The doubly reinforced section having d = 600 mm, reinforced with 6-25 T at bottom and 4-20 T at top, b = 350 mm, d' = 60 mm, Fe 415 steel grade and M 20 grade is

Concept: 

Moment of resistance for balanced section,

\({M_{u,lim}} = Q{f_{ck}}b{d^2}\)

Force from the compression side,

C = 0.36 × fck × b × xu

Force from the tension side,

T = 0.87 × fy × Ast

Calculation:

Given, B = 350 mm, d = 600 mm, fck = 20 MPa and fy = 415 MPa.

For a balanced section, neutral axis depth, x_u = x_u,lim

For Fe 415, xu (lim) = 0.48 × d = 0.48 × 600 = 288 mm

From the static equilibrium condition, equating the compressive force (C) due to concrete and tensile force (T) due to steel along the neutral axis:

C = T

C = 0.36 fck× B × xu, lim 

T = 0.87 fy × Ast, lim

\({{\rm{A}}_{{\rm{st}},{\rm{lim}}}} = \frac{{0.36 × {{\rm{f}}_{{\rm{ck}}}} × {\rm{\;B\;}} × {\rm{\;}}{{\rm{x}}_{{{\rm{u}}_{{\rm{lim}}}}}}}}{{{{\rm{f}}_{{\rm{ds}}}}}}\)

\({{\rm{A}}_{{\rm{st}},{\rm{lim}}}} = \frac{{0.36{\rm{\;}} × {\rm{\;}}20{\rm{\;}} × {\rm{\;}}350{\rm{\;}} × {\rm{\;}}288}}{{0.87{\rm{\;}} × {\rm{\;}}415}}\)

∴ A_{st lim} = 2010 mm²    ........ (1)

Now we have to calculate the actual area to be provided, for which we have to add the area of tensile and compressive reinforcement

A_{st pro} = A_st1 + A_st2

= 6 × π × 25² /4 +  4 × π × 202/4

A_{st pro} = 4201 mm² ........ (2)

From the above 2 equations, we can say that area of reinforcement provided in greater than limiting reinforcement, so the given section is over reinforced section