Question

An ISA $75\times 65\times 8$ having cross-sectional area, $A=976m{m}^2$ is used as a tie member in a truss. The connection of angle section with gusset plate is done using a single row of 4 - 20 mm diameter bolts of grade 4.6 at a pitch and end distances of 60 mm and 40 mm respectively.

The strength of angle governed by block shear considering Fe 410 grade steel is

Answer 1

$A_{vg}=8\times (3\times 60+40)=1760m{m}^2$

$A_{vn}=8\times (3\times 60+40-3.5\times 22)=1144m{m}^2$

$A_{tg}=8\times 40=320m{m}^2$

$A_{tn}=8\times (40-0.5\times 22)=232m{m}^2$

Strength governed by block shear

$T_{d{b}_1}=\frac{f_y{A}_{vg}}{\sqrt{3}{\gamma }_{m_0}}+\frac{0.9f_u{A}_{tn}}{\gamma m_1}$

$=\frac{250\times 1760}{\sqrt{3}\times 1.1}+\frac{0.9\times 410\times 232}{1.25}=299.43kN$

$T_{d{b}_2}=\frac{0.9f_u{A}_{vn}}{\sqrt{3}{\gamma }_{m_1}}+\frac{f_y{A}_{tg}}{{\gamma }_{m_0}}$

$=\frac{0.9\times 410\times 1144}{\sqrt{3}\times 1.25}+\frac{250\times 320}{1.1}=267.70kN$

$T_{db}=267.70kN$