Concrete structures miscellaneous
Direction: A singly reinforced rectangular concrete beam has a width of 150 mm and an effective depth of 330 mm. The characteristics compressive strength of concrete is 20 MPa and the characteristics tensile strength of steel is 415 MPa. Adopt the stress block for concrete as given in IS: 456-2000 and take limiting value of depth of neutral axis as 0.48 times the effective depth of the beam.
- The limiting value of the moment of resistance of the beam is kN.m is
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Limit values of moment of resistance = F × lever arm
= 0.36.fck .b.xlim (d - 0.42 × 0.48 × d)
= 0.36 × 20 × 150 × 0.48 × 330 (330 – 0.48 × 330 × 0.42)
= 45.07 × 106 Nmm0
= 45.07 kNmCorrect Option: C
Limit values of moment of resistance = F × lever arm
= 0.36.fck .b.xlim (d - 0.42 × 0.48 × d)
= 0.36 × 20 × 150 × 0.48 × 330 (330 – 0.48 × 330 × 0.42)
= 45.07 × 106 Nmm0
= 45.07 kNm
- The limiting area of tension steel in mm2 is
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0.87 fy Ast = 0.30 fck = b.xlimit
∴ Ast = .36 × 20 × 150 × .48 × 330 0.87 × 415
= 473.82 mm2Correct Option: A
0.87 fy Ast = 0.30 fck = b.xlimit
∴ Ast = .36 × 20 × 150 × .48 × 330 0.87 × 415
= 473.82 mm2
Direction: A doubly reinforced rectangular concrete beam has a width of 300 mm and an effective depth of 500 mm. The beam is reinforced with 2200 mm2 of steel in tension and 628 mm2 of steel in compression. The effective cover for compression steel is 50 mm. Assume that both tension and compression steel yield. The grades of concrete and steel used are M20 and Fe250 respectively. The stess lock parameters (rounded off to first two decimal places) for concrete shall be as per IS 456: 2000.
- The depth of neutral axis is
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Total compression force = Total tension force
36 fck b – 0.87 fy Asc = 0.87 fy = Ast∴ x = 0.87 fy(Ast - Asc) 0.36 × fy ∴ x = 0.87 × 250 × (2200 - 628) = 158.29 mm 0.36 × 300 × 20
≈ 160.91 mmCorrect Option: C
Total compression force = Total tension force
36 fck b – 0.87 fy Asc = 0.87 fy = Ast∴ x = 0.87 fy(Ast - Asc) 0.36 × fy ∴ x = 0.87 × 250 × (2200 - 628) = 158.29 mm 0.36 × 300 × 20
≈ 160.91 mm
- The moment of resistance of the section is
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MR = 0.36 × fck (d – 0.42 x) + fsc Asc × (d – d1)
= 0.36 × 20 × 300 × 160.91 (500 – 0.42 × 160.91) + 0.87 × 250 × 628 (500 – 50)
= 211.75 ×106 ≈ 209.2 kN–MCorrect Option: B
MR = 0.36 × fck (d – 0.42 x) + fsc Asc × (d – d1)
= 0.36 × 20 × 300 × 160.91 (500 – 0.42 × 160.91) + 0.87 × 250 × 628 (500 – 50)
= 211.75 ×106 ≈ 209.2 kN–M
Direction: 
The cross-section at mid-span of a beam at the edge of a slab is shown in the sketch. A portion of the slab is considered as the effective flange width for the beam. The grades of concrete and reinforcing steel are M25 and Fe415, respectively. The total area of reinforcing bars (As) is 4000 mm2. At the ultimate limit state, xu denotes the depth of the neutral axis from the top fibre. Treat the section as under-reinforced and flanged (xu > 100 mm).
- The value of xu (in mm) computed as per the Limit State Method of IS 456: 2000 is
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Cw = 0.362 fck
Take xu ≥ 7 × Df 3
Cf = 0.447 fck. (bf – bw) Φf.
T = 0.87fy Ast
Total compressive force = total tensile force
Cw + Cf = T
0.362 fck + 0.447. fck (bf – bw) Df = 0.87 fy Ast∴ xu = (0.87 × 415 × 4000) - (0.447 × 25 × 675 × 100) (0.362 × 25 × 325) = 1444200 - 754312.5 = 234.55 mm 2941.25 Correct Option: C
Cw = 0.362 fck
Take xu ≥ 7 × Df 3
Cf = 0.447 fck. (bf – bw) Φf.
T = 0.87fy Ast
Total compressive force = total tensile force
Cw + Cf = T
0.362 fck + 0.447. fck (bf – bw) Df = 0.87 fy Ast∴ xu = (0.87 × 415 × 4000) - (0.447 × 25 × 675 × 100) (0.362 × 25 × 325) = 1444200 - 754312.5 = 234.55 mm 2941.25
