RCC design B.C.Punmia
RETAINING WALL 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 β
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.
18.2 TYPE OF RETAINING WALLS 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) 1 Gravity walls 2 Cantilever retaining walls 3 Counterfort retainig walls. 4 Buttresssed walls.
a. T- shaped
b. L- shaped
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 va
pressure by way of beending of varios components acting as cantilever s.A coomon form of cantilever retaining waal
retaining waal
DESIGN OF CANTILEVER RETAINING WALL with horizontal back fill and traffic load. Height of Retaining wall = 6.00 m Super imposed load due to road traffic = 18.00 KN/m2 Unit weight of Earth = 18.00 KN/m3 Angle of repose = 30 Degree KN/m2 Safe Bearing capacity of soil = 160 Coffiecent of friction = 0.4 m Height of Parapet wall = 1.00 wt. Concrete M20 25000 N/m3 c σcbc N/mm2 13.33 7 m σst N/mm2 Steel 230 Fy 415 Nominal cover = 30 mm
200
Footing width
DESIGN SUMMARY At footing 600 mm At top Heel width 1800 mm Toe width 4200 mm Key 600 x 600
200 mm 1800 mm mm
Reinforcement Summary STEM:-
1000
Height of Parapet wall
10
mm Φ 300 @ c/c
10
6000 mm Φ 300 @ c/c 5540
2000
3100
mm Φ 160 @ c/c
10
20 20
600 260
Stem thickness
Q= 2 kN/m
200
Toe 900
460
8 mm Φ 120 @ c/c
1800
20 4200 600
10 mm Φ 300 @ c/c
20 mm Φ 520 @ c/c
20 mm Φ 260 @ c/c mm Φ 130 @ c/c mm Φ 180 @ c/c 8 mm Φ 120 @ c/c Heel 200
1800 mm Φ 130 @ c/c 600 10
mm Φ 160 @ c/c
Main
20 20 20
mm Φ@ mm Φ@ mm Φ@
130 260 520
mm c/c mm c/c mm c/c
10 mm Φ 300 @ c/c
Distribution Tamprecture
8 8
mm Φ@ mm Φ@
160 300
mm c/c mm c/c
5540
Main Distribution
20 8
mm Φ@ mm Φ@
130 120
mm c/c mm c/c
Main Distribution
20 8
mm Φ@ mm Φ@
120 120
mm c/c mm c/c
100% Reinforcement upto
mm
50% Reinforcement upto
mm
25% Reinforcement upto
3100 2000 Top
3.10
2.00
20 mm Φ 260 @ c/c
TOE:-
20 mm Φ 130 @ c/c
HEEL:-
[email protected]
20 mm Φ 520 @ c/c
All diamention in mm
Out side face
Earth side face
Not to scale
DESIGN OF T SHAPED CANTILEVER RETAINING WALL Height of Retaining wall Super imposed load due to road traffic Unit weight of Earth Angle of repose Safe Bearing capacity of soil Coffiecent of friction Height of Parapet wall Concrete 0 = 13.33 Steel fe = 415
= = γ = = q0 = µ = = = wt.c = σst = σcbc = =
Nominal cover 1 Design Constants:- For HYSD Bars
k=
m*c
m*c+σst j=1-k/3 = 1 = R=1/2xc x j x k 0.5
= x
Cocrete M =
6.00 18.00 18 30 160 0.4 1.00
m kN/m2 kN/m3 = Degree kN/m2 = m
18000 N/m2
25
N/mm2
M - 20 25000 N/m3 230 N/mm2 7 30 mm
20
13.33 13.33 x 0.289 7
x 7 = 0.289 7 + 230 / 3 = 0.904 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, in determining the valueof b and α etc. we will use a height he =
H2=H+he'= 6.00 + 1.00 = 7.00 m The ratio of the length of toe slab EF to the base width b may be determinined by eq. q0 160 1 = 1 = 0.423 α = 2.2 y H 2.2 x 18 x 7.00 …. Eq (1) 0.42 Keep α = The width of base is given by Eq. Ka 1 - 0.50 1-sin Φ b = 0.95 H x = = 0.33 Ka = (1- a)x(1+3 a) 1 + 0.5 1+sinΦ
b =
0.95
x
7.00
(
1
-
0.333 0.42 )x( 1
+ 1.26 )
=
3.35
The base width from the considration of sliding is given by Eq. 0.7 x 7.00 x 0.33 = 7.04 m ( 1 - 0.42 )x 0.4 (1-α) µ This width is excessive. Normal practice is to provide b between 0.5 to 0.6 H . Taking maximum value of H = 0.6 b = 0.60 x 7.00 = 4.20 m Hence Provided, b = 4.20 m b =
0.7HKa
=
The wall will be unsafe against sliding. This will be made safe by providing a shear Key at base . Width of toe slab = α x b = 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 for design purpose Hence width of heel slab = 4.20 - 0.60 - 1.80 = 1.80 m
[email protected]
3 Thickness of stem:Heigth AB = 6.00 - 0.60 = 5.40 m consider 1 m length of retaining wall Due to retained soil, the earth pressure diagram will be a triangle,having an ordinate equal to Ka.y.h at h below B.Due to surcgarge w, there will be a uniform horizontal pressure Ka.w = 0.33 x 18 = 6.00 kN/m2 throughout the height. The total bending moment at C will be due to moment of horizontal force Q= 2 kN acting at A, plus moment of rectangular pressure distribution, plus moment of triangular pressure distribution. H12 H13 Q(H1+he) = + Kaw x + Kay x ∴ M 2 6 2 0.33 x 18 x( 5.40 5.40 x + 1.00 + 0.33 x 18 x ∴ M 2 x ( 5.40 6 2 Kn-m M= 257.74 ∴ 257.74 x 10 6 ΒΜ Effective depth = = 531 mm required = Rxb 0.913 x 1000 Keep d = = 540 + 60 = 600 540 mm and total thickness Reduce the total thickness = 200 mm at top so that effective depth of = 140 mm to continue uniform thickness of 200 mm from B to A 4 Stability of wall:Full dimension wall is shown in fig 1a Length of heel slab CD = 4.20 - 1.80 - 0.60 = 1.80 m Let w1 = weight of rectangular portion of stem w2 = weight of triangular portion of stem w3 = weight of base slab w4 = weight of soil on heel slab. w5 = Total super imposed traffic load, over heel slab. The calculation are arrenged in Table Detail force(kN) lever arm Moment about toe (KN-m) w1 1 x 0.20 x 6.40 x 25 = 32.00 2.30 73.60 w2 1/2 x 0.40 x 6.40 x 25 = 32.00 2.00 64.00 w3 1 x 4.20 x 0.60 x 25 = 63.00 2.10 132.30 w4 1 x 1.80 x 5.40 x 18 = 175.00 3.30 577.37 w5 1 x 1.80 x 1.00 x 18.00 = 32.00 3.30 106.92 847.27 Σw = 302.00 total MR Total resisting moment = 847.27 kN-m Ka x y x H2
0.33
x
18 2
x( 6.00 )2
Over turning Over turning moment at Toe, due to horizontal force
=
2
x
Over turning moment at Toe, due to earth pressure
=
108
x
Earth pressure p
=
2
=
Over turning moment due to Horizontal pressure = caused by live load Total over turning moment
2
x( 6.00 )2
108
kN
=
14
=
216
=
108
338 kN/m 847.27 F.S. against over turning = = 2.51 > 2 'Hence safe' ∴ 338 2 + 108 +( 18 x 6.00 )= 146 Total horizontal pressure, Σp = 0.33 x 0.40 x 302 µΣW = 0.83 < 1.5 ∴ F.S. against Sliding = 146 Σp Special shear key will have to be designed to make the wall safe against sliding
[email protected]
=
0.33 x
7 6 3 18
=
Pressure distribution net moment ΣM = 847.27 - 338 = 509.27 kN-m ∴ Distance x of the point of application of resultant, from toe is b 4.20 ΣM 509.27 = = 1.69 m = = 0.70 m x = 302.00 6 6 Σw b 4.20 Eccenticity e = x = - 1.69 = 0.41 < 0.70 'Hence safe' 2 2 ΣW 6e 302.00 6x 0.41 kN -m2 Pressure p1 at = 1+ = x 1+ = 114.4 < 160 toe b b 4.20 4.20 Hence safe 6e 302.00 6x 0.41 kN -m2 ΣW Pressure p1 at = 1= x 1= 29.40 < 160 Heel b b 4.20 4.20 Hence safe Pressure p at the junction of stem with toe slab is 114.40 - 29.40 p = 114.40 x 1.80 = 77.97 kN-m2 4.20 Pressure p at the junction of stem with Heel slab is 114.40 - 29.40 p = 114.40 x 2.40 = 65.83 kN-m2 4.20 5 Design of toe slab: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 (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 kN-m2 Hence net pressure intensities will be = 114.40 - 15.00 = 99.40 kN-m2 under D = 77.97 - 15.00 = 62.97 kN-m2 under E Total force = S.F. at F = 0.50 x( 99.40 + 62.97 ) x 1.80 = 146.00 kN 62.97 + 2.00 x 99.40 1.80 C.G. of force from F = x = 0.97 m 62.97 + 99.40 3 B.M. at F = 146.00 x 0.97 = ∴ 141.23 kN-m 141.23 x 10 6 ΒΜ Effective depth required = = = 393 mm Rxb 0.913 x 1000 Keep effective depth d = = 400 + 60 = 460 400 mm and total thickness Reduce the total thickness to = say = 0.46 200 mm or 0.20 m at edge mm c/c giving A Bars available from stem reinforcemnet are 20 mm Φ @ 130 = 2416 st 146 x 1000 Tv = = 0.32 > 0.30 N/mm2 Hence un safe 1000 x 460 100As 100 x 2416 = % of reinforcement provided = = 0.53 % ∴ tc = 0.30 Rafer table 3.1 bd 1000 x 460 Hence unsafe, To make safe, either increase the reinforcment or increase the Depth. The reinforcement % required to get Tc >= = 0.32 N/mm2 is equal to 0.6 % reinforcement 0.60 x b x d x 1000 x 400 Rafer table 3.1 0.6 Ast = = = 2400 mm2 100 100 Π D2 3.14 x ( 20 )'2 Using 20 mm Φ bars, Area = = = 314 mm2 4 4 1000 x 314 = mm c/c 130 Hence Spacing = = 131 mm say 2400 Hence the slab just safe in shear 460 + 200 0.12 Distribution steel = x 1000 x = 396 mm2 2 100 Π D2 3.14 x ( 8 )'2 Using 8 mm Φ bars, Area = = = 50 mm2 4 4 1000 x 50 mm c/c = 127 mm say = 120 ∴ Spacing = 396
[email protected]
6 Design of heel slab :Four force act on it 1. down ward weight of soil = 5.40 m high 3 down ward weight of heel slab 4 upward soil pressure 2 Live traffic load Total weight of soil = 1.80 x 1 5.40 x x 18 = 175.0 KN acting at 0.90 m from C. Live Load = 18.00 x x 1.80 = 32.4 KN acting at 0.90 m from C. 1 Total weight of heel slab = 1.80 x 0.46 x 25 = 20.70 KN acting at 0.90 m from C. Total upward soil pressure = 1/2 x( 65.83 + 29.40 )x 1.80 = 85.71 kN 2 x 29.40 65.83 + 1.80 Acting at = x = 0.79 m from B + 29.40 65.83 3 ∴ Total force = S.F. at B= 175 + 32.40 + 20.70 - 85.71 = 142.35 kN =( ∴ B.M.at C 175 + 32.40 + 20.70 ) x 0.90 - 85.71 x 0.79 = B.M.at C = 138.0 x 10'6 N-mm2 137.95 x 10 6 ΒΜ Effective depth required = = = 389 mm Rxb 0.913 x 1000 Keep effective depth d = = 400 + 60 = 460 400 mm and total thickness Reduce the total thickness to = mm or m at edge (as that of toe slab) 200 0.20 6 BM 137.95 x 10 = Ast = = 1660 mm2 σst x j x D 230 x 0.904 x 400 Π D2 3.14 x ( 20 )'2 Using 20 mm Φ bars, Area = = = 314 mm2 4 4 1000 x 314 mm c/c = 189 mm say = 180 ∴ Spacing = 1660 142.35 x 1000 Tv = = 0.36 N/mm2 1000 x 400 The reinforcement % required to get Tc > =
Ast
=
Using 20
.b.d 100
0.82
mm Φ bars, Area
Hence Spacing = Hence provided
20
Distribution steel Using 8
∴
∴
1000
= x 2575
0.82 Π D2 4 314
N/mm2 is equal to
1000 x 100 3.14 x ( = 4
=
@
120
0.12 100
x
1000 x
=
mm Φ bars, Area Spacing =
1000
= x 396
Π D2 4 50
122
mm c/c,
=
=
460
20
= )'2
2575 =
mm2 314
Rafer table 3.1
mm2
say
120
+
200
=
2 3.14 x ( 4 127
% of reinforcemenrt
= mm c/c 120 Hence the slab just safe in shear 1000 x 314 Actual Ast = = ###
mm
8
)'2
mm say =
257.74 x 10 6 = 230 x 0.904 x 540 Π D2 3.14 x ( = = mm Φ bars, Area 4 4 1000 x 314 Spacing = = 137 mm 2297 314 Actual AS provided = 1000 x x 130
[email protected]
0.82
314
x
mm bars
7 Reinforcement in the stem:BM x1000 = Ast = σst x j x D Using 20
=
0.36
2297 20
)'2
say = 2416
396
mm2
=
50
120
mm c/c
mm2
mm2 =
314
130
mm c/c
mm2
mm2
Bent all bars bars in the toe slab to serve as tensile reinforcement there. However in order to make the toe slab safe in shear, steel required Ast = 2400 mm2 Give spacing = 130 mm c/c mm Φ bars @ 130 mm c/c . Due to this, sufficient bond length will be Hence provide 20 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 C =
2
+
0.33
x
18
x
5.40
+
0.33
x
18
5.40 2 Tv
=
2
###
Tc < 1000 = 0.23 < 0.28 N/mm2 540.00 Hence safe 100As 100 x 2400 ∴ tc = 0.28 Rafer table 3.1 = % of reinforcement provided = = 0.44 bd 1000 x 540 Let us curtail reinforcement between C and B. If there were no external force, except the earth pressure, and if the depth of stem were constant, half the bars could have been curtailed at a depth Shear stress at C
Tv
=
121.88 x 1000 x
and if the depth of stem were constant, half the bars could have been curtailed at a depth H1 H1 = D = 0.79 H1 Below the point B because of presence of other = (1/3) 2 2(1/3) 0.65 H1 = 0.65 x 5.40 = 3.51 m below B to see whether force, let us try at depth = = 3.51 m half bars could be curtailed there or not. Thus, depth of section below B = h H12 H13 Q(H1+he) = + Kaw x + Kay x ∴ Β. M. 2 6 2 0.33 x 18 x( 3.51 3.51 x ∴ B. M. = 2 x ( 3.51 + 1.00 + 0.33 x 18 x 6 2 Kn-m 89.22 ∴ Β.Μ. = 540 - 140 The effective depth d' at section is = 140 + x h (where h In meter) H1 - 140 x 4 = 400 mm 5.40 BM 89.22 x 10 6 = Ast = = 1074 mm2 < 1149 σst x j x D 230 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 However, the bars should be extented by a distance of =12 Φ = 400 mm beyond this point, whichever is more. or D = 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 H12 H13 Q(H1+he) = + Kaw x + Kay x ∴ Β. M. 2 6 2 0.33 x 18 x( 2.30 2.30 x + 1.00 + 0.33 x 18.00 x ∴ B. M. = 2 x ( 2.30 6 2 Kn-m 34.64 ∴ Β.Μ. = 2400 - 140 The effective depth d' at section is = 140 + x h (where h In meter) H1 d'
=
140
+
d'
=
140
+
540
540
-
140
x 2.30 = 310 mm 5.40 BM 34.64 x 10 6 = Ast = = mm2 < 574 537 σst x j x D 230 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 However, the bars should be extented by a distance of =12 Φ = 310 mm beyond this point, whichever is more. or D = Hence h = 2.30 0.31 = 2.00 m. Hence curtailed half bars at a height of 2.00
[email protected]
Distribution and temprechure reinforcement:Average thickness of stem
∴ Using 10
Distribution reinforcement mm Φ bars, Area
=
x 79 = 480 Hence provided 10 mm F bars @ for tempreture reinforcement Area = 480 mm2 provide ∴ spacing =
1000
=
600
=
0.12 100
Π D2 = 4 164
+ 2
200
=
x 1000 x
3.14 x ( 4
mm say =
10
)'2
400 400
mm =
=
mm2
480 78.50
mm2
160 mm c/c
160
mm c/c at the inner face of wall,along its length.
10
mm bars =
300 mm c/c both way in outer face
Design of shear key:The wall is in unsafe in sliding, Let us provide a shear keys of depth a below the stem 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 1 1 Where Kp = = = 3 3 ∴ Pp = 3.00 x 77.97 = 233.90 kN/m 0.33 Ka 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 ∴ total passive pressure Pp = Pp x a = 233.90 a.kN Keeping a = 600 mm (equal to stem width) Pp x a = 233.90 x 0.60 = 140 kN Total sliding force at the bottom of key is =
x
6.6
)+ 0.33 x
18
x
Σ P = 172.28 kN Weight of the soil between bottom of the base and JJ = 302.00 + 45.36 = ∴ ΣW = 347.36
4.20 kN
x
18
=
µ Σw+Pp
140
Hence F.S. against sliding is =
ΣP
=
2
0.4
x ( 6.00
x
347 + 172.28
0.6 x
6.60 2 45.36 (Approx)
= 1.62
1.5
Hence safe
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 Φ
=
a tan
x
45
+
Φ 2
= a √kp
where (45 + Φ/2) = shearing angle of passive resistance
a1 = 0.6 x ( 3.00 )1/2 ∴ a1 = 1.04 m .Actual length of the slab available GF = 1.80 m Hence satisfactory Let us keepthe width of key = 600 mm (Equal to stem width) Actual force to be resisted by the key = 1.5.Σ.P - µΣW = 1.5 x ### - 0.4 x 347.36 = kN 119 119 x 1000 Permissible shear stress = 0.20 N/mm2 < 1.8 ∴ shear stress = in M20 concrete 600 x 1000 119 x 300 x 1000 Hence safe ∴ Bending stress = 2 1/6 x 1000 x( 600 ) 2 = 0.59 N/mm < 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
[email protected]
[email protected]
minined by eq.
0.33
m
.
y at base . for design purpose
[email protected]
ng at A, plus sure distribution.
)3
mm
nt about toe (KN-m)
..(1)
..(2)
kN-m kN-m kN-m
'Hence safe' kN
ng
[email protected]
'Hence safe' kN -m2 Hence safe kN -m2 Hence safe
eight of soil above
mm m mm2
Hence un safe Rafer table 3.1
epth. Rafer table 3.1
st safe in shear
[email protected]
soil pressure m from C. m from C. m from C.
mm
Rafer table 3.1
st safe in shear mm2
[email protected]
kN
Rafer table 3.1
earth pressure, a depth
esence of other
)3
his depth.
m
h
)3
his depth.
m
[email protected]
g its length.
y in outer face
stem r key.this intencity
little increase kN/m2
ual to stem width) 2
kN (Approx)
Hence safe only when length
assive resistance
Hence satisfactory
sible shear stress M20 concrete
in M20 concrete
of stem with the
[email protected]
[email protected]
0.20 1.00
0.20 2
0.20
18.00 KN/m2 kN/m
A
A
A
H= 4.00 H1=5.40 m
6.00 m
H1=
6.00 m
W1
αb 1.80
5.40 m
W1
W1
2.40 W2
1.80
E
B 0.60
heel C
D
4.20 1.80
1.80
E
B
C
D
B
b=
2.40
m 29.40
m
C
a1 P= 65.83
4.20
E
1.80
0.60
Toe b=
W2
0.60
P= 77.97
Toe
W2
114.40
toe D
5.40 m
Kay(H+a)
a D1
e
Pp = Kpp
C1
2 Height of parapet wall
1.00
kN/m
77.97
114.40
P=
P= 0.20 18.00 KN/m2
A
Outer side face 10 mm Φ @ 300 c/c
Earth side Face ` 10 mm Φ @ 300 c/c
20 mm Φ@ 520 C/C
6.00 m
2.00
H=
3.10
10 mm Φ@ 160 C/C 5.54
10 mm Φ
10 mm Φ @ 160 c/c
@ 300 c/c
20 mm Φ@ 260 C/C 10 mm Φ @ 300 c/c
20 mm Φ@ 130 C/C N.S.L.
8
mm Φ @ 120 c/c
20 mm Φ@ 180 C/C 600
260
Toe
Heel
200
460
Earth side Face Reinforcement Detail
200 Foundation level
20 mm Φ
600
8
mm Φ
Outer side face Reinforcement Detail
@ 130 c/c
@ 120 c/c 600
`
mm Φ c/c
mm Φ c/c
Table 1.15. PERMISSIBLE DIRECT TENSILE STRESS Grade of concrete
Tensile stress N/mm2
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
Table 1.16.. Permissible stress in concrete (IS : 456-2000) 2 Grade of Permission stress in compression (N/mm ) Permissible stress in bond (Average) for plain bars in tention (N/mm2) Bending αcbc Direct (αcc) concrete M Kg/m2 (N/mm2) Kg/m2 (N/mm2) (N/mm2) in kg/m2 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 1600 1200 140 16.0 12.0 1.4
Table 1.18. MODULAR RATIO Grade of concrete
Modular ratio m
M-10 31 (31.11)
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 2.1. VALUES OF DESIGN CONSTANTS Grade of concrete Modular Ratio
σcbc N/mm m σcbc
2
k (a) σst = c jc 140 N/mm2 Rc (Fe 250) P (%) c kc (b) σst = j c 190 Rc N/mm2 Pc (%) kc (c ) σst = jc 230 Rc N/mm2 (Fe 415) P (%) c kc (d) σst = jc 275 Rc N/mm2 (Fe 500)
M-15 18.67 5 93.33 0.4 0.87 0.87 0.71 0.33 0.89 0.73 0.43 0.29 0.9 0.65 0.31 0.25 0.92 0.58
M-20 13.33 7 93.33 0.4 0.87 1.21 1 0.33 0.89 1.03 0.61 0.29 0.9 0.91 0.44 0.25 0.92 0.81
M-25 10.98 8.5 93.33 0.4 0.87 1.47 1.21 0.33 0.89 1.24 0.74 0.29 0.9 1.11 0.53 0.25 0.92 0.99
M-30 9.33 10 93.33 0.4 0.87 1.73 1.43 0.33 0.89 1.46 0.87 0.29 0.9 1.31 0.63 0.25 0.91 1.16
M-35 8.11 11.5 93.33 0.4 0.87 1.99 1.64 0.33 0.89 1.68 1 0.29 0.9 1.5 0.72 0.25 0.92 1.33
M-40 7.18 13 93.33 0.4 0.87 2.25 1.86 0.33 0.89 1.9 1.13 0.29 0.9 1.7 0.82 0.25 0.92 1.51
Grade of concrete
(d) σst = 275 N/mm2 (Fe 500)
Pc (%)
0.23
0.32
0.39
0.46
0.53
0.6
Table 3.1. Permissible shear stress Table τc in concrete (IS : 456-2000) 100As
Permissible shear stress in concrete tc N/mm2
bd
M-15 0.18 0.22 0.29 0.34 0.37 0.40 0.42 0.44 0.44 0.44 0.44 0.44 0.44
< 0.15 % 0.25 % 0.50 % 0.75 % 1.00 % 1.25 % 1.50 % 1.75 % 2.00 % 2.25 % 2.50 % 2.75 % 3.00 and above
M-20 0.18 0.22 0.30 0.35 0.39 0.42 0.45 0.47 0.49 0.51 0.51 0.51 0.51
M-25 0.19 0.23 0.31 0.36 0.40 0.44 0.46 0.49 0.51 0.53 0.55 0.56 0.57
M-30 0.20 0.23 0.31 0.37 0.41 0.45 0.48 0.50 0.53 0.55 0.57 0.58 0.6
M-35 0.20 0.23 0.31 0.37 0.42 0.45 0.49 0.52 0.54 0.56 0.58 0.60 0.62
M-40 0.20 0.23 0.32 0.38 0.42 0.46 0.49 0.52 0.55 0.57 0.60 0.62 0.63
200 1.20
175 150 or less 1.25 1.30
Table 3.2. Facor k Over all depth of slab
300 or more
κ
1.00
275 1.05
250 1.10
225 1.15
Table 3.3. Maximum shear stress τc.max in concrete (IS : 456-2000) M-15 1.6
Grade of concrete
τc.max
M-20 1.8
M-25 1.9
M-30 2.2
M-35 2.3
M-40 2.5
Table 3.4. Permissible Bond stress Table τbd in concrete (IS : 456-2000) Grade of concrete M-10 τbd (N / mm2) --
M-15 0.6
M-20 0.8
M-25 0.9
M-30 1
M-35 1.1
M-40 1.2
M-45 1.3
Table 3.5. Development Length in tension Grade of concrete M 15 M 20 M 25 M 30 M 35 M 40 M 45 M 50
τbd
Plain M.S. Bars (N / mm2) kd = Ld Φ 0.6 0.8 0.9 1 1.1 1.2 1.3 1.4
58 44 39 35 32 29 27 25
τbd
H.Y.S.D. Bars (N / mm2) kd = Ld Φ 0.96 1.28 1.44 1.6 1.76 1.92 2.08 2.24
60 45 40 36 33 30 28 26
M-50 1.4
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
