Clear Span = 7.450 mThickness of slab = 65 cm(assumed)Thickness of w.c = 75 mmClear cover = 4.00 cm(as per clause 304.3.1 and table 10 of IRC bridge code 21-2000section III )Main reinforcement = 20 mm dia. HYSD bars conforming to IS-1786

(Deformed bars)Concrete Mix is M 20 gradeBearing(assumed) = 49 cm

Effective depth = 65 - 4.00 - 1.00= 60 cm

Effective SpanAs per clause 305.1.2 of IRC code 21-2000Effective 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= 8.050 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 mCarriage way width = 4.25 m(clause 113.1 of IRC 5-1985)

Kerb width = 0.225 mWidth of slab = 4.25 + 0.225 x 2

= 4.700 mLoading IRC class 'A'

As per clause 303.1 and Table 6 of IRC code 21-2000, for M 20 grade RCCPermissible flexural compressive stresses 66.67

= 6.667 M.Pa= 66.67 kg/cm2

Modular ratio = Es = 10.00Ec

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

c allowable

As per clause 303.2.1 of IRC code 21-2000Permissible tensile stress in steel for combined bendingFor steel S 415 t = 200 M.Pa

m = 10.00n = mc / (mc+t)n = 10.00 x 66.67

66.67 x 10.00 + 2000= 0.250

j = 1 - n/3= 1 - 0.25 /3 Q = 1/2*c*n*j= 0.917 = 7.640

As per clause 211.2 of IRC code 6Impact factor fraction = 4.5

6+Lwhere L is Effective span= 7.940 mImpact factor fraction = 4.5

6 + 7.940= 0.323

Dead Load Bending Moment

Weight of slab = 0.65 * 2500 = 1625 kg/mWeight of w.c = 0.075 * 2400 = 180 kg/mTotal dead load 1805 kg/m

Bending Moment = wl2 = 1805 x 7.940 x 7.94due to dead load 8 8

= kg-m= kg-cm

Live Load Bending MomentAs per clause 305.13.2.1 of IRC code 21-2000Ratio b / lowhere b = width of slab = 4.70 m

lo = Effective span = 7.940 m

Ratio = 4.70 = 0.5927.94

for simply supported slab0.5 1.720.6 1.96

For b/lo = 1.72 + 0.24 * 0.0920.1

= 1.941

As per clause 305.1.3.2(1) IRC 21-2000Solid slabs spanning in one directionFor a single concentrated load, the effective width may be calculated in accordance with the following equation

bef = a (1 - a/lo) +b1where bef = The effective width of slab on which the load acts.

lo = The effective span as indicated in clause 305.1a = The distance of the centre of gravity of the concentrated load

from the nearer supportb1 = The breadth of concentration area of the load, i.e the dimension

b/lo

14224.212251422421.225

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

= a constant depending upon the ratiob/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

For maximum bending moment, the two loads of 11.40 t should be kept such that the resultant ofthe load system and the load under consideration should be equidistant from the centre of span.

Position of Loads2.7 11.4 11.4

3.2 1.2A C D E B

7.940CG of the load sytem from C

= (11.4 x 3.2)+(11.4 x 4.4) = 86.64 =(2.7 + 11.4 + 11.4) 25.5

CG from D = 0.1982.7 11.4 R 11.4

0.099 0.099 1.0023.2 1.2 2.869

A C D E B3.970 3.970

7.940

Dispersion Width Under 'C' 2.7 t loadbef = a ( 1 - a/lo) + b1

= 1.941 a from A = 0.671a = 0.671 a from B = 7.269

lo = 7.94b1 = 0.35 mbef = 1.543 m < 1.8 mDispersion Widths do not overlap

0.225 0.2 0.1 1.8

0.2 0.2

4.70Left Dispersion calculated (bef/2) = 0.7713

left dispersion means the possible dispersion on the left side from the centre

Class 'A' Train

3.398

0.671

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 <

here left is going beyond the slab edge, hence the left dispersion is limited to 0.48

Dispersion width under one wheel= 0.475 + /2 + 0 = 1.246 m

Intensity of load under 'C' = 1.35 (1.00 + 0.323 )1.246

= 1.433 tDispersion 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 = 4.069lo = 7.94b1 = 0.65 mbef = 4.500 m > 1.8 mDispersion Widths overlap

1.543

0.7713

0.225 0.2 0.25 1.8

0.5 0.5

4.70Left Dispersion calculated (bef/2) = 2.250

here left dispersion means the possible dispersion on the left side from the centreof the left wheel of the vehicle. = kerb width + kerb to wheel gap + tyre width / 2 =

= 0.225 + 0.15 + 0.5/2 0.625 <here left is going beyond the slab edge, hence the left dispersion is limited toCombined dispersion width = + 4.500 /2 + 1.8

= 4.675 mIntensity of load under 'D' = 11.4 ( 1.00 + 0.323 ) 4.675

= 3.226 tDispersion Width Under 'E' 11.4 t loadbef = a (1 - a/lo) +b1

= 1.941 a from A = 5.071a = 2.869 a from B = 2.869

lo = 7.940b1 = 0.65 mbef = 4.206 m > 1.8 mDispersion Widths overlap

0.225 0.2 0.25 1.8

0.5 0.5

4.70Left Dispersion calculated (bef/2) = 2.103

left dispersion means the possible dispersion on the left side from the centreof the left wheel of the vehicle. = kerb width + kerb to wheel gap + tyre width / 2 =

0.225 + 0.15 + 0.5/2 0.625 <here left dispersion is going beyond the slab edge, hence the left dispersion is limited Combined dispersion width = 0.625 + 4.206 /2 + 1.8

= 4.528 mIntensity of load under 'E' = 11.4 ( 1.00 + 0.323 ) 4.528

= 3.330 t

1.433 3.226 R 3.3300.099 0.099 1.002

3.2 1.2 2.869C D E3.97 3.970

A 7.94 B

0.62502.250

0.6250

0.62502.103

0.671

Taking moments about 'B'Ra x 7.9 = 3.330 * 2.869 + 3.226 *

1.433 * 7.269Ra x 7.9 =

Ra = 4.184 tRb = 1.433 + 3.226 + 3.330 -Rb = 3.805 t

4.07

33.219

4.184

Maximum live load Bending Moment= ( 4.184 * 3.871 ) - ( 1.433 * ### )

= 16.196 - 4.585= t-m= kg-cm

Total B.M = Dead load B.M + Live Load B.M= += kg-cm

Effective depth = M =required Q*b 7.640 x100

= 58.153 cm < 60 cmSteel OKsteel at bottomMain 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 = 0.12 % of total cross sectional areaOn each face = 0.12 * 100 * 65

* 100= 7.80 cm2 < cm2

provide 20 mm dia. HYSD bars at 130 mm c/c

Ast provided = 24.15 cm2 OKDistribution SteelAs per clause 305.18.1 of IRC code 21-2000Resisting moment

= 0.3 times the moment due to concentrated live loads plus 0.2 times the momentdue to other loads such as dead load, shrinkage, temperature etc.

= 0.3 * L.L B. Mom. + 0.2*D.L B. Mom.= 0.3x + 0.2x= kg-cm

Assuming the dia. Of distribution 12 mm HYSDEffective depth = 58.40 cmAst required = M =

t j d 2000 x 0.917 x 58.4= 5.910 cm2

As per clause 305.19 of IRC code 21-2000Minimum distribution reinforcem = 0.12 % of total cross section area

= 0.12 x 100 x 58.40100

= 7.008 cm2 > 5.910 cm2

2583499

2583499

2583499

632807.50

11.610781161078

1161077.51422421.2

23.486

1161077.5 1422421.23632807.4993

Hence provide minimun distribution reinforceme 7.008 cm2

provide 12 mm dia. HYSD bars at 130 mm c/cAst provided = 8.695 cm2 OK

Top ReinforcementMin. Reinforcement = 7.008 cm2

Provide 12 mm dia. HYSD bars at 130 mm c/c Also provide distribution steel of 12 mm dia. HYSD bars at 130 mm c/c

Check for shearAs per clause 305.13.3 of IRC code 21-2000Dispersion of loads along the spanLongitudinal dispersion =

The effect of contact of wheel or track load in the direction of spanlength 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 ofthe wearing surface.

Longitudinal dispersion = 0.25 + 2 ( 0.65 + 0.075 )= 1.70 m

so 11 t load may be k 0.85 m from the support to get max. shear11.4 11.4 6.8 t

0.85 1.2 4.3 1.590C D E

3.970 3.970A 7.940 B

The load of 11.40 t may be kept at 1.70 /2 = 0.85 m from the support to Dispersion Width Under 'C' 11.4 t get max. shearbef = a (1 - a/lo) +b1

= 1.941 a from A = 0.850a = 0.850 a from B = 7.090lo = 7.940b1 = 0.65 mbef = 2.123 m > 1.8 m

Dispersion Widths overlap

0.225 0.15 0.25 1.8

0.5 0.5

4.70

Left Dispersion (bef/2) = 1.0616left dispersion means the possible dispersion on the left side from the centreof the left wheel of the vehicle. = kerb width + kerb to wheel gap + tyre width / 2 =

= 0.225 + 0.15 + 0.5/2 = 0.625 <here left is going beyond the slab edge, hence the left dispersion is limited to 0.625

1.0616

Combined dispersion widt= 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 tDispersion Width Under 'D' 11.4 tbef = a (1 - a/lo) +b1

= 1.94 a from A = 2.05a = 2.050 a from B = 5.890lo = 7.94b1 = 0.65 mbef = 3.602 m > 1.8 m

Dispersion Widths overlapLeft calculated Dispersion (bef/2) = 1.8009

left dispersion means the possible dispersion on the left side from the centreof the left wheel of the vehicle. = kerb width + kerb to wheel gap + tyre width / 2 =

= 0.225 + 0.15 + 0.5/2 0.625 <here left is going beyond the slab edge, hence the left dispersion is limited toCombined dispersion width + 3.602 /2 + 1.80

= 4.226 mIntensity of load under 'D' = 11.4 ( 1.0 + 0.323 ) 4.226

= 3.569 tDispersion Width Under 'E' 6.8 tbef = a (1 - a/lo) +b1

= 1.941 a from A = 6.350a = 1.590 a from B = 1.590lo = 7.940b1 = 0.53 mbef = 2.998 m > 1.8 m

Dispersion Widths overlap

0.225 0.2 0.19 1.8

0.38 0.5

4.70

Left Dispersion calculated (bef/2) = 1.4991left dispersion means the possible dispersion on the left side from the centreof the left wheel of the vehicle. = kerb width + kerb to wheel gap + tyre width / 2 =

0.225 + 0.15 + 0.38/2 = 0.565 <here left is going beyond the slab edge, hence the left dispersion is limited to 0.565Combined dispersion width = 0.565 + 2.998 /2 + 1.8

= 3.864 mIntensity of load under 'E' = 6.8 ( 1.00 + 0.323 )

3.864

1.80090.6250

1.4991

0.6250

= 2.328 t

4.325 3.569 2.328

0.85 1.2 4.3 1.59C D E

3.970 3.970A 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 = 55.385Ra = 6.975 tRb = 3.246 t

Shear due to dead load = wl2

= 1805 x 7.9402

= 7165.9 kgsTotal shear due to dead load & live loa = +

kgsAs per clause 304.7.1 of IRC 21-2000Design shear stress τc = V

bdwhere V = The design shear across the section

d = The depth of the sectionb = The breadth of the rectangular beam or slab, or

the breadth of the rib in the case of flanged beamhere V = 14141 kgs

b = 100 cmsd = 60 cms

= 14141100x 60

= 2.36 kg/cm2As per clause 304.7.1.1 of IRC 21-2000 Max. permissible shear s 20 grade concrete

τcmax = 1.8 N/mm2

for solid slabs the permissible shear stress shall not exceed half the value ofτcmax given in Table 12 A

Hence τcmax = 0.5 x 1.8= 0.9 N/mm2

Hence max. permissible stress = 9.0 kg/cm2Ast provided = 24.154 cm2

100 Ast / b d = 0.403from Table 12B of IRC:21-2000

τc = 0.269 N/mm2

= 2.690 kg/cm2

For solid slabs the permissible shear in concrete k * τc

(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 * τc= 1.0 x 2.690 = 2.690 Kg/cm2 > 2.36 kg/cm2

14141.3016975.45 7165.850

Hence safe against shear.