Question

A single unequal angle 65 x 65 x 6 is connected to a 8 mm thick gusset plate using 4 m size fillet weld. Block shear capacity of the angle is_______. [Grade of plate is fe 410]

• A. 465.19 kN
• B. 420.6 kN
• C. 185.3 kN
• D. 658.1 kN

Answer 1

The correct option is B 420.6 kN

Since the member is welded to the gusset plate, no net areas are involved and hence

$A_{vn}and{A}_{tn}\text{in the equation for}{T}_{db}\text{should be taken to be the corresponding gross areas.}$

$A_{vn}=A_{vg}=8\times (56.5+198)=2036m{m}^2$
$A_{tn}=A_{tg}=65\times 8=520m{m}^2$

Block shear capacity
$T_{d{b}_1}=\frac{250}{\sqrt{3}\times 1.1}\times 2036+\frac{0.9\times 410}{1.25}\times 520=420.66kN$

$T_{d{b}_2}=\frac{0.9\times 410\times 2036}{\sqrt{3}\times 1.25}+\frac{250\times 520}{1.1}=465.19kN$

Hence,
$T_{db}=420.66kN$