Question

Two strips of metal are riveted together at their ends by four rivets, each of diameter $6.0\,{\text{mm}}$. What is the maximum tension that can be exerted by the riveted strip if the shearing stress is not to exceed $6.9 \times {10^7}\,{\text{Pa}}$ ?
Assume that each rivet is to carry one quarter of the load.

Answer 1

Find the radius. Calculate maximum load with the help of formula which relates load with stress and area of cross section. Multiply the load found with $4$, to find the tension.

Complete step by step answer:
According to the question, there are two metal strips which are riveted together, by four rivets.
Diameter $\left( D \right)$ of each rivet is given as $6.0\,{\text{mm}}$ .
So, the radius $\left( R \right)$ of each rivet is given by:

$$R = \dfrac{D}{2} \\\ \Rightarrow R = \dfrac{6}{2}\,{\text{mm}} \\\ \Rightarrow R= 3\,{\text{mm}} \\\$$

We know,
$1\,\,{\text{mm}} = {10^{ - 3}}\,{\text{m}}$
So,

$$R = 3\,{\text{mm}} \\\ \Rightarrow R = 3 \times {10^{ - 3}}\,{\text{m}} \\\$$

Given,
Maximum stress is $6.9 \times {10^7}\,{\text{Pa}}$ .
We now, calculate the maximum load that a rivet can hold, by the formula:
$W = \sigma \times A$ …… (1)
Where,
$W$ indicates maximum load on a rivet.
$\sigma$ indicates maximum stress.
$A$ indicates the area of the cross section of a rivet.
Now, substituting the values in equation (1), we get:

$$W = \sigma \times A \\\ \Rightarrow W = 6.9 \times {10^7} \times \pi \times {\left( {3 \times {{10}^{ - 3}}} \right)^2} \\\ \Rightarrow W = 6.9 \times {10^7} \times 3.14 \times 9 \times {10^{ - 6}} \\\ \Rightarrow W = 1949.94\,{\text{N}} \\\$$

The maximum load on a rivet comes out to be $1949.94\,{\text{N}}$ .
Since, one rivet can hold one quarter of the load. So, the maximum tension $\left( T \right)$
that can be exerted by the rivet strip is given by the following calculation:

$$T = 4 \times W \\\ \Rightarrow T= 4 \times 1949.94\,{\text{N}} \\\ \Rightarrow T= 7.799 \times {10^3}\,{\text{N}} \\\ \therefore T= 7.8 \times {10^3}\,{\text{N}} \\\$$

Hence, the maximum tension $\left( T \right)$ that can be exerted by the rivet strip is ${\text{7}}{\text{.8}} \times {10^3}\,{\text{N}}$.

Note: While solving this problem, you should have some understanding about load and tension. It is important to note that the diameter is given in millimetres, but you should convert it into metres, which is a S.I unit. After calculating the maximum load on a rivet, you should multiply this answer with $4$, since it is given in the question that one rivet can hold one quarter of the load. Failing to do this will affect the result.