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Design Code BS 8110
ANALYSIS FOR RETAINING WALL 1. DESIGN DATA. Unit weight of soil
=
=
19.0 kN/m3
Angle of internal friction of soil
=
=
32.0
Safe bearing capacity of soil
=
qb
=
230.0 kN/m2
Surcharge load
=
Qs
=
10.0 kN/m2
Co-efficient of friction
=
=
0.39
Total height of retaining wall
=
H
=
3.80 m
Depth of base slab below ground level
=
D
=
1.2 m
Width of base slab
=
B
=
3.0 m
Width of heel
=
C
=
2.40 m
Thickness of stem
=
t
=
0.40 m
Thickness of base slab
=
=
0.45 m
Grade of concrete
=
h fcu
=
30 N/mm2
Grade of reinforcement
=
fy
=
420 N/mm2
Density of concrete
=
c
=
25 kN/m3
Elastic modulus of concrete
=
Ec
=
2.60E+04 N/mm2
2. RETAINING WALL ARRANGEMENT +29.5
W1
2.6
W3
3.80
+26.9
0.40
0.20
2.40
1.2
W2 22.4 kN/m2
T 3.0 A
B
Earth pressure
C
3.10 Surcharge pressure
D
29.7
kN/m2
100.73 118.5 kN/m2
BASE PRESSURE DISTRIBUTION
3. CALCULATION FOR LATERAL FORCES / MOMENTS Ka Co-efficient of active earth pressure = Lateral pressure due to earth Lateral pressure due to surcharge
Retaining wall Prepd. By: Shanti Srinivasan email:
[email protected]
= =
=
p1
=
p1
=
p2
0.31 Kas H
(DBM-3.4.6.2)
kN/m2
=
22.4 Ka Qs
=
3.10
kN/m2
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Design Code BS 8110
Horizontal force due to earth pressure
=
P1
=
42.53
kN
Horizontal force due to surcharge
=
P2
=
11.78
kN
Total horizontal force
=
=
54.31
kN
Moment of earth pressure about toe
=
P M1
=
53.87
kN.m
Moment of surcharge pressure about toe
=
M2
=
22.38
kN.m
Total induced moment at base
=
M
=
76.25
kN.m
4. RESTORING FORCES / MOMENT CALCULATION SL.No
Load due to
Magnitude (kN)
Dist. of CG
Moment about 'T'
1
W1
33.5
0.40
13.4 kN.m
2
W2
33.8
1.50
50.6 kN.m
3
W3
155.0
279.1 kN.m
W
222.3
1.80 MR
Total
343.1 kN.m
5. CHECK FOR SLIDING Total horizontal force
=
Total vertical restoring force
=
Horizontal frictional resistance
=
Restoring force
=
P WR
=
54.31
kN
=
222.29
kN
WR WHR
=
86.69
kN
=
86.69
kN
(Passive resistance of soil above Toe slab is not considered conservatively) Factor of safety against sliding = WHR/P = 1.60 >
1.50
OK
=
76.2
kN.m
=
343.1
kN.m
=
4.50
>
1.80
6. CHECK FOR OVERTURNING Total induced moment at base
=
Total restoring moment at base
=
M MR
Factor of safety against overturning
OK
7. CHECK FOR BEARING PRESSURE Distance of the point of application of the resultant force from "A"
=
z
B/2
=
=
e
SUM (V)
=
1.20
=
1.50
B/6 The eccentricity
MR-M m
0.50 =
B/2 - z
=
0.30
m
< B/6 No Tension Gross bearing capacity of soil at fdn. Level Therefore, the extreme pressure Maximum soil pressure Minimum soil pressure
=
qs(gross)
=
=
f
=
=
f1
=
118.49
kN/m2
OK
=
f2
=
29.71
kN/m
OK
=
axmax
=
P1H3/15EcI
(Table-26
=
P2H /8EcI
Reynolds
252.8
kN/m2
(W) {1 + (6e/B)} B 2
8. CHECK FOR DEFLECTION (STEM) Deflection due earth pressure Deflection due surcharge pressure Retaining wall Prepd. By: Shanti Srinivasan email:
[email protected]
axmax
3
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Moment of Inertia for 1m wide wall
Retaining wall Prepd. By: Shanti Srinivasan email:
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Design Code BS 8110
=
I
=
5.3E+09
mm4
Handbook)
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Design Code BS 8110
P1
=
42525.8
N
=
P2
=
11780.0
N
Horizontal force due to earth pressure
=
Horizontal force due to surcharge Horizontal deflection due to P1
=
ax1
=
1.12
mm
Horizontal deflection due to P2
=
ax2
=
0.58
mm
Total Horizontal maximum deflection
=
axmax
=
1.70
mm
<
15.20
mm
H/250
9. CALCULATION OF DESIGN FORCES / MOMENTS 9.1 FOR VERTICAL STEM =
M1
=
53.9
kN.m
BM due to Surcharge pressure (p2)
=
M2
=
22.4
kN.m
Total Maximum moment
=
Ms
=
kN.m
Total Maximum shear force
=
Fs
=
76.2 P1+P2
=
54.3
kN
=
W3 x C / 2
=
186.0 Qs x C2/2
BM due to Earth pressure (p1)
9.2 FOR HEEL SLAB Moment due soil over heel slab
=
Mh1
Moment due to Surcharge pressure
=
Mh2
= =
Moment due to selfweight of toe slab
=
Mh3
=
28.8 kN.m (C x h x c)C/2
=
32.4
153.7 kN.m Mh1+Mh2+Mh3-Mh4
=
Mh4
=
Net moment for Heel slab
=
Mh
=
Shear due to soil over heel slab
=
Fh1
Moment due to upward soil pressure
= = = =
Fh2
Shear due to selfweight
=
Fh3
Shear due to upward earth pressure
=
Fh4
Total design SF for Heel slab
=
Fh
Shear due to surcharge
= =
10. ABSTRACT DESIGN VALUES (SERVICE) Thickness (mm) Component
kN.m
93.5 W3
kN.m
155.0 P2
kN
=
11.8 kN (C x E x c)
=
27.0
= =
156.5 kN Fh1+Fh2+Fh3-Fh4
=
37.3
kN
kN
Max.BM (kNm) Max.SF (kN)
Stem
400
76.2
54.3
Heel slab
450
93.5
37.3
Retaining wall Prepd. By: Shanti Srinivasan email:
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kN.m
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Design Code BS 8110
I. RETAININGWALL -BASESLAB - DESIGN FOR MAIN REINFORCEMENT 1. DESIGN DATA. Ms Design service moment = = 93.5 kN.m Design ultimate moment = M = 149.6 kN.m Design ultimate shear force = V = 37.3 kN Assumed width of slab = B = 1000.0 mm Thickness of slab = D = 450.0 mm f Grade of concrete = = 30 N/mm2 cu fy Grade of reinforcement = = 420 N/mm2 Ec Elastic modulus of concrete = = 2.6E+04 N/mm2 Es Elastic modulus of reinforcement = = 2.0E+05 N/mm2 Clear cover to main reinforcement = c = 70 mm Design allowable crack width = w = 0.30 mm 2.0 DESIGN 2.1 DESIGN FOR FLEXURE. Effective depth
=
d
k z z - Max. (As)req. (As)Prov.
=
Safety index
=
2.2 CHECK FOR SHEAR Design shear stress
=
0.8[Sqrt(fcu.)] 100As / bd (400/d) Therefore 400/d Design concrete shear stress
=
2.3 CHECK FOR CRACK WIDTH Rebar spacing Steel ratio
= =
Diameter Spacing 1340 1061
= = = = = = = = = = = =
D-c-(Dia/2) 372 mm 2 M / bd fcu
=
1.26
0.036 d [ 0.5 +( 0.25 - k/0.9)0.5] 356 mm 353 mm M / 0.95 fy z 1061 16 150 1340
mm2 mm mm mm2 > 1
= V / bvd = 0.10 N/mm2 = 4.4 N/mm2 < 0.8[sqrt(fcu.)] & 5 N/mm2 = 0.360 % = 1.08 = 1.08 vc = 0.49 N/mm2 > Design shear stress. Hence no shear reinforcement required
OK
v
S
= = =
150 mm (As)Prov. / bd 0.0036
Base slab of Retaining wall by: Shanti Srinivasan email:
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OK
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Design Code BS 8110
=
Modular ratio
e
= =
e * X /d Depth of neutral axis Lever arm
= =
X Z
Concrete compressive stress
=
fcb
Tensile stress on steel
=
2.3.1 Check stress levels Allowable comp. stress in concrete
Allowable tensile stress in steel
Elastic strain at surface
=
=
=
fs
fcb'
fs
'
E1
= = = = = = = = = = = = = > = = > = =
Stiffening effect of concrete (For bending only)
= = = = = =
Em Acr
Crack width
=
Es / (Ec/2) 2.0E+05 1.30E+04 15.38 0.055 e*{[(1+ (2 /e* - 1} 0.282 104.9 mm d - X/3 337.0 mm 2Ms / Z b X 5.3 Ms / Z As
N/mm2
207.0
N/mm2
0.45 fcu 13.5 fcb 0.8 fy
N/mm2
336.0 fs
N/mm2
OK
OK
(h-x) * fs (d-x) * Es 1.34E-03 bt * (h-x) * (a' - x) 3*Es * As * (d-x) 5.54E-04 E1 - Stiffening effect of concrete 7.828E-04 [(S/2)2+(c+ Dia./2)2]0.5 - Dia /2 100.2 mm 3 * Acr * Em 1+2[(Acr-Cmin.) / (h-x)]
= <
0.200 mm Allowable crack width.
Diameter Spacing
= = = =
585 12 150 754
mm2/m mm mm mm2
754 585
=
1.29
> 1
OK
2.4 DESIGN FOR DISTRIBUTION REINFORCEMENT Min. % of steel required (As)Prov.
Safety index
= = =
=
Astmin
Base slab of Retaining wall by: Shanti Srinivasan email:
[email protected]
OK
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Design Code BS 8110
I. RETAININGWALL -STEM - DESIGN FOR MAIN REINFORCEMENT 1.0 DESIGN DATA. Ms Design service moment = = 76.2 kN.m Design ultimate moment = M = 121.9 kN.m Design ultimate shear force = V = 54.3 kN Assumed width of slab = B = 1000.0 mm Thickness of slab = D = 400.0 mm f Grade of concrete = = 30 N/mm2 cu fy Grade of reinforcement = = 420 N/mm2 Ec Elastic modulus of concrete = = 2.6E+04 N/mm2 Es Elastic modulus of reinforcement = = 2.0E+05 N/mm2 Clear cover to main reinforcement = c = 70 mm Design allowable crack width = w = 0.30 mm 2.0 DESIGN 2.1-DESIGN FOR FLEXURE Effective depth
=
d
k z z - Max. (As)req. (As)Prov.
=
Safety index
=
2.2-CHECK FOR SHEAR Design shear stress
=
0.8[Sqrt(fcu.)] 100As / bd (400/d) Therefore 400/d Design concrete shear stress
=
2.3. CHECK FOR CRACK WIDTH Rebar spacing
=
Diameter Spacing 1340 999
= = = = = = = = = = = =
D-c-(Dia/2) 322 mm 2 M / bd fcu
=
1.34
0.039 d [ 0.5 +( 0.25 - k/0.9)0.5 ] 307 mm 306 mm M / 0.95 fy z 999 16 150 1340
mm2 mm mm mm2 > 1
= V / bvd = 0.17 N/mm2 = 4.4 N/mm2 < 0.8[sqrt(fcu.)] & 5 N/mm2 = 0.416 % = 1.24 = 1.24 vc = 0.53 N/mm2 > Design shear stress. Hence no shear reinforcement required
OK
v
S
=
150
Stem of Retaining wall by: Shanti Srinivasan email:
[email protected]
mm
OK
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Design Code BS 8110
Steel ratio Modular ratio
=
=
e
= = = = = = = = = = = = = = =
e * X /d Depth of neutral axis Lever arm
= =
X Z
Concrete compressive stress
=
fcb
Tensile stress on steel
=
2.3.1 Check stress levels:Allowable comp. stress in concrete
Allowable tensile stress in steel
Elastic strain at surface
=
=
=
fs
fcb'
fs
= = > = = >
'
E1
= =
Stiffening effect of concrete (For bending only)
= = = = = =
Em Acr
Crack width
= = <
(As)Prov. / bd 0.0042 Es / (Ec/2) 2.0E+05 1.30E+04 15.38 0.064 e*{[(1+ (2 /e* - 1} 0.300 96.4 mm d - X/3 289.9 mm 2Ms / Z b X 5.5 Ms / Z As
N/mm2
196.1
N/mm2
0.45 fcu 13.5 fcb 0.8 fy
N/mm2
336.0 fs
N/mm2
OK
OK
(h-x) * fs (d-x) * Es 1.32E-03 bt * (h-x) * (a' - x) 3*Es * As * (d-x) 5.08E-04 E1 - Stiffening effect of concrete 8.118E-04 [(S/2)2+(c+ Dia./2)2]0.5 - Dia /2 100.2 mm 3 * Acr * Em 1+2[(Acr-Cmin.) / (h-x)]
0.204 mm Allowable crack width.
2.4. DESIGN FOR DISTRIBUTION REINFORCEMENT Min. % of steel required (As)Prov.
= = =
Astmin Diameter Spacing
= = =
520 12 150
Stem of Retaining wall by: Shanti Srinivasan email:
[email protected]
mm2/m mm mm
OK
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Safety index
Design Code BS 8110
=
754 520
=
754
=
1.45
Stem of Retaining wall by: Shanti Srinivasan email:
[email protected]
mm2 > 1
OK
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Design Code BS8110
33400mm A C A New Retaining wall (Existing bund wall to be demolished)
Access stair over the existing bund PLAN Existing Bund (To be retained) D SECTION A-A 0.40
+29.5
3.80 +26.9 1.2
3
REINFORCEMENT DETAILS
16
@
150
12
@
12
@
150 (Both faces) 150 Earth side
12
@
150 (Top & Bottom)
By: Shanti Srinivasan emailto:
[email protected]
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Design Code BS8110
16
@ 150 (Top & Bottom)
By: Shanti Srinivasan emailto:
[email protected]
