A rectangle beam $230mm \times 450mm$ (effective depth) is reinforced with a bar of $16$ mm diameter out of which two bars are bent at $45^{\circ}$.

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written 8.8 years ago by

teamques10 ★ 69k

• modified 8.8 years ago

$b=230 mm$

$d=450mm \\ V_{uD}=300 KN\\ Ast_{support}=4\times \dfrac \pi4 \times 16^2=804mm^2 \\ Ast_{lentup}= 2\times \dfrac \pi4 \times 16^2=402mm^2\\ Z_{uv}=\dfrac {V_{UD}}{b\times d}=\dfrac {300\times 10^3}{230\times 450}=2.7N/mm^2 \lt 2.8 N/mm^2 \therefore safe\\ pt=\dfrac {100Ast}{b\times d}=\dfrac {100\times 804}{300\times 450}=0.77N/mm^2$

shear stress in concrete $(Z_{uc})$

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$V_{uc}=Z_{uc}bd=0.564\times 230\times 450=58.37KN$

$V_{u\space min}=0.4bd=0.4\times 230\times 450=41.4 KN$ Shear load taken by bentup bars:

$V_u \space bentup1=0.87f_yAst_{bentup} \times \sin\alpha$

$=0.87\times 415\times 402\times \sin 45\\ =102.63 KN\\ V_u \space bentup2= \dfrac {V_{UD}-V_{UC}}2=\dfrac {300-58.37}2=120.81 KN \\ V_u \space bentup =102.63 KN\\ V_{uc}+V_{u\space min}+ V_{u} \space bentup =58.37+41.4+102.63=202 KN\\ V_{UD}=300 KN\\ V_{UD} \gt V_{uc}+V_{u\space min} +V_{u\space bentup}$

Hence design & provide shear reinforcement.

$V_{US}=V_{UD}-V_{UC}-V_{u\space bentup}$

$V_{US}=300-58.37-102.63\ V_{US}=139KN$ Spacing:-

$S_1=\dfrac {0.87f_ya_{sv}d}{V_{us}}$

$S_1=\dfrac {0.81\times 415\times 100\times 450}{139\times 10^3}=116.88 mm \\ S_2=0.75d=0.75\times 450=337.5 mm \\ S_3=300 mm$

Hence $10$ mm $\phi2$ strumps at $110$ mm c/c

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