Download Box Culvert by B.C.punmia Example 30.2...
DESIGN OF BOX TYPE CULVERT 1 In side side diam diamen enti tion ons s
3.50
2 Supe Superr impo impose sed d load load
12000
N/m3
3 Live load
45000
N/m2
4 Wieg Wieght ht of so soil
18000
N/m2
5 Angl Angle e of of rep repos ose e
30
Degree
6 Nomi Nomina nall cove coverr top top / bot botto tom m
50
mm
6 Cocrete
3.50
Thickness of top slab Thickness of bottom slab
25000
N/m2
330 320 350
8 Thic Thicke kess ss of side side wall wall
50
Nominal cover side m
ter side σst
415
m
wt. of concrete
7
σcbc
9
x
20
M-
7 Steel
m
N/m2 σst
150
mm kg/m3
13 190
mm
thickness of side wall is OK
mm
O.K.
N/m2
mm
Reinforcement
Top slab
Bottom slab
mm c/c
mm Φ @
130 130 200 120 120 300 300
8
mm Φ @
130
mm c/c
260
mm c/c
Main
20
mm Φ @
Distribution
8
mm Φ @
At supports
mm Φ @
Vertical
8 20 8 8 20
Distribution
Main Distribution At supports
Side vertical wa wall
20 mm Φ @
mm Φ @ mm Φ @ mm Φ @
8 mm Φ @
200
mm c/c mm c/c mm c/c mm c/c mm c/c
Through out slab at bottom
mm c/c
Both side
mm C/C
8 mm Φ @
130
mm C/c
320 700
20 mm Φ @
20 mm Φ @
130 mm C/C
300 mm C/C 3.50
8 mm Φ @ 130 mm C/C 20 mm Φ @ 120 mm c/c 350 20 mm Φ @ 240 mm c/c
8 mm Φ @ 200
C/C
8 mm Φ @ 130
C/
O.K.
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DESIGN OF BOX TYPE CULVERT 1 In side diamentions
3.5
2 Su Super imposed load
12000
N/m3
3 Live Live load load
45000 45000
N/m2
4 Wieght of soil
18000
N/m2
5 Angle of repose
30
Degree
6 Nominal cover top/bottom
50
mm
6 Cocrete 7 Steel
M-
20
σcbc
7
Fy
1 Soluti Solution on
x
3.5
m
9800
N/m3
Nominal cover Side
50
mm
wt. of concrete
25000
N/m3
wt. of water
N/m2
m Out side σst
190
N/m2
water side side σst
150
N/m2
415
13
Genra Genrall For the purpose of design , one metre length of the box is considered.
The analysis is done for the f ollowing cases. (I) Live load, dead load and earth prssure acting , with no water pressure from inside. (II) Live and dead load on top and earth pressure acting from out side, and water pressure acting from ins with no live load on sides (III) Dead load and earth pressure acting from out side and water pressure from in side. Let the thicness of Horizontal slab
330 mm
=
0.33 m
Vertical wall thicness
320 mm
=
0.32 m
Effective slab span
3.5
+
0.33 =
3.83 m
Effective Height of wall
3.5
+
0.32 =
3.82 m
2 Case 1 : Dead and live load from out side of while no water pressure from inside. Self weight og top slab =
0.33 x
Live load and dead load =
45000 +
Weight of side wall = ∴
3.82 x
Upward soil reaction at base = ( 65250 x
25000 =
8250
N/m2
=
57000
N/m2
Total load on top =
65250
N/m2
25000
30560
N/m
1x
1x
12000 0.32 x 3.83 )+(
=
2x
30560 )=
3.83 Ka = ∴
1 - sin 30 1 + sin 30
=
1-
0.5
1+
0.5
0.5
=
1.5
81208.22 N/m2 1
=
= 0.33
3
Latral pressure due to dead load and live load = Pv x Ka p=
57000 x
Latral pressure due to soil Ka x w x h
=
0.33 x
Hence total pressure
=
19000 +
Latral presure intencity at top
=
19000
Latr Latral al press ressu ure inte ntenci ncity at botto ottom m
=
19000+ 9000+
0.33 =
19000
18000 h =
6000
h
6000 h N/m2 6000 000 x w =
Fig 1 show the box culvert frame ABC D, D, along with the external loads, Due to symmetry, half of frame (i.e. AEFD) of box culvert is considered for moment distribution. Since all the members have uniform thickness, and uniform diamentions, the relative stiffness K for AD will be equal to 1 while the relative stiffness for AE AE and DF will be 1/2.
N/m2
3.82 =
41920
N/m2
65250 N/m2
19000
19000 A
h
E
B
3.83
19000
3.82
6000 6000 h D
F
C
41920
19000 ### N/m2
22920
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1
Distribution factore for AD and DA=
=
1+1/2
MFAD = + MFAD = +
pL2 12 19000
###
12 3.82 2=
### x
=
12
=
1/3
N-m
98751.91 N - m
12
Where W is the total tringular earth pressure.
3.82
2
+ 22920 x
12
3.82
2
x
3.82 15
=
34254
N-m
WL
-
12
15
19000 x
MFDA = -
wL2
3.82 2=
65250 x
1+1/2
15
x
pL2
MFDA = -
+
WL
+
=
12
Mfdc=
1/2
Distribution factore for AB and DC=
wL2
MFAB=
Fix end moments will be as under :
2/3
3.82
2-
- 22920 x
12
3.82
2
x
3.82 10
= -23105
-16723
=
-39
The Moment distribution is carried out as illustrate in table Fixed End Moments
Member
DC
DA
AD
AB
98751.91
-39828
34254
-79346
55075
65250
46852
The moment distribution carried out as per table 1 for case 1 Joint
D
A
124627.5
1246
Member
DC
DA
AD
AB
Distribution factore
0.33
0.67
0.67
0.33
98751.91
-39828
34254
-79346
A
A
-19641
-39283
30061
15031
55075
1.91
15031
-19641
-10020
13094
6547
-5010
-4365
3340
1670
-2182
-1113
1455
727
-557
-485
371
186
-242
-124
162
81
-62
-54
41
21
-27
-7
-14
18
9
71023
-71023
55075
-55075
Fix end moment Balance Carry over balance
-5010
Carry over balance
-2182
Carry over balance
-557
Carry over balance
-242
Carry over balance
-62
Carry over balance
-27
Carry over balance Final moment
46852
6547
0.5
x
65250
1670
71023
727
41920
186
155514
D 69810
71023 155
69810
81 81208
21
Fig 2
x
65250
3.82 =
Vertical reaction at D and C =
3.83
81208.22 x
=
N/m2.
124627.5 N/m2
### N/m2
x
1.91
22920 D
Similarly, for the Bottom slab DC carrying U.D.L.loads @ 0.5
7197
3.82 m
For horizontal slab AB, carrying UDL @ Vertical reactionat a and B =
55075
19000
###
N
The body diagram for various members, including loading, B.M. And reactions are shown in fig.2 For the vertical member AD, the horizontal reaction at A is found by taking moments at D.Thus ( -ha x
3.83
)+ +
55075 - 71023 + 19000 x 1/2
22920
3.83 3.83
x
3.83 3 83
x
1/2 1/3
Hence , hd =( 19000 + 41920 )x
3.83 -
46852 =
69810 N
2
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65250
Free B.M. at mid point E =
x
3.83
2
=
Net B.M. at E =
119644 81208.22
Similarly, free B.M. at F =
-
55075
x
2
3.83
= =
148904.42
-
64569
###
8
Net B.M. at F =
N-m
119644
8
71023
N-m
N -m
=
77881
N-m
For vertical member AD , Simply supported B.M. At mid span 19000
imply supporetd at mid sapn
x
3.83
2
+
1/16
8 71023
Net B.M. =
+
55075
x 22920 x
=
2
63049
-
3.83
2
=
###
###
=
7197
N-m
3 Case 2 : Dead load and live load from out side and water pressure from inside. 9800 x
In this case , water pressure having an intensity of zero at A and At D, is acting, in addition to the pressure considered in case 1. The various pressures are marked in fig 3 .The vertical walls will thus be subjected to a net latral pressure of Itensity And
= 19000 = =
-
= 37436 N/m2
65250 N/m2
w = 19000
19000
N/m2 At the Top
41920
3.82
A
E
190
14516
B
37436
4484 N/m2 at the bottom
3.83 3.82
F
D
C
41920
41920 w
4484
### N/m2
=
Fig 3 Fix end moments will be as under :
wL2
MFAB=
12
MFAD = + MFAD = + MFDA = MFDA = -
pL2
+
12 4484
x
=
12
WL
-
12 4484 x
3.83 2=
### x 12
N-m
99269.61 N - m
Where W is the total tringular earth pressure.
10 3.83
2
+ 14516 x
12 pL2
###
12
wL2
Mfdc=
3.83 2=
= 65250 x
3.83
2
x
3.83 10
=
16128
N-m
=
-12579
N -m
WL 15 3.83
2
-
14516 x
12
3.83
2
x
3.83 15
The moment distribution is carrired out as illustred in table. Fixed End Moments
Member
DC
DA
AD
AB
99269.61
-12579
16128
-79762
45069
65250
23451
The moment distribution carried out as per table 1 for case 1 Joint
D
A
124627.5
Member
DC
DA
AD
AB
Distribution factore
0.33
0.67
0.67
0.33
19000
99269.61
-12579
16128
-79762
A
Fix end moment
1246
23451
45069 A
balance
-7070
-14141
19265
9632
-7070
-6422
4714
2357
-3211
-1571
2141
1070
-786
-714
524
262
-357
-175
238
119
-87
-79
58
29
-40
-10
-19
26
13
58813
-58813
45069
-45069
Carry over balance
-3211
Carry over balance
-786
Carry over balance
-357
Carry over balance
-87
Carry over balance
-40
Carry over balance Final moment
9632
3.82
2357
58813 4484 D
1070
0.5
x
65250
D 21404
262
58813
155108
1551
21404
119 81208
29
Fig 4
For horizontal slab AB, carrying UDL @ Vertical reactionat a and B =
1.91
x
N/m2.
65250
3.82 =
124627.5 N/m2
Similarly, for the Bottom slab DC carrying U.D.L.loads @ 81208 N/m2 Vertical reaction at D and C =
0.5
x
81208
x
3.82
=
###
N
The body diagram for various members, including loading, B.M. And reactions are shown in fig.3 For the vertical member AD, the horizontal reaction at A is found by taking moments at D.Thus ( -ha x
3.82
)+
45069 - 58813 + 4484
+ -ha x
3.82 +
1/2
x
-13744
+
From which,
ha =
Hence , hd =( 4484
14516
x
3.82
x
3.82
x
1/2
x
3.82
x
3.82
x
2/3
32716.16 + 70607.76
23451
+ 19000 )x
3.82 -
23451 =
21404 N
2 65250
Free B.M. at mid point E =
3.82
2
=
119020
8
Net B.M. at E =
119020
-
81208
x
Similarly, free B.M. at F = Net B.M. at F =
x
45069 3.82
=
-
73951
N-m
2
=
###
8 148127.86
N-m
58813
N -m
=
89315
N-m
For vertical member AD , Simply supported B.M. At mid span 4484
imply supporetd at mid sapn
x
3.82
2
+
1/16
8 58813
Net B.M. =
+
45069
x 14516 x
=
2
51941
3.82
2
= 21418
- 21418 =
30523 N-m
4 Case 3 : Dead load and live load on top water pressure from inside no live load on side. in this case, it is assume that there is no latral oressure due to live load . As before . The top slab is subjected to a load of '=
65250
N/m2
and the bottom slab is subjected to a load Itensity = 81208. N/m2 Lateral pressure due to dead load = 1/3 x
12000
Lateral pressure due to soil 1/3 x
18000
=
w = 4000
Hence earth pressure at depth h is =
4000 A
4000
E
B
N/m2
= =
65250 N/m2
3.83
6000
N/m2
3.82
4000
Earth pressure intensity at top
=
4000
37436 w=
N/m2
### N/
37436
Fig 5 Earth pressure intensity at Bottom=
### +
6000 x
3.82
=
26920
N/m2
In addition to these, the vertical wall lslab subjectednto water pressure of intensity ZERO at top and
374
N/m2 at Bottom, acting from inside . The lateral pressure on vertical walls Is shown in fig 5 and 6
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wL2
MFAB=
Fix end moments will be as under :
12 wL2
Mfdc= MFAD = + MFAD = + MFDA = MFDA = -
pL2 12 4000
4000
3.83 2=
### x 12
N-m
99269.61 N - m
Where W is the total tringular earth pressure.
3.83
2
-
14516 x
12
3.83
2 WL
+
12
###
12
15
x
pL2
3.83 2=
65250 x
=
12
WL
-
=
### -
x
3.83 15
=
-2209
N-m
=
5757
N -m
###
10
x
3.83
2
-
14516 x
12
3.83
2
x
3.83 10
The moment distribution is carrired out as illustred in table. Fixed End Moments
Member
DC
DA
AD
AB
99269.61
5757
-2209
-79762
35902
65250
3
=
The moment distribution carried out as per table 1 for case 1 Joint
D
A
124627.5
1246
Member
DC
DA
AD
AB
Distribution factore
0.33
0.67
0.67
0.33
4000
99269.61
5757
-2209
-79762
A
A
-35009
-70018
54647
27324
35902
1.91
27324
-35009
-18216
23339
11670
-9108
-7780
6072
3036
-3890
-2024
2593
1297
-1012
-864
675
337
-432
-225
288
144
-112
-96
75
37
-48
-12
-25
32
16
49646
-49646
35902
-35902
Fix end moment Balance Carry over balance
-9108
Carry over balance
-3890
Carry over balance
-1012
Carry over balance
-432
Carry over balance
-112
Carry over balance
-48
Carry over balance Final moment
11670
3.82
3036
0.5
x
65250
1.91
0D 1297
14516
337
155108
8
48748
49646
D 49646
144 81208 37
Fig 4
x
65250
3.82 =
N/m2.
124627.5 N
Similarly, for the Bottom slab DC carrying U.D.L.loads @ 81208 N/m2 Vertical reaction at D and C =
0.5
x
81208
x
3.82
=
###
9 155
5200
For horizontal slab AB, carrying UDL @ Vertical reactionat a and B =
35902
N
The body diagram for various members, including loading, B.M. And reactions are shown in fig.6
-ha x
3.82 +
1/2
x
14516
x
-13744
+
29184.8
-
From which,
ha =
3.82
x
3.82
x
1/3
-
5200 = 7245.56
35304
5200
Hence , hd =( 14516 x
3.82 )-
4000 x
3.82
119644
N-m
2 65250
Free B.M. at mid point E =
x
2
3.83
=
8
Net B.M. at E =
119644 81208
Similarly, free B.M. at F =
-
35902
x
2
3.83
= =
###
8
Net B.M. at F =
148904.42
-
83742
49646
N-m
N -m
=
99258
N-m
For vertical member AD , Simply supported B.M. At mid span 4000
Simply supporetd at mid sapn =
x
2
+
3.83
1/16
8 49646
Net B.M. =
+
35902
x 14516 x
=
2
42774
3.83
2
###
=
+
= 5973.91 48748 N-m
5 Design of top slab : Mid section The top slab is subjected to following values of B.M. and direct force Case
B.M. at Center (E)
B.M. at ends (A)
Direct force (ha)
(i)
64569
55075
46852
(II)
73951
45069
23451
(II)
83742
35902
5200
The section will be design for maximum B.M. =
83742
N -m
for water side force σst
= σcbc = m k=
=
150
N/mm2
=
7
N/mm2
=
13
wt of water = 9800 N/mm2 for water side f
m*c
m*c+σst
j=1-k/3 = R=1/2xc x j x k =
wt. of concrete = 25000 N/m3
1 0.5
13
= x
13 x
7
0.378 /
3
7
Provide over all thickness Mr = R . B .D
2
=
1.155 x
x
BMx100/σstxjxD=
using
20 mm Φ bars
Spacing of Bars = x1000/Ast Hence Provided Acual Ast provided
Ast using
= 8
0.24
mm Φ bars
=
0.3
= 0.874
J
=
0.8
x 0.378 = 1.155
R
=
1.1
=
320 mm so effective thicknesss 270
2
=
84216794
83742000 150 x A
=
314
x
20
0.874 x
=
0.1
320
x
A
=
10
3.83 /
x(
320 -
=
3.14xdia2 4 100
3.14 x
83742000
133
O.K.
x
20
say
=
100
759 mm2
130
mm
mm2
2415
100
=
= 314
mm c/c
5
=
20 4
130
130 =
= 450 -
mm
2365 mm
2365 =
mm Φ Bars @ 314 /
270
2
=
4 x100 1000 /
>
=
270
3.14xdia2
1000 x
0.3 x
K
150
0.87
Bend half bars up near support at distance of L/5 Area of distributionn steel =
= 0.378
+
x
1000 x
Ast =
7
=
0.80 m 0.24
%
area on each face= 3.14 x
8 4
x
8
# =
50
Section at supports :- Maximum B.M.=
55075
N-m. There is direct compression of
46852 N also.
But it effect is not considered because the slab is actually reinforced both at top and bottom . σst =
Since steel is at top k=
0.32
J=
concrete M
0.89
R=
55075000
=
Ast
∴
N/mm2
190
190 x
0.89 x
=
270
20
1.01 mm2
1204
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Area
available from the bars bentup from the middle section = 1204 <
###
2415
/
2
=
mm2
###
Hence these bars will serve the purpose. However, provide 8 mm dia. Additional bars @ 200 mm c/c
6 Design of bottom slab: The bottom slab has the following value of B.M. and direct force. Case
B.M. at Center (F)
B.M. at ends (D)
Direct force (ha)
(i)
77881
71023
69810
(II)
89315
58813
21404
(II)
99258
49646
7246
The section will be design for maximum B.M. =
99258
N -m
for water side force σst
= σcbc = m k=
=
150
N/mm2
=
7
N/mm2
=
13
m*c+σst 1 0.5
13
= -
7
0.378 /
3
x
1000 x
7
x
7
= 0.378
K
=
0.3
= 0.874
J
=
0.8
x 0.378 = 1.155
R
=
1.1
+
0.87
=
1.155
Provide thickness of bottom slab D=
x
13 x
99258416
d =
Ast =
wt of water = 9800 N/mm2 for water side f
m*c
j=1-k/3 = R=1/2xc x j x k =
wt. of concrete = 25000 N/m3
294
150
D
mm
350 mm so that d = 99258416
BMx100/σstxjxD=
150 x
300 =
0.874 x
20
mm bars
Spacing of Bars = x1000/Ast
A
=
314
x
20
Hence Provided Acual Ast provided
Area of distributionn steel = Ast
=
0.23
using 8
0.3
1000 / 314 /
x
3.83 /
x(
350 -
450 -
350
x
A
=
50
x
mm bars
Spacing of Bars = Ax1000/Ast =
0.1
=
8
Hence Provided
10
=
say
=
100
800 mm2
4 x100
400 =
mm Φ Bars @
20
=
= 314
120
mm
mm2
2617 5
=
x
mm c/c
100
3.14xdia2 1000 /
124
120
120 =
20 4
2523 =
mm Φ Bars @
-
mm
3.14 x
=
4 x100
Bend half bars up near support at distance of L/5
mm
300
3.14xdia
1000 x
344
2523 mm2
2
using
=
=
0.80 m 0.23 %
area on each face= 3.14 x
8
x
8
4 126
120
say
=
120
400 =
50 mm
mm c/c on each face
Section at supports :- Maximum B.M.=
71023
N-m. There is direct compression of
69810 N also.
But it effect is not considered because the slab is actually reinforced both at top and bottom . Si
t
l i
tt
σ
190
N/mm2
t M
20
190 x Area
0.89 x
300
available from the bars bentup from the middle section = 1397 >
###
mm bars
Spacing of Bars = Ax1000/Ast = Hence Provided
/
2
=
mm2
###
Fail , hence additional reinforcement will provided. Additional reinforcemet required
using 8
2617
8 mm Φ Bars @
A
=
50
x
3.14xdia2
=
3.14 x
=
4 x100 1000 /
89
mm2
88.67
8
x
8
=
4
=
567
say
560
=
50 mm
300 mm c/c throught out the slab, at its bottom.
7 Design of side wall: The side wall has the following value of B.M. and direct force. Case
B.M. at Center (F)
B.M. at ends (D)
Direct force (ha)
(i)
7197
71023
155514
(II)
30523
58813
155108
(II)
48748
49646
155108
The section will be design for maximum B.M. =
71023
71023
Eccentricity =
N -m,
x
1000
=
155514
proposed thickness of side wall '=
330 mm
∴
and direct force
e/D
=
457
mm
457 /
330 =
155514
1.38 <
1.5
thickness of side wall is OK
20
Let us reinforce the section with in fig xxx . With cover of Asc = Ast =
50 1000 300
mm
and D
=
20
x
3.14 x
x
300 mm c/c provided on both faces, as sho
mm Φ bars @
330 mm 20
4
=
mm2
1047
The depth of N.A. is computed from following expression: bn
D - dt -
3
n + (m - 1)Asc 1 (n - dc)(D - dt- dc) n
3
b n + (m -1) Asc
n - dc n
1000 n or
2 1000 n+ 2
500 n 500 n+
140000 n 500 n
- m Ast
330
-
12
x
280
-
12560 n
+
50 ### n n 3
=
e+
+
12
x
n
-
50
-
n
-
50
x
D - dt - n
- dt
2
n
x
+
n 3
x
n
-
###
n2
+ -1256000
12560
D
-
50
-
628000 n
### n
### n 13
x x
n ### n
-
50
x 330
x -
50
-1256000 n
x
280
= 457 + 115 -
n
-62800000
-
n
-
### n
= +
572
###
multiply by n 140000 n2
-
500 2
+
### n3 12560
+
-1256000 n 628000
-
-62800000 ###
+
###
=
572
286000 n2
+
14967333 n
-146000 n2
-
13711333 n
-876 n2
+
82268
Solwing this trial and error we get, n ∴
= (
c'
500
x
n =
-
-2475659733
= 167 n3
-
14853958
= n3
91.47 mm 12 x
91.47 +
x(
2538459733
###
( 91.47 -
91.47
330 -
50 -
91.47 )
137.32 x
41.47 -
### x
50
)-
13 x
###
91.47
[email protected]
or
### + ∴
c' =
Also stress in steel t =
=
155514 m c' n
6.65
=
23383
(D-dc-n)
178.21 N/mm2
= <
7
< 13 x
6.65
91.47
190 N/mm2
188.53 = N/mm2
x(
330
23383
Stress is less than permissi -
50
O.K.
Stress in steel is less than permissiable Hence section is O.K.
-
91.47
)
[email protected]
ide,
28
55075
27.5
64569
77881
14
00 m a r g a i d e r u s s e r p l a r t a l t e N
45069
27.5
73951
89315
08
m a r g a i d e r u s s e r p l a r t a
36
5902
27.5 3742
9258 08
orce 78 74 55
mm2
mm2 mm2
orce 78 74 55
mm2
mm2 mm2
mm2
N
OK n
-100 -n
ble
Box culverts
20 m Φ @
260
mm c/c
8 mm Φ @
200
mm C/C
8 mm Φ @
130
mm C/c
320 700
20 mm Φ @
20 mm Φ @
130 mm C/C
300 mm C/C 3.50
8 mm Φ @ 130 mm C/C 20 mm Φ @ 120 mm c/c 350 20 m Φ @ 240 mm c/c 330
8 mm Φ @
8 mm Φ @
200 mm C/C
130 mm C/c
3.50
330
Table 3.1
Table 1.15. PERMISSIBLE DIRECT TENSILE STRESS Grade of co
Tensile stress N/mm2
10
M-10
M-15
M-20
M-25
M-30
M-35
M-40
1.2
2.0
2.8
3.2
3.6
4.0
4.4
b < 0 0. 0. 0.
Table 1.16.. Permissible stress in concrete (IS : 456-2000) Permission stress in compression (N/mm2)
Grade of concrete M M M M M M M M M
Bending αcbc
(N/mm2) 3.0 5.0 7.0 8.5 10.0 11.5 13.0 14.5 16.0
10 15 20 25 30 35 40 45 50
Kg/m2 300 500 700 850 1000 1150 1300 1450 1600
Direct (αcc) (N/mm2) 2.5 4.0 5.0 6.0 8.0 9.0 10.0 11.0 12.0
Kg/m2 250 400 500 600 800 900 1000 1100 1200
1.
Permissible stress in bond (Average) for plain bars in tention (N/mm 2)
1. 1. 1. 2. 2. 2. 2.
(N/mm2) in kg/m2 --0.6 60 0.8 80 0.9 90 1.0 100 1.1 110 1.2 120 1.3 130 1.4 140
3.00 an
Over all de
Table 1.18. MODULAR RATIO M-10 31 (31.11)
Grade of co
Modular ra
M-15 19 (18.67)
M-20 13 (13.33)
M-25 11 (10.98)
M-30 9 (9.33)
M-35 8 (8.11)
M-40 7 (7.18)
Table Grade of τc.
Table 2.1. VALUES OF DESIGN CONSTANTS Grade of concrete
M-15
M-20
M-25
M-30
M-35
M-40
Modular Ratio σcbc N/mm2
18.67 5
13.33 7
10.98 8.5
9.33 10
8.11 11.5
7.18 13
m σcbc
93.33
93.33
93.33
93.33
93.33
93.33
0.4
0.4
0.4
0.4
0.4
0.4
0.87
0.87
0.87
0.87
0.87
0.87
0.87
1.21
1.47
1.73
1.99
2.25
0.71
1
1.21
1.43
1.64
1.86
0.33
0.33
0.33
0.33
0.33
0.33
M 15
0.89
0.89
0.89
0.89
0.89
0.89
M 20
0.73
1.03
1.24
1.46
1.68
1.9
M 25
0.43
0.61
0.74
0.87
1
1.13
M 30
kc
0.29
0.29
0.29
0.29
0.29
0.29
jc
0.9
0.9
0.9
0.9
0.9
0.9
M 40
0.65
0.91
1.11
1.31
1.5
1.7
M 45
k (a) σst = c jc 140 N/mm2 Rc (Fe 250) P (%) c kc (b) σst = j c 190 Rc N/mm2 Pc (%) (c ) σst = 230 N/mm2
Rc
Grade of concre τbd (N / mm2
Grade of concrete
M 35
e
Pc (%)
Shear stress tc 100As bd 0.15 0.16 0.17 0.18 0.19 0.2 0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.29 0.3 0.31 0.32 0.33 0.34 0.35 0.36 0.37 0.38 0.39 0.4 0.41 0.42 0.43 0.44 0.45 0.46 0.47 0.48 0.49 0.5 0.51 0.52
M-20 0.18 0.18 0.18 0.19 0.19 0.19 0.2 0.2 0.2 0.21 0.21 0.21 0.22 0.22 0.22 0.23 0.23 0.24 0.24 0.24 0.25 0.25 0.25 0.26 0.26 0.26 0.27 0.27 0.27 0.28 0.28 0.28 0.29 0.29 0.29 0.30 0.30 0.30
0.31
0.44
Reiforcement % 100As M-20 bd 0.18 0.15 0.19 0.18 0.2 0.21 0.21 0.24 0.22 0.27 0.23 0.3 0.24 0.32 0.25 0.35 0.26 0.38 0.27 0.41 0.28 0.44 0.29 0.47 0.30 0.5 0.31 0.55 0.32 0.6 0.33 0.65 0.34 0.7 0.35 0.75 0.36 0.82 0.37 0.88 0.38 0.94 0.39 1.00 0.4 1.08 0.41 1.16 0.42 1.25 0.43 1.33 0.44 1.41 0.45 1.50 0.46 1.63 0.46 1.64 0.47 1.75 0.48 1.88 0.49 2.00 0.50 2.13 0.51 2.25
0.53
0.63
Degree 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 30 35 40 45 50 55 60 65
0.72
0.82
Value of angle sin cos 0.17 0.98 0.19 0.98 0.21 0.98 0.23 0.97 0.24 0.97 0.26 0.97 0.28 0.96 0.29 0.96 0.31 0.95 0.33 0.95 0.34 0.94 0.36 0.93 0.37 0.93 0.39 0.92 0.41 0.92 0.42 0.91 0.50 0.87 0.57 0.82 0.64 0.77 0.71 0.71 0.77 0.64 0.82 0.57 0.87 0.50 0.91 0.42
M 50
tan 0.18 0.19 0.21 0.23 0.25 0.27 0.29 0.31 0.32 0.34 0.36 0.38 0.40 0.42 0.45 0.47 0.58 0.70 0.84 1.00 1.19 1.43 1.73 2.14
0.53 0.54 0.55 0.56 0.57 0.58 0.59 0.6 0.61 0.62 0.63 0.64 0.65 0.66 0.67 0.68 0.69 0.7 0.71 0.72 0.73 0.74 0.75 0.76 0.77 0.78 0.79 0.8 0.81 0.82 0.83 0.84 0.85 0.86 0.87 0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1.00 1.01 1.02
0.30 0.30 0.31 0.31 0.31 0.31 0.31 0.32 0.32 0.32 0.32 0.32 0.33 0.33 0.33 0.33 0.33 0.34 0.34 0.34 0.34 0.34 0.35 0.35 0.35 0.35 0.35 0.35 0.35 0.36 0.36 0.36 0.36 0.36 0.36 0.37 0.37 0.37 0.37 0.37 0.37 0.38 0.38 0.38 0.38 0.38 0.38 0.39 0.39 0.39
1.03 1.04 1.05 1.06 1.07 1.08 1.09 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20 1.21 1.22 1.23 1.24 1.25 1.26 1.27 1.28 1.29 1.30 1.31 1.32 1.33 1.34 1.35 1.36 1.37 1.38 1.39 1.40 1.41 1.42 1.43 1.44 1.45 1.46 1.47 1.48 1.49 1.50 1.51 1.52
0.39 0.39 0.39 0.39 0.39 0.4 0.4 0.4 0.4 0.4 0.4 0.4 0.4 0.41 0.41 0.41 0.41 0.41 0.41 0.41 0.41 0.41 0.42 0.42 0.42 0.42 0.42 0.42 0.42 0.42 0.43 0.43 0.43 0.43 0.43 0.43 0.43 0.43 0.44 0.44 0.44 0.44 0.44 0.44 0.44 0.44 0.44 0.45 0.45 0.45
1.53 1.54 1.55 1.56 1.57 1.58 1.59 1.60 1.61 1.62 1.63 1.64 1.65 1.66 1.67 1.68 1.69 1.70 1.71 1.72 1.73 1.74 1.75 1.76 1.77 1.78 1.79 1.80 1.81 1.82 1.83 1.84 1.85 1.86 1.87 1.88 1.89 1.90 1.91 1.92 1.93 1.94 1.95 1.96 1.97 1.98 1.99 2.00 2.01 2.02
0.45 0.45 0.45 0.45 0.45 0.45 0.45 0.45 0.45 0.45 0.46 0.46 0.46 0.46 0.46 0.46 0.46 0.46 0.46 0.46 0.46 0.46 0.47 0.47 0.47 0.47 0.47 0.47 0.47 0.47 0.47 0.47 0.47 0.47 0.47 0.48 0.48 0.48 0.48 0.48 0.48 0.48 0.48 0.48 0.48 0.48 0.48 0.49 0.49 0.49
2.03 2.04 2.05 2.06 2.07 2.08 2.09 2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18 2.19 2.20 2.21 2.22 2.23 2.24 2.25 2.26 2.27 2.28 2.29 2.30 2.31 2.32 2.33 2.34 2.35 2.36 2.37 2.38 2.39 2.40 2.41 2.42 2.43 2.44 2.45 2.46 2.47 2.48 2.49 2.50 2.51 2.52
0.49 0.49 0.49 0.49 0.49 0.49 0.49 0.49 0.49 0.49 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51
2.53 2.54 2.55 2.56 2.57 2.58 2.59 2.60 2.61 2.62 2.63 2.64 2.65 2.66 2.67 2.68 2.69 2.70 2.71 2.72 2.73 2.74 2.75 2.76 2.77 2.78 2.79 2.80 2.81 2.82 2.83 2.84 2.85 2.86 2.87 2.88 2.89 2.90 2.91 2.92 2.93 2.94 2.95 2.96 2.97 2.98 2.99 3.00 3.01 3.02
0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51
3.03 3.04 3.05 3.06 3.07 3.08 3.09 3.10 3.11 3.12 3.13 3.14 3.15
0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51 0.51
. Permissible shear stress Table
c
in concrete (IS : 456-2000)
Permissible shear stress in concrete tc N/mm2
As d .15 5 0 5
M-15 0.18 0.22 0.29 0.34
M-20 0.18 0.22 0.30 0.35
M-25 0.19 0.23 0.31 0.36
M-30 0.2 0.23 0.31 0.37
M-35 0.2 0.23 0.31 0.37
M-40 0.2 0.23 0.32 0.38
0
0.37
0.39
0.40
0.41
0.42
0.42
5
0.40
0.42
0.44
0.45
0.45
0.46
0 5 0 5 0 5
0.42 0.44 0.44 0.44 0.44 0.44 0.44
0.45 0.47 0.49 0.51 0.51 0.51 0.51
0.46 0.49 0.51 0.53 0.55 0.56 0.57
0.48 0.50 0.53 0.55 0.57 0.58 0.6
0.49 0.52 0.54 0.56 0.58 0.60 0.62
0.49 0.52 0.55 0.57 0.60 0.62 0.63
200 1.20
175 1.25
above
Table 3.2. Facor k pth of slab
300 or more
1.00
275 1.05
250 1.10
.3. Maximum shear stress M-15 1.6
concrete ax
M-20 1.8
τc.max
M-25 1.9
225 1.15
15 0.6
20 0.8
25 0.9
M-30 2.2
M-35 2.3
bd
M-40 2.5
in concrete (IS : 456-2000)
30 1
35 1.1
40 1.2
Table 3.5. Development Length in tension Plain M.S. Bars τbd
(N / mm2)
H.Y.S.D. Bars
kd = Ld Φ
1.30
in concrete (IS : 456-2000)
able 3.4. Permissible Bond stress Table 10 --
150 or less
τbd
(N / mm2)
kd = Ld Φ
0.6
58
0.96
60
0.8
44
1.28
45
0.9
39
1.44
40
1
35
1.6
36
1.1
32
1.76
33
1.2
29
1.92
30
1.3
27
2.08
28
45 1.3
50 1.4
1.4
tan 0.18 0.19 0.21 0.23 0.25 0.27 0.29 0.31 0.32 0.34 0.36 0.38 0.40 0.42 0.45 0.47 0.58 0.70 0.84 1.00 1.19 1.43 1.73 2.14
25
Value of angle Degree sin 10 0.17 11 0.19 12 0.21 13 0.23 14 0.24 15 0.26 16 0.28 17 0.29 18 0.31 19 0.33 20 0.34 21 0.36 22 0.37 23 0.39 24 0.41 25 0.42 30 0.50 35 0.57 40 0.64 45 0.71 50 0.77 55 0.82 60 0.87 65 0.91
2.24
cos 0.98 0.98 0.98 0.97 0.97 0.97 0.96 0.96 0.95 0.95 0.94 0.93 0.93 0.92 0.92 0.91 0.87 0.82 0.77 0.71 0.64 0.57 0.50 0.42
26
