Design the shear reinforcement for the rectangular beam of dimensions $300 \times 50$ mm (effective) provided with $4 -20$mm in tension zone. beam is subjected to a udl of $50$KN/m having span $4m.$

Use M20 concrete and $F_{e415}$ steel. Refer table from paper of Zuc values Adopt LSM

1 Answer

$$b= 300 mm \ d=500mm\ d_c=50mm \ l=4m\ Ast=4-20mm\space \phi=1256mm^2\ D=550 mm \ self \space wt=(0.3\times 0.55)\times25=4.125kn/m$$

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factored shear force $(V_{UD} )= 1.5 \times v \\ = 1.5 \times 81.18 \\ = 121.78 KN \\ Z_{uv}=\dfrac {V_{UD}}{b\times d}=\dfrac {121.78\times10^3}{300\times500}=0.81 N/mm^2 \lt 2.8 N/mm^2 \therefore safe\\ pt=\dfrac {100Ast}{b\times d}=\dfrac {100\times1256}{300\times500}= 0.83\%$

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

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$V_{u\space min} =0.4bd=0.4\times300\times500=60 KN\\ V_{uc}+V_{u\space min}=86.85 + 60 =46.85 KN\\ \therefore V_{UD} \lt V_{uc}+ V_{u\space min}$

Hence design & provide shear reinforcement.


$$\phi_s=10mm \ a_{sv}=2\times \dfrac \pi4 \times 10^2=157mm^2$$ Spacing:- $$S_1=\dfrac {0.87f_ya_{sv}d}{V_{us}}$$

$S_1=\dfrac {0.87\times 415\times 157\times 500}{60\times 10^3}=472.37 mm \\ S_2=0.75d=0.75\times 500=375 mm \S_3=300 mm $

Hence $10$ mm $\phi2$ strumps at $175$ mm c/c

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