Slrb Deck Slab Design

Clear Span = 7.450 m

Thickness of slab = 65 cm

(assumed)

Thickness of w.c = 75 mm

Clear cover = 4.00 cm

(as per clause 304.3.1 and table 10 of IRC bridge code 21-2000

section III )

Main reinforcement = 20 mm dia. HYSD bars conforming to IS-1786

(Deformed bars)

Concrete Mix is M 20 grade

Bearing(assumed) = 49 cm

Effective depth = 65 - 4.00 - 1.00

= 60 cm

Effective SpanAs per clause 305.1.2 of IRC code 21-2000

Effective span shall be the least of the following

i). Effective span = l1 + d

where l1 = clear span

= 7.45 m

d = effective depth

= 0.6 m

Effective span = 7.45 + 0.6000

= 8.0500 m

ii) l = distance from centre of supports

= 7.45 + 2x 0.490

2

= 7.940 m

Effective span shall be least of (i) and (ii)

Effective span = 7.940 m

Carriage way width = 4.25 m

width

(clause 113.1 of IRC 5-1985)

Kerb width = 0.225 m

Width of slab = 4.25 + 0.225 x 2

= 4.700 m

Loading IRC class 'A'

As per clause 303.1 and Table 6 of IRC code 21-2000, for M 20 grade RCC

Permissible flexural compressive stresses 66.67

= 6.667 M.Pa

= 66.67 kg/cm2

Modular ratio = Es = 10.00

Ec

As per clause 303.2.1 of IRC code 21-2000

Permissible tensile stress in steel for combined bending

For steel S 415 = 200 M.Pa

m = 10.00

n = mc/mc+t

n = 10.00 x 66.67

66.67 x 10.00 + 2000

= 0.250

j = 1 - n/3

= 1 - 0.250009 /3 Q = 1/2*c*n*j

= 0.917 = 7.640

D E S I G N O F D E C K S L A B

c allowable

As per clause 211.2 of IRC code 6

Impact factor fraction = 4.5

6+L

where L is Effective span= 7.940 m

Impact factor fraction = 4.5

6 + 7.940

= 0.323

Dead Load Bending Moment

Weight of slab = 0.65 * 2500 = 1625 kg/m

Weight of w.c = 0.075 * 2400 = 180 kg/m

Total dead load 1805 kg/m

Bending Moment = wl2

= 1805 x 7.940 x 7.94

due to dead load 8 8

= 14224.2 kg-m

= 1422421 kg-cm

Live Load Bending MomentAs per clause 305.13.2.1 of IRC code 21-2000

Ratio b / lo

where b = width of slab = 4.70 m

lo = Effective span = 7.940 m

Ratio = 4.70 = 0.592

7.94

for simply supported slab

0.5 1.72

0.6 1.96

For b/lo = 1.72 + 0.24 * 0.092

0.1

= 1.941

As per clause 305.1.3.2(1) IRC 21-2000

Solid slabs spanning in one direction

For a single concentrated load, the effective width may be calculated

in accordance with the following equation

bef = a (1 - a/lo) +b1

where bef = The effective width of slab on which the load acts.

lo = The effective span as indicated in clause 305.1

a = The distance of the centre of gravity of the concentrated load

from the nearer support

b1 = The breadth of concentration area of the load, i.e the dimension

of the tyre or track contact area over the road surface of the slab

in a direction at right angles to the span plus twice the thickness

of the wearing coat or surface finish above the structural slab.

= a constant having the following values depending upon the ratio

b/lo where b is the width of the slab

2.7 11.4 11.4 6.8 6.8

3.2 1.2 4.3 3

b/lo

Class 'A' Train

For maximum bending moment, the two loads of 11.40 t should be kept such that the resultant

of the load system and the load under consideration should be equidistant from the centre of span.

Position of Loads

2.7 11.4 11.4

3.2 1.2

A C D E B

7.940

CG of the load sytem from C

= (11.4 x 3.2)+(11.4 x 4.4) = 86.64 = 3.398

(2.7 + 11.4 + 11.4) 25.5

CG from D = 0.198

2.7 11.4 R 11.4

0.099 0.099 1.002

0.671 3.2 1.2 2.869

C D E

3.970 3.970

A 7.940 B

Dispersion Width Under 'C' 2.7 t load

bef = a ( 1 - a/lo) + b1

= 1.941 a from A = 0.671

a = 0.671 a from B = 7.269

lo = 7.94

b1 = 0.35 m

bef = 1.543 m < 1.8 m

Dispersion Widths do not overlap

0.225 0.15 0.1 1.8

0.2 0.2

4.70

Left Dispersion (bef/2) = 0.7713

here left dispersion means the possible dispersion on the left side from the centre of the left wheel of the vehicle.

= kerb width + kerb to wheel gap + tyre width / 2 = 0.225 + 0.15 + 0.2/2 = 0.475 < 0.7713

here left is going beyond the slab edge, hence the left dispersion is limited to 0.475

Dispersion width under one wheel

= 0.475 + 1.543 /2 + 0

= 1.246 m

Intensity of load under 'C' = 1.35 (1.00 + 0.323 )

1.246

= 1.433 t

Dispersion Width Under D 11.4 t load

bef = a (1 - a/lo) +b1

= 1.941 a from A = 3.871

a = 3.871 a from B = 4.069

lo = 7.94

b1 = 0.65 m

bef = 4.500 m > 1.8 m

Dispersion Widths overlap

0.225 0.15 0.25 1.8

0.5 0.5

4.70





Combined dispersion width

= 0.6250 + 4.500 /2 + 1.8

= 4.675 m

Intensity of load under 'D' = 11.4 ( 1.00 + 0.323 )

4.675

= 3.226 t

Dispersion Width Under 'E' 11.4 t load

bef = a (1 - a/lo) +b1

= 1.941 a from A = 5.071

a = 2.869 a from B = 2.869

lo = 7.940

b1 = 0.65 m

bef = 4.206 m > 1.8 m


0.225 0.15 0.25 1.8

0.5 0.5

4.70




here left dispersion is going beyond the slab edge, hence the left dispersion is limited to 0.6250


= 0.6250 + 4.206 /2 + 1.8

= 4.528 m

Intensity of load under 'E' = 11.4 ( 1.00 + 0.323 )

4.528

= 3.330 t

1.433 3.226 R 3.330

0.099 0.099 1.002

0.671 3.2 1.2 2.869

C D E

3.97 3.970

A 7.94 B

Taking moments about 'B'

Ra x 7.94 = 3.330 * 2.869 + 3.226 * 4.069

1.433 * 7.269

Ra x 7.94 = 33.219

Ra = 4.184 t

Rb = 1.433 + 3.226 + 3.330 - 4.184

Rb = 3.805 t

Maximum live load Bending Moment

= ( 4.184 * 3.871 ) - ( 1.433 * 3.20 )

= 16.196 - 4.585

= 11.61078 t-m

= 1161078 kg-cm

Total B.M = Dead load B.M + Live Load B.M

= 1422421 + 1161078

= kg-cm

Effective depth = M =

required Q*b 7.640 x100

= 58.153 cm < 60 cm

Steel OKsteel at bottom

Main steel reinforcement required = M =

t j d 2000 x 0.917 x 60

= 23.486 cm2

As per clause 305.19 of IRC code 21-2000

Minimum area of tension

reinforcement = 0.175 % of total cross sectional area

On each face = 0.35 * 100 * 60

2 * 100

= 11.375 cm2

< 23.486 cm2

provide 20 mm dia. HYSD bars at 120 mm c/c

Ast provided = 26.1667 cm2

OK

Distribution SteelAs per clause 305.15.1 of IRC code 21-2000

Resisting moment = 0.3 times the moment due to concentrated live loads plus 0.2 times the moment

due to other loads such as dead load, shrinkage, temperature etc.

= 0.3 * L.L B. Mom. + 0.2*D.L B. Mom.

= 0.3x 1161078 + 0.2x 1422421

= 632807.5 kg-cm

Assuming the dia. Of distribution bar 12 mm HYSD

Effective depth = 58.40 cm

Ast required = M =

t j d 2000 x 0.917 x 58.4

= 5.910 cm2

As per clause 305.16 of IRC code 21-1987

Minimum distribution reinforcement = 0.175 % of total cross section area

= 0.35 x 100 x 58.40

2 100

= 10.220 cm2

> 5.910 cm2

Hence provide minimun distribution reinforcement of 10.220 cm2

provide 12 mm dia. HYSD bars at 110 mm c/c


OK

2583499

632807.50

2583499

2583499

Top Reinforcement

Min. Reinforcement = 10.22 cm2

Provide 12 mm dia. HYSD bars at 110 mm c/c

Also provide distribution steel of 12 mm dia. HYSD bars at 110 mm c/c

Check for shear

As per clause 305.13.3 of IRC code 21-2000

Dispersion of loads along the span

Longitudinal dispersion = The effect of contact of wheel or track load in the direction of span

length shall be taken as equal to the dimension of the tyre contact

area over the wearing surface of the slab in the direction of the span

plus twice the overall depth of the slab inclusive of the thickness of

the wearing surface.

Longitudinal dispersion = 0.25 + 2 ( 0.65 + 0.075 )

= 1.70 m

so 11.4 t load may be kept 0.85 m from the support to get max. shear

11.4 11.4 6.8 t

0.85 1.2 4.3 1.590

C D E

3.970 3.970

A 7.940 B

The load of 11.40 t may be kept at 1.70 /2 = 0.85 m from the support to get max. shear

Dispersion Width Under 'C' 11.4 t

bef = a (1 - a/lo) +b1

= 1.941 a from A = 0.850

a = 0.850 a from B = 7.090

lo = 7.940

b1 = 0.65 m

bef = 2.123 m > 1.8 m


0.225 0.15 0.25 1.8

0.5 0.5

4.70






= 0.625 + 2.123 /2 + 1.8

= 3.487 m

Intensity of load under 'C' = 11.4 ( 1.00 + 0.323 )

3.487

= 4.325 t

Dispersion Width Under 'D' 11.4 t

bef = a (1 - a/lo) +b1

= 1.94 a from A = 2.05

a = 2.050 a from B = 5.890

lo = 7.94

b1 = 0.65 m

bef = 3.602 m > 1.8 m







= 0.6250 + 3.602 /2 + 1.80

= 4.226 m

Intensity of load under 'D' = 11.4 ( 1.00 + 0.323 )

4.226

= 3.569 t

Dispersion Width Under 'E' 6.8 t

bef = a (1 - a/lo) +b1

= 1.941 a from A = 6.350

a = 1.590 a from B = 1.590

lo = 7.940

b1 = 0.53 m

bef = 2.998 m > 1.8 m


0.225 0.15 0.19 1.8

0.38 0.5

4.70






= 0.565 + 2.998 /2 + 1.8

= 3.864 m

Intensity of load under 'E' = 6.8 ( 1.00 + 0.323 )

3.864

= 2.328 t

4.325 3.569 2.328

0.85 1.2 4.3 1.59

C D E

3.970 3.970

A 7.940 B

Taking moments about 'B'

Ra x 7.94 = 4.325 * 7.09 + 3.569 * 5.89

+ 2.328 * 1.59

Ra x 7.94 = 55.385

Ra = 6.975 t

Rb = 3.246 t

Shear due to dead load = wl

2

= 1805 x 7.940

2

= 7165.85 kgs

Total shear due to dead load & live load = 6975.45 + 7165.85

kgs

As per clause 304.7.1 of IRC 21-2000

Design shear stress c = Vbd

where V = The design shear across the section

d = The depth of the section

b = The breadth of the rectangular beam or slab, or the

breadth of the rib in the case of flanged beam

here V = 14141.3 kgs

b = 100 cms

d = 60 cms

= 14141.3

100x 60

= 2.36 kg/cm2

As per clause 304.7.1.1 of IRC 21-2000 Max. permissible shear stress for M 20 grade concrete

cmax = 1.8 N/mm2

for solid slabs the permissible shear stress shall not exceed half the value of cmax given in Table 12 AHence cmax = 0.5 x 1.8

= 0.9 N/mm2

Hence max. permissible stress = 9.0 kg/cm2


100 Ast / b d = 0.436

from Table 12B of IRC:21-2000

c = 0.280 N/mm2

= 2.800 kg/cm2

For solid slabs the permissible shear in concrete = k * c(vide cl. 304.7.1.3.2 of IRC:21-2000)

k for solid slab of 65 cm thick = 1.00 (Table 12C of IRC:21-2000)

k * c= 1.0 x 2.800 = 2.800 Kg/cm2 > 2.36 kg/cm2

Hence safe against shear.

14141.301

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Slrb Deck Slab Design