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Design the shear reinforcement for the rectangular beam of dimension $300 \times 500$ mm(effective) provided with $4-20 $ mm dia.

In tension zone the beam is subjected to udl of $50$ KN/m over the span of $7m$.Use M20 concrete and $F_{e415}$ steel. Adopt LSM.

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$$b = 300mm $$ $ d = 500mm \\ L = 7m. \\ \text{ Effective cover } (d_c ) = 50mm\\ Ast = 4 – 20mm \phi = 1256mm^2 \\ D = d + d_c =550mm \\ \text{ Self wf } = (0.3 \times 0.55) \times 25 = 4.125kn/m $

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$$Z_{uv}=\dfrac {V_{UD}}{b\times d}=\dfrac {243.55\times 10^3}{300\times 500}=1.62N/mm^2 < 2.8 N/mm^2 \therefore safe$$ $$ pt=\dfrac {100 Ast}{b\times d}=\dfrac {100\times 1256}{300\times 500}= 0.83\%$$

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

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$$V_{uc}=Z_{uc}bd=0.579\times 300\times 500=86.85KN$$ $$V_{u\space min}=0.4bd=0.4\times 300\times 500=60KN$$ $$ V_{UD}=243.55KN\ V_{UC}+V_{u\space min}=86.85 + 60 =146.85 KN$$ $$ V_{UD} > V_{UC} + V_{u\space min}$$ Hence design & provide shear reinforcement. $$V_{US}=V_{UD}-V_{UC}$$

$V_{US}=243.55-86.85\\ V_{US}=156.7 KN\\ Assume, \phi=10 mm\\ a_{sv}=2\times \dfrac \pi4 \times 10^2=157 mm^2$

Spacing :-

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

$S_1=\dfrac {0.87\times 415\times 157\times 500}{156.7\times 10^3}=180.87 \approx 175 mm \\ S_2=0.75d=0.75\times 500=375mm \\ S_3=300 mm$

Hence $10$ mm $\phi 2$ strumps at $110$ mm c/c

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