A 5m long simply supported beam carries a superimposed load of $20kN/m$.

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written 8.8 years ago by

teamques10 ★ 69k

• modified 8.8 years ago

Given: b= 250mm(Assumed)

d= 500mm

$w_L=20kN/m$

L=5m

To find

Ast = ? (ULM Method)
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

$Ast= \frac{0.5×20×250×500}{415}×\Big[1-\sqrt{1-\frac{4.6×93.75×10^3}{20×250×500^2}}\Big] \\ Ast=574.33 mm^2$

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