Question

A double riveted double strap butt joint has to sustain a force of 160 kN. Allowable shearing and crushing stresses of rivet material are 70 MPa and 175 MPa respectively. If the rivet diameter is 18 mm and plate thickness is 9 mm then the maximum number of rivets required per plate is

• A. 6
• B. 8
• C. 4
• D. 5

Answer 1

The correct option is A 6

Force, F = 160 kN
Allowable shearing stress, $\tau$ = 70 MPa
Crushing stress, ${\sigma }_c$ = 175 MPa
Rivet diameter, d = 18 mm
Thickness, t = 9 mm
Minimum no. of rivet, n = ?
For saftey,

$f\le f_a$

$f\le \tau n\left(2\frac{\pi d^2}{4}\right)$

$n\ge \frac{4f}{2\pi d^2\tau }$

$n\ge \frac{4\times 160\times {10}^3}{2\times \pi \times {18}^2\times 70}$

$n\ge 4.49$
Further, For safety

$f\le f_c$

$f\le {\sigma }_{c}ndt$

$n>\frac{f}{{\sigma }_{c}dt}$

$n\ge \frac{160\times {10}^3}{175\times 18\times 9}$

$n\ge 5.64$
From both the conditions, we see that minimum 6 rivets are requied per plate.