reatining wall with horzontal bach fill and trrafic load
TRANSCRIPT
RCC design B.C.Punmia
RETAINING WALL
18.2 TYPE OF RETAINING WALLS
1 Gravity walls2 Cantilever retaining walls a. T- shaped b. L- shaped3 Counterfort retainig walls.4 Buttresssed walls.
The cantilever retaining wall resist the horizontal earth pressure as well as other vertical pressure by way of beending of varios components acting as cantilever s.A coomon form of cantilever retaining waal
A retaining wall or retaining structure is used for maintaining the ground surfgaces at
defrent elevations on either side of it. Whenever embankments are involed in construction ,retaining
wall are usually necessary. In the construction of buildins having basements, retaining walls are
mandatory. Similsrly in bridge work, the wing walls and abutments etc. are designed as retaining
walls , to resist earth pressure along with superimposed loads. The material retained or supported by
a retaining wall is called backfill lying above the horizontal plane at the elevation of the top of a wall
is called the surcharge, and its inclination to horizontal is called the surcharge angle b
In the design of retaining walls or other retaining structures, it is necessary to compute the lateral earth pressure exerted bythe retaining mass of soil. The equation of finding out the lateral earth pressure against retaining wall is one of the oldest in Civil Engineering field. The plastic state of strees, when the failure is imminent, was invetigated by Rankine in1860. A Lot of theoretical experiment work has been done in this field and many theory and hypothesis heve benn proposed.
Retaining walls may be classified according to their mode of resisting the earth pressure,and according to their shape. Following are some of commen types of retaining walls (Fig)
A gravity retaining wall shown in fig 1 is the one in which the earth pressure exrted by the back fill is resisted by dead weight of wall, which is either made of masonry or of mass concrete . The stress devlop in the wall is very low ,These walls are no proportioned that no tension is devloped any where, and the resultant of forces remain withen the middle third of the base.
The cantilever retaining wall resist the horizontal earth pressure as well as other vertical pressure by way of beending of varios components acting as cantilever s.A coomon form of cantilever retaining waal
The cantilever retaining wall resist the horizontal earth pressure as well as other vertical pressure by way of beending of varios components acting as cantilever s.A coomon form of cantilever retaining waal
200 Q= 2 kN/m
Height of Parapet wall 1000
Height of Retaining wall = 6.00 m
Super imposed load due to road traffic = 18.00 10 300 @ c/c 20
Unit weight of Earth = 18.002000
520 @ c/c
Angle of repose = 30 Degree 6000
Safe Bearing capacity of soil = 160 10 300 @ c/c
Coffiecent of friction = 0.4 5540 20
Height of Parapet wall = 1.00 m3100
260 @ c/c
Concrete M- 20 25000 10 160 @ c/c
m 13.33 7 20 130 @ c/c
Steel Fy 415 230600
20 180 @ c/c
Nominal cover = 30 mm 8 120 @ c/c
260 Toe Heel
DESIGN SUMMARY200 460
900200
Stem thickness At footing 600 mm At top 200 mm
Heel width 1800 mm Toe width 1800 mm 8 1800
600Footing width 4200 mm Key 600 x 600 mm 120 @ c/c 1800 20 130
4200
Reinforcement Summary 10 600 10 160
300 @ c/c
Main
100% Reinforcement upto mm 3100 20 130 mm c/c 10 20
50% Reinforcement upto mm 2000 20 260 mm c/c 300 @ c/c3.10 2.00
520 @ c/c
25% Reinforcement upto Top 20 520 mm c/c
Distribution 8 160 mm c/c 5540
Tamprecture 8 300 mm c/c 20
260 @ c/c
Main 20 130 mm c/c
Distribution 8 120 mm c/c 20
130 @ c/c
Main 20 120 mm c/c All diamention in mm
Distribution 8 120 mm c/c Out side face Earth side face Not to scale
DESIGN OF CANTILEVER RETAINING WALL with horizontal back fill and traffic load.
KN/m2 mm F mm F
KN/m3
KN/m2 mm Fmm F
wt.c N/m3 mm Fscbc N/mm2 mm F
sst N/mm2 mm Fmm F
mm F
mm F @ c/c
mm F mm F @ c/c
STEM:-
mm F@ mm F mm F
mm F@
mm F@
mm F@
mm F@ mm F
TOE:-mm F@
mm F@ mm FHEEL:-
mm F@
mm F@
DESIGN OF T SHAPED CANTILEVER RETAINING WALL
Height of Retaining wall = 6.00 m
Super imposed load due to road traffic = 18.00Unit weight of Earth g = 18 = 18000Angle of repose = 30 Degree
Safe Bearing capacity of soil = 160Coffiecent of friction m = 0.4 = 25Height of Parapet wall = 1.00 m
Concrete = M - 20
0 = 13.33 wt.c = 25000 N/m3
Steel fe = 415 = 230
= 7
Nominal cover = 30 mm
1 Design Constants:- For HYSD Bars Cocrete M = 20
m*c=
13.33 x 7= 0.289
13.33 x 7 + 230
= 1 - 0.289 / 3 = 0.904
= 0.5 x 7 x 0.904 x 0.289 = 0.913
2 Diamension of base:-
Assume that a horizontal force Q= 2 kN/m length of parapet wall will be act because of person standing near the parapet
Due to surcharge eqivalent height of fill given by w=
18= 1.00 m
y 18.00
Hence, H2=H+he'= 6.00 + 1.00 = 7.00 m
a = 1 - = 1 -160
= 0.4232.2 y H 2.2 x 18 x 7.00
= 0.42 …. Eq (1)
The width of base is given by Eq.
b = 0.95 H x Ka = =1 - 0.50
= 0.33(1- a)x(1+3 a) 1 + 0.5
b = 0.95 x 7.000.333
= 3.35 ( 1 - 0.42 )x( 1 + 1.26 )
The base width from the considration of sliding is given by Eq.
b =0.7HKa
=0.7 x 7.00 x 0.333
= 7.04 m( 1 - 0.42 )x 0.4
Taking maximum value of H = 0.6 b = 0.60 x 7.00 = 4.20 m
Hence Provided, b = 4.20 m
The wall will be unsafe against sliding. This will be made safe by providing a shear Key at base .
Width of toe slab = = 0.42 x 4.20 = 1.76 m Provided toe slab = 1.80 m
Let the thickness of base be = H/12 = 7.00 / 12 = 0.58 or say = 0.60 m
Hence width of heel slab = 4.20 - 0.60 - 1.80 = 1.80 m
kN/m2
kN/m3 N/m2
q0 kN/m2
N/mm2
sst N/mm2
scbc
k=m*c+sst
j=1-k/3
R=1/2xc x j x k
he=
in determining the valueof b and a etc. we will use a height
The ratio of the length of toe slab EF to the base width b may be determinined by eq.
q0
Keep a
Ka 1-sin F1+sinF
(1-a) mThis width is excessive. Normal practice is to provide b between 0.5 to 0.6 H .
a x b
for design purpose
3 Thickness of stem:-Heigth AB = 6.00 - 0.60 = 5.40 m consider 1 m length of retaining wall
below B.Due to surcgarge w, there will be a uniform horizontal pressure
= 0.33 x 18 = 6.002
moment of rectangular pressure distribution, plus moment of triangular pressure distribution.
= + Kaw x +2 6
x ( 5.40 + 1.00 + 0.33 x 18 x5.40 0.33 x 18 x( 5.40
2 6\ M = 257.74 Kn-m
BM=
257.74 x= 531 mm
Rxb 0.913 x 1000
Keep d = 540 mm and total thickness = 540 + 60 = 600= 200 mm at top so that effective depth of = 140 mm
continue uniform thickness of 200
4 Stability of wall:-Full dimension wall is shown in fig 1a
= 4.20 - 1.80 - 0.60 = 1.80 m
= weight of rectangular portion of stem
= weight of triangular portion of stem
= weight of base slab
= weight of soil on heel slab.
= Total super imposed traffic load, over heel slab.
The calculation are arrenged in Table
Detail force(kN) lever arm Moment about toe (KN-m)
1 x 0.20 x 6.40 x 25 = 32.00 2.30 73.60
1/2 x 0.40 x 6.40 x 25 = 32.00 2.00 64.00
1 x 4.20 x 0.60 x 25 = 63.00 2.10 132.30
1 x 1.80 x 5.40 x 18 = 175.00 3.30 577.37
1 x 1.80 x 1.00 x 18.00 = 32.00 3.30 106.92
= 302.00 847.27
Total resisting moment = 847.27 kN-m
Earth pressure p = =0.33 x 18 x( 6.00
= 108 kN2 2
Over turning
Over turning moment at Toe, due to horizontal force = 2 x 7 = 14
Over turning moment at Toe, due to earth pressure = 108 x6
= 2163
=0.33 x 18 x( 6.00
= 1082
Total over turning moment = 338 kN/m
\ F.S. against over turning =847.27
= 2.51 > 2 'Hence safe'338
2 + 108 +( 0.333 x 18 x 6.00 )= 146
\ F.S. against Sliding = =0.40 x 302
0.83 < 1.5146
Special shear key will have to be designed to make the wall safe against sliding
Due to retained soil, the earth pressure diagram will be a triangle,having an ordinate equal to Ka.y.h at h
Ka.w kN/m2 throughout the height.
The total bending moment at C will be due to moment of horizontal force Q= kN acting at A, plus
\ M Q(H1+he)H1
2
Kay xH1
3
\ M 22x
Effective depth required =
10 6
Reduce the total thickness to
mm from B to A
Length of heel slab CD
Let w1
w2
w3
w4
w5
w1
w2
w3
w4
w5
Sw total MR
Ka x y x H2 )2
Over turning moment due to Horizontal pressure caused by live load
)2
Total horizontal pressure, Sp =
mSW
Sp
Pressure distribution 847.27 - 338 = 509.27 kN-m\ Distance x of the point of application of resultant, from toe is
x = =509.27
= 1.69 mb
=4.20
= 0.70 m302.00 6 6
Eccenticity e =b
- x =4.20
- 1.69 = 0.41 < 0.70 'Hence safe'2 2
= 1 +6 e
=302.00
x 1 +6x 0.41
= 114.4 < 160b b 4.20 4.20 Hence safe
= 1 -6 e
=302.00
x 1 -6x 0.41
= 29.40 < 160b b 4.20 4.20 Hence safe
Pressure p at the junction of stem with toe slab is
p = 114.40 -114.40 - 29.40
x 1.80 = 77.974.20
Pressure p at the junction of stem with Heel slab is
p = 114.40 -114.40 - 29.40
x 2.40 = 65.834.20
5 Design of toe slab:-
(1) Up ward soil pressure (2) Down ward weight of slab
Down ward weight of slab per unit area = 0.60 x 1 x 1.00 x 25 = 15.00
Hence net pressure intensities will be = 114.40 - 15.00 = 99.40
= 77.97 - 15.00 = 62.97
= 0.50 x( 99.40 + 62.97 ) x 1.80 = 146.00 kN
=62.97 2.00 x 99.40
x1.80
= 0.97 m62.97 + 99.40 3
\ = 146.00 x 0.97 = 141.23 kN-m
Effective depth required =BM
=141.23 x
= 393 mmRxb 0.913 x 1000
Keep effective depth d = 400 mm and total thickness = 400 + 60 = 460Reduce the total thickness to = 200 mm or 0.20 m at edge say = 0.46
Bars available from stem reinforcemnet are 20 130 = 2416
=146 x 1000
= 0.32 > 0.30 Hence un safe1000 x 460
% of reinforcement provided =100As
=100 x 2416
= 0.53 % = 0.30 Rafer table 3.1bd 1000 x 460
Hence unsafe, To make safe, either increase the reinforcment or increase the Depth.
= 0.32 0.6 reinforcement
Ast =0.60 x b x d
=0.6 x 1000 x 400
= 2400 Rafer table 3.1
100 100
Using 20 = =3.14 x ( 20
= 3144 4
Hence Spacing =1000 x 314
= 131 mm say= 130 mm c/c
2400 Hence the slab just safe in shear
Distribution steel =0.12
x 1000 x460 + 200
= 396100 2
Using 8 = =3.14 x ( 8
= 504 4
\ Spacing =1000 x 50
= 127 mm say = 120 mm c/c396
net moment SM =
SM
Sw
Pressure p1 at toe
SW kN -m2
Pressure p1 at Heel
SW kN -m2
kN-m2
kN-m2
The upward pressure distribution on the toe slab is shown in fig 1b .The weight of soil above the toe slab is neglicted . Thus two forces are acting on it
kN-m2
kN-m2 under D
kN-m2 under E
Total force = S.F. at F
C.G. of force from F+
B.M. at F
10 6
mm F @ mm c/c giving Ast
Tv N/mm2
\ tc
The reinforcement % required to get Tc >= N/mm2 is equal to %
mm2
mm F bars, Area P D2 )'2
mm2
mm2
mm F bars, Area P D2 )'2
mm2
6 Design of heel slab :-Four force act on it 1. down ward weight of soil = 5.40 m high
2 Live traffic load 3 down ward weight of heel slab 4 upward soil pressure
Total weight of soil = 1.80 x 5.40 x 1 x 18 = 175.0 KN acting at 0.90Live Load = 18.00 x 1 x 1.80 = 32.4 KN acting at 0.90
Total weight of heel slab = 1.80 x 0.46 x 25 = 20.70 KN acting at 0.90Total upward soil pressure = 1/2 x( 65.83 + 29.40 )x 1.80 = 85.71 kN
Acting at =65.83 + 2 x 29.40
x1.80
= 0.7965.83 + 29.40 3
\ Total force = 175 + 32.40 + 20.70 - 85.71 = 142.35 kN\ = ( 175 + 32.40 + 20.70 ) x 0.90 - 85.71 x 0.79 =
= 138.0 x
Effective depth required =BM
=137.95 x
= 389 mmRxb 0.913 x 1000
Keep effective depth d = 400 mm and total thickness = 400 + 60 = 460Reduce the total thickness to = 200 mm or 0.20 m at edge (as that of toe slab)
Ast = =137.95 x
= 1660230 x 0.904 x 400
Using 20 = =3.14 x ( 20
= 3144 4
\ Spacing =1000 x 314
= 189 mm say = 180 mm c/c1660
=142.35 x 1000
= 0.361000 x 400
= 0.36 0.82 % of reinforcemenrt
Ast =0.82 .b.d
=0.82 x 1000 x 314
= 2575Rafer table 3.1
100 100
Using 20 = =3.14 x ( 20
= 3144 4
Hence Spacing =1000 x 314
= 122 mm say = 120 mm c/c
2575 Hence the slab just safe in shear
Hence provided 20 mm bars @ 120 mm c/c, Actual Ast =1000 x 314
= 2617120
Distribution steel =0.12
x 1000 x460 + 200
= 396100 2
Using 8 = =3.14 x ( 8
= 504 4
\ Spacing =1000 x 50
= 127 mm say = 120 mm c/c396
7 Reinforcement in the stem:-
Ast =BM x1000
=257.74 x
= 2297230 x 0.904 x 540
Using 20 = =3.14 x ( 20
= 3144 4
\ Spacing =1000 x 314
= 137 mm say = 130 mm c/c2297
= 1000 x314
x 2416130
m from C.
m from C.
m from C.
m from B
S.F. at B=
B.M.at C
B.M.at C 10'6 N-mm2
10 6
BM 10 6
mm2
sst x j x D
mm F bars, Area P D2 )'2
mm2
Tv N/mm2
The reinforcement % required to get Tc > N/mm2 is equal to
mm2
mm F bars, Area P D2 )'2
mm2
mm2
mm F bars, Area P D2 )'2
mm2
10 6
mm2
sst x j x D
mm F bars, Area P D2 )'2
mm2
Actual AS provided mm2
Bent all bars bars in the toe slab to serve as tensile reinforcement there. However in order to make the
= 2400 = 130 mm c/c
Hence provide 20 130
= 2 + 0.33 x 18 x 5.40 + 0.33 x 185.40
1222
=121.88 x 1000
= 0.23 < 0.28<
1000 x 540.00 Hence safe
% of reinforcement provided =100As
=100 x 2400
= 0.44 = 0.28 Rafer table 3.1bd 1000 x 540
and if the depth of stem were constant, half the bars could have been curtailed at a depth
and if the depth of stem were constant, half the bars could have been curtailed at a depth
D = = = 0.79
force, let us try at depth = 0.65 H1 = 0.65 x 5.40 = 3.51
= 3.51 m
= + Kaw x +2 6
x ( 3.51 + 1.00 + 0.33 x 18 x3.51 0.33 x 18 x( 3.51
2 6\ = 89.22 Kn-m
= 140 +540 - 140
x h
d' = 140 +540 - 140
x 4 = 400 mm5.40
Ast = =89.22 x
= 1074 < 1149230 x 0.904 x 400
This is less than half of that provided at C. Hence half bars can be curtailed at this depth.
= 12 x 20 = 240 mm
or D = 400 mm beyond this point, whichever is more.
Hence h = 3.51 - 0.4 = 3.10 m. Hence curtailed half bars at a height of 3.10
If we wish to curtailed half of the remaning part, let us try it at a section at depth
h = 0.65 x 3.51 = 2.30 m
= + Kaw x +2 6
x ( 2.30 + 1.00 + 0.33 x 18.00 x2.30 0.33 x 18 x( 2.30
2 6\ = 34.64 Kn-m
= 140 +2400 - 140
x h
d' = 140 +540 - 140
x 2.30 = 310 mm5.40
Ast = =34.64 x
= 537 < 574230 x 0.904 x 310
This is less than half of that provided at C. Hence half bars can be curtailed at this depth.
= 12 x 20 = 240 mm
or D = 310 mm beyond this point, whichever is more.
Hence h = 2.30 - 0.3104 = 2.00 m. Hence curtailed half bars at a height of 2.00
toe slab safe in shear, steel required Ast mm2 Give spacing
mm F bars @ mm c/c . Due to this, sufficient bond length will be
available to the side of point C (point of maximum bending moment)
Total Shear force at C =Q+Ka.w.H1+KayH12/2
S.F. at C2=
Shear stress at C Tv N/mm2Tv Tc
\ tc
Let us curtail reinforcement between C and B. If there were no external force, except the earth pressure,
H1 H1H1 Below the point B because of presence of other
2(1/3) 2(1/3)
m below B to see whether
half bars could be curtailed there or not. Thus, depth of section below B = h
\ B. M. Q(H1+he)H1
2
Kay xH1
3
\ B. M. = 22x
The effective depth d' at section is (where h In meter)H1
BM 10 6
mm2
sst x j x D
However, the bars should be extented by a distance of =12 F
\ B. M. Q(H1+he)H1
2
Kay xH1
3
\ B. M. = 22x
The effective depth d' at section is (where h In meter)H1
BM 10 6
mm2
sst x j x D
However, the bars should be extented by a distance of =12 F
Distribution and temprechure reinforcement:-
Average thickness of stem = 600 + 200= 400 mm
2
\ Distribution reinforcement =0.12
x 1000 x 400 = 480100
Using 10 = =3.14 x ( 10
= 78.504 4
= 1000 x 79= 164 mm say = 160 mm c/c
480
Hence provided 10 mm F bars @ 160 mm c/c at the inner face of wall,along its length.
Area = 480 provide 10 mm bars = 300Design of shear key:-
The wall is in unsafe in sliding, Let us provide a shear keys of depth a below the stem
==
1=
1= 3
= 3.00 x 77.97 = 233.90 Ka 0.33
This intencity may be considered to be constant along the depth of key, through there will be little increase
in Pp , because of increase in pwith depth.Wewill, however,consider the constant value of Pp = 233.90
Pp x a = 233.90 a.kN Keeping a = 600Pp x a = 233.90 x 0.60 = 140 kN
Total sliding force at the bottom of key is = 2 x ( 6.00 x 6.6 )+ 0.33 x 18 x6.60
2
= 172.28 kN
Weight of the soil between bottom of the base and JJ = 4.20 x 0.6 x 18 = 45.36\ = 302.00 + 45.36 = 347.36 kN (Approx)
Hence F.S. against sliding is = =0.4 x 347 + 140
= 1.62 1.5 Hence safe172.28
= = a tan x 45 +F
=2 shearing angle of passive resistance
\ = 0.6 x ( 3.00
= 1.04 m .Actual length of the slab available GF = 1.80 m Hence satisfactory
Let us keepthe width of key = 600Actual force to be resisted by the key =
= 1.5 x 172 - 0.4 x 347.36
= 119 kN
=119 x 1000
= 0.20 < 1.8600 x 1000
=119 x 300 x 1000 Hence safe
1/6 x 1000 x( 600
= 0.59 < 7 Permissible Bending stress in M20 concrete
Hence safe
Since concrete can take this much of tensile stress no special reinforcement is necessary for key.
The key is to be cast monolithically with the base.
9 Construction Joint:-
A construction joint, in the form of key, is to be provided at the junction of stem with the
base slab. The width of key is kept equal to d/4 = 540 / 4 = 140 mm
mm2
mm F bars, Area P D2 )'2
mm2
\ spacing
for tempreture reinforcement
mm2 mm c/c both way in outer face
Let Pp be the intensity of passive pressure Pp devloped just in front of shear key.this intencity Pp depend upon the soil pressure P just in front of the key
Pp Kp x P Where Kp
\ Pp kN/m3
\ total passive pressure Pp = mm (equal to stem width)
S P
SW
m Sw+Pp
SP
it should be noted that passive pressure taken into account above will be devloped only when length a1 given below is avilable in front of key ;
a1 a tan F a Ökp
where (45 + F/2) =
a1 )1/2
a1
mm (Equal to stem width)
1.5.S.P - mSW
\ shear stress N/mm2 Permissible shear stress in M20 concrete
\ Bending stress )2
N/mm2
DESIGN OF T SHAPED CANTILEVER RETAINING WALL
Eq (1)
0.33
m
The base width from the considration of sliding is given by Eq.
The wall will be unsafe against sliding. This will be made safe by providing a shear Key at base .
may be determinined by eq.
H .
for design purpose
moment of rectangular pressure distribution, plus moment of triangular pressure distribution.
mm
Moment about toe (KN-m)
..(1)
..(2)
kN-m
kN-m
kN-m
'Hence safe'
kN
Special shear key will have to be designed to make the wall safe against sliding
kN acting at A, plus
)3
'Hence safe'
Hence safe
Hence safe
mm
m
Hence un safe
Rafer table 3.1
Hence unsafe, To make safe, either increase the reinforcment or increase the Depth.
Rafer table 3.1
Hence the slab just safe in shear
kN -m2
kN -m2
The upward pressure distribution on the toe slab is shown in fig 1b .The weight of soil above
mm2
upward soil pressure
mm
(as that of toe slab)
Rafer table 3.1
Hence the slab just safe in shear
m from C.
m from C.
m from C.
mm2
kN
Rafer table 3.1
and if the depth of stem were constant, half the bars could have been curtailed at a depth
This is less than half of that provided at C. Hence half bars can be curtailed at this depth.
m
If we wish to curtailed half of the remaning part, let us try it at a section at depth
This is less than half of that provided at C. Hence half bars can be curtailed at this depth.
m
If there were no external force, except the earth pressure,
because of presence of other
)3
)3
mm c/c at the inner face of wall,along its length.
The wall is in unsafe in sliding, Let us provide a shear keys of depth a below the stem
This intencity may be considered to be constant along the depth of key, through there will be little increase
kN
(Approx)
Hence safe
shearing angle of passive resistance
Hence satisfactory
Permissible Bending stress in M20 concrete
A construction joint, in the form of key, is to be provided at the junction of stem with the
both way in outer face
devloped just in front of shear key.this intencity
kN/m2
(equal to stem width)
2
it should be noted that passive pressure taken into account above will be devloped only when length
Permissible shear stress in M20 concrete
0.20 0.20 0.2018.00 KN/m2
1.002 kN/m
A A A
H= 4.00
6.00 m 5.40 m 6.00 m 5.40 m 5.40 m
W1 W1 W1
2.40 4.201.80 W2 1.80 W2 1.80 1.80 W2 1.80
toe heelD E B C D E B C D E B C
0.60 0.60 0.60
Toe Toeb = 4.20 m b = 2.40 m
##
#
##
#
##
# ##
# a
e
P=
P=
H1= H1=
a b
a1
Kay(H+a)
D1 C1
Pp = Kpp
P=
P=
##
#
##
#
0.20
2 kN/mHeight of parapet wall 1.00
18.00 KN/m2A
Outer side face10 Earth side Face
@ 300 c/c `
10@ 300 c/c
H= 6.00 m 20 520 C/C
##
#
10 160 C/C
##
#
5.54
1010 @ 300 c/c
@ 160 c/c
20 260 C/C
10@ 300 c/c
20 130 C/C
N.S.L.
8 20 180 C/C
@ 120 c/c600
260 Toe Heel Earth side Face Outer side faceReinforcement Detail Reinforcement Detail
200 460 200Foundation level
20 600 8
mm F
mm F
mm F@
mm F@
mm Fmm F
mm F@
mm F
mm F@
mm F mm F@
mm F mm F
@ 130 c/c @ 120 c/c `
600
c/c
c/c
mm F
mm F
Grade of concrete 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)
10 3.0 300 2.5 250 -- --
15 5.0 500 4.0 400 0.6 60
20 7.0 700 5.0 500 0.8 80
25 8.5 850 6.0 600 0.9 90
30 10.0 1000 8.0 800 1.0 100
35 11.5 1150 9.0 900 1.1 110
40 13.0 1300 10.0 1000 1.2 120
45 14.5 1450 11.0 1100 1.3 130
50 16.0 1600 12.0 1200 1.4 140
Table 1.18. MODULAR RATIO
Grade of concrete M-10 M-15 M-20 M-25 M-30 M-35 M-40
Modular ratio m
Table 2.1. VALUES OF DESIGN CONSTANTSGrade 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
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
0.253 0.253 0.253 0.253 0.253 0.253
0.916 0.916 0.916 0.914 0.916 0.916
0.579 0.811 0.985 1.159 1.332 1.506
Table 1.15. PERMISSIBLE DIRECT TENSILE STRESS
Tensile stress N/mm2
Table 1.16.. Permissible stress in concrete (IS : 456-2000)
Grade of concrete
M
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
31 (31.11)
19 (18.67)
13 (13.33)
11 (10.98)
9 (9.33)
8 (8.11)
7 (7.18)
scbc 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 (%)
(d) sst = 275 N/mm2 (Fe 500)
kc
jc
Rc
0.23 0.322 0.391 0.46 0.53 0.599
(d) sst = 275 N/mm2 (Fe 500)
Pc (%)
bd M-15 M-20 M-25 M-30 M-35 M-40
% 0.18 0.18 0.19 0.20 0.20 0.20
0.25 % 0.22 0.22 0.23 0.23 0.23 0.23
0.50 % 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 0.49
1.75 % 0.44 0.47 0.49 0.50 0.52 0.52
2.00 % 0.44 0.49 0.51 0.53 0.54 0.55
2.25 % 0.44 0.51 0.53 0.55 0.56 0.57
2.50 % 0.44 0.51 0.55 0.57 0.58 0.60
2.75 % 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 M-10 M-15 M-20 M-25 M-30 M-35 M-40 M-45 M-50
-- 0.6 0.8 0.9 1 1.1 1.2 1.3 1.4
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
Degree sin cos tan
10 0.174 0.985 0.176
15 0.259 0.966 0.268
16 0.276 0.961 0.287
17 0.292 0.956 0.306
18 0.309 0.951 0.325
19 0.326 0.946 0.344
20 0.342 0.940 0.364
21 0.358 0.934 0.384
22 0.375 0.927 0.404
23 0.391 0.921 0.424
24 0.407 0.924 0.445
25 0.422 0.906 0.466
30 0.500 0.866 0.577
35 0.573 0.819 0.700
40 0.643 0.766 0.839
45 0.707 0.707 1.000
50 0.766 0.643 1.192
55 0.819 0.574 1.428
60 0.866 0.500 1.732
65 0.906 0.423 2.145