unit i

DESIGN OF CONCRETE STRUCTURES - I

1

V.A.SHANMUGAVELU, Assistant Professor (Selection Grade ), Department of Civil Engineering, Periyar Maniammai University, Vallam – 613 403, Thanjavur

Problems

1 Design a reinforced concrete beam spaced at a clear distance of 5 m and

supported on two walls 230 mm thick. The beam carries a super-imposed load of

2 kN/m. The width of beam is half the depth of the beam. Use M20 concrete and

Fe415 steel.

Data Given : L = 5 m ; tw = 230 mm ; w l = 2 kN/m; b = 175 mm; M20 ; Fe415

Required : Design a reinforced concrete beam

Solution :

Effective span = 5 + 0.23 = 5.23

Effective depth = 20

5230=261.5 mm

Clear cover is 20 mm and use 16 mm dia bars

Overall depth ‘D’ = 261.5 + 20 + 8 = 289.5 mm

Provide Overall depth ‘D’ = 300 mm

Width of the beam ‘b’ = 150 mm

Effective depth d = 300 – 28 = 272 mm

Again Effective span

i. 5.23 m

ii. 5+ 0.272 = 5.272 m

Consider Le = 5.23 m

Calculation of Loads:

Dead Load = 0.15 x 0.30 x 25 = 1.125 kN/m

Live load = 1.750 kN/m

Total = 2.875 kN/m

Bending Moment = M = 8

223.5x875.2= 9.83 kN-m


2


Moment of Resistance = MR = 0.914 x 150 x 2722 = 10.14 kN-m

M < MR, the section is underreinforced

Area of steel Ast =272x904.0x230

610x83.9= 173.8 mm2

No. of bars = 216x

4x8.173

= 0.86

Provide 2 bars of 16 mm diameter (Ast = 402 mm2)

2. Design a reinforced concrete beam spaced at a clear distance of 7 m and

supported on two walls 230 mm thick. The beam carries a super-imposed load of

9 kN/m. The width of beam is restricted to 230 mm. Use M20 concrete and Fe415

steel.

Data Given : L = 7 m ; tw = 230 mm ; w l = 9 kN/m; b = D/2; M20 ; Fe415

Required : Design a reinforced concrete beam

Solution :

Effective span = 7 + 0.23 = 7.23

Effective depth = 20

7230=361.5 mm


Overall depth ‘D’ = 361.5 + 20 + 8 = 389.5 mm



Again Effective span (Le)

i. 7.23 m

ii. 7 + 0.372 = 7.372 m

Consider Le = 7.23 m


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Calculation of Loads:

Dead Load = 0.230 x 0.400 x 25 = 2.30 kN/m

Live load = 9.00 kN/m

Total = 11.30 kN/m

Bending Moment = M = 8

223.7x300.11

= 73.83 kN-m

Moment of Resistance (MR) = 0.914 x 230 x 3722

= 29091084.48N-mm or 29.10 kN-m

M > MR, the section is Overreinforced

Assume n = nc = 372x2307x33.13

7x33.13

= 107.40 mm

Stress in concrete at compression steel level /cbc

= 7x40.107

2840.107 = 5.175 N/mm2

Equating moment of resistance and find the area of steel in compression

61083.7328372175.5scA133.135.1

3

40.107372

2

740.107230

Asc = 1323.80 mm2

No. of bars = 216x

4x80.1323

= 6.58

Provide 7 bars of 16 mm diameter (Asc = 1407 mm2)

Total Compression = Total Tension, find the Ast

175.51407133.135.12

740.107230

= 230

stA

Ast = 977.23 mm2


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No. of bars = 2

16x

4x23.977

= 4.86


Alternative Solution:

Lever arm a =

3

40.107372 = 336.2 mm

M1 for singly reinforced beam =

3

40.107372

2

740.107230

= 29066843.4N-mm or 29.10 kN-m

For this moment, calculate Ast1

Ast1 =2.336230

61010.29

= 376.33 mm2

Balance Bending moment = M2 = (73.83-29.10) x 106 = 44.73 x 106 N-mm

Tensile Steel required for this BM Ast1 =

28372230

61073.44

= 565.34 mm2

Total Ast = 376.33 + 565.34 = 941.67mm2

No. of bars =

216x

467.941

= 4.68


Compression Steel Asc =

34.5652840.107133.135.1

40.10737233.13

=1322.11 mm2

No. of bars =

216x

4x11.1322

= 6.58

Provide 7 bars of 16 mm diameter (Asc = 1407 mm2)


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3. Design a slab for living room of the residential building having a dimension of

2.5m x 7m. The simply supported along the edges. The width of bearing of wall is

300 mm. The live load acting on the slab is 2 kN/m2. Adopt M20 grade of concrete

and Fe 415 steel.

Data Given : Size = 2.5 m x 7 m ; tw = 300 mm ; w l = 2 kN/m2 ; M20 ; Fe415;

Simply supported

Required : Design a reinforced slab

Solution :

Effective span

Short span = 2.5 + 0.3 = 2.8 m

Long span = 7.0 + 0.3 = 7.3 m

Span Ratio = 8.2

3.7 = 2.61 > 2, One way slab

Effective depth =

20

2800=140 mm


Overall depth ‘D’ = 140 + 15 + 5 = 160 mm



Again find the Effective span

i. 2.80 m

ii. 2.5+ 0.155 = 2.655 m

Le = 2.655 m

Calculation of Loads: Consider 1 m wide slab

Dead load = 0.175 x 25 = 4.375 kN/m2

Live load = 2.000 kN/m2

Total = 6.375 kN/m2


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Bending Moment = M =

8

2655.2x375.6= 5.618 kN-m

Moment of Resistance = MR = 0.914 x1000 x 1552 = 21.96 kN-m

M < MR , the section is designed as underreinforced section

Main Reinforcement:

Area of steel Ast =155x904.0x230

610x618.5= 174.3 mm2

Ast min = 175x1000x100

12.0= 210 mm2

Spacing of 10 dia bars =

210x4

210xx1000

= 374 mm

Maximum spacing : Whichever is lesser of the following

i. 374 mm

ii. 300 mm

iii. 3 x 130 = 390 mm

Spacing Provided = 300 mm

Provide 10 mm diameter @ 300 mm c/c (Ast = 378 mm2)

Distribution Reinforcement:

Asd = 175x1000x100

12.0= 210 mm2

Spacing of 8 dia bars =210x4

28xx1000

= 239 mm

Maximum spacing : Whichever is lesser of the following

i. 239 mm

ii. 450 mm

iii. 5 x 154 = 770 mm

Provide 8 mm diameter @ 225 mm c/c (Ast = 224 mm2)


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4. A simply supported beam of 300 mm wide and 500 mm effective depth carries a

uniformly distributed load of 50 kN/m including its own weight over an effective

span of 6m. Design the shear reinforcement by vertical stirrups. The tensile

reinforcement consists of 3 bars of 20 mm diameter. Adopt M20 grade concrete

and Fe415 steel.

Data Given : L = 6 m ; b=300 mm ; d = 500 mm ; w = 50 kN/m; Ast =3-20 mm; M20

; Fe415

Required : Design a vertical stirrups

Solution :

Ast =

4

203 2xx= 943 mm2

Shear force = V =

4

650x = 75 kN

Nominal shear stress = v =

500300

1075 3

x

x = 0.5 N/mm2

% p =

5003004

203100 2

xx

xxx = 0.63

Table 23 of IS 456,Permissible shear stress in concrete c = 0.326 N/mm2

Table 23 of IS 456, Maximum shear stress cmax = 1.8 N/mm2

Cl. B5.4 of IS 456 v > c < cmax , Shear reinforcement shall be provided.

Net shear force = 75000 – 0.326 x 300 x 500 = 26100 N

Use two legged 10 mm diameter stirrups

sv =

261004

500230102 2

x

xxxx = 692 mm


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Spacing of stirrups shall not exceed the following

i. 692 mm

ii. 450 mm

iii. 0.75 x 500 = 375 mm

Provide 2 legged 10mm dia stirrups @ 350 mm c/c

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