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8/12/2019 Design of Rectangular Water Tank (1)
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Capacity 80000 Litres (given)
Material M20 (given)
Fe 415 Grade HYSD reinforcement (given)
80400 Litres
L / B = 6 / 4 = 1.5 < 2 .
H / 4 = 3.5 / 4 = 0.875 m
2.5 x YW= 2.5 x 9.8
= 24.5 KN / m2
where Ywis unit weight of water = 9.8 KN / m3
6 m
6 m A E
F
3.5 m 24.5 KN / m2
D
34.3 KN / m2
Elevation Plan
MAB =
= 73.5 KNm
MAD =
= -32.66 KNm
- 32.66 73.5
0 -12.25 -8.17 0
0 -12.25 -8.17 0
D 0 A 0 B
-57.16 57.16
Fixed end moments :-
Design of Rectangular water tank CASE-1 ( L / B < 2 )
Grade Concrete
Volume = 6 x 4 x 3.35 x 103
=
Water pressure at 3.5 - 1 = 2.5 m height from top =
Solution :-
Provide 6 m x 4 m x 3.5 m tank with free board of 150 mm.
The top portion of side walls will be designed as a continuous frame.
bottom 1 m or H / 4 whichever is more is designed as cantilever.
bottom 1 m will be designed as cantilever.
Rotation factor at Joint A
To find moment in side walls, moment distribution or kani's method is used. As the
frame is symmetrical about both the axes, only one joint is solved
Kani's Method :-
24.5 x 62/ 12 =
24.5 x 42/ 12 =
w x l2/ 12 =
w x l2/ 12 =
-3/10 -2/1040.84
2.5 m
1 m
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Joint Member Relative
Stiffness( K ) K
Rotation Factor
u =(-1/2) k / K
AB I / 6 - 2 / 10
AD I / 4 - 3 / 10
MAF =
40.84 KNm
MAB=
=
= 57.16
MAD=
= (- 32.66 ) + 2 x (- 12.25 ) + 0
= -57.16
=
53.09 KNm
=
-8.16 KNm
= 49 KN
= 73.5 KN
M = 57.16 KNm
T = 49 KN
Q = 0.306
D = M / Q x b
= 57.16 x 106/ 0.306 x 1000
= 432.2 mm,
Take D = 450 mm d = 450 - 25 - 8
= 417 mm
Direct tension in long wall =
A 5 * I / 12
Sum of FEM
73.5-32.66
B.M. at centre of long span =
B.M. at centre of short span =
MABF+ 2 MAB' + MBA'
73.5 + 2 x (- 8.17 ) + 0
MADF+ 2 MAD' + MDA'
Direct tension in short wall = Yw( H - h ) x L / 2
24.5 x 6 / 2 =
Design of Long Walls :-
24.5 x 4 / 2 =
w x l2/ 8 - 57.16
w x l2/ 8 - 57.16
Yw( H - h ) x B / 2
24.5 x 42/ 8
- 57.16
24.5 x 62/ 8
- 57.16
Tension on liquid face.
Assuming d / D = 0.9
At support
From Table 9-6
From Table 9-5
Assuming d / D = 0.9
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M / stx j x d
=
= 1048 mm2
Ast2for direct tension = T / st
=
= 327 mm2
= 1375 mm2
=
= 146.15273 mm
Provide 16 mm O bar @ 130 mm C/Cmarked(a) = 1546 mm2/ m.
At centre
M = 53.09 KNm
T = 49 KN
e = M / T =
= 1.08 m
E = e + D / 2 - d b
=
= 888 mm
D
= 49 x 0.888d
= 43.51 KNm
d'
M / stx j x d
=
= 617 mm2
Ast2for direct tension = T / st
=
= 327 mm2
= 944 mm2
=
= 212.88136 mm
Total Ast1+ Ast2 = 617 + 327
Provide 16 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m2/ m
200.96 x 1000 / 944
E = e + D / 2 - d
1080 + 450 / 2 - 417
modified moment
Ast1for moment =
43.51 x 106
/ 190 x 0.89 x 417
49 x 103
/ 150
Larger steel area is provided to match with the steel of short walls.
49 x 103
/ 150
Total Ast1+ Ast2 = 1048 + 327
i.e.tension is small
Line of action of forces lies outsi
spacing of bar = Area of one bar x 1000 / required area in m2/ m
57.16 x 106
/ 150 x 0.872 x 417
Ast1for moment =
53.09 / 49
200.96 x 1000 / 1375
Provide 16 mm O bar
tension on remote face
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Provide 16 mm O bar @ 200 mm C/Cmarked(b) = 1005 mm2
= 720 mm2
360 mm2
=
= 139.55556 mm
Provide 8 mm O bar @ 130 mm C/Cmarked(d) = 385 mm2
=
= 218.05556 mm
Provide 10 mm O bar @ 200 mm C/Cmarked(c) = 392 mm2
Remote face ( b) - Provide 16 mm O @ 200 mm C/C = 1005 mm2
385 mm2
mm2
M = 57.16 KNm
T = 73.5 KN
M / stx j x d
== 1048 mm
2
Ast2for direct tension = T / st
=
= 490 mm2
= 1538 mm2
== 130.6632 mm
Provide 16 mm O bar @ 130 mm C/Cmarked(b) = 1546 mm2/ m.
At centre
M = 8.16 KNm
T = 73.5 KN
200.96 x 1000 / 1538
tension on liquid face
57.16 x 106
/ 150 x 0.872 x 417
73.5 x 103
/ 150
Total Ast1+ Ast2 = 1048 + 490
Provide 16 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m2/ m
Vertical Steel ( c) - Provide 10 mm O @ 200 mm C/C both faces = 392
Design of short walls :-
At support
tension on liquid face
From Table 9-5
Ast1for moment =
spacing of bar = Area of one bar x 1000 / required area in m2/ m
50.24 x 1000 /360
Horizontal steel :-
Liquid face ( d) - Provide 8 mm O @ 130 mm C/C =
Provide 10 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m2/ m
78.50 x 1000 /360
Vertical Steel ( c)
Distribution steel =
From Table 9-3 minimum reinforcement 0.16 %
0.16 / 100 x 1000 x 450
On each face =
Provide 8 mm O bar
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M / stx j x d
=
= 150 mm2
Ast2for direct tension = T / st
=
= 490 mm2
= 640 mm2
=
= 176.625 mm
Provide 12 mm O bar @ 130 mm C/Cmarked(e) = 869 mm2/ m.
= 720 mm2
360 mm2
=
= 139.55556 mm
Provide 8 mm O bar @ 130 mm C/Cmarked(d) = 385 mm2
=
= 218.05556 mm
Provide 10 mm O bar @ 200 mm C/Cmarked(c) = 392 mm2.
Remote face ( e) - Provide 12 mm O @ 130 mm C/C = 869 mm2
385 mm2
mm2
M = OR Ywx H / 6 , whichever is greater.
= = 9.8 x 3.5 / 6
= 5.72 KNm = 5.72
78.50 x 1000 /360
Horizontal steel :-
Liquid face ( d) - Provide 8 mm O @ 130 mm C/C =
Vertical Steel ( c) - Provide 10 mm O @ 200 mm C/C both faces = 392
spacing of bar = Area of one bar x 1000 / required area in m2/ m
50.24 x 1000 /360
Vertical Steel ( c)
Provide 10 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m2/ m
From Table 9-3 minimum reinforcement 0.16 %
Distribution steel = 0.16 / 100 x 1000 x 450
On each face =
Provide 8 mm O bar
Total Ast1+ Ast2 = 150 + 490
Provide 12 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m2/ m
113.04 x 1000 / 640
From Table 9-5
Ast1for moment =
8.16 x 106
/ 150 x 0.872 x 417
73.5 x 103
/ 150
Bottom 1 m will be designed as cantilever
,tension on liquid face.
Cantilever moment : -
Ywx H x h2/ 6
9.8 x 3.5 x 1 / 6
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M / stx j x d
=
= 105 mm2
= 720 mm2
360 mm2
=
= 218 mm
Provide 10 mm O bar @ 200 mm C/Cmarked(c) = 392 mm2.
0.229%
=
= 344 mm2
=
= 292 mm
Ast = 346 mm2
lx= 4 + 0.15 = 4.15 say 4.5 m
ly= 6 + 0.15 = 6.15 say 6.5 m
3.75 KN / m2
1.0 KN / m2
1.5 KN / m2
6.25 KN / m2
PU=
= 9.38 KN / m
ly/ lx= 6.5 / 4.5
= 1.4
From Table 9-3 minimum reinforcement 0.16 %
From Table 9-5
Astfor moment =
5.72 x 106
/ 150 x 0.872 x 417
spacing of bar = Area of one bar x 1000 / required area in m2/m
78.50 x 1000 /360
each face
Distribution steel = 0.16 / 100 x 1000 x 450
On each face =
Provide 10 mm O bar
Base slab :-
Base slab is resting on ground. For a water head 3.5 m, provide 150 mm thick slab.
From table 9-3
Top slab may be designed as two-way slab as usual for a live load of 1.5 KN / m2
Minimum steel =
0.229 / 100 x 1000 x 150
Provide 8 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m2/ m
50.24 x 1000 /172
Dead Load : self 0.15 x 25 =
floor finish =
1.5 x 6.25
,172 mm2bothway
Provide 8 mm O bar @ 290 mm C/C both ways, top and bottom.
Designed section,Elevation etc. are shown in fig.
Live load =
consider 1 m wide strip. Assume 150 mm thick slab.
For 1 m wide strip
AS Per IS-456-2000,Four Edges Discontinuous,positive moment at mid-span.
Top slab : -
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Table 26 x = 0.085
y = 0.056
Mx= My=
= =
= 16.15 KNm = 10.64 KNm
M / Q x b
= 16.15 x 106/ 2.76 x 1000
= 76.50 mm
dshort=
= 130 > 76.50 mm
dlong= 130 - 10 = 120 mm
= 0.96
Pt=
fy/ fck
=
415 / 20
=
= 0.29%
= 377 mm2
=
= 208 mm
Provide 10 mm O bar @ 210 mm c/c = 374 mm2.
= 0.74
Pt=fy/ fck
=
415 / 20
(O.K.)
Larger depth is provided due to deflection check.
y x w x lx2
0.085 x 9.38 x 4.52
0.056 x 9.38 x 4.52
x x w x lx2
From Table 6-3 ,Q = 2.76
Mu/ b x d2(short) = 16.15 x 10
6/ 1000 x 130 x 130
50 1-1-(4.6 / fck) x (Mu / b x d2)
Mu/ b x d2(long) = 10.64 x 10
6/ 1000 x 120 x 120
50 1-1-(4.6 / fck) x (Mu / b x d2
)
50 1-1-(4.6 / 20) x (0.74)
50 1-1-(4.6 / 20) x (0.96)
50 [(1-0.88) x 20 / 415 ]
78.50 x 1000 /377
Ast(short) = 0.29 x 1000 x 130 / 100
Provide 10 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m2/m
drequired=
150 - 15(cover) - 5
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=
= 0.22%
= 264 mm2
=
= 190 mm
= 264 mm2.
50 [(1-0.91) x 20 / 415 ]
Provide 8 mm O bar @ 190 mm c/c
Provide 8 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m2/ m
50.24 x 1000 /264
Ast(long) = 0.22 x 1000 x 120 / 100
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B
4 m
C
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Q = M / bD2
Pt Q = M / bD2
Pt
0.75 0.3 0.4 0.295 0.289
0.8 0.305 0.37 0.299 0.272
0.85 0.31 0.355 0.302 0.258
0.9 0.314 0.335 0.306 0.246
Balanced Design Factors for members in bending
For M20 Grade Concrete Mix
Mild steel HYSD barsd / D
TABLE 9-5
TABLE 9-6
Members in bending ( Cracked condition )
Coefficients for balanced design
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Grade of
concrete
Grade of
steelcbc
N / mm2
st
N / mm2 k j Q
M20 Fe250 7 115 0.445 0.851 1.33
Fe415 7 150 0.384 0.872 1.17
For members more than 225mm thickness and tension away from liquid face
M20 Fe250 7 125 0.427 0.858 1.28
Fe415 7 190 0.329 0.89 1.03
D / 2
e = M / T
0.214
0.24
0.2290.217
0.206
0.194
0.183
0.171
300
350
400
0.3
0.2860.271
0.257
0.243
0.229
Thickness, mm
100
150200
250
TABLE 9-3
Minimum Reinforcement for Liquid Retaining Structures
% of reinforcement
Mild Steel HYSD bars
ide the section
For members less than 225mm thickness and tension on liquid face
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0.2 0.16450 or more
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3500
150
1 : 4 : 8 P.C.C.
150 450 6000 450 150
Elevation
1500
450
1000
4000
1000
450
1500 1500
6000
450 450
Section A-A
Table 6-3
A15001500
150
150
8 O @ 290 c/c both ways top and bottom
10 O @ 200 c/c
10 O @ 200 c/c - shape
- shape
10 O @ 210 c/c 8 O @ 190 c/c
150 Free board
1000
16 O @ 130 c/c (a)
10 O @ 200 c/c both faces (c)
12 O @ 130 c/c (e)
16 O @ 200 c/c (b)
8 O @ 130 c/c (d)
( a ) ( a )( b ) ( d )
( c )
( d )
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250 415 500 550
15 2.22 2.07 2.00 1.94
20 2.96 2.76 2.66 2.5825 3.70 3.45 3.33 3.23
30 4.44 4.14 3.99 3.87
Limiting Moment of resistance factor Qlim, N / mm2
fy, N / mm2
fck N / mm2
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Pt,bal
1.36
0.98
1.2
0.61
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(given)
Material M20 (given)
Fe 415 (given)
86400 Litres
L / B = 8 / 3.6 = 2.22 > 2 .
H / 4 = 3.0 / 4 = 0.75 m
=
=
= 44.1 KNm.
=
=
= 26.13 KNm.
==
= 19.60 KNm.
For bottom portion
M = OR
= =
= 4.90 KNm = 4.90 KNm
=
= 35.28 KN
=
= 19.6 KN
Volume = 3.6 x 8 x 3.0 x 103
=
The short walls are designed as supported on long walls.
If thickness of long walls is 400 mm, the span of the short wall = 3.6 + 0.4 = 4.0 m.
The long walls are designed as vertical cantilevers from the base.
bottom 1 m or H / 4 whichever is more is designed as cantilever.
bottom h = 1 m will be designed as cantilever.
Design of Rectangular water tank CASE-2 ( L / B 2 )
Grade Concrete
Solution :-
Size of tank : 3.6 m x 8.0 m x 3.0 m high
Grade HYSD reinforcement
Size of tank : 3.6 m x 8.0 m x 3.0 m high
Maximum B.M. in long walls at the base
(1 / 6 ) x Ywx H3
( 1 / 6 ) x 9.8 x 33
Moments and tensions :
Maximum ( - ve ) B.M. in short walls at support
Ywx ( H - h ) x B2/ 12
9.8 x ( 3 - 1 ) x 42/ 12
Maximum ( + ve ) B.M. in short walls at centre
Direct tension in long wall = Yw x( H - h ) x B / 2
Direct tension in short wall= Yw( H - h ) x 1
Ywx ( H - h ) x B2
/ 169.8 x ( 3 - 1 ) x 4
2/ 16
Ywx H x h2/ 6
9.8 x 3.0 x 1 / 6
Ywx H / 6 , whichever is greater
9.8 x 3.0 / 6
9.8 x ( 3 - 1 ) x 1
9.8 x ( 3 - 1 ) x 3.6 / 2
It is assumed that end one metre width of long wall gives direct tension to short walls.
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M ( - ) = 44.1 KNm
T = 35.28 KN
D = M / Q x b
=
= 379.6 mm,
Take D = 400 mm d =
= 367 mm
Ast =
=
= 918.68 mm2
=
= 218.75 mm
= 1005 mm2.
As=
= 684 mm2.
342 mm2.
= T / st
=
= 235 mm2.
== 146.9 mm
= 357 mm2
Provide 16 mm O bar @ 200 mm c/c
44.1 x 10 6/ 0.306 x 1000
400 - 25 - 8
M / stx j x d
Design of long walls : -
( water face )
( perpendicular to moment steel )
Assume d / D = 0.9 Q = 0.306
200.96 x 1000 / 918.68
( 0.171 / 100 ) x 1000 x 400
From Table 9-6
From Table 9-5 ,
Provide 16 mm O bar
spacing of bar =
Astfor Moment
Provide 8 mm O bar
Note : The design is made at the base. The moment reduces from base to top.For ec
reinforcement can be curtailed or the thickness of wall can also be reduced as we hav
cantilever retaining wall.
Distribution steel = 0.171 % for 400 mm depth
From Table 9-3
Steel required for direct tension
35.28 x 103/ 150
( 2 )
44.1 x 106
/ 150 x 0.872 x 367
Area of one bar x 1000 / required area in m2/ m
From ( 1 ) and ( 2 ) , minimum steel is sufficient for resisting direct tension.
spacing of bar = Area of one bar x 1000 / required area in m2/ m
50.24 x 1000 / 342
on each face = ( 1 )
Provide 8 mm O bar @ 140 mm c/c on each face
on each face
Design of short walls :-
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M = 26.13 KNm
T = 19.6 KN
=
= 544 mm2
T / st
=
= 131 mm2
= 675 mm2
=
= 167.47 mm= 706 mm
2.
1000
203.56
400 367
163.44
= 13.33
=
= 203.56 mm
D - x = 196.44 mm
d - x = 163.44 mmAT=
=
= 408705 mm2
Ixx=
Provide 12 mm O bar@160 mm c/c
checking :
At support
From Table 9-5
Ast1for moment =
113.04 x 1000 / 675
19.6 x 103
/ 150
M / stx j x d
26.13 x 106
/ 150 x 0.872 x 367
Ast2for direct tension =
Area of one bar x 1000 / required area in m2/ m
Total Ast1+ Ast2 = 544 + 131
Provide 12 mm O bar
spacing of bar =
( 1000 x 4002/ 2 ) + ( 706 x ( 13.33 - 1 ) x 367 )
( 1000 x 400 ) + ( ( 13.33 - 1 ) x 706 )
b x D + ( m - 1 ) x Ast
1000 x 400 + (13.33 - 1 ) x 706
modular ratio m = 280 / 3 x cbc
x =b x D
2/ 2 + Ast( m - 1 ) x d
b x D + ( m - 1 ) x Ast
( 1 / 3 ) x b x ( x3+ ( D - x )
3) + ( m - 1 ) x Astx ( d - x )
2
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=
= 5.34E+09 + 2.33E+08
= 5.57E+09 mm4
fct = T / AT
=
= 0.048 N / mm2
fcbt =
=
= 0.767 N / mm2
check :
1
0.4912 1
M = 19.6 KNm
T = 19.6 KN
=
= 408 mm2
T / st
=
= 131 mm2
= 539 mm2
=
= 209.72 mm
= 565 mm2.
As=
= 684 mm2.
342 mm2.
26.13 x 106x 163.44 / 5.57 x 10
9
( fct/ ct ) + ( fcbt/ cbt )( 0.048 / 1.2 ) + ( 0.767 / 1.7 ) 1
( 1 / 3 ) x 1000 x ( 203.563+ 196.44
3) + ( 13.33 - 1 ) x 706 x 163.44
2
19.6 x 10 3/ 408705
M x ( d - x ) / Ixx
From Table 9-2
At centre :
From Table 9-5
Ast1for moment = M / stx j x d
0.04 + 0.4512 1
.. ( O. K. )
Provide 12 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m2/ m
113.04 x 1000 / 539
19.6 x 106
/ 150 x 0.872 x 367
Ast2
for direct tension =
19.6 x 103
/ 150
Total Ast1+ Ast2 = 408 + 131
on each face = ( 1 )
Provide 12 mm O bar @ 200 mm c/c
From Table 9-3
Distribution steel = 0.171 % for 400 mm depth
( 0.171 / 100 ) x 1000 x 400
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= T / st
=
= 131 mm2.
=
= 146.9 mm
= 357 mm2
M = 4.9 KNm
Ast =
=
= 102 mm2
= 357 mm2
0.229%
=
= 344 mm
2
=
= 292 mm
Ast = 346 mm2
lx= 3.6 + 0.4 = 4 say 4 mly= 8 + 0.15 = 8.15 say 8.5 m
3.75 KN / m2
1.0 KN / m2
1.5 KN / m2
6.25 KN / m2
Top slab may be designed as a one-way slab as usual for a live load of 1.5 KN / m2
Top slab : -
consider 1 m wide strip. Assume 150 mm thick slab.
Dead Load : self 0.15 x 25 =
floor finish =
Live load =
Provide 8 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m2/ m
50.24 x 1000 /172
Provide 8 mm O bar @ 290 mm C/C both ways, top and bottom.
Designed section,Elevation etc. are shown in fig.
Minimum steel =
0.229 / 100 x 1000 x 150
,172 mm
2
bothway
Base slab :-
Base slab is resting on ground. For a water head 3 m, provide 150 mm thick slab.
From table 9-3
Steel required for direct tension
19.6 x 103/ 150
50.24 x 1000 / 342
on each face
Bottom cantilever
Provide 8 mm O bar @ 140 mm c/c on each face
( 2 )From ( 1 ) and ( 2 ) , minimum steel is sufficient for resisting direct tension.
Provide 8 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m2/ m
Provide 8 mm O bar @ 140 mm c/c on each faces
on each face
From Table 9-5
M / stx j x d
4.9 x 106
/ 150 x 0.872 x 367
Minimum steel = 342 mm2on each face.
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PU=
= 9.38 KN / m
= 18.76 KNm
= 16.88 KN
drequired= M / Q x b
= 18.76 x 10 6/ 2.76 x 1000= 82.44 mm
dprovided=
=
= 1.13
Pt=
fy/ fck
=
415 / 20
== 0.34%
= 439 mm2
=
= 114 mm
= 457 mm2.
= 180 mm2
=
Provide 8 mm O bar @ 110 mm c/c
Provide 6 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m2/ m
Distribution steel = ( 0.12 / 100 ) x 1000 x 150
28.26 x 1000 /180
Maximum moment = 9.38 x 42/ 8
Maximum shear = 9.38 x 3.6 / 2
From Table 6-3 ,Q = 2.76
Minimum steel is 0.15 % for mild steel and 0.12 % for HYSD Fe415 reinforcement
Provide 8 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m2/m
50.24 x 1000 /439
50 1-1-(4.6 / 20) x (1.13)
50 [(1-0.86) x 20 / 415 ]
Ast = 0.34 x 1000 x 129 / 100
(O.K.)
Larger depth is provided due to deflection check.
Mu/ b x d2 = 18.76 x 10
6/ 1000 x 129 x 129
50 1-1-(4.6 / fck) x (Mu / b x d2)
129 > 82.44
Design for flexure :
150 - 15(cover) - 6
1.5 x 6.25
For 1 m wide strip
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= 157 mm
= 188 mm2.Provide 6 mm O bar @ 150 mm c/c
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nomy, the
e done for
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x
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M15 1.1 1.5 1.5
M20 1.2 1.7 1.7
M25 1.3 1.8 1.9
M30 1.5 2.0 2.2
M35 1.6 2.2 2.5
M40 1.7 2.4 2.7
Table 9-2
Grade of
concrete
Permissible stresses in N / mm2
Direct
tension ct
Tension due to
bending cbtShear stress
v = V / b j d
Permissible concrete stresses in calculations relating
to resistance to cracking
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150
3000
150
1 : 4 : 8 P.C.C. 150
150 400 8000 400
Section A-A
2000
400
900
3600
900
400
2000 2000
8000
400 Sectional plan 400
150
3000
Base details not
8 O @ 140 c/c (chipiya)
20002000
8 O @ 290 c/c both ways top and bottom
12 O @ 200 c/c
- shape
8 O @ 110 c/c 6 O @ 150 c/c
900
12 O @ 160 c/c(chipiya)16 O @ 200 c/c ( chipiya )
150
8 O @ 140 mm c/c
8 O @ 140 c/c
12 O @ 200 c/c
B
B
8 O @ 140 c/c
8 O @ 140 c/c
8 O @ 110 c/c 6 O @ 150 c/c
16 O @ 200 c/c ( chipiya )
8 O @ 140 c/c
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150
150
150 400 3600 400 150
Section B- B
shown for clarity900 900
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( given )
2.5 KN / m2 ( given )
1 KN / m2 ( given )
( given )
( given )
= 25.2
= 158.7 mm
D =
= 178.7 mm180 mm
4.5 KN / m2
1.0 KN / m2
2.5 KN / m2
Total 8.0 KN / m2
1.5 x 8 = 12 KN / m
w x l2/ 8
= 12 x 42/ 8
= 24 KNm
w x l / 2
=
= 24 KN
d =
= 160 mm
= 0.94
Pt= 50 1-1-(4.6 / fck) x (Mu/ b x d2)
fy/ fck
=
415 / 15
=
= 0.289%
M15 grade concrete
HYSD reinforcement grade Fe415
Design of simply supported one way slab
effective span = 4 m supported on masonry wall of 230 mm thickness
Live load =
( span / d ) ratio permissible = 1.26 x 20
IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement
for simply supported, basic span / effective depth ratio = 20
solution : -
Assume 0.4 % steel , a trial depth by deflection criteria
modification factor = 1.26
Floor finish =
material
DL = 0.18 x 25 =
Floor finish =
Live load =
Factored load =
drequired= 4000 / 25.2
158.7 + 15 ( cover ) + 5 ( assume 10 O bar )
Assume an overall depth =
Mu/ b x d2= 24 x 10
6/ 1000 x (160)
2
50 1-1-(4.6 / 15) x (0.94)
Maximum moment =
Maximum shear =
12 x 4 / 2
Design for flexure : -
Consider 1 m length of slab
50 [(1-0.84) x 15 / 415 ]
180 - 15 - 5
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= 462 mm2
== 170 mm
= 462 mm2.
= 0.128
= 216 mm2
=
= 233 mm
= 218 mm2.
Vu= 24 KN
=
= 0.150 N / mm2
d = 160 mm
As= 231 mm .
= 0.144
6 x =
= 0.8 x 15 / 6.89 x 0.144
= 12.1
6 x 12.1
= 0.277
N / mm2
spacing of bar = Area of one bar x 1000 / required area in m2/ m
78.50 x 1000 /462
Provide 10 mm O bar @ 170 mm c/c
Ast= 0.289 x 1000 x 160 / 100
Provide 10 mm O bar
> 0.12 % ( minimum steel for Fe415)
i.e. remaining bars provide minimum steel. Thus, half the bars may be bent up.
Distribution steel = ( 0.12 /100 ) x 1000 x 180
Half the bars are bent at 0.1 l = 400 mm , and
remaining bars provide 231 mm2area
100 x As/ ( b x D ) = 100 x 231 / ( 1000 x 180 )
spacing of bar =
50.24 x 1000 /216
Provide 8 mm O bar @ 230 mm c/c
Area of one bar x 1000 / required area in m2/m
Maximum spacing = 5 x 160 = 800 or 450 mm i.e. 450 mm
Provide 8 mm O bar
100 x As/ b x d = 100 x 231 / 1000 x 160
For bars at support
Check for shear : -
Actual Shear stress = Vu/ b x d
24 x 103/ 1000 x 160
for Pt = 0.144 c= 0.28
IS 456-2000 Table 19 from table 7-1
< ( C)N / mm2 ( too small )
Design shear strength c= 0.85 0.8 x fck( 1 + 5 x - 1 )
0.8 x fck/ 6.89 Pt , but not less than 1.0
Design shear strength c= 0.85 0.8 x 15 ( 1 + 5 x 12.1 - 1 )
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IS 456-2000 clause 40.2.1.1 -0.05
k = 1.24 ? -0.04
= 0.347 N / mm2 .( O.K.)
8 OAs= 231 mm
Mu1= OR
= { 1 - (415 x 231 / 15 x 1000 x 160 ) }
= 13.34 { 1- 0.0399 }
= 12.812 KNm
Vu = 24 KN
693.875 + 8 O 56 O48 O 693.88
O 14.46 mm .( O.K.)
20
Pt=
= 0.289IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement
= 28.4
= 25.00 < 28.4 .( O.K.)
= 3 x 160 = 480 mm170 mm
5 x effective depth of slab or
= 5 x 160
230 mm
Check for development length : -
Assuming L0=
for 180 mm slab depth
Design shear strength = 1.24 x 0.28
0.87 x 415 x 231 x 160
= 56 O ( from Table 7-6 )
Check for deflection : -
Basic ( span / d ) ratio =
100 x Ast/ b x d = 100 x 462 / 1000 x 160
modification factor = 1.42
For tying the bent bars at top , provide 8 mm O @ 230 mm c/c
( span / d ) ratio permissible = 1.42 x 20
Actual (span / d ) ratio = 4000 / 160
Main bars : maximum spacing permitted =
Distribution bars : maximum spacing permitted =
spacing provided =
25 difference
20 difference
(HYSD Fe415 steel ) For continuing bars
0.87 x fyx Astx d { 1 - ( fyx Ast/ fckx b x d ) }
The depth could be reduced
1.3 x ( 12.81 x 106/ 24 x 10
3) + 8 O 56 O
which gives
.( O.
= 800 mm
.( O.
spacing provided = < 300 mm
3 x effective depth of slab or 300 mm
or 300 m
Check for cracking : -
< 450 mm
IS 456-2000 , clause 26.3.3
Development length of bars Ld= O s/ 4 x bd1.3 x ( Mu1/ Vu ) + L0Ld
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NOTE : -
NOTE : -
If clear span = 3.77 m , effective span = 3.77 + 0.23 = 4 m OR
effective span = 3.77 + 0.16 ( effective depth ) = 3.93 m
0.6 % for mild steel reinforcement and 0.3 to 0.4 % for HYSD Fe415 grade
reinforcement
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For mild steel minimum reinforcement 0.15 %
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Pt=
=
= 0.14
Pt= 50 1-1-(4.6 / fck) x (Mu/ b x d2)fy/ fck
we get , 0.49
Mu1=
= 12.54 KNm
160
180
450 mm whichever is small
0.49x 1000 x 1602
x 10-6
400
For checking development lmay be assumed as 8 O for
bars ( usually end anchorag
provided ) and 12 O for mild
U hook is provided usually
anchorage length is 16 O.
100 x As/ b x d
100 x 231 / 1000 x 160
From equation
Mu1/ b x d2=
4000
400
.)
or 450 mm i.e. 450 mm
.)
whichever is small
i.e. 300 mm
8 O @ 230 c/c10 O @ 170 c/c( alternate bent )
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( c )Cantilever-The effective length of a cantilever
shall be taken as its length to the face of the
support plus half the effective depth except
centre to centre distance shall be used.
where it forms the end of a continuous beamwhere the length to the centre of support shall
be taken.
( d )Frames-In the analysis of a continuous frame,
of the beam or slab or the clear span plus
half the width of the discontinuous support,
whichever is less;
3) In the case of spans with roller or rocket
bearings, the effective span shall always be
the distance between the centres of bearings.
other continuous or for intermediate spans,
the effective span shall be the clear span
between supports;
2) For end span with one end free and the other
continuous, the effective span shall be equal
to the clear span plus half the effective depth
shall be taken as under:
( b )Continuous Beam or Slab - In the case of
continuous beam or slab, if the width of the
support is less than l/12 of the clear span, theeffective span shall be as in (a). If the
1) For end span with one end fixed and the
Effective Span
IS 456-2000 clause 22.2
( a ) Simply Supported Beam or Slab -
The effective span of a member that is not built integrally with its
supports shall be taken as clear span plus the effective depth of
slab or beam or centre to centre of supports , whichever is less.
supports are wider than I/12 of the clear span
or 600 mm whichever is less, the effective span
If ly/ lx 2 ,called one way slab provided thatit is supported on all four edges . Note that , if
all four edge is not supported and ly/ lx< 2 ,
then also it is one-way slab,If ly/ lx< 2 , called
two-way slab.provided that it is supported on
all four edges.
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ngth , l0HYSD
is not
steel (
hose
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material ( given )
( given )
( Balcony slab )
DL LL
For S2 3 0 KN / m2
1 0 KN / m2
0 2 KN / m2
Total 4 2 KN / m2
Pu= ( 6 + 3 ) KN / m2
DL LL
For S1 3 0 KN / m2
1 0 KN / m2
0 3 KN / m2
Total 4 3 KN / m2
Pu= (6 + 4.5) KN / m2
1.875 KN / m
Pu= 2.8 KN / m
9 KN/m 6 KN/m
A 3m B 1.2m C
9 KN/m 2.8 KN
A 3m B 1.2m C
considering fig (a)
wx l2/ 2
= 6 x 1.22/ 2
mild steel grade Fe250
1.5 ( 4 + 3 ) =
Weight of parapet 0.075 x 25 x 1 =
1.5 x 1.875 =
Consider 1 m long strip(1) To get maximum positive moment in slab S2only dead load on slab
S1and total load on slab S2shall be considered
(b) Loads for maximum negative
moment,maximum shear for cantilever span
and maximum reaction at support B
cantilever moment =
Solution : -
For slab S2live load = 2 KN /m2
For slab S1live load = 3 KN /m2
Assume 120 mm thick slab
self load = 0.12 x 25 =
floor finish =
live load =
1.5 ( 4 + 2 ) =
self load = 0.12 x 25 =
Live Load As per IS 875
ly=
(a) Loads for maximum positive moment
Design of Cantilever one way slab
floor finish =
live load =
10.5 KN/m
M15 grade concrete
used for residential purpose
at the free end of slab S1,concrete parapet of 75 mm thick and 1 m high.
S S2
S2 S1
S1
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= 4.32 KNm
shear =
=
= 12.06 KN
= 1.34 m
=
= 8.08 KNm
=
= 10.92 KNm
Vu,BA=
== 17.14 KN
Vu,BC=
=
= 15.4 KN
10.92 KNm
Q = 2.22
=
= 70.14 mm,
= 99 mm,
= 0.82
Pt= 50 1-1-(4.6 / fck) x (Mu/ b x d2)fy/ fck
=
250 / 15
=
= 0.405%
= 401 mm2
Moment steel :
50 1-1-(4.6 / 15) x (0.82)
50 [(1-0.865) x 15 / 250 ]
10.5 x 1.2 + 2.8
Maximum moment =
Mu/ b x d2( + ) =
120 - 15 - 6 ( assume 12 O bar )
Ast( + ) = 0.405 x 1000 x 99 / 100
.( O.K.)8.08 x 10
6/ 1000 x (99)
2
From Table 6-3
dprovided=
10.92 x 10 6/ 2.22 x 1000
Maximum shear at B
9 x 3 / 2 + 10.92 / 3
w x l / 2 + Moment @ B at distance 3 m
w x l + 2.8
the slab is loaded with full loads as shown in fig (b)
Maximum positive moment = 12.06 x 1.34 - W x l2/ 2
12.06 x 1.34 - 9 x 1.342/ 2
M / Q x b
9 x 3 / 2 - 4.32 / 3
w x l / 2 - Moment @ B at distance 3 mReaction at A =
2.8 x 1.2 + 10.5 x 1.22/ 2
Maximum negative moment = w x l + w x l2/ 2
(2) To get maximum negative moment and maximum shear at B,
Point of zero shear from A = 12.06 ( KN ) / 9 ( KN / m )
drequired=
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=
= 195.76 mm
= 462 mm2.
= 1.11
Pt= 50 1-1-(4.6 / fck) x (Mu/ b x d2)fy/ fck
=
250 / 15
=
= 0.564%
= 558 mm2
= 231 mm2. ( bent bars extended )
=
= 231 mm2
remaining area = 558 - 231
= 327 mm2
=
= 346 mm
= 332 mm2. ( Extra )
= 180 mm2.
=
= 157 mm
= 188 mm2.
For negative moment reinforcement
Total 231 + 332 mm2= 563 mm
2steel provided
( from Table 7-6 )
28.26 x 1000 / 180
Provide 6 mm O bar@150 mm c/c
Development length of bars Ld= O s/ 4 x bd
Distribution steel = ( 0.15 /100 ) x 1000 x 120
Provide 6 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m2/ m
For mild steel minimum reinforcement 0.15 %
Provide 12 mm O bar
Provide 12 mm O bar@340 mm c/c
spacing of bar = Area of one bar x 1000 / required area in m2/ m
Area provided = Area of one bar x 1000 / spacing of bar in m
Provide 10 mm O bar
Mu/ b x d2( - ) = 10.92 x 106
/ 1000 x (99)2
78.5 x 1000 / 401
113.04 x 1000 / 327
Note that at simple support , the bars are bent at 0.1 l whereas at continuity of slab it
is bent at 0.2 l
( alternate bent up )
78.5 x 1000 / 340
Provide 10 mm O bar@340 mm c/c
50 [(1-0.812) x 15 / 250 ]
Ast( - ) = 0.564 x 1000 x 99 / 100
Provide 10 mm O bar@170 mm c/c
Provide 10 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m2/ m
50 1-1-(4.6 / 15) x (1.11)
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=
= 54.4 O
Ld= 54.4 x ( 10 + 12 ) / 2
= 598 mm.
12 O (mild steel )
At A , Pt=
=
= 0.233
OR
Pt= 50 1-1-(4.6 / fck) x (Mu/ b x d2) Mu1=fy/ fck =
we get , 0.487 = 4.974
Mu1= = 4.78= 4.77 KNm
Vu = 12.06 KN
=54.4 O
514.18 + 12 O 54.4 O42.4 O 514.18
O 12.13 mm
At B , Pt=
== 0.233
OR
Pt= 50 1-1-(4.6 / fck) x (Mu/ b x d2) Mu1=fy/ fck =
we get , 0.487 = 4.974
Mu1= = 4.78
= 4.77 KNm
Vu = 13.09 KN
=54.4 O
473.72 + 12 O 54.4 O42.4 O 473.72
O 11.2 mmwhich gives
Mu1/ b x d2=
.( O.K.)
0.487 x 1000 x 992
x 10-6
Development length of bars Ld= O s/ 4 x bd = O x 0.87 x 250 / 4 x 1
Near point of contraflexure i.e. 0.15 x l from B
1.3 x ( Mu1/ Vu ) + L0Ld
1.3 x ( 4.77 x 106/ 13.09 x 10
3) + 12 O 54.4 O
100 x As/ b x d Half bars bent = 462 / 2 = 231 mm2)
100 x 231 / 1000 x 99
From equation
0.87 x fy
0.87 x 25
which gives .( O.K.)
Half bars bent = 462 / 2 = 231 mm2)
0.87 x fy
0.87 x 25
Development length of bars Ld= O s/ 4 x bd = O x 0.87 x 250 / 4 x 1
1.3 x ( Mu1/ Vu ) + L0Ld
100 x 231 / 1000 x 99
From equation
Mu1/ b x d2=
0.487 x 1000 x 992
x 10-6
1.3 x ( 4.77 x 106/ 12.06 x 10
3) + 12 O 54.4 O
Check for development length : -
Assuming L0=
100 x As/ b x d
O x 0.87 x 250 / 4 x 1
say 600 mmAs a thumb rule, a bar shall be given an anchorage equal to the length of
the cantilever.
17.14 - ( 0.15 x 3 ) x 9 =
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13.09 KN
=
= 0.132 N / mm < c
Pt=
= 0.233
6 x =
=
= 7.47
6 x 7.47
= 0.34
N / mm2
IS 456-2000 clause 40.2.1.1 0.07
k = 1.3 ? 0.014
= 0.442 N / mm2
> v
Vu=
=
= 0.173 N / mm2
< c
Pt=
= 0.569
6 x =
=
= 3.06
6 x 3.06
= 0.487
N / mm2
Span AB :
Check for shear : -
Use Vu= 13.09 KN
Shear stress v= Vu/ b x d
At A , Vu,AB= 12.06 KN ( for maximum loading )
At B , shear at point of contraflexure =
0.8 x fck/ 6.89 Pt , but not less than 1.0
Design shear strength c= 0.85 0.8 x 15 ( 1 + 5 x 7.47 - 1 )
IS 456-2000 Table 19 from table 7-1
13.09 x 103/ 1000 x 99
100 x As/ b x d = 100 x 231 / 1000 x 99
Design shear strength c= 0.85 0.8 x fck( 1 + 5 x - 1 )
.( O.K.)
0.8 x 15 / 6.89 x 0.233
.( O.K.)
for Pt = 0.233 c= 0.34
0.1 difference
for 120 mm slab depth 0.02 difference
Span BC :
17.14 KNShear stress v= Vu/ b x d
Design shear strength = 1.3 x 0.34
Design shear strength c= 0.85 0.8 x fck( 1 + 5 x - 1 )
0.8 x fck/ 6.89 Pt , but not less than 1.0
0.8 x 15 / 6.89 x 0.569
17.14 x 103/ 1000 x 99
.( O.K.)
100 x As/ b x d = 100 x 563 / 1000 x 99
Design shear strength c= 0.85 0.8 x 15 ( 1 + 5 x 3.06 - 1 )
IS 456-2000 Table 19 from table 7-1
for Pt = 0.569 c= 0.48
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IS 456-2000 clause 40.2.1.1 0.08
k = 1.3 ? 0.0579
= 0.624 N / mm2
> v
20
Pt=
= 0.467
IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement
= 40
= 30.30 < 40 .( O.K.)
7
Pt=
= 0.569
IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement
= 12.6
= 12.12 < 12.6 .( O.K.)
= 3 x 99 = 297 mm170 mm
5 x effective depth of slab or
= 5 x 99
150 mm
0.25 difference
for 120 mm slab depth 0.181 difference
Design shear strength = 1.3 x 0.48
modification factor = 2
( span / d ) ratio permissible = 2 x 20
Actual (span / d ) ratio = 3000 / 99
.( O.K.)Check for deflection : -
Basic ( span / d ) ratio =
100 x Ast/ b x d = 100 x 462 / 1000 x 99
For span AB :
(1) Main bars : maximum spacing permitted = 3 x effective depth of slab or 300 mm
( span / d ) ratio permissible = 1.8 x 7
Actual (span / d ) ratio = 1200 / 99
For span BC :
Basic ( span / d ) ratio =
100 x Ast/ b x d = 100 x 563 / 1000 x 99
modification factor = 1.8
spacing provided = < 450 mm .( O.K
or 300 mspacing provided = < 297 mm .( O.K
(2 )Distribution bars : maximum spacing permitted == 495 mm
Check for cracking : -
IS 456-2000 , clause 26.3.3
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1.2m
0.15m
0.15m
0.15m 0.15m
1.2mlx= 3m
6m S2 S1
S3
B1B2
B3
B4
1m high parapet
Column 300 x 300
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KNm
KNm
( 1 - 0.0389 )
Astx d ( 1 - fyx Ast/ b x d x fck)
0 x 231 x 99 ( 1 - 250 x 231 / 1000 x 99 x 15 ) x 10-6
Astx d ( 1 - fyx Ast/ b x d x fck)
0 x 231 x 99 ( 1 - 250 x 231 / 1000 x 99 x 15 ) x 10-6
( 1 - 0.0389 )
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1200 1200
150 3000 150 1200
450 mm whichever is small
hichever is small
.)
i.e. 297 mm .)
120 125
600300
or 450 mm i.e. 450 mm
10 O @ 340 c/c (bent)+ 12 O @ 340 c/c (extra)
10 O @ 170 c/c
6 O @ 150 c/c
6 O @ 150 c/c
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= 250 mm2
=
= 200.96 mm
= 251 mm2.
= 0.81
Pt= 50 1-1-(4.6 / fck) x (Mu/ b x d2)fy/ fck
=
415 / 15
== 0.240%
= 312 mm2
=
= 161 mm
= 335 mm2.
For HYSD Fe415
= 180 mm2
=
= 126 mm2
=
= 279.11 mm
= 193 mm
2
.
= 188 mm2.
Provide 8 mm O bar@ 260 mm c/cMore steel is provided to match with the torsion reinforcement.
In edge strip , minimum reinforcement is provided equal to 8 mm O @ 260 mm c/c.
Torsion steel :-
At corner A , steel required = ( 3/4 ) x 250
Provide 8 mm O bar
At discontinuous edges 4 and 5 , 50 % of the positive steel is required at top
This is less than minimum , therefore , use minimum steel at location 4 and 5 .
Provide 8 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m2/ m
50.24 x 1000 / 180
spacing of bar = Area of one bar x 1000 / required area in m2/ m
50.24 x 1000 / 312
Provide 8 mm O bar@ 150 mm c/c
Minimum steel = ( 0.12 / 100 ) x 1000 x 150
50.24 x 1000 / 250
Provide 8 mm O bar@ 200 mm c/c
Mu/ b x d2( - ) = 13.66 x 10
6/ 1000 x (130)
2
( 1 / 2 ) x 251
50 1-1-(4.6 / 15) x (0.81)
50 [(1-0.867) x 15 / 415 ]
Ast = 0.24 x 1000 x 130 / 100
Provide 8 mm O bar
Provide 8 mm O bar
Ast = 0.208 x 1000 x 120 / 100
spacing of bar = Area of one bar x 1000 / required area in m2/ m
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=
= 267.23 mm
= 193 mm2.
= 94 mm2.
=
= 279.11 mm
= 193 mm2.
S.F. =
=
= 31.795 KN
= 0.258
N / mm2
0.25 differen 0.11
IS 456-2000 clause 40.2.1.1 0.242 differe ? -0.1065
k = 1.3
= 0.460 N / mm2
OR
6 x =
= 6.75
6 x 6.75
= 0.356
IS 456-2000 clause 40.2.1.1
k = 1.3
= 0.463 N / mm2
=
= 0.245 N / mm2
< c .( O.K.)
Design shear strength = 1.3 x 0.356
.( O.K.)Actual shear stress = Vu/ b x d
31.795 x 103/ ( 1000 x 130 )
Design shear strength c= 0.85 0.8 x fck( 1 + 5 x - 1 )
0.8 x fck/ 6.89 Pt , but not less than 1.0
Design shear strength c= 0.85 0.8 x 15 ( 1 + 5 x 6.75 - 1 )
for 150 mm slab depth
from table 7-1
for Pt= 0.258 , c= 0.354
for 150 mm slab depth
Design shear strength = 1.3 x 0.354
.( O.K.)
Note that positive reinforcements are not curtailed because if they are curtailed , the
remaining bars do not provide minimum steel.
Check for shear : -
At point 1 or 3w x l / 2 + Moment @ point 1 or 3 in that span
11.625 x 5 / 2 + 13.66 / 5
100 x As/ b x d = 100 x 335 / ( 1000 x 130 )
Provide 8 mm O bar
spacing of bar = Area of one bar x 1000 / required area in m2/ m
50.24 x 1000 / 180
Provide 8 mm O bar@ 260 mm c/c
This will be provided by minimum steel .
spacing of bar = Area of one bar x 1000 / required area in m2/ m
50.24 x 1000 / 188
Provide 8 mm O bar@ 260 mm c/c
This will be provided by minimum steel of edge strip,
At corner B , steel required = ( 1/2 ) x 188
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S.F. = w x l / 2
=
= 29.06 KN
= 0.209
N / mm2
0.1 differenc 0.07
IS 456-2000 clause 40.2.1.1 0.041 differe ? -0.0287
k = 1.3
= 0.417 N / mm2
OR
6 x =
= 8.33
6 x 8.33
= 0.326
IS 456-2000 clause 40.2.1.1
k = 1.3
= 0.424 N / mm2
=
= 0.242 N / mm < c
Vu= 29.06 KN
8 O (HYSD Fe415 steel )
Pt=
=
= 0.209
OR
Pt= 50 1-1-(4.6 / fck) x (Mu/ b x d2) Mu1=
fy/ fck =
we get , 0.711 = 10.875
Mu1= = 10.246
= 10.24 KNm
=54.3 O
This is critical at point 4 or 5.
No bar is curtailed or bent up.
Mu1/ b x d2=
0.711 x 1000 x 1202
x 10-6
Development length of bars Ld= O s/ 4 x bd (From Table 7-6 )
Assuming L0=
0.87 x fy
Actual shear stress = Vu/ b x d
29.06 x 103/ ( 1000 x 120 )
.( O.K.)
0.87 x 41
0.85 0.8 x 15 ( 1 + 5 x 8.33 - 1 )
for 150 mm slab depth
Design shear strength = 1.3 x 0.326
100 x As/ b x d
100 x 251 / 1000 x 120
From equation
Check for development length : -
At point 4 or 5 ,
1.3 x 0.321
Design shear strength c= 0.85 0.8 x fck( 1 + 5 x - 1 )
0.8 x fck/ 6.89 Pt , but not less than 1.0
Design shear strength c=
At point 4 or 5
11.625 x 5 / 2
100 x As/ b x d = 100 x 251 / ( 1000 x 120 )
from table 7-1
for Pt= 0.209 , c= 0.321
for 150 mm slab depth
Design shear strength =
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458.36 + 8 O 54.3 O46.3 O 458.36
O 9.90 mm
Vu= = 25.31 KN
8 O (HYSD Fe415 steel )
Pt=
=
= 0.289
OR
Pt= 50 1-1-(4.6 / fck) x (Mu/ b x d2) Mu1=fy/ fck =
we get , 0.9595 = 26.689
Mu1
= = 24.554
= 24.56 KNm
=56 O
1261.5 + 8 O 56 O48 O 1261.5
O 26.28 mm
26
251 mm2.
= 123 mm.
Pt= 240.7
= 0.204
IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement
= 42.9
= 40.65 < 42.9
= 3 x 130 = 390 mm
200 mm
= 5 x 120 = 600 mm
< 300 mm .( O.(2) Distribu. bars : maximum spacing permitted = 5 x effective depth of slab or 4
.( O.K.)
IS 456-2000 , clause 26.3.3Check for cracking : -
3 x effective depth of slab or 3(1) Main bars : maximum spacing permitted =
spacing provided =
1.3 x ( Mu1/ Vu ) + L0Ld
1.3 x ( 24.56 x 106/ 25.31 x 10
3) + 8 O 56 O
which gives
Note that the bond is usually critical along long direction.
Actual (span / d ) ratio = 5000 / 123
.( O.K.)
0.9595 x 1000 x 1602
x 10-6
Development length of bars Ld= (From Table 7-6 )
.( O.K.)
Short span 11.25 x ( 4.5 / 2 )
0.87 x fy
0.87 x 41
1.3 x ( Mu1/ Vu ) + L0Ld
1.3 x ( 10.246 x 106/ 29.06 x 10
3) + 8 O 54.3 O
which gives
O s/ 4 x bd
( span / d ) ratio permissible = 1.65 x 26
Basic ( span / d ) ratio =
Mu1/ b x d2=
Assuming L0=
100 x As/ b x d
100 x 462 / 1000 x 160
From equation
Check for deflection : -
100 x Ast/ b x d = 100 x 251 / 1000 x 123
positive moment steel =actual d = 150 - 15 -8 - 4
modification factor = 1.65
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260 mm
This is 180 mm2
=
= 279.11 mm
= 193 mm2.
625 3750 625
150
625 3750 625
Section A-A
50.24 x 1000 / 180
Provide 8 mm O bar@ 260 mm c/c for uniformity in spacing.
For clarity , top and bottom reinforcements are shown separately.
3750
625
Note that the bottom reinforcements are both ways and therefore there is no
necessity of secondary reinforcements.However , top reinforcement in edge strip
requires the secondary steel for tying the bars.
minimum (0.12 / 100) x 150 x 1000 =
Provide 8 mm O barspacing of bar = Area of one bar x 1000 / required area in m
2/ m
spacing provided = < 450 mm .( O.
625
MiddleStrip
S1
Edgestrip
Edgestrip
Edge strip Edge stripMiddle Strip
8O@2
60c/c
8O@2
60c/c
8O@200c/c
8 O @ 260 c/c8 O @ 260 c/c8 O @ 200 c/c
A A
B
B
500
1500 1500
8 O @ 260 c/c
8 O @ 150 c/c
8 O @ 260 c/c
8 O @ 200 c/c
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5 m
625 3750 625 5 m
ly/8 (3/4)ly ly/8
5 m
5 m
Middle Strip
S1
S1
Edgestrip
Edgestrip
A B
B
1
2
3
4
50.0
35
-0.0
47
-
0.035
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KNm
( 1 - 0.0579 )
Astx d ( 1 - fyx Ast/ b x d x fck)
5 x 251 x 120 ( 1 - 415 x 251 / 1000 x 120 x 15 ) x 10-6
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KNm
or 300 mm i.e. 300 mm
or 450 mm i.e. 450 mm
.) 50 mm whichever is small
00 mm whichever is small
( 1 - 0.0799 )
Astx d ( 1 - fyx Ast/ b x d x fck)
5 x 462 x 160 ( 1 - 415 x 46
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1000
625 3750 625
150
625 3750 625
1000
.)
625
3750
625
MiddleStrip
S1
Edgestrip
Edgestrip
Edge strip Edge stripMiddle Strip
8O@2
40c/c
8O@2
40c/c
8O@140c/c
8 O @ 240 c/c8 O @ 240 c/c8 O @ 140 c/c
500
1500 1500
8 O @ 240 c/c
8 O @ 140 c/c
8 O @ 240 c/c
8 O @ 240 c/c
500
1500
1500
1500 1500
8 O @ 180 c/c
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0.047
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given
Material
DL LL
3 0 KN / m2
1 0 KN / m2
0 3 KN / m2
Total 4 3 KN / m2
=
3m 3m 3m 3m 3m
=
= 4.5 + 4.05
= 8.55 KNm
=
= 3.38 + 3.38
= 6.75 KNm
=
= 5.4 + 4.5
= 9.9 KNm
=
= 4.50 + 4.50
= 9 KNm
=
= 18.9 KN
KNm
Q = 2.07
0.6 x 6 x 3 + 0.6 x 4.5 x 3
Maximum moment is Mu3( - ) = 9.9
From Table 6-3
( 1 / 10 ) x 6 x 32+ ( 1 / 9 ) x 4.5 x 3
2
Mu4( - ) = ( 1 / 12 ) x w ( DL ) x l2+ ( 1 / 9 ) x w ( LL ) x l
2
( 1 / 12 ) x 6 x 32+ ( 1 / 9 ) x 4.5 x 3
2
Maximum shear is Vu(BA)= 0.6 x w x l + 0.6 x w x l
Mu3( - ) = ( 1 / 10 ) x w ( DL ) x l2+ ( 1 / 9 ) x w ( LL ) x l
2
( 1 / 12 ) x 6 x 32+ ( 1 / 10 ) x 4.5 x 3
2
Mu2( + ) = ( 1 / 16 ) x w ( DL ) x l2+ ( 1 / 12 ) x w ( LL ) x l
2
( 1 / 16 ) x 6 x 32+ ( 1 / 12 ) x 4.5 x 3
2
Design of Continuous One-way slab
A five span continuous one-way slab used as an office floor.
The centre-to-centre distance of supporting beams is 3 m
Live load 3 KN / m2and floor finish 1 KN / m
2
The factored moments at different points using the coefficients are as follows :
Mu1( + ) = ( 1 / 12 ) x w ( DL ) x l2+ ( 1 / 10 ) x w ( LL ) x l
2
Dead load 0.12 x 25 =
floor finish =
live load =
factored load =
HYSD reinforcement of grade Fe415
M15 grade concrete
Solution : -
Try 120 mm thick slab
1.5 ( 4 + 3 )
( 6 + 4.5 ) KN / m2
Consider 1 m wide strip of the slab.
A B C D E F
1 2 2 2 1
( 6 + 4.5 ) KN / m
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=
= 69.2 mm,
= 90 mm,
Pt= 50 1-1-(4.6 / fck) x (Mu/ b x d2)fy/ fck
= 132 mm2
= 165 mm2
=
= 171
=
= 344
306
Table
8 mm O @ 220 c/c
= 228 mm2
10 mm O @ 220 c/c
= 357 mm2
10 mm O @ 220 c/c
= 357 mm24 ( - ) 9.0 1.11 0.34
223
3 ( - ) 9.9 1.22 0.38 342
2 ( + ) 6.75
10 mm O @ 440 c/c + 8 mm O @ 440 c/c
= 178 +114 = 292 mm2(Half 10 O+half 8 O)290
0.83 0.248
1 ( + ) 8.55 1.06 0.322
Factored
moment
KNmpoint Mu/(b x d2) Pt
Ast
mm2
Steel Provided
dprovided= 110 - 15 - 5 ( assume 10 O bar ).( O.K.)
Try 110 mm overall depth
spacing of bar = Area of one bar x 1000 / required area in m2/ m
Ast = Ptx b x d / 100
drequired= M / Q x b9.9 x 10 6/ 2.07 x 1000
For Main steel ,HYSD Fe415 reinforcement
minimum steel area = ( 0.12 / 100 ) x 1000 x 110
For Distribution steel , mild steel Fe250 reinforcement
minimum steel area = ( 0.15 / 100 ) x 1000 x 110
Use 6 mm O @ 160 mm c/c = 177 mm2.
Note that the positive bars cannot be curtailed as the remaining bars in the internal
spans ( + ve moment ) will not provide minimum area.
Provide 50 % Ast at end support top bars i.e. 292 / 2 = 146 mm2.
Use 8 mm O
Use 6 mm O
spacing of bar = Area of one bar x 1000 / required area in m2/ m
28.26 x 1000 /165
Check for shear : -
spacing of bar = Area of one bar x 1000 / required area in m2/ m
50.24 x 1000 /146
Use 8 mm O @ 340 mm c/c = 148 mm2.
Maximum shear = 18.9 KN
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=
= 0.210 N / mm2
d = 90 mm
As= 357 mm2.
= 0.397
6 x =
=
= 4.4
6 x 4.4
= 0.42
N / mm2
IS 456-2000 clause 40.2.1.1 0.11
k = 1.3 ? 0.0453
= 0.546 N / mm2
At point of contraflexure i.e. 0.15 x l from B
Vu=
= 14.18 KN
=
= 0.158 N / mm2
d = 90 mm
As= 292 mm2.
= 0.324
6 x =
=
= 5.4
Actual Shear stress = Vu/ b x d
18.9 x 103/ 1000 x 90
for 110 mm slab depth 0.103 difference
Design shear strength c= 0.85 0.8 x fck( 1 + 5 x - 1 )
0.8 x fck/ 6.89 Pt , but not less than 1.0
Design shear strength c=0.85 0.8 x 15 ( 1 + 5 x 4.4 - 1 )
0.8 x 15 / 6.89 x 0.397
IS 456-2000 Table 19 from table 7-1
for Pt= 0.397 c= 0.42
< ( C)N / mm2
0.25 difference
( too small )
For bars at support
100 x As/ b x d = 100 x 357 / 1000 x 90
.( O.K.)
18.9 - 0.15 x 3 x 10.5
Actual Shear stress = Vu/ b x d
Design shear strength = 1.3 x 0.42
100 x As/ b x d = 100 x 292 / 1000 x 90
Design shear strength c= 0.85 0.8 x fck( 1 + 5 x - 1 )
14.18 x 103/ 1000 x 90
< ( C)N / mm2 ( too small )
For bars at support
with positive moment reinforcement ( 29
0.8 x fck/ 6.89 Pt , but not less than 1.0
0.8 x 15 / 6.89 x 0.324
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6 x 5.4
OR = 0.39
N / mm2
IS 456-2000 clause 40.2.1.1 0.11k = 1.3 ? 0.0774
= 0.494 N / mm2
Vu=
= 13.28 KN
At A , Pt=
=
= 0.324
OR
Pt= 50 1-1-(4.6 / fck) x (Mu/ b x d2) Mu1=fy/ fck =
we get , 1.064 = 9.4884
Mu1= = 8.64
= 8.62 KNm
8 O (HYSD Fe415 steel )
=56.4 O
845.783 + 8 O 56.4 O48.4 O 845.78
O 17.47 mmAt support B, point of contraflexure is assumed at 0.15 x l from B
Vu=
= 14.18 KN
Mu1= 8.64 KNm
L0= 12 O
792.102 + 12 O 56.4 O44.4 O 792.1
O 17.84 mm
IS 456-2000 Table 19 from table 7-1
for Pt= 0.324 c= 0.38
0.25 differencefor 110 mm slab depth 0.176 difference
Design shear strength c= 0.85 0.8 x 15 ( 1 + 5 x 5.4 - 1 )
0.4 x 6 x 3 + 0.45 x 4.5 x 3 =
100 x 292 / 1000 x 90
From equation
1.064 x 1000 x 902
x 10-6
Design shear strength = 1.3 x 0.38
.( O.K.)
0.87 x fy
0.87 x 41
Mu1/ b x d2=
Check for development length : -
Assuming L0=
100 x As/ b x d
Span AB is critical for checking this requirement
At support A
0.4 x w x l + 0.45 x w x l
as before
( actual anchorage is more than 12 O but L0is limited to 12 O or d
, i.e. 90 mm whichever is greater )
1.3 x ( Mu1/ Vu ) + L0Ld
Development length of bars Ld= O s/ 4 x bd = O x 0.67 x 415 / 4 x 1
1.3 x ( Mu1/ Vu ) + L0Ld1.3 x ( 8.64 x 10
6/ 13.28 x 10
3) + 8 O 56.4 O
which gives .( O.K.)
1.3 x ( 8.64 x 106/ 14.18 x 10
3) + 12 O 56.4 O
18.9 - 0.15 x 3 x 10.5
which gives .( O.K.)Check for deflection : -
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26
Pt=
= 0.324
IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement
= 34.84
= 33.33 < 34.84 .( O.K.)
26
Pt=
= 0.253
IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement
= 41.6
= 33.33 < 41.6 .( O.K.)
= 3 x 90 = 270 mm
220 mm
5 x effective depth of slab or
= 5 x 90
160 mm
.( O.
Basic ( span / d ) ratio =
modification factor = 1.34( span / d ) ratio permissible = 1.34 x 26
or 300 m
IS 456-2000 , clause 26.3.3
(1) Main bars : maximum spacing permitted = 3 x effective depth of slab or 300 mm
100 x Ast/ b x d = 100 x 292 / 1000 x 90
Maximum positive moment occurs in span AB. Therefore, this check is critical in span
Check for cracking : -
100 x Ast/ b x d = 100 x 228 / 1000 x 90
modification factor = 1.6
( span / d ) ratio permissible = 1.6 x 26
Actual (span / d ) ratio = 3000 / 90
For span BC :
Basic ( span / d ) ratio =
spacing provided = < 450 mm .( O.
Actual (span / d ) ratio = 3000 / 90
(2 )Distribution bars : maximum spacing permitted == 450 mm
spacing provided = < 270 mm
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Outer side Inner side
IS 456-2000 Clause -22.5
( 22.5.1 ) Unless more exact estimates are made, for
beams of uniform cross-section which support
substantially uniformly distributed loads over three or
more spans which do not differ by more than 15 percent
of the longest, the bending moments and shear forcesused in design may be obtained using the coefficients
Where coefficients given in Table 12 are used for
calculation of bending moments, redistribution referred
to in 22.7 shall not be permitted.
(22.5.2 ) Beams and Slabs Over Free End Supports
given in Table 12 and Table 13 respectively.
For moments at supports where two unequal spans
the average of the two values for the negative moment
at the support may be taken for design.
meet or in case where the spans are not equally loaded,
and I is the effective span, or such other restraining
moment as may be shown to be applicable. For such a
condition shear coefficient given in Table 13 at the
end support may be increased by 0.05.
Where a member is built into a masonry wall whichdevelops only partial restraint, the member shall be
designed to resist a negative moment at the face of the
support of Wl / 24 where W is the total design load
+ 1 / 12
+ 1 / 10
Table 12 Bending Moment coefficients
Near middle
of end span
At middle of
interior span
Support m
At support next to
the end support
Type of load
Span moments
NOTE -For obtaining the bending moment, the coefficient shall be multiplie
design load and effective span.
Table 13 Shear Force coefficients
+ 1 / 16
+ 1 / 12
-1 / 10
-1 / 9
Dead load and imposed
load ( fixed )
imposed load (
not fixed )
Type of load
Dead load and imposed
load ( fixed )
imposed load (
not fixed )
At support next to the end supportAt end
support
0.45 0.6 0.6
0.4 0.6 0.55
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90110
450 900 900 900
3000 3000
10 O @ 220 c/c8 O @ 340 c/c
6 O @ 160 c/c8 O @ 220 c/c8 O @ 440 c/c
+ 10 O @ 440 c/c
( 0.3 l1 ) ( 0.3 l1 ) ( 0.3 l1 )( 0.15 l1 )
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2 mm2)
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KNm
Astx d ( 1 - fyx Ast/ b x d x fck)
5 x 292 x 90 ( 1 - 415 x 292 / 1000 x 90 x 15 ) x 10-6
( 1 - 0.0898 )
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450 mm whichever is small
.) i.e. 270 mm
hichever is small
B
.)or 450 mm i.e. 450 mm
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oments
At other interior
supports
-1 / 12
-1 / 9
by the total
0.6
At all other
interior
supports
0.5
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900
3000
10 O @ 220 c/c
( 0.3 l1 )
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material
lx= say 4.5 m
ly = say 6.75 m
4.5 KN / m2
1.0 KN / m2
2.0 KN / m2
Total 7.5 KN / m
2
Pu= 1.5 x 7.5
= 11.25 KN / m
ly/ lx=
= 1.5
Mux = 23.7 KNm
Muy= 10.48 KNm
Q = 2.07
=
= 107 mm,
= 160 mm,
160 - 10
= 150 mm,
= 0.926
Pt= 50 1- 1-(4.6 / fck) x (Mu/ b x d2)
fy/ fck
=
Solution :
Consider 1 m wide strip. Assume 180 mm thick slab.
Design of Simply supported two way slab
residential building drawing room 4.3 m x 6.55 m
It is supported on 350 mm thick walls on all four sides.
M15 grade concrete
6.75 / 4.5
4.3 + 0.18 = 4.48
6.55+0.18 = 6.73
Live load ( residence ) =
For 1 m wide strip
Dead load : self 0.18 x 25 =
floor finish =
HYSD reinforcement of grade Fe415
given
IS 456-2000 Table -27
dshortprovided= 180 - 15 - 5 ( assume 10 O bar )
.( O.K.)> 107 mm
dlongprovided=
> 107 mm
xx w x lx2
=
yx w x lx2
= 0.046 x 11.25 x 4.52 =
0.104 x 11.25 x 4.52
=
From Table 6-3
drequired= M / Q x b23.7 x 10 6/ 2.07 x 1000
Larger depth is provided due to deflection check.
Mu/ b x d2( short ) = 23.7 x 10
6/ 1000 x (160)
2
50 1-1-(4.6 / 15) x (0.926)
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415 / 15
=
= 0.28%
= 448 mm2
=
= 175.22 mm
= 462 mm2.
= 0.466
Pt= 50 1- 1-(4.6 / fck) x (Mu/ b x d2)
fy/ fck
=
415 / 15
=
= 0.134%
= 201 mm2
= 216 mm2.
=
= 232.59 mm
= 218 mm2.
Vu= = 25.31 KN
8 O (HYSD Fe415 steel )
50 [(1-0.846) x 15 / 415 ]
Ast( short ) = 0.28 x 1000 x 160 / 100
Provide 10 mm O bar
Mu/ b x d2( long ) = 10.48 x 10
6/ 1000 x (150)
2
50 1-1-(4.6 / 15) x (0.466
50 [(1-0.926) x 15 / 415 ]
spacing of bar = Area of one bar x 1000 / required area in m2/ m
78.5 x 1000 / 448
Provide 10 mm O bar@170 mm c/c ( short span )
Ast( long ) = 0.134 x 1000 x 150 / 100
For HYSD Fe415 minimum reinforcement 0.12 %
Provide 8 mm O bar
Minimum steel = ( 0.12 /100 ) x 1000 x 180
spacing of bar = Area of one bar x 1000 / required area in m2/ m
Provide 8 mm O bar@ 230 mm c/c ( long span )
50.24 x 1000 / 216
The bars cannot be bent or curtailed because if 50 % of long span bars are curtailed ,the remaining bars will be less than minimum
At top on support , provide 50 % of bars of respective span to take into account
negative moment due to slab nature.
Check for development length : -
Assuming L0=
Long span 11.25 x ( 4.5 / 2 )
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Pt=
=
= 0.145
OR
Pt= 50 1- 1-(4.6 / fck) x (M
u/ b x d
2) M
u1=
fy/ fck =
we get , 0.5023 = 11.806
Mu1= = 11.331
= 11.30 KNm
=56 O
580.4 + 8 O 56 O
48 O 580.403O 12.09 mm
Vu= = 25.31 KN
8 O (HYSD Fe415 steel )
Pt=
=
= 0.289
OR
Pt= 50 1- 1-(4.6 / fck) x (Mu/ b x d2) Mu1=
fy/ fck =
we get , 0.9595 = 26.689
Mu1= = 24.554
= 24.56 KNm
=56 O
1261.5 + 8 O 56 O48 O 1261.48
O 26.28 mm
Vu= 25.31 KN
=
From equation
0.87 x fy
1.3 x ( Mu1/ Vu ) + L0Ld
0.9595 x 1000 x 1602x 10
-6
Check for shear : -
Shear stress = Vu/ b x d
Note that the bond is usually critical along long direction.
Development length of bars Ld= O s/ 4 x bd (From Table 7-6 )
Mu1/ b x d2=
0.87 x 41
Mu1/ b x d2=
0.5023 x 1000 x 1502x 10
-6
1.3 x ( 24.56 x 106/ 25.31 x 10
3) + 8 O 56 O
which gives .( O.K.)
From equation
0.87 x fy
0.87 x 41
100 x 462 / 1000 x 160
Short span 11.25 x ( 4.5 / 2 )
(From Table 7-6 )
100 x As/ b x d
100 x 218 / 1000 x 150
Development length of bars Ld= O s/ 4 x bd1.3 x ( Mu1/ Vu ) + L0Ld
1.3 x ( 11.30 x 106/ 25.31 x 10
3) + 8 O 56 O
which gives .( O.K.)
Assuming L0=
100 x As/ b x d
25.31 x 103/ 1000 x 150
This is critical along long span
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= 0.169 N / mm2
= 0.145
N / mm 25 difference -0.05IS 456-2000 clause 40.2.1.1 20 difference ? -0.04
k = 1.24
= 0.347 N / mm2
OR
6 x =
= 12.011
6 x 12.011
= 0.278
IS 456-2000 clause 40.2.1.1 25 difference -0.05
k = 1.24 20 difference ? -0.04
= 0.345 N / mm2
20
Pt=
= 0.289
IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement
= 28.8
= 28.00 < 28.8
= 3 x 160 = 480
170 mm
= 5 x 150 = 750
230 mm
for 180 mm slab depth
Design shear strength = 1.24 x 0.28
< ( C)N / mm2 ( too small )100 x As/ b x d = 100 x 218 / 1000 x 150
0.8 x fck/ 6.89 Pt , but not less than 1.0
Design shear strength c= 0.85 0.8 x 15 ( 1 + 5 x 12.01 - 1 )
for 180 mm slab depth
.( O.K.)
Design shear strength c= 0.85 0.8 x fck( 1 + 5 x - 1 )
from table 7-1
for Pt= 0.145 c= 0.28
IS 456-2000 , clause 26.3.3
Basic ( span / d ) ratio =
100 x Ast/ b x d = 100 x 462 / 1000 x 160
modification factor = 1.44
Design shear strength = 1.24 x 0.278
.( O.K.)Check for deflection : -
This check shall be done along short span
( span / d ) ratio permissible = 1.44 x 20
Check for cracking : -
Actual (span / d ) ratio = 4480 / 160
.( O.K.)
.( O.
3 x effective depth
spacing provided = < 300 mm .( O.
(1) Main bars : maximum spacing permitted for short span steel =
(2) Distribu. bars : maximum spacing permitted for long span steel = 5 x effective depth
spacing provided = < 450 mm
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1.0 1.1 1.2 1.3Interior Panels:
Negative moment at continuous edge 0.032 0.037 0.043 0.047
Positive moment at mid-span 0.024 0.028 0.032 0.036
One Short Edge Discontinuous:
Negative moment at continuous edge 0.037 0.043 0.048 0.051
Positive moment at mid-span 0.028 0.032 0.036 0.039
One long Edge Discontinuous:
Negative moment at continuous edge 0.037 0.044 0.052 0.057
Positive moment at mid-span 0.028 0.033 0.039 0.044Two Adjacent Edges Discontinuous:
Negative moment at continuous edge 0.047 0.053 0.060 0.065
Positive moment at mid-span 0.035 0.040 0.045 0.049
Two Short Edges Discontinuous:
Negative moment at continuous edge 0.045 0.049 0.052 0.056
Positive moment at mid-span 0.035 0.037 0.040 0.043
Two Long Edges Discontinuous:
Negative moment at continuous edge - - - -
Positive moment at mid-span 0.035 0.043 0.051 0.057
Three Edges Discontinuous
(One Long Edge Continuous):
Negative moment at continuous edge 0.057 0.064 0.071 0.076
Positive moment at mid-span 0.043 0.048 0.053 0.057
Three Edges Discrmntinuous
(One Short Edge Continuous) :
Negative moment at continuous edge - - - -
Positive moment at mid-span 0.043 0.051 0.059 0.065
Four-Edges Discontinuous:
Positive moment at mid-span 0.056 0.064 0.072 0.079
ly/ lx 1.0 1.1 1.2 1.3 1.4
Table - 26 Bending moment coefficients for rectangular panels supported on four sid
short span c
( Values
3
4
5
Type of Panel and Moments
consideredCase No.
1
2
6
7
8
9
Table - 27 Bending moment coefficients for slabs spanning in two directions at right a
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x 0.062 0.074 0.084 0.093 0.099
y 0.062 0.061 0.059 0.055 0.051
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KNm
KNm
350 6550 350
350
4300 4650
Astx d ( 1 - fyx Ast/ b x d x fck)
( 1 - 0.0799 )
5 x 462 x 160 ( 1 - 415 x 462 / 1000 x 160 x 15 ) x 10-6
Astx d ( 1 - f
yx A
st/ b x d x f
ck)
5 x 218 x 150 ( 1 - 415 x 218 / 1000 x 150 x 15 ) x 10-6
( 1 - 0.0402 )
460
690 690
A
8 O @ 230 c/c
10 O @ 170 c/c
10 O @ 340 c/c
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1.4 1.5 1.75 2.0
0.051 0.053 0.060 0.065 0.032
0.039 0.041 0.045 0.049 0.024
0.055 0.057 0.064 0.068 0.037
0.041 0.044 0.048 0.052 0.028
0.063 0.067 0.077 0.085 0.037
0.047 0.051 0.059 0.065 0.028
0.071 0.075 0.084 0.091 0.047
0.053 0.056 0.063 0.069 0.035
0.059 0.060 0.065 0.069 -
0.044 0.045 0.049 0.052 0.035
- - - - 0.045
0.063 0.068 0.080 0.088 0.035
0.080 0.084 0.091 0.097 -
0.060 0.064 0.069 0.073 0.043
- - - - 0.057
0.071 0.076 0.087 0.096 0.043
0.085 0.089 0.100 0.107 0.056
1.5 1.75 2.0 2.5 3.0
s with provision for torsion at corners
oefficient x
of ly/ lx)
Long span
coefficient yfor
all values of ly/ lx
ngles , simply supported on Four sides
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0.104 0.113 0.118 0.122 0.124
0.046 0.037 0.029 0.020 0.014
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Material
say
= == 648 = 648
( 2 / 3 ) x 6 x 162 ( 2 / 3 ) x 6 x 162
6MA+ 24MB+ 6MC= - [ 1944 ] - [ 1944 ]
MA( 6 / I ) + 2MB( 6 / I + 6 / I ) + MC( 6 / I ) = - [ 6 x 648 x 3 / ( Ix 6 ) ] - [ 6 x 648 x 3 / ( Ix
4MB+ MC= - 648 As MA= 0 .( 1 )
Using three moment equation for span ABC
MA( L1/ I1) + 2MB( L1/ I1+ L2/ I2) + MC( L2/ I2) = - 6 A1a1/ ( I1L1) -6 A2a2/ ( I2L2)
A1= ( 2 / 3 ) x Base x h1
a1= 3 m
A2= ( 2 / 3 ) x Base x h1
a2= 3 m
Live load = 3 KN / m2
Floor finish = 1 KN / m2
Rib size = 230 mm x 450 mm
Main beams = 300 mm x 570 mm overall .
Design of Continuous Beam
An R.C.C. floor is used as a banking hall
Design the beams B10-B11-B12.
The Slab thickness is 120 mm .
Floor finish = 1 + 0 KN /m2
Total 4 + 3 KN / m2
Load on beam = 3 ( 4 + 3 ) = 12 + 9 KN / m
(Given )
M15 grade concrete
Slab 120 mm thick 0.12 x 25 = 3 + 0 KN / m2
HYSD reinforcement of grade Fe415 .
Column = 300 mm x 300 mm
Solution : -
( a ) Load calculations and analysis :
Self wt. = 0.23 x 0.45 x 25 = 2.58 + 0 KN / m
Total 14.58 + 9 KN / m
Factored Load = 1.5 ( 14.58 + 9 )
= 21.87 + 13.5
Live load = 0 + 3 KN / m2
( 22 + 14 ) KN /m .
Case ( a ) Maximum moment at B
B10
6 m
6 m
36 KN / m 36 KN / m 22 KN / m
6 m 6 m 6 mA B C D
36 KN / m
162
36 KN / m
162
A B B C
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= =
= 648 = 396
6 m Distance 138 KNm3 m Distance ( ? ) 69
6 m Distance 96 KNm
3 m Distance ( ? ) 48
6 m Distance 42 KNm
3 m Distance ( ? ) 21
- 2088 - 16 MC+ MC= - 648
15 MC= - 1440
MC= - 96 KNm
4MB+ ( - 96 ) = - 648
4 MB= - 552MB= - 138 KNm
By putting Value of MCinto Equation ( 1 )
4MB+ MC = - 648MB+ 4MC = - 522
.( 1 ).( 2 )
By putting Value of MBfrom Equation ( 2 ) into Equation ( 1 )
4(- 522 - 4MC) + MC= - 648
MB+ 4MC= - 522 As MD= 0 .( 2 )
( 2 / 3 ) x 6 x 162 ( 2 / 3 ) x 6 x 99
a2= 3 m a3= 3 m
MB( 6 / I ) + 2MC( 6 / I + 6 / I ) + MD( 6 / I ) = - [ 6 x 648 x 3 / ( Ix 6 ) ] - [ 6 x 396 x 3 / ( Ix
6MB+ 24MC+ 6MD= - [ 1944 ] - [ 1188 ]
Using three moment equation for span BCD
MB( L2/ I2) + 2MC( L2/ I2+ L3/ I3) + MD( L3/ I3) = - 6 A2a2/ ( I2L2) -6 A3a3/ ( I3L3)
A2= ( 2 / 3 ) x Base x h2 A3= ( 2 / 3 ) x Base x h3
36 KN / m
162
22 KN / m
99B
CC D
36 KN / m 36 KN / m 22 KN / m
6 m 6 m 6 mA B C D138
96162
162 99( - )
( + )( - ) ( + )
96
( - ) ( - )
( + ) ( + ) ( + )
93 5145
138
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X dist.
6 - X dist.
Span AB or CD
= 1650 mm > 3000 mm
BC ( + ) = 58 KNmBC ( minimum - ve ) = 5 KNm
Factored maximum moments :
B or C , negative moment = 138 KNm
AB ( + ) = CD ( + ) = 110 KNm
Three cases are considered for getting maximum values of moments. Case ( a ) gives ma
negative moment at B.The same moment shall be used at C also because of symmetry. C
gives the maximum positive moment in span BC while the case ( c ) gives maximum positi
moments in span AB and CD.
85 x 12 + VBx 6 - 36 x 6 x 9 - 36 x 6 x 3 = - 96
VC= 101 KN
50 x 12 + VCx 6 - 36 x 6 x 3 - 22 x 6 x 9 = -138
VB= 246 KN
VB+ 131 = 246
VB= 115 KN
VC= 132 - 50
VC= 82 KN MC= - 96 KNm
VDx 6 -22 x 6 x 3 = -96
VD= 50 KN
VC+ VD= 22 x 6
VB= 216 - 85VB= 131 KN
MB= - 138 KNm
Vc= 183 KN
VC+ 82 = 183
MB= - 138 KNm
VAx 6 -36 x 6 x 3 = -138
VA= 85 KN
MC= - 96 KNm
VA+ VB= 36 x 6 131 X =8
216 X =X = 2.36
BC ( - ) = 11.6 KNm
BC ( + ) = 51.6 KNm > 0.7 x 58 KNm.
The moment redistribution shall be now carried out. Maximum negative moment = 138 KN
reduce it by 20 % . Then Mu = 0.8 x 138 = 110.4 KNm. For all the cases , redistribute ( inc
decrease ) the moment at B or C = 110.4 KNm ( Hogging ).
Maximum design moments :
The beam acts as a flanged beamFor T-beams , bf= ( l0/ 6 ) + bw+ 6 Dfbf= ( 0.7 x 6000 / 6 ) + 230 + 6 x 120
( As per I
( As per IS456-2000 ,Clause 23.1.2 , Note
Note that after redistribution , the design positive moments also have been reduced.
( b ) Design for flexure :
Mu( + ) = 111.6 KNm.
Support B or C = 110.4 KNm > 0.7 x 138 KNm ( O.K.)AB or CD ( + ) = 111.6 KNm > 0.7 x 110 KNm
85
131
2.36 m50
82
115
101
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For bf /bw = 7
0.01 diff. 0.014 0.016
0.006 diff. ? ?
-0.0084 -0.0096
= 662.6 For 0.224 0.6566 0.743
1 diff 0.0864
0.83 diff ? 0.0717
= 651 mm2.
Span BC :
= 301 mm .
fy/ fck
415 / 15
= 0.549
Ast= 603 135
Assuming one layer of 20 mm diameter bars
d = 450 + 120 -25 - 10 = 535 mm .
Minimum Ast= ( 0.205 / 100 ) x 230 x 535 = 252 mm2
( As per IS456-2000 ,Cl
If Mu< Mu,lim : design as under-reinforced section (singly reinforced beam) as explaine
Ast= Mu/ 0.87 x fyx lever arm
KNm > 111.6 KNm
bf/ bw= 1650 / 230 = 7.17
Df/ d = 120 / 535 = 0.224
Mu,lim/ fckbwd2= 0.671 ( As per SP:16 ,Table 58 )
For bf /
Mu.lim= 0.671 x 15 x 230 x 5352x 10
-6
Mu( + ) = 51.6 KNm
Ast= 51.6 x 106/ 0.87 x 415 x ( 535 - 60 )
Provide 3 - 12 mm O = 339 mm2.
In span AB , curtail 3 - 12 O at 300 mm ( 0.05 ) from A and at 900 mm ( 0.15 ) from B.
where lever arm = d - D f/ 2 = 535 - 120 / 2 = 535 - 60
Ast= 111.6 x 106/ 0.87 x 415 x ( 535 - 60 )
Provide 6 - 12 mm O = 678 mm2.
Continue 3 - 12 O in span BC as required for flexure.
Support B or C :Mu( - ) = 110.4 KNm
Mu/ bd2= 110.4 x 10
6/ 230 x 535
2= 1.68 < 2.07. ( From Table 6-3 )
The section is under-reinforced.
Pt= 501 - 1 - ( 4.6 / fck) x ( Mu / bd2 )
Provide 2- 10 mm O anchor bars = 157 mm2. At support , provide 3 - 16 mm O extra at to
one of which may be curtailed at 0.15 = 900 mm from centre of support B and remainingat 0.25 = 1500 mm from B .
Pt= 501 - 1 - ( 4.6 / 15 ) x ( 1.68)
( 0.549 / 100 ) x 230 x 535 = 676 mm
2
20 % of steel should be carried through the span = 0.2 x 676 = 135 mm2
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= == 396 = 648
6MA+ 24MB+ 6MC= - [ 1188 ] - [ 1944 ]
4MB+ MC= - 522 As MA= 0
A2= ( 2 / 3 ) x Bas
( 2 / 3 ) x 6 x 99 ( 2 / 3 ) x
a1= 3 m a2= 3 m
MA( 6 / I ) + 2MB( 6 / I + 6 / I ) + MC( 6 / I ) = -6) ]
Using three moment equation for span ABC
MA( L1/ I1) + 2MB( L1/ I1+ L2/ I2) + MC( L2/ I
A1= ( 2 / 3 ) x Base x h1
Case ( b ) Maximum positive moment in span B
B11 B12
6 m 6 m
3 m
3 m
22 KN / m 36 KN / m 22 KN / m
6 m 6 m 6 mA B C D
22 KN / m
99
36 K
162
A B B
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= =
= 648 = 396
MB= - 104 KNm
MC= - 104 KNm
MB+ 4MC = - 522 .
By putting Value of MCinto Equation ( 1 )
4MB+ ( - 104 ) = - 522
4 MB= - 418
By putting Value of MBfrom Equation ( 2 ) into
4(- 522 - 4MC) + MC = - 522
- 2088 - 16 MC+ MC= - 522
15 MC= - 1566
MB( 6 / I ) + 2MC( 6 / I + 6 / I ) + MD( 6 / I ) = -
6MB+ 24MC+ 6MD= - [ 1944 ] - [ 1188 ]
MB+ 4MC= - 522 As MD= 0
4MB+ MC = - 522 .
A2= ( 2 / 3 ) x Base x h2 A3= ( 2 / 3 ) x Bas
( 2 / 3 ) x 6 x 162 ( 2 / 3 ) x
a2= 3 m a3= 3 m
Using three moment equation for span BCD
MB( L2/ I2) + 2MC( L2/ I2+ L3/ I3) + MD( L3/ I
( 162 - ( 96 + 21 ) = 45 KNm )
( 99 - 48 = 51 KNm )
( 162 - 69 = 93 KNm )
6) ]
36 KN / m
162
22 K
99B
CC
22 KN / m 36 KN / m 22 KN / m
6 m 6 m 6 mA B C D104 104
99162
99( - )
( - ) ( + )
( - ) ( - )
( + ) ( + ) ( + )
47 4758
104 104
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85
131
imum
ase ( b )
ve
VC= 132 - 48.67 VC= 108 KN
VC= 83.33 KN MC= - 104 KNm
VD= 48.67 KN Vc= 191.33 KN
VB= 191.33 KN
VB= 108 KN
48.67 x 12 + VBx
VB+ 83.33 = 191.3
VC+ VD= 22 x 6 VC+ 83.33 = 191.
MC= - 104 KNm MB= - 104 KNm
VDx 6 -22 x 6 x 3 = -104 48.67 x 12 + VCx
5 ( 6 - X )
10 m
m.
rease or
appropriate combinations of loads.
( c ) The elastic moment at any sec
particular combination of loads shallthan 30 % of the numerically largest
the elastic maximum moments di
member , covering all appropriate co
IS 456-2000 ,Clause 23.1.2 EffectiveIn the absence of more accurate dete
width of flange may be taken as the f
IS 456-2000 ,Clause 37.1.1 Redistri
Continuous Beams and Frames -
( b ) The ultimate moment of resistan
member is not less than 70 percent of
obtained from an elastic maximum m
S 456-2000 , Clause 37.1.1 )
)
48.67
83.33
2.21 m48.67
83.33
108
108
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Where ,
0.671
26.5.1 Beams
26.5.1.1 Tension Reinforcement
where ,
For mild steel
greater than the breadth of the web pl
distances to the adjacent beams on ei
( a ) For T-beams , bf= ( l0/ 6 ) + bw+
( b ) For L-beams , bf= ( l0/ 12 ) + bw
NOTE - For continuous beams and fr
assumed as 0.7 times the effective sp
bf= effective width of flange ,
l0= distance between points of zero
bw= breadth of the web ,
Df= thickness of flange , and
b = actual width of the flange.
a ) Minimum reinforcement -The mi
shall not be less than that given by th
As/ b d = 0.85 / fy
lause 26.5( a ) )
IS 456-2000 Clause 26.5 Requireme
below.
b ) Maximum reinforcement -The m
reinforcement shall not exceed 0.04 b
Minimum steel %
100 As/ b d = 100 x 0.85 / 250 = 0.34
For HYSD steel , Fe415 grade
As= minimum area of tension reinforc
b = breadth of the beam or the breadt
d = effective depth , andfy= characteristic strength of reinforce
Structural Members
bw = 8
For singly reinforced rectangular sections
100 As/ b d = 100 x 0.85 / 415 = 0.20
For HYSD steel , Fe500 grade100 As/ b d = 100 x 0.85 / 500 = 0.17
Table 6-3
Limiting Moment of resistance factor Q lim, N /
p = 603 mm2,
2 - 16 mm O
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250 415 500
15 2.22 2.07 2.00
20 2.96 2.76 2.66
25 3.70 3.45 3.33
30 4.44 4.14 3.99
fck
N / mm2
fy, N / mm2
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== 648
Using three mome
MA( L1/ I1) + 2MB
A1= ( 2 / 3 ) x Bas
( 2 / 3 ) x
.( 1 )
x h1
6 x 162
[ 6 x 396 x 3 / ( Ix 6 ) ] - [ 6 x 648 x 3 / ( Ix 6) ]
Case ( c ) Maximu
2) = - 6 A1a1/ ( I1L1) -6 A2a2/ ( I2L2)
6MA+ 24MB+ 6MC
4MB+ MC = - 522
MA( 6 / I ) + 2MB(
C
a1= 3 m
N / m
C
36 KN