Question

The plates of a 6 mm thick tank are connected by a single bolted lap joint with 20 mm diameter bolts at 60 mm pitch and 35 mm edge distance. Take $f_u$ for plate as 410 MPa and assume 4.6 grade bolts. Shear plane passes through threaded portion. The efficiency of joint per pitch is_______.

• A. 56.5%
• B. 53%
• C. 67.5%
• D. 55.4%

Answer 1

The correct option is D 55.4%

Strength of bolt in single shear
$\left(V_{dsb}\right)=\frac{f_{ub}}{\sqrt{3{\gamma }_{m_b}}}(n_n{A}_{nb}+n_s{A}_{sb})$

Given $n_s=0,V_{dsb}=\frac{400}{\sqrt{3}\times 1.25}\times (0.78\times \frac{\pi }{4}\times {20}^2)=45.3kN$

Bearing capacity of bolt
$V_{dpb}=2.5k_b.d.t.f_u/{\gamma }_{m_1}$

$k=min[\frac{e}{3d_0},\frac{P}{3d_0}-0.25,\frac{f_{ub}}{f_u},1]$

$=min[\frac{35}{3\times 22},\frac{60}{3\times 22}-0.25,\frac{400}{410},1]=0.53$

$V_{dpb}=\frac{2.5\times 0.53\times 20\times 6\times 410}{1.25}=52.15kN$

$\text{Strength of bolt = min}[V_{dsb},V_{dpd}]=45.3kN$

Strength of net section per pitch length

$=\frac{0.9f_u{A}_n}{{\gamma }_{m_1}}$

$=\frac{0.9\times 410\times 6\times (60-22)}{1.25}=67.3kN$

Strength of joint per pitch length
$=min[45.3,67.3]=45.3$

Strength of solid plate per pitch length
$=\frac{250\times 6\times 60}{1.10}=81.82kN$

Joint efficiency
$=\frac{45.3}{81.82}\times 100=55.36\%$