1. DESIGN OF CONCRETE STRUCTURES - I
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 =
5230
=261.5 mm
20
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
kN/m
= 1.750
kN/m
Total = 2.875
Live load
= 1.125
kN/m
2.875x5.232
Bending Moment = M =
= 9.83 kN-m
8
1
V.A.SHANMUGAVELU, Assistant Professor (Selection Grade ), Department of Civil Engineering,
Periyar Maniammai University, Vallam – 613 403, Thanjavur

2. DESIGN OF CONCRETE STRUCTURES - I
Moment of Resistance = MR = 0.914 x 150 x 2722 = 10.14 kN-m
M < MR, the section is underreinforced
Area of steel Ast =
No. of bars =
9.83x106
= 173.8 mm2
230x0.904x272
173.8x4
= 0.86
2
x16
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 =
7230
=361.5 mm
20
Clear cover is 20 mm and use 16 mm dia bars
Overall depth ‘D’ = 361.5 + 20 + 8 = 389.5 mm
Provide Overall depth ‘D’ = 400 mm
Effective depth d = 400 – 28 = 372 mm
Again Effective span (Le)
i. 7.23 m
ii. 7 + 0.372 = 7.372 m
Consider Le = 7.23 m
2
V.A.SHANMUGAVELU, Assistant Professor (Selection Grade ), Department of Civil Engineering,
Periyar Maniammai University, Vallam – 613 403, Thanjavur

3. DESIGN OF CONCRETE STRUCTURES - I
Calculation of Loads:
Dead Load = 0.230 x 0.400 x 25
=
2.30 kN/m
=
9.00 kN/m
Total =
11.30 kN/m
Live load
Bending Moment = M =
11.300x7.232
= 73.83 kN-m
8
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 =
13.33x7
x372= 107.40 mm
13.33x7  230
107.40  28
Stress in concrete at compression steel level  / =
x7 = 5.175 N/mm2
cbc
107.40
Equating moment of resistance and find the area of steel in compression
230 107.40 7 
107.40 
6
 372
  1.5 13.33  1 A sc  5.175 372 28  73.83 10
2
3 

Asc = 1323.80 mm2
No. of bars =
1323.80x4
= 6.58
x162
Provide 7 bars of 16 mm diameter (Asc = 1407 mm2)
Total Compression = Total Tension, find the Ast
230 107.40 7
 1.5 13.33  1 1407 5.175= A st  230
2
Ast = 977.23 mm2
3
V.A.SHANMUGAVELU, Assistant Professor (Selection Grade ), Department of Civil Engineering,
Periyar Maniammai University, Vallam – 613 403, Thanjavur

4. DESIGN OF CONCRETE STRUCTURES - I
No. of bars =
977.23x4
= 4.86
2
x16
Provide 5 bars of 16 mm diameter (Ast = 1005 mm2)
Alternative Solution:
Lever arm a = 372
107.40
= 336.2 mm
3
M1 for singly reinforced beam =
230 107.40 7 
107.40 
 372

2
3 

= 29066843.4N-mm or 29.10 kN-m
For this moment, calculate Ast1
Ast1 =
6
29.10 10
= 376.33 mm2
230 336.2
Balance Bending moment = M2 = (73.83-29.10) x 106 = 44.73 x 106 N-mm
Tensile Steel required for this BM Ast1 =
6
44.73 10
= 565.34 mm2
230 372 28
Total Ast = 376.33 + 565.34 = 941.67mm2
No. of bars =
941.67  4
= 4.68
2
x16
Provide 5 bars of 16 mm diameter (Ast = 1005 mm2)
Compression Steel Asc =
No. of bars =
13.33  372  107.40
 565.34 =1322.11 mm2
1.5  13.33  1  107.40  28
1322.11x4
= 6.58
x162
Provide 7 bars of 16 mm diameter (Asc = 1407 mm2)
4
V.A.SHANMUGAVELU, Assistant Professor (Selection Grade ), Department of Civil Engineering,
Periyar Maniammai University, Vallam – 613 403, Thanjavur

5. DESIGN OF CONCRETE STRUCTURES - I
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 =
7 .3
= 2.61 > 2, One way slab
2 .8
Effective depth =
2800
=140 mm
20
Clear cover is 15 mm and use 10 mm dia bars
Overall depth ‘D’ = 140 + 15 + 5 = 160 mm
Provide Overall depth ‘D’ = 175 mm
Effective depth d = 175 – 20 = 155 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 =
=
4.375
kN/m2
=
2.000
kN/m2
Total =
Live load
0.175 x 25
6.375
kN/m2
5
V.A.SHANMUGAVELU, Assistant Professor (Selection Grade ), Department of Civil Engineering,
Periyar Maniammai University, Vallam – 613 403, Thanjavur

6. DESIGN OF CONCRETE STRUCTURES - I
Bending Moment = M =
6.375x2.6552
= 5.618 kN-m
8
Moment of Resistance = MR = 0.914 x1000 x 1552 = 21.96 kN-m
M < MR , the section is designed as underreinforced section
Main Reinforcement:
5.618x106
Area of steel Ast =
= 174.3 mm2
230x0.904x155
Ast min =
0.12
x1000x175= 210 mm2
100
Spacing of 10 dia bars =
1000xx102
= 374 mm
4x210
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 =
0.12
x1000x175= 210 mm2
100
Spacing of 8 dia bars =
1000xx82
= 239 mm
4x210
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)
6
V.A.SHANMUGAVELU, Assistant Professor (Selection Grade ), Department of Civil Engineering,
Periyar Maniammai University, Vallam – 613 403, Thanjavur

7. DESIGN OF CONCRETE STRUCTURES - I
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 =
:
3xx 20 2
= 943 mm2
4
Shear force = V =
50x6
= 75 kN
4
Nominal shear stress = v =
%p=
75 x10 3
= 0.5 N/mm2
300 x500
100 x3xx 20 2
= 0.63
4 x300 x500
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 =
2 xx10 2 x 230 x500
= 692 mm
4 x 26100
7
V.A.SHANMUGAVELU, Assistant Professor (Selection Grade ), Department of Civil Engineering,
Periyar Maniammai University, Vallam – 613 403, Thanjavur

8. DESIGN OF CONCRETE STRUCTURES - I
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
8
V.A.SHANMUGAVELU, Assistant Professor (Selection Grade ), Department of Civil Engineering,
Periyar Maniammai University, Vallam – 613 403, Thanjavur