Question

Two metal strips are riveted together at their ends by four rivets, each of diameter $6 \, \text{mm}$ . Assuming that each of the rivets carries one-quarter of the load, the maximum tension that can be produced in the riveted strip is [Breaking shear stress of the rivet = $6.9\times 10^{7} \, Pa$ ]

• A. $7.8 \times 10^{3} \, \text{N}$
• B. $6.9 \times 10^{3} \, \text{N}$
• C. $3.14 \times 10^{3} \, \text{N}$
• D. none of these

Answer 1

Answer: (A)

Diameter of each rivet $\left(\right. \text{D} \left.\right) = 6 \, \, \text{mm}$
$\therefore \text{Radius (r)} = \frac{\text{D}}{2} = \text{3 mm} = 3 \times \left(\text{10}\right)^{- 3} \text{m}$
Maximum shearing stress on each rivet = 6.9 × 10⁷ Pa
Let w be the maximum load that can be subjected to the riveted strip.
As each rivet carry one-quarter of the load, therefore
$\text{Load on each rivet} = \frac{\text{w}}{4}$
$\text{Maximum shearing stress} = \frac{\text{Maximum shearing force}}{\text{Area}} \, \, \, $
$\therefore \text{6.9} \times \text{10}^{7} = \frac{\text{w / 4}}{\text{\pi } \text{r}^{2}}$
or $\text{w} = 6.9 \times 10^{7} \, 4 \text{\pi r}^{2}$
​or $\text{w} = 6.9 \times \left(10\right)^{7} \times 4 \times 3.1 \, 4 \times \left(\right. 3 \times \left(10\right)^{- 3} \left.\right)^{2}$
$= 6.9 \times 4 \times 3.14 \times 9 \times 10$
$= 7.8 \times 10^{3} \, \text{N}$