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22 - Reinforced Concrete Retaining Walls
01: Design a reinforced concrete wall for a sloped backfill.
02: Design a reinforced concrete bridge abutment wall.
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**RC Retaining Walls01:Design a RC wall for a sloped backfill.(Revision: Oct-08)
1'
2'
15'
16'-6"
1'-6"
18
.33'
1.8
3'
PvPh
3
1
1
23
4
5
66
.11'
3' 1'-6" 5'-6"
10'
Design a reinforced concrete wall with a backfill = 125 pcf, an allowable soil bearing
capacity ofqall= 3 ksf, and a friction at the base of = 30. Design the wall and check for
its stability under working loads. (Note: All loads, shears and moments are per linear ft. of retaining wall).
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Solution:
( )2 2 2
Step 1: Find the lateral pressures and forces.
The active lateral earth pressure is,
30tan 45 tan 45 tan 30 0.33
2 2
The presure at the base (invert) of the wall footing is,
a
a
b a
K
K
p HK
= = = =
= ( ) ( ) ( )0.120 18.33 0.33 0.733
The forces on the wall (per unit length) are,
(12)(18.33) 2.0
(39)(18.33) 6.6
Step 2: Stability analysis for sliding and overturning.
v
h
kcf ft ksf
P kips
P kips
= =
= =
= =
Moment About A
Area Area Force(kip)
Arm(ft)
Moment(kip-ft)
1 x 5.5x 1.83 = 5.03 x 0.125 = 0.63 1.83 1.22 5.5 x 15.0 = 82.5 x 0.125 = 10.31 2.75 28.4
3 1.0 x 15.0 = 15.0 x 0.150 = 2.25 6.00 13.5
4 x 0.5 x 15.0 = 3.75 x 0.150 = 0.56 6.67 3.7
5 10.0 x 1.5 = 15.0 x 0.150 = 2.25 5.00 11.3
6 3.0 x 2.0 = 6.0 x 0.125 = 0.75 8.50 6.4
Pv 2.00
Ph H = 6.60 6.11 40.4
V=18.75 M=104.9
Location of Resultant
From point A, 104.9 = 5.6 ft
18.75
then e = 5.6 10 = 0.6 ft o.k. < B
2 6
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Soil Pressure at Toe of Base
qmax = 18.75 (1 + 6 x 0.6 ) = 1.875 (1 + 0.36) = 2.55 ksf OK < 3 ksf
10 10
Check F against SlidingShear available along base = 18.75 kips x 0.58 = 10.9 kips
Passive force at toe
Use S = 2/3 (30) = 20 , Pp = ( cos ) = 5.8 kips H
Pp = 5.8 (0.125) (3.5) = 4.7 kips2 (0.940)
Min. F = 10.9 = 1.7 kips , Max F = 10.9 + 4.7 = 15.6 = 2.4 kips OK without Key6.6 6.6 6.6
Step 2: Design parameters.
Load Factors
Stem Use 1.7 Ph
Base (toe and heel) distribute V uniformly over front B/3
Concrete and Steel Data
Capacity reduction factors: 0.90(flexure); 0.85(shear)
562
Fc = 3,000psi x 0.85 = 2,550 psi (for stress block)
Vc = 2 3,000 = 110 psi
fy = 40,000 psi ; min = 0.005 ; max = 0.0278 ; shrinkage
= 0.002
ld = 0.04 Ab (40,000) = 29.2 Ab (bottom bars) x 1.4 =
40.9 (top bars)3,000
Step 3: Design the stem of the wall.
Vertical Reinforcement
Ph = 1.7 39 (15) = 7.46
M = 7.46 x 5 x 12 = 448 kip-in
Use 6 batter on front, then t = 12 + 6 = 18
P v
P h
14"
5'
15
'
12"
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Use d = 18 4 = 14
Assume arm = d a/2 = 13
T = 448/13 = 34.5 kip
As = 34.5 / (40 x 0.90) = 0.96 in/ft
At bottom of wall Use #6 @ 5 ctrs
As = 1.06 in/ft
=1.06/(14 x 12) = 0.0063 >0.005 and
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Compute level of cutoff for #6 @ 10 ctrs. :
As = 0.53 in/ft ; = 0.53 / (11 x 12) = 0.004 > 0.002 < 0.005
Therefore, cutoff and develop bars above level where:
M = 0.53 (40) (11) (0.90) = 158 kip-in
1.33
M = 158kip-ft @ 10-6 level
Developmental length = 1-6
Level of Cutoff = 9-0
Step 4: Design the toe of the base of the wall.
Distribute V over front B/3. Assume t = 18 , d =14
V = 18.75 kips = qmax = 5.63 ksf(10 / 3)
wt. of soil over toe = 2.0 (0.125) = 0.25 ksf
wt of concrete base = 1.5 (0.150) = 0.23 ksf
Net toe pressure for design = qmax - wt. of soil over toe - wt of concrete base
= 5.63ksf - 0.25ksf - 0.23 ksf = 5.15 ksf
5.15 kip
1'-6"
3'
d =14"
front face of stem
V = 5.15 x (3.0) = 15.45 kips
M = 15.45 x (3/2) x 12 =278 kip-in
Assume arm = 13 in
T = 278/13 = 21.4 kips = 0.59 in/ft
(40 x 0.9)
Check = 0.59 = 0.0035 > 002, shrinkage OK14 x 12
< 0.005, so increase As by 1/3 , then As = 0.59 x 1.33 = 0.80 in/ftuse #6 @ 6 ctrs. As = 0.88 in/ft
Compressive stress block and shear OK by inspection after stem computations
Development length : Extend full base width, therefore ld OK
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Step 5:Design the heel of the base of the wall.
At Stem:
Weight of soil above heel at back face of stem = 15.0 (0.125) = 1.88 ksf
Weight of concrete base = 1.5 (0.150) = 0.23 ksf
Net pressure for design = 1.88 + 0.23 = 2.11 ksf
At Back:
Weight of soil above hell at back = 16.83 (0.125) = 2.10 ksf
Weight of concrete base = 0.23 ksf
Net pressure for design = 2.33 ksf
1'-6"
5'-6"
d =14"
2.33 ksf
2.11 ksf
V1 = (2.11) (5.5) = 5.80 kips
V2 = (2.33) (5.5) = 6.41 kips
Total V = 5.80 + 6.41 = 12.21 kips
M1= 5.8 (1/3) (5.5) (12) = 128 kip-in
M2 = 6.41 (2/3) (5.5) (12)=282 kip-in
Total M = 128 + 282 = 410 kip-in
Assume arm = 13
T= 410/13 = 31.5 kipAs = 31.5kip = 0.88 in/ft
(40 x 0.9)
Use #6 @ 6 ctrs. As = 0.88 in/ft > 0.002 and >0.005 and < 0.0278 OK
Compressive stress block and shear OK
Development length: Extend full base width, therefore ld OK
Horizontal shrinkage steel in stem:
Required: 0.002 (15) (12) = 0.36 in/ft of height
Use #4 @ 9 ctrs. front As = 0.27 in/ft
Use #4 @ 18 ctrs back As = 0.13 in/ft
Total As = 0.27 + 0.13 = 0.40 in/ft
Horizontal shrinkage steel in base:
Use #4 @ 12 ctrs. top and bottom As = 0.40 in/ft
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Step 6:Finished sketch of the wall.
3
1
12"
16'-6"
2'
1'-6"
3' 1'-6" 5'-6"
Ver
t.ba
rs,
Ver
t.ba
rs,
bac
kface
bac
kface
#6@
5"
#6@
10"
Provide 3" clear to
all bars except 112 "clear to bars infront face of stem
#4 @ 2'-0"
6" drains
@ 10'-0"
#6 @ 6" 2"x6" key
#6 @ 6"
#4 @ 12"
#4@ 9" front face
#4@18" back face
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**RC Retaining Walls02: Design a RC bridge abutment wall.
Design a bridge abutment for a backfill of = 110 lb/ft3
, an equivalent surcharge of 3
feet, an allowable soil pressure of 3.5kips/ft2, and an allowable shear between soil and
base of 45% of the vertical shear.
12'-0"
5'-6"
12"12"
3'-0"
2'-6" 12" 3'-6"
1'-6"
7'-0"
4'-0
" 6'-0"
9"
12"
A
5
72
8
4
1
3
6
3.88'
L =5.26 k
L =0.24 k
v
h
N
ote:
All loads and moments are per lineal foot of abutment.
Solution:
Phs = 30(3.0)(12) = 1.08 kips
Ph = 1/2(30)(122) = 2.16 kips
Stability Computations:
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Moments about A
Area Force Arm Moment
1 2.5 x 10.5 x 0.110 = 2.89 1.25 3.61
2 1.0 x 5.5 x 0.110 = 0.61 3.00 1.83
3 1/2 x 1.0 x 1.0 x 0.110 = 0.06 2.83 0.15
4 0.75 x 3.0 x 0.150 = 0.34 2.88 0.98
5 2.0 x 1.0 x 0.150 = 0.30 3.50 1.05
6 1/2 x 1.0 x 1.0 x 0.150 = 0.08 3.17 0.25
7 1.0 x 6.5 x 0.150 = 0.97 4.00 3.88
8 7.0 x 1.5 x 0.150 = 1.58 3.50 5.53
Lv 5.26 3.88 20.40
Lh 0.24 9.00 2.16
Phs 1.08 6.00 6.48
Ph 2.16 4.00 8.64
SH = 3.48 kips SV = 12.09 kips SMA= 54.96 ft-kips
Location of Resultant
From point A,09.12
96.54= 4.55 then e = 4.55 -
2
0.7= 1.05 1.5 ok
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**RC Retaining Walls02:Design a RC bridge abutment wall.
Design a bridge abutment for a backfill of_= 110 lb/ft3, an equivalent surcharge offeet, an allowable soil pressure of 3.5kips/ft2, and an allowable shear between soil
and base of 45% of the vertical shear.
All loads and moments are per lineal foot of abutment.
Solution:
Phs = 30(3.0)(12) = 1.08 kips
Ph = 1/2(30)(122) = 2.16 kips
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Stability Computations:
Moments about A
Area Force Arm Moment
1 2.5x10.5x0.11 = 2.89 1.25 3.61
2 1x5.5x0.11 = 0.61 3 1.83
3 1/2x1x1x0.11 = 0.06 2.83 0.15
4 0.75x3x0.15 = 0.34 2.88 0.98
5 2x1x0.15 = 0.3 3.5 1.05
6 .5x1x1x0.15 = 0.08 3.17 0.25
7 1x6.5x.15 = 0.97 4 3.88
8 7x1.5x0.15 = 1.58 3.5 5.53
Lv 5.26 3.88 20.4
Lh 0.24 9 2.16
Phs 1.08 6 6.48
Ph 2.16 4 8.64
H 3.48kips
V 12.09kips
MA 54.96kips
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Location of Resultant
Soil Pressure at Base:
Sliding:
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Design the stem:
Vertical Reinforcement
Ph + Phs = 8.64 +6.48 = 15.12
M = 15.12 x 12/3 x 12 = 726 kip in
Use 6 batter on front, then t = 9 + 6 = 15
Assume arm 15 4 = 11T = 726/11 = 66 kipAs = 66/ 40 x 0.9 = 1.83 in^2/ft
At bottom of wall Use #5 @ 2in =1.86in^2/ft or #9 at 6 c.c. = 2.00in^2/ft
p = 2.00/ 14 x 9 = 0.015 >.005 < 0.0278 OK
Check compressive stress block
C = T = 66kip / 2.55 x 12 x 0.9 = 2.39 >2 NOT GOOD
Check Shear
15.12 / 14x9 x .75 = 160psi > 110 psi. Need Shear Design for stirrups
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Resisting Moment
Top : 2 x 40 x .9 x 9 = 648 kip in
Bottom : 2 x 40 x.9 x 14 = 1008 kip in
Horizontal shrinkage in stem
Required = 0.002 x 14 x 9 = 0.336in^2/ft
Use #4@9 c.c front As = 0.27in^2/ft
Use #4@18 c.c back As = 0.13in^2/ft
Total As = 0.27 + 0.13 = 0.40