Design of Rectangular water tank CASE-1 ( L / B < 2 )Capacity 80000 Litres (given)Material M20 Grade Concrete (given)
Fe 415 Grade HYSD reinforcement (given)Solution :-
Provide 6 m x 4 m x 3.5 m tank with free board of 150 mm.
80400 Litres
L / B = 6 / 4 = 1.5 < 2 .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.
H / 4 = 3.5 / 4 = 0.875 mbottom 1 m will be designed as cantilever.
Water pressure at 3.5 - 1 = 2.5 m height from top = 2.5 x 9.8
= 24.5
6 m
6 m A E
F
3.5 m
D
Elevation Plan
Fixed end moments :-
= 73.5 KNm
= -32.66 KNm
Kani's Method :-
- 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
Volume = 6 x 4 x 3.35 x 10 3 =
2.5 x YW
=
KN / m2
where Yw is unit weight of water = 9.8 KN / m3
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
24.5 KN / m2
34.3 KN / m2
MAB
= w x l2 / 12 =
24.5 x 62 / 12 =
MAD
= w x l2 / 12 =
24.5 x 42 / 12 =
-3/10 -2/1040.84
2.5 m
1 m
Rotation factor at Joint A
Joint Member ∑ K
AAB I / 6
5 * I / 12- 2 / 10
AD I / 4 - 3 / 10
Sum of FEM
73.5-32.6640.84 KNm
= 73.5 + 2 x (- 8.17 ) + 0= 57.16
= (- 32.66 ) + 2 x (- 12.25 ) + 0= -57.16
B.M. at centre of long span =
=53.09 KNm
B.M. at centre of short span =
=-8.16 KNm
Direct tension in long wall = = 24.5 x 4 / 2 = 49 KN
Direct tension in short wall = = 24.5 x 6 / 2 = 73.5 KN
Design of Long Walls :-At support
M = 57.16 KNm
T = 49 KN Tension on liquid face.From Table 9-6 Q = 0.306 Assuming d / D = 0.9
D =
== 432.2 mm, Assuming d / D = 0.9
Take D = 450 mm d = 450 - 25 - 8
Relative Stiffness( K )
Rotation Factor u =(-1/2) k / ∑ K
MA
F =
MAB
= MAB
F + 2 MAB
' + MBA
'
MAD
= MAD
F + 2 MAD
' + MDA
'
w x l 2 / 8 - 57.16
24.5 x 62 / 8 - 57.16
w x l 2 / 8 - 57.16
24.5 x 42 / 8 - 57.16
Yw ( H - h ) x B / 2
Yw ( H - h ) x L / 2
√M / Q x b
√57.16 x 10 6 / 0.306 x 1000
= 417 mmFrom Table 9-5
=
= 1048
=
= 327
1048 + 327
= 1375Provide 16 mm O bar
spacing of bar == 200.96 x 1000 / 1375= 146.152727 mm
Provide 16 mm O bar @ 130 mm C/C…marked(a) = 1546Larger steel area is provided to match with the steel of short walls.At centre
M = 53.09 KNmT = 49 KN tension on remote facee = M / T = 53.09 / 49
= 1.08 m Line of action of forces lies outside the sectioni.e.tension is small
E = e + D / 2 - d b
= 1080 + 450 / 2 - 417 = 888 mm
D
modified moment = 49 x 0.888 d
= 43.51 KNmd'
E = e + D / 2 - d
=
= 617
=
= 327
617 + 327
= 944Provide 16 mm O bar
Ast1
for moment = M / σst x j x d
57.16 x 10 6 / 150 x 0.872 x 417
mm2
Ast2
for direct tension = T / σst
49 x 10 3 / 150
mm2
Total Ast1
+ Ast2
=
mm2
Area of one bar x 1000 / required area in m2 / m
mm2 / m.
Ast1
for moment = M / σst x j x d
43.51 x 10 6 / 190 x 0.89 x 417
mm2
Ast2
for direct tension = T / σst
49 x 10 3 / 150
mm2
Total Ast1
+ Ast2
=
mm2
spacing of bar == 200.96 x 1000 / 944= 212.881356 mm
Provide 16 mm O bar @ 200 mm C/C…marked(b) = 1005From Table 9-3 minimum reinforcement 0.16 %Distribution steel = 0.16 / 100 x 1000 x 450
= 720
On each face = 360Provide 8 mm O bar
spacing of bar == 50.24 x 1000 /360= 139.555556 mm
Provide 8 mm O bar @ 130 mm C/C…marked(d) = 385Vertical Steel ( c)Provide 10 mm O bar
spacing of bar == 78.50 x 1000 /360= 218.055556 mm
Provide 10 mm O bar @ 200 mm C/C…marked(c) = 392Horizontal steel :-
Remote face ( b) - Provide 16 mm O @ 200 mm C/C = 1005
Liquid face ( d) - Provide 8 mm O @ 130 mm C/C = 385
Vertical Steel ( c) - Provide 10 mm O @ 200 mm C/C both faces = 392
Design of short walls :-At support M = 57.16 KNm
T = 73.5 KN tension on liquid faceFrom Table 9-5
=
= 1048
=
= 490
1048 + 490
= 1538Provide 16 mm O bar
spacing of bar == 200.96 x 1000 / 1538= 130.663199 mm
Provide 16 mm O bar @ 130 mm C/C…marked(b) = 1546
Area of one bar x 1000 / required area in m2 / m
mm2
mm2
mm2
Area of one bar x 1000 / required area in m2 / m
mm2
Area of one bar x 1000 / required area in m2 / m
mm2
mm2
mm2
mm2
Ast1
for moment = M / σst x j x d
57.16 x 10 6 / 150 x 0.872 x 417
mm2
Ast2
for direct tension = T / σst
73.5 x 10 3 / 150
mm2
Total Ast1
+ Ast2
=
mm2
Area of one bar x 1000 / required area in m2 / m
mm2 / m.
At centreM = 8.16 KNmT = 73.5 KN tension on liquid face
From Table 9-5
=
= 150
=
= 490
150 + 490
= 640Provide 12 mm O bar
spacing of bar == 113.04 x 1000 / 640= 176.625 mm
Provide 12 mm O bar @ 130 mm C/C…marked(e) = 869
From Table 9-3 minimum reinforcement 0.16 %Distribution steel = 0.16 / 100 x 1000 x 450
= 720
On each face = 360Provide 8 mm O bar
spacing of bar == 50.24 x 1000 /360= 139.555556 mm
Provide 8 mm O bar @ 130 mm C/C…marked(d) = 385Vertical Steel ( c)Provide 10 mm O bar
spacing of bar == 78.50 x 1000 /360= 218.055556 mm
Provide 10 mm O bar @ 200 mm C/C…marked(c) = 392
Horizontal steel :-
Remote face ( e) - Provide 12 mm O @ 130 mm C/C = 869
Liquid face ( d) - Provide 8 mm O @ 130 mm C/C = 385
Vertical Steel ( c) - Provide 10 mm O @ 200 mm C/C both faces = 392
Bottom 1 m will be designed as cantilever
Ast1
for moment = M / σst x j x d
8.16 x 10 6 / 150 x 0.872 x 417
mm2
Ast2
for direct tension = T / σst
73.5 x 10 3 / 150
mm2
Total Ast1
+ Ast2
=
mm2
Area of one bar x 1000 / required area in m2 / m
mm2 / m.
mm2
mm2
Area of one bar x 1000 / required area in m2 / m
mm2
Area of one bar x 1000 / required area in m2 / m
mm2 .
mm2
mm2
mm2
Cantilever moment : -
M = OR , whichever is greater.= 9.8 x 3.5 x 1 / 6 = 9.8 x 3.5 / 6= 5.72 KNm = 5.72 ,tension on liquid face.
From Table 9-5
=
= 105From Table 9-3 minimum reinforcement 0.16 %Distribution steel = 0.16 / 100 x 1000 x 450
= 720
On each face = 360Provide 10 mm O bar
spacing of bar == 78.50 x 1000 /360= 218 mm
Provide 10 mm O bar @ 200 mm C/C…marked(c) = 392each face
Base slab :-Base slab is resting on ground. For a water head 3.5 m, provide 150 mm thick slab.From table 9-3Minimum steel = 0.229%
= 0.229 / 100 x 1000 x 150
= 344Provide 8 mm O bar
spacing of bar == 50.24 x 1000 /172= 292 mm
Provide 8 mm O bar @ 290 mm C/C both ways, top and bottom.
346Designed section,Elevation etc. are shown in fig.
Top slab : - consider 1 m wide strip. Assume 150 mm thick slab.
4 + 0.15 = 4.15 say 4.5 m
6 + 0.15 = 6.15 say 6.5 m
Dead Load : self 0.15 x 25 = 3.75
floor finish = 1.0
Live load = 1.5
6.25For 1 m wide strip
Yw x H x h2 / 6 Y
w x H / 6
Ast for moment = M / σ
st x j x d
5.72 x 10 6 / 150 x 0.872 x 417
mm2
mm2
mm2
Area of one bar x 1000 / required area in m2 /m
mm2 .
mm2 ,172 mm2 bothway
Area of one bar x 1000 / required area in m2 / m
Ast = mm2
Top slab may be designed as two-way slab as usual for a live load of 1.5 KN / m2
lx =
ly =
KN / m2
KN / m2
KN / m2
KN / m2
1.5 x 6.25 = 9.38 KN / m
6.5 / 4.5= 1.4
AS Per IS-456-2000,Four Edges Discontinuous,positive moment at mid-span.
Table 26 0.085
0.056
= == 16.15 KNm = 10.64 KNm
From Table 6-3 ,Q = 2.76
== 76.50 mm
150 - 15(cover) - 5= 130 > 76.50 mm …………(O.K.)
130 - 10 = 120 mmLarger depth is provided due to deflection check.
= 0.96
=415 / 20
= 50 [(1-0.88) x 20 / 415 ]= 0.29%
0.29 x 1000 x 130 / 100
= 377Provide 10 mm O bar
spacing of bar == 78.50 x 1000 /377= 208 mm
Provide 10 mm O bar @ 210 mm c/c = 374
= 0.74
PU =
ly / l
x =
αx =
αy =
Mx = α
x x w x l
x2 M
y = α
y x w x l
x2
0.085 x 9.38 x 4.52 0.056 x 9.38 x 4.52
drequired
= √M / Q x b
√16.15 x 10 6 / 2.76 x 1000
dshort
=
dlong
=
Mu / b x d2 (short) = 16.15 x 106 / 1000 x 130 x 130
Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2)
fy / f
ck
50 1-√1-(4.6 / 20) x (0.96)
Ast (short) =
mm2
Area of one bar x 1000 / required area in m2 /m
mm2 .
Mu / b x d2 (long) = 10.64 x 106 / 1000 x 120 x 120
=415 / 20
= 50 [(1-0.91) x 20 / 415 ]= 0.22%
0.22 x 1000 x 120 / 100
= 264
Provide 8 mm O bar
spacing of bar == 50.24 x 1000 /264= 190 mm
Provide 8 mm O bar @ 190 mm c/c = 264
Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2)
fy / f
ck
50 1-√1-(4.6 / 20) x (0.74)
Ast (long) =
mm2
Area of one bar x 1000 / required area in m2 / m
mm2 .
B
4 m
C
TABLE 9-6Balanced Design Factors for members in bending
For M20 Grade Concrete Mix
d / DMild steel HYSD bars
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 0.2580.9 0.314 0.335 0.306 0.246
TABLE 9-5
Q = M / bD2 Pt Q = M / bD2 P
t
Members in bending ( Cracked condition )Coefficients for balanced design
k j Q
For members less than 225mm thickness and tension on liquid face
M20 Fe250 7 115 0.445 0.851 1.33Fe415 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
Line of action of forces lies outside the section
D / 2
e = M / T
TABLE 9-3Minimum Reinforcement for Liquid Retaining Structures
Thickness, mm% of reinforcement
Mild Steel HYSD bars
100 0.3 0.24
150 0.286 0.229
200 0.271 0.217250 0.257 0.206
Grade of concrete
Grade of steel
σcbc
N / mm2
σst
N / mm2
300 0.243 0.194350 0.229 0.183400 0.214 0.171
450 or more 0.2 0.16
3500
150
1 : 4 : 8 P.C.C.
150 450 6000 450 150
Elevation1500
450
1000
4000
1000
450
1500 1500
6000
450 450
Section A-A
A 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
vv vv1000
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 )
Table 6-3
250 415 500 550
15 2.22 2.07 2.00 1.9420 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 Q lim
, N / mm2
fck
N / mm2f
y, N / mm2
TABLE 9-5
Members in bending ( Cracked condition )Coefficients for balanced design
1.360.98
1.2
0.61
Pt,bal
Design of Rectangular water tank CASE-2 ( L / B ≥ 2 )Size of tank : 3.6 m x 8.0 m x 3.0 m high (given)Material M20 Grade Concrete (given)
Fe 415 Grade HYSD reinforcement (given)Solution :-Size of tank : 3.6 m x 8.0 m x 3.0 m high
86400 Litres
L / B = 8 / 3.6 = 2.22 > 2 .The long walls are designed as vertical cantilevers from the base. 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.bottom 1 m or H / 4 whichever is more is designed as cantilever.
H / 4 = 3.0 / 4 = 0.75 mbottom h = 1 m will be designed as cantilever.
Moments and tensions :Maximum B.M. in long walls at the base
=
== 44.1 KNm.
Maximum ( - ve ) B.M. in short walls at support
=
== 26.13 KNm.
Maximum ( + ve ) B.M. in short walls at centre
=
== 19.60 KNm.
For bottom portion
M = OR = 9.8 x 3.0 x 1 / 6 = 9.8 x 3.0 / 6= 4.90 KNm = 4.90 KNm
Direct tension in long wall = = 9.8 x ( 3 - 1 ) x 3.6 / 2 = 35.28 KN
Direct tension in short wall= = 9.8 x ( 3 - 1 ) x 1= 19.6 KN
Volume = 3.6 x 8 x 3.0 x 10 3 =
(1 / 6 ) x Yw x H3
( 1 / 6 ) x 9.8 x 33
Yw x ( H - h ) x B2 / 12
9.8 x ( 3 - 1 ) x 42 / 12
Yw x ( H - h ) x B2 / 16
9.8 x ( 3 - 1 ) x 42 / 16
Yw x H x h2 / 6 Y
w x H / 6 , whichever is greater
Yw
x ( H - h ) x B / 2
Yw ( H - h ) x 1
It is assumed that end one metre width of long wall gives direct tension to short walls.
Design of long walls : -M ( - ) = 44.1 KNm ( water face )
T = 35.28 KN ( perpendicular to moment steel )From Table 9-6 Assume d / D = 0.9 Q = 0.306
D =
== 379.6 mm,
Take D = 400 mm d = 400 - 25 - 8= 367 mm
From Table 9-5 ,
=
= 918.68Provide 16 mm O bar
spacing of bar == 200.96 x 1000 / 918.68= 218.749 mm
Provide 16 mm O bar @ 200 mm c/c = 1005
From Table 9-3Distribution steel = 0.171 % for 400 mm depth
( 0.171 / 100 ) x 1000 x 400
= 684
on each face = 342 …………………… ( 1 )
Steel required for direct tension
=
=
= 235 …………………… ( 2 )From ( 1 ) and ( 2 ) , minimum steel is sufficient for resisting direct tension.Provide 8 mm O bar
spacing of bar == 50.24 x 1000 / 342= 146.901 mm
Provide 8 mm O bar @ 140 mm c/c on each face = 357on each face
√M / Q x b
√44.1 x 10 6 / 0.306 x 1000
Ast for Moment
Ast = M / σ
st x j x d
44.1 x 10 6 / 150 x 0.872 x 367
mm2
Area of one bar x 1000 / required area in m2 / m
mm2 .
Note : The design is made at the base. The moment reduces from base to top.For economy, the reinforcement can be curtailed or the thickness of wall can also be reduced as we have done for cantilever retaining wall.
As =
mm2 .
mm2 .
T / σst
35.28 x 103 / 150
mm2 .
Area of one bar x 1000 / required area in m2 / m
mm2
Design of short walls :-At support M = 26.13 KNm
T = 19.6 KNFrom Table 9-5
=
= 544
T / σst
=
= 131
544 + 131
= 675Provide 12 mm O bar
spacing of bar == 113.04 x 1000 / 675= 167.467 mm
Provide 12 mm O bar@160 mm c/c = 706
1000
203.56
400 367
163.44
checking :
modular ratio m = = 13.33
x =
=( 1000 x 400 ) + ( ( 13.33 - 1 ) x 706 )
= 203.56 mm
D - x = 196.44 mmd - x = 163.44 mm
= 1000 x 400 + (13.33 - 1 ) x 706
Ast1
for moment = M / σst x j x d
26.13 x 10 6 / 150 x 0.872 x 367
mm2
Ast2
for direct tension =
19.6 x 10 3 / 150
mm2
Total Ast1
+ Ast2
=
mm2
Area of one bar x 1000 / required area in m2 / m
mm2.
280 / 3 x σcbc
b x D2 / 2 + Ast ( m - 1 ) x d
b x D + ( m - 1 ) x Ast
( 1000 x 4002 / 2 ) + ( 706 x ( 13.33 - 1 ) x 367 )
AT = b x D + ( m - 1 ) x A
st
= 408705
== 5.34E+09 + 2.33E+08
= 5.57E+09
=
= 0.048
=
= 0.767check :
( 0.048 / 1.2 ) + ( 0.767 / 1.7 ) ≤ 1 From Table 9-2
0.4912 ≤ 1 ………………….. ( O. K. )
At centre :M = 19.6 KNmT = 19.6 KN
From Table 9-5
=
= 408
T / σst
=
= 131
408 + 131
= 539Provide 12 mm O bar
spacing of bar == 113.04 x 1000 / 539= 209.722 mm
Provide 12 mm O bar @ 200 mm c/c = 565
From Table 9-3Distribution steel = 0.171 % for 400 mm depth
mm2
Ixx
= ( 1 / 3 ) x b x ( x3 + ( D - x )3 ) + ( m - 1 ) x Ast x ( d - x )2
( 1 / 3 ) x 1000 x ( 203.563 + 196.443 ) + ( 13.33 - 1 ) x 706 x 163.442
mm4
fct
= T / AT
19.6 x 10 3 / 408705
N / mm2
fcbt
= M x ( d - x ) / Ixx
26.13 x 106 x 163.44 / 5.57 x 10 9
N / mm2
( fct / σ
ct ) + ( f
cbt / σ
cbt ) ≤ 1
0.04 + 0.4512 ≤ 1
Ast1
for moment = M / σst x j x d
19.6 x 10 6 / 150 x 0.872 x 367
mm2
Ast2
for direct tension =
19.6 x 10 3 / 150
mm2
Total Ast1
+ Ast2
=
mm2
Area of one bar x 1000 / required area in m2 / m
mm2.
( 0.171 / 100 ) x 1000 x 400
= 684
on each face = 342 …………………… ( 1 )
Steel required for direct tension
=
=
= 131 …………………… ( 2 )From ( 1 ) and ( 2 ) , minimum steel is sufficient for resisting direct tension.Provide 8 mm O bar
spacing of bar == 50.24 x 1000 / 342= 146.901 mm
Provide 8 mm O bar @ 140 mm c/c on each face = 357on each face
Bottom cantileverM = 4.9 KNm
From Table 9-5
=
= 102
Provide 8 mm O bar @ 140 mm c/c on each faces = 357on each face
Base slab :-Base slab is resting on ground. For a water head 3 m, provide 150 mm thick slab.From table 9-3Minimum steel = 0.229%
= 0.229 / 100 x 1000 x 150
= 344Provide 8 mm O bar
spacing of bar == 50.24 x 1000 /172= 292 mm
Provide 8 mm O bar @ 290 mm C/C both ways, top and bottom.
346Designed section,Elevation etc. are shown in fig.
Top slab : - consider 1 m wide strip. Assume 150 mm thick slab.
3.6 + 0.4 = 4 say 4 m
As =
mm2 .
mm2 .
T / σst
19.6 x 103 / 150
mm2 .
Area of one bar x 1000 / required area in m2 / m
mm2
Ast = M / σ
st x j x d
4.9 x 10 6 / 150 x 0.872 x 367
mm2
Minimum steel = 342 mm2 on each face.
mm2
mm2 ,172 mm2 bothway
Area of one bar x 1000 / required area in m2 / m
Ast = mm2
Top slab may be designed as a one-way slab as usual for a live load of 1.5 KN / m2
lx =
8 + 0.15 = 8.15 say 8.5 m
Dead Load : self 0.15 x 25 = 3.75
floor finish = 1.0
Live load = 1.5
6.25For 1 m wide strip
1.5 x 6.25 = 9.38 KN / m
Maximum moment = = 18.76 KNm
Maximum shear = 9.38 x 3.6 / 2= 16.88 KN
From Table 6-3 ,Q = 2.76
== 82.44 mm
150 - 15(cover) - 6= 129 > 82.44 …………(O.K.)
Larger depth is provided due to deflection check.Design for flexure :
= 1.13
=415 / 20
= 50 [(1-0.86) x 20 / 415 ]= 0.34%
0.34 x 1000 x 129 / 100
= 439Provide 8 mm O bar
spacing of bar == 50.24 x 1000 /439= 114 mm
Provide 8 mm O bar @ 110 mm c/c = 457
ly =
KN / m2
KN / m2
KN / m2
KN / m2
PU =
9.38 x 42 / 8
drequired
= √M / Q x b
√18.76 x 10 6 / 2.76 x 1000
dprovided
=
Mu / b x d2 = 18.76 x 106 / 1000 x 129 x 129
Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2)
fy / f
ck
50 1-√1-(4.6 / 20) x (1.13)
Ast =
mm2
Area of one bar x 1000 / required area in m2 /m
mm2 .
Minimum steel is 0.15 % for mild steel and 0.12 % for HYSD Fe415 reinforcementDistribution steel = ( 0.12 / 100 ) x 1000 x 150
= 180
Provide 6 mm O bar
spacing of bar == 28.26 x 1000 /180= 157 mm
Provide 6 mm O bar @ 150 mm c/c = 188
mm2
Area of one bar x 1000 / required area in m2 / m
mm2 .
If thickness of long walls is 400 mm, the span of the short wall = 3.6 + 0.4 = 4.0 m.
It is assumed that end one metre width of long wall gives direct tension to short walls.
Note : The design is made at the base. The moment reduces from base to top.For economy, the reinforcement can be curtailed or the thickness of wall can also be reduced as we have done for
x
Table 9-2
M15 1.1 1.5 1.5M20 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 2.5M40 1.7 2.4 2.7
) + ( 13.33 - 1 ) x 706 x 163.442
Permissible concrete stresses in calculations relating to resistance to cracking
Grade of concrete
Permissible stresses in N / mm2
Direct tension σct
Tension due to bending σcbt
Shear stress ح
v = V / b j d
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
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
vv vv900
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 A A
B
B
8 O @ 140 c/c
8 O @ 140 c/c
8 O @ 110 c/c 6 O @ 150 c/c
3000
150
150
150 400 3600 400 150
Section B- B
Base details not shown for clarity
16 O @ 200 c/c ( chipiya )
900 900
8 O @ 140 c/c
Design of simply supported one way slabeffective span = 4 m supported on masonry wall of 230 mm thickness ( given )
Live load = 2.5 ( given )
Floor finish = 1 ( given )material M15 grade concrete ( given )
HYSD reinforcement grade Fe415 ( given )solution : - Assume 0.4 % steel , a trial depth by deflection criteria IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementmodification factor = 1.26 for simply supported, basic span / effective depth ratio = 20( span / d ) ratio permissible = 1.26 x 20
= 25.2
4000 / 25.2
= 158.7 mmD = 158.7 + 15 ( cover ) + 5 ( assume 10 O bar )
= 178.7 mmAssume an overall depth = 180 mmDL = 0.18 x 25 = 4.5
Floor finish = 1.0
Live load = 2.5
Total 8.0Factored load = 1.5 x 8 = 12 KN / m
Consider 1 m length of slabMaximum moment =
== 24 KNm
Maximum shear = w x l / 2= 12 x 4 / 2= 24 KN
Design for flexure : -d = 180 - 15 - 5
= 160 mm
= 0.94
=415 / 15
= 50 [(1-0.84) x 15 / 415 ]
KN / m2
KN / m2
drequired
=
KN / m2
KN / m2
KN / m2
KN / m2
w x l2 / 8
12 x 42 / 8
Mu / b x d2 = 24 x 10 6 / 1000 x (160)2
Pt = 50 1-√1-(4.6 / fck) x (M
u / b x d2)
fy / f
ck
50 1-√1-(4.6 / 15) x (0.94)
= 0.289%
0.289 x 1000 x 160 / 100
= 462Provide 10 mm O bar
spacing of bar == 78.50 x 1000 /462= 170 mm
Provide 10 mm O bar @ 170 mm c/c = 462
100 x 231 / ( 1000 x 180 )
= 0.128 > 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 = 216
Maximum spacing = 5 x 160 = 800 or 450 mm i.e. 450 mmProvide 8 mm O bar
spacing of bar == 50.24 x 1000 /216= 233 mm
Provide 8 mm O bar @ 230 mm c/c = 218Check for shear : -
24 KN
Actual Shear stress =
=
= 0.150 ( too small )
For bars at support
d = 160 mm
231
100 x 231 / 1000 x 160= 0.144
6 x β
β == 0.8 x 15 / 6.89 x 0.144= 12.1
6 x 12.1= 0.277
IS 456-2000 Table 19 from table 7-1
Ast =
mm2
Area of one bar x 1000 / required area in m2 / m
mm2 .
Half the bars are bent at 0.1 l = 400 mm , and remaining bars provide 231 mm2 area
100 x As / ( b x D ) =
mm2
Area of one bar x 1000 / required area in m2 /m
mm2 .
Vu =
Vu / b x d
24 x 103 / 1000 x 160
N / mm2 ح ) >C )
N / mm2
As = mm2 .
100 x As / b x d =
Design shear strength حc = 0.85 √ 0.8 x f
ck ( √ 1 + 5 x β - 1 )
0.8 x fck
/ 6.89 Pt , but not less than 1.0
Design shear strength حc = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 12.1 - 1 )
IS 456-2000 clause 40.2.1.1 25 difference -0.05k = 1.24 for 180 mm slab depth 20 difference ? -0.04
Design shear strength = 1.24 x 0.28
= 0.347 ……………….( O.K.)Check for development length : -
8 O (HYSD Fe415 steel ) For continuing bars
231
OR= 0.87 x 415 x 231 x 160 { 1 - (415 x 231 / 15 x 1000 x 160 ) }= 13.34 { 1- 0.0399 }= 12.812 KNm
24 KN
= 56 O ( from Table 7-6 )
693.875 + 8 O693.88
which gives 14.46 mm ……………….( O.K.)Check for deflection : -
Basic ( span / d ) ratio = 20
100 x 462 / 1000 x 160= 0.289
IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementmodification factor = 1.42( span / d ) ratio permissible = 1.42 x 20
= 28.4Actual (span / d ) ratio = 4000 / 160
= 25.00 < 28.4 ……………….( O.K.)The depth could be reducedCheck for cracking : -IS 456-2000 , clause 26.3.3
Main bars : maximum spacing permitted = 3 x effective depth of slab or 300 mm whichever is small= 3 x 160 = 480 mm or 300 mm i.e. 300 mm
spacing provided = 170 mm < 300 mm ……………….( O.K.)Distribution bars : maximum spacing permitted = 5 x effective depth of slab or 450 mm whichever is small
= 5 x 160 = 800 mm spacing provided = 230 mm < 450 mm ……………….( O.K.)
for Pt = 0.144 حc = 0.28 N / mm2
N / mm2
Assuming L0 =
As = mm2
Mu1
= 0.87 x fy x A
st x d { 1 - ( f
y x A
st / f
ck x b x d ) }
Vu =
Development length of bars Ld = O σ
s / 4 x ح
bd
1.3 x ( Mu1
/ Vu ) + L
0 ≥ L
d
1.3 x ( 12.81 x 106 / 24 x 103 ) + 8 O ≥ 56 O≥ 56 O
48 O ≤O ≤
Pt = 100 x A
st / b x d =
For tying the bent bars at top , provide 8 mm O @ 230 mm c/c
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
For mild steel minimum reinforcement 0.15 %
= 100 x 231 / 1000 x 160= 0.14
From equation
we get , 0.49
= 12.54 KNm
160
180
400 400
4000
3 x effective depth of slab or 300 mm whichever is smallor 300 mm i.e. 300 mm
……………….( O.K.)5 x effective depth of slab or 450 mm whichever is small
or 450 mm i.e. 450 mm……………….( O.K.)
For checking development length , l0
may be assumed as 8 O for HYSD bars ( usually end anchorage is not provided ) and 12 O for mild steel ( U hook is provided usually whose anchorage length is 16 O.
Pt = 100 x A
s / b x d
Pt = 50 1-√1-(4.6 / fck) x (M
u / b x d2)
fy / f
ck
Mu1
/ b x d2 =
Mu1
= 0.49x 1000 x 1602 x 10-6
vv
vv
vvvv
vv
8 O @ 230 c/c10 O @ 170 c/c ( alternate bent )
IS 456-2000 clause 22.2Effective Span( a ) Simply Supported Beam or Slab -
( b )Continuous Beam or Slab - In the case ofcontinuous 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
supports are wider than I/12 of the clear span
or 600 mm whichever is less, the effective span
shall be taken as under:
1) For end span with one end fixed and theother continuous or for intermediate spans,the effective span shall be the clear spanbetween supports;
2) For end span with one end free and the othercontinuous, the effective span shall be equalto the clear span plus half the effective depthof the beam or slab or the clear span plushalf the width of the discontinuous support,whichever is less;3) In the case of spans with roller or rocketbearings, the effective span shall always bethe distance between the centres of bearings.
( 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
where it forms the end of a continuous beam
where the length to the centre of support shall
be taken.
( d )Frames-In the analysis of a continuous frame,centre to centre distance shall be used.
If ly / l
x ≥ 2 ,called one way slab provided that it
is supported on all four edges . Note that , if all four edge is not supported and l
y / l
x < 2 , then
also it is one-way slab,If ly / l
x < 2 , called two-
way slab.provided that it is supported on all four edges.
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.
For checking development length , l0
may be assumed as 8 O for HYSD bars ( usually end anchorage is not provided ) and 12 O for mild steel ( U hook is provided usually whose anchorage length is 16 O.
Design of Cantilever one way slabused for residential purpose
material M15 grade concrete ( given )mild steel grade Fe250 ( given )
Live Load As per IS 875Solution : -
( Balcony slab )Assume 120 mm thick slab DL LL
self load = 0.12 x 25 = 3 0
floor finish = 1 0
live load = 0 2
Total 4 2
1.5 ( 4 + 2 ) = ( 6 + 3 )DL LL
self load = 0.12 x 25 = 3 0
floor finish = 1 0
live load = 0 3
Total 4 3
1.5 ( 4 + 3 ) = (6 + 4.5)Weight of parapet 0.075 x 25 x 1 = 1.875 KN / m
1.5 x 1.875 = 2.8 KN / mConsider 1 m long strip
9 KN/m 6 KN/m
A 3m B 1.2m C
(a) Loads for maximum positive moment9 KN/m 10.5 KN/m 2.8 KN
A 3m B 1.2m C
considering fig (a)
cantilever moment =
at the free end of slab S1 ,concrete parapet of 75 mm thick and 1 m high.
For slab S2 live load = 2 KN /m2
For slab S1 live load = 3 KN /m2
ly = 6m
For S2 KN / m2
KN / m2
KN / m2
KN / m2
Pu = KN / m2
For S1 KN / m2
KN / m2
KN / m2
KN / m2
Pu = KN / m2
Pu =
(1) To get maximum positive moment in slab S2 only dead load on slab
S1 and total load on slab S
2 shall be considered
(b) Loads for maximum negative moment,maximum shear for cantilever span and maximum reaction at support B
wx l2 / 2
S2S2
S2
S1
S1
== 4.32 KNm
shear = Reaction at A = w x l / 2 - Moment @ B at distance 3 m= 9 x 3 / 2 - 4.32 / 3= 12.06 KN
Point of zero shear from A = 12.06 ( KN ) / 9 ( KN / m )= 1.34 m
Maximum positive moment =
== 8.08 KNm
(2) To get maximum negative moment and maximum shear at B,the slab is loaded with full loads as shown in fig (b)
Maximum negative moment =
== 10.92 KNm
Maximum shear at B
w x l / 2 + Moment @ B at distance 3 m= 9 x 3 / 2 + 10.92 / 3= 17.14 KN
w x l + 2.8= 10.5 x 1.2 + 2.8= 15.4 KN
Moment steel :Maximum moment = 10.92 KNm
From Table 6-3 Q = 2.22
== 70.14 mm,
120 - 15 - 6 ( assume 12 O bar )= 99 mm, ……………….( O.K.)
= 0.82
=250 / 15
= 50 [(1-0.865) x 15 / 250 ]= 0.405%
6 x 1.22 / 2
12.06 x 1.34 - W x l2 / 2
12.06 x 1.34 - 9 x 1.342 / 2
w x l + w x l2 / 2
2.8 x 1.2 + 10.5 x 1.22 / 2
Vu,BA
=
Vu,BC
=
drequired
= √M / Q x b
√10.92 x 10 6 / 2.22 x 1000
dprovided
=
Mu / b x d2 ( + ) = 8.08 x 10 6 / 1000 x (99)2
Pt = 50 1-√1-(4.6 / fck) x (M
u / b x d2)
fy / f
ck
50 1-√1-(4.6 / 15) x (0.82)
0.405 x 1000 x 99 / 100
= 401Provide 10 mm O bar
spacing of bar == 78.5 x 1000 / 401= 195.761 mm
Provide 10 mm O bar@170 mm c/c = 462 ( alternate bent up )
= 1.11
=250 / 15
= 50 [(1-0.812) x 15 / 250 ]= 0.564%
0.564 x 1000 x 99 / 100
= 558Provide 10 mm O bar
Provide 10 mm O bar@340 mm c/c = 231 ( bent bars extended )Area provided = Area of one bar x 1000 / spacing of bar in m
= 78.5 x 1000 / 340
= 231remaining area = 558 - 231
= 327Provide 12 mm O bar
spacing of bar == 113.04 x 1000 / 327= 346 mm
Provide 12 mm O bar@340 mm c/c = 332 ( Extra )
For mild steel minimum reinforcement 0.15 %Distribution steel = ( 0.15 /100 ) x 1000 x 120
= 180Provide 6 mm O bar
spacing of bar == 28.26 x 1000 / 180= 157 mm
Ast ( + ) =
mm2
Area of one bar x 1000 / required area in m2 / m
mm2.
Mu / b x d2 ( - ) = 10.92 x 10 6 / 1000 x (99)2
Pt = 50 1-√1-(4.6 / fck) x (M
u / b x d2)
fy / f
ck
50 1-√1-(4.6 / 15) x (1.11)
Ast ( - ) =
mm2
mm2.
mm2
mm2
Area of one bar x 1000 / required area in m2 / m
mm2.
Total 231 + 332 mm2 = 563 mm2 steel provided
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
mm2.
Area of one bar x 1000 / required area in m2 / m
Provide 6 mm O bar@150 mm c/c = 188For negative moment reinforcement
( from Table 7-6 )= O x 0.87 x 250 / 4 x 1= 54.4 O
54.4 x ( 10 + 12 ) / 2= 598 mm. say 600 mm
Check for development length : -
12 O (mild steel )
At A ,= 100 x 231 / 1000 x 99= 0.233
From equation OR
=
we get , 0.487 = 4.97401
= 4.78= 4.77 KNm
12.06 KN
= O x 0.87 x 250 / 4 x 1 =54.4 O
514.179 + 12 O514.179
which gives 12.13 mm ……………….( O.K.)
At B ,= 100 x 231 / 1000 x 99= 0.233
From equation OR
=
we get , 0.487 = 4.974
= 4.78= 4.77 KNm
Near point of contraflexure i.e. 0.15 x l from B
17.14 - ( 0.15 x 3 ) x 9 = 13.09 KN
= O x 0.87 x 250 / 4 x 1 =54.4 O
mm2.
Development length of bars Ld = O σ
s / 4 x ح
bd
Ld =
As a thumb rule, a bar shall be given an anchorage equal to the length of the cantilever.
Assuming L0 =
Pt = 100 x A
s / b x d Half bars bent = 462 / 2 = 231 mm2 )
Pt = 50 1-√1-(4.6 / fck) x (M
u / b x d2) M
u1 = 0.87 x f
y x A
st x d ( 1 - f
y x A
st / b x d x f
ck )
fy / f
ck 0.87 x 250 x 231 x 99 ( 1 - 250 x 231 / 1000 x 99 x 15 ) x 10 -6
Mu1
/ b x d2 =
Mu1
= 0.487 x 1000 x 992 x 10-6
Vu =
Development length of bars Ld = O σ
s / 4 x ح
bd
1.3 x ( Mu1
/ Vu ) + L
0 ≥ L
d
1.3 x ( 4.77 x 106 / 12.06 x 103 ) + 12 O ≥ 54.4 O≥ 54.4 O
42.4 O ≤O ≤
Pt = 100 x A
s / b x d Half bars bent = 462 / 2 = 231 mm2 )
Pt = 50 1-√1-(4.6 / fck) x (M
u / b x d2) M
u1 = 0.87 x f
y x A
st x d ( 1 - f
y x A
st / b x d x f
ck )
fy / f
ck 0.87 x 250 x 231 x 99 ( 1 - 250 x 231 / 1000 x 99 x 15 ) x 10 -6
Mu1
/ b x d2 =
Mu1
= 0.487 x 1000 x 992 x 10-6
Vu =
Development length of bars Ld = O σ
s / 4 x ح
bd
473.72 + 12 O473.72
which gives 11.2 mm ……………….( O.K.)Check for shear : - Span AB :
12.06 KN ( for maximum loading )At B , shear at point of contraflexure = 13.09 KN
13.09 KN
=
= 0.132 ……………….( O.K.)
100 x 231 / 1000 x 99= 0.233
6 x β
β == 0.8 x 15 / 6.89 x 0.233= 7.47
6 x 7.47= 0.34
IS 456-2000 Table 19 from table 7-1
IS 456-2000 clause 40.2.1.1 0.1 difference 0.07k = 1.3 for 120 mm slab depth 0.02 difference ? 0.014
Design shear strength = 1.3 x 0.34
= 0.442 ……………….( O.K.)Span BC :
17.14 KN
=
= 0.173 ……………….( O.K.)
100 x 563 / 1000 x 99= 0.569
6 x β
β == 0.8 x 15 / 6.89 x 0.569
1.3 x ( Mu1
/ Vu ) + L
0 ≥ L
d
1.3 x ( 4.77 x 106 / 13.09 x 103 ) + 12 O ≥ 54.4 O≥ 54.4 O
42.4 O ≤O ≤
At A , Vu,AB
=
Use Vu =
Shear stress حv = V
u / b x d
13.09 x 10 3 / 1000 x 99
N / mm2 ح >c
Pt = 100 x A
s / b x d =
Design shear strength حc = 0.85 √ 0.8 x f
ck ( √ 1 + 5 x β - 1 )
0.8 x fck
/ 6.89 Pt , but not less than 1.0
Design shear strength حc = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 7.47 - 1 )
for Pt = 0.233 حc = 0.34 N / mm2
N / mm2 ح <v
Vu =
Shear stress حv = V
u / b x d
17.14 x 10 3 / 1000 x 99
N / mm2 ح >c
Pt = 100 x A
s / b x d =
Design shear strength حc = 0.85 √ 0.8 x f
ck ( √ 1 + 5 x β - 1 )
0.8 x fck
/ 6.89 Pt , but not less than 1.0
= 3.06
6 x 3.06= 0.487
IS 456-2000 Table 19 from table 7-1
IS 456-2000 clause 40.2.1.1 0.25 difference 0.08k = 1.3 for 120 mm slab depth 0.181 difference ? 0.05792
Design shear strength = 1.3 x 0.48
= 0.624 ……………….( O.K.)Check for deflection : -For span AB :
Basic ( span / d ) ratio = 20
100 x 462 / 1000 x 99= 0.467
IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementmodification factor = 2( span / d ) ratio permissible = 2 x 20
= 40Actual (span / d ) ratio = 3000 / 99
= 30.30 < 40 ……………….( O.K.)For span BC :
Basic ( span / d ) ratio = 7
100 x 563 / 1000 x 99= 0.569
IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementmodification factor = 1.8( span / d ) ratio permissible = 1.8 x 7
= 12.6Actual (span / d ) ratio = 1200 / 99
= 12.12 < 12.6 ……………….( O.K.)Check for cracking : -IS 456-2000 , clause 26.3.3(1) Main bars : maximum spacing permitted = 3 x effective depth of slab or 300 mm whichever is small
= 3 x 99 = 297 mm or 300 mm i.e. 297 mmspacing provided = 170 mm < 297 mm ……………….( O.K.)
(2 )Distribution bars : maximum spacing permitted = 5 x effective depth of slab or 450 mm whichever is small= 5 x 99 = 495 mm
spacing provided = 150 mm < 450 mm ……………….( O.K.)
Design shear strength حc = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 3.06 - 1 )
for Pt = 0.569 حc = 0.48 N / mm2
N / mm2 ح <v
Pt = 100 x A
st / b x d =
Pt = 100 x A
st / b x d =
1.2m
0.15m
0.15m
0.15m 0.15m
1.2m
ly = 6m
lx = 3m
S2 S1
S3
B1B2
B3
B4
1m high parapet
Column 300 x 300
( 1 - 0.0389 )
KNm
( 1 - 0.0389 )
KNm
0.87 x fy x A
st x d ( 1 - f
y x A
st / b x d x f
ck )
0.87 x 250 x 231 x 99 ( 1 - 250 x 231 / 1000 x 99 x 15 ) x 10 -6
0.87 x fy x A
st x d ( 1 - f
y x A
st / b x d x f
ck )
0.87 x 250 x 231 x 99 ( 1 - 250 x 231 / 1000 x 99 x 15 ) x 10 -6
1200 1200
120 125
300 600
150 3000 150 1200
3 x effective depth of slab or 300 mm whichever is smallor 300 mm i.e. 297 mm
……………….( O.K.)5 x effective depth of slab or 450 mm whichever is small
or 450 mm i.e. 450 mm……………….( O.K.)
10 O @ 340 c/c (bent)+ 12 O @ 340 c/c (extra)
10 O @ 170 c/c
6 O @ 150 c/c6 O @ 150 c/c
Design of Continuous Two-way slabBanking hallslab is restrained with edge beams
givenMaterial M15 grade concreteHYSD reinforcement of grade Fe415
Solution : - Assume 150 mm thick slab
Self load 0.15 x 25 = 3.75
floor finish = 1.00
Live Load = 3.00
Total 7.75
1.5 x 7.75
= 11.625
5 / 5= 1 < 2 i.e. two-way slab.
Middle strip : IS 456-2000 Table -26 Two adjacent edges Discontinuous
13.66 KNm
10.17 KNmFrom Table 6-3
For M15 mix and Fe415 steel, Q = 2.07
== 81.23 mm,
= 150 - 15 - 10 - 5 ( assume 10 O bar )= 120 mm, > 81.23 mm ……………….( O.K.)
= 150 - 15 - 5 = 130 mm, > 81.23 mm ……………….( O.K.)
= 0.706
=415 / 15
= 50 [(1-0.885) x 15 / 415 ]= 0.208%
KN / m2
KN / m2
KN / m2
KN / m2
Pu =
KN / m2
ly / l
x =
Mu1
, Mu3
( - ) = αx x w x l
x2 = 0.047 x 11.625 x 52 =
Mu2
( + ) = αy x w x l
x2 = 0.035 x 11.625 x 52 =
drequired
= √M / Q x b
√13.66 x 10 6 / 2.07 x 1000
dprovided
for positive moment reinforcement
dprovided
for negative moment reinforcement
Mu / b x d2 ( + ) = 10.17 x 10 6 / 1000 x (120)2
Pt = 50 1-√1-(4.6 / fck) x (M
u / b x d2)
fy / f
ck
50 1-√1-(4.6 / 15) x (0.706)
0.208 x 1000 x 120 / 100
= 250Provide 8 mm O bar
spacing of bar == 50.24 x 1000 / 250= 200.96 mm
Provide 8 mm O bar@ 200 mm c/c = 251
= 0.81
=415 / 15
= 50 [(1-0.867) x 15 / 415 ]= 0.240%
0.24 x 1000 x 130 / 100
= 312Provide 8 mm O bar
spacing of bar == 50.24 x 1000 / 312= 161 mm
Provide 8 mm O bar@ 150 mm c/c = 335For HYSD Fe415
Minimum steel = ( 0.12 / 100 ) x 1000 x 150
= 180At discontinuous edges 4 and 5 , 50 % of the positive steel is required at top
= ( 1 / 2 ) x 251
= 126This is less than minimum , therefore , use minimum steel at location 4 and 5 .Provide 8 mm O bar
spacing of bar == 50.24 x 1000 / 180= 279.111 mm
Provide 8 mm O bar@ 260 mm c/c = 193More 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
Ast =
mm2
Area of one bar x 1000 / required area in m2 / m
mm2.
Mu / b x d2 ( - ) = 13.66 x 10 6 / 1000 x (130)2
Pt = 50 1-√1-(4.6 / fck) x (M
u / b x d2)
fy / f
ck
50 1-√1-(4.6 / 15) x (0.81)
Ast =
mm2
Area of one bar x 1000 / required area in m2 / m
mm2.
mm2
mm2
Area of one bar x 1000 / required area in m2 / m
mm2.
= 188Provide 8 mm O bar
spacing of bar == 50.24 x 1000 / 188= 267.234 mm
Provide 8 mm O bar@ 260 mm c/c = 193This will be provided by minimum steel of edge strip,At corner B , steel required = ( 1/2 ) x 188
= 94Provide 8 mm O bar
spacing of bar == 50.24 x 1000 / 180= 279.111 mm
Provide 8 mm O bar@ 260 mm c/c = 193This will be provided by minimum steel .
Check for shear : -At point 1 or 3
S.F. = w x l / 2 + Moment @ point 1 or 3 in that span= 11.625 x 5 / 2 + 13.66 / 5= 31.795 KN
100 x 335 / ( 1000 x 130 )= 0.258
from table 7-1
0.25 differen 0.11IS 456-2000 clause 40.2.1.1 0.242 differe ? -0.1065
k = 1.3 for 150 mm slab depthDesign shear strength = 1.3 x 0.354
= 0.460 ……………….( O.K.)OR
6 x β
β == 6.75
6 x 6.75= 0.356
IS 456-2000 clause 40.2.1.1k = 1.3 for 150 mm slab depth
Design shear strength = 1.3 x 0.356
= 0.463 ……………….( O.K.)
Actual shear stress =
mm2.
Area of one bar x 1000 / required area in m2 / m
mm2.
mm2.
Area of one bar x 1000 / required area in m2 / m
mm2.
Note that positive reinforcements are not curtailed because if they are curtailed , the remaining bars do not provide minimum steel.
100 x As / b x d =
for Ptح , 0.258 =
c = 0.354 N / mm2
N / mm2
Design shear strength حc = 0.85 √ 0.8 x f
ck ( √ 1 + 5 x β - 1 )
0.8 x fck
/ 6.89 Pt , but not less than 1.0
Design shear strength حc = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 6.75 - 1 )
N / mm2
Vu / b x d
=
= 0.245 ……………….( O.K.)At point 4 or 5
S.F. = w x l / 2 = 11.625 x 5 / 2 = 29.06 KN
100 x 251 / ( 1000 x 120 )= 0.209
from table 7-1
0.1 differenc 0.07IS 456-2000 clause 40.2.1.1 0.041 differe ? -0.0287
k = 1.3 for 150 mm slab depthDesign shear strength = 1.3 x 0.321
= 0.417OR
6 x β
β == 8.33
6 x 8.33= 0.326
IS 456-2000 clause 40.2.1.1k = 1.3 for 150 mm slab depth
Design shear strength = 1.3 x 0.326
= 0.424
Actual shear stress =
=
= 0.242 ……………….( O.K.)Check for development length : -This is critical at point 4 or 5.
At point 4 or 5 , 29.06 KN No bar is curtailed or bent up.
8 O (HYSD Fe415 steel )
= 100 x 251 / 1000 x 120= 0.209
From equation OR
=
we get , 0.711 = 10.8748
31.795 x 103 / ( 1000 x 130 )
N / mm2 ح > c
100 x As / b x d =
for Ptح , 0.209 =
c = 0.321 N / mm2
N / mm2
Design shear strength حc = 0.85 √ 0.8 x f
ck ( √ 1 + 5 x β - 1 )
0.8 x fck
/ 6.89 Pt , but not less than 1.0
Design shear strength حc = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 8.33 - 1 )
N / mm2
Vu / b x d
29.06 x 103 / ( 1000 x 120 )
N / mm2 ح > c
Vu =
Assuming L0 =
Pt = 100 x A
s / b x d
Pt = 50 1-√1-(4.6 / fck) x (M
u / b x d2) M
u1 = 0.87 x f
y x A
st x d ( 1 - f
y x A
st / b x d x f
ck )
fy / f
ck 0.87 x 415 x 251 x 120 ( 1 - 415 x 251 / 1000 x 120 x 15 ) x 10 -6
Mu1
/ b x d2 =
= 10.246= 10.24 KNm
=54.3 O (From Table 7-6 )
458.355 + 8 O458.355
which gives 9.90 mm ……………….( O.K.)
Short span 11.25 x ( 4.5 / 2 ) = 25.31 KN
8 O (HYSD Fe415 steel )
= 100 x 462 / 1000 x 160= 0.289
From equation OR
=
we get , 0.9595 = 26.6888
= 24.5539= 24.56 KNm
=56 O (From Table 7-6 )
1261.48 + 8 O1261.48
which gives 26.28 mm ……………….( O.K.)Note that the bond is usually critical along long direction.Check for deflection : -
Basic ( span / d ) ratio = 26
positive moment steel = 251actual d = 150 - 15 -8 - 4
= 123 mm.
100 x 251 / 1000 x 123 240.7= 0.204
IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementmodification factor = 1.65( span / d ) ratio permissible = 1.65 x 26
= 42.9Actual (span / d ) ratio = 5000 / 123
= 40.65 < 42.9 ……………….( O.K.)Check for cracking : -IS 456-2000 , clause 26.3.3
Mu1
= 0.711 x 1000 x 1202 x 10-6
Development length of bars Ld = O σs / 4 x ح
bd
1.3 x ( Mu1
/ Vu ) + L
0 ≥ L
d
1.3 x ( 10.246 x 106 / 29.06 x 103 ) + 8 O ≥ 54.3 O≥ 54.3 O
46.3 O ≤O ≤
Vu =
Assuming L0 =
Pt = 100 x A
s / b x d
Pt = 50 1-√1-(4.6 / fck) x (M
u / b x d2) M
u1 = 0.87 x f
y x A
st x d ( 1 - f
y x A
st / b x d x f
ck )
fy / f
ck 0.87 x 415 x 462 x 160 ( 1 - 415 x 462 / 1000 x 160 x 15 ) x 10 -6
Mu1
/ b x d2 =
Mu1
= 0.9595 x 1000 x 1602 x 10-6
Development length of bars Ld = O σ
s / 4 x ح
bd
1.3 x ( Mu1
/ Vu ) + L
0 ≥ L
d
1.3 x ( 24.56 x 106 / 25.31 x 103 ) + 8 O ≥ 56 O≥ 56 O
48 O ≤O ≤
mm2.
Pt = 100 x A
st / b x d =
(1) Main bars : maximum spacing permitted = 3 x effective depth of slab or 300 mm whichever is small= 3 x 130 = 390 mm
spacing provided = 200 mm < 300 mm ……………….( O.K.)(2) Distribu. bars : maximum spacing permitted = 5 x effective depth of slab or 450 mm whichever is small
= 5 x 120 = 600 mmspacing provided = 260 mm < 450 mm ……………….( O.K.)
This is minimum (0.12 / 100) x 150 x 1000 = 180Provide 8 mm O bar
spacing of bar == 50.24 x 1000 / 180= 279.111 mm
Provide 8 mm O bar@ 260 mm c/c = 193 for uniformity in spacing.For clarity , top and bottom reinforcements are shown separately.
625 3750 625
625
3750
625
150
625 3750 625
Section A-A
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.
mm2
Area of one bar x 1000 / required area in m2 / m
mm2.
Mid
dle
Str
ip
S1
Edg
e s
trip
Ed
ge
stri
p
Edge strip Edge stripMiddle Strip
8 O
@ 2
60
c/c
8 O
@ 2
60
c/c
8 O
@ 2
00
c/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
Design of Continuous Two-way slab 5 m
5 m
5 m
625 3750 625 5 m
ly/8 (3/4)l
yly/8
Middle Strip
S1
S1
Ed
ge s
trip
Ed
ge
str
ip
A B
B
1
2
3
4
5
0.0
35
-0.0
47
-0.047
0.035
( 1 - 0.0579 )
0.87 x fy x A
st x d ( 1 - f
y x A
st / b x d x f
ck )
0.87 x 415 x 251 x 120 ( 1 - 415 x 251 / 1000 x 120 x 15 ) x 10 -6
KNm
( 1 - 0.0799 )
KNm
0.87 x fy x A
st x d ( 1 - f
y x A
st / b x d x f
ck )
0.87 x 415 x 462 x 160 ( 1 - 415 x 462 / 1000 x 160 x 15 ) x 10 -6
3 x effective depth of slab or 300 mm whichever is smallor 300 mm i.e. 300 mm
……………….( O.K.)5 x effective depth of slab or 450 mm whichever is small
or 450 mm i.e. 450 mm……………….( O.K.)
1000
625 3750 625
1000 625
3750
625
150
625 3750 625
Mid
dle
Str
ip
S1
Ed
ge s
trip
Ed
ge
str
ip
Edge strip Edge stripMiddle Strip
8 O
@ 2
40
c/c
8 O
@ 2
40
c/c
8 O
@ 1
40
c/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
15
001
500
1500 1500
8 O @ 180 c/c
3
-0.047
Design of Continuous One-way slabA five span continuous one-way slab used as an office floor.The centre-to-centre distance of supporting beams is 3 m
givenMaterial M15 grade concrete
HYSD reinforcement of grade Fe415Solution : -Try 120 mm thick slab DL LL
Dead load 0.12 x 25 = 3 0
floor finish = 1 0
live load = 0 3
Total 4 3factored load = 1.5 ( 4 + 3 )
=Consider 1 m wide strip of the slab.
3m 3m 3m 3m 3m
The factored moments at different points using the coefficients are as follows :
== 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
0.6 x w x l + 0.6 x w x l= 0.6 x 6 x 3 + 0.6 x 4.5 x 3= 18.9 KN
KNm
Live load 3 KN / m2 and floor finish 1 KN / m2
KN / m2
KN / m2
KN / m2
KN / m2
( 6 + 4.5 ) KN / m2
Mu1
( + ) = ( 1 / 12 ) x w ( DL ) x l2 + ( 1 / 10 ) x w ( LL ) x l2
( 1 / 12 ) x 6 x 32 + ( 1 / 10 ) x 4.5 x 32
Mu2
( + ) = ( 1 / 16 ) x w ( DL ) x l2 + ( 1 / 12 ) x w ( LL ) x l2
( 1 / 16 ) x 6 x 32 + ( 1 / 12 ) x 4.5 x 32
Mu3
( - ) = ( 1 / 10 ) x w ( DL ) x l2 + ( 1 / 9 ) x w ( LL ) x l2
( 1 / 10 ) x 6 x 32 + ( 1 / 9 ) x 4.5 x 32
Mu4
( - ) = ( 1 / 12 ) x w ( DL ) x l2 + ( 1 / 9 ) x w ( LL ) x l2
( 1 / 12 ) x 6 x 32 + ( 1 / 9 ) x 4.5 x 32
Maximum shear is Vu(BA)
=
Maximum moment is Mu3
( - ) = 9.9
A B C D E F
1 2 2 2 1
( 6 + 4.5 ) KN / m
From Table 6-3 Q = 2.07
== 69.2 mm,
Try 110 mm overall depth
110 - 15 - 5 ( assume 10 O bar )= 90 mm, ……………….( O.K.)
spacing of bar =Table
point Steel Provided
1 ( + ) 8.55 1.06 0.322 29010 mm O @ 440 c/c + 8 mm O @ 440 c/c
2 ( + ) 6.75 0.83 0.248 2238 mm O @ 220 c/c
3 ( - ) 9.9 1.22 0.38 34210 mm O @ 220 c/c
4 ( - ) 9.0 1.11 0.34 30610 mm O @ 220 c/c
For Main steel ,HYSD Fe415 reinforcementminimum steel area = ( 0.12 / 100 ) x 1000 x 110
= 132For Distribution steel , mild steel Fe250 reinforcement
minimum steel area = ( 0.15 / 100 ) x 1000 x 110
= 165Use 6 mm O
spacing of bar == 28.26 x 1000 /165= 171
Use 8 mm O
spacing of bar == 50.24 x 1000 /146= 344
drequired
= √M / Q x b
√9.9 x 10 6 / 2.07 x 1000
dprovided
=
Pt = 50 1-√1-(4.6 / fck) x (M
u / b x d2)
fy / f
ck
Ast = P
t x b x d / 100
Area of one bar x 1000 / required area in m2 / m
Factored moment
KNmMu/(b x d2 ) P
t
Ast
mm2
= 178 +114 = 292 mm2 (Half 10 O+half 8 O)
= 228 mm2
= 357 mm2
= 357 mm2
mm2
mm2
Area of one bar x 1000 / required area in m2 / m
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 .
Area of one bar x 1000 / required area in m2 / m
Use 8 mm O @ 340 mm c/c = 148 mm2.
Check for shear : -Maximum shear = 18.9 KN
Actual Shear stress =
=
= 0.210 ( too small )
For bars at support
d = 90 mm
357
100 x 357 / 1000 x 90= 0.397
6 x β
β == 0.8 x 15 / 6.89 x 0.397= 4.4
6 x 4.4= 0.42
IS 456-2000 Table 19 from table 7-1
IS 456-2000 clause 40.2.1.1 0.25 difference 0.11k = 1.3 for 110 mm slab depth 0.103 difference ? 0.04532Design shear strength = 1.3 x 0.42
= 0.546 ……………….( O.K.)At point of contraflexure i.e. 0.15 x l from B
18.9 - 0.15 x 3 x 10.5= 14.18 KN
Actual Shear stress =
=
= 0.158 ( too small )
For bars at support
d = 90 mm
292
100 x 292 / 1000 x 90= 0.324
6 x β
β =
Vu / b x d
18.9 x 103 / 1000 x 90
N / mm2 ح ) >C )
N / mm2
As = mm2 .
100 x As / b x d =
Design shear strength حc = 0.85 √ 0.8 x f
ck ( √ 1 + 5 x β - 1 )
0.8 x fck
/ 6.89 Pt , but not less than 1.0
Design shear strength حc = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 4.4 - 1 )
for Ptح 0.397 =
c = 0.42 N / mm2
N / mm2
with positive moment reinforcement ( 292 mm2 )
Vu =
Vu / b x d
14.18 x 103 / 1000 x 90
N / mm2 ح ) >C )
N / mm2
As = mm2 .
100 x As / b x d =
Design shear strength حc = 0.85 √ 0.8 x f
ck ( √ 1 + 5 x β - 1 )
0.8 x fck
/ 6.89 Pt , but not less than 1.0
= 0.8 x 15 / 6.89 x 0.324= 5.4
6 x 5.4OR = 0.39IS 456-2000 Table 19 from table 7-1
IS 456-2000 clause 40.2.1.1 0.25 difference 0.11k = 1.3 for 110 mm slab depth 0.176 difference ? 0.07744Design shear strength = 1.3 x 0.38
= 0.494 ……………….( O.K.)Check for development length : -Span AB is critical for checking this requirementAt support A
0.4 x w x l + 0.45 x w x l= 0.4 x 6 x 3 + 0.45 x 4.5 x 3 = 13.28 KN
At A ,= 100 x 292 / 1000 x 90= 0.324
From equation OR
=
we get , 1.064 = 9.48839
= 8.64= 8.62 KNm
8 O (HYSD Fe415 steel )
= O x 0.67 x 415 / 4 x 1 =56.4 O
845.7831 + 8 O845.783
which gives 17.47 mm ……………….( O.K.)At support B, point of contraflexure is assumed at 0.15 x l from B
18.9 - 0.15 x 3 x 10.5= 14.18 KN
8.64 KNm as before
12 O
792.1016 + 12 O
Design shear strength حc = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 5.4 - 1 )
for Ptح 0.324 =
c = 0.38 N / mm2
N / mm2
Vu =
Pt = 100 x A
s / b x d
Pt = 50 1-√1-(4.6 / fck) x (M
u / b x d2) M
u1 = 0.87 x f
y x A
st x d ( 1 - f
y x A
st / b x d x f
ck )
fy / f
ck 0.87 x 415 x 292 x 90 ( 1 - 415 x 292 / 1000 x 90 x 15 ) x 10 -6
Mu1
/ b x d2 =
Mu1
= 1.064 x 1000 x 902 x 10-6
Assuming L0 =
Development length of bars Ld = O σ
s / 4 x ح
bd
1.3 x ( Mu1
/ Vu ) + L
0 ≥ L
d
1.3 x ( 8.64 x 106 / 13.28 x 103 ) + 8 O ≥ 56.4 O≥ 56.4 O
48.4 O ≤O ≤
Vu =
Mu1
=
L0 = ( actual anchorage is more than 12 O but L
0 is limited to 12 O or d ,
i.e. 90 mm whichever is greater )
1.3 x ( Mu1
/ Vu ) + L
0 ≥ L
d
1.3 x ( 8.64 x 106 / 14.18 x 103 ) + 12 O ≥ 56.4 O≥ 56.4 O
792.102which gives 17.84 mm ……………….( O.K.)
Check for deflection : -Maximum positive moment occurs in span AB. Therefore, this check is critical in span AB
Basic ( span / d ) ratio = 26
100 x 292 / 1000 x 90= 0.324
IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementmodification factor = 1.34( span / d ) ratio permissible = 1.34 x 26
= 34.84Actual (span / d ) ratio = 3000 / 90
= 33.33 < 34.84 ……………….( O.K.)For span BC :
Basic ( span / d ) ratio = 26
100 x 228 / 1000 x 90= 0.253
IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementmodification factor = 1.6( span / d ) ratio permissible = 1.6 x 26
= 41.6Actual (span / d ) ratio = 3000 / 90
= 33.33 < 41.6 ……………….( O.K.)Check for cracking : -IS 456-2000 , clause 26.3.3
(1) Main bars : maximum spacing permitted = 3 x effective depth of slab or 300 mm whichever is small= 3 x 90 = 270 mm or 300 mm i.e. 270 mm
spacing provided = 220 mm < 270 mm ……………….( O.K.)(2 )Distribution bars : maximum spacing permitted = 5 x effective depth of slab or 450 mm whichever is small
= 5 x 90 = 450 mm spacing provided = 160 mm < 450 mm ……………….( O.K.)
44.4 O ≤O ≤
Pt = 100 x A
st / b x d =
Pt = 100 x A
st / b x d =
IS 456-2000 Clause -22.5( 22.5.1 ) Unless more exact estimates are made, forbeams of uniform cross-section which support
substantially uniformly distributed loads over three ormore spans which do not differ by more than 15 percentof the longest, the bending moments and shear forcesused in design may be obtained using the coefficientsgiven in Table 12 and Table 13 respectively.
For moments at supports where two unequal spans
meet or in case where the spans are not equally loaded,
the average of the two values for the negative moment
at the support may be taken for design.Where coefficients given in Table 12 are used for
calculation of bending moments, redistribution referredto in 22.7 shall not be permitted. (22.5.2 ) Beams and Slabs Over Free End SupportsWhere a member is built into a masonry wall whichdevelops only partial restraint, the member shall bedesigned to resist a negative moment at the face of thesupport of Wl / 24 where W is the total design loadand I is the effective span, or such other restrainingmoment 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.Table 12 Bending Moment coefficients
Type of load
Span moments Support moments
+ 1 / 12 + 1 / 16 -1 / 10
+ 1 / 10 + 1 / 12 -1 / 9
Table 13 Shear Force coefficients
Type of loadAt support next to the end support
Outer side Inner side
0.4 0.6 0.55
0.45 0.6 0.6
Near middle of end span
At middle of interior span
At support next to the end support
Dead load and imposed load ( fixed )
imposed load ( not fixed )
NOTE -For obtaining the bending moment, the coefficient shall be multiplied by the total design load and effective span.
At end support
Dead load and imposed load ( fixed )
imposed load ( not fixed )
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 )
with positive moment reinforcement ( 292 mm2 )
( 1 - 0.0898 )
KNm
0.87 x fy x A
st x d ( 1 - f
y x A
st / b x d x f
ck )
0.87 x 415 x 292 x 90 ( 1 - 415 x 292 / 1000 x 90 x 15 ) x 10 -6
Maximum positive moment occurs in span AB. Therefore, this check is critical in span AB
3 x effective depth of slab or 300 mm whichever is smallor 300 mm i.e. 270 mm
……………….( O.K.)5 x effective depth of slab or 450 mm whichever is small
or 450 mm i.e. 450 mm……………….( O.K.)
( 22.5.1 ) Unless more exact estimates are made, forbeams of uniform cross-section which support
substantially uniformly distributed loads over three ormore spans which do not differ by more than 15 percentof the longest, the bending moments and shear forcesused in design may be obtained using the coefficientsgiven in Table 12 and Table 13 respectively.
For moments at supports where two unequal spans
meet or in case where the spans are not equally loaded,
the average of the two values for the negative moment
at the support may be taken for design.Where coefficients given in Table 12 are used for
calculation of bending moments, redistribution referredto in 22.7 shall not be permitted. (22.5.2 ) Beams and Slabs Over Free End SupportsWhere a member is built into a masonry wall whichdevelops only partial restraint, the member shall bedesigned to resist a negative moment at the face of thesupport of Wl / 24 where W is the total design loadand I is the effective span, or such other restrainingmoment 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.Table 12 Bending Moment coefficients
Support moments
-1 / 12
-1 / 9
Table 13 Shear Force coefficients
0.5
0.6
At other interior supports
NOTE -For obtaining the bending moment, the coefficient shall be multiplied by the total
At all other interior
supports
900
3000
10 O @ 220 c/c
( 0.3 l1 )
Design of Simply supported two way slabresidential building drawing room 4.3 m x 6.55 m
It is supported on 350 mm thick walls on all four sides.given
material M15 grade concreteHYSD reinforcement of grade Fe415
Solution :Consider 1 m wide strip. Assume 180 mm thick slab.
4.3 + 0.18 = 4.48 say 4.5 m
6.55+0.18 = 6.73 say 6.75 m
Dead load : self 0.18 x 25 = 4.5
floor finish = 1.0
Live load ( residence ) = 2.0
Total 7.5For 1 m wide strip
1.5 x 7.5= 11.25 KN / m
6.75 / 4.5 = 1.5
IS 456-2000 Table -27
23.7 KNm
10.48 KNmFrom Table 6-3 Q = 2.07
== 107 mm,
180 - 15 - 5 ( assume 10 O bar )= 160 mm, > 107 mm ……………….( O.K.)
160 - 10 = 150 mm, > 107 mm
Larger depth is provided due to deflection check.
= 0.926
lx =
ly =
KN / m2
KN / m2
KN / m2
KN / m2
Pu =
ly / l
x =
Mux
= αx x w x l
x2 = 0.104 x 11.25 x 4.52 =
Muy
= αy x w x l
x2 = 0.046 x 11.25 x 4.52 =
drequired
= √M / Q x b
√23.7 x 10 6 / 2.07 x 1000
dshort
provided
=
dlong
provided
=
Mu / b x d2 ( short ) = 23.7 x 10 6 / 1000 x (160)2
Pt = 50 1-√1-(4.6 / fck) x (M
u / b x d2)
fy / f
ck
=
415 / 15
= 50 [(1-0.846) x 15 / 415 ]= 0.28%
0.28 x 1000 x 160 / 100
= 448Provide 10 mm O bar
spacing of bar == 78.5 x 1000 / 448= 175.223 mm
Provide 10 mm O bar@170 mm c/c = 462 ( short span )
= 0.466
=415 / 15
= 50 [(1-0.926) x 15 / 415 ]= 0.134%
0.134 x 1000 x 150 / 100
= 201For HYSD Fe415 minimum reinforcement 0.12 %
Minimum steel = ( 0.12 /100 ) x 1000 x 180
= 216Provide 8 mm O bar
spacing of bar == 50.24 x 1000 / 216= 232.593 mm
Provide 8 mm O bar@ 230 mm c/c = 218 ( long span )The bars cannot be bent or curtailed because if 50 % of long span bars are curtailed , the remaining bars will be less than minimum
50 1-√1-(4.6 / 15) x (0.926)
Ast ( short ) =
mm2
Area of one bar x 1000 / required area in m2 / m
mm2.
Mu / b x d2 ( long ) = 10.48 x 10 6 / 1000 x (150)2
Pt = 50 1-√1-(4.6 / fck) x (M
u / b x d2)
fy / f
ck
50 1-√1-(4.6 / 15) x (0.466
Ast ( long ) =
mm2
mm2.
Area of one bar x 1000 / required area in m2 / m
mm2.
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 : -
Long span 11.25 x ( 4.5 / 2 ) = 25.31 KN
8 O (HYSD Fe415 steel )
= 100 x 218 / 1000 x 150= 0.145
From equation OR
=
we get , 0.5023 = 11.8063
= 11.3313= 11.30 KNm
=56 O (From Table 7-6 )
580.403 + 8 O580.403
which gives 12.09 mm ……………….( O.K.)
Short span 11.25 x ( 4.5 / 2 ) = 25.31 KN
8 O (HYSD Fe415 steel )
= 100 x 462 / 1000 x 160= 0.289
From equation OR
=
we get , 0.9595 = 26.6888
= 24.5539= 24.56 KNm
=56 O (From Table 7-6 )
1261.48 + 8 O1261.48
which gives 26.28 mm ……………….( O.K.)Note that the bond is usually critical along long direction.
Vu =
Assuming L0 =
Pt = 100 x A
s / b x d
Pt = 50 1-√1-(4.6 / fck) x (M
u / b x d2) M
u1 = 0.87 x f
y x A
st x d ( 1 - f
y x A
st / b x d x f
ck )
fy / f
ck 0.87 x 415 x 218 x 150 ( 1 - 415 x 218 / 1000 x 150 x 15 ) x 10 -6
Mu1
/ b x d2 =
Mu1
= 0.5023 x 1000 x 1502 x 10-6
Development length of bars Ld = O σ
s / 4 x ح
bd
1.3 x ( Mu1
/ Vu ) + L
0 ≥ L
d
1.3 x ( 11.30 x 106 / 25.31 x 103 ) + 8 O ≥ 56 O≥ 56 O
48 O ≤O ≤
Vu =
Assuming L0 =
Pt = 100 x A
s / b x d
Pt = 50 1-√1-(4.6 / fck) x (M
u / b x d2) M
u1 = 0.87 x f
y x A
st x d ( 1 - f
y x A
st / b x d x f
ck )
fy / f
ck 0.87 x 415 x 462 x 160 ( 1 - 415 x 462 / 1000 x 160 x 15 ) x 10 -6
Mu1
/ b x d2 =
Mu1
= 0.9595 x 1000 x 1602 x 10-6
Development length of bars Ld = O σs / 4 x ح
bd
1.3 x ( Mu1
/ Vu ) + L
0 ≥ L
d
1.3 x ( 24.56 x 106 / 25.31 x 103 ) + 8 O ≥ 56 O≥ 56 O
48 O ≤O ≤
Check for shear : -This is critical along long span
25.31 KN
Shear stress =
=
= 0.169 ( too small )
100 x 218 / 1000 x 150= 0.145
from table 7-1
25 difference -0.05IS 456-2000 clause 40.2.1.1 20 difference ? -0.04
k = 1.24 for 180 mm slab depthDesign shear strength = 1.24 x 0.28
= 0.347 ……………….( O.K.)OR
6 x β
β == 12.0114
6 x 12.011= 0.278
IS 456-2000 clause 40.2.1.1 25 difference -0.05k = 1.24 for 180 mm slab depth 20 difference ? -0.04
Design shear strength = 1.24 x 0.278
= 0.345 ……………….( O.K.)Check for deflection : -This check shall be done along short span
Basic ( span / d ) ratio = 20
100 x 462 / 1000 x 160= 0.289
IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementmodification factor = 1.44( span / d ) ratio permissible = 1.44 x 20
= 28.8Actual (span / d ) ratio = 4480 / 160
= 28.00 < 28.8 ……………….( O.K.)Check for cracking : -IS 456-2000 , clause 26.3.3
Vu =
Vu / b x d
25.31 x 103 / 1000 x 150
N / mm2 ح ) >C )
N / mm2
100 x As / b x d =
for Ptح 0.145 =
c = 0.28 N / mm2
N / mm2
Design shear strength حc = 0.85 √ 0.8 x f
ck ( √ 1 + 5 x β - 1 )
0.8 x fck
/ 6.89 Pt , but not less than 1.0
Design shear strength حc = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 12.01 - 1 )
N / mm2
Pt = 100 x A
st / b x d =
(1) Main bars : maximum spacing permitted for short span steel = 3 x effective depth of slab or 300 mm whichever is small= 3 x 160 = 480
spacing provided = 170 mm < 300 mm ……………….( O.K.)(2) Distribu. bars : maximum spacing permitted for long span steel = 5 x effective depth of slab or 450 mm whichever is small
= 5 x 150 = 750spacing provided = 230 mm < 450 mm ……………….( O.K.)
Table - 26 Bending moment coefficients for rectangular panels supported on four sides with provision for torsion at corners
Case No. Type of Panel and Moments considered
1.0 1.1 1.2 1.3
10.032 0.037 0.043 0.047
Positive moment at mid-span 0.024 0.028 0.032 0.036
2 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
3 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 0.044
4 Two Adjacent Edges Discontinuous:
Negative moment at continuous edge 0.047 0.053 0.060 0.065Positive moment at mid-span 0.035 0.040 0.045 0.049
5 Two Short Edges Discontinuous:Negative moment at continuous edge 0.045 0.049 0.052 0.056Positive moment at mid-span 0.035 0.037 0.040 0.043
6 Two Long Edges Discontinuous:
Negative moment at continuous edge - - - -Positive moment at mid-span 0.035 0.043 0.051 0.057
7Three 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
8 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
9 Four-Edges Discontinuous:Positive moment at mid-span 0.056 0.064 0.072 0.079
Table - 27 Bending moment coefficients for slabs spanning in two directions at right angles , simply supported on Four sides
short span coefficient αx
( Values of l
Interior Panels:Negative moment at continuous edge
1.0 1.1 1.2 1.3 1.4
0.062 0.074 0.084 0.093 0.099
0.062 0.061 0.059 0.055 0.051
ly / l
x
αx
αy
( 1 - 0.0402 )
KNm
( 1 - 0.0799 )
KNm
350 6550 350
350
0.87 x fy x A
st x d ( 1 - f
y x A
st / b x d x f
ck )
0.87 x 415 x 218 x 150 ( 1 - 415 x 218 / 1000 x 150 x 15 ) x 10 -6
0.87 x fy x A
st x d ( 1 - f
y x A
st / b x d x f
ck )
0.87 x 415 x 462 x 160 ( 1 - 415 x 462 / 1000 x 160 x 15 ) x 10 -6
A
4300 4650
350
6900
PLAN460 460
180
350 4300 350
Section A-A
460
460
690 690
A
8 O @ 230 c/c
8 O @ 460 c/c
10 O @ 170 c/c
10 O @ 340 c/c
vv
vv
vv
vv
vv
vv
10 O @ 340 c/c
10 O @ 170 c/c
8 O @ 230 c/c
3 x effective depth of slab or 300 mm whichever is smallmm or 300 mm i.e. 300 mm
……………….( O.K.)5 x effective depth of slab or 450 mm whichever is small
mm or 450 mm i.e. 450 mm……………….( O.K.)
Table - 26 Bending moment coefficients for rectangular panels supported on four sides with provision for torsion at corners
1.4 1.5 1.75 2.0
0.051 0.053 0.060 0.065 0.0320.039 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 0.0470.053 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
- - - - 0.0450.063 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
- - - - 0.0570.071 0.076 0.087 0.096 0.043
0.085 0.089 0.100 0.107 0.056
Table - 27 Bending moment coefficients for slabs spanning in two directions at right angles , simply supported on Four sides
short span coefficient αx Long span
coefficient αy for all
values of ly / lx
( Values of ly / l
x )
1.5 1.75 2.0 2.5 3.0
0.104 0.113 0.118 0.122 0.124
0.046 0.037 0.029 0.020 0.014
Design of Continuous BeamAn R.C.C. floor is used as a banking hall
The Slab thickness is 120 mm .
(Given )Rib size = 230 mm x 450 mmColumn = 300 mm x 300 mmMain beams = 300 mm x 570 mm overall .Material M15 grade concrete
HYSD reinforcement of grade Fe415 .Solution : -( a ) Load calculations and analysis :
Load on beam = 3 ( 4 + 3 ) = 12 + 9 KN / m Self wt. = 0.23 x 0.45 x 25 = 2.58 + 0 KN / m
Total 14.58 + 9 KN / mFactored Load = 1.5 ( 14.58 + 9 )
= 21.87 + 13.5say ( 22 + 14 ) KN /m .
Case ( a ) Maximum moment at B
Using three moment equation for span ABC
= ( 2 / 3 ) x 6 x 162 = ( 2 / 3 ) x 6 x 162= 648 = 648
Design the beams B10
-B11
-B12
.
Live load = 3 KN / m2
Floor finish = 1 KN / m2
Slab 120 mm thick 0.12 x 25 = 3 + 0 KN / m2
Floor finish = 1 + 0 KN /m2
Live load = 0 + 3 KN / m2
Total 4 + 3 KN / m2
MA ( L
1 / I
1 ) + 2M
B ( L
1 / I
1 + L
2 / I
2 ) + M
C ( L
2 / I
2 ) = - 6 A
1a
1 / ( I
1 L
1 ) -6 A
2a
2 / ( I
2L
2)
A1 = ( 2 / 3 ) x Base x h
1A
2 = ( 2 / 3 ) x Base x h
1
a1 = 3 m a
2 = 3 m
MA ( 6 / I ) + 2M
B ( 6 / I + 6 / I ) + M
C ( 6 / I ) = - [ 6 x 648 x 3 / ( I
x
6 ) ] - [ 6 x 648 x 3 / ( I
x 6) ]
6MA + 24M
B + 6M
C = - [ 1944 ] - [ 1944 ]
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
…………………….( 1 )Using three moment equation for span BCD
= ( 2 / 3 ) x 6 x 162 = ( 2 / 3 ) x 6 x 99= 648 = 396
…………………….( 2 )
…………………….( 1 )
…………………….( 2 )
6 m Distance 138 KNm3 m Distance ( ? ) 696 m Distance 96 KNm3 m Distance ( ? ) 486 m Distance 42 KNm
4MB + M
C = - 648 As M
A = 0
MB ( L
2 / I
2 ) + 2M
C ( L
2 / I
2 + L
3 / I
3 ) + M
D ( L
3 / I
3 ) = - 6 A
2a
2 / ( I
2 L
2 ) -6 A
3a
3 / ( I
3L
3)
A2 = ( 2 / 3 ) x Base x h
2A
3 = ( 2 / 3 ) x Base x h
3
a2 = 3 m a
3 = 3 m
MB ( 6 / I ) + 2M
C ( 6 / I + 6 / I ) + M
D ( 6 / I ) = - [ 6 x 648 x 3 / ( I
x
6 ) ] - [ 6 x 396 x 3 / ( I
x 6) ]
6MB + 24M
C + 6M
D = - [ 1944 ] - [ 1188 ]
MB + 4M
C = - 522 As M
D = 0
4MB + M
C = - 648
MB + 4M
C = - 522
By putting Value of MB from Equation ( 2 ) into Equation ( 1 )
4(- 522 - 4MC ) + M
C = - 648
- 2088 - 16 MC + M
C = - 648
15 MC = - 1440
MC = - 96 KNm
By putting Value of MC into Equation ( 1 )
4MB + ( - 96 ) = - 648
4 MB = - 552
MB = - 138 KNm
36 KN / m
162
22 KN / m
99
B C C D
36 KN / m 36 KN / m 22 KN / m
6 m 6 m 6 mA B C D138
96162162 99
( - )( + ) ( - ) ( + )
( - ) ( - )
( + ) ( + ) ( + )
93 5145
3 m Distance ( ? ) 21
X dist.
6 - X dist.
131 X =85 ( 6 - X )
216 X = 510
X = 2.36 m
Three cases are considered for getting maximum values of moments. Case ( a ) gives maximumnegative moment at B.The same moment shall be used at C also because of symmetry. Case ( b ) gives the maximum positive moment in span BC while the case ( c ) gives maximum positive moments in span AB and CD. Factored maximum moments :B or C , negative moment = 138 KNmAB ( + ) = CD ( + ) = 110 KNmBC ( + ) = 58 KNmBC ( minimum - ve ) = 5 KNmThe moment redistribution shall be now carried out. Maximum negative moment = 138 KNm.reduce it by 20 % . Then Mu = 0.8 x 138 = 110.4 KNm. For all the cases , redistribute ( increase ordecrease ) the moment at B or C = 110.4 KNm ( Hogging ).Maximum design moments :Support B or C = 110.4 KNm > 0.7 x 138 KNm ………………( O.K.) ( As per IS 456-2000 , Clause 37.1.1 )AB or CD ( + ) = 111.6 KNm > 0.7 x 110 KNm BC ( - ) = 11.6 KNmBC ( + ) = 51.6 KNm > 0.7 x 58 KNm.Note that after redistribution , the design positive moments also have been reduced.( b ) Design for flexure :Span AB or CD
The beam acts as a flanged beam
MB = - 138 KNm
VA x 6 -36 x 6 x 3 = -138
VA = 85 KN
VA + V
B = 36 x 6
VB = 216 - 85
VB = 131 KN
MC = - 96 KNm M
B = - 138 KNm
VD x 6 -22 x 6 x 3 = -96 50 x 12 + V
C x 6 - 36 x 6 x 3 - 22 x 6 x 9 = -138
VD = 50 KN V
c = 183 KN
VC + V
D = 22 x 6 V
C + 82 = 183
VC = 132 - 50 V
C = 101 KN
VC = 82 KN M
C = - 96 KNm
85 x 12 + VB x 6 - 36 x 6 x 9 - 36 x 6 x 3 = - 96
VB = 246 KN
VB + 131 = 246
VB = 115 KN
Mu ( + ) = 111.6 KNm.
96138
85
131
2.36 m50
82
115
101
( As per IS456-2000 ,Clause 23.1.2 , Note )= 1650 mm > 3000 mm
Assuming one layer of 20 mm diameter barsd = 450 + 120 -25 - 10 = 535 mm .
( As per IS456-2000 ,Clause 26.5( a ) )
For bf /bw = 7 For bf /bw = 8
0.01 diff. 0.014 0.016
( As per SP:16 ,Table 58 ) 0.006 diff. ? ?
-0.0084 -0.0096
= 662.597 KNm > 111.6 KNm For 0.224 0.6566 0.743
1 diff 0.08640.83 diff ? 0.07171
= 535 - 120 / 2 = 535 - 60
= 651
Span BC :
= 301
Continue 3 - 12 O in span BC as required for flexure.Support B or C :
( From Table 6-3 )The section is under-reinforced.
415 / 15
= 0.549
603 135
For T-beams , bf = ( l
0 / 6 ) + b
w + 6 D
f
bf = ( 0.7 x 6000 / 6 ) + 230 + 6 x 120
Minimum Ast = ( 0.205 / 100 ) x 230 x 535 = 252 mm2
bf / b
w = 1650 / 230 = 7.17
Df / d = 120 / 535 = 0.224
Mu,lim
/ fck
bw d2 = 0.671
Mu.lim
= 0.671 x 15 x 230 x 5352 x 10-6
If Mu < M
u,lim
: design as under-reinforced section (singly reinforced beam) as explained below.
Ast = M
u / 0.87 x f
y x lever arm
where lever arm = d - Df / 2
Ast = 111.6 x 106 / 0.87 x 415 x ( 535 - 60 )
mm2 .
Provide 6 - 12 mm O = 678 mm2 .
Mu ( + ) = 51.6 KNm
Ast = 51.6 x 106 / 0.87 x 415 x ( 535 - 60 )
mm2 .
Provide 3 - 12 mm O = 339 mm2 .
In span AB , curtail 3 - 12 O at 300 mm ( 0.05 ℓ ) from A and at 900 mm ( 0.15 ℓ ) from B.
Mu ( - ) = 110.4 KNm
Mu / bd2 = 110.4 x 106 / 230 x 5352 = 1.68 < 2.07.
Pt = 50
1 - √ 1 - ( 4.6 / fck
) x ( Mu / bd2 )
fy / f
ck
Pt = 50 1 - √ 1 - ( 4.6 / 15 ) x ( 1.68 )
Ast = ( 0.549 / 100 ) x 230 x 535 = 676 mm2
20 % of steel should be carried through the span = 0.2 x 676 = 135 mm2
Provide 2- 10 mm O anchor bars = 157 mm2 . At support , provide 3 - 16 mm O extra at top = 603 mm2 , one of which may be curtailed at 0.15 ℓ = 900 mm from centre of support B and remaining 2 - 16 mm Oat 0.25 ℓ = 1500 mm from B .
Case ( b ) Maximum positive moment in span BC
Using three moment equation for span ABC
= ( 2 / 3 ) x 6 x 99 = ( 2 / 3 ) x 6 x 162= 396 = 648
MA ( L
1 / I
1 ) + 2M
B ( L
1 / I
1 + L
2 / I
2 ) + M
C ( L
2 / I
2 ) = - 6 A
A1 = ( 2 / 3 ) x Base x h
1A
2 = ( 2 / 3 ) x Base x h
a1 = 3 m a
2 = 3 m
6 ) ] - [ 6 x 648 x 3 / ( I x 6) ] M
A ( 6 / I ) + 2M
B ( 6 / I + 6 / I ) + M
C ( 6 / I ) = - [ 6 x 396 x 3 / ( I
6MA + 24M
B + 6M
C = - [ 1188 ] - [ 1944 ]
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 KN / m
162
A B B
Using three moment equation for span BCD
= ( 2 / 3 ) x 6 x 162 = ( 2 / 3 ) x 6 x 99= 648 = 396
…………………….( 1 )
…………………….( 2 )
( 162 - 69 = 93 KNm )
( 99 - 48 = 51 KNm )
4MB + M
C = - 522 As M
A = 0
MB ( L
2 / I
2 ) + 2M
C ( L
2 / I
2 + L
3 / I
3 ) + M
D ( L
3 / I
3 ) = - 6 A
A2 = ( 2 / 3 ) x Base x h
2A
3 = ( 2 / 3 ) x Base x h
a2 = 3 m a
3 = 3 m
6 ) ] - [ 6 x 396 x 3 / ( I x 6) ] M
B ( 6 / I ) + 2M
C ( 6 / I + 6 / I ) + M
D ( 6 / I ) = - [ 6 x 648 x 3 / ( I
6MB + 24M
C + 6M
D = - [ 1944 ] - [ 1188 ]
MB + 4M
C = - 522 As M
D = 0
4MB + M
C = - 522
MB + 4M
C = - 522
By putting Value of MB from Equation ( 2 ) into Equation ( 1 )
4(- 522 - 4MC ) + M
C = - 522
- 2088 - 16 MC + M
C = - 522
15 MC = - 1566
MC = - 104 KNm
By putting Value of MC into Equation ( 1 )
4MB + ( - 104 ) = - 522
4 MB = - 418
MB = - 104 KNm
36 KN / m
162
22 KN / m
99
B C C
22 KN / m 36 KN / m 22 KN / m
6 m 6 m 6 mA B C D104 104
99162
99( - )
( - ) ( + )
( - ) ( - )
( + ) ( + ) ( + )
47 4758
( 162 - ( 96 + 21 ) = 45 KNm )
85
131
131 X =85 ( 6 - X )
216 X = 510
X = 2.36 m
Three cases are considered for getting maximum values of moments. Case ( a ) gives maximumnegative moment at B.The same moment shall be used at C also because of symmetry. Case ( b ) gives the maximum positive moment in span BC while the case ( c ) gives maximum positive
The moment redistribution shall be now carried out. Maximum negative moment = 138 KNm.reduce it by 20 % . Then Mu = 0.8 x 138 = 110.4 KNm. For all the cases , redistribute ( increase or
( As per IS 456-2000 , Clause 37.1.1 ) ( b ) The ultimate moment of resistance provided at any section of a member is not less than 70 percent of the moment at that sectionobtained from an elastic maximum moment diagram covering all appropriate combinations of loads.
MC = - 104 KNm M
B = - 104 KNm
VD x 6 -22 x 6 x 3 = -104 48.67 x 12 + V
C x 6 - 36 x 6 x 3 - 22 x 6 x 9 = -104
VD = 48.67 KN V
c = 191.33 KN
VC + V
D = 22 x 6 V
C + 83.33 = 191.33
VC = 132 - 48.67 V
C = 108 KN
VC = 83.33 KN M
C = - 104 KNm
48.67 x 12 + VB x 6 - 36 x 6 x 3 - 22 x 6 x 9 = - 104
VB = 191.33 KN
VB + 83.33 = 191.33
VB = 108 KN
IS 456-2000 ,Clause 37.1.1 Redistribution of moments in Continuous Beams and Frames -
( c ) The elastic moment at any section in a member due to a particular combination of loads shall not be reduced by more than 30 % of the numerically largest moment given anywhere by the elastic maximum moments diagram for the particular member , covering all appropriate combination of loads.
104
48.67
83.33
2.21 m48.67
83.33
108
108
104
IS 456-2000 ,Clause 23.1.2 Effective width of flange - ( As per IS456-2000 ,Clause 23.1.2 , Note )
( As per IS456-2000 ,Clause 26.5( a ) )
For bf /bw = 8
Where ,
0.671 b = actual width of the flange.
IS 456-2000 Clause 26.5 Requirements of Reinforcement for
Structural Members
26.5.1 Beams26.5.1.1 Tension Reinforcement
shall not be less than that given by the following :
where ,
b = breadth of the beam or the breadth of the web of T- beam ,d = effective depth , and
Minimum steel %
For mild steel
For HYSD steel , Fe415 grade
For HYSD steel , Fe500 grade
In the absence of more accurate determination , the effective width of flange may be taken as the following but in no case greater than the breadth of the web plus half the sum of the clear distances to the adjacent beams on either side :
( a ) For T-beams , bf = ( l
0 / 6 ) + b
w + 6 D
f
( b ) For L-beams , bf = ( l
0 / 12 ) + b
w + 3 D
bf = effective width of flange ,
l0 = distance between points of zero moments in the beam ,
bw = breadth of the web ,
Df = thickness of flange , and
: design as under-reinforced section (singly reinforced beam) as explained below. NOTE - For continuous beams and frames , ' l0 ' may be assumed
as 0.7 times the effective span.
a ) Minimum reinforcement - The minimum area of tension reinforcement
As / b d = 0.85 / f
y
As = minimum area of tension reinforcement ,
fy = characteristic strength of reinforcement in N / mm2 .
b ) Maximum reinforcement - The maximum area of tension reinforcement shall not exceed 0.04 b D.
100 As / b d = 100 x 0.85 / 250 = 0.34
100 As / b d = 100 x 0.85 / 415 = 0.205
100 As / b d = 100 x 0.85 / 500 = 0.17
Table 6-3
For singly reinforced rectangular sections
250 415 50015 2.22 2.07 2.0020 2.96 2.76 2.6625 3.70 3.45 3.3330 4.44 4.14 3.99
. At support , provide 3 - 16 mm O extra at top = 603 mm2 , = 900 mm from centre of support B and remaining 2 - 16 mm O
Limiting Moment of resistance factor Q lim
, N / mm
fck
N / mm2
fy, N / mm2
Case ( b ) Maximum positive moment in span BC Case ( c ) Maximum positive moment in span AB and CD
Using three moment equation for span ABC
( 2 / 3 ) x 6 x 162 = ( 2 / 3 ) x 6 x 162= 648
) + MC ( L
2 / I
2 ) = - 6 A
1a
1 / ( I
1 L
1 ) -6 A
2a
2 / ( I
2L
2) M
A ( L
1 / I
1 ) + 2M
B ( L
= ( 2 / 3 ) x Base x h1
A1 = ( 2 / 3 ) x Base x h
a1 = 3 m
( 6 / I ) = - [ 6 x 396 x 3 / ( I x
6 ) ] - [ 6 x 648 x 3 / ( I
x 6) ] M
A ( 6 / I ) + 2M
B ( 6 / I + 6 / I ) + M
6MA + 24M
B + 6M
C = - [ 1944 ] - [ 1188 ]
36 KN / m
C
36 KN / m
6 mA B
36 KN / m
162
A
…………………….( 1 )Using three moment equation for span BCD
( 2 / 3 ) x 6 x 99 = ( 2 / 3 ) x 6 x 99= 396
…………………….( 2 )
…………………….( 1 )
…………………….( 2 )
6 m Distance 104 KNm3 m Distance ( ? ) 52 ( 99 - 52 = 47 KNm ) Moment in span BC = ( 162 - 104 ) = 58 KNm
4MB + M
C = - 522
) + MD ( L
3 / I
3 ) = - 6 A
2a
2 / ( I
2 L
2 ) -6 A
3a
3 / ( I
3L
3) M
B ( L
2 / I
2 ) + 2M
C ( L
= ( 2 / 3 ) x Base x h3
A2 = ( 2 / 3 ) x Base x h
a2 = 3 m
( 6 / I ) = - [ 6 x 648 x 3 / ( I x
6 ) ] - [ 6 x 396 x 3 / ( I
x 6) ] M
B ( 6 / I ) + 2M
C ( 6 / I + 6 / I ) + M
= - [ 1944 ] - [ 1188 ] 6MB + 24M
C + 6M
D = - [ 1188 ] - [ 1944 ]
MB + 4M
C = - 522
4MB + M
C = - 522
MB + 4M
C = - 522
from Equation ( 2 ) into Equation ( 1 ) By putting Value of M
4(- 522 - 4MC ) + M
- 2088 - 16 MC + M
15 MC = - 1566
MC = - 104.4 KNm
By putting Value of M
4MB + ( - 104.4 ) = - 522
4 MB = - 417.6
MB = - 104.4 KNm
22 KN / m
D
22 KN / m
99
B
36 KN / m
6 mA B104162 ( - )
( + )
( - )
( + )
110
X dist. 48.67
6 - X dist. 83.33
83.33 X = 48.67 ( 6 - X )
132 X = 292.02
X = 2.21 m
( b ) The ultimate moment of resistance provided at any section of a member is not less than 70 percent of the moment at that sectionobtained from an elastic maximum moment diagram covering all
MB = - 104 KNm
VA x 6 -22 x 6 x 3 = -104
VA = 48.67 KN
VA + V
B = 22 x 6
VB = 132 - 48.67
VB = 83.33 KN
MC = - 104 KNm
48.67 x 12 + VC x 6 - 36 x 6 x 3 - 22 x 6 x 9 = -104 V
D x 6 -36 x 6 x 3 = -104
VD = 90.67 KN
+ 83.33 = 191.33 VC + V
D = 36 x 6
VC = 216 - 90.67
VC = 125.33 KN
48.67 x 12 + VB x 6 - 36 x 6 x 3 - 22 x 6 x 9 = - 104
+ 83.33 = 191.33
IS 456-2000 ,Clause 37.1.1 Redistribution of moments in Continuous Beams and Frames -
( c ) The elastic moment at any section in a member due to a particular combination of loads shall not be reduced by more than 30 % of the numerically largest moment given anywhere by the elastic maximum moments diagram for the particular member , covering all appropriate combination of loads.
104
90.67
125.33
2.52 m
66
IS 456-2000 ,Clause 23.1.2 Effective width of flange -
IS 456-2000 Clause 26.5 Requirements of Reinforcement for
Structural Members
shall not be less than that given by the following :
b = breadth of the beam or the breadth of the web of T- beam ,
In the absence of more accurate determination , the effective width of flange may be taken as the following but in no case greater than the breadth of the web plus half the sum of the clear distances to the adjacent beams on either side :
/ 6 ) + bw + 6 D
f
/ 12 ) + bw + 3 D
f
= distance between points of zero moments in the beam ,
NOTE - For continuous beams and frames , ' l0 ' may be assumed
as 0.7 times the effective span.
a ) Minimum reinforcement - The minimum area of tension reinforcement
= minimum area of tension reinforcement ,
= characteristic strength of reinforcement in N / mm2 .
b ) Maximum reinforcement - The maximum area of tension reinforcement shall not exceed 0.04 b D.
/ b d = 100 x 0.85 / 250 = 0.34
/ b d = 100 x 0.85 / 415 = 0.205
/ b d = 100 x 0.85 / 500 = 0.17
Table 6-3
For singly reinforced rectangular sections
5501.942.583.233.87
Limiting Moment of resistance factor Q lim
, N / mm2
, N / mm2
Case ( c ) Maximum positive moment in span AB and CD
Using three moment equation for span ABC
( 2 / 3 ) x 6 x 162 = ( 2 / 3 ) x 6 x 99= 396
/ I1 ) + 2M
B ( L
1 / I
1 + L
2 / I
2 ) + M
C ( L
2 / I
2 ) = - 6 A
1a
1 / ( I
1 L
1 ) -6 A
2a
2 / ( I
2L
2)
= ( 2 / 3 ) x Base x h1
A2 = ( 2 / 3 ) x Base x h
1
a2 = 3 m
( 6 / I ) + 2MB ( 6 / I + 6 / I ) + M
C ( 6 / I ) = - [ 6 x 648 x 3 / ( I
x
6 ) ] - [ 6 x 396 x 3 / ( I
x 6) ]
+ 24MB + 6M
C = - [ 1944 ] - [ 1188 ]
22 KN / m 36 KN / m
6 m 6 mC D
36 KN / m 22 KN / m
99
B B C
…………………….( 1 )Using three moment equation for span BCD
( 2 / 3 ) x 6 x 99 = ( 2 / 3 ) x 6 x 162= 648
…………………….( 2 )
…………………….( 1 )
…………………….( 2 )
6 m Distance 104 KNm3 m Distance ( ? ) 52 ( 162 - 52 = 110 KNm )
Moment in span BC = ( 104 - 99 ) = 5 KNm
C = - 522 As M
A = 0
/ I2 ) + 2M
C ( L
2 / I
2 + L
3 / I
3 ) + M
D ( L
3 / I
3 ) = - 6 A
2a
2 / ( I
2 L
2 ) -6 A
3a
3 / ( I
3L
3)
= ( 2 / 3 ) x Base x h2
A3 = ( 2 / 3 ) x Base x h
3
a3 = 3 m
( 6 / I ) + 2MC ( 6 / I + 6 / I ) + M
D ( 6 / I ) = - [ 6 x 396 x 3 / ( I
x
6 ) ] - [ 6 x 648 x 3 / ( I
x 6) ]
+ 24MC + 6M
D = - [ 1188 ] - [ 1944 ]
C = - 522 As M
D = 0
C = - 522
C = - 522
By putting Value of MB from Equation ( 2 ) into Equation ( 1 )
(- 522 - 4MC ) + M
C = - 522
- 2088 - 16 MC + M
C = - 522
= - 104.4 KNm
By putting Value of MC into Equation ( 1 )
+ ( - 104.4 ) = - 522
= - 104.4 KNm
22 KN / m 36 KN / m
162
C C D
22 KN / m 36 KN / m
6 m 6 mC D104 104
99
162
( - ) ( - )( + )
( - )
( + )
110
5.0
X dist. 90.67
6 - X dist. 125.33
125.33 X =90.67 ( 6 - X )
216 X = 544.02
X = 2.52 m
MB = - 104 KNm
VA x 6 -36 x 6 x 3 = -104
VA = 90.67 KN
VA + V
B = 36 x 6
VB = 216 - 90.67
VB = 125.33 KN
MB = - 104 KNm
x 6 -36 x 6 x 3 = -104 90.67 x 12 + VC x 6 - 22 x 6 x 3 - 36 x 6 x 9 = -104
Vc = 191.33 KN
VC + 125.33 = 191.33
VC = 66 KN
MC = - 104 KNm
90.67 x 12 + VB x 6 - 36 x 6 x 9 - 22 x 6 x 3 = - 104
VB = 191.33 KN
VB + 125.33 = 191.33
VB = 66 KN
104
125.33
90.67
125.33
66
( 162 - 52 = 110 KNm )
Design of Cantilever BeamSpan = 3 mCharacteristic U.D.L. = 12 KN /mAssume that sufficient safety against overturning is there
( Given )and reinforcement anchorages are also available.Material M15 grade concrete
mild steel reinforcement.Solution : -Assume an initial trial section of 230 mm x 600 mm overall depth to consider self-weight.The self-weight = 0.23 x 0.6 x 25 = 3.45 KN / m
Total load = 12 + 3.45 = 15.45 KN / mFactored load = 1.5 x 15.45 = 23.2 KN / mFactored S.F. =
Factored B.M. =
Depth required = √M / Q b ( From Table 6-3 )= 452 mm
Using one layer of 20 mm O bars and overall depth of 550 mmd = 550 - 25 - 10 = 515 mm OR
= 1.71 < 2.22 ( Table 6-3 ) = 135.424 KNm
The section is singly reinforced ( under-reinforced )
250 / 15= 0.93
Check for development length :
( From Table 7-6 )IS 456-2000 clause 26.2.1 ( From Table 7-5 )
== 47.71 O= 47.71 x 20= 954 mm
The bar shall extend into the support for a straight length of 954 mm . Provide anchorage of 1200 mm . If in some case the bars are to be bent e.g. anchored in column , the bearing stress around the bend has to be checked as discussed
in the next example . Check for shear :
w x ℓ = 23.2 x 3 = 69.6 KN
w x ℓ2 / 2 = 23.2 x 32 / 2 = 104.4 KNm
= √( 104.4 x 106 ) / ( 2.22 x 230 )
Mu / b d2 == 104.4 x 106 / ( 230 x 5152) M
u,lim = 2.22 x 230 x 5152 x 10-6
Mu < M
u,lim
Pt = 50
1 - √ 1 - ( 4.6 / fck
) x ( Mu / bd2 )
fy / f
ck
Pt = 50 1 - √ 1 - ( 4.6 / 15 ) x ( 1.71 )
Ast = ( 0.93 / 100 ) x 230 x 515 = 1102 mm2
Provide 4 - 20 mm O giving Ast = 4 x 314 = 1256 mm2 .
Ld = 1102 / 1256 = 55 O
Development length of bars Ld = O σ
s / 4 x ح
bd
O ( 0.87 x 250 x 1102 ) / ( 1256 x 4 x 1.0 )
104.4 KNm
=
= 0.588
( From IS 456-2000 , table 19 table 7-1 )= 1.06 0.25 difference
0.61 0.19 differenceProvide minimum shear reinforcement . For 230 mm wide beam , From IS 456-2000 clause 26.5.1.6Spacing of minimum shear reinforcement using 6 O stirrups
== 0.87 x 56 x 250 / 0.4 x 230= 132.4 mm
spacing should not exceed( i ) 450 mm
( ii ) 0.75 d = 0.75 x 515 = 386 mm( iii ) ≤132.4 mm ( minimum )
( iv ) 569.2 mm ( designed )Provide 6 mm O two-legged stirrups @ 130 mm c/cCheck for deflection :Basic span / d ratio = 7
modification factor = 1.4 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementSpan / d permissible = 7 x 1.4 = 9.8Actual span / d = 3000 / 515 = 5.83 < 9.8 ………………………( safe )Check for cracking (spacing of bars ) :Clear distance between bars
= ( 230 - 50 - 4 x 20 ) / 3 = 10 mmMaximum clear distance permitted
= 300 mm ( cracking - table 8-1 , IS 456-2000 , table 15 ) ………………………( safe )The design beam is shown in fig.
Vu = 69.6 KN
Actual shear strength حv = V
u / bd
69.6 x 103 / ( 230 x 515 )
N / mm2
100 x As / b d = ( 100 x 1256 ) / ( 230 x 515 )
design ( permissible ) shear strength حc = N / mm2 ح <
v
0.87 Asv
fy / 0.4 b
100 Ast / b d = 100 x 1256 / 230 x 515 = 1.06
6 O @ 130 c/c
Fixed end support
1200 3000
2-12 O
4 -20 O
v v v v
vv
230
550
Stirrups 6 O @ 130 c/c
Design of Cantilever BeamA cantilever rectangular bracket projects from a column of size 230 mm x 500 mmin the direction of 500 mm for a length of 3 mFactored load of 20 KN / m inclusive of self - weightMaterial M15 grade concrete
HYSD reinforcement of grade Fe415 .Solution : -
( a ) Moment steel :Take size of Beam 230 mm x 550 mm overall.Assuming 16 mm diameter bars in one layerd = 550 - 25 ( Cover ) -8 OR
= 517 mm
= 127.256 KNm
= 1.46 < 2.07 ( Table 6-3 )The section is singly reinforced ( under-reinforced ) Beam
415 / 15= 0.464
Use 2- 10 mm O as bottom anchor bars .( b ) Check for development length :IS 456-2000 clause 26.2.1
= 0.87 x 415 x 552 / 603 = 330.5
IS 456-2000 clause 26.2.1 ( From Table 7-5 )
== 51.64 O= 51.64 x 16= 826 mm
The arrangement of anchoring the bars is shown in fig. and is equivalent to an anchorageof 1127 mm. The bearing stress inside the bend is now checked .O = 16 mma = 82 mm for internal bar ( centre to centre distance between bar ) ( = ( 230 - 50 - 3 x 16 ) / 2 + 16 = 82 mm )
= 25 + 16 = 41 mm for external bar . ( = cover + dia of bar )
Vu = w x ℓ = 20 x 3 = 60 KN
Mu = w x ℓ2 / 2 = 20 x 32 / 2 = 90 KNm
Mu,lim
= 2.07 x 230 x 5172 x 10-6
Mu / b d2 =90 x 106 / ( 230 x 5172)
Mu < M
u,lim
Pt = 50
1 - √ 1 - ( 4.6 / fck
) x ( Mu / bd2 )
fy / f
ck
Pt = 50 1 - √ 1 - ( 4.6 / 15 ) x ( 1.46 )
Ast = ( 0.464 / 100 ) x 230 x 517 = 552 mm2
Provide 16 mm O - 3 No. = 603 mm2.
Development length of bars Ld = O σ
s / 4 x ح
bd
Stress in bar σs = 0.87 x f
y N / mm2
ح ) bd
= 1.6 x 1 = 1.6 N / mm2
Development length of bars Ld = O σ
s / 4 x ح
bd
O 330.5 / ( 4 x 1.6 )
33
The bearing stress is critical in external bar . Check for this stress for a = 41 mm.
= 1.5 x 15 / [ 1 + ( 2 x 16 / 41 ) ]= 22.5 / 1.78
= 12.6404At the centre of bend , the anchorage available = 279 + 147 = 426 mm
Stress in bar at centre of the bend = 330.5 x ( 826 - 426 ) / 826
= 160
=
= ……………………..( safe )The arrangement is thus satisfactory. ( c )Check for shea
=
= 0.505
( From IS 456-2000 , table 19 table 7-1 )= 0.507 0.25 difference
0.46 0.243 differenceshear design is necessary .
At support , OR
= = 5.35= 60 - 54.6986= 5.3 KN
== 1188 mm
Provide minimum shear reinforcement . For 230 mm wide beam , From IS 456-2000 clause 26.5.1.6Spacing of minimum shear reinforcement using 6 O stirrups
== 0.87 x 56 x 250 / 0.4 x 230= 132.4 mm
spacing should not exceed( i ) 450 mm ( ii ) 0.75 d = 0.75 x 517 = 388 mm
Design bearing stress = 1.5 x fck
/ [1 + ( 2O/a ) ]
N / mm2
N / mm2
Fbt = 160 x 201 x 10-3 = 32.16 KN ( area of 16 O bar = 201 mm2 )
Bearing stress = Fbt / r O
32.16 x 103 / 180 x 16
11.16 N / mm2 < 12.64 N /mm2
Vu = 60 KN
Actual shear strength حv = V
u / bd
60 x 103 / ( 230 x 517 )
N / mm2
100 x As / b d = ( 100 x 603 ) / ( 230 x 517 )
design ( permissible ) shear strength حc = N / mm2 ح >v
Vus
= Vuح -
c b d V
us = ( 0.505 - 0.46 ) x 230 x 517 x10-3
60 - 0.46 x 230 x 517 x 10-3
Using 6 mm O ( mild steel ) Two legged stirrups , Asv
= 28 x 2 = 56 mm2 .
fy = 250 N / mm2
Sv = 0.87 f
y A
sv d / V
us
0.87 x 250 x 56 x 517 / 5.3 x 103
0.87 Asv
fy / 0.4 b
550
33
( iii ) ≤132.4 mm ( minimum )( iv ) 1188 mm ( designed )Provide 6 mm O two-legged stirrups @ 130 mm c/c( d ) Check for deflection :Basic span / d ratio = 7
modification factor = 1.2 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementSpan / d permissible = 7 x 1.2 = 8.4Actual span / d = 3000 / 517 = 5.8 < 8.4 ………………………( safe )( e ) Check for cracking (spacing of bars ) :Clear distance between bars
= ( 230 - 50 - 3 x 16 ) / 2 = 66 mmMinimum clear distance permitted
= 20 + 5 = 25 mm or 16 mm ( O of bar ) i.e. 25 mm .Maximum clear distance permitted
= 180 mm ( cracking - table 8-1 , IS 456-2000 , table 15 ) ………………………( safe )The design beam is shown in fig.
100 Ast / b d = 100 x 603 / 230 x 517 = 0.507
hagg
+ 5 mm =
279295
295
108
150
internal radiusr = 180
r
r
500 3000 mm
2-10 O
3-16 O50
550
Stirrups 6 mm O @ 130 mm c/c
Assume an initial trial section of 230 mm x 600 mm overall depth to consider self-weight.
IS 456-2000 , Table 19Table 7-1
Concrete grade
2.22 x 230 x 5152 x 10-6
Design shear strength of concrete , حC, N / mm2P
t = 100
x As
3 m
23.2 KN /m
69.6 KN
-104.4 KNm
M15 M20 M25 M30 M35
0.28 0.28 0.29 0.29 0.29
0.25 0.35 0.36 0.36 0.37 0.37
0.50 0.46 0.48 0.49 0.50 0.50
( From IS 456-2000 , table 19 table 7-1 ) 0.75 0.54 0.56 0.57 0.59 0.590.04 1.00 0.60 0.62 0.64 0.66 0.67
? -0.0304 1.25 0.64 0.67 0.70 0.71 0.731.50 0.68 0.72 0.74 0.76 0.781.75 0.71 0.75 0.78 0.80 0.822.00 0.71 0.79 0.82 0.84 0.86
2.25 0.71 0.81 0.85 0.88 0.902.50 0.71 0.82 0.88 0.91 0.932.75 0.71 0.82 0.90 0.94 0.963.00 0.71 0.82 0.92 0.96 0.99
The above given table is based on the following formula
6 x β
β =
IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement
………………………( safe )
x As
b x d
≤ 0.15
Design shear strength حc = 0.85 √ 0.8 x f
ck ( √ 1 + 5 x β - 1 )
0.8 x fck
/ 6.89 Pt , but not less than 1.0
v
v
4-20 O
2-12 O
( given )
The arrangement of anchoring the bars is shown in fig. and is equivalent to an anchorage
( = ( 230 - 50 - 3 x 16 ) / 2 + 16 = 82 mm )
2.07 x 230 x 5172 x 10-6
33 3382 82
IS 456-2000 Clause 26.2.2.5 Bearing stresses at bends
Where ,
r = internal radius of the bend , and
O = size of the bar or , in bundle , the size of bar of equivalent area
Where,
( From IS 456-2000 , table 19 table 7-1 )0.08
? -0.0778
KN
The bearing stress in concrete for bends and hooks described in IS : 2502-1963 need not be checked . The bearing stress inside a bend in any other bend shall be calculated as given below :
Bearing stress = Fbt / r O
Fbt = tensile force due to design loads in a bar or group of bars,
For Limit state method of design , this stress shall not exceed 1.5 f
fck
is the characteristic strength of concrete and
a is the centre to centre distance between bars or group of bars , for a bar or group of bars adjacent to the face of the member a shall be taken as the cover plus size of bar ( O )
= ( 0.505 - 0.46 ) x 230 x 517 x10-3
230
33 3382 82
16 25
IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement
………………………( safe )
230
2-10 O
3-16 O
IS 456-2000 , Table 19Table 7-1
Concrete grade
حC, N / mm2
M40
0.30
0.38
0.51
0.600.68
0.740.790.840.88
0.920.950.981.01
1 + 5 x β - 1 )
, but not less than 1.0
IS 456-2000 Clause 26.2.2.5 Bearing stresses at bends
O = size of the bar or , in bundle , the size of bar of equivalent area
The bearing stress in concrete for bends and hooks described in IS : 2502-1963 need not be checked . The bearing stress inside a bend in any other bend shall be
= tensile force due to design loads in a bar or group of bars,
For Limit state method of design , this stress shall not exceed 1.5 fck
/ [ ( 1+ 2O/a ) ]
a is the centre to centre distance between bars or group of bars , for a bar or group of bars adjacent to the face of the member a shall be taken as the cover plus size of
Design of Simply supported Doubly reinforced BeamSpan = 5 m ( simply supported rectangular beam )super-imposed load = 40 KN / m Beam section = 230 mm x 600 mm ( Given )Material M15 grade concrete
HYSD reinforcement of grade Fe415Solution : -
DL of beam 0.23 x 0.60 x 25 = 3.45 KN / m super-imposed load = 40 KN / m
Total 43.45 KN / mFactored load = 1.5 x 43.45 = 65.18 KN / m
== 203.688 KNm
w x ℓ / 2= 65.18 x 5 / 2= 162.95 KN
( a ) Moment steel :Assuming 20 mm diameter bars in two layerd = 600 - 25 ( Cover ) - 20 - 10 OR
= 545 mm
= 141.414 KNm
= 2.98 > 2.07 ( Table 6-3 )The section is Doubly reinforced ( over-reinforced )
= 141.414 KNm
= 203.69 - 141.41 = 62.28 KNmLet the compression reinforcement be provided at an effective cover of 40 mm .
d' / d = 40 / 545= 0.07 consider d' / d = 0.1 .
=
= 349Corresponding tension steel
= 349 x 353 / ( 0.87 x 415 )
= 341.2
=
=
Mu = w x ℓ2 / 8
65.18 x 52 / 8
Vu =
Mu,lim
= 2.07 x 230 x 5452 x 10-6
Mu / b d2 = 203.69 x 106 / ( 230 x 5452)
Mu > M
u,lim
Mu,lim
= 2.07 x 230 x 5452 x 10-6
Mu2
= Mu - M
u,lim
Stress in compression steel , fsc
= 353 N / mm2 ( refer to table 6-6 )
Asc
= Mu2
/ ( fsc
x ( d - d' ) )
62.28 x 106 / 353 ( 545 - 40 )
mm2
Ast2
= Asc
fsc
/ 0.87 fy
mm2
Ast,lim
= Mu,lim
/ ( 0.87 fy ( d - 0.42 X
u,max ) )
141.41 x 106 / ( 0.87 x 415 ( 545 - 0.42 x 0.48 x 545 ) )
141.46 x 106 / 361.05 ( 435.13 )
= 900
=
= 1241.2Provide
2-16 O = 402
4-20 O = 1256 ( all straight )( b ) Check for development length :
1625.02 ≥ 44 OO ≤ 36.93 mm
( c ) Check for shear
162.95 KN
=
= 1.3
( From IS 456-2000 , table 19 table 7-1 )= 1.0
0.6As the ends are confined with compressive reaction , shear at distance d will be used forchecking shear at support . At 545 mm , shear is equal to
162.95 - 0.545 x 43.45 = 139.27 KN
=
== 307 mm
From IS 456-2000 clause 26.5.1.6Spacing of minimum shear reinforcement using 8 mm O stirrups
== 0.87 x 100 x 415 / 0.4 x 230= 392.4 mm
spacing should not exceed( i ) 450 mm ( ii ) 0.75 d = 0.75 x 545 = 408 mm
mm2
Ast = A
st,lim + A
st2 .
900 + 341.2 mm2
mm2
Asc
= mm2
Ast = mm2
As all the bars are taken into support , Mu1
may be taken as Mu .
Assume L0 = 12 O L
d = 56 O ( From Table 7-6 )
1.3 x M1 / V + L
0 ≥ L
d
1.3 x 203.69 x 106 / 162.95 x 103 +12 O ≥ 56 O
Oprovided
= 20 mm ……………………..( safe )
Vu =
Actual shear strength حv = V
u / bd
162.95 x 103 / ( 230 x 545 )
N / mm2
100 x As / b d = ( 100 x 1256 ) / ( 230 x 545 )
design ( permissible ) shear strength حc = N / mm2 ح >v
Vu =
Vus
= Vuح -
c b d
139.27 - 0.6 x 230 x 545 x 10-3 = 64.06 KN
Assuming 8 mm O two - legged stirrups , Asv
= 100 mm2 , fy = 415 N / mm2 .
Sv = 0.87 f
y A
sv d / V
us
0.87 x 415 x 100 x 545 / 64.06 x 103
0.87 Asv
fy / 0.4 b
( iii ) ≤392.4 mm ( minimum )( iv ) 307 mm ( designed )Minimum shear reinforcement of 8 mm O @ 390 mm c/c will be used .At support provide 8 mm O @ 300 mm c/c .Shear resistance of beam with minimum shear reinforcement
=
== 50.45 + 75.21= 125.66 KN
Designed shear reinforcement is required from face of the support upto the distance equal to
162.95 - 125.66 / 43.45 = 0.858 mNo. of stirrups = ( 858 / 300 ) + 1 = 3.86 say 4 Provide 8 mm O two-legged stirrups @ 300 mm c/c upto 4 no. from face of the supportand 8 mm O @ 390 mm c/c in remaining central portion .( d ) Check for deflection :Basic span / d ratio = 20
modification factor = 0.97 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement
modification factor = 1.08 IS 456-2000 clause 23.2.1 fig-5 ,for compression reinforcementSpan / d permissible = 20 x 0.97 x 1.08 = 20.95Actual span / d = 5000 / 545 = 9.17 ………………………( safe )( d ) Check for cracking (spacing of bars ) :Clear distance between bars
= ( 230 - 50 - 4 x 20 ) / 3 = 10 mmMinimum clear distance permitted
= 20 + 5 = 25 mm or 20 mm ( O of bar ) i.e. 25 mm .Maximum clear distance permitted
= 180 mm ( cracking - table 8-1 , IS 456-2000 , table 15 ) ………………………( safe )The design beam is shown in fig.
( 0.87 fy A
sv d / S
vح +(
c b d
(0.87 x 415 x 100 x 545 x 10-3 / 390 ) + 0.6 x 230 x 545 x 10-3
100 Ast / b d = 100 x 1256 / 230 x 545 = 1.0
100 Asc
/ b d = 100 x 402 / 230 x 545 = 0.32
hagg
+ 5 mm =
230
600545
2-16 O
4- 20 O
600
5000 c/c
2-16 O
4- 20 O ( straight ) 600
DIANO.
SPA.
8 O 8 O 8 O4 4rest
300 300390
bearing
Table 6-6
d ' / d0.05 0.1 0.15 0.2
250 217 217 217 217415 355 353 342 329500 424 412 395 370
550 458 441 419 380
where ,
STRESS IN COMPRESSION REINFORCEMENT fsc , N / mm2 IN DOUBLY REINFORCED BEAMS
fy N / mm2
2.07 x 230 x 5452 x 10-6
If Mu > M
u,lim
: design the section either increasing the dimensions of section or deign as Over reinforced section ( doubly-reinforced beam ).The additional moment of resistance M
u2
needed is obtained by providing compression( top ) reinforcement and additional tensile reinforcement. M
u2 = M
u - M
u,lim
as explained below .
Ast,lim
= Mu,lim
/ ( 0.87 fy ( d - 0.42 X
u,max ) )
Asc
= Mu2
/ ( fsc
x ( d - d' ) )A
st2 = A
sc f
sc / 0.87 f
y
Ast = A
st,lim + A
st2 .
If Xu < X
u,max the section is under-reinforced ( singly reinforced )
If Xu = X
u,max the section is balanced
If Xu > X
u,max the section is over-reinforced ( doubly reinforced )
Xu,max
= 0.53 x d ( for Fe250 mild steel )
Xu,max
= 0.48 x d ( for Fe415 HYSD steel )
Xu,max
= 0.46 x d ( for Fe500 HYSD steel )
Xu,max
= 0.44 x d ( for Fe550 HYSD steel )
Xu =( 0.87 f
y A
st ) / ( 0.36 f
ck b )
( From IS 456-2000 , table 19 table 7-1 )
As the ends are confined with compressive reaction , shear at distance d will be used for
Provide 8 mm O two-legged stirrups @ 300 mm c/c upto 4 no. from face of the support
IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement
IS 456-2000 clause 23.2.1 fig-5 ,for compression reinforcement
………………………( safe )
: design the section either increasing the dimensions of section or deign as Over reinforced section ( doubly-reinforced beam ).The additional moment of resistance M
u2
needed is obtained by providing compression( top ) reinforcement and additional tensile as explained below .
the section is under-reinforced ( singly reinforced )
the section is over-reinforced ( doubly reinforced )
Design of Simply supported Singly reinforced BeamSpan = 6 m ( simply supported rectangular beam )characteristic load = 20 KN / m inclusive of its self-weightBeam section = 230 mm x 600 mmMaterial M15 grade concrete ( Given )
HYSD reinforcement of grade Fe415
The beam is resting on R.C.C. columns.Solution : -Factored load = 1.5 x 20 = 30 KN /m.
=
= 135 KNm
w x ℓ / 2
= 30 x 6 / 2
= 90 KN
( a ) Moment steel :
Assuming 20 mm diameter bars in one layer
d = 600 - 25 ( Cover ) -10 OR= 565 mm
= 151.983 KNm
= 1.84 < 2.07 ( Table 6-3 )
The section is singly reinforced ( under-reinforced )
415 / 15
= 0.614
As per IS 456-2000 clause 26.5.1.1 ( a )
From Table 6-4
Let 2 bars are bent at 1.25 D = 1.25 x 600 = 750 mm , from the face of the support . ( b ) Check for development length :
Mu = w x ℓ2 / 8
30 x 62 / 8
Vu =
Mu,lim
= 2.07 x 230 x 5652 x 10-6
Mu / b d2 = 135 x 106 / ( 230 x 5652)
Mu < M
u,lim
Pt = 50
1 - √ 1 - ( 4.6 / fck
) x ( Mu / bd2 )
fy / f
ck
Pt = 50
1 - √ 1 - ( 4.6 / 15 ) x ( 1.84 )
Ast = ( 0.614 / 100 ) x 230 x 565 = 798 mm2
Minimum steel , As = ( 0.205 / 100 ) x 230 x 565 = 266 mm2
Ast,lim
= 0.72 / 100 x 230 x 565 = 936 mm2.
Provide 16 mm O - 4 No. = 804 mm2.
( 1 ) A bar can be bent up at a distance greater than Ld = 56 O ( Table 7-6 )
From the centre of the support , i.e. 56 x 16 = 896 mm .
in this case , the distance is ( 3000 - 750 ) = 2250 mm ……………….( safe )
== 82.01 x 0.91= 74.99 KNm
90 KN ,
As the reinforcement is confined by compressive reaction
1083.19 ≥ 44 OO ≤ 24.62 mm
( c ) Check for shear :
As the ends of the reinforcement are confined with compressive reaction , shear at distance dwill be used for checking shear at support.
=
= 0.562
( From IS 456-2000 , table 19 table 7-1 )= 0.309 0.25 difference
0.376 0.191 differenceshear design is necessary .
At support , OR
= = 24.17= 73.05 - 48.86= 24.19 KN
Capacity of bent bars to resist shear
== 102.6 KN
Bent bars share 50 % = 12.09 KNStirrups share 50 % = 12.09 KN
== 569.2 mm
( 2 ) For the remaining bars , Ast = 402 mm2
Mu1
= 0.87 fy A
st d ( 1 - [( f
y A
st ) / ( f
ck b d ) ] )
0.87 x 415 x 402 x 565 ( 1 - [ ( 415 x 402 ) / ( 15 x 230 x 565 ) ] ) x 10-6
Vu = L
0 = 12 O ( assume )
1.3 x M1 / V + L
0 ≥ L
d
1.3 x 74.99 x 106 / 90 x 103 +12 O ≥ 56 O
Oprovided
= 16 mm ……………………..( safe )
At support , Vu = 90 KN
Vu = 90 - 0.565 x w = 90 - 0.565 x 30 = 73.05 KN
Actual shear strength حv = V
u / bd
73.05 x 103 / ( 230 x 565 )
N / mm2
100 x As / b d = ( 100 x 402 ) / ( 230 x 565 )
design ( permissible ) shear strength حc = N / mm2 ح >
v
Vus
= Vuح -
c b d V
us = ( 0.562 - 0.376 ) x 230 x 565 x10-3
73.05 - 0.376 x 230 x 565 x 10-3
2 x 201 x 0.87 x 415 x sin 45º x 10-3 ( 0.87 fy A
sv sin α )
Using 6 mm O ( mild steel ) Two legged stirrups , Asv
= 28 x 2 = 56 mm2 .
Sv = 0.87 f
y A
sv d / V
us
0.87 x 250 x 56 x 565 / 12.09 x 103
From IS 456-2000 clause 26.5.1.6Spacing of minimum shear reinforcement using 6 O stirrups
== 0.87 x 56 x 250 / 0.4 x 230
= 132.4 mmspacing should not exceed
( i ) 450 mm ( ii ) 0.75 d = 0.75 x 565 = 423 mm( iii ) ≤132.4 mm ( minimum )( iv ) 569.2 mm ( designed )Provide 6 mm O two-legged stirrups @ 130 mm c/cAt 1.25 D = 750 mm from face of the support where contribution of bent bars is not available
== 67.5 - 48.86= 18.64 KN
Provide minimum 6 mm O M.S. two -legged stirrups @ 130 mm c/c throught the beam.
( d ) Check for deflection :Basic span / d ratio = 20
modification factor = 1.1 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement
Span / d permissible = 20 x 1.1 = 22
Actual span / d = 6000 / 565 = 10.62 ………………………( safe )
( d ) Check for cracking (spacing of bars ) :Clear distance between bars
= 230 - 50 - 2 x 16 = 148 mm
Minimum clear distance permitted
= 20 + 5 = 25 mm or 16 mm ( O of bar ) i.e. 25 mm .Maximum clear distance permitted
= 180 mm ( cracking - table 8-1 , IS 456-200 , table 15 ) ………………………( safe )
The design beam is shown in fig.
0.87 Asv
fy / 0.4 b
Vu = 90 - 0.75 x w = 90 - 0.75 x 30 = 67.5 KN
Vus
= Vuح -
c b d
67.5 - 0.376 x 230 x 565 x 10-3
100 Ast / b d = 100 x 804 / 230 x 565 = 0.62
hagg
+ 5 mm =
230
600565
2-10 O
4- 16 O
750
6000 c/c
2-10 O
4- 16 O ( 2 straight +2 bent )
6 mm O @ 130 mm c/c
Consider width of the beam equal to 230 mm. The depth may be assumedas 1 / 10 to 1 / 8 of the span.To find steel area( 1 ) For a given ultimate moment ( also known as factored moment )and assumed width of section , find out d from equation
( 2 ) For a given factored moment ,width and depth of section .
where ,
Table 6-2
Limiting Moment of Resistance and Reinforcement
d = √ Mu / Q
lim b
This is a balanced section and steel area may be found out from table P
Obtain Mu,lim
= Qlim
bd2 .
If Mu < M
u,lim
: design as under-reinforced section (singly reinforced beam) as explained below.
Pt = 50
1 - √ 1 - ( 4.6 / fck
) x ( Mu / bd2 )
fy / f
ck
If Mu = M
u,lim
: design as balanced section as explained in ( 1 ).
If Mu > M
u,lim
: design the section either increasing the dimensions of section or deign as Over reinforced section ( doubly-reinforced beam ).The additional moment of resistance M
u2
needed is obtained by providing compression( top ) reinforcement and additional tensile reinforcement. M
u2 = M
u - M
u,lim
as explained below .2.07 x 230 x 5652 x 10-6
Ast,lim
= Mu,lim
/ ( 0.87 fy ( d - 0.42 X
u,max ) )
Asc
= Mu2
/ ( fsc
x ( d - d' ) )
Ast2
= Asc
fsc
/ 0.87 fy
Ast = A
st,lim + A
st2 .
If Xu < X
u,max the section is under-reinforced ( singly reinforced )
If Xu = X
u,max the section is balanced
If Xu > X
u,max the section is over-reinforced ( doubly reinforced )
Xu,max
= 0.53 x d ( for Fe250 mild steel )
Xu,max
= 0.48 x d ( for Fe415 HYSD steel )
Xu,max
= 0.46 x d ( for Fe500 HYSD steel )
Xu,max
= 0.44 x d ( for Fe550 HYSD steel )
Xu =( 0.87 f
y A
st ) / ( 0.36 f
ck b )
Index for Singly Reinforced Rectangular Sections
250 415 500 550
0.148 0.138 0.133 0.129
21.93 19.86 19.03 18.2
Table 6-3
For singly reinforced rectangular sections
250 415 500 550
15 2.22 2.07 2.00 1.9420 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
Table 6-4As the ends of the reinforcement are confined with compressive reaction , shear at distance d
For singly reinforced rectangular sections
250 415 500 550
15 1.32 0.72 0.57 0.50
20 1.75 0.96 0.76 0.66
( From IS 456-2000 , table 19 table 7-1 ) 25 2.19 1.20 0.95 0.830.11 30 2.63 1.44 1.14 0.99
? -0.084IS 456-2000 Clause 26.5 Requirements of Reinforcement for
Structural Members
KN 26.5.1 Beams26.5.1.1 Tension Reinforcement
shall not be less than that given by the following :
where ,
b = breadth of the beam or the breadth of the web of T- beam ,
d = effective depth , and
fy , N / mm2
Mu,lim
/ fck
b d2
Pt,lim
fy / f
ck
Limiting Moment of resistance factor Q lim
, N / mm2
fck
N / mm2
fy, N / mm2
Limiting Percentage of Reinforcement Pt,lim
fck
N / mm2
fy, N / mm2
= ( 0.562 - 0.376 ) x 230 x 565 x10-3
a ) Minimum reinforcement - The minimum area of tension reinforcement
As / b d = 0.85 / f
y
As = minimum area of tension reinforcement ,
fy = characteristic strength of reinforcement in N / mm2 .
b ) Maximum reinforcement - The maximum area of tension reinforcement shall not exceed 0.04 b D.
Minimum steel %For mild steel
For HYSD steel , Fe415 grade
For HYSD steel , Fe500 grade
At 1.25 D = 750 mm from face of the support where contribution of bent bars is not available
check for development lengthIS 456-2000 clause 26.2.1
IS 456-200 clause 26.2.1.1
Table 7-5
IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement
M15 M20 M25 M30 M35 M40
1.0 1.2 1.4 1.5 1.7 1.9
For mild steel Fe250
For Fe415IS 456-2000 clause 26.2.3.3
………………………( safe )
Table 7-6Development length for single mild steel bars
Tension bars Compression bars
M15 M20 M15 M20250 55 O 26 O 44 O 37 O415 56 O 47 O 45 O 38 O
100 As / b d = 100 x 0.85 / 250 = 0.34
100 As / b d = 100 x 0.85 / 415 = 0.205
100 As / b d = 100 x 0.85 / 500 = 0.17
For checking development length , l0
may be assumed as 8 O for HYSD bars ( usually end anchorage is not
provided ) and 12 O for mild steel ( U hook is provided usually whose
anchorage length is 16 O.
Development length of bars Ld = O σ
s / 4 x ح
bd
Design bond stress (حbd
) for plain bars in tension
Concrete grade
ح ) bd
N / mm
Note-1 : حbd
shall be increased by 25 % for bars in compressionNote-2 : In case of deformed bars confirming to IS : 1786-1985 , the value of ح
bd shall be increased by 60 %.
σs = 0.87 x f
y
σs = 0.67 x f
y
Ld ≤ M
1 / V + L
0
L0 = effective depth of the members or 12 O , whichever is greater
if ends of the reinforcement are confined by a compressive reaction M1 / V increased by 30 % ( 1.3 M1 / V )
fy N /
mm2
500 69 O 58 O 54 O 46 O
check for shearIS 456-2000 , Table 19
Table 7-1
Concrete gradeM15 M20 M25 M30 M35 M400.28 0.28 0.29 0.29 0.29 0.30
0.25 0.35 0.36 0.36 0.37 0.37 0.380.50 0.46 0.48 0.49 0.50 0.50 0.510.75 0.54 0.56 0.57 0.59 0.59 0.601.00 0.60 0.62 0.64 0.66 0.67 0.681.25 0.64 0.67 0.70 0.71 0.73 0.741.50 0.68 0.72 0.74 0.76 0.78 0.791.75 0.71 0.75 0.78 0.80 0.82 0.842.00 0.71 0.79 0.82 0.84 0.86 0.882.25 0.71 0.81 0.85 0.88 0.90 0.922.50 0.71 0.82 0.88 0.91 0.93 0.952.75 0.71 0.82 0.90 0.94 0.96 0.983.00 0.71 0.82 0.92 0.96 0.99 1.01
The above given table is based on the following formula
6 x β
β =
IS 456-2000 , Table 20Table 7-2
M15 M20 M25 M30 M35 M40
2.5 2.8 3.1 3.5 3.7 4.0
IS 456-2000 Clause 26.5.1.6 Minimum shear reinforcementMinimum shear reinforcement in the form of stirrups shall be provided
such that :where,
total cross-sectional area of stirrup legs effective in shear ,
stirrup spacing along the length of the member , b = breadth of the beam or breadth of the web of flanged beam , and
Design shear strength of concrete , حC, N / mm2
Pt = 100 x A
s
b x d≤ 0.15
Design shear strength حc = 0.85 √ 0.8 x f
ck ( √ 1 + 5 x β - 1 )
0.8 x fck
/ 6.89 Pt , but not less than 1.0
Maximum shear stress , حC, N / mm2
Concrete grade
ح )c )
max N/mm2
Asv
/ b sv ≥ 0.4 / 0.87 f
y
Asv
=
Sv =
fy = characteristic strength of the stirrup reinforcement in N / mm2
which shall not be taken greater than 415 N / mm2 .
check for deflectionBasic values of span to effective depth ratios for spans upto 10 m :
cantilever 7simply supported 20
continuous 26
check for crackingIS 456-2000 26.3.3 Maximum distance between bars in tension
% redistribution to or from section considered-30 -15 0 +15 +30
Clear distance between bars
mm mm mm mm mm250 215 260 300 300 300415 125 155 180 210 235500 105 130 150 175 195
For spans above 10 m, the values in (a) may be multiplied by 10 / span in metres , except for cantilever in which case deflection calculations should be made.
Table 15 Clear distance Between Bars ( Clause 26.3.3 )
fy
N / mm2
Consider width of the beam equal to 230 mm. The depth may be assumedas 1 / 10 to 1 / 8 of the span.
( 1 ) For a given ultimate moment ( also known as factored moment )
This is a balanced section and steel area may be found out from table Pt,lim
, SP : 16 ,2.3
: design as under-reinforced section (singly reinforced beam) as explained below.
: design as balanced section as explained in ( 1 ).
: design the section either increasing the dimensions of section or deign as Over reinforced section ( doubly-reinforced beam ).The additional moment of resistance M
u2
needed is obtained by providing compression( top ) reinforcement and additional tensile
The minimum area of tension reinforcement
The maximum area of tension
= effective depth of the members or 12 O , whichever is greater
Minimum shear reinforcement in the form of stirrups shall be provided
total cross-sectional area of stirrup legs effective in shear ,
breadth of the beam or breadth of the web of flanged beam , and
characteristic strength of the stirrup reinforcement in N / mm2
Design of slender ( Long ) columns ( with biaxial bending )Size of column 400 x 300 mm Column is restrained against sway.Concrete grade M 30
Unsupported length = 7.0 m
Factored load 1500kN
Factored.moment in the direction of larger dimension = 40 kNm at top and 22.5 KNm at bottom.
Factored.moment in the direction of shorter dimension = 30 kNm at top and 20 KNm at bottom.
Solution : -About x axis :
For Beam :
= 0.7 x 5000 / 6 + 230 + 6 x 120= 1533.33333
2.07Beam stiffness
=
= 2836699Column stiffness
= 175238.095
= 2 x 175238 / 2 ( 175238 + 2836699 )= 0.0582
as per IS 456-2000 fig. 27
1.035 < 1.2 ……………….consider 1.2
1.2 x unsupported length= 1.2 ( 7000 - 620 )= 7656 mm
7656/400= 19.14 > 12
The column is long about x direction.About Y axis :
Characteristic strength of reinforcement 415 N/mm2
Effective length for bending parallel to larger dimension ℓex
= 6.0 m
Effective length for bending parallel to shorter dimension ℓey
= 5.0 m
β1 and β
2 are the same
bf = l
0 / 6 + b
w + 6 D
f
bf / b
w = 1533.3 / 230 = 6.67
Df / D = 120 / 600 = 0.2
Kt from chart 88 , SP : 16 =
Kb = 1.5 x I
b / l
1.5 x ( 2.07 x ( 1 / 12 ) x 230 x 6203 ) / 5000
mm3
Kc = I
c / l = 1/12 x 230 x 4003 / 7000
mm3
β1 = β
2 = ∑ K
c / ( ∑K
c + ∑K
b )
lef / l =
lex
=
lex
/ D =
Beam stiffness
=
= 426006Column stiffness
= 57938.0952
= 2 x 57938.1 / 2 ( 57938.1 + 426006 )= 0.1197
as per IS 456-2000 fig. 27
1.06 < 1.2 ……………….consider 1.2
1.2 x unsupported length= 1.2 ( 7000 -420 )= 7896 mm
7896/300= 26.32 > 12
The column is long about Y direction.The column is bent in double curvature. Reinforcement will be distributed equally on four sides.
6000 / 400= 15.0 > 12
5000 / 300= 16.7 > 12
Therefore the column is slender about both the axes.Additional moments
== 67.5 KNm
== 62.75 KNm
The above moments will have to be reduced in accordance with clause 39.7.1.1 of the IS 456-2000but multiplication factors can be evaluated only if the reinforcement is known. For first trial , assume p = 3.0 ( with reinforcement equally on all the four sides ).
400 x 300
= 120000 400 x 300 - 120000*3/100
OR = 116400
= 2700 KN = 1571.4 +
= 2692 KN
Kb = 1.5 x I
b / l
1.5 x ( ( 1 / 12 ) x 230 x 4203 ) / 5000
mm3
Kc = I
c / l = 1/12 x 400 x 2303 / 7000
mm3
β1 = β
2 = ∑ K
c / ( ∑K
c + ∑K
b )
lef / l =
ley
=
ley
/ b =
ℓex
/ D =
ℓey
/ b =
Max
= ( Pu D / 2000 ) x ( ℓ
ex / D )2
( 1500 x 400 / 2000 ) x (15)2 x 10-3
May
= ( Pu b / 2000 ) x ( ℓ
ey / b )2
( 1500 x 300 / 2000 ) x (16.7)2 x 10-3
Ag = P
uz = 0.45 f
ck A
c + 0.75 f
mm2 Ac =
From chart 63 , puz
/ Ag = 22.5 N/mm2
mm2
Puz
= 22.5 x 120000 x 10-3 Puz
= 0.45 x 30 x 116400 x 10-3 + 0.75 x 415 x 3600 x 10-3
Calculation of Pb :
Assuming 25 mm dia bars with 40 mm cover
d' / D ( about xx-axis ) = 52.5 / 400 = 0.13 ……………use d' / D = 0.15
d' / D ( about yy-axis ) = 52.5 / 300 = 0.18 ……………use d' / D = 0.2
From Table 60 , SP 16
= 779 KN
= 672 KN
= ( 2700 - 1500 ) / ( 2700 - 779 )= 1200 / 1921= 0.625
= ( 2700 - 1500 ) / ( 2700 - 672 )= 1200 / 2028= 0.592
The additional moments calculated earlier , will now be multiplied by the above values of k .
67.5 x 0.625 = 42.2 KNm
62.75 x 0.592 = 37.15 KNmThe additional moments due to slenderness effects should be added to the initial moments
after modifying the initial moments as follows ( see note 1 under 39.7.1 of the code IS 456-2000 )
( 0.6 x 40 - 0.4 x 22.5 ) = 15.0 KNm
( 0.6 x 30 - 0.4 x 20 ) = 10.0 KNmThe above actual moments should be compared with those calculated from minimum eccentricity
consideration ( see 25.4 of the code ) and greater value is to be taken as the initial moment for
adding the additional moments.
( ℓ / 500 ) + ( D / 30 ) = ( 7000 / 500 ) + ( 400 / 30) = 27.3333
( ℓ / 500 ) + ( b / 30 ) = ( 7000 / 500 ) + ( 300 / 30) = 24Moments due to minimum eccentricity :
Total moments for which the column is to be designed are :
41.0 + 42.2 = 83.2 KNm
36.0 + 37.15 = 73.15 KNmThe section is to be checked for biaxial bending
Pb ( about xx-axis ) = ( k
1 + k
2 p / f
ck ) f
ck b D
Pbx
= ( 0.196 + 0.203 x 3 /30 ) 30 x 300 x 400 x10-3
Pb ( about yy-axis ) = ( k
1 + k
2 p / f
ck ) f
ck b D
Pby
= ( 0.184 + 0.028 x 3 /30 ) 30 x 300 x 400 x10-3
Kx = ( P
uz - P
u ) / ( P
uz - P
bx )
Ky = ( P
uz - P
u ) / ( P
uz - P
by )
Max
=
May
=
Mux
=
Muy
=
ex =
ey =
Mux
= 1500 x 27.33 x 10-3 = 41.0 KNm > 15.0 KNm
Muy
= 1500 x 24 x 10-3 = 36.0 KNm > 10.0 KNm
Mux
=
Muy
=
Pu / f
ck b D = 1500 x 103 / ( 30 x 300 x 400 )
= 0.417
3 / 30 = 0.10
referring to chart 45 (d' / D = 0.15 ) ,
0.104
= 149.8 KNm
referring to chart 46 (d' / D = 0.2 ) ,
0.096
= 103.7 KNm
83.2 / 149.8 = 0.56
73.15 / 103.7 = 0.71
referring to chart 64 , the maximum allowable value of Mux / Mux1 corresponding to the above values of Muy / Muy1 and Pu / Puz is 0.58 which is slightly higher than the actual
value of 0.56 . The assumed reinforcement of 3.0 % is therefore satisfactory.
From , IS 456-2000 , Clause 39.6
1.602 0.2 difference 0.34
check : 0.04 difference ?
+
( 0.56 ) + ( 0.71 )
0.395 + 0.577
= 0.972 ……………………..( O.K.)
p x b x D / 100 = 3.0 x 300 x 400 / 100
= 3600
Ties : -
25 / 4= 6.25
Provide 8 mm O M.S. ties.Spacing should not exceed lesser of( i ) 300 mm( ii ) 16 x 25 = 400 mm
p / fck
=
Mu / f
ck b D2 =
Mux1
= 0.104 x 30 x 300 x 4002
Mu / f
ck b D2 =
Muy1
= 0.096 x 30 x 400 x 3002
Mux
/ Mux1
=
Muy
/ Muy1
=
Pu / P
uz = 1500 / 2700 = 0.56
for Pu / P
uz = 0.56 , α
n = 1.602
αn =
Mux
Muy ≤ 1M
ux1M
uy1
≤ 1
As =
mm2
Provide 25 mm diameter bar ( 3600 / 491 ) = 8 no. = 3928 mm2
O min
=
αn
αn
1.602 1.602
500
300
( iii ) 48 x 8 = 384 mmProvide 8 mm O M.S. ties @ 300 mm c/c
( 500 - 80 -25 = 395 > 48 x 8 = 384 ). Therefore two sets of closed ties shall be used .
Design of slender ( Long ) columns ( with Uniaxial bending )Size of column 230 x 450 mm Column of a braced frameConcrete grade M 20
Unsupported length in both the direction = 5.0 m
The column is bent in double curvature and is slender about both the axis.
Solution : -Assume adjustment factor k = 0.8 for the first trial .Additional moments
== 39.2 KNm
== 28 KNm
About XX
1000 KN
= 0.6 x 80 - 0.4 x 60= 48 - 24= 24 KNm < 0.4 x 80 = 32 KNm
Take , 32 KNm
32 + 0.8 x 39.2= 63.36 KNm. < 80 KNm
Note that the distance between corner bars in one face is more than 48 Otr
Note that if this distance would be less than ( 48 x 8 ) mm , open ties for internal bars would be sufficient.
Characteristic strength of reinforcement 415 N/mm2 HYSD reinforcement
Factored load Pu = 1000 kN
Factored.moment in the direction of larger dimension Muxx
= 80 kNm at top and 60 KNm at bottom.
Factored.moment in the direction of shorter dimension Muyy
= 40 kNm at top and 30 KNm at bottom.
The slenderness ratios ℓex
/ Ixx
and ℓey
/ Iyy are respectively 13.2 and 15.6
Assume that the moments due to minimum eccentricities about both the axes are less than applied moments.
Max
= ( Pu D / 2000 ) x ( ℓ
ex / D )2
( 1000 x 450 / 2000 ) x (13.2 )2 x 10-3
May
= ( Pu b / 2000 ) x ( ℓ
ey / b )2
( 1000 x 230 / 2000 ) x (15.6 )2 x 10-3
Pu =
Muxx
= Mi + k x M
ax
Mi = 0.6 M
u2 + 0.4 M
u1
Mi =
Note that Mu1
is considered negative as the column bends in double curvature .
Mu,xx
=
300
Take , 80 KNmAbout YY
1000 KN
= 0.6 x 40 - 0.4 x 30= 24 - 12= 12 KNm < 0.4 x 40 = 16 KNm
Take , 16 KNm
16 + 0.8 x 28= 38.4 KNm. < 40 KNm
Take , 40 KNmFinally design the column for
1000 KN
80 KNm
40 KNmFor the first trial , assume uniaxial bending about y axis for the following values .
1000 KN
( 230 / 450 ) ( 80 + 40 ) = 61.33 KNmd' / D = 50 / 230 = 0.22 Say 0.2
from chart 46 , SP : 16 = 0.162p = 0.162 x 20
= 3.24
( 3.24 /100 ) x 230 x 450
= 3353Assumptions made above for k = 0.8 and for uniaxial moment to find out first trial steel are
Now check the assumed section as follows :
230 x 450 - 3220
= 100280
= 902.52 + 1002.225= 1904.7 KN
( p / 100 ) x b x D
( 3220 x 100 ) / ( 230 x 450 x 20 )
Mu,xx
=
Pu =
Muyy
= Mi + k x M
ay
Mi = 0.6 M
u2 + 0.4 M
u1
Mi =
Mu,yy
=
Mu,yy
=
Pu =
Mu,xx
=
Mu,yy
=
P'u =
M'uy
=
P'u / f
ck b D = 1000 x 103 / 20 x 230 x 450 = 0.48
M'uy
/ fck
b D2 = 61.33 x 106 / 20 x 450 x 2302 = 0.129p / f
ck =
Asc
=
mm2
usually conservative . Let us try 4 - 25 O + 4 - 20 O = 3220 mm2
Puz
= 0.45 fck
Ac + 0.75 f
y A
sc
Ac =
mm2
Puz
= 0.45 x 20 x 100280 x 10-3 + 0.75 x 415 x 3220 x 10-3
Pb = ( k
1 + k
2 p / f
ck ) f
ck b D
Asc
=
p / fck
=
= 0.156
= 0.181 x 2070= 373.8 KN
k == ( 1904.7 - 1000 ) / ( 1904.7 - 373.8 )= 904.7 / 1530.9= 0.59
Design for
1000 KN
32 + 0.59 x 39.2 = 55.13 KNm < 80
16 + 0.59 x 28 = 32.52 KNm < 40
the moment capacities can be found out as follows :About XXd' / D = 50 / 450 = 0.11 ≈ 0.15From chart 45 , SP : 16
= 135.07 KNmAbout YYd' / D = 50 / 230 = 0.22 ≈ 0.2From chart 46 , SP : 16
= 61.89 KNmCheck :
From , IS 456-2000 , Clause 39.6
1.542 0.2 difference 0.34check : 0.075 difference ?
+
80 + 40135.07 61.89
0.446 + 0.51
0.956 …………….( O.K.)
For d' / D = 0.2 , k1 = 0.184 and k
2 = -0.022 from table 60 ,SP : 16
Pb = ( 0.184 - 0.022 x 0.156 ) x 20 x 230 x 450 x 10-3
( Puz
- Pu ) / ( P
uz - P
b )
Pu =
Mux
= Take , Mux
= 80 KNm
Muy
= Take , Muy
= 40 KNm
For p / fck
= 0.162 and Pu / f
ck b D = 0.48 , the reinforcement being equally distributed ,
Mux1
/ fck
b D2 = 0.145
Mux1
= 0.145 x 20 x 230 x 4502 x 10-6
Muy1
/ fck
b D2 = 0.13
Muy1
= 0.13 x 20 x 450 x 2302 x 10-6
Pu / P
uz = 1000 / 1904.7 = 0.525
αn =
Mux
Muy ≤ 1M
ux1M
uy1
≤ 1
αn
αn
1.542 1.542
230
Provide 4 - 25 O + 4 - 20 O equally distributed.
= 6.25 mmUse 8 mm diameter M.S. ties.Spacing should not exceed lesser of( i ) 230 mm ( least lateral dimension )( ii ) 16 x 25 = 400 mm ( 16 times least longitudinal diameter of bar )( iii ) 48 x 8 = 384 mm ( 48 times diameter of tie )Provide 8 mm O M.S. ties @ 230 mm c/c
( 450 - 80 -25 = 345 < 48 x 8 = 384 ). Therefore two sets of Opened ties shall be used .
Ties : Minimum diameter Otr = 25 / 4
Note that the distance between corner bars in one face is less than 48 Otr
Note that if this distance would be more than ( 48 x 8 ) mm , closed ties for internal bars would be sufficient.
230
Design of slender ( Long ) columns ( with biaxial bending ) IS 456-2000 clause 25
( Given ) D = depth in respect of the major axis
Factored.moment in the direction of larger dimension = 40 kNm at top and 22.5 KNm at bottom.b = width of the member
Factored.moment in the direction of shorter dimension = 30 kNm at top and 20 KNm at bottom.
IS 456 : 2000Table 28 Effective Length of Compression Members ( Clause E-3 )
Symbol
Note :- A column may be considered as short when both slenderness ratios lex
/ D and ley
/ b ≤ 12 OR lex
/ ixx
< 40 where, For Circular column lex
/ D ≤ 10 .
lex
= effective length in respect of the major axis
ley
= effective length in respect of the minor axis
ixx
= radius of gyration in respect of the major axis.
iyy = radius of gyration in respect of the minor axis.
Degree of End Restraint of Compression members
Effectively held in position and restrained against rotation in both ends
Effectively held in position at both ends , restrained against rotation at one ends
Effectively held in position at both ends , but not restrained against rotation .
Effectively held in position and restrained against rotation at one end , and at the other end restrained against rotation but not held in position
Effectively held in position and restrained against rotation at one end , and at the other partially restrained against rotation but not held in position
Effectively held in position at one end but not restrained against rotation and at the other end restrained against rotation but not held in position
Effectively held in position and restrained against rotation at one end but not held in position nor restrained against rotation at the other end.
NOTE - ℓ is the unsupported length of compression member.
The column is bent in double curvature. Reinforcement will be distributed equally on four sides.
The above moments will have to be reduced in accordance with clause 39.7.1.1 of the IS 456-2000
400 x 300 - 120000*3/100
1120.5
Effectively held in position and restrained against rotation at one end but not held in position nor restrained against rotation at the other end.
Effective length of column ( ℓef ) : It is the distance between the points of zero moment (contraflexure ) along the
column height .
Ac + 0.75 f
y A
sc
= 0.45 x 30 x 116400 x 10-3 + 0.75 x 415 x 3600 x 10-3
( i + 1 ) th floor
( i ) th floor
slab
Beam
ℓ
SP : 16 ,Table 60
Section d' / D0.05 0.10 0.15
Rectangular 0.219 0.207 0.196
Circular 0.172 0.160 0.149
Section d' / D0.05 0.10
250 -0.045 -0.045415 0.096 0.082
500 0.213 0.173250 0.215 0.146415 0.424 0.328500 0.545 0.425
The additional moments calculated earlier , will now be multiplied by the above values of k .
Circular
250 0.193 0.148
415 0.410 0.323
500 0.543 0.443The additional moments due to slenderness effects should be added to the initial moments IS 456-2000 clause 39.7.1
after modifying the initial moments as follows ( see note 1 under 39.7.1 of the code IS 456-2000 )
The above actual moments should be compared with those calculated from minimum eccentricity Where ,
consideration ( see 25.4 of the code ) and greater value is to be taken as the initial moment for axial load on the member,
effective length in respect of the major axis ,
mm > 20 mm effective length in respect of the minor axis ,
mm > 20 mm D =
b = width of the member
NOTES : - 1) A column may be considered braced in a given plane if lateral
stability to the structure as a whole is provided by walls or
bracing or buttressing designed to resist all lateral forces inthat plane. It should otherwise be considered unbraced.
2 ) In the case of a braced column without any transverse loads
Slender compression members- Values of P
Rectangular sections : Pb / f
ck b D = k
1 + k
2 . p / f
Circular sections : Pb / f
ck D2 = k
1 + k
2 . p / f
ck
Values of k1
Values of k2
fy N/ mm2
Rectangular ; equal reinforcement on two
opposite sides
Rectangular ; equal reinforcement on four
sides
The additional moments Max
and May
shall be calculated by the following formulae :
Max
= ( Pu D / 2000 ) x ( ℓ
ex / D )2
May
= ( Pu b / 2000 ) x ( ℓ
ey / b )2
Pu =
ℓex
=
ℓey
=
depth of the cross - section at right angles to the major axis , and
occurring in its height, the additional moment shall be added
end moment (assumed negative if the column is bent in double
curvature). In no case shall the initial moment be less than
shall be added to the end moments.
3 ) Unbraced compression members, at any given level or storey,subject to lateral load are usually constrained to deflect
equally. In such cases slenderness ratio for each column may
be taken as the average for all columns acting in the same
direction.IS 456-2000 clause 39.7.1.1The values given by equation 39.7.1 may be multiplied by the following factor :
where ,
axial load on compression member,
as defined in 39.6,
-0.068 axial load corresponding to the condition
of maximum compressive strain of
0.0035 in concrete and tensile strain of0.002 in outer most layer of tension steel.
to an initial moment equal to sum of 0.4 Mu1
, and 0.6 Mu2
,
where Mu2
is the larger end moment and Mu1
is the smaller
0.4 Mu2
nor the total moment including the initial moment be
less than Mu2
. For unbraced columns, the additional moment
K = ( Puz
- Pu ) / ( P
uz - P
b ) ≤ 1
Pu =
Puz
= Puz
= 0.45 fck
Ac + 0.75 f
Pb =
8-25 O
( 500 - 80 -25 = 395 > 48 x 8 = 384 ). Therefore two sets of closed ties shall be used .
Design of slender ( Long ) columns ( with Uniaxial bending )
( Given )
Note that if this distance would be less than ( 48 x 8 ) mm , open ties for internal bars would be
= 80 kNm at top and 60 KNm at bottom.
= 40 kNm at top and 30 KNm at
Assume that the moments due to minimum eccentricities about both the axes are less than
8 O @300 c/c(two sets )
Assumptions made above for k = 0.8 and for uniaxial moment to find out first trial steel are
-0.1275
450
4-25 O + 4 -20 O
( 16 times least longitudinal diameter of bar )
( 450 - 80 -25 = 345 < 48 x 8 = 384 ). Therefore two sets of Opened ties shall be used .
Note that if this distance would be more than ( 48 x 8 ) mm , closed ties for internal bars would be
8 O @ 230 mm c/c( three sets )
Table 28 Effective Length of Compression Members ( Clause E-3 )
Symbol
0.5 ℓ 0.65 ℓ
0.7 ℓ 0.8 ℓ
1.0 ℓ 1.0 ℓ
1.0 ℓ 1.2 ℓ
- 1.5 ℓ
2.0 ℓ 2.0 ℓ
2.0 ℓ 2.0 ℓ
Note :- A column may be considered as short when both slenderness ratios < 40 where, For Circular column l
ex / D ≤
= effective length in respect of the major axis
= effective length in respect of the minor axis
= radius of gyration in respect of the major axis.
= radius of gyration in respect of the minor axis.
Theoretical Value of effective length
Recommended Value of effective length
2.0 ℓ 2.0 ℓ
NOTE - ℓ is the unsupported length of compression member.
) : It is the distance between the points of zero moment (contraflexure ) along the
slab
Beam
SP : 16 ,Table 60
d' / D0.20
0.184
0.138
d' / D0.15 0.20
-0.045 -0.0450.046 -0.022
0.104 -0.0010.061 -0.0110.203 0.0280.256 0.0400.077 -0.020
0.201 0.036
0.291 0.056
effective length in respect of the major axis ,
effective length in respect of the minor axis ,
1) A column may be considered braced in a given plane if lateral
stability to the structure as a whole is provided by walls or
bracing or buttressing designed to resist all lateral forces inthat plane. It should otherwise be considered unbraced.
2 ) In the case of a braced column without any transverse loads
Slender compression members- Values of Pb
b D = k1 + k
2 . p / f
ck
= k1 + k
2 . p / f
ck
shall be calculated by the following formulae :
depth of the cross - section at right angles to
occurring in its height, the additional moment shall be added
end moment (assumed negative if the column is bent in double
curvature). In no case shall the initial moment be less than
shall be added to the end moments.
3 ) Unbraced compression members, at any given level or storey,subject to lateral load are usually constrained to deflect
equally. In such cases slenderness ratio for each column may
be taken as the average for all columns acting in the same
The values given by equation 39.7.1 may be multiplied by the following factor :
, and 0.6 Mu2
,
is the smaller
nor the total moment including the initial moment be
. For unbraced columns, the additional moment
= 0.45 fck
Ac + 0.75 f
y A
sc
Design of short eccentrically loaded square columns - Biaxial bending.Size 500 mm x 500 mmAxial factored load 1500 KN
( Given )
moment due to minimum eccentricity is less than the applied moment.Material M15 grade concrete
HYSD reinforcement of grade Fe415Solution :-
= 0.4
= 0.112d' = 40 + 10 = 50 mmd' / D = 50 / 500 = 0.1From chart 32 , SP-16
p = 0.078 x 15 = 1.17
= 2925
p = 3220 x 100 / ( 500 x 500 )p = 1.288
= 0.086The assumed section is now checked.
The reinforcement being equally distributed , the moment capacities from
Chart - 44 , SP-16
= 202.5 KNm
500 x 500 - 3220
= 246780
= 1665.77 + 1002.23
= 2667.99 KN
Factored moment Mux
= 90 KNm
Muy
= 120 Knm
Assume an axial load P'u of 1500 KN and a uniaxial moment M'
ux = 90 + 120 = 210 KNm .
P'u / ( f
ck x b x D ) = 1500 x 103 / ( 15 x 500 x 500 )
M'ux
/ ( fck
x b x D2 ) = 210 x 106 / ( 15 x 500 x 5002 )
p / fck
= 0.078
As = 1.17 x b x D = 1.17 x 500 x 500 / 100
mm2
Provide 4-25 mm O + 4-20 mm O = 3220 mm2 , equally distributed
p / fck
= 1.288 / 15
For p / fck
= 0.086 and Pu / ( f
ck x b x D ) = 0.4 ,
Mux1
/ ( fck
x b x D2 ) = Muy1
/ ( fck
x b x D2 ) = 0.108
Mux1 = Muy1 = 0.108 x 15 x 500 x 5002 x 10-6
Puz
= 0.45 fck
Ac + 0.75 f
y A
sc
Ac =
mm2
Puz
= 0.45 x 15 x 246780 x 10-3 + 0.75 x 415 x 3220 x 10-3
1500 / 2668= 0.56
From , IS 456-2000 , Clause 39.6
1.602 0.2 difference 0.34check : 0.04 difference ? -0.068
+
90+
120202.5 202.5
0.273 + 0.433
0.706
= 6.25 mmUse 8 mm diameter M.S. ties.Spacing should not exceed lesser of( i ) 500 mm( ii ) 16 x 25 = 400 mm( iii ) 48 x 8 = 384 mmProvide 8 mm O M.S. ties @ 350 mm c/c
Design of short eccentrically loaded rectangle columns - Biaxial bending.Size 300 mm x 500 mmAxial factored load 1500 KN
( Given )
moment due to minimum eccentricity is less than the applied moment.Material M15 grade concrete
HYSD reinforcement of grade Fe415Solution :-
= 0.67
= 0.107d' = 40 + 10 = 50 mmd' / D = 50 / 300 = 0.167 ………………..use 0.2From chart 34 , SP-16
p = 0.167 x 15 = 2.5
Pu / P
uz =
αn =
Mux
Muy ≤ 1M
ux1M
uy1
≤ 1Ties : Minimum diameter O
tr = 25 / 4
Factored moment Mux
= 60 KNm
Muy
= 60 KNm
Assume an axial load P'u of 1500 KN and a uniaxial moment M'
uy = 300 / 500 ( 60 + 60 ) = 72 KNm .
P'u / ( f
ck x b x D ) = 1500 x 103 / ( 15 x 300 x 500 )
M'u / ( f
ck x b x D2 ) = 72 x 106 / ( 15 x 500 x 3002 )
p / fck
= 0.167
αn
αn
1.602 1.602
500
500
4-25 O +4-20 O
8 O @350 c/c
= 3750
p = 3928 x 100 / ( 300 x 500 )p = 2.62
= 0.175The assumed section is now checked.About X
Chart - 44 , SP-16
= 146.25 KNmAbout Y
Chart - 46 , SP-16
= 69.53 KNmPure axial load capacity
300 x 500 - 3928
= 146072
= 985.986 + 1222.59= 2208.58 KN
1500 / 2208.6= 0.68
From , IS 456-2000 , Clause 39.6
1.802 0.2 difference 0.33check : 0.12 difference ? -0.198
+
60 + 60146.25 69.53
0.2 + 0.77
As = 2.5 x b x D = 2.5 x 300 x 500 / 100
mm2
Provide 8-25 mm O = 3928 mm2 , equally distributed
p / fck
= 2.62 / 15
d' / D = 50 / 500 = 0.1 , For p / fck
= 0.175 and Pu / ( f
ck x b x D ) = 0.67 ,
Mux1
/ ( fck
x b x D2 ) = 0.13
Mux1
= 0.13 x 15 x 300 x 5002 x 10-6
d' / D = 50 / 300 = 0.167 say 0.2 , For p / fck
= 0.175 and Pu / ( f
ck x b x D ) = 0.67 ,
Muy1
/ ( fck
x b x D2 ) = 0.103
Muy1
= 0.103 x 15 x 500 x 3002 x 10-6
Puz
= 0.45 fck
Ac + 0.75 f
y A
sc
Ac =
mm2
Puz
= 0.45 x 15 x 146072 x 10-3 + 0.75 x 415 x 3928 x 10-3
Pu / P
uz =
αn =
Mux
Muy ≤ 1M
ux1M
uy1
αn
αn
1.802 1.802
0.97
= 6.25 mmUse 8 mm diameter M.S. ties.Spacing should not exceed lesser of( i ) 300 mm( ii ) 16 x 25 = 400 mm( iii ) 48 x 8 = 384 mmProvide 8 mm O M.S. ties @ 300 mm c/c
( 500 - 80 -25 = 395 > 48 x 8 = 384 ). Therefore two sets of closed ties shall be used .
≤ 1Ties : Minimum diameter O
tr = 25 / 4
Note that the distance between corner bars in one face is more than 48 Otr
Note that if this distance would be less than ( 48 x 8 ) mm , open ties for internal bars would be sufficient.
500
300
39.6 Members Subjected to Combined Axial Loadand Biaxial Bending where ,
+
= 90 + 120 = 210 KNm .
M
Mux
Muy ≤ 1M
ux1M
uy1M
ux1, M
uy1 =
αn is related to P
u/P
uz
where Puz
= 0.45 fck
Ac + 0.75 f
y A
sc
For values of Pu / P
uz = 0.2 to 0.8, ( 0.2 , 0.4 , 0.6 , 0.8 )the values of
linearly from 1 .0 to 2.0.( 1.0 , 1.33 , 1.67 , 2.0 ) For values less than 0.2, α
αn
αn
( least lateral dimension )( 16 times least longitudinal diameter of bar )
( 48 times diameter of tie )
1 .O; for values greater than 0.8, αn is 2.0.
= 300 / 500 ( 60 + 60 ) = 72 KNm .
4-25 O +4-20 O
8 O @350 c/c
( least lateral dimension )( 16 times least longitudinal diameter of bar )
( 48 times diameter of tie )
( 500 - 80 -25 = 395 > 48 x 8 = 384 ). Therefore two sets of closed ties shall be used .
8-25 O
8 O @300 c/c(two sets )
moments about x and y axes
due to design loads,
maximum uniaxial moment
bending about x and y axes
respectively, and
Mux
, Muy
=
Mux1
, Muy1
=
capacity for an axial load of Pu,
= 0.2 to 0.8, ( 0.2 , 0.4 , 0.6 , 0.8 )the values of αn vary
linearly from 1 .0 to 2.0.( 1.0 , 1.33 , 1.67 , 2.0 ) For values less than 0.2, αn is
Design of short eccentrically loaded columns - uniaxial bending.Size 300 mm x 600 mmAxial factored load 600 KNFactored moment 300 KNm
( Given )moment due to minimum eccentricity is less than the applied moment.Material M15 grade concrete
HYSD reinforcement of grade Fe415Solution :-Using 25 mm diameter bars with 40 mm clear coverd' = 40 + 12.5 = 52.5 mmd' / D = 52.5 / 600 = 0.088 ,say 0.1
= 0.222
= 0.185From chart 32 , SP : 16
p = 0.1 x 15 = 1.5
= 2700
As the distance between two opposite corner bars is more than 300 mm ,provide 2- 12 mm O at centre of long sides of the column, which may be tied by open ties.
= 6.25 mmUse 8 mm diameter M.S. ties.Spacing should not exceed lesser of( i ) 300 mm( ii ) 16 x 25 = 400 mm( iii ) 48 x 8 = 384 mmProvide 8 mm O M.S. ties @ 300 mm c/c ( two sets )
( 600 - 80 -25 = 495 > 48 x 8 = 384 ). Therefore two sets of closed ties shall be used .
Pu / ( f
ck x b x D ) = 600 x 103 / ( 15 x 300 x 600 )
Mu / ( f
ck x b x D2 ) = 300 x 106 / ( 15 x 300 x 6002 )
p / fck
= 0.1
As = 1.5 x b x D = 1.5 x 300 x 600 / 100
mm2
Provide 22 mm O ( 2700 / 380 = 8 no.) = 3041 mm2
Provide 8- 25 O + 2 - 12 O = 4151 mm2
Ties : Minimum diameter Otr = 25 / 4
Note that the distance between corner bars in one face is more than 48 Otr
Note that if this distance would be less than ( 48 x 8 ) mm , open ties for internal bars would be sufficient.
600
300
provide 2- 12 mm O at centre of long sides of the column, which may be tied by open ties.
( least lateral dimension )( 16 times least longitudinal diameter of bar )
( 48 times diameter of tie )
( 600 - 80 -25 = 495 > 48 x 8 = 384 ). Therefore two sets of closed ties shall be used .
8-25 O+ 2-12 O
8 mm O @300 mm c/c( two sets )
Design of short circular columnWorking load = 1200 KN
Given( a ) lateral ties & ( b ) helical reinforcementMaterial M20 grade concrete
HYSD steel Fe415For Lateral reinforcement mild steel Fe250Solution : -
Factored load = 1.5 x 1200 = 1800 KN
( a ) lateral ties : -
Assume 0.8 % minimum steel.
Then ,
=Substituting , we have
7.936 2.2244
10.1604
177165If D is the diameter of the column
177165D = √ 225688D = 475 mm
Use 475 mm diameter column.
0.008 x 177165
= 1417Minimum 6 bars shall be used.Provide 16 mm diameter bars ( 1417 / 201 ) = 8 no.
8 x 201
= 1608Use 6 mm O lateral ties , Spacing shall be lesser of( i ) 475 mm ( least lateral dimension )( ii ) 16 x 16 = 256 mm ( 16 times least longitudinal diameter of bar )( iii ) 48 x 6 = 288 mm ( 48 times diameter of tie )Provide 6 mm O lateral ties @ 250 mm c/c .( b ) Helical reinforcement : -The column with helical reinforcement can support 1.05 times the load of a similarmember with lateral ties. Therefore
Assume emin
< 0.05 D
Pu = 0.4 f
ck A
c + 0.67 f
y A
sc
Asc
= 0.008 Ag
Ac = A
g - A
sc
0.992 Ag
1800 x 103 = 0.4 x 20 x 0.992 Ag + 0.67 x 415 x 0.008 A
g
1800 x 103 = Ag + A
g
1800 x 103 = Ag
Ag = mm2
( ¶ / 4 ) x D2 =
Asc
=
mm2
Asc
=
mm2
Pu = 1.05 [ 0.4 f
ck A
c + 0.67 f
y A
sc ]
8.3328 2.33562
10.66842
168729If D is the diameter of the column
168729D = √ 214941D = 463 mm
Use 450 mm diameter column.
Then ,
= 158962.5
= 1335289 - 8.4 + 291.953
464711 = 283.55
1638.903provide 20 mm diameter bars ( 1638.9 / 314 =) 6 No.
6 x 314
= 1884Assume 8 mm O M.S. bars for helix at 40 mm clear cover .
450 - 40 - 40= 370 mm
= 50
== 0.36 x 0.4792 x 20 / 250= 0.013801
Now0.0138 = 4 x 50 / p x 370
p = 39.17 mm …………………….. ( 1 )As per IS 456-2000 clause 26.5.3.2 ( d )The pitch < 75 mm
> 25 mm> 3 x dia of helix bar = 3 x 8 = 24 mm ……………( 2 )
From ( 1 ) and ( 2 ) , provide 8 mm O helix @ 35 mm pitch.
1800 x 103 = 1.05 [0.4 x 20 x 0.992 Ag + 0.67 x 415 x 0.008 A
g ]
1800 x 103 = Ag + A
g
1800 x 103 = Ag
Ag = mm2
( ¶ / 4 ) x D2 =
Ag = ( ¶ / 4 ) x 4502
mm2
1800 x 103 = 1.05 [0.4 x 20 x (158963 - Asc
) + 0.67 x 415 x Asc
]
Asc
Asc
Asc
Asc
= mm2
Asc
=
mm2
Dc =
asp
= ( ¶ / 4 ) x 82
mm2
Minimum ρs = 0.36 ( (A
g / A
cr) - 1 ) f
ck / f
y
0.36 ( ( 4502 / 3702 ) - 1 ) x 20 / 250
ρs = 4 x a
sp / p x D
c
< Dc / 6 ( = 370 / 6 = 61.67 mm )
( 16 times least longitudinal diameter of bar )
450
6 - 20 O
6 - 20 O
35
8 mm O @35 mm c/c
Design of short columnFactored load = 1500 KN
GivenMaterial M15 grade concrete
mild steel Fe250
Solution : -
Therefore , size of column shall be minimum ( 20 / 0.05 = 400 ) 400 mm x 400 mm
Assume 0.8 % minimum steel.Then ,
=
=5.952 1.34
7.292
205705
If the column is to be a square , the side of column = √205705 = 453 mm
Adopt 450 mm x 450 mm size column .
Then ,
1215000 + 167.5
285000 = 161.5
1765Provide 20 mm diameter bars = ( 1765 / 314 ) = 6 No.
Note that the distance between the bars exceeds 300 mm on two parallel sides .the arrangement of reinforcement should be changed.Provide then 4 no. 20 mm diameter bars plus 4 no. 16 mm diameter bars giving
4 x 314 + 4 x 201= 1256 + 804
= 2060Lateral ties :Use 6 mm O lateral ties.Spacing should be lesser of :( i ) 450 mm ( least lateral dimension )( ii ) 16 x 16 = 256 mm ( 16 times least longitudinal diameter of bar )(iii) 48 x 6 = 288 mm . ( 48 times diameter of tie )Provide 6 mm O ties about 250 mm c/c .
Assume emin
< 0.05 D
Here , emin
< 0.05 D , But emin
= 20 mm
Asc
= 0.008 Ag
Ac = A
g - A
sc
0.992 Ag
Pu = 0.4 f
ck A
c + 0.67 f
y A
sc
0.4 x 15 x 0.992 Ag + 0.67 x 250 x 0.008 A
g
1500 x 103 = Ag + A
g
1500 x 103 = Ag
Ag = mm2
1500 x 103 = 0.4 x 15 x ( 450 x 450 - Asc
)+ 0.67 x 250 x Asc
1500 x 103 = - 6 Asc
Asc
Asc
Asc
= mm2
giving , Asc
= 6 x 314 = 1884 mm2
Asc
=
mm2
( 450 - 80 -20 = 350 > 48 x 6 = 288 ). Therefore two sets of closed ties shall be used .
wrong arrangement correct arrangement
Note that the distance between corner bars in one face is more than 48 Otr
Note that if this distance would be less than ( 48 x 6 ) mm , open ties for internal bars would be sufficient.
450
450
450
450
6 - 20 O4 - 20 O + 4 - 16 O
6 mm O @ 250 mm c/c(double ties )
IS 456-2000 clause 25
D = depth in respect of the major axis
b = width of the member
400 mm x 400 mm
IS 456-2000 clause 39.3
IS 456-2000 clause 26.5.3.1
NOTE - The use of 6 percent reinforcement may involve
practical difficulties in placing and compacting of concrete;hence lower percentage is recommended. Where bars fromthe columns below have to be lapped with those in thecolumn under consideration, the percentage of steel shall
usually not exceed 4 percent26.5.3 columns
26.5.3.1 Longitudinal reinforcementa) The cross-sectional area of longitudinalreinforcement, shall be not less than 0.8 percentnor more than 6 percent of the gross crosssectionalarea of the column.NOTE - The use of 6 percent reinforcement may involvepractical difficulties in placing and compacting of concrete;hence lower percentage is recommended. Where bars from
Note :- A column may be considered as short when both slenderness ratios lex
/ D and ley
/ b ≤ 12 OR lex
/ ixx
< 40 where,
lex
= effective length in respect of the major axis
ley
= effective length in respect of the minor axis
ixx
= radius of gyration in respect of the major axis.
iyy = radius of gyration in respect of the minor axis.
All columns shall be designed for minimum eccentricity equal to the unsupported length of column / 500 plus lateral dimension / 30 , subject to a minimum of 20 mm. For bi-axial bending this eccentricity exceeds the minimum about one axis at a time.
When emin
≤ 0.05 D , the column shall be designed by the following equation:
Pu = 0.4 f
ck A
c + 0.67 f
y A
sc
Ac = Area of concrete ,
Asc
= Area of longitudinal reinforcement for columns.
If emin
> 0.05 D , the column shall be designed for moment also.
The cross-sectional area of longitudinal reinforcement , shall be not less than 0.8 % nor more than 6 % of the gross cross-sectional area of the column.
the columns below have to be lapped with those in the( 450 - 80 -20 = 350 > 48 x 6 = 288 ). Therefore two sets of closed ties shall be used . column under consideration, the percentage of steel shall
usually not exceed 4 percent.b) In any column that has a larger cross-sectionalarea than that required to support the load,the minimum percentage of steel shall bebased upon the area of concrete required toresist the direct stress and not upon the actualarea.c)The minimum number of longitudinal barsprovided in a column shall be four in rectangularcolumns and six in circular columns.d )The bars shall not be less than 12 mm indiameter.e) A reinforced concrete column having helicalreinforcement shall have at least six bars oflongitudinal reinforcement within the helicalreinforcement.f )In a helically reinforced column, the longitudinalbars shall be in contact with the helicalreinforcement and equidistant around its innercircumference.g )Spacing of longitudinal bars measured alongthe periphery of the column shall not exceed300 mm.h )In case of pedestals in which the longitudinalreinforcement is not taken in account in strengthcalculations, nominal longitudinal reinforcementnot less than 0.15 percent of the cross-sectionalarea shall be provided.
4 - 20 O + 4 - 16 O
6 mm O @ 250 mm c/c(double ties )
Note :- A column may be considered as short when both slenderness ratios
All columns shall be designed for minimum eccentricity equal to the unsupported length of column / 500 plus lateral dimension / 30 , subject to a minimum of 20 mm. For bi-axial bending this eccentricity exceeds the
≤ 0.05 D , the column shall be designed by the following
The cross-sectional area of longitudinal reinforcement , shall be not less than 0.8 % nor more than 6 % of the gross cross-sectional area of the
Design of dog-legged staircase
230
UP 900
Floor A B 150 1950
900
230300 900 2250 900 230
Rise of step = 160 mm Rise 175 mm to 200 mm
Tread = 250 mm Tread 250 mm to 280 mmNosing is not provided ( given )Material M15 Grade concretemild steel reinforcement Fe250Solution : - Assume 150 mm thick waist slab. Both the landings can span on walls.Landing A or B
Self-load 0.15 x 25 = 3.75
Floor finish = 1.00
Live load ( residence ) = 3.00
Total 7.75
11.63Span = 1950 + 150 = 2100 i.e. 2.1 m
Consider 1 m length of slab
M =
== 6.41 KNm
Reinforcement will be in second layer, Assuming 12 mm O barsd = 150 - 15 (cover ) -12 - 6
= 117 mm
= 0.468
=
> 200 mm
< 230 mm
KN / m2
KN / m2
KN / m2
KN / m2
PU = 1.5 x 7.75 = KN / m2
w x l2 / 8
11.63 x 2.12 / 8
Mu / b x d2 = 6.41 x 106 / 1000 x 1172
Pt = 50 1-√1-(4.6 / fck) x (M
u / b x d2)
fy / f
ck
50 1-√1-(4.6 / 15) x (0.468)
1 5 9 10
11151920
vvv
vv
250 / 15
= 50 [(1-0.93) x 15 / 250 ]= 0.224%
0.224 x 1000 x 117 / 100
= 262
Minimum steel = ( 0.15 / 100 ) x 1000 x 150
= 225Provide 10 mm O bar
spacing of bar == 78.50 x 1000 /262= 300 mm
Provide 10 mm O bar @ 280 mm c/c = 280Maximum spacing = 3 x d = 3 x 117
= 351 mm ……………….( O.K.)Check for shear : -
w x l / 2= 11.63 x 2.1 / 2= 12.21 KN
Shear stress =
=
= 0.104 ( too small )
( from table 7-1 )……………….( O.K.)
Check for development length : -
12 O (mild steel )
= 100 x 280 / 1000 x 117= 0.239
From equation
0.499
= 6.83 KNm
12.21 KN
Ast =
mm2
According to IS 456-200 clause 26.5.2.1 Minimum steel 0.15 % for Fe250 and 0.12 % for Fe415
mm2
Area of one bar x 1000 / required area in m2 /m
mm2 .
Vu =
Vu / b x d
12.21 x 103 / 1000 x 117
N / mm2 ح ) >C )
N / mm2
for Pt = 0.224 حc = 0.28 < 0.28
Assuming L0 =
Pt = 100 x A
s / b x d
Pt = 50 1-√1-(4.6 / fck) x (M
u / b x d2)
fy / f
ck
Mu1
/ b x d2 =
Mu1
= 0.499 x 1000 x 1172 x 10-6
Vu =
1.3 x ( Mu1
/ Vu ) + L
0 ≥ L
d
727.191 + 12 O
727.19which gives 16.91 mm ……………….( O.K.)
Check for deflection : -
Basic ( span / d ) ratio = 20
100 x 280 / 1000 x 117= 0.239
IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement
modification factor = 2
( span / d ) ratio permissible = 2 x 20
= 40Actual (span / d ) ratio = 2100 / 117
= 17.95 < 40 ……………….( O.K.)
Design of flight : -
Loads :
Inclined length of waist slab for one step = = 296.8 mm
Assuming 150 mm thick waist slab
self-load in plan = ( 296.8 / 250 ) x 0.15 x 25 = 4.45Floor finish length for one step = 160 + 250
= 410 mm
floor finish = ( 410 / 250 ) x 1 = 1.64 250
Weight of step ( ( 0 + 160 ) / 2000 ) x 25 = 2.00 160 296.8
Live load = 3.00Total = 4.45 + 1.64 + 2.00 + 3.00
= 11.09
16.64IS 456-2000 , clause 33.2
16.64 KN / m
11.63 KN / m 11.63 KN / m
525 2250 525 525 = 900 / 2 +75
1.3 x ( 6.83 x 106 / 12.21 x 103 ) + 12 O ≥ 55 O≥ 55 O
43 O ≤O ≤
Pt = 100 x A
st / b x d =
Span and loading on both the flights are the same. Therefore same design will be adopted.
√ 2502 + 1602
KN / m2
KN / m2
KN / m2
KN / m2
KN / m2
Pu = 1.5 x 11.09 = KN / m2
Landing is spanning in transverse direction. The span of stair according to discussion made in art 13-1 and loading for 1 m width of stair is shown in fig below.
RA = 24.83 KN RB = 24.83 KN
3300
0.525 x 11.63 + 16.64 x 2.25 / 2 = 6.11 + 18.72= 24.83 KN
= 40.97 - 15.83 - 3.17= 21.97 KNm
From Table 6-3 Q = 2.22
== 99.48 mm,
150 - 15 - 6= 129 mm, ……………….( O.K.)
= 1.32
=250 / 15
= 50 [(1-0.77) x 15 / 250 ]= 0.687%
0.687 x 1000 x 129 / 100
= 886Provide 12 mm O bar
spacing of bar == 113.04 x 1000 / 886= 127.585 mm
Provide 12 mm O bar@125 mm c/c = 904Minimum steel is 0.15 % for mild steel and 0.12 % for HYSD Fe415 reinforcement
Distribution steel = ( 0.15 / 100 ) x 1000 x 150
= 225Provide 8 mm O bar
spacing of bar == 50.24 x 1000 /225= 223 mm
Provide 8 mm O bar @ 220 mm c/c = 228
RA = R
B =
Mu = 24.83 x 3.3/2 - 11.63 x (3.3/2)2 / 2 - ( ( 16.64 - 11.63 ) x ( 2.25/2 )2 ) / 2
drequired
= √M / Q x b
√21.97 x 10 6 / 2.22 x 1000
dprovided
=
Mu / b x d2 = 21.97 x 10 6 / 1000 x (129)2
Pt = 50 1-√1-(4.6 / fck) x (M
u / b x d2)
fy / f
ck
50 1-√1-(4.6 / 15) x (1.32)
Ast =
mm2
Area of one bar x 1000 / required area in m2 / m
mm2.
mm2
Area of one bar x 1000 / required area in m2 /m
mm2 .
Check for shear : -
24.83 KN
Shear stress =
=
= 0.192 ( too small )
100 x 904 / 1000 x 129= 0.70
from table 7-1
IS 456-2000 clause 40.2.1.1k = 1.3 for 150 mm slab depth
Design shear strength = 1.3 x 0.524
= 0.681 ……………….( O.K.)Check for development length : -
12 O (mild steel )
= 100 x 904 / 1000 x 129= 0.70
From equation
we get , 1.344
= 22.37 KNm
24.83 KN
= O x 0.87 x 250 / 4 x 1 =54.4 O
1171.2 + 12 O1171.2
which gives 27.62 mm ……………….( O.K.)From Crossing of bars, the bars must extend upto 54.4 x 12 = 653
say 700 mm.Check for deflection : -
Basic ( span / d ) ratio = 20
100 x 904 / 1000 x 129= 0.70
IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementmodification factor = 1.62( span / d ) ratio permissible = 1.62 x 20
Vu =
Vu / b x d
24.83 x 103 / 1000 x 129
N / mm2 ح ) >C )
N / mm2
100 x As / b x d =
for Pt = 0.7 حc = 0.524 N / mm2
N / mm2
Assuming L0 =
Pt = 100 x A
s / b x d
Pt = 50 1-√1-(4.6 / fck) x (M
u / b x d2)
fy / f
ck
Mu1
/ b x d2 =
Mu1
= 1.344 x 1000 x 1292 x 10-6
Vu =
Development length of bars Ld = O σ
s / 4 x ح
bd
1.3 x ( Mu1
/ Vu ) + L
0 ≥ L
d
1.3 x ( 22.37 x 106 / 24.83 x 103 ) + 12 O ≥ 54.4 O≥ 54.4 O
42.4 O ≤O ≤
Pt = 100 x A
st / b x d =
= 32.4Actual (span / d ) ratio = 3300 / 129
= 25.58 < 32.4 ……………….( O.K.)The depth could be reducedCheck for cracking : -IS 456-2000 , clause 26.3.3
Main bars : maximum spacing permitted = 3 x effective depth of slab or 300 mm whichever is small= 3 x 129 = 387 mm or 300 mm
spacing provided = 125 mm < 300 mm ……………….( O.K.)Distribution bars : maximum spacing permitted = 5 x effective depth of slab or 450 mm whichever is small
= 5 x 129 = 645 mmspacing provided = 220 mm < 450 mm ……………….( O.K.)
175 mm to 200 mm
250 mm to 280 mm
check for shear
IS 456-2000 , Table 19Table 7-1
Concrete grade
M15 M20 M25 M30 M350.28 0.28 0.29 0.29 0.29
0.25 0.35 0.36 0.36 0.37 0.37
0.50 0.46 0.48 0.49 0.50 0.50
0.75 0.54 0.56 0.57 0.59 0.59
1.00 0.60 0.62 0.64 0.66 0.67
1.25 0.64 0.67 0.70 0.71 0.731.50 0.68 0.72 0.74 0.76 0.781.75 0.71 0.75 0.78 0.80 0.82
2.00 0.71 0.79 0.82 0.84 0.86
2.25 0.71 0.81 0.85 0.88 0.902.50 0.71 0.82 0.88 0.91 0.932.75 0.71 0.82 0.90 0.94 0.963.00 0.71 0.82 0.92 0.96 0.99
The above given table is based on the following formula
6 x β
β =
IS 456-2000 , Table 20
Table 7-2
Design shear strength of concrete , حC, N / mm2
Pt = 100 x A
s
b x d≤ 0.15
Design shear strength حc = 0.85 √ 0.8 x f
ck ( √ 1 + 5 x β - 1 )
0.8 x fck
/ 6.89 Pt , but not less than 1.0
Concrete grade M15 M20 M25 M30 M35
2.5 2.8 3.1 3.5 3.7check for development lengthIS 456-2000 clause 26.2.1
IS 456-200 clause 26.2.1.1Table 7-5
Concrete grade M15 M20 M25 M30 M35
1.0 1.2 1.4 1.5 1.7
For mild steel Fe250
For Fe415IS 456-2000 clause 26.2.3.3
Table 7-6Development length for single mild steel bars
Tension bars Compression bars
M15 M20 M15 M20
250 55 O 26 O 44 O 37 O
415 56 O 47 O 45 O 38 O500 69 O 58 O 54 O 46 O
check for deflectionBasic values of span to effective depth ratios for spans upto 10 m :
cantilever 7simply supported 20
continuous 26
check for cracking(a)The horizontal distance between parallel main
Maximum shear stress , حC, N / mm2
ح ) c )
max N / mm2
Development length of bars Ld = O σ
s / 4 x ح
bd
Design bond stress (حbd
) for plain bars in tension
ح ) bd
N / mm2
Note-1 : حbd
shall be increased by 25 % for bars in compressionNote-2 : In case of deformed bars confirming to IS : 1786-1985 , the value of ح
bd shall be increased by 60 %.
σs = 0.87 x f
y
σs = 0.67 x f
y
Ld ≤ M1 / V + L
0
L0 = effective depth of the members or 12 O , whichever is greater
if ends of the reinforcement are confined by a compressive reaction M1 / V increased by 30 % ( 1.3 M1 / V )
fy N / mm2
For spans above 10 m, the values in (a) may be multiplied by 10 / span in metres , except for cantilever in which case deflection calculations should be made.
reinforcement bars ( spacing )shall not be more than ttimes the effective depth of solid slab or
300 mm whichever is smaller.(b)The horizontal distance between parallelreinforcement bars ( spacing ) provided against
shrinkage and temperature (distribution bar) shall not be morethan five times the effective depth of a solidslab or 450 mm whichever is smaller.
IS 456-2000 Clause 40.2.1.1
275 250 225 200k 1.00 1.05 1.10 1.15 1.20
sketch :The reinforcements for both the flights are shown in fig.
700
700
Overall depth of slab , mm
300 or more
v vv v
v
10 O @ 280 c/c ( Landing )
Flight-B
Flight-A
v vv
vv
v vv
v vv
10 O @ 280 c/c ( Landing )
12 O @ 125 c/c150 mm thick waist slab
8 O @ 220 c/c
12 O @ 125 c/c
vv
vv
vv
vv
v
v
vv v v v v
8 O @ 220 c/c
12 O @ 125 c/c
10 O @ 280 c/c ( Landing )
150 mm thick waist slab
3 x effective depth of slab or 300 mm whichever is smallor 300 mm i.e. 300 mm
……………….( O.K.)5 x effective depth of slab or 450 mm whichever is small
or 450 mm i.e. 450 mm……………….( O.K.)
IS 456-2000 , Table 19Table 7-1
Concrete grade
M400.300.38
0.51
0.60
0.68
0.740.790.84
0.88
0.920.950.981.01
IS 456-2000 , Table 20
Table 7-2
, N / mm2
1 + 5 x β - 1 )
, but not less than 1.0
M40
4.0
IS 456-200 clause 26.2.1.1Table 7-5
M40
1.9
&
Basic values of span to effective depth ratios for spans upto 10 m :
, N / mm2
) for plain bars in tension
shall be increased by 25 % for bars in compressionNote-2 : In case of deformed bars confirming to IS : 1786-1985 , the
= effective depth of the members or 12 O , whichever is greater For checking development length , l
0
may be assumed as 8 O for HYSD bars ( usually end anchorage is not provided ) and 12 O for mild steel ( U hook is provided usually whose anchorage length is 16 O.
if ends of the reinforcement are confined by a compressive reaction M1 /
For spans above 10 m, the values in (a) may be multiplied by 10 / span in metres , except for cantilever in which case deflection
1751.25 1.30
150 or less
150
v vv
v vv
v v vvv v v
vv
vv
vv
vv
10 O @ 280 c/c ( Landing )
12 O @ 125 c/c
8 O @ 220 c/c
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