Question

A 20 mm diameter bolt of grade 4.6 is used for double cover butt joint . Cover plates and main plates to be jointed are 8 mm and 12 mm thick respectively . The strength of 20 mm diameter bolt is _______. [ Assume thread area is at shear plane]

• A. 93.44 kN
• B. 88.60 kN
• C. 90.53 kN
• D. 99.83 kN

Answer 1

The correct option is C 90.53 kN

The strength of bolt in double shear

$V_{mb}=2\times A_{mb}\frac{f_{mb}}{\sqrt{3}\times y_{mb}}$
$=2\times (0.78\times \frac{\pi }{4}\times {20}^2)\times \frac{400}{\sqrt{3}\times 1.25}\times {10}^{-3}kN$
$=2\times 245\times \frac{400}{\sqrt{3}\times 1.25}\times {10}^{-3}kN$
=90.53 kN

The strength of the bolt in bearing

$V_{pb}=2.5K_b.d.t.\frac{f_u}{y_{mb}}$

Where $K_b$ is least of
$\frac{e}{3d_0},\frac{p}{3d_0}-0.25,\frac{f_{ub}}{f_u}$ and 1

$\Rightarrow k_{b_1}=\{\begin{array}{c}\frac{e}{3d_0}=\frac{33}{3\times 22}=0.5\\ \frac{P}{3d_0}-0.25=\frac{50}{3\times 22}=0.25=0.5\\ \frac{f_{ub}}{f_u}=\frac{400}{410}=0.975\\ and1.0\end{array}$

$\Rightarrow K_b=0.5$

$\Rightarrow V_{pb}=2.5K_b.d.t.\frac{f_u}{Y_{mb}}$

$=2.5\times 0.5\times 20\times 12\times \frac{410}{1.25}\times {10}^{-3}=98.4kN$

Therefore, the strength of the bolt will be the minimum of the strength in shearing and bearing and that is 90.53 kN.