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. 205.3 mm
2. 184.56 mm
3. 160.91 mm
4. 145.30 mm

Answer 1

Correct Answer - Option 3 : 160.91 mm

Concept:

Resultant compressive force in concrete is given by,

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

Resultant compressive force in the compressive steel is given by,

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

Given compression and tension steel yield

∴ fsc = 0.87 fy

\({C_S} = {A_{sc}} \times \left( {0.87 \times {f_y} - 0.45 \times {f_{ck}}} \right)\)

Total Tensile force T = 0.87 fy × Ast

To find the Actual Neutral Axis, Equate Tensile and Compressive forces

Total compressive force = Total tensile force

i.e. Cc + Cs = T

Calculation

Given:

b = 300 mm, d = 500 mm,

A_st = 2200 mm2, A_sc = 628 mm²

M 20,∴ f_ck = 20 N/mm², Fe 250, ∴ f_y = 250 N/mm²

Given compression and tension steel yield

∴ fsc = 0.87 fy

To find the neutral axis, we equate the Total compressive and tensile forces

\(0.36 \times {f_{ck}} \times b \times {x_u} + {A_{sc}} \times \left( {0.87 \times {f_y} - 0.45 \times {f_{ck}}} \right) = 0.87 \times {f_y} \times {A_{st}}\)

\({x_u} = \frac{{0.87 \times {f_y} \times {A_{st}} - \;{A_{sc}} \times \left( {0.87 \times {f_y} - 0.45 \times {f_{ck}}} \right)}}{{0.36 \times {f_{ck}} \times b}}\)

\({x_u} = \frac{{0.87 \times 250 \times 2200 - \;628 \times \left( {0.87 \times 250 - 0.45 \times 20} \right)}}{{0.36 \times 20 \times 300}}\;\)

= 160.91 mm

∴ The actual Neutral axis depth is 160.91 mm