Question

A rectangular beam of 230 mm width and effective depth = 450 mm, is reinforced with 4 bars of 12 mm diameter. The grade of concrete is M 20, grade of steel is Fe500. Given that for M20 grade of concrete, the ultimate shear strength ${\tau }_{uc}=0.36N/m{m}^2$ for steel percentage of p = 0.25, and ${\tau }_{uc}=0.48N/m{m}^2$ for steel percentage p = 0.5. For a factored shear force of 45 kN, the diameter (mm) of Fe500 steel 2 legged stirrups to be used at spacing of 375 mm should be

• A. 8
• B. 10
• C. 12
• D. 16

Answer 1

The correct option is A 8


${\tau }_{uc}=0.36N/m{m}^2\left[ForM20\right]for\frac{A_{st}}{bd}\times 100=0.25$

${\tau }_{uc}=0.48N/m{m}^2\left[ForM20\right]for\frac{A_{st}}{bd}\times 100=0.5$

$FactoredSF=45kN=V_u$

We have to calculate the dia of Fe500 2-legged stirrup to be used at a spacing of 325 mm c/c.

${\tau }_v=\frac{V_u}{bd}=\frac{45\times 1000}{230\times 450}=0.4348N/m{m}^2$

$\%tensilesteel=\frac{4\times \frac{\pi }{4}(12{)}^2}{230\times 450}\times 100=0.437\%$

${\tau }_c=0.36+\frac{0.12}{0.25}\times (0.437-0.25)$

$=0.45N/m{m}^2$

Since ${\tau }_v-{\tau }_c<0$

$\Rightarrow$ Min shear reinforcement is required
$\Rightarrow$ Min shear reinforcement is given by

$\frac{A_{sv}}{b{S}_v}=\frac{0.4}{0.87fy}$

$A_{sv}=\frac{0.4\times \left(S_v\right)\left(b\right)}{0.87fy}$

$sincewelimit{f}_{y}to415N/m{m}^{2}hence,$

$A_{sv}=2\times \frac{\pi }{4}(\varphi {)}^2=\frac{0.4\times 375\times 230}{0.87\times 415}$

$\varphi =7.79mm$

$weadopt\varphi =8mm$