Question

Point E у tw 15 kips 1 bf Point F

The dimensions for the W 21x62 beam cross section (shown below) are in inches: t_f = .615, t_w = .4, b_f = 8.24, d = 21. What is the maximum transverse shear stress on the beam cross section if the shear force acting on it is 5.2 kip?

0.706 ksi

0.25 ksi

0.365 ksi

1.023 ksi

Answer 1

Moment of inertia about neutral axis;

I bfd (bf - tw)(d – 2t :) 12 12

$\Rightarrow I = \frac{8.24(21)^3}{12} - \frac{7.84(19.77)^3}{12}$

$\Rightarrow I = 1310.808\;\;in^4$

First moment of area about neutral axis;

$Q _{max}= \sum Ay$

$\Rightarrow Q_{max} = \left [ t_w\left ( \frac{d}{2}-t_f \right ) \right ]\left [ \frac{d-2t_f}{4} \right ]+(b_f\times t_f)\left [ \frac{d-t_f}{2} \right ]$

$\Rightarrow Q_{max} = \left [ 0.4\left ( 9.885 \right ) \right ]\left [ 4.9425 \right ]+(8.24\times 0.615)\left [ 10.1925 \right ]$

$\Rightarrow Q_{max} =71.194\;\;in^3$

Maximum shear stress will be;

$\tau _{max} = \frac{VQ_{max}}{It}$

$\Rightarrow \tau_{max} = \frac{5.2(71.194)}{( 1310.808)(0.4)}$

$\Rightarrow \tau_{max} =0.706\;\;ksi$

...(Answer)