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A simply supported reinforced concrete beam of size $230mm x 600$ deep (effective) carries a super imposed load of $40$KN/m over a span of $10$ m.

The beam is reinforced with $6$ Nos. $25$mm dia. an tension fau. Design the shear reinforcement using vertical stirrups.

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$$b=230 mm , d=600 mm ,l=10m, d_c=50 mm , Ast=6-25mm$$

$ =2946mm^2\\ D=d+d_c=650mm \\ self \space ast=(0.23\times0.65)\times 23= 3.73 kn/m$

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factored shear force $(V_{UD} )= 1.5 x v \\ = 1.5 x 192.41 \\ = 288.61KNm \\ Z_{uv}=\dfrac {V_{UD}}{b\times d}=\dfrac {288.6\times10^3}{230\times600}=1.92 N/mm^2 \lt 2.8 N/mm^2 \therefore safe \\ pt= \dfrac {100Ast}{b\times d}=\dfrac {100\times6\times491}{230\times 600}=2.13 \% $

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

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$$Z_{uc} =0.8N/mm^2$$

$ V_{uc} = Z_{uc}bd= 0.8\times 230\times 600=110.4KN\\ V_{u\space min} =0.4bd=0.4\times 230\times 600=55.2 KN\\ V_{uc}+V_{u\space min}=110.4+55.2=165.6KN \\ \therefore V_{UD} \gt V_{uc}+V_{u\space min}$

Hence design & provide shear reinforcement.

$$V_{us}=V_{uD}-V_{uc}=288.61-110.6=178.21 KN\ \phi^2=10 mm $$ Assume $$a_{sv}=2\times \dfrac \pi4 \times 10^2=157mm^2$$ Spacing:- $$S_1=\dfrac {0.87f_ya_{sv}d}{V_{us}}$$ $ S_1=\dfrac {0.87\times 415\times 157\times 600}{178.21\times 10^3}=190.84\\ S_2=0.75d=0.75\times 600=450mm \\ S_3=300 mm$

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

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