pcf,E_c=60,625\sqrt{f{\text{'}}_c}forW=150pcf$

751.40.8.14.1.2 Loads

1. Dead Loads
2. Live Load
   • As specified on the Bridge Memorandum.
   • Impact of 30% is to be used for design of the beam. No impact is to be used for design of any other portion of bent including the piles.
3. Temperature, Wind and Frictional Loads

751.40.8.14.1.3 Distribution of Loads

1. Dead Loads
   • Loads from stringers, girders, etc. shall be concentrated loads applied at the intersection of centerline of stringer and centerline of bearing. Loads from concrete slab spans shall be applied as uniformly, distributed loads along the centerline of bearing.
2. Live Load
   • Loads from stringers, girders, etc. shall be applied as concentrated loads at the intersection of centerline of stringer and centerline of bearing. For concrete slab spans distribute two wheel lines over 10'-0" (normal to centerline of roadway) of substructure beam. This distribution shall be positioned on the beam on the same basis as used for wheel lines in Traffic Lanes for Substructure Design.
3. Wing with Detached Wing Wall
   • When wing length, L, is greater than 17 feet, use maximum length of 10 feet rectangular wing wall combined with a detached wing wall. When detached wing walls are used, no portion of the bridge live load shall be assumed distributed to the detached wing walls. Design detached wing wall as a retaining wall. (The weight of barrier or railing on top of the wall shall be included in Dead Load.)

751.40.8.14.1.4 Design Examples

Design H-bar and F-bar of an intermediate wing as shown in the figures below (wing length = 12.5', wing thickness = 24", wing height = 8'-4"), a Seismic Force of   $\omega$ = 12.21 kips/ft. is applied on the wall.

Section Near Intermediate Wing	Section B-B
	Interior Wing Design
Section C-C

Solve: Assume #6 V bar, #8 H bar, #6 F bar

1.)	Design H-bar for bending
	$d=24in.-2in.(clr.)-0.75in.(VBar)-0.5\times 1in.(Hbar)=20.75in.$.
	$\ell =11ft.,$   $\omega =12.21kips/ft.,$   $b=8ft.4in.=100in.$
	At Section A-A:
	$Mu=(1.0)(\omega {\ell }^2/2)=12.21\times 11^2/2=738.705kip-ft.$
	$Ru=Mu/(\varphi b{d}^2)=738.705\times 12,000/(0.9\times 100in.\times (20.75{)}^2)=228.85psi$
	Use $f-c=3kisi,$   $fy=60ksi$
	$m=fy/(0.85f^{\prime }c)=60/(0.85\times 3)=23.53$
	$\rho =(1/m)[1-\sqrt{1-2Rum/fy}]=(1-\sqrt{1-2\times 228.85\times 23.53/60000})/23.53=0.004003$
	As (Req'd) = $\rho bd=0.004003\times 100in.\times 20.75in.=8.31sq.in.$
	Try No. 8 @ 9", USE   $\frac{100in-3in.(clr.)-2in.(clr.)-1in(No.8bar)}{9in}=10.44spacing$
	Say 11 spacings, 12 bars (Each Face)
	Total Area = $12(0.7854)=9.42sq.in.>8.31sq.in.,$   USE 12-No. 8 H-bar (Each Face)

2.)	Design F-bar for shear
	$Vu\le \varphi (Vc=Vs),\varphi =0.85$   (AASHTO Article 8.16.6.1.1)
	At Section A-A:
	$Vu=1.0\times (\omega \ell )=(12.21kips/ft.)(11ft.)=134.11kips$
	$Vc=bd(\vartheta c)=bd(2\sqrt{f^{\prime }c}=(100in.\times 20.75in.)(2\times \sqrt{3000})/1000=227.30kips$
	$\varphi Vc=0.85Vc=0.85\times 227.30kips=193.20kips$
	$\varphi Vc=193.20kips>Vu=134.11kips,$   No   $Vs$   needed by AASHTO Article 8.16.6.3.1.
	Minimum shear reinforcement is required by AASHTO Article 8.19.1.1(a).(ACI 318-95 11.5.5.1)
	F-bar is a single group of parallel bars, all bent up at the same distance from support (no "spacing" along the "L" direction of the wing).
	Try No. 6 @ 12" F-bar (each face).
	Try $(100in.-3in.-2in.-1in.)/12in=7.83,$ say 8 spacing, 9 bars (each face).
	Since seismic force is a cyclic loading, assume one bar works at any instance.
	$Av(provided)=1\times 9\times (0.4418sq.in.)=3.98sq.in.$
	$Vs=Av(FySin45^{\circ })=(3.98sq.in.)(60ksi)(Sin45^{\circ })=168.7kips$
	Check   $3\sqrt{f^{\prime }c}{b}_{\omega }d=3\sqrt{3000}\times 100in.\times 20.75in./1000=341.0kips$
	$Vs=Av(fySin45^{\circ })\le 3\sqrt{f^{\prime }c}{b}_{\omega }d,$   O.K. by AASHTO Article 8.16.6.3.4.
	USE 9 No. 6 F-bars (each face).

751.40.8.14.2 Reinforcement

751.40.8.14.2.1 Earthquake Loads at End Bent – Intermediate Wing (Seismic Shear Wall)

Section Near Intermediate Wing	Section B-B
Section A-A

*	Use 1.25 x development length for seismic design.
**	Additional reinforcing steel by design if required.
Note:	Make sure reinforcement does not interfere with girders.

751.40.8.15 Cast-In-Place Concrete Retaining Walls

751.40.8.15.1 Loads

Dead Loads

Dead loads shall be determined from the unit weights in EPG 751.2.1.1 Dead Load.

Equivalent Fluid Pressure (Earth Pressures)

Additional Information

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.

Rankine Formula: $P_a=\frac{1}{2}{C}_a{\gamma }_s{H}^2$ where:

C_a = $\cos \delta \left[\frac{\cos \delta -\sqrt{{\cos }^2\delta -{\cos }^2\varphi }}{\cos \delta +\sqrt{{\cos }^2\delta -{\cos }^2\varphi }}\right]$ = coefficient of active earth pressure

P_a = equivalent active earth pressure

H = height of the soil face at the vertical plane of interest

${{\gamma }}_{\mathit{s}}$ = unit weight of soil

${\delta }$= slope of fill in degrees

${\varphi }$ = angle of internal friction of soil in degrees

Example

Given:

δ = 3:1 (H:V) slope

ϕ = 25°

γ_s = 0.120 kcf

H = 10 ft

δ = arctan$\left[\frac{1}{3}\right]$ = 18.4°

C_a = $\cos (18.4^{\circ })\left[\frac{\cos (18.4^{\circ })-\sqrt{{\cos }^2(18.4^{\circ })-{\cos }^2(25^{\circ })}}{\cos (18.4^{\circ })+\sqrt{{\cos }^2(18.4^{\circ })-{\cos }^2(25^{\circ })}}\right]$ = 0.515

P_a = (1/2)(0.515)(0.120 kips/ft³)(10 ft)² = 3.090 kips per foot of wall length

The ϕ angle shall be determined by the Materials Division from soil tests. If the ϕ angle cannot be provided by the Construction and Materials Division a ϕ angle of 27 degrees shall be used.

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.

Surcharge Due to Point, Line and Strip Loads

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.

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.

Retaining Wall in front of an Abutment

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.

The following variables are shown in the figure below:

β = slope of the active failure plane in degrees

δ = slope of fill in degrees

H = height of the pseudo-wall (fom the bottom of the footing).

L₁ = distance from back of stem to back of footing heel

L₂ = distance from footing heel to intersection of failure plane with ground surface

Determination of Active Wedges

In order to determine β, the following equation which has been derived from Rankine's active earth pressure theory must be solved by iteration:

$\tan (-\beta )+\frac{1}{\tan (\beta -\varphi )}-\frac{1}{\tan (\beta -\delta )}+\frac{1}{\tan (90^{\circ }+\varphi +\delta -\beta )}=0$

ϕ = angle of internal friction of soil in degrees

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 L₁ is needed to determine L₂. From the geometry of the variables shown in the above figure:

$L_2=H\frac{\cos \delta \cos \beta }{\sin (\beta -\delta )}$

The resultant pressure due to the strip load surcharge and its location are then determined. The following variables are shown in the figure below:

q = load per unit area

P_s = resultant pressure on wall due only to surcharge earth pressure

$\overline{z}$ = location of P_s measured from the bottom of the footing

L₃ = distance from back of stem to where surcharge pressure begins

Surcharge Pressure on Retaining Wall

From the figure:

P_s = $\frac{q}{90}[H({\theta }_2-{\theta }_1)]$ where

${\theta }_1=\arctan \left[\frac{L_3}{H}\right]and{\theta }_2=\arctan \left[\frac{L_2}{H}\right]$

$\overline{z}=\frac{H^2({\theta }_2-{\theta }_1)-(R-Q)+57.03L_{4}H}{2H({\theta }_2-{\theta }_1)}$ where

$R=(L_2{)}^2(90^{\circ }-{\theta }_2)andQ=(L_3{)}^2(90^{\circ }-{\theta }_1)$

When applicable, P_s is applied to the wall in addition to other earth pressures. The wall is then designed as usual.

Live Load Surcharge

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.

Live Load Surcharge

P_LLS = (2 ft.) γ_s C_a H = pressure due to live load surcharge only

γ_s = 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 EPG 751.2.1.1 Dead Load.)

C_a = coefficient of active earth pressure

H = height of the soil face at the vertical plane of interest

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.

Live Load Wheel Lines

Live load wheel lines shall be applied to the footing when the footing is used as a riding or parking surface.

Additional Information

AASHTO 3.24.5.1.1 & 5.5.6.1

Distribute a LL_WL equal to 16 kips as a strip load on the footing in the following manner.

P = LL_WL/E

where E = 0.8X + 3.75

X = distance in ft. from the load to the front face of the wall

Additional Information

AASHTO 3.24.2 & 3.30

Two separate placements of wheel lines shall be considered, one foot from the barrier or wall and one foot from the toe of the footing.

Collision Forces

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.

Section

Distribute the force to the wall in the following manner:

Force per ft of wall = (10 kips)/2L

Profile

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.

Wind and Temperature Forces

These forces shall be disregarded except for special cases, consult the Structural Project Manager.

When walls are longer than 84 ft., an expansion joint shall be provided.

Contraction joint spacing shall not exceed 28 feet.

Seismic Loads

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 AASHTO specifications for the corresponding SPC.

In seismic category B, C and D determine equivalent fluid pressure from Mononobe-Okabe static method.

Additional Information

1992 AASHTO Div. IA Eqns. C6-3 and C6-4

P_AE = equivalent active earth pressure during an earthquake

P_AE = 0.5 γ_sH²(1 - k_v)K_AE where

K_AE = seismic active pressure coefficient

$K_{AE}=\frac{{\cos }^2(\varphi -\theta -\beta )}{\cos \theta {\cos }^{}\beta \cos (\delta +\beta +\theta )\{1+\sqrt{\frac{\sin (\varphi +\delta )\sin (\varphi -\theta -i)}{\cos (\delta +\beta +\theta )\cos (i-\beta )}}{\}}^2}$

γ_s = unit weight of soil

Additional Information

AASHTO 5.2.2.3 & Div. IA 6.4.3

k_v = vertical acceleration coefficient

k_h = horizontal acceleration coefficient which is equal to 0.5A for all walls,

but 1.5A for walls with battered piles where

A = seismic acceleration coefficient

The following variables are shown in the figure below:

ϕ = angle of internal friction of soil

θ = $\arctan \left(\frac{k_h}{1-k_v}\right)$

β = slope of soil face

δ = angle of friction between soil and wall in degrees

i = backfill slope angle in degrees

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.

Active Soil Wedge

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.

Additional Information

AASHTO Table 3.22.1A

AASHTO Group I Load Factors for Load Factor Design of concrete: γ = 1.3

β_D = 1.0 for concrete weight

β_D = 1.0 for flexural member

β_E = 1.3 for lateral earth pressure for retaining walls

β_E = 1.0 for vertical earth pressure

β_LL = 1.67 for live load wheel lines

β_LL = 1.67 for collision forces

Additional Information

AASHTO 5.14.2

β_E = 1.67 for vertical earth pressure resulting from live load surcharge

β_E = 1.3 for horizontal earth pressure resulting from live load surcharge

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.

Additional Information

AASHTO Div. IA 4.7.3

When seismic loads are considered, load factor for all loads = 1.0.

751.40.8.15.3 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 Standard Detailing Notes.

Spread Footing

For foundation material capacity, see Foundation Investigation Geotechnical Report.

751.40.8.15.4 Design

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. 40.8.15.4.1

Additional Information

AASHTO 5.5.5

751.40.8.15.4.1 Spread Footings

Location of Resultant

The resultant of the footing pressure must be within the section of the footing specified in the following table.

When Retaining Wall is Built on:	AASHTO Group Loads I-VI	For Seismic Loads
Soila	Middle 1/3	Middle 1/2 b
Rockc	Middle 1/2	Middle 2/3
a Soil is defined as clay, clay and boulders, cemented gravel, soft shale, etc. with allowable bearing values less than 6 tons/sq. ft.
b MoDOT is more conservative than AASHTO in this requirement.
c 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.

Factor of Safety Against Overturning

Additional Information

AASHTO 5.5.5

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.

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.

For collision forces:

• F.S. for overturning ≥ 1.2.

Factor of Safety Against Sliding

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.

• 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:

• 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

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.

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.

Rankine Formula: $P_p=\frac{1}{2}{C}_p{\gamma }_s[H^2-H_1^2]$ where thefollowing variables are defined in the figure below

C_p = $\tan (45^{\circ }+\frac{\varphi }{2})$

y₁ = $\frac{H_1{y}_2^2+\frac{2}{3}{y}_2^3}{H^2-H_1^2}$

P_p = passive force at shear key in pounds per foot of wall length

C_p = coefficient of passive earth pressure

${{\gamma }}_{\mathit{s}}$ = unit weight of soil

H = height of the front face fill less than 1 ft. min. for erosion

H₁ = H minus depth of shear key

y₁ = location of P_p from bottom of footing

${\varphi }$ = angle of internal friction of soil

Additional Information

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.

Additional Information

MoDOT Materials Division

Frost line is set at 36 in. at the north border of Missouri and at 18 in. at the south border.

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_PE = equivalent passive earth pressure during an earthquake

Additional Information

1992 AASHTO Div. IA Eqns. C6-5 and C6-6

$P_{PE}=\frac{1}{2}{\gamma }_s{H}^2(1-k_v)K_{PE}$ where:

K_PE = seismic passive pressure coefficient

$K_{PE}=\frac{{\cos }^2(\varphi -\theta -\beta )}{\cos \theta {\cos }^{}\beta \cos (\delta +\beta +\theta )[1+\sqrt{\frac{\sin (\varphi +\delta )\sin (\varphi -\theta -i)}{\cos (\delta +\beta +\theta )\cos (i-\beta )}}{]}^2}$

${{\gamma }}_s$ = unit weight of soil

H = height of soil at the location where the earth pressure is to be found

k_V = vertical acceleration coefficient

${\varphi }$ = angle of internal friction of soil

${\theta }=arctan\left[\frac{k_h}{1-k_V}\right]$

k_H = horizontal acceleration coefficient

${\beta }$ = slope of soil face in degrees

i = backfill slope angle in degrees

${\delta }$ = angle of friction between soil and wall

Special Soil Conditions

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 (${\varphi }$) less than 22°, or a soil shear less than 0.8 ksf. Soil shear is determined from a standard penetration test.

Soil Shear $\left(\frac{k}{f{t}^2}\right)=\frac{blowsper12in.}{10}$

Friction

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.

When A Shear Key Is Not Used

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 is not readily available or is inconsistent, the friction factor (f) can be taken as

f =$tan\left(\frac{2\varphi }{3}\right)$ where ${\varphi }$ is the angle of internal friction of the soil (Civil Engineering Reference Manual by Michael R. Lindeburg, 6th ed., 1992).

Coefficient of Friction Values for Soil Against Concrete
Soil Typea	Coefficient of Friction
coarse-grained soil without silt	0.55
coarse-grained soil with silt	0.45
silt (only)	0.35
clay	0.30b
a It is not necessary to check rock or shale for sliding due to embedment.
b Caution should be used with soils with ${\varphi }$ < 22° or soil shear < 0.8 k/sq.ft. (soft clay soils). Removal and replacement of such soil with suitable material should be considered.

When A Shear Key Is Used

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.

Bearing Pressure

Additional Information

AASHTO 4.4.7.1.2 & 4.4.8.1.3

Group Loads I - VI

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 bearing pressure on the supporting soil shall not be greater than the allowable bearing pressure given on the "Design Layout Sheet".

Seismic Loads

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".

Stem Design

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

Footing Design

Additional Information

AASHTO 5.5.6.1

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.

Heel

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.

Shear Key Design

The shear key shall be designed as a cantilever supported at the bottom of the footing.

751.40.8.15.4.2 Pile Footings

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.

Pile Arrangement

For retaining walls subject to moderate horizontal loads (walls 15 to 20 ft. tall), the following layout is suggested.

Section

Plan

For higher walls and more extreme conditions of loading, it may be necessary to:

• use the same number of piles along all rows

• use three rows of piles

• provide batter piles in more than one row

Loading Combinations for Stability and Bearing

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

DL = dead load weight of the wall elements

SUR = two feet of live load surcharge

E = earth weight

EP = equivalent fluid earth pressure

COL = collision force

EQ = earthquake inertial force of failure wedge

Loading Case	Vertical Loads	Horizontal Loads	Overturning Factor of Safety	Sliding Factor of Safety
				Battered Toe Piles	Vertical Toe Piles
Ia	DL+SUR+E	EP+SUR	1.5	1.5	2.0
II	DL+SUR+E	EP+SUR+COL	1.2	1.2	1.2
III	DL+E	EP	1.5	1.5	2.0
IVb	DL+E	None	-	-	-
Vc	DL+E	EP+EQ	1.125	1.125	1.5
a Load Case I should be checked with and without the vertical surcharge.
b A 25% overstress is allowed on the heel pile in Load Case IV.
c 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

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.

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.

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.

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.

Neutral Axis of Pile Group

Locate the neutral axis of the pile group in the repetitive strip from the toe of the footing at the bottom of the footing.

Moment of Inertia of Pile Group

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_A) + Σ(Ad²) where :

I_A = moment of inertia of a pile about its neutral axis

A = area of a pile

d = distance from a pile's neutral axis to pile group's neutral axis

I_A may be neglected so the equation reduces to:

I = Σ(Ad²)

Resistance To Sliding

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.

Friction or Bearing Piles With Batter (Case 1)

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.

Friction or Bearing Piles Without Batter (Case 2)

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.

Concrete Pedestal Piles or Drilled Shafts (Case 3)

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.

Resistance Due to Passive Pressure Against Pile

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.

F = the lateral force due to passive pressure on pile

$F=\frac{1}{2}{\gamma }_s{C}_P{H}^{2}B$ , where: $C_P=ta{n}^2[45+\frac{\varphi }{2}]$

${{\gamma }}_{\mathit{s}}$ = unit weight of soil

H = depth of pile considered for lateral resistance (H_max= 6B)

C_P = coefficient of active earth pressure

B = width of pile

${\varphi }$ = angle of internal friction of soil

Resistance Due to Pile Batter

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.)

b = the amount of batter per 12 inches.

$c=\sqrt{(12in.{)}^2+b^2}$

$P_{HBatter}=P_T\left(\frac{b}{c}\right)$ (# of battered piles) where:

P_HBatter = the horizontal force due to the battered piles

P_T = the allowable pile load

Maximum batter is 4" per 12".

Resistance Due to Shear Keys

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.

Resistance to 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.

Pile Properties

Location of Resultant

The location of the resultant shall be evaluated at the bottom of the footing and can be determined by the equation below:

$e=\frac{\mathrm{\Sigma }M}{\mathrm{\Sigma }V}$ where:

e = the distance between the resultant and the neutral axis of the pile group

ΣM = the sum of the moments taken about the neutral axis of the pile group at the bottom of the footing

ΣV = the sum of the vertical loads used in calculating the moment

Pile Loads

The loads on the pile can be determined as follows:

$P=\frac{\mathrm{\Sigma }V}{A}\pm \frac{Mc}{I}$ where:

P = the force on the pile

A = the areas of all the piles being considered

M = the moment of the resultant about the neutral axis

c = distance from the neutral axis to the centerline of the pile being investigated

I = the moment of inertia of the pile group

Additional Information

AASHTO 5.5.6.2

Stem Design

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

Footing Design

Toe

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.

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.

Shear Key Design

The shear key shall be designed as a cantilever supported at the bottom of the footing.

751.40.8.15.4.3 Counterfort Walls

Assumptions:

(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

$P=C_a{{\gamma }}_s$

where:

C_a = coefficient of active earth pressure

${{\gamma }}_s$ = unit weigt 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.

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.

Maximum concrete stress shall be assumed as the greater of the two thus obtained.

The application of these horizontal pressures shall be as follows:

Counterfort Wall Section 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.

(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.

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.

(4) Footing

The footing of the counterfort walls shall be designed as a continuous beam of spans equal to the distance between the counterforts.

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.

Divide footing (transversely) into four (4) equal sections for design footing pressures.

Counterfort walls on pile are very rare and are to be treated as special cases. See Structural Project Manager.

(5) Sign-Board type walls

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.

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:

Wall

Footing

The individual footings shall be designed transversely as cantilevers supported by the wall. Refer to AASHTO Section 5.

751.40.8.15.5 Example 1: Spread Footing Cantilever Wall

Typical Section thru Wall(Spread Footing)

f'_c = 3,000 psi

f_y = 60,000 psi

φ = 24 in.

γ_s = 120 pcf (unit wgt of soil)

Allowable soil pressure = 2 tsf

γ_c = 150 pcf (unit wgt of concr.)

Retaining wall is located in Seismic Performance Category (SPC) B.

A = 0.1 (A = seismic acceleration coefficient)

$P_a=\frac{1}{2}{\gamma }_s{C}_a{H}^2$

$P_p=\frac{1}{2}{\gamma }_s{C}_p{H}_2^2-H_1^2$

Assumptions

• 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.

• Design is for a unit length (1 ft.) of wall.

• Sum moments about the toe at the bottom of the footing for overturning.

• 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.

• 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.

• Base of footing is below the frost line.

• 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.

• Soil cohesion along failure plane is neglected.

• 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.

• Neglect soil weight above toe of footing in design of the toe.

• The wall is designed as a cantilever supported by the footing.

• Load factors for AASHTO Groups I - VI for design of concrete:

• γ = 1.3.
• β_E = 1.3 for horizontal earth pressure on retaining walls.
• β_E = 1.0 for vertical earth pressure.

• Load factor for earthquake loads = 1.0.

Lateral Pressures Without Earthquake

C_a = $\cos \delta \left[\frac{\cos \delta -\sqrt{{\cos }^2\delta -{\cos }^2\varphi }}{\cos \delta +\sqrt{{\cos }^2\delta -{\cos }^2\varphi }}\right]$

C_a = $\cos 18.435^{\circ }\left[\frac{\cos 18.435^{\circ }-\sqrt{{\cos }^{2}18.435^{\circ }-{\cos }^{2}24^{\circ }}}{\cos 18.435^{\circ }+\sqrt{{\cos }^{2}18.435^{\circ }-{\cos }^{2}24^{\circ }}}\right]$ = 0.546

$C_p=ta{n}^2(45^{\circ }+\frac{\varphi }{2})=ta{n}^2(45^{\circ }+\frac{24^{\circ }}{2})=2.371$

$P_A=\frac{1}{2}[0.120\frac{k}{f{t}^3}](1ft)(0.546)(10.667ft{)}^2=3.726k$

$P_P=\frac{1}{2}[0.120\frac{k}{f{t}^3}](1ft)(2.371)[(5.0{)}^2-(2.5{)}^2]=2.668k$

$P_{AV}=P_A(sin\delta )=3.726k(sin18.435^{\circ })=1.178k$

$P_{AH}=P_A(cos\delta )=3.726k(cos18.435^{\circ })=3.534k$

Load	Area (ft2)	Force (k) = (Unit Wgt.)(Area)	Arm (ft.)	Moment (ft-k)
(1)	(0.5)(6.667ft)(2.222ft) = 7.407	0.889	7.278	6.469
(2)	(6.667ft)(6.944ft) = 46.296	5.556	6.167	34.259
(3)	(0.833ft)(8.000ft) + (0.5)(0.083ft)(8.000ft) = 7.000	1.050	2.396	2.515
(4)	(1.500ft)(9.500ft) = 14.250	2.138	4.750	10.153
(5)	(2.500ft)(1.000ft) = 2.500	0.375	2.500	0.938
(6)	(1.000ft)(1.917ft)+(0.5)(0.010ft)(1.000ft) = 1.922	0.231	0.961	0.222
Σ	-	ΣV = 10.239	-	ΣMR = 54.556
PAV	-	1.178	9.500	11.192
Σ resisting	-	ΣV = 11.417	-	ΣMR = 65.748
PAH	-	3.534	3.556	12.567
PP	-	2.668	1.3891	-
1 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.

$\overline{y}=\frac{H_1{y}^2+\frac{2}{3}{y}^3}{H_2^2-H_1^2}=\frac{(2.5ft)(2.5ft{)}^2+\frac{2}{3}(2.5ft{)}^3}{(5.0ft{)}^2-(2.5ft{)}^2}$ = 1.389 ft.

Overturning

F.S. = $\frac{M_R}{M_{OT}}=\frac{65.748(ft-k)}{12.567(ft-k)}=5.232\ge 2.0$ o.k.

where: M_OT = overturning moment; M_R = resisting moment

Resultant Eccentricity

$\overline{x}=\frac{(65.748-12.567)(ft-k)}{11.417k}$ = 4.658 ft.

$e=\frac{9.500ft}{2}-4.658ft.=0.092ft.$

$\frac{L}{6}=\frac{9.500ft}{6}=1.583ft>e$ o.k.

Sliding

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

F.S. = $\frac{\mathrm{\Sigma }V(tan{\varphi }_{s-c})}{P_{AH}}=\frac{11.042k(tan\frac{2}{3}(24^{\circ })}{3.534k}$= 0.896 no good - shear key req'd

where: φ_s-c = angle of friction between soil and concrete = (2/3)φ_s-s

F.S. = $\frac{P_P+(\mathrm{\Sigma }V)(\frac{L_2}{L_1}tan{\varphi }_{s-s}+\frac{L_3}{L_1}tan{\varphi }_{s-c})}{P_{AH}}$

where: φ_s-s = angle of internal friction of soil

F.S. = $\frac{2.668k+(11.417k)\left[\right(\frac{2ft}{9.50ft})tan24^{\circ }+(\frac{7.50ft}{9.50ft}tan(\frac{2}{3}(24^{\circ }))]}{3.534k}$ = 1.789 ≥ 1.5 o.k.

Footing Pressure

$P=\frac{\mathrm{\Sigma }V}{bL}[1\pm \frac{6e}{L}]$

P_H = pressure at heel $P_H=\frac{11.417k}{(1ft)9.50ft}[1-\frac{6(0.092ft)}{9.50ft}]$ = 1.132 k/ft²

P_T = pressure at toe $P_T=\frac{11.417k}{(1ft)9.50ft}[1+\frac{6(0.092ft)}{9.50ft}]$ = 1.272 k/ft²

Allowable pressure = 2 tons/ft² = 4 k/ft² ≥ 1.272 k/ft² o.k.

Lateral Pressures With Earthquake

k_h = 0.5A = 0.5 (0.1) = 0.05

k_v = 0

$\theta =arctan\left[\frac{k_h}{1-k_v}\right]=arctan\left[\frac{0.05}{1-0}\right]=2.862^{\circ }$

Active Pressure on Psuedo-Wall

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.)

i = 18.435°

β = 0°

$K_{AE}=\frac{co{s}^2(\varphi -\theta -\beta )}{cos\theta co{s}^2\beta cos(\delta +\beta +\theta )(1+\sqrt{\frac{sin(\varphi +\delta )sin(\varphi -\theta -i)}{cos(\delta +\beta +\theta )cos(I-\beta )}}{)}^2}$

$K_{AE}=\frac{co{s}^2(24^{\circ }-2.862^{\circ }-0^{\circ })}{cos(2.862^{\circ })co{s}^2(0^{\circ })cos(24^{\circ }+0^{\circ }+2.862^{\circ })(1+\sqrt{\frac{sin(24^{\circ }+24^{\circ })sin(24^{\circ }-2.862^{\circ }-18.435^{\circ })}{cos(24^{\circ }+0^{\circ }+2.862^{\circ })cos(18.435^{\circ }-0^{\circ })}}{)}^2}$

K_AE = 0.674

P_AE = ½γ_sH²(1 − k_v)K_AE

P_AE = ½[0.120 k/ft³](10.667 ft)²(1 ft.)(1 - 0)(0.674) = 4.602k

P_AEV = P_AE(sinδ) = 4.602k(sin24°) = 1.872k

P_AEH = P_AE(cosδ) = 4.602k(cos 24°) = 4.204k

P'_AH = P_AEH − P_AH = 4.204k − 3.534k = 0.670k

P'_AV = P_AEV − P_AV = 1.872k − 1.178k = 0.694k

where: P'_AH and P'_AV are the seismic components of the active force.

Passive Pressure on Shear Key

δ = φ = 24° (δ = φ because the soil wedge considered is assumed to form in front of the footing.)

i = 0

β = 0

$K_{PE}=\frac{co{s}^2(\varphi -\theta +\beta )}{cos\theta co{s}^2\beta cos(\delta -\beta +\theta )(1-\sqrt{\frac{sin(\varphi -\delta )sin(\varphi -\theta +i)}{cos(\delta -\beta +\theta )cos(I-\beta )}}{)}^2}$

$K_{PE}=\frac{co{s}^2(24^{\circ }-2.862^{\circ }+0^{\circ })}{cos(2.862^{\circ })co{s}^2(0^{\circ })cos(24^{\circ }-0^{\circ }+2.862^{\circ })(1-\sqrt{\frac{sin(24^{\circ }-24^{\circ })sin(24^{\circ }-2.862^{\circ }+0^{\circ })}{cos(24^{\circ }-0^{\circ }+2.862^{\circ })cos(0^{\circ }-0^{\circ })}}{)}^2}$

K_PE = 0.976

P_PE = ½γ_sH²(1 − k_v)K_PE

P_PE = ½[0.120 k/ft³][(5.0 ft)² - (2.5 ft²)](1 ft.)(1 - 0)(0.976) = 1.098k

Load	Force (k)	Arm (ft)	Moment (ft-k)
Σ (1) thru (6)	10.239	-	54.556
PAV	1.178	9.500	11.192
P'AV	0.694	9.500	6.593
Σresisting	ΣV = 12.111	-	ΣMR = 72.341
PAH	3.534	3.556	12.567
P'AH	0.670	6.400a	4.288
PPEV	0.447b	0.000	0.000
PPEH	1.003b	1.389c	0.000
-	-	-	ΣMOT = 16.855
a P'AH acts at 0.6H of the wedge face (1992 AASHTO Div. IA Commentary).
b PPEH and PPEH are the components of PPE with respect to δ (the friction angle). PPE does not contribute to overturning.
c The line of action of PPEH can be located as was done for PP.

Overturning

$F.S{.}_{OT}=\frac{72.341ft-k}{16.855ft-k}=4.292>1.5$ o.k.

Resultant Eccentricity

$\overline{x}=\frac{72.341ft-k-16.855ft-k}{12.111k}=4.581ft.$

$e=\frac{9.5ft.}{2}-4.581ft.=0.169ft.$

$\frac{L}{4}=\frac{9.5ft.}{4}=2.375ft.>e$ o.k.

Sliding

$F.S.=\frac{1.003k+12.111k[(\frac{2}{9.5})tan24^{\circ }+(\frac{7.5}{9.5})tan(\frac{2}{3}(24^{\circ })\left)\right]}{4.204k}=1.161>1.125$ o.k.

Footing Pressure

for e ≤ L/6:

$P=\frac{\mathrm{\Sigma }V}{bL}[1\pm \frac{6e}{L}]$

$P_H=pressureatheel{P}_H=\frac{12.111k}{(1ft.)9.50ft.}[1-\frac{6(0.169ft.)}{9.50ft}]$ = 1.139 k/ft²

$P_{T}H=pressureattoe{P}_T=\frac{12.111k}{(1ft.)9.50ft.}[1+\frac{6(0.169ft.)}{9.50ft}]$ = 1.411 k/ft²

Allowable soil pressure for earthquake = 2 (allowable soil pressure)

(2)[4 k/ft²] = 8 k/ft² > 1.411 k/ft² o.k.

Reinforcement-Stem

d = 11" - 2" - (1/2)(0.5") = 8.75"

b = 12"

f'_c = 3,000 psi

Without Earthquake

P_AH = ½ [0.120 k/ft³](0.546)(6.944 ft.)²(1 ft.)(cos 18.435°) = 1.499k

γ = 1.3

β_E = 1.3 (active lateral earth pressure)

M_u = (1.3)(1.3)(1.499k)(2.315ft) = 5.865 (ft-k)

With Earthquake

k_h = 0.05

k_v = 0

Additional Information

1992 AASHTO Div. IA Commentary

θ = 2.862°

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.

φ = 24°

i = 18.435°

β = 0°

K_AE = 0.654

P_AEH = 1/2 γ_sK_AEH²cosδ

P_AEH = 1/2 [0.120k/ft](0.654)(6.944 ft.)²(1 ft.) cos(12°) = 1.851k

M_u = (1.499k)(2.315 ft.) + (1.851k − 1.499k)(0.6(6.944 ft.)) = 4.936(ft−k)

The moment without earthquake controls:

$R_n=\frac{M_u}{\varphi b{d}^2}=\frac{5.865(ft-k)}{0.9(1ft.)(8.75in.{)}^2}\left(1000\frac{lb}{k}\right)$ = 85.116 psi

ρ = $\frac{0.85f{\text{'}}_c}{f_y}[1-\sqrt{1-\frac{2R_n}{0.85f{\text{'}}_c}}]$

ρ = $\frac{0.85(3.000psi}{60,000psi}[1-\sqrt{1-\frac{2(85.116psi}{0.85(3000psi)}}]$ = 0.00144

Additional Information

AASHTO 8.17.1.1 & 8.15.2.1.1

ρ_min = $1.7[\frac{h}{d}{]}^2\frac{\sqrt{f{\text{'}}_c}}{f_y}=1.7[{\frac{11in.}{8.75in.}}^2\frac{\sqrt{3000psi}}{60,000psi}]$ = 0.00245

Use ρ = 4/3 ρ = 4/3 (0.00144) = 0.00192

A_SReq = ρbd = 0.00192 (12 in.)(8.75 in.) = 0.202 in.²/ft

One #4 bar has A_S = 0.196 in²

$\frac{s}{0.196in{.}^2}=\frac{12in.}{0.202in{.}^2}$

s = 11.64 in.

Use #4's @ 10" cts.

Check Shear

V_u ≥ φ V_n

Without Earthquake

V_u, = (1.3)(1.3)(1.499k) = 2.533k

With Earthquake

V_u = 1.851k

The shear force without earthquake controls.

$\frac{{\nu }_u}{\varphi }=\frac{2.533k}{0.85(12in.)(8.75in.)}(1000lb/k)$ = 28.4 psi

${\nu }_c=2\sqrt{3,000psi}$ = 109.5 psi > 28.4 psi o.k.

Reinforcement-Footing-Heel

Note: Earthquake will not control and will not be checked.

β_E = 1.0 (vertical earth pressure)

d = 18" - 3" - (1/2)(0.750") = 14.625"

b = 12"

f'_c = 3,000 psi

M_u = 1.3 [(5.556k + 1.500k)(3.333ft) + 0.889k(4.444ft) + 1.178k(6.667ft)]

M_u = 45.919(ft−k)

$R_n=\frac{45.919(ft-k)}{0.9(1ft.)(14.625in.{)}^2}(1000\frac{lb}{k})$ = 238.5 psi

ρ = $\frac{0.85(3000)psi}{60,000psi}[1-\sqrt{1-\frac{2(238.5psi)}{0.85(3000psi)}}]$ = 0.00418

ρ_min = $1.7[\frac{18in.}{14.625in.}{]}^2\frac{\sqrt{3000psi}}{60,000psi}$ = 0.00235

A_SReq = 0.00418 (12 in.) (14.625 in.) = 0.734 in²/ft.

Use #6's @ 7" cts.

Check Shear

Shear shall be checked at back face of stem.

V_u = 1.3 (5.556k + 1.500k + 0.889k + 1.178k) = 11.860k

$\frac{{\nu }_u}{\varphi }=\frac{11.860k}{0.85(12in.)(14.625in.)}(1000\frac{lb}{k})=79.5psi<2\sqrt{3,000psi}$ = 109.5 psi o.k.

Reinforcement-Footing-Toe

d = 18" - 4" = 14"

b = 12"

Without Earthquake

Apply Load Factors

load 4 (weight) = 0.431k(1.3)(1.0) = 0.560k

β_E = 1.3 for lateral earth pressure for retaining walls.

β_E = 1.0 for vertical earth pressure.

ΣM_OT = 12.567(ft−k)(1.3)(1.3) = 21.238(ft−k)

ΣM_R = [54.556(ft−k) + 11.192(ft−k)](1.3)(1.0) = 85.472(ft−k)

ΣV = 11.417k(1.3)(1.0) = 14.842k

$\overline{x}=\frac{85.472(ft-k)-21.238(ft-k)}{14.842k}$ = 4.328 ft.

e = (9.5 ft./2) − 4.328 ft. = 0.422 ft.

$P_H=\frac{14.842k}{(1ft.)(9.5ft.)}[1-\frac{6(0.422ft.)}{9.5ft.}]$ = 1.146k/ft²

$P_T=\frac{14.842k}{(1ft.)(9.5ft.)}[1+\frac{6(0.422ft.)}{9.5ft.}]$ = 1.979k/ft²

$P=\left[\frac{1.979\frac{k}{ft.}-1.146\frac{k}{ft.}}{9.5ft.}\right](7.583ft.)+1.146\frac{k}{ft.}$ = 1.811k/ft.

$M_u=1.811\frac{k}{ft.}\frac{(1.917ft.{)}^2}{2}+\frac{1}{2}(1.917ft.{)}^2[1.979\frac{k}{ft.}-1.811\frac{k}{ft.}]\frac{2}{3}-0.560k(0.958ft.)$

M_u = 2.997(ft−k)

With Earthquake

P_H = 1.139 k/ft

P_T = 1.411 k/ft

$P=\left[\frac{1.411\frac{k}{ft.}-1.139\frac{k}{ft.}}{9.5ft.}\right](7.583ft.)+1.139\frac{k}{ft.}$ = 1.356 k/ft

$M_u=1.356\frac{k}{ft.}\frac{(1.917ft.{)}^2}{2}+\frac{1}{2}(1.917ft.{)}^2[1.411\frac{k}{ft.}-1.356\frac{k}{ft.}]\frac{2}{3}-0.431k(0.958ft.)$

M_u = 2.146(ft−k)

The moment without earthquake controls.

$R_n=\frac{2.997(ft-k)}{0.9(1ft.)(14.0in.{)}^2}(1000\frac{lb}{k})$ = 16.990 psi

ρ = $\frac{0.85(3000psi)}{60,000psi}[1-\sqrt{1-\frac{2(16.990psi)}{0.85(3000psi)}}]$ = 0.000284

ρ_min = $1.7[\frac{18in.}{14.0in.}{]}^2\frac{\sqrt{3,000psi}}{60,000psi}$ = 0.00257

Use ρ = 4/3 ρ = $\frac{4}{3}(0.000284)$ = 0.000379

A_SReq = 0.000379 (12 in.)(14.0 in.) = 0.064 in.²/ft.

$\frac{12in.}{0.064in{.}^2}=\frac{s}{0.196in{.}^2}$

s = 36.8 in.

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.

Use # 4's @ 10" cts.

Check Shear

Shear shall be checked at a distance "d" from the face of the stem.

Without Earthquake

$P_d=\left[\frac{1.979\frac{k}{ft.}-1.146\frac{k}{ft.}}{9.5ft.}\right](8.750ft.)+1.146\frac{k}{ft.}$ = 1.913k/ft.

$V_u=\frac{1.979\frac{k}{ft.}+1.913\frac{k}{ft.}}{2}(0.750ft.)-1.3[0.225\frac{k}{ft.}](0.750ft.)$ = 1.240k

With Earthquake

$P_d=\left[\frac{1.411\frac{k}{ft.}-1.139\frac{k}{ft.}}{9.5ft.}\right](8.750ft.)+1.139\frac{k}{ft.}$ = 1390k/ft.

$V_u=\frac{1.411\frac{k}{ft.}+1.139\frac{k}{ft.}}{2}(0.750ft.)-[0.225\frac{k}{ft.}](0.750ft.)$ = 0.788k

Shear without earthquake controls.

$\frac{{\nu }_u}{\varphi }=\frac{1.240k}{0.85(12in.)(14.0in.)}(1000\frac{lb}{k})=8.7psi<2\sqrt{3000psi}$ = 109.5 psi o.k.

Reinforcement-Shear Key

The passive pressure is higher without earthquake loads.

γ = 1.3

β_E = 1.3 (lateral earth pressure)

d = 12"-3"-(1/2)(0.5") = 8.75"

b = 12"

M_u = (3.379k)(1.360 ft.)(1.3)(1.3) = 7.764(ft−k)

$R_n=\frac{7.764(ft-k)}{0.9(1ft.)(8.75in.{)}^2}(1000\frac{lb}{k})$ = 112.677 psi

ρ = $\frac{0.85(3000psi)}{60,000psi}[1-\sqrt{1-\frac{2(112.677psi)}{0.85(3000psi)}}]$ = 0.00192

ρ_min = $1.7[\frac{12in.}{8.75in.}{]}^2\frac{\sqrt{3000psi}}{60,000psi}$ = 0.00292

Use ρ = 4/3 ρ = 4/3 (0.00192) = 0.00256

A_SReq = 0.00256(12 in.)(8.75 in.) = 0.269 in.²/ft.

Use # 4 @ 8.5 in cts.

Check Shear

$\frac{{\nu }_u}{\varphi }=\frac{1.3(3.379k)(1.3)}{0.85(12in.)(8.75.)}(1000\frac{lb}{k})=64.0psi<2\sqrt{3000psi}$ = 109.5 psi o.k.

Reinforcement Summary

751.40.8.15.6 Example 2: L-Shaped Cantilever Wall

Typical Section thru Wall(Spread Footing)

f'_c = 4000 psi

f_y = 60,000 psi

φ = 29°

γ_s = 120 pcf

Allowable soil pressure = 1.5 tsf = 3.0 ksf

Retaining wall is located in Seismic Performance Category (SPC) A.

$\delta =ta{n}^{-1}\frac{1}{2.5}$ = 21.801°

$C_a=cos\delta \left[\frac{cos\delta -\sqrt{co{s}^2\delta -co{s}^2\varphi }}{cos\delta +\sqrt{co{s}^2\delta -co{s}^2\varphi }}\right]$ = 0.462

$C_p=ta{n}^2[45+\frac{\varphi }{2}]$ = 2.882

P_A = 1/2 γ_s C_aH² = 1/2 (0.120 k/ft³)(0.462)(4.958 ft.)² = 0.681k

For sliding, P_P 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.

P_P = 1/2 (0.120 k/ft³)(2.882)[(2.458 ft.)² − (1.500 ft.)²] = 0.656k

Assumptions

• Design is for a unit length (1 ft.) of wall.

• Sum moments about the toe at the bottom of the footing for overturning.

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

• F.S. for sliding ≥ 1.5 for footings on soil.

• Resultant of dead load and earth pressure to be in back half of the middle third of the footing if subjected to frost heave.

• For all loading combinations the resultant must be in the middle third of the footing except for collision loads.

• The top 12 in. of the soil is not neglected in determining the passive pressure because the soil there will be maintained.

• Frost line is set at 18 in. at the south border for Missouri.

• Portions of shear key which are above the frost line are assumed not to resist sliding by passive pressure.

• 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.

• Soil cohesion along the failure plane is neglected.

• Live loads can move to within 1 ft. of the stem face and 1 ft. from the toe.

• The wall is designed as a cantilever supported by the footing.

• 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.

• Load factors for AASHTO Groups I-VI for design of concrete are:

• γ = 1.3.

• β_E = 1.3 for horizontal earth pressure on retaining walls.

• β_E = 1.0 for vertical earth pressure.

• β_LL = 1.67 for live loads and collision loads.

Dead Load and Earth Pressure - Stabilty and Pressure Checks

Dead Load and Earth Pressure - Stabilty and Pressure Checks
Load	Force (k)	Arm (in.)	Moment (ft-k)
(1)	(0.833 ft.)(5.167 ft.)(0.150k/ft3) = 0.646	5.333	3.444
(2)	(0.958ft)(5.750ft)(0.150k/ft3) = 0.827	2.875	2.376
(3)	(1.000ft)(1.500ft)(0.150k/ft3) = 0.22534.259	4.250	0.956
ΣV = 1.698			ΣMR = 6.776
PAV	0.253	5.750	1.455
ΣV = 1.951			ΣMR = 8.231
PAH	0.633	1.653	1.045
PP	0.656	1.061	-
ΣMOT = 1.045
1 The passive pressure at the shear key is ignored in overturning checks.

Overturning

$F.S.=\frac{\mathrm{\Sigma }{M}_R}{\mathrm{\Sigma }{M}_{OT}}=\frac{8.231(ft-k)}{1.045(ft-k)}$ = 7.877 ≥ 2.0 o.k.

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:

$[\frac{5.750ft.}{2}=2.875ft.]\le \overline{x}\le \left[\right(\frac{5.750ft.}{2}+\frac{5.750ft.}{6})=3.833ft.]$

$\overline{x}=\frac{M_{NET}}{\mathrm{\Sigma }V}=\frac{8.231(ft-k)-1.045(ft-k)}{1.951k}$ = 3.683 ft. o.k.

Sliding

$F.S.=\frac{P_P+\mathrm{\Sigma }V\left[\right(\frac{L_2}{L_1})tan{\varphi }_{s-s}+(\frac{L_3}{L_1}\left)tan{\varphi }_{s-c}\right]}{P_{AH}}$

where:

φ_s-s = angle of internal friction of soil

φ_s-c = angle of friction between soil and concrete = (2/3)φ_s-s

$F.S.=\frac{0.656k+(1.951k)\left[\right(\frac{3.75ft.}{5.75ft.})tan29^{\circ }+(\frac{1ft.}{5.75ft.}\left)tan\right(\frac{2}{3}(29^{\circ })\left)\right]}{0.633k}$ = 2.339 ≥ 1.5 o.k.

Footing Pressure

$P=\frac{\mathrm{\Sigma }V}{bL}[1\pm \frac{6e}{L}]$

$e=\overline{x}-\frac{L}{2}=3.683ft.-\frac{5.75ft.}{2}$ = 0.808 ft.

Heel: $P_H=\frac{1.951k}{(1ft.)(5.75ft.)}[1+\frac{6(0.808ft.)}{5.75ft.}]$ = 0.625 ksf < 3.0 ksf o.k.

Toe: $P_T=\frac{1.951k}{(1ft.)(5.75ft.)}[1-\frac{6(0.808ft.)}{5.75ft.}]$ = 0.053 ksf < 3.0 ksf o.k.

Dead Load, Earth Pressure, and Live Load - Stability and Pressure Checks

Stability is not an issue because the live load resists overturning and increases the sliding friction force.

The live load will be distributed as:

$F_{LL}=\frac{L{L}_{WL}}{E}$

where E = 0.8X + 3.75

X = distance in feet from the load to the front face of wall

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.

Live Load 1 ft From Stem Face

Resultant Eccentricity

X = 1 ft.

E = 0.8(1 ft.) + 3.75 = 4.55 ft.

$F_{LL}=\frac{16k}{4.55ft.}(1ft.)$ = 3.516k

$\overline{x}=\frac{M_{NET}}{\mathrm{\Sigma }V}=\frac{8.231(ft-k)+(3.516k)(3.917ft.)-1.045(ft-k)}{1.951k+3.516k}$ = 3.834 ft.

$e=\overline{x}-\frac{L}{2}=3.834ft.-\frac{5.75ft.}{2}=0.959ft.\le \frac{L}{6}$ = 5.75 ft. o.k.

Footing Pressure

$P=\frac{\mathrm{\Sigma }V}{bL}[1\pm \frac{6e}{L}]$

Allowable Pressure = 3.0 ksf

Heel: $P_H=\frac{5.467k}{(1ft.)(5.75ft.)}[1+\frac{6(0.959ft.)}{5.75ft.}]$ = 1.902 ksf

Toe: $P_T=\frac{5.467k}{(1ft.)(5.75ft.)}[1-\frac{6(0.959ft.)}{5.75ft.}]$ = 0.000ksf

Live Load 1 ft From Toe

Resultant Eccentricity

X = 3.917 ft.

E = 0.8(3.917 ft.) + 3.75 = 6.883 ft.

$F_{LL}=\frac{16k}{6.883ft}(1ft.)$ = 2.324k

$x=\frac{8.231(ft-k)+(2.324k)(1ft.)-1.045(ft-k)}{1.951k+2.324k}$ = 2.225 ft.

$e=\frac{L}{2}-\overline{x}=\frac{5.75ft.}{2}-2.225ft.=0.650ft.\le \frac{L}{6}=\frac{5.75ft.}{6}$ = 0.958 ft. o.k.

Footing Pressure

Allowable Pressure = 3.0ksf

Heel: $P_H=\frac{4.275k}{(1ft.)(5.75ft.}[1-\frac{6(0.650ft.)}{5.75ft.}]$ = 0.239ksf o.k.

Toe: $P_T=\frac{4.275k}{(1ft.)(5.75ft.}[1+\frac{6(0.650ft.)}{5.75ft.}]$ = 1.248ksf o.k.

Dead Load, Earth Pressure, Collision Load, and Live Load - Stability and Pressure Checks

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_LL = 3.516k

F_COLL = $\frac{10k}{2(3ft.)}(1ft.)$ = 1.667k

$C_P=cos\delta \left[\frac{cos\delta +\sqrt{co{s}^2\delta -co{s}^2\varphi }}{cos\delta -\sqrt{co{s}^2\delta -co{s}^2\varphi }}\right]$ = 1.867

$P_{PH}=\frac{1}{2}{\gamma }_s{C}_P{H}^{2}cos\delta =\frac{1}{2}(0.120kcf)(1.867)(4.958ft{)}^{2}cos(21.801^{\circ })$

P_PH = 2.556k > F_COLL Thus the soil will develop an equal but opp. force.

Overturning About the Heel

F.S. = $\frac{(0.646k)(0.417ft.)+(0.827k)(2.875ft.)+(0.225k)(1.500ft.)+(3.516k)(1.833ft.)+(1.667k)\left(\frac{4.958ft.}{3}\right)}{(1.667k)(3.958ft.)}$

F.S. = $\frac{12.184(ft-k)}{6.598(ft-k)}$ = 1.847 ≥ 1.2 o.k.

Footing Pressure

$\overline{x}=\frac{12.184(ft-k)-6.598(ft-k)}{1.951k+3.516k}$ = 1.022 ft. from heel

e = $\frac{5.75ft.}{2}-1.022ft.$ = 1.853 ft.

Allowable Pressure = (1.25)(3.0ksf) = 3.75ksf

Heel: $P_H=\frac{2(\mathrm{\Sigma }V)}{3b[\frac{L}{2}-e]}=\frac{2(5.467k)}{3(1ft.)[\frac{5.75ft.}{2}-1.853ft.]}$ = 3.566ksf o.k.

Stem Design-Steel in Rear Face

γ = 1.3

β_E = 1.3 (active lateral earth pressure)

d = 10 in. − 2 in. − (0.5 in./2) = 7.75 in.

$P_{AH}=\frac{1}{2}{\gamma }_s{C}_a{H}^{2}cos\delta =\frac{1}{2}[0.120\frac{k}{f{t}^3}](0.462)(4ft.{)}^2(1ft.)cos21.801^{\circ }$

P_AH = 0.412k

M_u = (1.333 ft.)(0.412k)(1.3)(1.3) = 0.928(ft−k)

$R_n=\frac{M_u}{\varphi b{d}^2}=\frac{0.928(ft-k)}{(0.9)(1ft.)(7.75in.{)}^2}\left(1000\frac{lb}{k}\right)$ = 17.160psi

$\rho =\frac{0.85f_c}{f_y}[1-\sqrt{1-\frac{2R_n}{0.85f_c}}]$

$\rho =\frac{4000psi}{60,000psi}[1-\sqrt{1-\frac{2(17.160psi)}{0.85(4000psi)}}]$ = 0.000287

${\rho }_{min}=1.7[\frac{h}{d}{]}^2\frac{\sqrt{f_c}}{f_y}$

${\rho }_{min}=1.7[\frac{10in.}{7.75in.}{]}^2\frac{\sqrt{4000psi}}{60000psi}$ = 0.00298

Use ρ = (4/3)ρ = (4/3)(0.000287) = 0.000382

$A_{S_{Req}}=\rho bd=0.000382(12in.)(7.75in.)=0.036\frac{i{n}^2}{ft.}$

One #4 bar has A_S = 0.196 in², so the required minimum of one #4 bar every 12 in. controls.

Use #4's @ 12 in. (min)

(These bars are also the bars in the bottom of the footing so the smaller of the two required spacings will be used.)

Check Shear

$\frac{{\nu }_u}{\varphi }\le V_n$

$\frac{{\nu }_u}{\varphi }=\frac{(1.3)(1.3)(0.412k)}{0.85(12in.)(7.75in.)}(1000\frac{lb}{k})$ = 8.8 psi

${\nu }_c=2\sqrt{f{\text{'}}_c}$

${\nu }_c=2\sqrt{4,000psi}$ = 126.5 psi > 8.8 psi o.k.

Stem Design-Steel in Front Face (Collision Loads)

The soil pressure on the back of the stem becomes passive soil pressure during a collision, however this pressure is ignored for reinforcement design.

γ = 1.3

β_LL = 1.67

$d=10in.-1.5in.-0.5in.-\frac{0.5in.}{2}$ = 7.75 in.

$F_{COLL}=\frac{10k}{2L}=\frac{10k}{(2)(3ft.)}$ = 1.667 k/ft.

M_u = 1.667k/ft. (1 ft.)(3 ft.)(1.3)(1.67) = 10.855(ft−k)

$R_n=\frac{10.855(ft-k)}{0.9(1ft.)(7.75in.{)}^2}(1000\frac{lb}{k})$ = 200.809 psi

$\rho =\frac{0.85(4000psi)}{60,000psi}[1-\sqrt{1-\frac{2(200.809psi)}{0.85(4000psi)}}]$ = 0.00345

${\rho }_{min}=1.7[\frac{10in.}{7.75in.}{]}^2\frac{\sqrt{4000psi}}{60,000psi}$ = 0.00298

$A_{S_{Req}}=0.00345(12in.)(7.75in.)=0.321\frac{in{.}^2}{ft.}$

One #4 bar has A_S = 0.196 in².

$\frac{s}{0.196in{.}^2}=\frac{12in.}{0.321in{.}^2}$

s = 7.3 in.

Use #4's @ 7 in.

Check Shear

$\frac{{\nu }_u}{\varphi }=\frac{(1.3)(1.67)(1.667k)}{(0.85)(12in.)(7.75in.)}(1000\frac{lb}{k})$ = 45.8 psi < 126.5 psi o.k.

Footing Design - Bottom Steel

It is not considered necessary to design footing reinforcement based upon a load case which includes collision loads.

Dead Load and Earth Pressure Only

Footing wt. = $[\frac{11.5}{12}ft.](4.917ft.)[0.150\frac{k}{ft{.}^3}](1ft.)$ = 0.707k

β_E = 1.3 (lateral earth pressure)

γ = 1.3

Apply Load Factors:

ΣV = 1.951k (1.3) = 2.536k

ΣM_R = 8.231(ft−k)(1.3) = 10.700(ft−k)

ΣM_OT = 1.045(ft−k)(1.3)(1.3) = 1.766(ft−k)

Footing wt. = 0.707k (1.3) = 0.919k

$\overline{x}=\frac{10.700(ft-k)-1.766(ft-k)}{2.536k}$ = 3.523 ft.

$e=3.523ft.-\frac{5.75ft}{2}$ = 0.648 ft.

$P_H=\frac{2.536k}{(1ft.)(5.75ft.)}[1+\frac{6(0.648ft.)}{5.75ft.}]$ = 0.739 ksf

$P_T=\frac{2.536k}{(1ft.)(5.75ft.)}[1-\frac{6(0.648ft.)}{5.75ft.}]$ = 0.143ksf

$P_W=0.143ksf+[0.739ksf-0.143ksf]\left[\frac{4.917ft.}{5.75ft.}\right]$ = 0.653 ksf

Moment at Wall Face:

$M_W=[0.143\frac{k}{ft.}]\left[\frac{(4.917ft.{)}^2}{2}\right]+\frac{1}{3}(4.917ft.{)}^2[0.653\frac{k}{ft.}-0.143\frac{k}{ft.}]\frac{1}{2}-0.919k[\frac{4.917ft.}{2}]$ = 1.524(ft−k)

Dead Load, Earth Pressure, and Live Load

Live Load 1 ft. From Stem Face

β_E = 1.3 (lateral earth pressure)

β_LL = 1.67

γ = 1.3

Apply Load Factors:

F_LL = 3.516k(1.3)(1.67) = 7.633k

ΣV = 7.633k + 1.951k(1.3) = 10.169k

ΣM_OT = 1.045(ft−k)(1.3)(1.3) = 1.766(ft−k)

ΣM_R = 8.231(ft−k)(1.3) + 3.917 ft.(7.633k) = 40.599(ft−k)

$\overline{x}=\frac{40.599(ft-k)-1.766(ft-k)}{10.169k}$ = 3.819 ft.

e = 3.819 ft. − (5.75 ft./2) = 0.944 ft.

$P_T=\left[\frac{10.169k}{(1ft.)(5.75ft.)}\right]\left[1-\frac{6(0.944ft.)}{5.75ft.}\right]$ = 0.026 ksf

$P_H=\left[\frac{10.169k}{(1ft.)(5.75ft.)}\right]\left[1+\frac{6(0.944ft.)}{5.75ft.}\right]$ = 3.511 ksf

$P_W=0.026ksf+[3.511ksf-0.026ksf]\left[\frac{4.917ft.}{5.75ft.}\right]$ = 3.006 ksf

$P_{LL}=0.026ksf+[3.511ksf-0.026ksf]\left[\frac{3.917ft.}{5.75ft.}\right]$ = 2.400 ksf

Footing wt. from face of wall to toe:

Footing wt. = $1.3[\frac{11.5}{12}ft.](4.917ft.)[0.150\frac{k}{f{t}^3}](1ft.)$ = 0.919k

Footing wt. from LL_WL to toe:

Footing wt. = $1.3[\frac{11.5}{12}ft.](3.917ft.)[0.150\frac{k}{f{t}^3}](1ft.)$ = 0.732k

Moment at Wall Face:

M_W = $0.026\frac{k}{ft}\frac{(4.917ft.{)}^2}{2}-7.633k(1ft.)+\frac{1}{2}[3.006\frac{k}{ft}-0.026\frac{k}{ft}](4.917ft.{)}^2[\frac{1}{3}]-0.919k\frac{(4.917ft.)}{2}$

M_W = 2.430(ft−k)

Moment at LL_WL:

M_LL = $0.026\frac{k}{ft}\frac{(3.917ft.{)}^2}{2}-0.732k\frac{(3.917ft.)}{2}+\frac{1}{2}[2.400\frac{k}{ft}-0.026\frac{k}{ft}](3.917ft.{)}^2[\frac{1}{3}]$ = 4.837(ft−k)

Live Load 1 ft. From Toe

Apply Load Factors:

F_LL = 2.324k(1.3)(1.67) = 5.045k

ΣV = 5.045k + 1.951k(1.3) = 7.581k

ΣM_OT = 1.045(ft−k)(1.3)(1.3) = 1.766(ft−k)

ΣM_R = 8.231(ft−k)(1.3) + 5.045k(1ft.) = 15.745(ft−k)

$\overline{x}=\frac{15.745(ft-k)-1.766(ft-k)}{7.581k}$ = 1.844 ft.

$e=\frac{5.75ft.}{2}-1.844ft.$ = 1.031 ft.

P_H = 0 ksf

$P_T=\frac{2(7.581k)}{3(1ft.)[\frac{5.75ft.}{2}-1.031ft.]}$ = 2.741 ksf

L₁ = 3[(L/2)− e]

L₁ = 3[(5.75 ft./2)− 1.031 ft.] = 5.532 ft.

$P_W=2.741ksf\left[\frac{0.615ft.}{5.532ft.}\right]$ = 0.305 ksf

$P_{LL}=2.741ksf\left[\frac{4.432ft.}{5.532ft.}\right]$ = 2.196 ksf

Moment at Wall Face:

M_W = $-5.045k(3.917ft.)-0.919k\left[\frac{4.917ft.}{2}\right]+\frac{1}{2}(0.305\frac{k}{ft.})(4.917ft.{)}^2+\frac{1}{2}(4.917ft.{)}^2[2.741\frac{k}{ft.}-0.305\frac{k}{ft.}]\left[\frac{2}{3}\right]$ = 1.298(ft−k)

Moment at LL_WL:

M_LL = $-0.187k(0.5ft.)+2.196\frac{k}{ft.}\frac{(1ft.{)}^2}{2}+\frac{1}{2}(1ft.)[2.741\frac{k}{ft.}-2.196\frac{k}{ft.}]\left[\frac{2}{3}\right](1ft.)$ = 1.186(ft−k)

Design Flexural Steel in Bottom of Footing

d = 11.5 in. − 4 in. = 7.500 in.

M_u = 4.837(ft−k) (controlling moment)

$R_n=\frac{4.837(ft-k)}{0.9(1ft.)(7.5in.{)}^2}$ = 0.096 ksi

$\rho =\frac{0.85(4000psi)}{60,000psi}[1-\sqrt{1-\frac{2(0.096ksi)}{0.85(4ksi)}}]$ = 0.00162

${\rho }_{min}=1.7[\frac{11.5in.}{7.5in.}{]}^2\frac{\sqrt{4000psi}}{60,000psi}$ = 0.00421

Use ρ = (4/3)ρ = (4/3)(0.00162) = 0.00216

A_SReq = 0.00216(12 in.)(7.5 in.) = 0.194 in²/ft.

$\frac{s}{0.196i{n}^2}=\frac{12in.}{0.194i{n}^2}$

s = 12.1 in.

Use #4's @ 12 in. cts. (Also use this spacing in the back of the stem.)

Check Shear

Dead Load and Earth Pressure Only

$V_W=0.143\frac{k}{ft.}(4.917ft.)+\frac{1}{2}(4.917ft.)[0.653\frac{k}{ft.}-0.143\frac{k}{ft.}]-0.919k$

V_W = 1.038k

Live Load 1 ft. From Stem Face

Shear at the wall can be neglected for this loading case.

$V_{LL}=0.026\frac{k}{ft.}(3.917ft.)+\frac{1}{2}(3.917ft.)[2.400\frac{k}{ft.}-0.026\frac{k}{ft.}]-0.732k$

V_LL = 4.019k

Live Load 1 ft. From Toe

$V_W=0.305\frac{k}{ft.}(4.917ft.)+\frac{1}{2}(4.917ft.)[2.741\frac{k}{ft.}-0.305\frac{k}{ft.}]-0.919k-5.045k$

V_W = 1.525k

$V_{LL}=2.196\frac{k}{ft.}(1ft)+\frac{1}{2}(1ft)[2.741\frac{k}{ft.}-2.196\frac{k}{ft.}]-0.187k$

V_LL = 2.282k

Use V_U = 4.019k

$\frac{{\nu }_u}{\varphi }=\frac{4019(lbs)}{0.85(12in.)(7.5in.)}=52.5psi<2\sqrt{4000psi}$ = 126.5 psi

Shear Key Design

For concrete cast against and permanently exposed to earth, minimum cover for reinforcement is 3 inches.

$d=12in.-3in.-\frac{1}{2}[\frac{1}{2}in.]$ = 8.75 in.

$P_1=0.120\frac{k}{f{t}^3}(1ft.)(2.882)[\frac{11.5}{12}ft.]$ = 0.331 k/ft.

$P_2=0.120\frac{k}{f{t}^3}(1ft.)(2.882)[\frac{29.5}{12}ft.]$ = 0.850 k/ft.

$M_u=(1.3)(1.3)\{0.331\frac{k}{ft.}\frac{(1.5ft.{)}^2}{2}+\frac{1}{2}(1.5ft.)[0.850\frac{k}{ft.}-0.331\frac{k}{ft}\left]\right[\frac{2}{3}](1.5ft.)\}$

M_u = 1.287(ft−k)

$R_n=\frac{1.287(ft-k)}{0.9(1ft.)(8.75in.{)}^2}$ = 0.0187 ksi

$\rho =\frac{0.85(4000psi)}{60,000psi}[1-\sqrt{1-\frac{2(0.0187ksi)}{0.85(4ksi)}}]$ = 0.000312

${\rho }_{min}=1.7[\frac{12in.}{8.75in.}{]}^2\frac{\sqrt{4000psi}}{60,000psi}$ = 0.00337

Use ρ = (4/3)ρ = (4/3)(0.000312) = 0.000416

A_SReq = 0.000416 (12 in.)(8.75 in.) = 0.0437 in²/ft.

$\frac{s}{0.196in{.}^2}=\frac{12in.}{0.0437in{.}^2}$

s = 53.8 in.

Use #4's @ 18 in. cts. (min)

Check Shear

V = 0.886k

$\frac{{\nu }_u}{\varphi }=\frac{(1.3)(1.3)(886lbs)}{0.85(12in.)(8.75in.)}$ = 16.8 psi < 126.5 psi o.k.

Reinforcement Summary

751.40.8.15.7 Example 3: Pile Footing Cantilever Wall

f’_c = 3,000 psi

f_y = 60,000 psi

φ = 27°

γ_s = 120 pcf

Pile type: HP 10 x 42

Allowable pile bearing = 56 tons

Pile width = 10 inches

Toe pile batter = 1:3

See EPG 751.12 Barriers, Railings, Curbs and Fences for weight and centroid of barrier.

Assumptions

• Retaining wall is located such that traffic can come within half of the wall height to the plane where earth pressure is applied.

• Reinforcement design is for one foot of wall length.

• 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.

• Neglect top one foot of fill over toe in determining soil weight and passive pressure on shear key.

• Neglect all fill over toe in designing stem reinforcement.

• The wall is designed as a cantilever supported by the footing.

• Footing is designed as a cantilever 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.

• For load factors for design of concrete, see EPG 751.24.1.2 Group Loads.

$C_A=cos\delta \left[\frac{cos\delta -\sqrt{co{s}^2\delta -co{s}^2\varphi }}{cos\delta +\sqrt{co{s}^2\delta -co{s}^2\varphi }}\right]$

δ = 0, ϕ = 27° so C_A reduces to:

$C_A=\frac{1-sin\varphi }{1+sin\varphi }=\frac{1-sin27^{\circ }}{1+sin27^{\circ }}$ = 0.376

$C_P=ta{n}^2[45^{\circ }+\frac{\varphi }{2}]=ta{n}^2[45^{\circ }+\frac{27^{\circ }}{2}]$ = 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).

Table 751.24.3.5.1

Load		Force (kips/ft)	Arm about C.L. of toe pile at 6B below footing (ft.)	Moment (ft-kips) per foot of wall length
Dead Load	(1)	0.340	2.542	0.864
	(2)	(1.333 ft.)(7.000 ft.)(0.150k/ft3) = 1.400	2.833	3.966
	(3)	(3.000 ft.)(8.500 ft.)(0.150k/ft3) = 3.825	4.417	16.895
	(4)	(1.000 ft.)(1.750 ft.)(0.150k/ft3) = 0.263	4.417	1.162
	Σ	ΣV = 5.828	-	ΣMR = 22.887
Earth Load	(5)	(7.000 ft.)(5.167 ft.)(0.120k/ft3) = 4.340	6.083	26.400
	(6)	(2.000 ft.)(2.000 ft.)(0.120k/ft3) = 0.480	1.167	0.560
	Σ	ΣV = 4.820	-	ΣMR = 26.960
Live Load Surcharge	PSV	(2.000 ft.)(5.167 ft.)(0.120k/ft3) = 1.240	6.083	MR = 7.543
	PSH	(2.000 ft.)(0.376)(10.000 ft.)(0.120k/ft3) = 0.902	10.000	MOT = 9.020
Earth Pressure	PA	2.2561	8.333	MOT = 18.799
	PP	3.2852	-	-
Collision Force (FCOL)		(10.000k)/[2(7.000 ft.)] = 0.714	18.000	MOT = 12.852
Heel Pile Tension (PHV)		(3.000 tons)(2 k/ton)(1 pile)/(12.000 ft.) = 0.500	7.167	MR = 3.584
Toe Pile Batter (PBH)		5.9033	-	-
Passive Pile Pressure (Ppp)		0.8324	-	-
1 $P_A=\frac{1}{2}{\gamma }_S{C}_A{H}^2=\frac{1}{2}[0.120\frac{k}{f{t}^3}](0.376)(10ft.{)}^3=2.256\frac{k}{ft}$
2 $P_P=\frac{1}{2}{\gamma }_S{C}_A[H_2^2-H_1^2]=\frac{1}{2}[0.120\frac{k}{f{t}^3}](2.663)[(6.75ft.{)}^2-(5ft.{)}^2]=3.285\frac{k}{ft}$
3 $P_{BH}=\left(56\frac{tons}{pile}\right)\left(2\frac{k}{ton}\right)(2piles)\left(\frac{4in.}{\sqrt{(12in.{)}^2+(4in.{)}^2}}\right)\left(\frac{1}{12ft.}\right)=5.903\frac{k}{ft}$
4 $P_{PP}=\frac{1}{2}(2.663)(5ft.{)}^2(0.120\frac{k}{f{t}^3})(0.833ft.)(3piles)(\frac{1}{12ft.})=0.832\frac{k}{ft}$

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).

Table 751.24.3.5.2

Load		Force (kips/ft)	Arm about C.L. of toe pile at 6B below footing (ft.)	Moment (ft-kips) per foot of wall length
Dead Load	(1)	0.340	0.875	0.298
	(2)	(1.333 ft.)(7.000 ft.)(0.150k/ft3) = 1.400	1.167	1.634
	(3)	(3.000 ft.)(8.500 ft.)(0.150k/ft3) = 3.825	2.750	10.519
	(4)	(1.000 ft.)(1.750 ft.)(0.150k/ft3) = 0.263	2.750	0.723
	Σ	ΣV = 5.828	-	ΣMR = 13.174
Earth Load	(5)	(7.000 ft.)(5.167 ft.)(0.120k/ft3) = 4.340	4.417	19.170
	(6)	(2.000 ft.)(2.000 ft.)(0.120k/ft3) = 0.480	-0.500	-0.240
	Σ	ΣV = 4.820	-	ΣMR = 18.930
Live Load Surcharge	PSV	(2.000 ft.)(5.167 ft.)(0.120k/ft3) = 1.240	4.417	MR = 5.477
	PSH	(2.000 ft.)(0.376)(10.000 ft.)(0.120k/ft3) = 0.902	5.000	MOT = 4.510
Earth Pressure	PA	2.256	3.333	MOT = 7.519
	PP	3.285	-	-
Collision Force (FCOL)		(10.000k)/[2(7.000 ft.)] = 0.714	13.000	MOT = 9.282
Heel Pile Tension (PHV)		(3.000 tons)(2 k/ton)(1 pile)/(12.000 ft.) = 0.500	5.500	MR = 2.750
Toe Pile Batter (PBH)		5.903	-	-
Passive Pile Pressure (Ppp)		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.

Neutral Axis Location = [2piles(1.5 ft.) + 1pile(7 ft.)] / (3 piles) = 3.333 ft. from the toe.

I = Ad²

For repetitive 12 ft. strip:

Total pile area = 3A

I = 2A(1.833 ft.)² + A(3.667 ft.)² = 20.167(A)ft.²

For a 1 ft. unit strip:

$I=\frac{20.167(A)ft{.}^2}{12ft.}=1.681(A)ft{.}^2$

Total pile area = (3A/12 ft.) = 0.250A

Case I

F.S. for overturning ≥ 1.5

F.S. for sliding ≥ 1.5

Check Overturning

Neglect resisting moment due to P_SV for this check.

ΣM_R = 22.887(ft−k) + 26.960(ft−k) + 3.584(ft−k)

ΣM_R = 53.431(ft−k)

ΣM_OT = 9.020(ft−k) + 18.799(ft−k) = 27.819(ft−k)

F.S._OT = $\frac{\mathrm{\Sigma }{M}_R}{\mathrm{\Sigma }{M}_{OT}}=\frac{53.431(ft-k)}{27.819(ft-k)}$ = 1.921 > 1.5 o.k.

Check Pile Bearing

Without P_SV :

ΣV = 5.828k + 4.820k = 10.648k

e = $\frac{\mathrm{\Sigma }M}{\mathrm{\Sigma }V}=\frac{(13.174+18.930)(ft-k)-(4.510+7.519)(ft-k)}{10.648k}$ = 1.885 ft.

Moment arm = 1.885 ft. - 1.833 ft. = 0.052 ft.

$P_T=\frac{\mathrm{\Sigma }V}{A}-\frac{M_c}{I}=\frac{10.648k}{0.250A}-\frac{10.648k(0.052ft.)(1.833ft.)}{1.681(A)f{t}^2}$

$P_T=\frac{41.988}{A}k$

$P_H=\frac{10.648k}{0.250A}+\frac{10.648k(0.052ft.)(3.667ft.)}{1.681(A)f{t}^2}$

$P_H=\frac{43.800}{A}k$

Allowable pile load = 56 tons/pile. Each pile has area A, so:

$P_T=41.988\frac{k}{pile}=20.944\frac{tons}{pile}$ o.k.

$P_H=43.800\frac{k}{pile}=21.900\frac{tons}{pile}$ o.k.

With P_SV:

ΣV = 5.828k + 4.820k + 1.240k = 11.888k

$e=\frac{(13.174+18.930+5.477)(ft-k)-(4.510+7.519)(ft-k)}{11.888k}$ = 2.149 ft.

Moment arm = 2.149 ft. - 1.833 ft. = 0.316 ft.

$P_T=\frac{11.888k}{0.250A}-\frac{11.888k(0.316ft.)(1.833ft.)}{1.681(A)f{t}^2}=43.456k=21.728\frac{tons}{pile}$ o.k.

$P_H=\frac{11.888k}{0.250A}+\frac{11.888k(0.316ft.)(3.667ft.)}{1.681(A)f{t}^2}=55.747k=27.874\frac{tons}{pile}$ o.k.

Check Sliding

$F.S{.}_{Sliding}=\frac{3.285k+5.903k+0.832k}{0.902k+2.256k}$ = 3.173 ≥ 1.5 o.k.

Case II

F.S. for overturning ≥ 1.2

F.S. for sliding ≥ 1.2

Check Overturning

ΣM_R = (22.887 + 26.960 + 7.543 + 3.584)(ft−k) = 60.974(ft−k)

ΣM_OT = (9.020 + 18.799 + 12.852)(ft−k) = 40.671(ft−k)

$F.S{.}_{OT}=\frac{\mathrm{\Sigma }{M}_R}{\mathrm{\Sigma }{M}_{OT}}=\frac{60.974(ft-k)}{40.671(ft-k)}$ = 1.499 ≥ 1.2 o.k.

Check Pile Bearing

$e=\frac{\mathrm{\Sigma }M}{\mathrm{\Sigma }V}=\frac{(13.174+18.930+5.477)(ft-k)-(4.510+7.519+9.282)(ft-k)}{(5.828+4.820+1.240)k}$ = 1.369 ft.

Moment arm = 1.833 ft. - 1.369 ft. = 0.464 ft.

$P_T=\frac{\mathrm{\Sigma }V}{A}+\frac{M_c}{I}=\frac{11.888k}{0.250A}+\frac{11.888k(0.464ft.)(1.833ft.)}{1.681(A)f{t}^2}$

$P_T=53.567\frac{k}{pile}=26.783\frac{tons}{pile}\le 56\frac{tons}{pile}$ o.k.

$P_H=\frac{11.888k}{0.250A}-\frac{11.888k(0.464ft.)(3.667ft.)}{1.681(A)f{t}^2}$ = 35.519k

$P_H=17.760\frac{tons}{pile}\le 56\frac{tons}{pile}$ o.k.

Check Sliding

$F.S{.}_{Sliding}=\frac{3.285k+5.903k+0.832k}{0.902k+2.256k+0.714k}$ = 2.588 ≥ 1.2 o.k.

Case III

F.S. for overturning ≥ 1.5

F.S. for sliding ≥ 1.5

Check Overturning

ΣM_R = (22.887 + 26.960 + 3.584)(ft−k) = 53.431(ft−k)

ΣM_OT = 18.799(ft−k)

$F.S{.}_{OT}=\frac{\mathrm{\Sigma }{M}_R}{\mathrm{\Sigma }{M}_{OT}}=\frac{53.431(ft-k)}{18.799(ft-k)}$ = 2.842 ≥ 1.5 o.k.

Check Pile Bearing

$e=\frac{\mathrm{\Sigma }M}{\mathrm{\Sigma }V}=\frac{(13.174+18.930)(ft-k)-7.519(ft-k)}{(5.828+4.820)k}$ = 2.309 ft.

Moment arm = 2.309 ft. - 1.833 ft. = 0.476 ft.

$P_T=\frac{10.648k}{0.250A}-\frac{10.648k(0.476ft.)(1.833ft.)}{1.681(A)f{t}^2}$ = 37.065k

$P_T=18.532\frac{tons}{pile}\le 56\frac{tons}{pile}$ o.k.

$P_H=\frac{10.648k}{0.250A}+\frac{10.648k(0.476ft.)(3.667ft.)}{1.681(A)f{t}^2}$ = 53.649k

$P_H=26.825\frac{tons}{pile}\le 56\frac{tons}{pile}$ o.k.

Check Sliding

$F.S{.}_{Sliding}=\frac{3.285k+5.903k+0.832k}{2.256k}$ = 4.441 ≥ 1.5 o.k.

Case IV

Check Pile Bearing

$e=\frac{\mathrm{\Sigma }M}{\mathrm{\Sigma }V}=\frac{(13.174+18.930)(ft-k)}{5.828k+4.820k}$ = 3.015 ft.

Moment arm = 3.015 ft. - 1.833 ft. = 1.182 ft.

$P_H=\frac{\mathrm{\Sigma }V}{A}+\frac{M_c}{I}=\frac{10.648k}{0.250A}+\frac{10.648k(1.182ft.)(3.667ft.)}{1.681(A)f{t}^2}$

$P_H=70.047k=35.024\frac{tons}{pile}$

25% overstress is allowed on the heel pile:

$P_H=35.024\frac{tons}{pile}\le 1.25(56\frac{tons}{pile})=70\frac{tons}{pile}$ o.k.

$P_T=\frac{10.648k}{0.250A}-\frac{10.648k(1.182ft.)(1.833ft.)}{1.681(A)f{t}^2}$ = 28.868k

$P_T=14.434\frac{tons}{pile}\le 56\frac{tons}{pile}$ o.k.

Reinforcement - Stem

b = 12 in.

cover = 2 in.

h = 16 in.

d = 16 in. - 2 in. - 0.5(0.625 in.) = 13.688 in.

F_Collision = 0.714k/ft

$P_{LL}={\gamma }_s{C}_{A}H(2.000ft.)=(2.000ft.)(0.376)(7.000ft.)(0.120\frac{k}{f{t}^3})=0.632\frac{k}{ft}$

$P_{A_{Stem}}=\frac{1}{2}{\gamma }_s{C}_A{H}^2=\frac{1}{2}[0.120\frac{k}{f{t}^3}](0.376)(7.000ft.{)}^2=1.105\frac{k}{ft}$

Apply Load Factors

F_Col. = γβ_LL(0.714k) = (1.3)(1.67)(0.714k) = 1.550k

P_LL = γβ_E (0.632k) = (1.3)(1.67)(0.632k) = 1.372k

P_AStem = γβ_E (1.105k) = (1.3)(1.3)(1.105k) = 1.867k

M_u = (10.00 ft.)(1.550k) + (3.500 ft.)(1.372k) + (2.333 ft.)(1.867k)

M_u = 24.658(ft−k)

$R_n=\frac{M_u}{\varphi b{d}^2}=\frac{24.658(ft-k)}{(0.9)(1ft.)(13.688in.{)}^2}$ = 0.146ksi

$\rho =\frac{0.85f{\text{'}}_c}{f_y}[1-\sqrt{1-\frac{2R_n}{0.85f{\text{'}}_c}}]=\frac{0.85(3ksi)}{60ksi}[1-\sqrt{1-\frac{2(0.146ksi)}{0.85(3ksi)}}]$ = 0.00251

${\rho }_{min}=1.7[\frac{h}{d}{]}^2\frac{\sqrt{f{\text{'}}_c}}{f_y}=1.7[\frac{16in.}{13.688in.}{]}^2\frac{\sqrt{3000psi}}{60,000psi}$ = 0.00212

ρ = 0.00251

$A_{S_{Req.}}=\rho bd=(0.00251)(12in.)(13.688in.)=0.412\frac{i{n}^2}{ft.}$

One #5 bar has A_S = 0.307 in²

$\frac{s}{0.307i{n}^2}=\frac{12in.}{0.412i{n}^2}$

s = 8.9 in.

Use # 5 bars @ 8.5 in. cts.

Check Shear

V_u ≤ φV_n

V_u = F_Collision + P_LL + P_AStem = 1.550k + 1.372k + 1.867k = 4.789k

$\frac{{\nu }_u}{\varphi }=\frac{v_u}{\varphi bd}=\frac{4789lbs}{0.85(12in.)(13.688in.)}$ = 34.301 psi

${\nu }_n={\nu }_c=2\sqrt{f{\text{'}}_c}=2\sqrt{3000psi}$ = 109.5 psi > 34.3 psi o.k.

Reinforcement - Footing - Top Steel

b = 12 in.

cover = 3 in.

h = 36 in.

d = 36 in. - 3 in. - 0.5(0.5 in.) = 32.750 in.

Design the heel to support the entire weight of the superimposed materials.

Soil(1) = 4.340k/ft.

LL_s = 1.240k/ft.

$Slabwt.=(3.000ft.)[0.150\frac{k}{f{t}^3}](5.167ft.)$ = 2.325k/ft.

Apply Load Factors

Soil(1) = γβ_E(4.340k) = (1.3)(1.0)(4.340k) = 5.642k

LL_s = γβ_E(1.240k) = (1.3)(1.67)(1.240k) = 2.692k

Slab wt. = γβ_D(2.325k) = (1.3)(1.0)(2.325k) = 3.023k

M_u = (2.583 ft.)(5.642k + 2.692k + 3.023k) = 29.335(ft−k)

$R_n=\frac{M_u}{\varphi b{d}^2}=\frac{29.335(ft-k)}{(0.9)(1ft.)(32.750in.{)}^2}$ = 0.0304 ksi

$\rho =\frac{0.85(3ksi)}{60ksi}[1-\sqrt{1-\frac{2(0.0304ksi)}{0.85(3ksi)}}]$ = 0.000510

${\rho }_{min}=1.7[\frac{36in.}{32.750in.}{]}^2\frac{\sqrt{3000psi}}{60,000psi}$ = 0.00188

Use ρ = 4/3 ρ = 4/3 (0.000510) = 0.000680

$A_{S_{Req}}=\rho bd=(0.000680)(12in.)(32.750in.)=0.267\frac{i{n}^2}{ft.}$

One #4 bar has A_s = 0.196 in.²

$\frac{s}{0.196i{n}^2}=\frac{12in}{0.267in{.}^2}$

s = 8.8 in.

Use #4 bars @ 8.5 in. cts.

Check Shear

$V_u=Soil(1)+L{L}_s+Slabwt.=5.642k+2.692k+3.023k=11.357k$

$\frac{{\nu }_u}{\varphi }=\frac{V_u}{\varphi bd}=\frac{11357lbs}{(0.85)(12in.)(32.750in.)}$ = 33.998 psi ≤ 109.5 psi = ν_c o.k.

Reinforcement - Footing - Bottom Steel

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.

Pile is embedded into footing 12 inches.

b = 12 in.

h = 36 in.

d = 36 in. - 4 in. = 32 in.

Apply Load Factors to Case IV Loads

$\mathrm{\Sigma }V=\gamma {\beta }_D[5.828\frac{k}{ft.}]+\gamma {\beta }_E[4.820\frac{k}{ft.}]$

$\mathrm{\Sigma }V=1.3(1.0)[5.828\frac{k}{ft.}]+1.3(1.0)[4.820\frac{k}{ft.}]$

ΣV = 13.842 k/ft.

$\mathrm{\Sigma }M=\gamma {\beta }_D[13.174\frac{(ft-k)}{ft.}]+\gamma {\beta }_E[18.930\frac{(ft-k)}{ft.}]$

$\mathrm{\Sigma }M=(1.3)(1.0)[13.174\frac{(ft-k)}{ft.}]+(1.3)(1.0)[18.930\frac{(ft-k)}{ft.}]$

ΣM = 41.735 (ft−k)/ft.

e = $\frac{\mathrm{\Sigma }M}{\mathrm{\Sigma }V}=\frac{41.735(ft-k)}{13.842k}$ = 3.015 ft.

Moment arm = 3.015 ft. - 1.833 ft. = 1.182 ft.

$P_H=\frac{\mathrm{\Sigma }V}{A}+\frac{M_c}{I}=\frac{13.842k}{0.250A}+\frac{13.842k(1.182ft.)(3.667ft.)}{1.681(A)f{t}^2}$

$P_H=91.059\frac{k}{pile}\left(\frac{1}{12ft.}\right)$ = 7.588 k/ft.

$M_u=(7.588\frac{k}{ft.})(3.667ft.)$ = 27.825(ft−k)/ft.

$R_n=\frac{M_u}{\varphi b{d}^2}=\frac{27.825(ft-k)}{(0.9)(1ft.)(32in.{)}^2}$ = 0.0301 ksi

$\rho =\frac{0.85(3ksi)}{60ksi}[1-\sqrt{1-\frac{2(0.0301ksi)}{0.85(3ksi)}}]$ = 0.000505

${\rho }_{min}=1.7[\frac{36in.}{32in.}{]}^2\frac{\sqrt{3000psi}}{60,000psi}$ = 0.00196

Use ρ = 4/3 ρ = 4/3 (0.000505) = 0.000673

A_SReq = ρbd = (0.000673)(12 in.)(32 in.) = 0.258 in²/ft.

One #4 bar has A_s = 0.196 in².

$\frac{s}{0.196in{.}^2}=\frac{12in.}{0.258in{.}^2}$

s = 9.1 in.

Use #4 bars @ 9 in. cts.

Check Shear

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.

V_u = 7.588k

$\frac{{\nu }_u}{\varphi }=\frac{V_u}{\varphi bd}=7588lbs0.85(12in.)(32in.)$ = 23.248 psi ≤ 109.5 psi = ν_c o.k.

Reinforcement - Shear Key

b = 12 in.

h = 12 in.

cover = 3 in.

d = 12 in. - 3 in. - 0.5(0.5 in.) = 8.75 in.

Apply Load Factors

P_P = γβ_E (3.845k) = (1.3)(1.3)(3.845k) = 6.498k

M_u = (0.912 ft.)(6.498k) = 5.926(ft−k)

$R_n=\frac{M_u}{\varphi b{d}^2}=\frac{5.926(ft-k)}{(0.9)(1ft.)(8.75in.{)}^2}$ = 0.0860 ksi

$\rho =\frac{0.85(3ksi)}{60ksi}[1-\sqrt{1-\frac{2(0.0860ksi)}{0.85(3ksi)}}]$ = 0.00146

${\rho }_{min}=1.7[\frac{12in.}{8.75in}{]}^2\frac{\sqrt{3000psi}}{60,000psi}$ = 0.00292

Use ρ = 4/3 ρ = 4/3(0.00146) = 0.00195

A_SReq = ρbd = (0.00195)(12 in.)(8.75 in.) = 0.205 in.²/ft.

One #4 bar has A_s = 0.196 in²

$\frac{s}{0.196in{.}^2}=\frac{12in.}{0.205in{.}^2}$

s = 11.5 in.

Use #4 bars @ 11 in. cts.

Check Shear

$\frac{{\nu }_u}{\varphi }=\frac{V_u}{\varphi bd}=\frac{6498lbs}{0.85(12in.)(8.75in.)}$ = 72.807 psi < 109.5 psi = ν_c

Reinforcement Summary

751.40.8.15.8 Dimensions

Cantilever Walls

Each section of wall shall be in increments of 4 ft. with a maximum length of 28'-0".

Each section of wall shall be in increments of 4 ft. with a maximum length of 28'-0".

Cantilever Walls - L-Shaped

Each section of wall shall be in increments of 4 ft. with a maximum length of 28'-0".

Counterfort Walls

Notes:

Dimension "A"

• Maximum length = 28'-0".

• Each section to be in 4'-0" increments.

• (See rustication recess details.)

Dimensions "B" & "C"

• As required by the design to balance the negative and positive moments. (See the design assumptions).

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

Notes:

Dimension "A"

* Maximum length = 28'-0".

* Each section to be in 4'-0" increments.

* (See 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.

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".

751.40.8.15.9 Reinforcement

Cantilever Walls

(*) 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 EPG 751.5.9.2.8.1 Development and Lap Splice General.)

(***) Theo. cut-off for bending + development length. (Wall height over 10' only.)

(*) 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 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

(*) 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 EPG 751.5.9.2.8.1 Development and Lap Splice General.)

Counterfort Walls

Wall and Stem

(For footing reinforcement, see the "Footing" diagram, below)

(*) Use development length or standard hook in accordance with EPG 751.5.9.2.8.1 Development and Lap Splice General.

(**) See lap splices Class B. (See EPG 751.5.9.2.8.1 Development and Lap Splice General.)

Footing

(*) 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

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.

751.40.8.15.10 Details

Non-Keyed Joints

Each section of wall shall be in increments of 4 ft. with a maximum length of 28'-0".

See EPG 751.50 Standard Detailing Notes for appropriate notes.

Keyed Joints

See EPG 751.50 Standard Detailing Notes for appropriate notes.

Rustication Recess

Drains

Note: French drains shall be used on all retaining walls, unless otherwise specified on the Design Layout.

Construction Joint Keys:

Cantilever Walls

Counterfort Walls

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

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).