When the riveted strip is subjected to a stretching load W, the tensile force (i.e. tension) in each strip (equal to W) provides the shearing force on the four rivets. Since the load is sheared uniformly, i.e. each rivet is under a shearing force equal to W$\frac{}{4}$.Maximum shearing stress on each rivet $= 6.9 \times 10^{7}$Pa.Let A = area of each rivet on which the shearing force acts.$\therefore$ Shearing stress on each rivet $= \frac{\text{Shearing force}}{\text{area}}$$= \frac{\frac{W}{4}}{A} = \frac{W}{4A}$If $W_{\text{max}}$ be the maximum permissible tension or load (i.e. force) exerted by the riveted strip, then $\frac{W_{\text{max}}}{4A}=6.9 \times 10^{7}$Here, diameter of each rivet, $D = 6$ mm $= 6 \times 10^{-3} \text{m}$$\therefore A = \frac{\pi D^{2}}{4}$$= \frac{3.142 \times (6 \times 10^{-3})^{2}}{4} \dots (ii)$From (i) and (ii), we get$= 7804.73 \text{N} = 7804.73 \text{N}$$= 7.8 \times 10^{3} \text{N}$