Question

A 200 ISF 16 plate is connected to 2 -200 ISF 10 plates as shown in the figure through 20 mm dia bolts. Bolt value will be _______.
[Assume both shear plane passes through shank portion of bolt]
$[Take{f}_u=400MPaand{k}_b=1]$

• A. 228 kN
• B. 116.08 kN
• C. 106.08 kN
• D. 256 kN

Answer 1

The correct option is B 116.08 kN

$\text{Bolt value = Min. of [ Shear strength, Bearing strength ]}$

Shear strength:
$V_{dsb}=\frac{f_{ub}}{\sqrt{3}\times {\gamma }_{m_0}}\times n\times \frac{\pi }{4}\times d^2$

$=\frac{400}{\sqrt{3}\times 1.25}\times 2\times \frac{\pi }{4}\times {20}^2[\because n=2,doubleshear]$

$=116.08kN$

Bearing strength:
$V_{dpb}=2.5k_b\times f_u\times \frac{d\times t}{1.25}$

$=2.5\times 1\times \frac{410\times 16\times 20}{1.25}=262.4kN$

Minimum strength in bearing and shear
$=116.08kN$

So, Bolt value
$=116.08kN$