Question

An I-section is built up by welding 250 mm x 15 mm web plate to two 150 mm x 15 mm flange plates by 8 mm fillet welds. If mean shearing stress in the web and the maximum shear stress in weld are not to exceed 100 MPa, then maximum shear force which can be permitted is____ kN.

Answer 1

$I_{xx}=\frac{1}{12}[150\times {280}^3-135\times {250}^3]=98.62\times {10}^{6}m{m}^4$

Shear stress, 'q' at the section passing through welds is given by,

$q=\frac{SA\overline{y}}{I_{xx}b}$

Here, b = effective thickness
= 2 x throat thickness
= 2 x 8 x 0.7 = 11.2 mm

$A\overline{y}$ = Moment of the area about x-x
= 150 x 15 (125 + 7.5) = 298125 mm${}^3$

q = Allowable shear stress=100 N/mm${}^2$

$100=\frac{S}{98.62\times {10}^6\times 11.2}$ $\times 298125$

S= 370496.9 N

Maximum shear force limited on the web
= 250 x 15 x 100 = 375000 N

Hence maximum allowable shearing force = lesser of the above two values

= 370496.9 N = 370497 N
= 370.497 kN
$\simeq$ 370.5 kN