Mumbai University > CIVIL > Sem 7 > Limit State Method

Marks: 10 M

Year: May 2013

Given: b=300mm

d = 450mm.

$$σ_{cu}=f_{ck}=20N/mm^2 \\ σ_{sy}=f_y=420N/mm^2 \\ M_u=240kNm$$

To find = Ast = ?

$$M_{umax}=0.25f_{ck} bd^2=0.25×20×300×450^2=303.75 kNm$$

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

$$M_u=C_u × l_a=\frac{2}{3}×f_{ck} b×a×\Big(d-\frac{a}{2}\Big) \\ 240×10^6=\frac{2}{3}×20×300×a×\Big(450-\frac{a}{2}\Big)$$

a = 162.77mm

$$M_u= T_u×l_a=f_y×Ast×\Big(d-\frac{a}{2}\Big) \\ 224×10^6= 420×Ast×\Big(450-\frac{162.77}{2}\Big) \\ Astr = 1550.2 mm^2$$

$Provide \ 5-20 \ mm∅ \\ Therefore \ Astp = 1570mm^2%$