Question

A plate is attached using 4 rivets as shown in figure:

If the maximum allowable shear stress is 80 MPa, then what will be the minimum diameter of rivets required (in mm)?

• A. None of the above
• B. 17.84
• C. 6.97
• D. 9.48

Answer 1

The correct option is B 17.84

$\text{Direct shear force},F_D=\frac{50}{4}kN=12.5kN$

Due to moment,

$F_1{r}_1+F_2{r}_2+F_3{r}_3+F_4{r}_4=P.e$

$\Rightarrow \frac{F_1}{r_1}(r_1^2+r_2^2+r_3^2+r_4^2)=P.e$

$\Rightarrow \frac{F_1}{150}({150}^2+{50}^2+{50}^2+{150}^2)=50\times 50$

$\Rightarrow F_1=7.5kN$

$\text{Maximum shear force}=F_1+F_D=7.5+12.5$

$=20kN$

As, $\frac{F_{max}}{A}={\tau }_{all}$

$\Rightarrow \frac{20\times {10}^3}{\pi d^2/4}=80\times {10}^6$

$\Rightarrow d^2=\frac{1}{\pi \times {10}^3}$

$d=17.84mm$