A Doubly reinforced rectangular beam has a width of 300 mm and an

SOLUTION

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

∴ f_sc = 0.87 f_y

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

Total Tensile force T = 0.87 f_y × A_st

For Actual Neutral Axis

Total compressive force = Total tensile force

i.e. C_c + C_s = T

Limiting NA for Fe250 grade is given by x_u,lim =  0.53 × d

Calculation

Given,

b = 300 mm, d = 500 mm,

Area of steel in tension zone = 2200 mm²,

Area of steel in Compression zone = 628 mm²

Given compression and tension steel yield

∴ f_sc = 0.87 f_y

For limiting depth of NA

We know for Fe250 grade of steel

\({x_{u,lm}} = 0.53 \times d\)

= 0.53 × 500 = 265 mm

For actual neutral axis

Compressive force = Tensile force.

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

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

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

= 160.91 mm

∴ Actual Neutral axis depth is 160.91 mm < 265 mm

i.e. \({x_u} < {x_{u,lim}}\)

∴ Beam is under reinforced