Question

Comprehension

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 stress lock parameters (rounded off to first two decimal places) for concrete should be as per IS: 456-2000

1. 206.00 kN - m
2. 209.20 kN - m
3. 237.80 kN - m
4. 251.90 kN - m

Answer 1

Correct Answer - Option 2 : 209.20 kN - m

Concept:

The moment of resistance(M_u) of the doubly reinforced beam is given by

M_u = C_C × (d - 0.42 × x_u) + C_S × (d - d')

Resultant compressive force in concrete (CC) is given by,

\({C_C} = 0.36 × {f_{ck}} × b × {x_u}\)

Resultant compressive force in the compressive steel(C_S) is given by,

\({C_S} = {A_{sc}} × \left( {{f_{sc}} - 0.45 × {f_{ck}}} \right)\)

Calculation:

Given:

b = 300 mm, d = 500 mm,

Ast = 2200 mm2, Asc = 628 mm2

M 20,∴ fck = 20 N/mm², Fe 250, ∴ fy = 250 N/mm2

Given compression and tension steel yield

∴ fsc = 0.87 fy

Resultant compressive force in concrete (CC) is given by,

\({C_C} = 0.36 × {{20}} × 300 × {160.91}\)(∵ x_u = 160.91 mm)

\({C_S} = {628} × \left( {{250×0.87} - 0.45 × {20}} \right)\)

Mu = 0.36 × 20 × 300 × 160.91 × (500 - 0.42 × 160.91) + 628 × (0.87× 250 - 0.45 × 20)  × (500 - 50)

M_u = 209.21 kN - m.