Design the mid span section of the beam if its effective depth is kept constant at 500mm.

Using

1) Ultimate load theory and

2) Limit state theory Neglect self weight of the beam. Use $M_{20}/F_{e415}$ steel. Comment on the results obtained using either theory.

Given: b= 250mm(Assumed)

d= 500mm

$$w_L=20kN/m$$

L=5m

To find

1. Ast = ? (ULM Method)
2. Ast = ? (LSM Method)

1) ULM Method

$$w_d=1.5 D.L+2.2L.L \\ w_d=0+(2.2×20) \\ w_d=44kN/m$$

Now

$$M_u=\frac{w_d×l^2}{8}=\frac{44×5^2}{8}=137.5 kNm$$ Also

$$M_{umax}=0.25f_{ck} bd^2=0.25×20×250×500^2=312.5 kNm$$

$M_u \lt M_{umax}→$ Singly reinforced section

$$M_u=C_u × l_a=\frac{2}{3}×f_{ck} b×a×\Big(d-\frac{a}{2}\Big) \\ 137.5 ×10^6=\frac{2}{3}×20×250×a×(500-0.5a) \\ a=90.73mm \\ M_u= T_u×l_a=f_y×Ast×\Big(d-\frac{a}{2}\Big) \\ 137.5 ×10^6=415×Ast×\Big(500-\frac{90.73}{2}\Big) \\ Ast = 728.77 mm^2$$

Case 2: LSM Method

$$w_d=1.5×20=30kN/m \\ M_u=\frac{w_d×l^2}{8}=\frac{30×5^2}{8}=93.75 kNm \\ M_{umax}=0.138f_{ck} bd^2=0.138×20×250×500^2=172.5 kNm$$

$M_u \lt M_{umax}→$ Singly reinforced section $$Ast= \frac{0.5×20×250×500}{415}×\Big[1-\sqrt{1-\frac{4.6×93.75×10^3}{20×250×500^2}}\Big] \\ Ast=574.33 mm^2$$