Box Culvert by B.C.punmia Example 30.2
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Transcript of Box Culvert by B.C.punmia Example 30.2
DESIGN OF BOX TYPE CULVERT
1 In side diamentions 3.50 m x 3.50 m
2 Super imposed load 12000
3 Live load 45000
4 Wieght of soil 18000
5 Angle of repose 30 Degree
6 Nominal cover top / bottom 50 mm Nominal cover side 50 mm
6 Cocrete M- 20 wt. of concrete 25000
7 m 13
7 Steel 415 150 190
8 Thickess of side wall 330 mm thickness of side wall is OK
Thickness of top slab 320 mm O.K.
Thickness of bottom slab 350 mm
9 Reinforcement
Top slab Main 20 130 mm c/c
Distribution 8 130 mm c/c
At supports 8 200 mm c/c
Bottom slab Main 20 120 mm c/c
Distribution 8 120 mm c/c
At supports 8 300 mm c/c Through out slab at bottom
Side vertical wall Vertical 20 300 mm c/c Both side O.K.
Distribution 8 130 mm c/c
20 260 mm c/c
8 200 mm C/C
8 130 mm C/c
320
700 20
20 130 mm C/C
300 mm C/C
3.50
8
130 mm C/C
20
120 mm c/c
350
20 8 8
240 mm c/c 200 mm C/C 130 mm C/c
330 3.50 330
N/m3
N/m2
N/m2
kg/m3
scbc N/m2
Water side sst N/m2 sst N/m2
mm F @
mm F @
mm F @
mm F @
mm F @
mm F @
mm F @
mm F @
mm F @
mm F @
mm F @
mm F @
mm F @
mm F @
mm F @
mm F @ mm F @ mm F @
DESIGN OF BOX TYPE CULVERT
1 In side diamentions 3.5 x 3.5 m
2 Super imposed load 12000
3 Live load 45000
4 Wieght of soil 18000 wt. of water 9800
5 Angle of repose 30 Degree
6 Nominal cover top/bottom 50 mm Nominal cover Side 50 mm
6 Cocrete M - 20 wt. of concrete 25000
7 m 13
7 Steel Fy 415 190
150
1 Solution Genral
For the purpose of design , one metre length of the box is considered.
The analysis is done for the following 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 inside,
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 1 x 1 x ### = 8250
Live load and dead load = 45000 + ### = 57000
Total load on top = 65250
Weight of side wall = 3.82 x 0.32 x ### = 30560 N/m
65250 x 3.83 )+( 2 x ### )=81208.2
3.83
Ka =1 - sin 30
=1 - 0.5
=0.5
=1
= 0.331 + sin 30 1 + 0.5 1.5 3
p = ### x 0.333 = 19000
Latral pressure due to soil Ka x w x h = 0.333 x ### h = 6000 h
Hence total pressure = ### + 6000 h
Latral presure intencity at top = 19000
Latral pressure intencity at bottom = ### + 6000 x 3.82 = 41920
w = 65250
19000 19000
A E B
h 3.83
19000 3.82
6000 h
D F C
41920 19000 22920
w = 81208
Fig 1
N/m3
N/m2
N/m2 N/m3
N/m3
scbc N/m2
Out side sst N/m2
water side side sst N/m2
N/m2
N/m2
N/m2
\ Upward soil reaction at base = (N/m2
\ Latral pressure due to dead load and live load = Pv x Ka
N/m2
N/m2
N/m2
Fig 1 show the box culvert frame ABCD, 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 and DF will be 1/2.
N/m2
N/m2
1= 2/3
1/2= 1/3
1+1/2 1+1/2
Fix end moments will be as under : =### x 3.82
-79346 N - m12 12
+ =### x 3.82
98751.9 N - m12 12
+ +WL Where W is the total tringular earth pressure.
12 15
+19000 x 3.82 ### x 3.82
x3.82
= 34254 N-m12 2 15
- -WL
12 15
-19000 x 3.82 ### x 3.82
x3.82
= -23105 -16723 = -3982812 2 10
The Moment distribution is carried out as illustrate in table
Fixed End MomentsMember DC DA AD AB 55075 65250 55075
98751.91 -39828 34254 -79346 46852
Joint D A 124627.5 124627.5
Member DC DA AD AB 4685255075
Distribution factore 0.33 0.67 0.67 0.33 19000 64569
Fix end moment 98751.91 -39828 34254 -79346 A A
Balance -19641 -39283 30061 15031 55075 1.91
Carry over 15031 -196417197
balance -5010 -10020 13094 6547 3.82 m
Carry over 6547 -5010
balance -2182 -4365 3340 1670 71023 1.91
Carry over 1670 -2182 22920 D D
balance -557 -1113 1455 727 41920 6981071023
77881
Carry over 727 -557
balance -242 -485 371 186 155514 155514
Carry over 186 -242 69810
balance -62 -124 162 81
Carry over 81 -62 81208
balance -27 -54 41 21 Fig 2
Carry over 21 -27
balance -7 -14 18 9
Final moment 71023 -71023 55075 -55075
For horizontal slab AB, carrying UDL @ 65250
Vertical reactionat a and B = 0.5 x 65250 x 3.82 = 124628 N/m2
Similarly, for the Bottom slab DC carrying U.D.L.loads @ ###
Vertical reaction at D and C = 0.5 x 81208.22 x 3.83 = 155514 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 - ### + ### x 3.83 x 3.83 x 1/2
+ 1/2 x 22920 x 3.83 x 3.83 x 1/3
-ha x 3.83 + -15948 + 139355 + 56035.2
From which, ha = 46852
Distribution factore for AD and DA= Distribution factore for AB and DC=
MFAB=wL2 2=
Mfdc=wL2 2=
MFAD =pL2
MFAD =2+
MFDA =pL2
MFDA =2--
The moment distribution carried out as per table 1 for case 1
N/m2.
N/m2
Hence , hd =( 19000 + ### )x 3.83 - ### = 69810 N
2
Free B.M. at mid point E =65250 x 3.83
119644 N-m8
Net B.M. at E = 119644 - 55075 = 64569 N-m
Similarly, free B.M. at F =81208.2245 x 3.83
148904 N -m8
Net B.M. at F = 148904.416 - 71023 = 77881 N-m
For vertical member AD , Simply supported B.M. At mid span
Simply supporetd at mid sapn =19000 x 3.83
1/16 x ### x 3.83 558528
Net B.M. =71023 + 55075
= 63049 - ### = 7197 N-m2
3 Case 2 : Dead load and live load from out side and water pressure from inside.
In this case , water pressure having an intensity of zero at A and 9800 x 3.82 = 37436
w = 65250
19000 19000 19000
Itensity = 19000 A E B 14516
Net
latr
al p
ress
ure
diag
ram
And = 41920 - 37436
= 4484 3.83
3.82
D F C
41920 41920 4484
w = 81208
Fig 3
Fix end moments will be as under := ### x 3.83
-79762 N - m12 12
=### x 3.83
99269.6 N - m12 12
+ +WL Where W is the total tringular earth pressure.
12 10
+4484 x 3.83 ### x 3.83
x3.83
= 16128 N-m12 2 10
- -WL
12 15
-4484 x 3.83 ### x 3.83
x3.83
= -12579 N -m12 2 15
The moment distribution is carrired out as illustred in table.
Fixed End MomentsMember DC DA AD AB 45069 65250 45069
99269.61 -12579 16128 -79762 23451
Joint D A 124627.5 124627.5
Member DC DA AD AB45069
Distribution factore 0.33 0.67 0.67 0.33 19000 23451 73951
Fix end moment 99269.61 -12579 16128 -79762 A A
Balance -28897 -57794 42423 21211 45069 1.91
Carry over 21211 -2889730523
2=
2=
2+2=
N/m2
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
N/m2
N/m2 At the Top
N/m2 at the bottom
N/m2
MFAB=wL2 2=
Mfdc=wL2 2=
MFAD =pL2
MFAD =2+
MFDA =pL2
MFDA =2-
The moment distribution carried out as per table 1 for case 1
balance -7070 -14141 19265 9632 3.8230523
Carry over 9632 -7070
balance -3211 -6422 4714 2357 58813 1.91
Carry over 2357 -3211 4484 D D
balance -786 -1571 2141 1070 2140458813
89315
Carry over 1070 -786
balance -357 -714 524 262 155108 155108
Carry over 262 -357 21404
balance -87 -175 238 119
Carry over 119 -87 81208
balance -40 -79 58 29 Fig 4
Carry over 29 -40
balance -10 -19 26 13
Final moment 58813 -58813 45069 -45069
For horizontal slab AB, carrying UDL @ 65250
Vertical reactionat a and B = 0.5 x 65250 x 3.82 = 124628 N/m2
Similarly, for the Bottom slab DC carrying U.D.L.loads @ ###
Vertical reaction at D and C = 0.5 x 81208 x 3.82 = 155108 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 - ### + 4484 x 3.82 x 3.82 x 1/2
+ 1/2 x 14516 x 3.82 x 3.82 x 2/3
-ha x 3.82 + -13744 + 32716.16 + 70607.8
From which, ha = 23451
Hence , hd =( 4484 + ### )x 3.82 - ### = 21404 N
2
Free B.M. at mid point E =65250 x 3.82
119020 N-m8
Net B.M. at E = 119020 - 45069 = 73951 N-m
Similarly, free B.M. at F =81208 x 3.82
148128 N -m8
Net B.M. at F = 148127.862 - 58813 = 89315 N-m
For vertical member AD , Simply supported B.M. At mid span
Simply supporetd at mid sapn =4484 x 3.82
1/16 x ### x 3.82 214188
Net B.M. =58813 + 45069
= 51941 - ### = 30523 N-m2
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
and the bottom slab is subjected to a load w = 65250
Itensity = 81208 4000 4000
Net
latr
al p
ress
ure
diag
ram
Lateral pressure due to dead load = A E B 4000
1/3 x 12000 = 4000
Lateral pressure due to soil = 3.83
1/3 x 18000 = 6000 3.82
Hence earth pressure at depth h is =
4000 + 6000 h D F C
26920 26920 14516
N/m2.
N/m2
2=
2=
2+2=
N/m2
N/m2
N/m2
N/m2
N/m2
Earth pressure intensity at top = 4000 37436 w= 81208 37436
Fig 5
Earth pressure intensity at Bottom= 4000 + 6000 x 3.82 = 26920
In addition to these, the vertical wall lslab subjectednto water pressure of intensity ZERO at top and 37436
N/m2 at Bottom, acting from inside . The lateral pressure on vertical walls Is shown in fig 5 and 6
Fix end moments will be as under : =### x 3.83
-79762 N - m12 12
=### x 3.83
99269.6 N - m12 12
+ -WL Where W is the total tringular earth pressure.
12 15
+4000 x 3.83 ### x 3.83
x3.83
= -2209 N-m12 2 15
- +WL 4890 - 7098
12 10
-4000 x 3.83 ### x 3.83
x3.83
= 5757 N -m12 2 10
The moment distribution is carrired out as illustred in table.
Fixed End MomentsMember DC DA AD AB 35902 65250 35902
99269.61 5757 -2209 -79762 =
Joint D A 124627.5 124627.5
Member DC DA AD AB35902
Distribution factore 0.33 0.67 0.67 0.33 4000 83742
Fix end moment 99269.61 5757 -2209 -79762 A A
Balance -35009 -70018 54647 27324 35902 1.91
Carry over 27324 -3500948748
balance -9108 -18216 23339 11670 3.82
Carry over 11670 -9108
balance -3890 -7780 6072 3036 49646 1.91
Carry over 3036 -3890 0 D D
balance -1012 -2024 2593 1297 1451649646
99258
Carry over 1297 -1012
balance -432 -864 675 337 155108 155108
Carry over 337 -432 5200
balance -112 -225 288 144
Carry over 144 -112 81208
balance -48 -96 75 37 Fig 4
Carry over 37 -48
balance -12 -25 32 16
Final moment 49646 -49646 35902 -35902
For horizontal slab AB, carrying UDL @ 65250
Vertical reactionat a and B = 0.5 x 65250 x 3.82 = 124628 N
Similarly, for the Bottom slab DC carrying U.D.L.loads @ ###
Vertical reaction at D and C = 0.5 x 81208 x 3.82 = 155108 N
The body diagram for various members, including loading, B.M. And reactions are shown in fig.6
For the vertical member AD, the horizontal reaction at A is found by taking moments at D.Thus
( ha x 3.82 ) + 35902 - ### + 4000 x 3.82 x 3.82 x 1/2
N/m2 N/m2
N/m2
MFAB=wL2 2=
Mfdc=wL2 2=
MFAD =pL2
MFAD =2-
MFDA =pL2
MFDA =2-
The moment distribution carried out as per table 1 for case 1
N/m2.
N/m2
- 1/2 x 14516 x 3.82 x 3.82 x 1/3
-ha x 3.82 + -13744 + 29184.8 - 35304
From which, ha = 5200
Hence , hd =( 14516 x 3.82 )- 4000 x 3.82 - 5200 = 7245.6
2
Free B.M. at mid point E =65250 x 3.83
119644 N-m8
Net B.M. at E = 119644 - 35902 = 83742 N-m
Similarly, free B.M. at F =81208 x 3.83
148904 N -m8
Net B.M. at F = 148904.416 - 49646 = 99258 N-m
For vertical member AD , Simply supported B.M. At mid span
Simply supporetd at mid sapn =4000 x 3.83
1/16 x ### x 3.83 5973.98
Net B.M. =49646 + 35902
= 42774 + 5974 = 48748 N-m2
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
= 150 wt. of concrete = ###
= 7 wt of water = 9800
m = 13 for water side force
m*c=
13 x 7= 0.378 K = 0.378
13 x 7 + 150
= 1 - 0.378 / 3 = 0.874 J = 0.874
= 0.5 x 7 x 0.87 x 0.378 = 1.155 R = 1.155
Provide over all thickness = 320 mm so effective thicknesss = 270 mm
= 1.155 x 1000 x 270 84216794 > 83742000 O.K.
Ast =83742000 = 2365
150 x 0.874 x 270
using ### A = =3.14 x 20 x 20
= 3144 x100 4
Spacing of Bars =Ax1000/Ast 314 x 1000 / 2365 = 133 say = 130 mm
Hence Provided 20 130 mm c/c
Acual Ast provided 1000 x 314 / 130 = 2415
Bend half bars up near support at distance of L/5 = 3.83 / 5 = 0.80 m
Area of distributionn steel = 0.3 -0.1 x( 320 - 100
= 0.24%
450 - 100
= 0.24 x 320 x 10 = 759 ###
using 8 A = =3.14 x 8 x 8
= 504 x100 4
Spacing of Bars = Ax1000/Ast = 50 x 1000 / 380 = 132 say = 130 mm
Hence Provided 8 130 mm c/c on each face
2=
2=
2+2=
sst = N/mm2 N/m3
scbc = N/mm2 N/mm2
k=m*c+sst
j=1-k/3
R=1/2xc x j x k
Mr = R . B .D2 2=
BMx100/sstxjxD=mm2
mm F bars 3.14xdia2
mm2
mm F Bars @
mm2
Ast mm2 area on each face= mm2
mm F bars 3.14xdia2
mm2
mm F Bars @
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 .
Since steel is at top = 190 concrete M 20
k = 0.3238 J = 0.892 R = 1.011
=55075000
= 1204190 x 0.892 x 270
Area available from the bars bentup from the middle section = 2415 / 2 = 1208
1204 < 1208
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
= 150 wt. of concrete = ###
= 7 wt of water = 9800
m = 13 for water side force
m*c=
13 x 7= 0.378 K = 0.378
13 x 7 + 150
= 1 - 0.378 / 3 = 0.874 J = 0.874
= 0.5 x 7 x 0.87 x 0.378 = 1.155 R = 1.155
=99258416
= 294 mm D = 344 mm1000 x 1.155
Provide thickness of bottom slab D= 350 mm so that d = 300 mm
Ast =99258416 = 2523
150 x 0.874 x 300
using ### mm bars A = =3.14 x 20 x 20
= 3144 x100 4
Spacing of Bars =Ax1000/Ast 314 x 1000 / 2523 = 124 say = 120 mm
Hence Provided 20 120 mm c/c
Acual Ast provided 1000 x 314 / 120 = 2617
Bend half bars up near support at distance of L/5 = 3.83 / 5 = 0.80 m
Area of distributionn steel = 0.3 -0.1 x( 350 - 100
= 0.23 %450 - 100
= 0.23 x 350 x 10 = 800 400
using 8 mm bars A = =3.14 x 8 x 8
= 504 x100 4
Spacing of Bars = Ax1000/Ast = 50 x 1000 / 400 = 126 say = 120 mm
Hence Provided 8 120 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 .
Since steel is at top = 190 concrete M 20
k = 0.3238 J = 0.892 R = 1.011
=71023000
= 1397
sst N/mm2
\ Ast mm2
mm2
Hence these bars will serve the purpose. However, provide 8 mm dia. Additional bars @ 200 mm c/c
sst = N/mm2 N/m3
scbc = N/mm2 N/mm2
k=m*c+sst
j=1-k/3
R=1/2xc x j x k
\ d
BMx100/sstxjxD=mm2
3.14xdia2
mm2
mm F Bars @
mm2
Ast mm2 area on each face= mm2
3.14xdia2
mm2
mm F Bars @
sst N/mm2
\ Ast mm2
=190 x 0.892 x 300
= 1397
Area available from the bars bentup from the middle section = 2617 / 2 = 1308
1397 > 1308 Fail , hence additional reinforcement will provided.
Additional reinforcemet required = 88.67
using 8 mm bars A = =3.14 x 8 x 8
= 504 x100 4
Spacing of Bars = Ax1000/Ast = 50 x 1000 / 89 = 567 say = 560 mm
Hence Provided 8 300 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 N -m, and direct force = 155514 N
Eccentricity =71023 x 1000
= 457 mm155514
proposed thickness of side wall '= 330 mm \ e / D 457 / 330 = 1.38 < 1.5 OK
thickness of side wall is OK
Let us reinforce the section with 20 300 mm c/c provided on both faces, as shown
in fig xxx . With cover of 50 mm and D = 330 mm
Asc = Ast =1000
x3.14 x 20 x 20
= 1047300 4
The depth of N.A. is computed from following expression:
n
3 3 n= e +
D- dt
b n + (m -1) Asc n - dc- m Ast
D - dt - n 2
n n
or
1000 n330 - 50 -
n+ 12 x
1047x n - 50 x -100
2 3 n
1000 n+12 x
1047x n - 50 - 13 x
1047x 330 - 50 - n
2 n n
500 n 280 -n
+ n - 50 x-1E+06
3 n= 457 + 115
500 n+12560
x n - 50 -###
x 280 - nn n
140000 n - 167 + -1256000 --62800000
n= 572
500 n + 12560 -628000
-3809867
+ ###n n
multiply by n
140000 n2 - 167 n3 + -1256000 n - -62800000= 572
500 n2 + 12560 n- 628000 - 3809867 + ### n
140000 n2 - 167 n3 + -1256000 n - -62800000= 0
\ Ast mm2
mm2
mm2
3.14xdia2
mm2
mm F Bars @
mm F bars @
mm2
b n D - dt - + (m - 1)Asc 1 (n - dc)(D - dt- dc)
n2
286000 n2 + 14967333 n - 2538459733= 0
-146000 n2 - 13711333 n - -2475659733 = 167
-876 n2 + 82268 n - 14853958 =
Solwing this trial and error we get, n = 91.47 mm
( 500 x 91.47 +12 x 1047
( 91.47 - 50 ) -13 x 1047
91.47 91.47
x ( 330 - 50 - 91.47 )
or 45734 + 137.3 x 41.47 - 149 x 188.53 = 23383
=155514
= 6.65 < 7 Stress is less than permissiable 23383
Also stress in steel t =m c'
(D-dc-n) =13 x 6.65
x ( 330 - 50 - 91.47 )n 91.47
= 178.2 N/mm2 < 190 N/mm2 O.K.
Stress in steel is less than permissiable Hence section is O.K.
n3
n3
\ c'
\ c' N/mm2
Box culverts
20 260 mm c/c
8 200 mm C/C
8 130 mm C/c
320
700 20
20 130 mm C/C
300 mm C/C
3.50
8
130 mm C/C
20
120 mm c/c
350
20 8 8
240 mm c/c 200 mm C/C 130 mm C/c
330 3.50 330
mm F @
mm F @
mm F @
mm F @
mm F @
mm F @
mm F @
mm F @ mm F @ mm F @
Grade of co 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
(N/mm2) (N/mm2) (N/mm2)M 10 3.0 300 2.5 250 -- --M 15 5.0 500 4.0 400 0.6 60M 20 7.0 700 5.0 500 0.8 80M 25 8.5 850 6.0 600 0.9 90M 30 10.0 1000 8.0 800 1.0 100M 35 11.5 1150 9.0 900 1.1 110M 40 13.0 1300 10.0 1000 1.2 120M 45 14.5 1450 11.0 1100 1.3 130M 50 16.0 1600 12.0 1200 1.4 140
Grade of co M-10 M-15 M-20 M-25 M-30 M-35 M-40Modular ra
Grade of concrete M-15 M-20 M-25 M-30 M-35 M-40
Modular Ratio 18.67 13.33 10.98 9.33 8.11 7.18 Grade of concrete M
5 7 8.5 10 11.5 13
93.33 93.33 93.33 93.33 93.33 93.33
0.4 0.4 0.4 0.4 0.4 0.4
0.867 0.867 0.867 0.867 0.867 0.867
0.867 1.214 1.474 1.734 1.994 2.254
0.714 1 1.214 1.429 1.643 1.857
0.329 0.329 0.329 0.329 0.329 0.329
0.89 0.89 0.89 0.89 0.89 0.89
0.732 1.025 1.244 1.464 1.684 1.903
0.433 0.606 0.736 0.866 0.997 1.127
0.289 0.289 0.289 0.289 0.289 0.289
0.904 0.904 0.904 0.904 0.904 0.904
0.653 0.914 1.11 1.306 1.502 1.698
0.314 0.44 0.534 0.628 0.722 0.816
Table 1.15. PERMISSIBLE DIRECT TENSILE STRESS
Tensile stress N/mm2
Table 1.16.. Permissible stress in concrete (IS : 456-2000)
Grade of concrete
Permission stress in compression (N/mm2) Permissible stress in bond (Average) for plain bars in tention (N/mm2)Bending acbc Direct (acc)
Kg/m2 Kg/m2 in kg/m2
Table 1.18. MODULAR RATIO
31 (31.11)
19 (18.67)
13 (13.33)
11 (10.98)
9 (9.33)
8 (8.11)
7 (7.18)
Table 2.1. VALUES OF DESIGN CONSTANTS
scbc N/mm2 tbd (N / mm2)
m scbc
(a) sst = 140
N/mm2 (Fe 250)
kc
jc
Rc
Pc (%)
(b) sst = 190
N/mm2
kc
jc
Rc
Pc (%)
(c ) sst = 230 N/mm2 (Fe 415)
kc
jc
Rc
Pc (%)
Reiforcement % Value of angle
M-20 M-20Degree sin cos tan
bd bd 10 0.17 0.98 0.180.15 0.18 0.18 0.15 11 0.19 0.98 0.190.16 0.18 0.19 0.18 12 0.21 0.98 0.210.17 0.18 0.2 0.21 13 0.23 0.97 0.230.18 0.19 0.21 0.24 14 0.24 0.97 0.250.19 0.19 0.22 0.27 15 0.26 0.97 0.270.2 0.19 0.23 0.3 16 0.28 0.96 0.29
0.21 0.2 0.24 0.32 17 0.29 0.96 0.310.22 0.2 0.25 0.35 18 0.31 0.95 0.320.23 0.2 0.26 0.38 19 0.33 0.95 0.340.24 0.21 0.27 0.41 20 0.34 0.94 0.360.25 0.21 0.28 0.44 21 0.36 0.93 0.380.26 0.21 0.29 0.47 22 0.37 0.93 0.400.27 0.22 0.30 0.5 23 0.39 0.92 0.420.28 0.22 0.31 0.55 24 0.41 0.92 0.450.29 0.22 0.32 0.6 25 0.42 0.91 0.470.3 0.23 0.33 0.65 30 0.50 0.87 0.58
0.31 0.23 0.34 0.7 35 0.57 0.82 0.700.32 0.24 0.35 0.75 40 0.64 0.77 0.840.33 0.24 0.36 0.82 45 0.71 0.71 1.000.34 0.24 0.37 0.88 50 0.77 0.64 1.190.35 0.25 0.38 0.94 55 0.82 0.57 1.430.36 0.25 0.39 1.00 60 0.87 0.50 1.730.37 0.25 0.4 1.08 65 0.91 0.42 2.140.38 0.26 0.41 1.160.39 0.26 0.42 1.250.4 0.26 0.43 1.33
0.41 0.27 0.44 1.410.42 0.27 0.45 1.500.43 0.27 0.46 1.630.44 0.28 0.46 1.640.45 0.28 0.47 1.750.46 0.28 0.48 1.880.47 0.29 0.49 2.000.48 0.29 0.50 2.130.49 0.29 0.51 2.250.5 0.30
0.51 0.300.52 0.300.53 0.300.54 0.300.55 0.310.56 0.310.57 0.310.58 0.310.59 0.310.6 0.32
Shear stress tc
100A s 100A s
0.61 0.320.62 0.320.63 0.320.64 0.320.65 0.330.66 0.330.67 0.330.68 0.330.69 0.330.7 0.34
0.71 0.340.72 0.340.73 0.340.74 0.340.75 0.350.76 0.350.77 0.350.78 0.350.79 0.350.8 0.35
0.81 0.350.82 0.360.83 0.360.84 0.360.85 0.360.86 0.360.87 0.360.88 0.370.89 0.370.9 0.37
0.91 0.370.92 0.370.93 0.370.94 0.380.95 0.380.96 0.380.97 0.380.98 0.380.99 0.381.00 0.391.01 0.391.02 0.391.03 0.391.04 0.391.05 0.391.06 0.391.07 0.391.08 0.41.09 0.41.10 0.41.11 0.41.12 0.41.13 0.41.14 0.4
1.15 0.41.16 0.411.17 0.411.18 0.411.19 0.411.20 0.411.21 0.411.22 0.411.23 0.411.24 0.411.25 0.421.26 0.421.27 0.421.28 0.421.29 0.421.30 0.421.31 0.421.32 0.421.33 0.431.34 0.431.35 0.431.36 0.431.37 0.431.38 0.431.39 0.431.40 0.431.41 0.441.42 0.441.43 0.441.44 0.441.45 0.441.46 0.441.47 0.441.48 0.441.49 0.441.50 0.451.51 0.451.52 0.451.53 0.451.54 0.451.55 0.451.56 0.451.57 0.451.58 0.451.59 0.451.60 0.451.61 0.451.62 0.451.63 0.461.64 0.461.65 0.461.66 0.461.67 0.461.68 0.46
1.69 0.461.70 0.461.71 0.461.72 0.461.73 0.461.74 0.461.75 0.471.76 0.471.77 0.471.78 0.471.79 0.471.80 0.471.81 0.471.82 0.471.83 0.471.84 0.471.85 0.471.86 0.471.87 0.471.88 0.481.89 0.481.90 0.481.91 0.481.92 0.481.93 0.481.94 0.481.95 0.481.96 0.481.97 0.481.98 0.481.99 0.482.00 0.492.01 0.492.02 0.492.03 0.492.04 0.492.05 0.492.06 0.492.07 0.492.08 0.492.09 0.492.10 0.492.11 0.492.12 0.492.13 0.502.14 0.502.15 0.502.16 0.502.17 0.502.18 0.502.19 0.502.20 0.502.21 0.502.22 0.50
2.23 0.502.24 0.502.25 0.512.26 0.512.27 0.512.28 0.512.29 0.512.30 0.512.31 0.512.32 0.512.33 0.512.34 0.512.35 0.512.36 0.512.37 0.512.38 0.512.39 0.512.40 0.512.41 0.512.42 0.512.43 0.512.44 0.512.45 0.512.46 0.512.47 0.512.48 0.512.49 0.512.50 0.512.51 0.512.52 0.512.53 0.512.54 0.512.55 0.512.56 0.512.57 0.512.58 0.512.59 0.512.60 0.512.61 0.512.62 0.512.63 0.512.64 0.512.65 0.512.66 0.512.67 0.512.68 0.512.69 0.512.70 0.512.71 0.512.72 0.512.73 0.512.74 0.512.75 0.512.76 0.51
2.77 0.512.78 0.512.79 0.512.80 0.512.81 0.512.82 0.512.83 0.512.84 0.512.85 0.512.86 0.512.87 0.512.88 0.512.89 0.512.90 0.512.91 0.512.92 0.512.93 0.512.94 0.512.95 0.512.96 0.512.97 0.512.98 0.512.99 0.513.00 0.513.01 0.513.02 0.513.03 0.513.04 0.513.05 0.513.06 0.513.07 0.513.08 0.513.09 0.513.10 0.513.11 0.513.12 0.513.13 0.513.14 0.513.15 0.51
bd M-15 M-20 M-25 M-30 M-35 M-400.18 0.18 0.19 0.2 0.2 0.2
0.25 0.22 0.22 0.23 0.23 0.23 0.230.50 0.29 0.30 0.31 0.31 0.31 0.32
0.75 0.34 0.35 0.36 0.37 0.37 0.38
1.00 0.37 0.39 0.40 0.41 0.42 0.42
1.25 0.40 0.42 0.44 0.45 0.45 0.46
1.50 0.42 0.45 0.46 0.48 0.49 0.491.75 0.44 0.47 0.49 0.50 0.52 0.522.00 0.44 0.49 0.51 0.53 0.54 0.552.25 0.44 0.51 0.53 0.55 0.56 0.572.50 0.44 0.51 0.55 0.57 0.58 0.602.75 0.44 0.51 0.56 0.58 0.60 0.62
3.00 and above 0.44 0.51 0.57 0.6 0.62 0.63
Over all depth of slab 300 or more 275 250 225 200 175 150 or less
k 1.00 1.05 1.10 1.15 1.20 1.25 1.30
Grade of concrete M-15 M-20 M-25 M-30 M-35 M-40
1.6 1.8 1.9 2.2 2.3 2.5
Grade of concrete M10 15 20 25 30 35 40 45
-- 0.6 0.8 0.9 1 1.1 1.2 1.3
Plain M.S. Bars H.Y.S.D. Bars
M 15 0.6 58 0.96 60
M 20 0.8 44 1.28 45
M 25 0.9 39 1.44 40
M 30 1 35 1.6 36
M 35 1.1 32 1.76 33
M 40 1.2 29 1.92 30
M 45 1.3 27 2.08 28
M 50 1.4 25 2.24 26
Table 3.1. Permissible shear stress Table tc in concrete (IS : 456-2000)100A s Permissible shear stress in concrete tc N/mm2
< 0.15
Table 3.2. Facor k
Table 3.3. Maximum shear stress tc.max in concrete (IS : 456-2000)
tc.max
Table 3.4. Permissible Bond stress Table tbd in concrete (IS : 456-2000)
tbd (N / mm2)
Table 3.5. Development Length in tension
Grade of concrete tbd (N / mm2) kd = Ld F tbd (N / mm2) kd = Ld F
Value of angle
tan Degree sin cos 0.18 10 0.17 0.980.19 11 0.19 0.980.21 12 0.21 0.980.23 13 0.23 0.970.25 14 0.24 0.970.27 15 0.26 0.970.29 16 0.28 0.960.31 17 0.29 0.960.32 18 0.31 0.950.34 19 0.33 0.950.36 20 0.34 0.940.38 21 0.36 0.930.40 22 0.37 0.930.42 23 0.39 0.920.45 24 0.41 0.920.47 25 0.42 0.910.58 30 0.50 0.870.70 35 0.57 0.820.84 40 0.64 0.771.00 45 0.71 0.711.19 50 0.77 0.641.43 55 0.82 0.571.73 60 0.87 0.502.14 65 0.91 0.42
150 or less
50
1.4
in concrete (IS : 456-2000)
