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7/31/2019 Retaining Wall Design_1
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Design of Retaing wall
Input
Height above GL = 3 m
Density of soil, = 18 kN/m3
Denity of Concrete = 25 kN/m3
= 30
= 0.5 radians
= 0 = 0 radians
= 18
= 0.3 radians
Grade of Concrete, fck = 20 Mpa
Grade of Steel, fy = 415 Mpa
SBC of foundation strata, q0 = 100 kN/m2
Coefficient of friction, = 0.5
b5
b3
b6
H4
H3
H1
H2
b2 b4
b1
W1
W2
W3
W4
D E B C
A
TOE HEEL
SHEAR
KEY
STEM
H6
H5
H7
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Solution
1 Design constants
Xu, max = 0.48
d
2 Depth of foundation
Density of soil, = 18 kN/m3
= 18000 N/m3
ymin = (q0/) x ((1-sin)/(1+sin))2
= 0.61728 m
Keep, H3 = 1 m
Height of wall above its base
Hence, H1 = 4 m
3 Dimensions of base
"D-E" Ratio =
b
= 1-(q0/(2.2 H1))
= 0.37
ka = (1-sin)/(1+sin)
ka = 0.33333
b1 = 0.95 H Sqrt{(Ka/((1-)(1+))}
b1 = 1.90288 m
Base width from sliding condition
b1 = 0.7 H Ka/(1-)
b1 = 2.96296 m
Base width from from normal practice
b1 = 0.6 H or 0.4 H
b1 = 2.4 m
Keep, b1 = 2.4 m
Width of Toe slab, b1 = 0.9 m
Let the thickness of base, H4 = H/12
= 0.3 m
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4 Thickness of stem
Height, H5 "A-B" = 3.7 m
Consider 1m length retaining wall
Maximum Bending moment @ "B"
= Ka H53/6
= 50.653 kN-m
M @ "B" = 5.1E+07 N-mm
Mu = 7.6E+07 N-mm
Hence effective depth, b3eff = Sqrt(Mu/Ru b)
= 159.73 mm
Keep, b3eff = 240 mm
Total thickness, b3 = 300 mm
Using 16 dia bars
b5 = 200 mm
b5eff = 140 mm
Width of Heel, b4 = 1.2 m
5 Stability of wall
Designation Force, kNLever arm,
m
Moment
@ toe,
kN-m
W1 18.5 1.1 20.35
W2 4.625 0.97 4.47083
W3 18 1.2 21.6
W4 79.92 1.8 143.856
Total W = 121.045 MR = 190.30
Total resisting moment MR = 190.3 kN-m
Earth pressure, P = Ka H12/2
= 48 kN
Check for overturning
Overturning moment, M0 = 64 kN-m
F.S. againt overturning = = 190.30
64
= 3.0 should be > 2
Hence Safe
Check for Sliding
F.S. againt Sliding = W
P
= 1.3 should be > 1.5
Hence Unsafe, Provide shear key
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Pressure distribution
Net moment, M = 190.3 - 64
M = 126.30 kN-m
Distance X of the pont of application of the resultant, from toe
X = MW
X = = 126.30
121.045
X = = 1.04 m
Eccentricity, e = (b1/2) - X
e = 0.16 m Should be < b1/6
b/6 = 0.4 m
Hence No tension
Pressure p1 at toe = (W/b) x (1+(6e/b))
= 70.2 kN/m2 Should be less than SBC
= 100 kN/m2
Hence Safe
Pressure p2 at heel = (W/b) x (1-(6e/b))
= 30.6917 kN/m2
Pressure at the junction of stem with toe slab,
p = 55.4 kN/m2
Pressure at the junction of stem with heel slab,
p = 50.4458 kN/m2
b3b2 b4
b1
D E B C
A
H7
p1
p2P
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6 Design of toe slab
Downward weight of slab per unit area = 7.5 kN/m2
Hence net pressure intensity under "D" = 70.2 - 7.5 = 62.7 kN/m2
Hence net pressure intensity under "E" = 55.4 - 7.5 = 47.9 kN/m2
Total force = Shear force at E = 50 kN
X from "E" = 0.47 m
Bending moment @ "E" = 23.3955 kN-m
= 2.3E+07 N-mm
Mu @ "E" = 3.5E+07 N-mm
H4eff = Sqrt(Mu/Ru b)
= 108.555 mm
Provide, H4 = 260 mm
H4eff = 200 mm
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Steel
Ast = ((0.5 x fck x b x d)/fy)x(1-SQRT(1-((4.6xMu)/(fckxbfxd^2)))
Ast = 513.599 mm2
This reinforcement to be provided at bottom face.
If alternate bars of stem reinforcement are to be bent & continued in the toe slab,
area available = 565.2 mm2 See step 8
Bar dia = 12 # bars
Spacing = 200 mm
Provide 12 # bars @ = 200 C/C at top of heel slab
Check for development length, Ld = 47
= 564 mm
Providing 50mm clear side cover ,
actual length available = 850 mm
Hence safe
Distribution reinforcement = 0.12 x 1000 x A /100
= 276 mm2
Bar dia= = 8 # bars
Spacing = 182 mm
Provide 8 # bars @ = 180 C/C
7 Design of heel slab
1) Total weight of soil = 79.92 kN
Lever arm @ "B" = 0.60 m
2) Total weight of heel slab = 7.80 kN
Lever arm @ "B" = 0.60 m
3) Total upward soil reaction = 48.68 kN
Lever arm @ "B" = 0.55 mTotal force = S.F. at "B" = 39.04 kN
B.M. @ "B" = 25.793 kN-m
= 2.6E+07 N-mm
Mu = 3.9E+07 N-mm
H7eff = Sqrt(Mu/Ru b)
= 113.981 mm
Provide, H6 = 260 mm
H7eff = 200 mm
Steel
Ast = ((0.5 x fck x b x d)/fy)x(1-SQRT(1-((4.6xMu)/(fckxbfxd^2)))
= 569.737 mm2
Bar dia = 10 # bars
Spacing = 138 mm
Provide 10 # bars @ 130 C/C at top of heel slab
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Development length, Ld = 47
to the left of "B" = 470 mm
= 500 mm
Distribution reinforcement = 0.12 x 1000 x A /100
= 276 mm2
Bar dia= = 8 # barsSpacing = 182 mm
Provide 8 # bars @ = 180 C/C
Shear stress, v = 1.5 x 39.04 x 1000/(1000 x 200)
= 0.2928 N/mm2
pt Ast x 100
1000 x d
pt c = 0.28487
0.25 0.36
0.5 0.48
0.28486828 0.38674Corresponding c = 0.38674 N/mm
2
Hence Safe
8 Reinforcement in stem
Revised H5 = 3.74 m
M = Ka H53/6
= 52.3136 kN-m
= 5.2E+07 N-mm
Mu = 7.8E+07 N-mm
b3eff = Sqrt(Mu/Ru b)
= 162.327 mm
Provide, b3eff = 250 mm
b3 = 310 mm
b5 = 200 mm
b5eff = 140 mm
Ast = = ((0.5 x fck x b x d)/fy)x(1-SQRT(1-((4.6xMu)/(fckxbfxd^2)))
= 943.711939 mm2
Bar dia= = 12 # bars
Spacing = 120 mm
Provide 12 # bars @ 100 C/C
Astact = 1130.4 mm2
Continue alternate bars in the toe slab to serve as tensile reinforcement there.
Discontinue the remaining half bars after a distance of
Ld = 47
= 564 mm
= 600 mm
beyond "B" in the toe slab.
=
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Between "A" & "B" some bars can be curtiled.
Consider a section at depth h below the top of the stem.
Effective depth at that section is
d' = 140 + 29.4117647 h mm (where h in m)
Now Ast (H13/d)
or H1 = (Ast d)1/3
Hence h/H1 = [(Ast' d')/(Ast d)]1/3
where Ast' = reinforcement at depth h
Ast = reinforcement at depth H1
d' = effective depth at depth h
d = effective depth at depth H1
If Ast' = Ast/2, Ast'/Ast = 1/2
Hence h/H1 = [d'/2d]1/3
Substituting d = 250
d' = 140 + 29.4117647 h
h = H1 [d'/2d]1/3
h = H1/(2)1/3
h = 2.96844 m
Bars should be extended by a distance of 12
= 144 mm
or b3 = 250 mm
Hence h = 2.70 m
Curtail bars at this height below the top
Check for shear
Shear force = P
Ka H52/2 = 41.9628 kN
Fu = 62.9442 kN
Shear stress, v = 62.9442 x 1000/(1000 x 250)
= 0.25178 N/mm2
pt Ast x 1001000 x d
pt c = 0.37748
0.25 0.36
0.5 0.48
0.37748478 0.43119
Corresponding c = 0.43119 N/mm2
Hence Safe
Distribition & temperature reinforcement
Average thk of stem = 255 mm
Distribution reinforcement = 0.12 x 1000 x A /100
= 306 mm2
Bar dia= = 8 # bars
Spacing = 164 mm
Provide 8 # bars @ 160 C/C
at inner face of the wall, along its length.
For temperature reinforcement, provide 8 mm # bars @ 300 C/C both ways, in outer face.
=
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9 Design of shear key
Let depth of key = a
Intensity of passive pressure pp developed in front of the key depends on the soil
pp = Kp p = 166.2 kN/m2
Total passive pressure pp = pp a = 166.2 a
Dsliding force at level D1C1 = (1/3) x (/2) x (H+a)2
or pH = (1/3) x (/2) x (H+a)2
Weigth of soil between bottom of base and D1C1 = b a x
= 43.2 a
W = 121.045 + 43.2 a
F.S. against sliding = 1.5
W + pp
pH
1.5 = 1.5127695
a = 0.08 m..Adjust this value till above value = 1.5
However provide minimum
a = 0.3 m
a = H6 = 300 mm
Keep H6 = 300 mm
Keep width, b6 = 300 mm
8# @160c/c
8# @300c/c
2.7m
8# @300c/c
12# @200c/c
12# @100c/c
10# @130c/c
8# @180c/c
12# @200c/c 8# @180c/c
1.5 =
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Design of Retaing wall
Input
Height above GL = 3 m
Density of soil, = 18 kN/m3
Denity of Concrete = 25 kN/m3
= 30
= 0.5 radians
= 0 = 0 radians
= 18
= 0.3 radians
Grade of Concrete, fck = 20 Mpa
Grade of Steel, fy = 415 Mpa
SBC of foundation strata, q0 = 100 kN/m2
Coefficient of friction, = 0.5
b5
b3
b6
H4
H3
H1
H2
b2 b4
b1
W1
W2
W3
W4
D E B C
A
TOE HEEL
SHEAR
KEY
STEM
H6
H5
H7
H8 H
PH
PV
H/3
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Solution
1 Design constants
Xu, max = 0.48
d
2 Depth of foundation
Density of soil, = 18 kN/m3
= 18000 N/m
3
ymin = (q0/) x ((1-sin)/(1+sin))2
= 0.61728395 m
Keep, H3 = 1 m
Height of wall above its base
Hence, H1 = 4 m
3 Dimensions of base
ka = cos (cos -sqrt(cos2
- cos2
))
(cos (cos +sqrt(cos2
- cos2
))
ka = 0.39480588
"D-E" Ratio =
b
= 1-(q0/(2.7 H1))
= 0.49
b1 = H1
b1
= 2.18384292 m
Base width from sliding condition
b1 = 0.7 H Ka/(1-)
b1 = 4.33512342 mBase width from from normal practice
b1 = 0.6 H or 0.4 H
b1 = 2.4 m
Keep, b1 = 2.4 m
Width of Toe slab, b1 = 1.2 m
Let the thickness of base, H4 = H/12
= 0.3 m
sqrt (Ka cos )
sqrt ((1-)(1+3))
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4 Thickness of stem
Height, H5 "A-B" = 3.7 m
Consider 1m length retaining wall
Earth pressure on stem is
p = Ka H52/2
= 48.6440328 kN acting at 18 to Horiz.
Hence horizontal earth pressure isPH = P cos
= 46.26 kN
M @ "B" = PH H1/3 N-mm
M @ "B" = 57.054 kN-m
= 57054000 N-mm
Mu = 85581000 N-mm
Shear force at "B" = PH
= 46.26 kN
FU = 69.39 kN
Hence effective depth, b3eff = Sqrt(Mu/Ru b)
= 169.522103 mm
Keep, b3eff = 240 mm
Total thickness, b3 = 300 mm
b3eff = 240 mm
Using 16 dia bars
b5 = 200 mm
b5eff = 140 mm
Width of Heel, b4 = 0.9 m
Shear stress, v = 1.5 x 46.26 x 1000/(1000 x 240)
= 0.289125 N/mm2
pt = 0.25 minimum
pt c
0.25 0.36
0.5 0.48
0.25 0.37
Corresponding c = 0.37 N/mm2
Hence Safe
5 Stability of wall
Height, H8 = H5 + b4 tan
Height, H8 = 3.99 m
Height, H = 4.29 m
Earth pressure, P = Ka H2/2
= 65.3944225 kN
Horizontal component
PH = P cos
= 62.1937917 kN
Vertical component
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PV = P sin
= 20.2079879 kN
Designation Force, kNLever arm,
m
Moment
@ toe,
kN-m
W1 18.5 1.4 25.9
W2 4.625 0.97 4.47083W3 18 1.2 21.6
W4 62.289 1.95 121.464
W5 = PV 20.207988 2.4 48.4992
Total W = 123.62199 MR = 221.90
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Total resisting moment MR = 221.9 kN-m
Check for overturning
Overturning moment, M0 = 88.9371221 kN-m
F.S. againt overturning = = 221.90
88.9371221
= 2.5 should be > 2Hence Safe
Check for Sliding
F.S. againt Sliding = W
P
= 1.0 should be > 1.5
Hence Unsafe, Provide shear key
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Pressure distribution
Net moment, M = 221.9 - 88.9371220988394
M = 132.96 kN-m
Distance X of the pont of application of the resultant, from toe
X = M
W
X = = 132.96
123.621988
X = = 1.08 m
Eccentricity, e = (b1/2) - X
e = 0.12 m Should be < b1/6
b/6 = 0.4 m
Hence No tension
Pressure p1 at toe = (W/b) x (1+(6e/b))
= 67.5 kN/m2 Should be less than SBC
= 100 kN/m2
Hence Safe
Pressure p2 at heel = (W/b) x (1-(6e/b))
= 35.4846746 kN/m2
Pressure at the junction of stem with toe slab,
p = 51.5 kN/m2
Pressure at the junction of stem with heel slab,
p = 47.4904216 kN/m2
b3b2
b1
D E B C
A
H7
p1
p2P
Ka H5
Ka H8
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6 Design of toe slab
Downward weight of slab per unit area = 7.5 kN/m2
Hence net pressure intensity under "D" = 67.5 - 7.5 = 60 kN/m2
Hence net pressure intensity under "E" = 51.5 - 7.5 = 44 kN/m2
Total force = Shear force at E = 62 kN
Fu = 93.6 kN
X from "E" = 0.63 mBending moment @ "E" = 39.36 kN-m
= 39360000 N-mm
Mu @ "E" = 59040000 N-mm
H4eff = Sqrt(Mu/Ru b)
= 140.80265 mm
Provide, H4 = 300 mm
H4eff = 240 mm
Shear stress, v = 93.6 x 1000/(1000 x 240)
= 0.39 N/mm2
pt = 0.3 minimum
pt c
0.25 0.36
0.5 0.48
0.3 0.394
Corresponding c = 0.394 N/mm2
Hence Safe
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Steel
Ast = ((0.5 x fck x b x d)/fy)x(1-SQRT(1-((4.6xMu)/(fckxbfxd^2)))
Ast = 727.437498 mm2
This reinforcement to be provided at bottom face.
If alternate bars of stem reinforcement are to be bent & continued in the toe slab,
area available = 1130.4 mm2 See step 8
Bar dia = 12 # bars
Spacing = 100 mm
Provide 12 # bars @ = 100 C/C at top of heel slab
Check for development length, Ld = 47
= 564 mm
Providing 50mm clear side cover ,
actual length available = 1150 mm
Hence safe
Distribution reinforcement = 0.12 x 1000 x A /100
= 324mm
2
Bar dia= = 8 # bars
Spacing = 155 mm
Provide 8 # bars @ = 150 C/C
7 Design of heel slab
1) Total weight of soil = 62.29 kN
Lever arm @ "B" = 0.45 m
2) Total weight of heel slab = 6.75 kN
Lever arm @ "B" = 0.45 m
3) Total force due to vertical
component of earth pressure = Ka (H5+H8) b4 tan sin
2
= 2.47 kNLever arm @ "B" = 0.46 m
4) Total upward soil reaction = 37.34 kN
Lever arm @ "B" = 0.43 m
Total force = S.F. at "B" = 34.17 kN
Fu = 51.26 kN
B.M. @ "B" = 16.200581 kN-m
= 16200581 N-mm
Mu = 24300871.5 N-mm
H7eff = Sqrt(Mu/Ru b)
= 90.3334459 mm
Provide, H6 = 350 mm
H7eff = 290 mm
Steel
Ast = ((0.5 x fck x b x d)/fy)x(1-SQRT(1-((4.6xMu)/(fckxbfxd^2)))
= 236.197916 mm2
Astmin = 360 mm2
Bar dia = 8 # bars
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Spacing = 140 mm
Provide 8 # bars @ 130 C/C at top of heel slab
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Development length, Ld = 47
to the left of "B" = 376 mm
= 400 mm
Distribution reinforcement = 0.12 x 1000 x A /100
= 384 mm2
Bar dia= = 8 # barsSpacing = 131 mm
Provide 8 # bars @ = 130 C/C
Shear stress, v = 1.5 x 34.17 x 1000/(1000 x 290)
= 0.17674138 N/mm2
pt Ast x 100
1000 x d
pt c = 0.12413793
0.15 0.18
0.15 0.18
0.12413793 #DIV/0!
Corresponding c = 0.18 N/mm
2
Hence Safe
8 Reinforcement in stem
Revised H5 = 3.65 m
Shear force at "B" = PH = Ka H52/2
= 48.6440328 kN
bending moment @ "B" = 59.9943071 kN-m
Mu = 89.9914607 kN-m
= 89991460.7 N-mm
b3eff = Sqrt(Mu/Ru b)
= 0.17383543 mm
Provide, b3eff = 260 mm
b3 = 320 mmb5 = 210 mm
b5eff = 150 mm
Ast = = ((0.5 x fck x b x d)/fy)x(1-SQRT(1-((4.6xMu)/(fckxbfxd^2)))
= 1046.53945 mm2
Bar dia= = 12 # bars
Spacing = 108 mm
Provide 12 # bars @ 100 C/C
Astact = 1130.4 mm2
Continue alternate bars in the toe slab to serve as tensile reinforcement there.
Discontinue the remaining half bars after a distance of
Ld = 47
= 564 mm
= 600 mm
beyond "B" in the toe slab.
=
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Between "A" & "B" some bars can be curtiled.
Consider a section at depth h below the top of the stem.
Effective depth at that section is
d' = 150 + 30.1369863 h mm (where h in m)
Now Ast (H3/d)
or H = (Ast d)1/3
Hence h/H5 = [(Ast' d')/(Ast d)]1/3
where Ast' = reinforcement at depth hAst = reinforcement at depth H5
d' = effective depth at depth h
d = effective depth at depth H5
If Ast' = Ast/2, Ast'/Ast = 1/2
Hence h/H5 = [d'/2d]1/3
Substituting d = 260
d' = 150 + 30.1369863 h
h = H5 [d'/2d]1/3
h = H5/(2)1/3
h = 2.89701 m
Bars should be extended by a distance of 12
= 144 mmor b3 = 260 mm
Hence h = 2.60 m
Curtail bars at this height below the top
Check for shear
Shear force = P
Ka H52/2 = 48.6440328 kN
Fu = 72.97 kN
Shear stress, v = 72.97 x 1000/(1000 x 260)
= 0.28065385 N/mm2
pt Ast x 100
1000 x d
pt c = 0.4030.25 0.36
0.5 0.48
0.40 0.44
Corresponding c = 0.44320728 N/mm2
Hence Safe
Distribition & temperature reinforcement
Average thk of stem = 265 mm
Distribution reinforcement = 0.12 x 1000 x A /100
= 318 mm2
Bar dia= = 8 # bars
Spacing = 158 mmProvide 8 # bars @ 150 C/C
at inner face of the wall, along its length.
For temperature reinforcement, provide 8 mm # bars @ 300 C/C both ways, in outer face.
=
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9 Design of shear key
Let depth of key = a
Intensity of passive pressure pp developed in front of the key depends on the soil
pp = Kp p = 130.443852 kN/m2
Total passive pressure pp = pp a = 130.443852 a
Dsliding force at level D1C1 = (1/3) x (/2) x (H+a)2
or pH = (1/3) x (/2) x (H+a)2
Weigth of soil between bottom of base and D1C1 = b a x
= 43.2 a
W = 103.414 + 43.2 a
F.S. against sliding = 1.5
W + pp
pH
1.5 = 1.5009279
a = 0.1765 m..Adjust this value till above value = 1.5
However provide minimum
a = 0.3 m
a = H6 = 300 mm
Keep H6 = 300 mm
Keep width, b6 = 300 mm
8# @150c/c
8# @300c/c
2.6m
8# @300c/c12# @200c/c
12# @100c/c
8# @130c/c
8# @150c/c
12# @100c/c 8# @130c/c
1.5 =
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M-15 M-20 M-25 M-30 M-35 M-40
0.18 0.18 0.19 0.2 0.2 0.2
0.22 0.22 0.23 0.23 0.23 0.23
0.29 0.30 0.31 0.31 0.31 0.32
0.34 0.35 0.36 0.37 0.37 0.380.37 0.39 0.40 0.41 0.42 0.42
0.40 0.42 0.44 0.45 0.45 0.46
0.42 0.45 0.46 0.48 0.49 0.49
0.44 0.47 0.49 0.50 0.52 0.52
0.44 0.49 0.51 0.53 0.54 0.55
0.44 0.51 0.53 0.55 0.56 0.57
0.44 0.51 0.55 0.57 0.58 0.60
0.44 0.51 0.56 0.58 0.60 0.62
0.44 0.51 0.57 0.6 0.62 0.63
Refer IS 456-2000
fy Xu, max
d
250 0.53
415 0.48
500 0.46
2.00
2.25
2.50
2.75
3.00 and above
1.75
Permissible shear stress Table v in concrete (IS : 456-2000)
100As Permissible shear stress in concrete v N/mm2
bd
< 0.15
0.25
0.50
0.751.00
1.25
1.50