Question

A single ISA $75\times 50\times 8$ is connected (longer leg) with gusset plate using 4 bolts of 20 mm diameter in one line at a pitch of 50 mm and edge distance of 30 mm. The design tensile strength due to block shear failure will be (Assume gauge distance = 35 mm) (Assume steel section is Fe410 grade steel)

• A. 213.16 kN
• B. 312.24 kN
• C. 176.45 kN
• D. 257.44 kN

Answer 1

The correct option is A 213.16 kN


$A_{vg}=8\times (3\times 50+30)=1440m{m}^2$
$A_{vn}=8\times (3\times 50+30-3.5\times 22)=824m{m}^2$
$A_{tg}=8\times 40=320m{m}^2$
Gauge, g = 35 mm for 75 mm leg
$A_{tn}=8\times (40-0.5\times 22)=232m{m}^2$
$T_{d{b}_1}=\frac{0.9A_{vn}{f}_u}{\sqrt{3}{\gamma }_{m_1}}+\frac{f_y{A}_{tg}}{{\gamma }_{m_0}}$
$=(\frac{0.9\times 410\times 824}{\sqrt{3}\times 1.25}+\frac{250\times 320}{1.1})N=213.16kN$
$T_{d{b}_2}=\frac{A_{vg}{f}_y}{\sqrt{3}{\gamma }_{m_0}}+\frac{0.9f_u{A}_{tn}}{{\gamma }_{m_1}}$
$=(\frac{1440\times 250}{\sqrt{3}\times 1.1}+\frac{0.9\times 410\times 232}{1.25})N$
$=257.44kN$
Minimum of $\{T_{d{b}_1},T_{d{b}_2}\}$
So, $T_{db}=213.16kN$