Question 8.26: A riveted joint, consisting of four identical rivets, is sub...

\text { Given } P=5 kN \quad e=200 mm \quad \tau=60 N / mm ^{2} .

Step I Primary shear force
From Eq. (8.63),

P_{1}^{\prime}=P_{2}^{\prime}=P_{3}^{\prime}=P_{4}^{\prime}=\frac{P}{(\text { No. of rivets })}              (8.63).

P_{1}^{\prime}=P_{2}^{\prime}=P_{3}^{\prime}=P_{4}^{\prime}=\frac{P}{4}=\frac{5 \times 10^{3}}{4}=1250 N .

Step II Secondary shear force

r_{1}=r_{2}=r_{3}=r_{4}=100 mm .

From Eq. (8.65),

C=\frac{P e}{\left(r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r_{4}^{2}\right)}                (8.65).

C=\frac{P e}{\left(r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r_{4}^{2}\right)}=\frac{\left(5 \times 10^{3}\right)(200)}{4(100)^{2}}=25 .

Therefore,

P_{1}^{\prime \prime}=P_{2}^{\prime \prime}=P_{3}^{\prime \prime}=P_{4}^{\prime \prime}=C r=25(100)=2500 N .

Step III Resultant shear force
The primary and secondary shear forces are shown in Fig. 8.69(b) and (c). It is observed from the figure that rivet-2 is subjected to maximum resultant force. At rivet-2, the primary and secondary shear forces are additive. Therefore,

P_{2}=P_{2}^{\prime}+P_{2}^{\prime \prime}=1250+2500=3750 N .

Step IV Diameter of Rivets

P_{2}=\frac{\pi}{4} d^{2} \tau \quad \text { or } \quad 3750=\frac{\pi}{4} d^{2}(60) .

\therefore \quad d=8.92 \text { or } 9 mm .