Question 8.27: A bracket is attached to a steel channel by means of nine id...

\text { Given } P=50 kN \quad e=300 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_{3}^{\prime}=P_{6}^{\prime}=P_{9}^{\prime}=\frac{P}{\text { (No. of rivets) }} .

=\frac{\left(50 \times 10^{3}\right)}{9}=5555.56 N .

Step II Secondary shear force
By symmetry, the centre of gravity G is located at the center of rivet-5. The radial distances of rivet centres from the centre of gravity G are as follows:

r_{5}=0 .

r_{2}=r_{6}=r_{8}=r_{4}=100 mm .

r_{3}=r_{9}=r_{7}=r_{1}=\sqrt{100^{2}+100^{2}}=141.42 mm .

The primary and secondary shear forces acting on rivets 3, 6 and 9 are shown in Fig. 8.71.

\tan \theta=\frac{100}{100} \quad \text { or } \quad \theta=45^{\circ} .

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}+\cdots+r_{9}^{2}\right)} .

=\frac{\left(50 \times 10^{3}\right)(300)}{\left(4 \times 100^{2}+4 \times 141.42^{2}+1 \times 0^{2}\right)}=125 .

Therefore,

P_{3}^{\prime \prime}=C r_{3}=125(141.42)=17677.5 N .

P_{6}^{\prime \prime}=C r_{6}=125(100)=12500 N .

P_{9}^{\prime \prime}=C r_{9}=125(141.42)=17677.5 N .

Step III Resultant shear force
The resultant force P_{3} is given by,

P_{3}=\sqrt{\left(P_{3}^{\prime \prime} \sin \theta\right)^{2}+\left(P_{3}^{\prime \prime} \cos \theta+P_{3}^{\prime}\right)^{2}} .

=\sqrt{\left[17677.5 \sin \left(45^{\circ}\right)\right]^{2}+\left[17677.5 \cos (45)^{\circ}+5555.56\right]^{2}} .

= 21 960.1 N
The resultant force P_{9} is equal to the resultant force P_{3} .

P_{6}=P_{6}^{\prime}+P_{6}^{\prime \prime}=5555.56+12500=18055.56 N .

Therefore, rivet-3 or rivet-9 is subjected to maximum shear force.
Step IV Diameter of rivets
Equating the maximum shear force to the shear strength of the rivet,

P_{3}=\frac{\pi}{4} d^{2} \tau \quad \text { or } \quad 21960.1=\frac{\pi}{4} d^{2}(60) .

\therefore \quad d=21.59 \text { or } 22 mm .