Retaining wall design_example
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CE 437/537, Spring 2011 Retaining Wall Design Example 1 / 8
Design a reinforced concrete retaining wall for the following conditions. f'c = 3000 psi fy = 60 ksi 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.
11
'
'
003.0003.0,003.0
/003.0:
]2[85.0
85.0,
]1[2
2
βε
εβ
φ
φ
+=
+=
=
==
⎟⎠⎞
⎜⎝⎛ −=
⎟⎠⎞⎜
⎝⎛ −=
=
s
s
y
cs
ysc
ysu
ysn
nu
da
daitycompatibilstrain
EqnabffA
fAbafTC
EqnadfAM
adfAM
MM
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,
]3[235.0,235.0 Eqndada ==
Fill: φ = 32o Unit wt = 100 pcf
Natural Soil: φ = 32o allowable bearing pressure = 5000psf
wtoe tstem wheel
HT = 18 ft
tf
surcharge = qs = 400psf
CE 437/537, Spring 2011 Retaining Wall Design Example 2 / 8
Substituting Eqn. 2 into Eqn. 1:
⎟⎠⎞⎜
⎝⎛ −⎟
⎟⎠
⎞⎜⎜⎝
⎛=
285.0
' adfabffM yy
cu φ
And substituting Eqn. 3 into the above:
d
ddfbdffM yy
cu
883.0
2235.0235.085.0
'
⎟⎠⎞⎜
⎝⎛ −= φ
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).
22 71.5)883.0)(235.0)(12(3)85.0(90.0 dMdM ink
uinksi
u ==
Equation for area of reinforcement, As. The area of reinforcement required is calculated from Eqn. 1:
dAMdAdfAM s
ksiu
ksisysu 7.47883.06090.0883.0 === φ
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"
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. calc. d:
lbftftftsurasur
lbftftfill
o
o
a
ftafill
psfhqkP
pcfP
k
pageofouthhkP
2070)1)(67.16)(400(31.0)1(
4310)1()67.16)(100)(31.0(21
31.0)32sin(1)32sin(1
sin1sin1
)1()(21
2
===
==
=+−=
+−=
=
φφ
γ
inininin
instem
ininininstem
inink
ftk
ink
u
ftkft
lbft
lbu
fillu
d
tusebarsassumet
ddftin
dM
M
hhPM
5.125.0215
15)8#(,3.14)0.1(21cover28.11
8.11,71.5)12(9.65
71.5
9.65)267.16)(2070)(6.1()
367.16)(4310)(6.1(
)2
)()(PLoadFactorLive()3
)()(LoadFactorPressureEarth(
2
2
sur
=−−=
==++=
==
=
=+=
+=
−
−
wtoe tstem wheel
h = 8ft – 16in/12in/ft
h =16.67ft
tf = 16in
ka γ h ka qs
h
CE 437/537, Spring 2011 Retaining Wall Design Example 4 / 8
calc. As:
inusebarin
ftin
wallofftinbarin
inbaroneofA
inAAftin
dAM
s
sin
sksiftk
sksi
u
6@8#,13.71233.1
79.0
79.08#
33.1),5.12(7.47)12(9.65
7.47
2
2
2
2
=
=
==
=
−
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). Neglect soil resistance in front of the wall.
foundstemfillT
osoilnatural
soilnaturalT
slidingresist
WWWW
W
FFSF
set
++=
==
==
==
=
62.0)32tan(tan
)(tanFfriction)offficientForce)(coe(VerticalF
slidingfor 1.5 Safety ofFactor FS
resist
resist
φ
φ
850200)1)(31215)(
1216)(150(
2810)1()12
121512)(67.16)(150(
1670)1)()(67.16)(100(
+=++=
=+=
==
heelftft
heelfound
lbftin
ftininft
stem
heelft
heelft
fill
wplfftwftpcfW
pcfW
wftlbwpcfW
lblblbsliding
lbftftsur
lbftftfill
surfillsliding
F
psfP
pcfP
PPF
725022305020
2230)1)(18)(40031.0(
5020)1()18)(10031.0(21 2
=+=
=×=
=×=
+=
18ft
tf = 16in
15in
12in
CE 437/537, Spring 2011 Retaining Wall Design Example 5 / 8
ftheel
ftheelheel
lblb
lbheel
lbheel
lb
wusewwftlb
wftlbw
ftlb
5.7,42.7,1870366062.05.17250
5.1
)62.0(850200281016707250
==+=
⎥⎦
⎤⎢⎣
⎡+++
=
5. Check Overturning.
)275.11(
)225.13(
3),31215
25.7(
2.50)9(23.2)6(02.5
)218()
318(
ft
found
ftft
stem
fttoe
ftft
fillresist
ftkftkftkover
ft
sur
ft
fillover
W
W
wassumeftWM
M
PPM
+
++
=++=
=+=
+=
−
OKFSMM
M
M
overftk
ftk
over
resist
ftkresist
ftkftklfftkftftklfresist
,0.247.22.502.124
2.124
)875.5)(85.05.720.0()625.3)(81.2()8)(5.767.1(
=>==
=
+×++×=
−
−
−
6. Check Bearing.
ftkftkftkftk
foundft
ftft
stem
ftft
fillover
kkkkT
foundstemfillT
Tv
M
WWWMM
W
WWWW
bLM
bLWtoeofendat
−− =+−=
+−++−−=
=++=
++=
<+=
9.29)25.2(81.2)125.2(53.122.50
)0()875.52
25.15.7()25.7875.5(
69.1735.281.245.12
6L e ifonly validis equation,
6
2σ
Check that e < L/6:
OKLeLWme ft
ftft
k
ftk
T,6
,96.1675.11
6,68.1
69.179.29 <∴=====
−
12.53k 2.35k
18ft
tf = 16in
12in
3' 15" 7.5'
CE 437/537, Spring 2011 Retaining Wall Design Example 6 / 8
OK capacity,bearingallowable0.580.2)75.11)(1(
61
9.29)75.11)(1(
69.172
=<=+=−
ksfksf
ftft
ftk
ftft
k
vσ
7. Heel Design. Max. load on heel is due to the weight of heel + fill + surcharge as the wall tries to tip over. Flexure:
surfillheel WWWW ++=
ftkftklf
uu
klf
ftft
ft
LwM
W
plfpcf
ftpcfW
−===
=
++
=
0.812
)5.7(88.22
88.2
)400(6.1)1)(67.16)(100(2.1
)1)(1216)(150(2.1
22
flexureforddink
ftin
dinkM
inftk
u
0.13,71.5)12(0.81
71.5
2
2
==
=
−
Shear:
controlsshearforddpsiVsetV
dpsidbfV
wV
ininlbcu
inwcc
kftklfftuu
,9.21,)12(30002)75.0(600,21,
)12(30002)75.0(2)75.0(
6.21)5.7(88.2)5.7(
'
===
==
===
φ
φ
Shear controls the thickness of the heel.
inheel
inininheel tusebarassumeint 5.21),8#(4.24
21cover29.21 ==++=
Reinforcement in heel:
"8@8#,83.8)12(07.1
79.0
07.1),9.21(7.47)12(0.81
7.47
2
2
2
useftin
ftinbarin
inAAftin
dAM
in
sin
sksiftk
sksi
u
=
==
=
−
7.5ft
16in Mu
wu
Vu
CE 437/537, Spring 2011 Retaining Wall Design Example 7 / 8
8. Toe Design. Earth Pressure at Tip of Toe:
change) neglible b.c. wt foundation recalcnot (did,7.32)1)(18)(4.0(6.1)35.281.253.12(2.1
)(6.1)(2.161 2
kftftksfkkku
surfoundstemfillu
uuv
W
WWWWW
bL
MbLW
=+++=
+++=
±=σ
[ ]
ksfftft
ksfksfksf
v
ksfksfksfv
ksfksfksfv
ftft
ftk
ftft
k
v
ftkftkftkftku
ftstem
ftsoiloveru
B
C
A
M
WWMM
74.3)75.8(75.1182.074.482.0
82.096.178.2
74.496.178.2
)75.11)(1(61
0.45)75.11)(1(
7.32
0.45)1(81.2)125.2(53.122.1)2.50(6.1
)0.1125.2(2.16.1
2
=−+=
=−=
=+=
+=
=+−=
×+×−=
−
−−
σ
σ
σ
σ
d for flexure:
flexureforddink
ftin
dinkM
M
inftk
u
ftkftftftksfft
ftftksfu
5.6,71.5)12(8.19
71.5
8.19)332)(1)(3)(00.1(
21)
23)(1)(3)(74.3(
2
2
==
=
=+=
−
−
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"
controlsflexurefordOKVpsiV
ftV
ft
ulbinin
c
kftftksfksfu
ksfftft
ksfksfksf
v tioncritical
,,750,17)18)(12(30002)75(.
74.6)1)(12183)(24.474.4(
21
24.4)121875.8(
75.1182.074.482.0
sec
>==
=−+=
=+−+=
φ
σ
A B C
3' 7.5' 1.25'
CE 437/537, Spring 2011 Retaining Wall Design Example 8 / 8
Reinforcement in toe:
"8@4#6.8)12(28.0
20.0
4#,,33)12(28.0
79.0
28.0),18(7.47)12(8.19
7.47
2
2
2
2
2
useftin
ftinbarin
saybarssmallertryftin
ftinbarin
inAAftin
dAM
in
in
sin
sksiftk
sksi
u
=
=
==
=
−