rc element structures
TRANSCRIPT
IS 3370 (Part II) 1984
Determine the suitable dimension of the cantilever retaining wall
which is required to support the bank of earth 4m height above. G.L. on
the toe side of the wall considered the back fill surface to the inclined at
an angle of 15o Assume good soil for foundation at a depth of 1.25m
below G.L SBC – 160 KN/m2. Further assume the back fill to comprise of
granular soil with unit wt 16KN/m3 and an ansle of shearing resistance of
30o. Assume co-efficient of friction b/w. soil and concrete is 0.5. Pa is the
active earth pressure exerted by the retain earth on the wall.
Solution : (both wan & the earth move in the same direction)
= 30o, SBC = 160 KN/m2,
= 16 KN/m3, μ= 0.5
Rankine’s min depth of foundation,
dmin = 2
sin1
sin1SBC
φ+φ−
γ
= 2
o
o
30sin1
30sin1
16
160
+−
= 1.11m 1.25m
Thickness of base slab h/12
14
IS 3370 (Part II) 1984
=
12
25.5
= 0.4375
Assume the top tk of stem as 150mm thickness of stem tappers from
450mm from bottom & 150mm to top.
Lmin =
R
1h
3
Ca
α
Co-efficient of active earth pressure
Ca =
θ
φ−θ+θ
φ−θ−θCos
CosCosCos
CosCosCos22
22
=
1
o2o2o
o2o2o
Cos30Cos15Cos15Cos
30Cos15Cos15Cos
−+
−−
= Ca = 0.373
Cp =
o
o
30sin1
30sin1
sin1
sin1
−+=
φ−φ+
= 3
Assuming surcharge height of 0.4m at the end of heel slab.
h′= 5.25 + 0.4
15
IS 3370 (Part II) 1984
h′= 5.65 m
Assuming the trapezoidal below the base slab.
α R = 0.67
Lmin =
Lmin = 2.97 m
Provide base slab width of 3m
Width of heel slab is Xmin = Lmin x α R
= 3 x 0.67 = 2m
The preliminary proportion is shown in fissure
For the assumed proportions the retaining wall is check for stability
against overturning and sliding
Force
ID
Force (KN) Distance from
heel (m)
Moment
KNm
W1 ½ x 1.85 x 0.5 x 16 = 7.44 1/3 x 1.85 =
0.616
4.583
W2 1.85x 5 x 25 – 0.45 x 16 =
142.08
½ (1.85) =
0.925
131.424
W3 0.15 x 4.8 x 25 = 18 1.85 + 0.15/2 =
1.925
34.65
14
IS 3370 (Part II) 1984
W4 ½ x (0.450 – 0.15) x 4.8 x
(25 – 16) = 6.48
1.85 – 1/3
(0.30) = 1.75
11.34
Pa sin 98.65 sin 15o = 25.53
Pa = Ca γ e x 4/2
= 0.373 x 16 x 2
2
75.5
Pa = 98.65KN
The resultant of the vertical of lies at a distance of Xw from the h
end
X4 =
W
MW
=
28.283
62.232
= 0.99m
Check for over turning moment
FoSo = > 1.4
For retaining wall with sloping back fi
Mo = Pa cos θ x h1/3 = 98.65 Cos 15o x 5.75 / 5
Mo = 182.63 KNm
14
IS 3370 (Part II) 1984
Mr = W (L. x W) + Pa Sin θ (L)
= 233.28 (3 – 0.99) + 98.65 sin 15o
= 545.5
Foso =
63.182
5.545x9.0
=2.68 > 1.4
Check for sliding
Fos sliding =
θCosPa
F9.0
F = μR = 0.5 x 232.28
= 116.64 KN
Fos sliding =
o15CosPa
64.116x9.0
= 1.10 < 1.4
Hence the section is not safe is sliding
The shear key is required to resist sliding
Assume shear key of size 300 x 300mm at a distance of 1.3m from the
too as shown in figure.
Figue***
15
IS 3370 (Part II) 1984
Tan 30o =
3.1
x
x = 0.75m
h1 = 1.25 - 0.3 + 0.8 = 1.25m
h2 = 1.25 + 0.75 = 2m
Ppe = Cp pe
( )21
22 hh −
= 3 x
2
))25.1()2((16 22 −
Pps = 58.5KN
Fos s =
o
ps
15cos65.98
PF9.0 +
=
o15cos65.98
5.58)64.116(F9.0 +
= 1.71 > 1.4
Hence the section is safe against sliding.
Soil pressure from the supporting soil on base slab :
15
IS 3370 (Part II) 1984
The distribution of soil pressure from the base is trapezoida in
nature. Maximum pressure at one end is
q max =
L
R
+
L
e61
and min-pressure is
qmin =
−
L
e61
L
R
(where qmax is direct stress + Bending stress
qmin is direct stress – bending stress)
here the eccentricity ‘e’ is given by
e = LR – L/2
where LR = The distance of R from the heel slab
LR = (Mo + Mw) /R
=
38.233
)62.23263.182( +
LR = 1.78m
E = 1.78 – (3/2) = 0.28m
14
IS 3370 (Part II) 1984
qmax =
+
3
)28.0(61
3
28.233
= 121.30KN/m2 α SBC
qmin =
−
3
)28.0(61
3
28.233
34.21 > 0 KN/m2
Hence pressure is within the limits.
Figure***
Design of Toe slab :
Fig***
The toe slab is design for a UD of 92.36
The Toe slab is design as cantilever
Assume the clear cover as 75mm for base slab and bar dia of 16mm
Eff. Depth, d = D – clear cover - φ /2
= 450 – 75 – 16/2
d = 367mm
The max. B.M at the rear face of the stem is
M = 92.36 x 1 x ½ + (½ x 1 x 28.96)
15
IS 3370 (Part II) 1984
M = 55.82 KNm
The ma. Shear force occurs at the distance of ‘d’ from the face of the
stear.
V = ½ x 1 (92.36 + 121.3) x (1-0.367)
V = 67.62 KN
Factored B.M = 1.5 x 55.82
Mu = 83.73 KNm
Factored shear force, Vu = 1.5 x 67.62
Vu = 101.43 KN
Mu = 0.87fy Ast (d – 0.47
fck36.0
Astfy87.0
)
83.73 x 106 = 0.87 x 415 x Ast (367 - 0.42
1000x20x36.0
Astfy87.0
Ast = 656.44mm2
S =
Ast
ast1000
14
IS 3370 (Part II) 1984
=
Ast4
)16(xx1000
2π
S = 306.19 mm
Refer table 2 in slab for M20 p fe 415 stal
Provide 16mm @ 300mm c/c.
Check for shear
Design of heel slab :
M = 34.3 x 1.55 x
2
55.1
½ x 1.55 x 47.89 x 1/3 x 1.55
M = 60.37 KNm
V = ½ (82.19 + 34.3) (1.55 – 0.367)
V = 68.90KN
Mu = 1.5 x 60.37 = 90.55 KNm
Vu = 1.5 x 68.90 = 103.35 KN
Mu = 0.87fy Ast d
−bdfck36.0
Astfy1
15
IS 3370 (Part II) 1984
90.55 x 106 = 0.87 x 415 x Ast x 367
−367x1000x20
Ast4151
Ast = 712.03mm2
S =
08.712416x
x10002π
S = 282.37mm
Provide 16mm @ 280mm c/c
Development length :
The main reinforcement to be developed into the fixed support for
a length of
Ld =
bd
s
4 τσφ
=
6.1x2.1x4
415x87.0x16
Ld = 752.18 mm
Check for shear
15
IS 3370 (Part II) 1984
τ v =
367x1000
10x35.103
bd
vu 3
=
= 0.281 N/mm2
Pt =
367x1000
03.712x100
bd
Ast100 =
= 0.194%
c = 0.315 N/mm2
τ v < τ c
Hence section is safe in shear
Distribution steel :
Ast min = 0.12 c/s is provided along the transverse direction of the
base slab.
Ast min = 0.12 bd = 0.12 x 1000 x 150
= 540.0 mm2
S =
5404
10xx1000
2π
= 145.44mm
Provide 10mm @ 140mm c/c
Design of stem
15
IS 3370 (Part II) 1984
Stem is designed as a cantilever slab for a height of 4800mm (5250
- ) The max. moment on cantilever slab in head stem is
M = C a h3/6
Assume clear cover of 50mm, bar of 16mm
d = 450 – 50 – 16/2 = 392mm
M = 0.373 x 16 x = 110KNm
The maximum shear force at ‘d’ from compound base slab is
V = Caγ Z2/2 where Z = 4.8 – 0.45 = 4.35m
V = 0.373 x 16 x (4.35/2)2
V = 56.46. KN
Mu = 1.5 x 110 = 165KNm
Vu = 1.5 x 56.46 = 84.69 KN
Mu = 0.87 fy Ast d
− bdfck
Astfy1
165 x 106 = 0.87 x 415 x Ast x 312
−312x1000x20
Ast4151
Ast = 1248.30mm2
14
IS 3370 (Part II) 1984
S = 161.06mm
Provide 16mm φ @ 160mm c/c.
Check for shear
v =
312x1000
10x67.84
bd
Vu 3
=
= 0.216 N/mm2
Pt =
392x1000
3.1248x100
bd
Ast100 =
= 0.318
C = 0.392 N/mm2
v < c
Hence section is safe in slab.
14
IS 3370 (Part II) 1984
Design of heel slab :
Wt from backfill = 16(5.25 – 0.45) = 76.8
Self wt of heel slab 25 (0.45) = 11.25 Kn
= 88.05 Kn
M = 5.86 x 1.55 x 1.55/2 +
½ (47.89) x 1.55 x 2/3 (1.55)
M = 45.39 KNm
V = ½ (5.86 + 53.75) (1.53 – 0.367) 5.86
V = 35.25 KN
Mu = 1.5 (45.39) = 68.08 KNm
Vu = 1.5 x 56.463 = 84.69 KN
Mu = 0.87 fy Ast. d (1-
3x1000x20
Ast415
)
S =
65.529412x
x10002π
S = 213.53mm
15
IS 3370 (Part II) 1984
Provide 12mm @ 210mm c/c.
Provide distribution steel 10mm @ 160mm c/c. in the base slab as well as
along the transverse direction for the steam in the rear face. Also provide
distasted for the front face of the stem along both direction as 10mm @
16mm c/c.
Reinforcement Details :
The main reinforcement in the stem can be curtailed at two places.
At ½ height (4/3m) half the reinforcement is curtail.
Provide 16mm @ 320mm c/c.
At 2/8 h
8.4x8
2
the reinforcement is reduced to half.
Provide 16mm @ 640mm c/c.
The distribution steel is also curtail in the similar method.
Design the cantilever retaining wall to retain a level difference of 4m.
Good soil is available at a depth of 1.25m below G.L. The unit wt of soil
16KN/m3 and SBC of soil is 160KN/m2. The backfill is leveled one with
angle of internal friction φ = 30o.
Soln :
14
IS 3370 (Part II) 1984
Height of wall above G.L= 4m
Good soil depth below G.L. = 1.25m
Unit wt of soil = 16 Kn/m3
SBC = 160KN/m2
= 30
Rankine’s min depth of foundation.
dmin = 2
sin1
sin1SBC
θ+θ−
γ
= 2
o
o
30sin1
30sin1
16
160
+−
= 1.11m 1.25
Earth pressure co-efficient
Ca =
θ+θ−
sin1
sin1
=
o
o
30sin1
30sin1
+−
= 0.333
Cp =
θ+θ−
s in1
sin1
=
o
o
30sin1
30sin1
+−
= 3
15
IS 3370 (Part II) 1984
Thickness of base slab =
12
25.5
12
h =
= 0.4375 = 0.45m
Assume top width of wall as 150mm & tk of stem tappers from 450mm to
150mm
Lmin =
3
CaR
h
α
Assuming the trapezoidal sters below the base slab
R = 0.67
Lmin =
67.0
25.5
3
333.0
= 2.610m
Provide base slab width of 3m
Width of heel slab is Xmin = Lmin x α R
= 3.0 x 0.67
= 2.01
= 2m
The preliminary proportions is shown in figure.
15
IS 3370 (Part II) 1984
For the assumed proportions the retaining is check for stability against
overturning (s)
Pa = Ca e x2
2
)h(
= 0.333 x 16 x
2
)25.5( 2
Pa = 73.43 Kn
Force
ID
Force (KN) Distance from
heel (m)
Moment
KNm
W1 ½ x 1.85 x 0.45 x 16 =
142.08
1/2 x 1.85 =
0.925
181.42
W2 0.15 x (5.25 – 0.45) x 25 =
18
1.85 +
2
15.0
=
1.925
34.65
W3 ½ x (0.45 – 0.15) x 4.8 x
(25-16) = 6.48
1.85 – 1/3(0.3)
= 1.75 3/2
11.35
W4 3 x 0.45 x 25 = 33.75 3/2 = 1.5 50.625
W = 200.31KN M = 228.045 KNm
The resistant of vertical forces lies at a the of x 01 from the heel end.
X4 =
31.200
045.228
W
Mw =
= 1.188m
14
IS 3370 (Part II) 1984
Check for overturning moment :
Foso =
Mo
Mr9.0
Mo = Pa cos θ x h1/3 = 73.43
3
25.5
Mo = 128.50 KNm
Mr = W (L. x W) + Pa Sin (L)
= 200.31 (3 – 1.138)
= 372.97KNm
Foso =
50.128
97.372x9.0
=2.61 > 1.4
Check for sliding
Fos sliding =
θCosPa
F9.0
F = μR = 0.5 x 200.31
= 100.155 KN
14
IS 3370 (Part II) 1984
Fos sliding =
43.73
155.100x9.0
= 1.22 < 1.4
Hence the section is not safe is sliding
The shear key is required to resist sliding
Assume shear key of size 300 x 300mm at a distance of 1.3m from the
too as shown in figure.
Figue***
Tan 30o =
3.1
x
x = 0.75m
h1 = 1.25 - 0.3 + 0.8 = 1.25m
h2 = 1.25 - 0.3 + 0.3 + 0.75 = 1.25m
Ppe = Cp e
( )21
22 hh −
= 3 x
2
))25.1()2((16 22 −
Pps = 58.5KN
15
IS 3370 (Part II) 1984
Fos s =
θ+cos65.98
PF9.0 ps
=
43.73
5.58)155.100(F9.0 +
= 2.02 > 1.4
Hence the section is safe against sliding.
Soil pressure from the supporting soil on base slab :
The distribution of soil pressure from the base is trapezoida in
nature. Maximum pressure at one end is
q max =
L
R
+
L
e61
and min-pressure is
qmin =
−
L
e61
L
R
here the eccentricity ‘e’ is given by
e = LR – L/2
where LR = The distance of R from the heel slab
15
IS 3370 (Part II) 1984
LR = (Mo + Mw) /R
=
31.200
)04.2285.128( +
LR = 1.78m
e = 1.78 – (3/2) = 0.28m
qmax =
+
3
)28.0(61
3
31.200
= 104.16KN/m2 α 160
qmin =
−
3
)28.0(61
3
28.233
29.37 Kn/m2 > 0
Hence pressure is within the limits.
Figure***
Design of Toe slab :
Fig***
2
x
3
79.74 1=
X = 49.86KN/m2
15
IS 3370 (Part II) 1984
55.1
x
3
79.74 =
X = 38.64 KN/m2
The Toe slab is design for a UDL ofas cantilever beam
Assume the clear cover as 75mm for base slab and bar dia of 16mm
Eff. Depth, d = D – clear cover - φ /2
= 450 – 75 – 16/2
d = 367mm
The max. B.M at the rear face of the stem is
M = 79.23 x 1 x ½ + (½ x 1 x 21.93) x 2/3
M = 47.925 KNm
The ma. Shear force occurs at the distance of ‘d’ from the face of the
stear.
V = ½ x 1 (79.23 + 104.16) x (1-0.367)
V = 58.04 KN
Factored B.M = 1.5 x 47.92
Mu = 71.85 KNm
Factored shear force, Vu = 1.5 x 58.04
14
IS 3370 (Part II) 1984
Vu = 87.06 KN
Mu = 0.87fy Ast (1 -
fck
Astfy
)
71.88 x 106 = 0.87 x 415 x Ast (367)
−367x1000x20
Ast4151
Ast = 560.21mm2
S =
Ast
ast1000
=
21.5604
)12(xx1000
2π
S = 201.80 mm
Provide 12mm @ 200mm c/c.
Check for shear
v =
367x1000
10x06.87
bd
Vu 3
=
= 0.237 N/mm2
16
IS 3370 (Part II) 1984
Pf =
367x1000
21.560x100
bd
Ast100 =
= 0.152%
c = 0.2816N/mm2
τ v = c.
Hence section is safe in shear.
Design of heel slab :
Wt from back fill = 16(5.25 – 0.45) = 76.8 Kn
Self wt of heel slab = 25 (0.45) = 11.25 KN/m2
88.05 Kn/m2
M = (20.04 x 1.55) x
2
55.1
½ x 1.55 x 38.68 x 2/3 x 1.55
M = 55.04 KNm
Mu = 1.5 (55.04) = 82.56 KNm
V = ½ (20.04 + 58.68) (1.55 – 0.367)
V = 46.56KN
Vu = 1.5 x (46.56) = 69.84 KNm
16
IS 3370 (Part II) 1984
Mu = 0.87fy Ast d
−bdfck36.0
Astfy1
82.56 x 106 = 0.87 x 415 x Ast x 367
−367x1000x20
Ast4151
Ast = 646.71mm2
S =
71.6464
16xx1000
2π
S = 174.85mm
Provide 12mm @ 170mm c/c
Distribution steel
Ast(min) = 0.12% bd
=
450x1000x100
12.0
= 540mm2
S =
5404
)0(xx1000
2π = 145.44mm
16
IS 3370 (Part II) 1984
Provide 10mm @ 140mm c/c.
Design of Stem :
Stem is designed as cantilever slab for a height of 5.25 – 0.45 =
4.8m
The max. moment on cantilever slab in the head stem is
M = Caγ h3/6
= 0.833 x 16 x
4
)8.4( 3
M = 98.20 Knm
Assume 50mm cover & 16mm bar
d = 450 – 50 – 16/2 = 392mm
Max. shear force at ‘d’ from compound slab is
V = Caγ Z2/2 Z = 4.8 – 0.45 – 4.3
V = 0.833 x 16 x
2
)35.4( 2
V = 50.40 KN
Mu = 1.5 x 98.20 = 147.3 Knm
14
IS 3370 (Part II) 1984
Vu = 1.5(50.40) = 75.6 KN
147.3 x 106 = 0.87 x 415 x Ast x 392
1000x20
Ast4151 −
Ast = 1105.44mm2
S =
44.1105416x
x10002π
= 181.88mm
Provide 16mm @ 180mm c/c.
Check for shear
τ v =
392x1000
10x6.75
bd
vu 3
=
= 0.192 N/mm2
Pt =
392x1000
44.1105x100
bd
Ast100 =
= 0.282%
τ c = 0.375 N/mm2
τ v < τ c
Hence section is safe in shear
Reinforcement details :
14
IS 3370 (Part II) 1984
Figure***
Ld =
6.1x2.1x4
415x87.0x16
4
0
bd
=τσφ
= 752mm = 750mm
Design a suitable counterfort retaining wall to support a leveled back fill
of height 7.5m above g.l on the toe side. Assume good soil for the
foundation at a design of 1.5m below G.L. The SBC of soil is 170KN/m2
with unit weight as 16KN/m3. The angle of internal friction is = 30o. The
co-efficient of friction below the soil and concrete is 0.5 use M25
concrete and Fe415 steel.
Figure ****
Soln :
Minimum depth of foundation = 2
sin1
sin1P
θ+θ−
γ
= 2
o
o
30sin1
30sin1
16
170
+−
= 1.180m < 1.5m
Depth of foundation = 1.5m
Height of wall = 7.5 + 1.5 = 9m
14
IS 3370 (Part II) 1984
Thickness of heel & stem = 5% of 9m = 0.45m = 0.5m
Thickness of toe slab = 6% of 9m = 0.54m
Stability Conditions :
Earth pressure calculations :
Force
ID
Force (KN) Distance from
heel (m)
Moment
KNm
W1 16(7.5+1.5 – 0.5) x 2.5 =
340
(3-0.5)/2 =
1.25
425
W2 25(0.5)(9-0.5) = 106.25 0.5/2 + 2.5 =
2.75
292.18
W3 25(0.5)x3 = 37.5 1.5 56.25
W4 25(1.5) (0.72) = 27 1.5/2+3=3.75 101.25
Total W=510.75 KN MH = 874.69
The resistant of vertical forces lies at a the of x 01 from the heel end.
XW =
75.510
69.874
W
Mw =
= 1.713m
Check for overturning moment :
Foso =
Mo
Mr9.0
14
IS 3370 (Part II) 1984
Mo = Pa cos θ x h1/3 = 0.333 x 16 x
6
93
Mo = 647.35 KNm
Mr = W (L. x W) = W = 510.75 (4.5 – 1.718)
= 1423.6KNm
Foso =
35.647
6.1423x9.0
=1.98 > 1.4
Hence, section is safe against overturning.
Check for sliding
Fos sliding =
θCosPa
F9.0
F = μR = 0.5 x 510.75
= 255.87 KN
Pa = Ca e x2
2
)h(
= 0.333 x 16 x
2
)9( 2
Pa = 215.78 Kn
16
IS 3370 (Part II) 1984
Fos sliding =
78.215
)37.255(x9.0
= 1.065 < 1.4
Hence the section is not safe is sliding
The shear key is required to resist sliding
Base pressure calculation:
q max =
L
R
+
L
e61
=
+
5.4
)73(61
4.5
510.75
= 223.97 KN/m2 > SBC un safe
qmin =
L
R
+
L
e61
=
+
5.4
)73(61
4.5
510.75
= 3.02 Kn/m2 > 0
Safe
16
IS 3370 (Part II) 1984
where the eccentricity ‘e’ is given by
e = LR – L/2
where LR = The distance of R from the heel slab
LR = (Mo + Mw) /R
=
75.510
)352.64768.874( +
LR = 2.98m
e = 2.98 – (4.5/2) = 0.72 x 1/6 (0.75m)
Since maximum earth pressure is greater than SBC of soil, the
length of base slab has to increased preferably along the toe side. Increase
the toe slab by 0.5m in length.
EH = 510.75 + 0.5 x 25 x 0.72 = 519.75 KN
LR =
75.519
352.647438.917
R
Mw) (Mo +=+ = 3.01m
∑M
Design of Toe slab :
16
IS 3370 (Part II) 1984
Fig***
2
x
3
79.74 1=
X = 49.86KN/m2
55.1
x
3
79.74 =
X = 38.64 KN/m2
The Toe slab is design for a UDL ofas cantilever beam
Assume the clear cover as 75mm for base slab and bar dia of 16mm
Eff. Depth, d = D – clear cover - φ /2
= 450 – 75 – 16/2
d = 367mm
The max. B.M at the rear face of the stem is
M = 79.23 x 1 x ½ + (½ x 1 x 21.93) x 2/3
M = 47.925 KNm
The ma. Shear force occurs at the distance of ‘d’ from the face of the
stear.
V = ½ x 1 (79.23 + 104.16) x (1-0.367)
14
IS 3370 (Part II) 1984
V = 58.04 KN
Factored B.M = 1.5 x 47.92
Mu = 71.85 KNm
Factored shear force, Vu = 1.5 x 58.04
Vu = 87.06 KN
Mu = 0.87fy Ast (1 -
fck
Astfy
)
71.88 x 106 = 0.87 x 415 x Ast (367)
−367x1000x20
Ast4151
Ast = 560.21mm2
S =
Ast
ast1000
=
21.5604
)12(xx1000
2π
S = 201.80 mm
Provide 12mm @ 200mm c/c.
Check for shear
16
IS 3370 (Part II) 1984
v =
367x1000
10x06.87
bd
Vu 3
=
= 0.237 N/mm2
Pf =
367x1000
21.560x100
bd
Ast100 =
= 0.152%
τ c = 0.2816N/mm2
τ v = c.
Hence section is safe in shear.
Design of heel slab :
Wt from back fill = 16(5.25 – 0.45) = 76.8 Kn
Self wt of heel slab = 25 (0.45) = 11.25 KN/m2
88.05 Kn/m2
M = (20.04 x 1.55) x
2
55.1
½ x 1.55 x 38.68 x 2/3 x 1.55
M = 55.04 KNm
Mu = 1.5 (55.04) = 82.56 KNm
V = ½ (20.04 + 58.68) (1.55 – 0.367)
16
IS 3370 (Part II) 1984
V = 46.56KN
Vu = 1.5 x (46.56) = 69.84 KNm
Mu = 0.87fy Ast d
−bdfck36.0
Astfy1
82.56 x 106 = 0.87 x 415 x Ast x 367
−367x1000x20
Ast4151
Ast = 646.71mm2
S =
71.6464
16xx1000
2π
S = 174.85mm
Provide 12mm @ 170mm c/c
Distribution steel
Ast(min) = 0.12% bd
=
450x1000x100
12.0
= 540mm2
14
IS 3370 (Part II) 1984
S =
5404
)0(xx1000
2π = 145.44mm
Provide 10mm @ 140mm c/c.
Design of Stem :
Stem is designed as cantilever slab for a height of 5.25 – 0.45 =
4.8m
The max. moment on cantilever slab in the head stem is
M = Caγ h3/6
= 0.833 x 16 x
4
)8.4( 3
M = 98.20 KNm
Assume 50mm cover & 16mm φ bar
d = 450 – 50 – 16/2 = 392mm
Max. shear force at ‘d’ from compound slab is
V = Ca Z2/2 Z = 4.8 – 0.45 – 4.3
V = 0.833 x 16 x
2
)35.4( 2
16
IS 3370 (Part II) 1984
V = 50.40 KN
Mu = 1.5 x 98.20 = 147.3 Knm
Vu = 1.5(50.40) = 75.6 KN
147.3 x 106 = 0.87 x 415 x Ast x 392
1000x20
Ast4151 −
Ast = 1105.44mm2
S =
44.1105416x
x10002π
= 181.88mm
Provide 16mm @ 180mm c/c.
Check for shear
v =
392x1000
10x6.75
bd
vu 3
=
= 0.192 N/mm2
Pt =
392x1000
44.1105x100
bd
Ast100 =
= 0.282%
c = 0.375 N/mm2
v < τ c
14
IS 3370 (Part II) 1984
Hence section is safe in shear
Reinforcement details :
Figure***
Ld =
6.1x2.1x4
415x87.0x16
4
0
bd
=τσφ
= 752mm = 750mm
Design a suitable counterturn retaining way to support a leveled back fill
of height 7.5 m above g. L on the toe side. Assume good soil for the
foundation at a deim of 1.5 m below G. L. The SBC of soil is 170 KN/m2
with unit weight as 16 KN/m3. The angle of internal friction is φ = 300
the co-efficient of friction b/w the soil of concrete is 0.5 use M25
concrete of Fe 415 steel.
Soln:-
Figure *********
Minimum depth of foundation = 2
sin1
sin1P
φ+φ−
γ
=
m5.1m180.130sin1
30Sin1
16
1700
0
<=
+−
Depth of foundation = 1.5 m
14
IS 3370 (Part II) 1984
Height of wall = 7.5 + 1.5 = 9 m
Thickness of heal & stem = 5% of 9m =
Thickness of heal & stem = 5% of 9m = 0.45 m = 0.5
Thickness of toe slab – 8% of 9m = 0.72 m.
h
m in 3
CaX
=
=
m0.393
333.0 =<
Lmin = 1.5 x 3 = 4.5 m
Thickness of countertort = 6% of 9 m = 0.54 m
Stabiling conditions:
Earth pressure calculations:
Force IDforce (KN) Distance from heal (m) Moment KNm
W1 16 ( 7.5 + 1.5 – 0.5) x 2.5
= 340
(3 – 0.5)/2 = 1.25 425
W2 25 (0.5) (9 – 0.5) =
106.25
0.5/2 + 2.5 = 2.75 292.18
W3 25(0.5)x 3 = 37.5 1.5 56.25
14
IS 3370 (Part II) 1984
W4 25 (1.5) (0.72) = 27 1.5/2 + 3 = 3.75 101.25
Total W = 510.75 KN M4 =
874.69
W = 25(0.3) x 0.75 = 9.00
XW =
m713.175.510
69.874 =
= 1.713 m
Fos(overturnis) =
0M
M9.0
Mo = Pa. h.3 = Caγ e. h3/6 = 0.333 x 16 x (9)3/6
= 647. 35 KNm
Mr = (L – Xw) w = 510.75 (4.5 – 1.713)
= 1423.6 KNm
Fos =
4.198.135.647
)6.1423(9.0 >=
Hence section is safe against, overturnins.
Silding:-
Fos(sliding) =
θPaCos
F9.0
14
IS 3370 (Part II) 1984
F = R = 0.5 x 510.75 = 255.37
Pa = Ca. γ e. h2/2 = 0.383 x 16 x (9)2/2
= 215.78
Fos sliding = 0.9 (255.37) / 215.75 = 1.065m < 1.4
Hence the section is not safe against sliding.
Base pressure calculation:
9 mas =
+=
+
5.4
)73.0(61
5.4
75.510
L
601
L
R
= 223.97 KN/m2 > SBC unsafe
Qmin =
+=
5.4
)73.0(61
5.4
75.510
L
e614
L
R
= 3.02 w/m2 > 0 safe
Where, LR =
where2/LLRCR
MM 0h −==+
LR =
m98.275.510
352.647688.874 =+
e = 2.98 – 4.5/2 = 0.78 < L/L (0.75mm)
16
IS 3370 (Part II) 1984
Since maximum earth pressure is stresses then SBC of sol, the
length of base slab has to b. increased praberables along the toe side.
Increase the toe slab by 0.5m in length.
w = 510.75 + 0.5 x 25 x 0.72 = 519.75 KN
Figure **********
Additional load due to increase in toe slab by 0.5.
Moment = 0.5/2 + 4.5 = 4.75m
Σ m = 874.69 + 42.75 = 917.44 KNm
LR=
m01.375.519
352.647x438.417
R
)MM( W0 ==+
e = LR – L/2 = 30.11 – 5/2 = 0.511 m < L/6 (0.833m)
q max =
+=
+
0.5
m5.0x61
0.5
75.519
L
e61
L
R
= 167.69 KN/m2 < SBC
Qmin =
−
L
e61
L
R
16
IS 3370 (Part II) 1984
=
−
5
)511.0(61
5
75.519
Qmin = 40.20 KN/m2 > 0
FOS siliding =
784.215
75.519x5.0x9.0
Pa
F9.0 =
= 1.082 < 1.4
Hence the section is not slab against sliding.
Shear they is provided to resist sliding. Assume shear they of size
300 x 300 mm
Figure *********
tan 300 =
2400
L
x = 1.385 = 1.39 m
PPb = Cp. Re
−2
hh2
1
2
2
h2 = 1.39 + 1.2 = 2.89m
h1 = 1.2 + 0 = 1.2m
15
IS 3370 (Part II) 1984
Meridonial thrust T =
30.11
WR
+
=
724.01
06.9x5.4
+
T = 23.64 KN
Meridonial stress
100x1000
10x64.23
s/c
T 3
=
= 0.2364 N/mm2 x 7 N/mm2
Hence the section is safe against meridonial stress.
Hoop stress:-
Hoop Stress =
ϑ+−θ
cos1
1Cos
T
WR
=
+−
721.01
1724.0
64.23
06.9x5.4
= 0.248 N/mm2 < 7 N/mm2
Since the section is safe to resist meridonial thrust of hoop stress
the provided 100 mm thickness of sufficient.
15
IS 3370 (Part II) 1984
Ast min = 0.3 % of C/S
=
2mm300100x1000x100
3.0 =
For 8mm φ spacing =
mm5.167300
48x
x10002
=
π
∴Provide 8 mm @ 160 mm c/c both way.
Design of Rin beam
The area of concrete received for the ring beam is bound for the
hoop stress and area of steel required is bound for the horizontal
component of meridmial thrust.
HZ component of Meridonial thrust = T Cos θ x D/2
= 23.64 x 0.724 x 12.5/2
= 106.97 KN
Ast =
23
mm13.713150
10x97.106 =
16
IS 3370 (Part II) 1984
For 16 mm φ , no. 08 bars =
416x13.713
2π
= 3.54 say 4 nos
Size of rins beam is based on tensile stress of concrete
PP8 = 3 x 16
−
2
)2.1()89.2( 22
PP8 = 165.89 KN/m2
FOS sliding =
−
2
)2.1()89.2( 22
= 1.852 > 1.4
Hence section is safe against sliding.
Design of Toe slab:
Eff.Cover = 75 + 20/2 = 85 mm
Toe slab is designed similer to cantition and with maximum moment at
trust face of the size a maximum shear at ‘d’ for face it beam.
D = 720 – 85 = 635 mm
16
IS 3370 (Part II) 1984
M =
2x3
2x94.49x2x
2
1
2
2x38.80
2
+
= 227.35 KNm
S.F at 0.635 m =
KN85.15635.0x2
94.49 =
Area of tralizium = ½ x (a + b) h = ½ (130.82 + 95.98) (2 – 0.635)
= 154.44 KN
Factored S.F = 1.5 (154.44)
S.F Vu = 231.66 KN
Factored B.M, Mu = 1.5 (227.35) = 341.02 KNm
K =
845.0845.0635x1000
02.341
bd
M22
u ===
Pt = 0.2445
Ct =
100
635x1000x245.0Ast
bd
Ast100 == >
Ast = 1552.575 mm2
16
IS 3370 (Part II) 1984
Spacing =
mm50.129575.1552
416x
x10002
=
π
Provide 16 mm @ 125 mm C/C
Transerve Reinforcement = 0.12% of C/S
=
2mm864720x1000x100
12.0 =
Spacing =
mm90.90864
410x
x10002
=
π
∴ Provide 10 mm @ 100 mm c/C
Check for shear :-
τ v =
23
u mm/N364.0635x1000
10x66.231
bd
V ==
τ c \ 0.36 N/mm2 τ v ~ τ c
∴ The section is safe in shear
Design of heel slab
16
IS 3370 (Part II) 1984
The had slab is designed by countention at regular interval. The
counterbant act as support and makes the had slab as one – way
continuous slab.
The heal slab is designed for a moment Wl2/l2 at the support &
Wl2/16 at the mid span. The maximum shear at the support is W **** the
maximum pressure at the best slab is consider for me design.
Moment at the support, MSuf =
12
2.3x92.106
12
Wl 2
=
= 111.65 KNm
Moment at the mid span, Mmid =
KNm72.8316
Wl 2
=
The max pressure acting on the had slab is the as ‘W’ for which the Ast
required at mid span and support one found.
Factored Msup = 167.47 KNm Ast = 1172.69 mm2
Factored Mmidspan = 125.58 Ast = 868.27mm
Using 16 mm φ bar, spacing =
02.170Ast
Ast1000 =
∴ Provide 16 mm @ 170 mm c/c.
16
IS 3370 (Part II) 1984
At mid span, spacing = 231.72 mm Provide 16mm @ 230 mm C/C.
Transverse reinforcement = 0.12% of Bd
=
2mm600500x1000x100
12.0 =
For 8mm bar, spacing = 83.775 mm
∴Provided 8 mm @ 80mm c/c.
Check for shear:-
Maximum shear
−=
− 415.0
2
5.2107d
2
lW
Factored S.F = 217.47 KN = Pt = 0.282
τ v = 0.624 N/mm2, τ c = 0.376 N/mm2, τ cmax = 3.1 N/mm2
τ v > c
∴Depth has to be increased.
Design of stem
16
IS 3370 (Part II) 1984
The stem is also designed as one – way continuous slab with
support moment
12
Wl 2
and midspan moment
16
Wl 2
For the negative moment at the support, reinforcement is provided at the
rear side & for positive moment at mid span, reinforcement is provided at
front base of the stem.
The maximum moment various beam a base
Intensitus of Ca γ e. h = 1/3 x 16 x (9 – 0.5)
= 45.33 KN/m
Mmid =
16
Wl 2
=
16
54.3x33.45x5.1 2
= 53.25 KNm (or) 307.04 KNm
Effective depth d = 500 – (50 + 20/2) = 440 mm
M support =
KNm7112
54.3x33.45x5.1
12
Wl 22
==
Ast at support = 454.73 spacing = 69 mm
16
IS 3370 (Part II) 1984
∴Provide 16 mm @ 300 mm C/C.
Ast at mid span = 339.54 mm2
For 16 mm , spacing = 592.54 mm
Provide 16 mm @ 300 mm C.C
Max S.F = W
− d
2
l
= 60.28 KN
Factored S.F = 90.42 KN
Transverse reinforcement = 0.12 % of 6D
= 0.12/100 x 1000 x 440 = 528 m
For 8 mm , spacing = 95
8mm @ 90 mm C/C 8t = 0.134.
τ v = 0.205 N/mm2, τ c = 0.29 N/mm2
v < τ c ∴
Shear of safe.
Design of countertors:-
16
IS 3370 (Part II) 1984
The countertort is desined as a cautilever beam whose depth is
equal to the length of the heal slab of the base is reduces to the thickness
of the stem at the top. Maximum moment at the base of coutnerfort.
Mmax = Ca γ e. h3/6 x Le
Where Le C/C distance from counterforb
Mmax = 1932.5 KNm
Factored Mmax = 2898.75 KNm
Ast = 2755.5 mm2
No. of bars required =
425
x
5.2755
ast
Ast2
π=
= 5.61
Say n = 6 nos
The main reinforcement is provided along the slanting face of the
counterbort.
Cartailment of reinforcement:
Not all the 6 bars need to be taken to the for end
Three bars are taken straight to the entire span of the beam upto me hop
of the stem.
14
IS 3370 (Part II) 1984
One bar is Cut al a distance of
2
2
1
5.8
h
n
1n =−, where n is the total no of
h1 distance from top .
n = 6, γ = 6-1/6 = h12/8.52
h1 = 7.72th (from bottom )
m94.65.8
h
6
26
5.8
h
n
2n2
2
2
2
2
2 ==−=>=−
The third part is cut at a distance of
)bottomFrom(m01.63h,5.8
h
n
3n2
2
3 ==−
Vertical ties and horizontal ties are provided to connect the counter fort
with the item is the heal slab.
Design of horizontal ties:-
Closed stirrups are provided t, the vertical stem is the countertort.
Considering 1m strip, the tension resisted by reinforcement is given by
lateral pressure on the wall multiplied by contributing area.
T = Cα.γ e.h
14
IS 3370 (Part II) 1984
Where Ast =
fy87.0
T
T = 1/3 x 16 x (9-0.5) x 3.54 = 160.48 KN
Ast =
23
mm72.666)48.444(5.1415x87.0
5.1x10x48.160 ==
For 10 mm φ spacing = 110 mm
Provide 10 mm @ 110 mm c/c closed stirrups as horizontal ties.
Design of vertical ties:-
The vertical stirrups connects the countertort and the heel slab
considering 1m strip, the tensile force is the product of the average
download pressure & the spacing between the counterforts.
T = Avg 943.56 + 167) + Le = 266.49 KN
Factored T = 399.74 KN
Ast = 1107.15 mm2. For 10mm , spacing = 70.93
Provide 10mm @ 70m c/c.
C/s of counter fort wall of midspan:
Reinforcement details of stem, toe slab of heel slab.
15
IS 3370 (Part II) 1984
Isotropicall reinforced – Squave slab – fixed on all edges – udl:-
External work done = W. S
W – load, S – virtual displacement
Internal work done = Σ Mθ = Σ m.lθ =
L
22.L2.m ¬
I.W.D work done by positive yield line (ab, bc, cd, da) for
L
2
2/L
1 ==θ
Mθ = 4[W x L x 2/L] = 8m
Total I.W.D = 16m
E.W.D, Σ W.S = ½ x L x L/2 x W x ½ x (1) x 4 = Wl2/3
I.W.D = E.W.D
16m = WL2/3
M =
b4
WL2
M – Moment per metre length alms the rived line.
14
IS 3370 (Part II) 1984
Design a square slab fixed along as the bour edges, with side 5m. the slab
hase, to support a service load of hase to be support a service load of 4
KN.m2. use M20 concrete and Fe 415 steel.
Soln:-
Side = L = 5m, All four edges fixed
Given
Service load = 4 KN/m2
Fcx = 20 N/mm2
Fy = 415 N/mm2
As per yield line theory for isotrophic reinforced square slab fixed on all
four edges,
M = *****
Step 1:-
As per IS 456 : 2000 L/d ratio for simply supported slab using Fe 415
steel.
L/D = 0.8 (35) = 28
D = 5000/28 = 178.57m
∴Provide effective depth d as 180 mm
15
IS 3370 (Part II) 1984
Solve the above problem using the co-efficient stam in IS 456:2000
Soln:-
Given Lens = l2 = ly = 5 m ly/l2 = 1
∴The slab is two way slab.
Assume all four edges discontinuous, rebearing annex d, take 20
αu = αy = 0.056
Mx = α x Wdx2 = 0.056 x 15 x 52 = 21 KN
M2 = My = 21 KNm
Mu =
−
bdFck
Astfy1dAstfy87.0
21 x 106 = 0.87 fy Ast d
−
180x1000x20
Ast4151
Ast = 336.15 mm2
S =
mm59.149336
48x
x10002
=
π
14
IS 3370 (Part II) 1984
∴ Provide 8 mm φ @ 140 mm C/C
Design the above problem for simply support condition.
Soln:-
M =
KNm625.1524
5x15
24
WL 22
==
Mu = 0.87 fy Ast d
−
bdfck
Astfy1
15.625 x 106 = 0.87 x 415 x Ast x 180
−
180x100x20
Ast4151
Ast = 247.48 mm2
S =
mm50.203247
48x
x10002
=
π
∴Provide 8 mm bar @ 220 mm c/C
Check for shear:-
16
IS 3370 (Part II) 1984
τ v =
2
d
u
mm/N208.0180x100
25x15
b
V ==
Pt =
%137.0180x1000
247x100
bd
Ast100 ==
τ c = 0.28 N/mm2
c = 1.2 x 0.28 = 0.336 N/mm2
τ v < c
∴Hence the section is safe is shear.
Vu =
KN272
4x5.13
2
lW xu ==
τ v =
23
mm/N18.0150x1000
10x27 =
Pt =
%184.0150x1000
277x100 =
τ c = 0.307 N/mm2
τ c = K c = 1.26 x 0.307 = 0.386 N/mm2
16
IS 3370 (Part II) 1984
∴Hence the section is safe is shear.
In the above problem design the slab if all support are fixed
Soln:-
Based on yield line theory, for rectangular slab fixed on all four edges,
subjects to UDL throushour
M =
µφα
22
u tan
24
LW
tan =
αµ−
αµ+µ
245.1
2
2
=
−
+
)67.0(2
7.0
)67.0(4
)7.0()7.0(5.1
2
2
tan φ = 0.8000
m =
7.0
8.0x8.0
48
6x5.13 2
= 9.26 KNm
18
IS 3370 (Part II) 1984
M = 0.87 fy Ast d
−
bdfck
Astfy1
9.26 x 106 = 0.87 x 415 x Ast x 150
−
150x1000x20
Ast4151
Ast = 175 mm2
Ast min =
2mm204170x1000x100
12.0 =
∴Ast provided on shorter direction is 204 mm2
Assume 80 mm φ , S =
2044
8xx1000
2π
S = 246.39 mm
∴Provide 8 mm @ 240 mm c/c both ways
Check for shear
v =
23
mm/N18.0150x1000
10x27
bd
Vu ==
16
IS 3370 (Part II) 1984
Pt =
%136.0150x1000
204x100 =
τ c = 0.28 N/mm2
τ c1 = K c = 1.26 x 0.28 = 0.352 N/mm2
τ v < c
∴Hence the section is sale in shear
Solve the above problem for two long edges fixed boundarks condition
Soln:-
M =
µφ22
xu tan
24
LW
=
7.0
8.0
24
4x5.13 22
M = 8.28 Knm
8.23 x 106 = 0.87 x 415 x Ast x 150
−
150x1000x20
Ast4151
Ast = 155.3 mm2
16
IS 3370 (Part II) 1984
Ast min =
2mm204150x1000x100
12.0 =
∴Ast provide on shorter direction is 204 mm2
Assume 8mm φ , S =
mm39.246204
48x
x10002
=
π
∴Provide 8 mm @ 240 mm C/C bothways
Check for shear
v =
23
mm/N18.0150x1000
10x27 =
Pt =
%136.0150x1000
204x100 =
τ c = 0.28 N/mm2
c1 = Kc = 1.26 x 0.28 = 0.352 N/mm2
τ v < c
Hence the section is safe in shear.
16
IS 3370 (Part II) 1984
Design an equilateral triangular slab of side 5m, isotropically reinforced
and simply supported along its edges. The span is subjected to a super
imposed load of 3 KN/m2 use M20 concrete of Fe – 1415
Soln:-
Based on yield line theory, the ultimate moment for an etuilateral
triangular slab simply supported alms all edges of subjected to VDL
through out.
M =
72
LW 2
u
Assume L/d = 28
D = 5000/28 = 178.54 mm
Assume eff. Depth d = 180 mm
Assume in eff.cover of 20 mm
∴overall depth, D = 180 + 20 = 200 mm
Load calculation:
Self wt of slab = 1 x 0.2 x 25 = 5 KN/m2
Live load = 3 KN/m2
16
IS 3370 (Part II) 1984
Floor finish = 1 KN/ m2
W = 9 KN/m2
Factored load, Wk = 1.5 x 9 = 43.5 KN/m for 1m strip.
M =
KNm69.472
5x5.13
72
LW 22
u ==
4.69 x 106 = 0.87 x 415 x Ast x 180
−
10x1000x20
Ast4151
Ast = 72. 77 mm2
8mm φ Transverse reinforce =
2mm240200x1000x100
12.0 =
Spacing =
mm209240
48x
x10002
=
π
∴Provide 8 mm @ 200 mm C/C.
Check for shear
τ v =
bd
Vu
16
IS 3370 (Part II) 1984
V =
75.332
5x5.13
2
WL ==
Vu = 33.75 = 33.75
τ v =
23
mm/N187.0180x1000
10x75.33 =
Pt
%133.0180x1000
240x100 =
τ c = 0.28 N/mm2 , τ c max = 2.5 N/mm2
τ c1 = K c = 1.2 x 0.28 = 0.336 N/mm2
τ v < c < τ c max
∴Section is safe in shear
A right angled triangular slab simply support along all the edges it has
sides AB = BC = 4m. the slab is isotrofically reinforced with 10 mm @
100 mm C/C both way use M20 and Fe 415 (HYDS) bars. Find the safe
permissible service load (Live load) that can be apply on the slab.
Soln:
16
IS 3370 (Part II) 1984
For A1m slab S.S along all the edges which is right angle isotuplication
reinforced subjected to UDL through out.
Mu =
)therelineyieldonBased(6
LxW 2
u
α = 1
4
4
S =
Ast
ast1000
Ast =
2
2
mm4.785100
410x
x1000=
π
L/4 = 28 d =
28
4000D = 170 mm
For Ast available moment resistance the section.
Mu = 0.87 fy Ast d
−
bdfck
Astfy1
14
IS 3370 (Part II) 1984
Mu = 0.87 x 415 x 785.4 x 150
−
150x1000x20
4.785x4151
Mu = 37.9 KN m
37.9 x 106 =
6
4x1xWu 2
Wu = 14.2 KN/m2
Factored load = 14.2 KN/m2
∴Working load = 14.2 / 1.5 = 9.46 KN/m2
Self wt = 0.17 x 25 = 4.25 KN/m2
Floor finish =
95.4
m/KN71.0 2
∴The service live load that can be safly apply on slab is 9.46 – 4.96 =
4.5 KN/m2
Design a circular slab of diameter 5m, S.S along the edges and subjected
to live load of 4 KN/m3 use M20 & Fe 415 steel.
Soln:-
16
IS 3370 (Part II) 1984
Based on yield line theory the ultimate moment for circular slab
isotopicus reinforcement Sr. S. along the edges subjectedto VDO through.
M =
6
rW 2
u
Assume L/d = 28
D = 5000 / 28 = 178.07mm
∴eff. Depth, d = 180 mm
Adopt eff. Cover = 20 mm
∴ overall depth, D = 200 mm
Load calculation
Self wt on slab = 0.2 x 25 = 5 KN/m2
L.L = 4 KN/m2
Floor finish = KN/m2
W = 10 KN/m2
Factored load, Wu = 1.5 x 10 = 15 KN/m
14
IS 3370 (Part II) 1984
M =
KNm625.156
)5.2(x15
6
rW 22
u ==
15.625 x 106 = 0.87 x 415 x Ast x 180
−
10x1000x20
Ast4151
Ast = 247.48 mm2
Transverse reinforcement = 0.12% of D
=
200x1000x100
12.0
= 240 mm2
For 8 mm φ S =
mm10.20348.247
48x
x10002
=
π
Provide 8 mm φ @ 200 mmC/C both ways
Check for shear:-
τ v =
208.0180x1000
210x5x15
bd2
lW
bd
Vu
3
u
===
16
IS 3370 (Part II) 1984
Pt =
%187.0180x1000
48.247x100 =
τ c = 0.28 N/mm2
τ v < τ c
Hence the section is safe in shear
Design a simply supported rectangular slab of sie 5m x 8m which is
orthotopilr reinforced with co-efficient of orthorooly µ = 0.75 the
service live load on the slab is 5 KN/m2 use M20 concrete and Fe 415
steel
Soln : L = 8m, α = 5m LL = 5 KN/m2
Based on yield line memory the rectangular slab orthotropically
reinforced, S.S. along with edges of subjected to VDL through out.
Α = 5/8 = 0.625
M =
[ ]222
2324
Wl µα−µ α+α
Assume L/D = 28
5000 / 28 = d, d = 178. 57 mm
Provide d = 180 mm
15
IS 3370 (Part II) 1984
Eff. Cover = 20 mm, overall depth , D = 200 mm
Load calculation
Self wt on slab = 0.2 x 25 = 5 KN/m2
L. Load = 5 KN/m2
Floor finish = l KN/m2
Total load, W = 11 KN/m2
For 1 m strip , 4 = 11 KN/m
Factored load, Wu = 1.5 x 11 = 16.5 KN/m
Mu =
[ ]222
2324
Wl µα−µ α+α
=
[ ]22
22
75.0625.0625.0x75.0324
8x625.0x5.16 −+
Mu = 27.86 KNm
27.86 x 106 = 0.87 x 415 x Ast x 180
−
180x1000x20
Ast4151
Ast = 452.26 mm2
Transverse reinforcement ast min
14
IS 3370 (Part II) 1984
For 10mm φ mm bar
Provide 10mm φ @ 170 mm C.C lonser direction Ast µ Ast = 0.75 x
452.26 = 339.19 mm2
Check for shear: provide 10 mm @ 230 mm C/C
τ v = Vu/ bd
Vu = WL/2 = 16.5 x 5 / 2 = 41.25 KN
v = 41.25 x 103 / 1000 x 180 = 0.229 N/mm2
Pt = 100Ast / bd = 100 x 452.26/1000x 180 = 0.25 %
τ c = 0.36 N/mm2
c1 = K τ c = 1.2 x 0.86 = 0.432 N/mm2
Hence section is safe in shear.
Design a rectangular slab of size 5m x 4m which is fixed along the long
edges of S.S along the two short edges. The slab is subjected to the
distributed live load of 4 KN/m2 design the slab for orthotopically
reinforced condition with co-efficient orthotropy µ = 0.7
Soln:
Based on yield line theory for orthotropically reinforced slab fixed along
long edges of S. S along the short edges of subjected to VDL through out,
the ultimate mum M is
16
IS 3370 (Part II) 1984
M = Wulx2 / 24 =
µ
2tan
tan φ =
αµ−
αµ+µ
245.1
2
2
Assume L/d = 28
4000 / d = 28
D = 142.85
Provide eff. Depth d = 150 mm
Effective cover = 20 mm
Overall depth D = 170 mm
Load calculation:
Self wt of slab = .0.17 x 25 = 4.25 KN/m2
L.L = 4 KN/m2
Floor finish = 0.75 KN/m2
W = 9 KN/m2
For 1 m strip, W = 9 KN/m
Wu = 1.5 x 9 = 13.5 KN/m2
16
IS 3370 (Part II) 1984
Α = 4/5 = 0.8
Tan φ = 1.5 (0.7)
Tan = 0.804
M = 8.311 KNm
M = 0.87 fy Ast x d
8.311 x 103 = 0.87 x 415 x Ast x 150
−
150x1000x20
Ast4151
Ast = 156.86 mm2
Ast provided in Longer direction = µ Ast = 0.7 (156.86) = 109.8
Ast min = 0.12% to D = 0.12 / 100 x 1000 x 170
Ast = 204 mm2
Provide Ast = 204 mm2
For 8 mm φ , spacking =
246204
410x
x10002
=
π
Provide 8 mm @ 240 mm c/c.
Check for shear
14
IS 3370 (Part II) 1984
v = Vu/bd
Vu = WuL/ 2 = 13.5 x 4 /2 = 27KN
τ v =
bd
Vu
Rf =
KN272
4x5.13
2
WuL ==
τ c = 0.28 N/mm2
c1 = K τ c = 1.26 x 0.28 = 0.3528 N/mm2
v < c1
Hence section is safe in shear
Design a doom for a cylindered water tank of diameter 12.5 m use M20
concrete of Fe 415 steel
Soln:
D = 12.5 m
Rise = 12.5/5 = 2.5m
R2 = (R – r)2 + (D/2)2
R2 = (R – 2.5)2 + (12.5/2)2
16
IS 3370 (Part II) 1984
R = 9.06 m
Cos θ = 0.724
The done is subjected to meridonial thorust and hoof force for which the
stress should be within permissible compressive strength of concrete
Assume thickness of slab, t = 100 mm
Loadings on slab:
Self wt = 25 x 0.1 = 2.5 KN/m2
Live load = 2
W = 4.5 KN/m2
Ast =
33
st
o mm38.603150
10x5.90compHZ ==σ
Assume 16mm , no of bars =
416
x
3.6032
π
Provide 4 nos of 16 mm bar
Size of ring beam is based on tensile stress of concrete
ct = Ft / Act (m-1) Ast
2.8 = 90.50 x 103 / AC + (13.33-1) 603.3
14
IS 3370 (Part II) 1984
Ac = 24886.4 mm2
Provide ring beam of size 160 x 160 mm
Design a water tank for field base condition for a capacities of 4 lakh
litres height of tanks is 4m. permissible stresses σ st = 150 N/mm2 for M20
came σ cbc = 7 N/mm2, j = 0.84, R = 1.16
Soln:-
Capacity 4 x 105 lr = 400 m3
Ht = 4m
Volume
4x4
πd 2
2d4x
4x400 =π
d = 11.28
Provide diameter d = 115.m
Thickness of tank based on emtiral relation.
T min = 30 (4) + 50 = 170 mm
Provide a thickness of 170 mm
Non – dimensional parameter =
18.879.0x5.11
)4(
Dt
4 22
==
Hoop tension devolor on the 10 all =
T = co-efficient WH D/2
16
IS 3370 (Part II) 1984
BM develop along the W911, Co-eff x WH3
Reborins tasie 9 of SI = 3370 (Pert IV)
Co-efficient for m x m hoof tension = 0.577
ett.CDt
H 2
=
8 = 0.575
10 – 0.608
8.18 – 0.577
For
Dt
H2
of 8.18 the co-efficient for hoof tension is actions it 0.64
∴mass – hoop tension, T = 0.577 x 9.81 x 4 x 11.5/2
T = 130.18 KN
Co – efficient for max B.M is taken from table to of IS 3370 (Part IV)
8 0.0146
10 0.0123
8.18 0.0148
Co- efficient of max B.M – 0.0143 actions at 1.04 (at Base)
Moment = Co. efficient x WH3
= -0.0143 x 9.81 x 43
B.M = -8.97 KNm
Transverse Reinforcement:
16
IS 3370 (Part II) 1984
Ast =
150
10x18.130H 8
st
t =σ
Ast = 867.86 mm2
For 16 mm φ S =
mm69.2318.867
416x
x10002
=
π
∴Provide 16 mm φ @ 230 mm c/c
Adopt 12 mm φ @ 130 mm c.c
Vertical Steel:- The vertical reinforcement tension for the max. BM 0.8
8.97 KN m by working shows mif
Ast =
sd
M
stσ
Min – depth, d =
m127mm93.871000x16.1
10x97.8
Rb
M 6
<<==
Overall depth of wan, thickness = 170 mm
Adopiting a clear cover of 25 mm
D = 170 – 25 – 12/2 – 12
D = 127 mm
Ast =
26
mm55.560127x84.0x150
10x97.8 =
15
IS 3370 (Part II) 1984
For 12mm, S =
mm77.2015.560
412x
x10002
=
π
∴Provide 12 mm @ 200 mm c/c.
Min. reinforcement Ast min = 0.3 % C.S
= *****
Ast min = 510 mm2 < Ast rerd.
Design of base slab:
Assume 150 mm thick base slab, provide 0.3% of C/S as
reinforcement along both the faces.
Ast =
2mm450150x1000x100
3.0 =
For 8 mm, S =
mm70.111450
48x
x10002
=
π
Provide 8 mm @ 200 mm c/c both base along both faces.
Provide haunch of side 150 x 150 mm with min. steel of 8 mm @ 200
mm C/C along the face of the haunch
Figure ****
TUTORIAL
Design a rectangular water tanks of size 4m x 7m with height 3.5m
use M20 concrete of Fe – 415 steel. Design constant j = 0.053 R = 1.32
15
IS 3370 (Part II) 1984
Soln:-
L x B 7m x 4m
L/B = 7/4 = 1.75 x2
H = 3.5 m
h = H/4 or 1
m1or875.04
5.3 ==
h = 1 m
P = p (H-h) = 9.81 (3.5 – 7)
P = 24.53 KN/m2
To find the final moment at the junction of lons will & short will based
on the FEM & D.F. Moment distribution is done
For any joint, D.F =
2
2
1
1
11
LI
LI
L/IF.D
+=
FEM MFAB =
KNm70.3212
4x53.24
12
PB 22
==
MFAD =
KNm16.10012
7x53.24
12
82 22
==
ADAB
MemberA
intJo
F.D
14
IS 3370 (Part II) 1984
2
2
1
1
22
2
2
1
1
11
LI
LI
L/I
LI
LI
L/IF.D
++
36.0
71
41
7/1611.0
71
41
4/1 =+
=+
Joint Member
A AB AD
D.F 0.64 0.36
FIM -32.70 100.16
B.M -43.17 -24.28
Total -75.87 KNm 75.88 KNm
Fixed B.M is 75.88 KNm
Free B.M (For B.M)
a) Along shorter direction =
KNm06.498
4x53.24
8
PB 22
==
b) Longer direction =
KNm24.1508
7x53.24
8
PL 22
==
resultant B.M at the mid span is
a) Short will == -75.88 to 49.06 = -26.82 KNm
b) Long will = -75.88 + 150.24 = 74.36 KNm
At support = 75.88 KNm
The will is design for me max B.M of for
14
IS 3370 (Part II) 1984
M max = 75.88 KNm
Reinforcement details:
Ast L =
st
2
st
L P
id
xPM
σ+
σ−
Ast B =
st
B
st
b P
id
xPM
σ+
σ−
PL = P x B/2 = 24.53 x 4/2 = 49.06 KN
PB = P x L/2 = 24.53 x 7/2 = 85.86 KN
X = D/2 – eff.cover
Providing a clear cover of 25 mm of bar 10 φ mm
Eff.cover = 25 x 10/2 = 30 mm
d =
mm75.2391000x32.1
1088.75
Rb
M 6
==
provide d = 240 mm
overall depth, D = 240 + 30 = 270 mm
∴thickness of will is 270 mm
X =
302
270 −
X = 105 mm
14
IS 3370 (Part II) 1984
Ast L =
150
10x06.49
240x853.0x150
105x10x86.8510x88.75 336
+−
Ast L = 2630.33 mm2
Ast B =
150
10x86.85
240x853.0x150
105x10x86.8510x88.75 336
+−
Ast 8 = 2749.83 mm2
Provide 20mm φ bar, S =
mm43.11933.2630
420x
x10002
=
π
∴Provide 20 mm φ @ 110 mm c/c
Since the thickness is greater than 200mm, the reinforcement is placed
alogn both forces.
∴Along transverse direction (HZ) provide
20 mm φ @ 220mm c/c along both faces in the short well & long wall.
Vertical Reinforcement
For the cantilever action, moment develop is
6
2x5.3x21.9
6
PHh 22
=
M = 5.723 KN-m
16
IS 3370 (Part II) 1984
Ast min:-
0.3 % c/s for 100 mm of
0.2% C/S for 450 mm
∴for 270 mm 0.251% C/S
Ast min =
2701000x100
251.0
= 677.7 mm2
For 10mm φ , S =
mm87.1157.677
410x
x10002
=
π
∴Provide 10mm @ 220 mm C/C as vertical reinforcement along both
the faces the taks for long will and short will.
Design of Base slab:-
Assuming a thickness of 150 mm min. slab is provided for the base
slab. Since it is resting on firm ground.
Ast min =
2mm450150x100x1000
3.0 =
= 450mm2
For 8mm φ S =
mm7.111450
48x
x10002
=
π
14
IS 3370 (Part II) 1984
∴ Provide 8 mm φ @ 220 mm C/C along both ways along both faces
Provide a layer of lean mix (M10) for a thickness of 75 mm below base
slab.
Figure **************
A straight staircase is made of structures independent tread slab
cantilevered from the R.C wall given riser is 150 mm and trend is
300mm. width if flight is 1.5m. design a typical cantilever tread slab. Any
live load for over crowing. Use M20 concrete of Fe. 250 steel.
Soln:-
Given
R = 159 mm, T= 300 mm, w = 1.5 m M20 9 Fe 250
Loading on tread slab (0.3m)
i) self wt = 0.15 x 25 x 0.3 = 1.125 KN/m
ii) floor finish = 0.6 x 0.3 = 1.125 KN/m
dead load = 1.305 KN/m
B.M D.L =
KNm46.12
5.1x5.13 2
=
Live load moment is a minimum of
i) UDL due to live load on stair (for over providing case) WL = 5 KN/m2
ii) For cantilever tread slab the live load condition, is check for it load of
1.3 KN at tread.
16
IS 3370 (Part II) 1984
Figure *************
B.M LL1 =
2
5.1x5.1 2
B.M LL2 = 1.3 x 1.5 = 1.95 KN
Maximum of B.M LL = 1.9 KN m
Total BM = BM DL + B.D LL = 1.46 + 19.5
= 3.41 KNm.
Factored B.M = 1.5 x 3.41 = 5.12 KNm
Assume clear covet of 15 mm & 12mm bar.
d = 150 – 15 – 12/2 = 129 mm
Mu = 0.87 fy Ast d
−tdFck
Astf1 y
5.12 x 106 – 0.87 x 250 x Ast x 129
−
129x300x20
Ast2501
Ast = 194.73 mm2
n =
nos3say47.2
410x
194373
ast
Ast2 =
π=
provide 3 nos of 10 mm bar
Distribution steel:-
14
IS 3370 (Part II) 1984
Ast min = 0.15% & D =
150x300x100
15.0
= 67.5 mm2
S =
mm7445.67
48x
x10002
=
π
Provide 8 mm @ 300 mm C/C
Check for shear:-
d
uv b
V=τ
Dead load
V DL = 1.3.5 x 1.5 = 1.959 KN
V LL1 = 1.5 x 1.5 = 2.25 KN
VLL2 = 1.3 KN
Max VLL 2.25 KN
Total Vu = 2.25 + 1.958 = 4.208 KN
Factored Vu = 1.5 x 4.208 = 6.31 KN
v =
2mm/N168..0129x300
10x31.6 =
∴τ c = 0.48 N/mm2
K = 1.3 from table
16
IS 3370 (Part II) 1984
Tc1 = K Tc = 1.3 x 0.48 = 0.624 N/mm2
τ c 2.8/2 = 1.4 N/mm2
τ v x c x τ c mm
Hence safe in shear
Ld =
mm15.4532.1x4
250x87.0x10
bd4
3
==τσφ
Design a suitable countertor reintaining way to support a leveled back till
of height 7.5m above g.L on the toe side. Assume good soil for the
foundation at a desm of 1.5m below G.L the SBC of soil is 170 KN/m2
with unit weight as 16KN/m3. The angle of internal friction is = 300
the co-efficient of friction b/w the soil and concrete is 0.5 use M25
concrete and Fe 415 steel.
Soln:-
Figure **************
Minimum depth of foundation = ***
= ***
Depth of foundation = 1.5 m
Height of wall = 7.5 + 1.5 = 9m
Thickness of had and stem = 5% of 9m = 0.45m ** 0.5.
Thickness of toe slab = 8% of 9 m = 0.72 m.
16