Question

A beam with four 32 mm bars as main tension steel has two of its bars symmetrically bent at the ends of the beam at ${45}^{\circ }$. The design shear force (in kN) to be resisted by the stirrups, if the factored shear force at critical section is 420 kN, is Assume
$b=350mm,d=550,f_{ck}=25and{f}_y=415N/m{m}^2,{\tau }_{cmax}=3.1N/m{m}^2,{\tau }_c=0.6N/m{m}^2$

• A. 152.075

Answer 1

The correct option is A 152.075
Normal shear stress,

${\tau }_v=\frac{V_u}{bd}=\frac{420\times {10}^3}{350\times 550}=2.18N/m{m}^2$

Here, ${\tau }_v<{\tau }_{cmax}(=3.1N/m{m}^2)$
and ${\tau }_v>{\tau }_c(=0.6N/m{m}^2)$

Hence, the shear stirrups is required.
$\therefore$ Shear to be carried by stirrups,

$V_m=({\tau }_v-{\tau }_c)bd$
$=(2.18-0.6)\times 350\times 550\times {10}^{-3}$
= 304.15 kN

$\because$ Shear resistance of bent-up bars

= 0.87 $f_y{A}_{sv}sin\alpha$

$=0.87\times 415\times (2\times \frac{\pi }{4}\times {32}^2)\times sin45$

= 410.65 kN

But bent up bars are assumed to take maximum of half of shear force to be carried by stirrups.
Hence , shear to be resisted by bent up bars

= Minimum of $\left[410.65kNor\frac{304.15}{2}kN\right]$

$=\frac{304.15}{2}=152.075kN$