Question 8.25: A riveted joint, consisting of two identical rivets, is subj...

\text { Given } P=15 kN \quad \tau=60 N / mm ^{2} .

Step I Primary shear force
The primary shear force on each rivet is shown in Fig. 8.68(b). It is given by,

P_{1}^{\prime}=P_{2}^{\prime}=\frac{P}{2}=\frac{15 \times 10^{3}}{2}=7500 N .

Step II Secondary shear force
By symmetry, the centre of gravity is located midway between the centres of two rivets.
Therefore,

e = 50 mm

r_{1}=r_{2}=50 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}\right)}=\frac{\left(15 \times 10^{3}\right)(50)}{\left[(50)^{2}+(50)^{2}\right]}=150 .

P_{1}^{\prime \prime}=P_{2}^{\prime \prime}=C r_{1}=150(50)=7500 N .

Step III Resultant shear force
As shown in Fig. 8.68(b) and (c), the primary and secondary shear forces at rivet-1 act in the same direction or are additive. On the other hand, the primary and secondary shear forces at rivet-4 act in the opposite direction or are subtractive. Therefore, rivet-1 is subjected to maximum shear force.

P_{1}=P_{1}^{\prime}+P_{1}^{\prime \prime}=7500+7500=15000 N .

Step IV Diameter of rivets
Equating the resultant shear force to the shear strength of the rivet,

P_{1}=\frac{\pi}{4} d^{2} \tau \quad \text { or } \quad 15000=\frac{\pi}{4} d^{2}(60) .

\therefore \quad d=17.84 \text { or } 18 mm .