Design of Retaing wall Input Height above GL
Density of soil, γ
= =
Denity of Concrete
=
φ˚
= = = = = = = =
30 0 .5 0 0 18 0 .3 20 415
=
100 kN/m2 0 .5
α α Grade of Concrete, fck Grade of Steel, fy SBC of foundation strata, q 0 Coefficient of friction, μ
3 m 18 kN/m3 25 kN/m3
=
˚ radians ˚ radians ˚ radians Mpa Mpa
b5 A
STEM
H2 H5 H1
W1
W4
W2 b3
b2
b4
H3 D
H4
B
E
TOE
C HEEL
W3 H6
SHEAR KEY
b6 b1
H7
Solution 1 Design Design constants constants Xu, max d
=
0 .4 8
2 Depth of founda foundation tion
Density of soil, γ
18 kN/m3 18000 N/m3
= =
ymin Keep, H3 Height of wall above its base Hence, H1 3 Dimens Dimension ionss of base "D-E" Ratio b
= = =
(q0/γ) x ((1-sinφ)/(1+sinφ)) ((1-sinφ)/(1+sinφ)) 0 .6 1 7 2 8 m 1 m
=
2
4 m
=
α
α
=
1-(q0/(2.2 γH1))
α
=
0 .3 7
ka
=
ka
=
b1
=
b1 Base width from sliding condition
=
b1
=
b1 Base width from from normal practice b1 b1 Keep, b1
=
2 .9 6 2 9 6 m
= = =
0.6 H or 0.4 H 2 .4 m 2.4 m
Width of Toe slab, α b1
=
0.9 m
Let the thickness of base, H4
= =
H/12 0 .3 m
(1-sinφ)/(1+sinφ) 0 .3 3 3 3 3 0.95 H Sqrt{(K a/((1-α)(1+α))} 1 .9 0 2 8 8 m 0.7 H Ka/(1-α)μ
4 Thickness of stem Height, H5
"A-B"
=
3.7 m
Consider 1m length retaining wall Maximum Bending moment @ "B" =
3
= = =
Ka γ H5 /6 50.653 kN-m 5.1E+07 N-mm 7.6E+07 N-mm
=
Sqrt(Mu/Ru b)
=
159.73 mm
Keep, b3eff
=
240 mm
Total thickness, b3 Using 16 dia bars b5
=
300 mm
=
200 mm
b5eff
=
140 mm
M @ "B" Mu Hence effective depth, b3eff
Width of Heel, b4
=
1.2 m
5 Stability of wall
W1
18.5
1.1
Moment @ toe, kN-m 20.35
W2
4.625
0.97
4.47083
W3
18
1.2
21.6
W4
79.92
1.8
143.856
Designation
Total
Force, kN
∑W =
Lever arm, m
MR =
121.045
Total resisting moment MR Earth pressure, P
= =
190.3
= Check for overturning Overturning moment, M0
=
64
F.S. againt overturning
= =
Check for Sliding F.S. againt Sliding
= = Hence
kN-m
2
Ka γ H1 /2 48
=
190.30
kN
kN-m
190.30 64 3.0 should be > 2 Hence Safe
μ ∑W P 1.3 should be > 1.5 Unsafe, Provide shear key
Pressure distribution Net moment, ∑M ∑M
= =
190.3 - 64 126.30 kN-m
Distance X of the pont of application of the resultant, from toe X
=
X=
=
X=
=
∑M ∑W 126.30 121.045 1.04 m
= = = Hence
(b1/2) - X 0.16 m 0.4 m No tension
Eccentricity, e e b/6
Pressure p1 at toe
= =
(∑W/b) x (1+(6e/b)) 70.2 kN/m2 = Hence
Pressure p2 at heel
Should be < b1/6
Should be less than SBC 100 kN/m2
Safe
=
(∑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
A
b3
b2 D
b4 B
E
C
H7 p2
p1 b1
P
6 Design of toe slab
Downward weight of slab per unit area
=
Hence net pressure intensity under "D" =
70.2 - 7.5
=
Hence net pressure intensity under "E" =
55.4 - 7.5
=
Total force = Shear force at E X from "E" Bending moment @ "E" Mu @ "E" H4eff Provide, H4 H4eff
= = = = =
0.47 m 23.3955 kN-m 2.3E+07 N-mm 3.5E+07 N-mm
=
Sqrt(Mu/Ru b)
= =
108.555 mm 260 mm
=
200 mm
7.5 kN/m2 62.7 kN/m2 47.9 kN/m2 50 kN
Steel Ast Ast
=
((0.5 x fck x b x d)/fy)x(1-SQRT(1-((4.6xMu)/(fckxbfxd^2))) = 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, 565.2 mm2
area available
=
Bar dia Spacing Provide 12 # bars @
= = =
12 # bars 200 mm 200 C/C at top of heel slab
Check for development length, Ld
=
47 ø
=
564 mm
Providing 50mm clear side cover , actual length available
=
850 mm Hence safe
Distribution reinforcement
=
0.12 x 1000 x A /100
= Bar dia= Spacing Provide 8 # bars @
= = =
7 Design of heel slab 1) Total weight of soil Lever arm @ "B" 2) Total weight of heel slab Lever arm @ "B" 3) Total upward soil reaction Lever arm @ "B" Total force = S.F. at "B" B.M. @ "B"
Mu H7eff
=
Provide, H6 H7eff Steel Ast
= = = = = = = = = =
=
276 mm2 8 # bars 182 mm 180 C/C
79.92 0.60 7.80 0.60 48.68 0.55 39.04 25.793 2.6E+07 3.9E+07
kN m kN m kN m kN kN-m N-mm N-mm
Sqrt(Mu/Ru b)
= =
113.981 260
mm mm
=
200
mm
((0.5 x fck x b x d)/fy)x(1-SQRT(1-((4.6xMu)/(fckxbfxd^2))) =
Bar dia Spacing Provide 10 # bars @
See step 8
569.737 mm2
= 10 # bars = 138 mm 130 C/C at top of heel slab
Development length, Ld
=
47 ø
to the left of "B"
= =
470 500
Distribution reinforcement
= =
mm mm
0.12 x 1000 x A /100 276
2
mm 8 # bars 182 mm 180 C/C
Bar dia= Spacing Provide 8 # bars @
= = =
Shear stress, τv
=
1.5 x 39.04 x 1000/(1000 x 200)
=
0.2928 N/mm2 Ast x 100 1000 x d 0.28487
pt pt
= τc
=
0.25 0.36 0.5 0.48 0.28486828 0.38674 Corresponding τc
= Hence
0.38674 N/mm2 Safe
8 Reinforcement in stem
Revised H5
=
3.74 m
M
= = = =
Ka γ H5 /6 52.3136 kN-m 5.2E+07 N-mm 7.8E+07 N-mm
=
Sqrt(Mu/Ru b)
=
162.327 mm
Provide, b3eff
=
250 mm
b3 b5 b5eff
= =
310 mm 200 mm
=
140 mm
Mu b3eff
Ast =
= =
Bar dia= Spacing Provide 12 # bars @ Astact
3
((0.5 x fck x b x d)/fy)x(1-SQRT(1-((4.6xMu)/(fckxbfxd^2))) 943.711939 mm2 = =
12 # bars 120 mm
100 C/C
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 ø =
= =
564 mm 600 mm beyond "B" in the toe slab.
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)
3
Now
Ast α (H1 /d)
or
H1 = (Ast d)
Hence
h/H1 = [(Ast' d')/(Ast d)] Ast' = reinforcement at depth h Ast = reinforcement at depth H1
1/3 1/3
where
d' = effective depth at depth h d = effective depth at depth H 1 If Ast' = Ast/2,
Ast'/Ast = 1/2 1/3
Hence Substituting d' =
h/H1 = [d'/2d] d= 250 140 + 29.4117647
h
1/3
h = H1 [d'/2d]
h = H1/(2) 1/3 h = 2.96844 m Bars should be extended by a distance of = = =
or b3 Hence h
12 ø 144 mm 250 mm 2.70 m Curtail bars at this height below the top
Check for shear Shear force = P 2
Ka γ H5 /2 Fu Shear stress, τv pt pt
=
41.9628
kN
=
62.9442
kN
=
62.9442 x 1000/(1000 x 250)
=
0.25178 N/mm2 Ast x 100 1000 x d 0.37748
= τc
0.25 0.36 0.5 0.48 0.37748478 0.43119 Corresponding τc
=
0.43119 N/mm2 Hence Safe
=
Distribition & temperature reinforcement Average thk of stem = Distribution reinforcement = =
255 mm 0.12 x 1000 x A /100 306
2
mm 8 # bars 164 mm
Bar dia= = Spacing = 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.
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 =
Dsliding force at level D1C1
=
(1/3) x (γ/2) x (H+a)
=
(1/3) x (γ/2) x (H+a)
pH
or
166.2
2 2
Weigth of soil between bottom of base and D1C1 = ∑W =
121.045
F.S. against sliding =
1.5 =
+
a
43.2
baxγ = a
43.2 a
1.5
μ ∑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
Design of Retaing wall Input Height above GL
= =
Density of soil, γ Denity of Concrete φ˚
3 m 18 kN/m3 25 kN/m3 30 ˚ 0.5 radians 0 ˚ 0 radians 18 ˚ 0.3 radians 20 Mpa 415 Mpa
= = = = = = = = =
β β Grade of Concrete, fck Grade of Steel, fy SBC of foundation strata, q 0
100 kN/m2 0.5
=
Coefficient of friction, μ
=
b5
β A
STEM
H2
PV
H5 H1
W1
PH
W4
W2
H/3 b3
b2
b4
H3 D
H4
B
E
TOE
C HEEL
W3 H6
SHEAR KEY
b6 b1
H7
H8
H
Solution 1 Design constants Xu, max d
=
0.48
2 Depth of foundation
Density of soil, γ
18 kN/m3 18000 N/m3
= =
ymin
=
Keep, H3 Height of wall above its base Hence, H1
= =
(q0/γ) x ((1-sinφ)/(1+sinφ)) 0.61728395 m 1 m
=
2
4 m
3 Dimensions of base
ka
=
2
cos β
2
(cos β -sqrt(cos β- cos φ))
(cos (cos β +sqrt(cos 2 β- cos2 φ)) ka
= "D-E" b
Ratio
0.39480588
=
α
α
=
1-(q0/(2.7 γH1))
α
=
b1 b1
=
0.49 H1
=
2.18384292
b1
=
0.7 H Ka/(1-α)μ
b1 Base width from from normal practice b1 b1 Keep, b1
=
4.33512342 m
= = =
0.6 H or 0.4 H 2.4 m 2.4 m
Width of Toe slab, α b1
=
1.2 m
Let the thickness of base, H4
= =
H/12 0.3 m
Base width from sliding condition
sqrt (Ka cos β) sqrt ((1-α)(1+3α)) m
4 Thickness of stem Height, H5
"A-B"
=
3.7 m
Consider 1m length retaining wall Earth pressure on stem is p
=
Hence horizontal earth pressure is PH
2
=
Ka γ H5 /2 48.6440328 kN
=
P cos β
= M @ "B"
=
M @ "B" Mu
= = =
Shear force at "B"
=
46.26 kN PH H1/3
N-mm
57.054 kN-m 57054000 N-mm 85581000 N-mm PH
=
46.26 kN
FU
=
69.39 kN
Hence effective depth, b3 eff
=
Sqrt(Mu/Ru b)
=
169.522103 mm
Keep, b3 eff
=
240 mm
Total thickness, b3
=
300 mm
b3eff
=
240 mm
Using 16 dia bars b5 b5eff
=
200 mm
=
140 mm
Width of Heel, b4
=
Shear stress, τ v
= =
pt pt
acting at 18˚ to Horiz.
0.9 m
1.5 x 46.26 x 1000/(1000 x 240) 0.289125
2
=
N/mm 0.25 minimum
=
0.37 N/mm2
τc 0.25 0.5 0.25
0.36 0.48 0.37
Corresponding τc
Hence
Safe
5 Stability of wall
Height, H8 Height, H8 Height, H
= = =
H5 + b4 tan β 3.99 m 4.29 m
Earth pressure, P
= =
Ka γ H /2 65.3944225 kN
=
P cos β
Horizontal component PH
= Vertical component
2
62.1937917 kN
PV
= =
P sin β 20.2079879 kN
W1
18.5
1.4
Moment @ toe, kN-m 25.9
W2
4.625
0.97
4.47083
W3
18
1.2
21.6
W4
62.289
1.95
121.464
20.207988
2.4
48.4992
Designation
W5 = PV Total
Force, kN
∑W = 123.62199
Lever arm, m
MR =
221.90
Total resisting moment M R
=
221.9
kN-m
Check for overturning Overturning moment, M 0
=
88.9371221
kN-m
F.S. againt overturning
=
=
=
Check for Sliding F.S. againt Sliding
= = Hence
221.90 88.9371221 2.5 should be > 2 Hence Safe
μ ∑W P 1.0 should be > 1.5 Unsafe, Provide shear key
Pressure distribution Net moment, ∑M ∑M
= =
221.9 - 88.9371220988394 132.96 kN-m
Distance X of the pont of application of the resultant, from toe X
=
X=
=
X=
=
∑M ∑W 132.96 123.621988 1.08 m
= = = Hence
(b1/2) - X 0.12 m 0.4 m No tension
Eccentricity, e e b/6
Pressure p 1 at toe
=
Should be < b1/6
(∑W/b) x (1+(6e/b)) 67.5 kN/m2
=
= Hence Pressure p 2 at heel
=
Should be less than SBC 100 kN/m2
Safe
(∑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
=
β A
D
Ka γ H8
Ka γ H5
b3
b2
β B
E
β C
H7 p2
p1 b1
P
6 Design of toe slab
Hence net pressure intensity under "D" =
67.5 - 7.5
=
7.5 kN/m2 60 kN/m2
Hence net pressure intensity under "E" = Total force = Shear force at E Fu
51.5 - 7.5
= = =
44 kN/m2 62 kN 93.6 kN
Downward weight of slab per unit area
X from "E" Bending moment @ "E" Mu @ "E" H4eff
=
= = = = =
Provide, H4 H4eff
Sqrt(Mu/Ru b)
= =
140.80265 mm 300 mm
=
240 mm
Shear stress, τ v
=
pt
= =
pt
0.63 m 39.36 kN-m 39360000 N-mm 59040000 N-mm
93.6 x 1000/(1000 x 240) 0.39 N/mm2 0.3 minimum
τc 0.25 0.5 0.3
Corresponding τc
0.36 0.48 0.394 0.394 N/mm2
= Hence
Safe
Steel Ast Ast
=
((0.5 x fck x b x d)/fy)x(1-SQRT(1-((4.6xMu)/(fckxbfxd^2))) = 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 Spacing Provide 12 # bars @
= = =
12 # bars 100 mm 100 C/C at top of heel slab
Check for development length, L d
=
47 ø
=
564 mm
Providing 50mm clear side cover , actual length available
= Hence
Distribution reinforcement
= =
Bar dia= Spacing Provide 8 # bars @
= = =
7 Design of heel slab 1) Total weight of soil Lever arm @ "B" 2) Total weight of heel slab Lever arm @ "B" 3) Total force due to vertical component of earth pressure
Lever arm @ "B" 4) Total upward soil reaction Lever arm @ "B" Total force = S.F. at "B" Fu B.M. @ "B" Mu H7eff Provide, H6 H7eff Steel Ast
=
= = = = =
1150 mm safe
0.12 x 1000 x A /100 324 mm2 8 # bars 155 mm 150 C/C
62.29 0.45 6.75 0.45
kN m kN m
= = = = = = = = =
Ka γ (H5+H8) b4 tan β sin β 2 2.47 kN 0.46 m 37.34 kN 0.43 m 34.17 kN 51.26 kN 16.200581 kN-m 16200581 N-mm 24300871.5 N-mm
=
Sqrt(Mu/Ru b)
= =
90.3334459 350
mm mm
=
290
mm
((0.5 x fck x b x d)/fy)x(1-SQRT(1-((4.6xMu)/(fckxbfxd^2))) =
Astmin
=
Bar dia
=
236.197916 mm2 360
2
mm
8 # bars
Spacing Provide 8 # bars @
= 140 mm 130 C/C at top of heel slab
Development length, L d
=
47 ø
to the left of "B"
= =
376 400
Distribution reinforcement
=
Bar dia= Spacing Provide 8 # bars @
= = = =
Shear stress, τ v
= =
pt pt
= τc
=
mm mm
0.12 x 1000 x A /100 384
2
mm 8 # bars 131 mm 130 C/C
1.5 x 34.17 x 1000/(1000 x 290) 0.17674138 N/mm2 Ast x 100 1000 x d 0.12413793
0.15 0.18 0.15 0.18 0.12413793 #DIV/0! Corresponding τc
= Hence
0.18 N/mm2 Safe
8 Reinforcement in stem
Revised H5
=
3.65 m 2
Shear force at "B" = PH
=
bending moment @ "B" Mu
= = = =
Ka γ H5 /2 48.6440328 59.9943071 89.9914607 89991460.7
b3eff
=
Sqrt(Mu/Ru b)
kN kN-m kN-m N-mm
=
0.17383543 mm
Provide, b3eff
=
260 mm
b3 b5
= =
320 mm 210 mm
b5eff
=
150 mm
Ast =
= =
Bar dia= Spacing Provide 12 # bars @ Astact
((0.5 x fck x b x d)/fy)x(1-SQRT(1-((4.6xMu)/(fckxbfxd^2))) 1046.53945 mm2 = =
12 # bars 108 mm
100 C/C
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 ø =
= =
564 mm 600 mm beyond "B" in the toe slab.
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 3 Now Ast α (H /d) or
H = (Ast d)
mm (where h in m)
1/3 1/3
Hence
h/H5 = [(Ast' d')/(Ast d)] where Ast' = reinforcement at depth h Ast = reinforcement at depth H5 d' = effective depth at depth h d = effective depth at depth H5 If Ast' = Ast/2, Ast'/Ast = 1/2 1/3
Hence Substituting d' =
h/H5 = [d'/2d] d= 260 150 + 30.1369863
h
1/3
h = H5 [d'/2d]
h = H5/(2)1/3 h = 2.89701 m Bars should be extended by a distance of = = =
or b3 Hence h
12 ø 144 260
mm mm 2.60 m Curtail bars at this height below the top
Check for shear Shear force = P 2
Ka γ H5 /2 Fu Shear stress, τ v
=
48.6440328
kN
=
72.97
kN
= =
pt
=
pt
τc 0.25 0.5 0.40
Corresponding τc
=
72.97 x 1000/(1000 x 260) 0.28065385 N/mm2 Ast x 100 1000 x d 0.403
0.36 0.48 0.44 0.44320728 N/mm2 Hence Safe
=
Distribition & temperature reinforcement Average thk of stem = Distribution reinforcement = =
265 mm 0.12 x 1000 x A /100 318
2
mm 8 # bars 158 mm
Bar dia= = Spacing = Provide 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.
9 Design of shear key Let depth of key =
a Intensity of passive pressure p p developed in front of the key depends on the soil pp =
Kp p
=
130.443852 kN/m2
Total passive pressure p p =
pp a =
Dsliding force at level D 1C1
=
(1/3) x (γ/2) x (H+a)
2
=
(1/3) x (γ/2) x (H+a)
2
pH
or
130.443852 a
Weigth of soil between bottom of base and D 1C1 = ∑W =
103.414
F.S. against sliding =
1.5 =
+
baxγ = a
43.2 1.5
μ ∑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/c 12# @200c/c
12# @100c/c 8# @130c/c 8# @150c/c
12# @100c/c
8# @130c/c
43.2 a
Permissible shear stress Table
< 0.15
0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 and above
Refer IS 456-2000 fy Xu, max d 250 0.53 415 0.48 500 0.46
in concrete (IS : 456-2000)
Permissible shear stress in concrete
100As
bd
v
M-15 0.18 0.22 0.29 0.34 0.37 0.40 0.42 0.44 0.44 0.44 0.44 0.44 0.44
M-20 0.18 0.22 0.30 0.35 0.39 0.42 0.45 0.47 0.49 0.51 0.51 0.51 0.51
M-25 0.19 0.23 0.31 0.36 0.40 0.44 0.46 0.49 0.51 0.53 0.55 0.56 0.57
M-30 0.2 0.23 0.31 0.37 0.41 0.45 0.48 0.50 0.53 0.55 0.57 0.58 0.6
τv
N/mm2
M-35 0.2 0.23 0.31 0.37 0.42 0.45 0.49 0.52 0.54 0.56 0.58 0.60 0.62
M-40 0.2 0.23 0.32 0.38 0.42 0.46 0.49 0.52 0.55 0.57 0.60 0.62 0.63