The document provides a design example for a reinforced concrete retaining wall with the following conditions: 1. The wall must retain a backfill with a unit weight of 100 pcf and a surcharge of 400 psf. 2. The wall stem is designed as a vertical cantilever beam to resist lateral earth pressures. 3. The base thickness is selected as 16 inches and the stem thickness as 15 inches with #8 reinforcing bars at 6 inches. 4. The heel width is selected as 7.5 feet to prevent sliding failure based on resisting and driving forces.

1. CE 437/537, Spring 2011 Retaining Wall Design Example 1/8
Design a reinforced concrete retaining wall for the
psf
following conditions. surcharge = qs = 400
f'c = 3000 psi
Fill: φ = 32o
fy = 60 ksi Unit wt = 100 pcf
HT = 18 ft
tf
wtoe tstem wheel
Natural Soil: φ = 32o
allowable bearing pressure = 5000psf
Development of Structural Design Equations. In this example, the structural design of the
three retaining wall components is performed by hand. Two equations are developed in this
section for determining the thickness & reinforcement required to resist the bending moment
in the retaining wall components (stem, toe and heel).
Equation to calculate effective depth, d: Three basic equations will be used to develop an
equation for d.
M u = φM n
⎛ a⎞
M n = As f y ⎜ d − ⎟
⎝ 2⎠
⎛ a⎞
M u = φAs f y ⎜ d − ⎟ [ Eqn 1]
⎝ 2⎠
C = T , 0.85 f c' a b = As f y
f c'
As = 0.85 ab [ Eqn 2]
fy
0.003 ε s + 0.003 a 0.003
strain compatibility : = , = β1
a / β1 d d ε s + 0.003
Assuming β1 = 0.85,
εs a/d
0.005 0.319
0.00785 0.235
0.010 0.196
and choosing a value for εs in about the middle of the practical design range,
a
= 0.235, a = 0.235 d [ Eqn 3]
d

2. CE 437/537, Spring 2011 Retaining Wall Design Example 2/8
Substituting Eqn. 2 into Eqn. 1:
⎛ f' ⎞ ⎛ a⎞
M u = φ ⎜ 0.85 c ab ⎟ f y ⎜ d − ⎟
⎜ fy ⎟ ⎝ 2⎠
⎝ ⎠
And substituting Eqn. 3 into the above:
f c' ⎛ 0.235 d ⎞
M u = φ 0.85 0.235d b f y ⎜ d − ⎟
fy ⎝ 2 ⎠
0.883d
Inserting the material properties: f'c = 3 ksi and fy = 60 ksi, and b = 12in (1-foot-wide
strip of wall, in the direction out of the paper).
k
M u = 0.90(0.85) 3ksi (12 in )(0.235)(0.883)d 2 M u = 5.71 in d 2
Equation for area of reinforcement, As. The area of reinforcement required is calculated
from Eqn. 1:
M u = φ As f y 0.883d = 0.90 As 60 ksi 0.883 d M u = 47.7 ksi As d
Design Procedure (after Phil Ferguson, Univ. Texas)
1. Determine HT. Usually, the top-of-wall elevation is determined by the client. The
bottom-of-wall elevation is determined by foundation conditions. HT = 18 feet.
2. Estimate thickness of base. tf ≈7% to 10% HT (12" minimum)
Tf = 0.07 (18' x 12"/') = 15.1" use tf = 16"

3. CE 437/537, Spring 2011 Retaining Wall Design Example 3/8
3. Design stem (tstem, Asstem). The stem is a vertical cantilever beam, acted on by the
horizontal earth pressure.
ft in in/ft
h = 8 – 16 /12
ft
h =16.67
h
tf = 16in
ka γ h ka qs
wtoe tstem wheel
calc. d:
1
Pfill = ( k a γ h ) h (1 ft out of page)
2
1 − sin φ 1 − sin( 32 o )
ka = = = 0.31
1 + sin φ 1 + sin( 32 o )
1
Pfill = (0.31)(100 pcf )(16.67 ft ) 2 (1 ft ) = 4310 lb
2
Psur = k a q sur h (1 ft ) = 0.31 ( 400 psf )(16.67 ft )(1 ft ) = 2070 lb
h h
M u = ( Earth Pressure LoadFactor)( Pfill )( ) + ( Live LoadFactor)(Psur )( )
3 2
16.67 ft 16.67 ft
M u = (1.6)(4310 lb )( ) + (1.6)(2070 lb )( ) = 65.9 k − ft
3 2
k
Mu = 5.71 in d2
k
in
65.9 k − ft (12 ) = 5.71 in d 2 , d = 11.8in
ft
1
t stem = 11.8in + 2 in cover + (1.0 in ) = 14.3in , ( assume #8 bars ) use t stem = 15in
2
d = 15 − 2 − 0.5 = 12.5in
in in in

4. CE 437/537, Spring 2011 Retaining Wall Design Example 4/8
calc. As:
M u = 47.7 ksi As d
in
65.9 k − ft (12 ) = 47.7 ksi As (12.5in ), As = 1.33 in 2
ft
As of one #8 bar = 0.79 in 2
in 2
0.79
bar in in
2
12 = 7.13 , use #8 @ 6 in
in ft bar
1.33
ft of wall
4. Choose Heel Width, wheel Select wheel to prevent sliding. Use a key to force sliding
failure to occur in the soil (soil-to-soil has higher friction angle than soil-to-concrete).
12in
Neglect soil resistance in front of the
wall.
Fresist 18
ft
set = Fsliding
FS
FS = Factor of Safety = 1.5 for sliding
Fresist = (Vertical Force)(coefficient of friction)
Fresist = WT (tan φnatural soil )
tf = 16in
tan φnatural soil = tan(32 o ) = 0.62
WT = W fill + Wstem + W found 15in
lb
W fill = (100 pcf )(16.67 ft )( wheel )(1 ft ) = 1670 wheel
ft
12 in + 15in 1 ft
Wstem = (150 pcf )(16.67 ft )( in
(1 ft ) = 2810 lb
2 12 )
16 15
W found = (150 pcf )( ft )( wheel + ft + 3 ft )(1 ft ) = 200 plf wheel + 850
12 12
Fsliding = Pfill + Psur
1
Pfill = (0.31 × 100 pcf )(18 ft ) 2 (1 ft ) = 5020 lb
2
Psur = (0.31 × 400 psf )(18 ft )(1 ft ) = 2230 lb
Fsliding = 5020 lb + 2230 lb = 7250 lb

5. CE 437/537, Spring 2011 Retaining Wall Design Example 5/8
⎡ lb lb lb lb ⎤
⎢1670 ft wheel + 2810 + 200 ft wheel + 850 ⎥(0.62)
7250lb = ⎣ ⎦
1.5
1.5 lb
7250lb = 3660lb + 1870 wheel , wheel = 7.42 ft , use wheel = 7.5 ft
0.62 ft
5. Check Overturning. 12in
18 ft 18 ft
M over = P fill ( ) + Psur ( )
3 2
M over = 5.02 k (6 ft ) + 2.23k (9 ft ) = 50.2 k − ft 18
ft
7.5 ft 15
M resist = W fill ( + ft + 3 ft ), assume wtoe = 3 ft
2 12
1.25 ft tf = 16in
+ Wstem (3 ft + )
2
11.75 ft 3' 15" 7.5'
+ W found ( )
2
M resist = (1.67 klf × 7.5 ft )(8 ft ) + ( 2.81k )(3.625 ft ) + (0.20klf × 7.5 ft + 0.85k )(5.875 ft )
12.53k 2.35k
M resist = 124.2 k − ft
M resist 124.2 k − ft
= = 2.47 > 2.0 = FSover , OK
M over 50.2 k − ft
6. Check Bearing.
WT M L
σ v at end of toe = + , equation is valid only if e <
bL bL2 6
6
WT = W fill + Wstem + W found
WT = 12.45 k + 2.81k + 2.35 k = 17.69 k
7.5 ft 1.25 ft
M = M over − W fill (5.875 ft − ) + Wstem (7.5 ft + − 5.875 ft ) + W found (0)
2 2
M = 50.2 k − ft − 12.53k ( 2.125 ft ) + 2.81k ( 2.25 ft ) = 29.9 k − ft
Check that e < L/6:
m 29.9 k − ft L 11.75 ft L
e= = k
= 1.68 ft , = = 1.96 ft , ∴ e < , OK
WT 17.69 6 6 6

6. CE 437/537, Spring 2011 Retaining Wall Design Example 6/8
17.69 k 29.9 k − ft
σv = + = 2.80 ksf < 5.0 ksf = allowable bearing capacity, OK
ft
(1 )(11.75 ) ft 1 ft
(1 )(11.75 ft ) 2
6
7. Heel Design.
Max. load on heel is due to the weight of heel + fill + surcharge as the wall tries to tip over.
Flexure:
W = Wheel + W fill + Wsur
16
W = 1.2(150 pcf )( ft )(1 ft ) wu
12 Vu
+ 1.2(100 pcf )(16.67 ft )(1 ft ) 16
in
Mu
+ 1.6( 400 plf ) 7.5
ft
klf
W = 2.88
wu L2 2.88klf (7.5 ft ) 2
Mu = = = 81.0 k − ft
2 2
k 2
M u = 5.71 d
in
in k
81.0 k − ft (12 ) = 5.71 d 2 , d = 13.0in for flexure
ft in
Shear:
Vu = wu (7.5 ft ) = 2.88 klf (7.5 ft ) = 21.6 k
φVc = (0.75) 2 f c' bw d = (0.75) 2 3000 psi (12 in ) d
setVu = φVc , 21,600lb = (0.75) 2 3000 psi (12 in ) d , d = 21.9 in for shear, controls
Shear controls the thickness of the heel.
1
t heel = 21.9 in + 2 in cover + in = 24.4 in (assume #8 bar ), use t heel = 21.5 in
2
Reinforcement in heel:
M u = 47.7 ksi As d
in
81.0 k − ft (12 ) = 47.7 ksi As (21.9 in ), As = 1.07in 2
ft
in 2
0.79
bar (12 in ) = 8.83 in , use #8 @ 8"
in 2 ft
1.07
ft

7. CE 437/537, Spring 2011 Retaining Wall Design Example 7/8
8. Toe Design.
Earth Pressure at Tip of Toe:
Wu Mu
σv = ±
bL 1 2
bL
6
Wu = 1.2(W fill + Wstem + W found ) + 1.6 (Wsur )
Wu = 1.2(12.53k + 2.81k + 2.35k ) + 1.6(0.4 ksf )(18 ft )(1 ft ) = 32.7 k , (did not recalc foundation wt b.c. neglible change)
M u = 1.6 M over − 1.2(Wsoil × 2.125 ft + Wstem × 1.0 ft ) 3' 1.25' 7.5'
[ ]
M u = 1.6(50.2 k − ft ) − 1.2 12.53k ( 2.125 ft ) + 2.81k (1 ft ) = 45.0k − ft
32.7 k 45.0k − ft
σv = +
(1 ft )(11.75 ft ) 1 (1 ft )(11.75 ft ) 2
6
ksf
σ vA = 2.78 + 1.96ksf = 4.74 ksf
A B C
σ vC = 2.78ksf − 1.96ksf = 0.82 ksf
4.74 ksf − 0.82 ksf
σ vB = 0.82 ksf + (8.75 ft ) = 3.74 ksf
11.75 ft
d for flexure:
3 ft 1 2
M u = (3.74 ksf )(3 ft )(1 ft )( ) + (1.00 ksf )(3 ft )(1 ft )( 3 ft ) = 19.8k − ft
2 2 3
k 2
M u = 5.71 d
in
in k
19.8k − ft (12 ) = 5.71 d 2 , d = 6.5in for flexure
ft in
d for shear:
Assume theel = ttoe = 21.5in
Critical section for shear occurs at "d" from face of stem, d = 21.5" – 3"cover-1/2"=18"
4.74 ksf − 0.82 ksf 18
σ vcritical sec tion = 0.82 ksf + ft
(8.75 ft + ft ) = 4.24 ksf
11.75 12
1 18
Vu = ( 4.74 ksf + 4.24 ksf )(3 ft − ft )(1 ft ) = 6.74 k
2 12
φVc = (.75)2 3000 psi (12in )(18in ) = 17,750lb > Vu , OK , d for flexure controls

8. CE 437/537, Spring 2011 Retaining Wall Design Example 8/8
Reinforcement in toe:
M u = 47.7 ksi As d
in
19.8k − ft (12 ) = 47.7 ksi As (18in ), As = 0.28 in 2
ft
in 2
0.79
bar (12 in ) = 33in , try smaller bars , say #4
in 2 ft
0.28
ft
in 2
0.20
bar (12 in ) = 8.6in use #4 @ 8"
in 2 ft
0.28
ft