rcc r-w examples
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CE 437/537, Spring 2011 Retaining Wall Design Example
Design a reinforced concrete retaining wall for thefollowing conditions.
f'c = 3000 psi
fy = 60 ksi
Development of Structural Design Equations. In this example, the structural three retaining wall components is performed by hand. Two equations are deve
section for determining the thickness & reinforcement required to resist the bendin the retaining wall components (stem, toe and heel).
Equation to calculate effective depth, d: Three basic equations will be used to d
equation for d.
11
'
'
003.0
003.0,
003.0
/
003.0:
]2[85.0
85.0,
]1[2
2
! "
"
!
#
#
+
=+
=
=
==
$ % &
'( ) *=
$
%
& '
(
) *=
=
s
s
y
c s
y sc
y su
y sn
nu
d
a
d aitycompatibil strain
Eqnab f
f A
f Aba f T C
Eqna
d f A M
ad f A M
M M
Assuming !1 = 0.85,
"s a/d
Fill: # = 32o Unit wt =
Natural Soil: # = 32o allowable bearing pre
wtoe tstem wheel
HT = 18 ft
tf
surcharge = qs = 4
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CE 437/537, Spring 2011 Retaining Wall Design Example
Substituting Eqn. 2 into Eqn. 1:
! " #
$% & '!
! "
# $$%
& =
285.0
'a
d f ab f
f M y
y
cu (
And substituting Eqn. 3 into the above:
d
d d f bd
f
f M y
y
cu
883.0
2
235.0235.085.0
'
! " # $
% & '= (
Inserting the material properties: f'c = 3 ksi and fy = 60 ksi, and b = 12in
(1
strip of wall, in the direction out of the paper).
2)883.0)(235.0)(12(3)85.0(90.0 d M inksiu
=
Equation for area of reinforcement, As. The area of reinforcement required is ca
from Eqn. 1:
d Ad f A M ksi s y su 883.06090.0883.0 ==!
Design Procedure (after Phil Ferguson, Univ. Texas)
1. Determine HT. Usually, the top-of-wall elevation is determined by the clien bottom-of-wall elevation is determined by foundation conditions.
2. Estimate thickness of base. tf !7% to 10% HT (12" minimum)
Tf = 0.07 (18' x 12"/') = 15.1"
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CE 437/537, Spring 2011 Retaining Wall Design Example
3. Design stem (tstem, Asstem). The stem is a vertical cantilever beam, acted on
horizontal earth pressure.
calc. d:
lb ft ft ft sur a sur
lb ft ft fill
o
o
a
ft a fill
psf hqk P
pcf P
k
pageof out hhk P
2070)1)(67.16)(400(31.0)1(
4310)1()67.16)(100)(31.0(2
1
31.0)32sin(1
)32sin(1
sin1
sin1
)1()(2
1
2
===
==
=+
!
=+
!
=
=
"
"
#
inininin
inininin stem
inin
k
ft k
in
k
u
ft k ft
lb ft
lbu
fill u
d
barsassumet
d d
ft
in
d M
M
hh
P M
5.125.0215
)8#(,3.14)0.1(2
1cover 28.11
8.11,71.5)12(9.65
71.5
9.65)2
67.16)(2070)(6.1()
3
67.16)(4310)(6.1(
)2)()(PLoadFactor Live()3)()(LoadFactor PressureEarth(
2
2
sur
=!!=
=++=
==
=
=+=
+=
!
!
wtoe tstem wheel
h = 8 – 16n/12
n
h =16.67ft
tf = 16in
ka $
h
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CE 437/537, Spring 2011 Retaining Wall Design Example
calc. As:
bar
in
ft
in
wall of ft in
bar
in
inbar oneof A
in A A ft
in
d A M
s
sin
sksi ft k
sksiu
,13.712
33.1
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 forc
failure to occur in the soil (soil-to-soil has higher friction angle than soil-to-conc
Neglect soil resistance in front of thewall.
found stem fill T
o soil natural
soil natural T
sliding resist
W W W W
W
F FS
F set
++=
==
=
=
==
=
62.0)32tan(tan
)(tanF
friction)of fficientForce)(coe(VerticalF
slidingfor1.5Safetyof FactorFS
resist
resist
!
!
850200)1)(312
15)(
12
16)(150(
2810)1()12
1
2
1512)(67.16)(150(
1670)1)()(67.16)(100(
+=++=
=+
=
==
heel ft ft
heel found
lb ft
in
ft inin ft
stem
heel ft
heel ft
fill
w plf ft w ft pcf W
pcf W
w ft
lbw pcf W
lb ft ft
lb ft ft fill
sur fill sliding
psfP
pcf P
P P F
2230)1)(18)(40031.0(
5020)1()18)(10031.0(2
1 2
=!=
=!=
+=
18t
tf = 16n
15 n
12n
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CE 437/537, Spring 2011 Retaining Wall Design Example
ft heel heel
lblb
lbheel
lbheel
lb
uww ft
lb
w ft
lbw
ft
lb
,42.7,1870366062.0
5.17250
5.1
)62.0(85020028101670
7250
=+=
!"
#$%
&+++
=
5. Check Overturning.
)2
75.11(
)2
25.13(
3),312
15
2
5.7(
2.50)9(23.2)6(02.5
)2
18()3
18(
ft
found
ft ft
stem
ft toe
ft ft
fill resist
ft k ft k ft k over
ft
sur
ft
fill over
W
W
wassume ft W M
M
P P M
+
++
=++=
=+=
+=
!
OK FS M
M
M
M
over ft k
ft k
over
resist
ft k resist
ft k ft klf ft k ft ft klf resist
,0.247.22.50
2.124
2.124
)875.5)(85.05.720.0()625.3)(81.2()8)(5.767.1(
=>==
=
+!++!=
"
"
"
6. Check Bearing.
found ft
ft ft
stem
ft ft
fillover
k k k k T
found stem fill T
T v
W W W M M
W
W W W W
bL
M
bL
W toeof end at
+!++!!=
=++=
++=
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CE 437/537, Spring 2011 Retaining Wall Design Example
OKcapacity, bearingallowable0.580.2
)75.11)(1(6
1
9.29
)75.11)(1(
69.17
2
=
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CE 437/537, Spring 2011 Retaining Wall Design Example
8. Toe Design. Earth Pressure at Tip of Toe:
foundat recalcnot(did,7.32)1)(18)(4.0(6.1)35.281.253.12(2.1
)(6.1)(2.1
6
1 2
k ft ft ksf k k k u
sur found stem fill u
uuv
W
W W W W W
bL
M
bL
W
=+++=
+++=
±=!
[ ]
ksf ft
ft
ksf ksf ksf
v
ksf ksf ksf v
ksf ksf ksf
v
ft ft
ft k
ft ft
k
v
ft k ft k ft k ft k u
ft stem
ft soil over u
B
C
A
M
W W M M
74.3)75.8(75.11
82.074.482.0
82.096.178.2
74.496.178.2
)75.11)(1(6
1
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
=!
+=
=!=
=+=
+=
=+!=
"+"!=
!
!!
#
#
#
#
d for flexure:
flexure for d d in
k
ft
in
d in
k M
M
in ft k
u
ft k ft ft ft ksf ft
ft ft ksf u
5.6,71.5)12(8.19
71.5
8.19)33
2)(1)(3)(00.1(
2
1)
2
3)(1)(3)(74.3(
2
2
==
=
=+=
!
!
d for shear:Assume theel = ttoe = 21.5
in
Critical section for shear occurs at "d" from face of stem, d = 21.5" – 3"cove
ft
kftftksfksf
ksf ft
ft
ksf ksf ksf
v tioncritical
181
24.4)12
1875.8(
75.11
82.074.482.0
sec=+
!
+=#
A B
3' 1.25'
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CE 437/537, Spring 2011 Retaining Wall Design Example
Reinforcement in toe:
#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
use ft
in
ft
inbar
in
saybars smaller try ft
in
ft
in
bar
in
in A A ft in
d A M
in
in
sin
sksi ft k
sksi
u
=
=
==
=
!
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