Question 8.24: A bracket, attached to a vertical column by means of four id...

\text { Given } P=25 kN \quad e=100 mm \quad \tau=60 N / mm ^{2} .

Step I Primary shear force
The primary shear force on each rivet is shown in Fig. 8.67(c). It is given by,

P_{1}^{\prime}=P_{2}^{\prime}=P_{3}^{\prime}=P_{4}^{\prime}=\frac{P}{4}=\frac{25 \times 10^{3}}{4}=6250 N .

Step II Secondary shear force
By symmetry, the centre of gravity of four rivets is located midway between the centres of rivets 2 and 3. The radial distances of the rivet centre from this centre of gravity are as follows:

r_{1}=r_{4}=150 mm .

r_{2}=r_{3}=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}+r_{3}^{2}+r_{4}^{2}\right)} .

=\frac{\left(25 \times 10^{3}\right)(100)}{\left[2(150)^{2}+2(50)^{2}\right]}=50 .

P_{1}^{\prime \prime}=P_{4}^{\prime \prime}=C r_{1}=50(150)=7500 N .

P_{2}^{\prime \prime}=P_{3}^{\prime \prime}=C r_{2}=50(50)=2500 N .

Step III Resultant shear force
Rivets 1 and 4, which are located at the farthest distance from the centre of gravity, are subjected to maximum shear force. As shown in Fig. 8.67(e), the primary and secondary shear forces acting on rivets 1 or 4 are at right angles to each other. The resultant shear force P_{1} is given by,

P_{1}=\sqrt{\left(P_{1}^{\prime}\right)^{2}+\left(P_{1}^{\prime \prime}\right)^{2}}=\sqrt{(6250)^{2}+(7500)^{2}} .

=9762.81 N .

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

P_{1}=\frac{\pi}{4} d^{2} \tau \quad \text { or } \quad 9762.81=\frac{\pi}{4} d^{2}(60) .

\therefore \quad d=14.39 \text { or } 15 mm .