Question

The bracket used in a crane to connect the tie rod is shown in the figure. The maximum force in the tie rod is 6 kN. The bracket is fastened by means of four identical bolts. Two at P and two at O. The bolts are made of steel of $S_{yt}$ 350 MPa and the factor of safety is 5. Assume Guests’ theory. What would be the size of bolts?

• A. M9
• B. M6
• C. M7
• D. M8

Answer 1

The correct option is D M8

The horizontal component produces direct shear force on each bolt.

$P_P^{\prime }=P_Q^{\prime }=\frac{6\times {10}^3}{4}=1500N$

The bolts at B are located at the farthest distance from the tilting edge R.

$P_Q^{\prime \prime }=\frac{Pe{l}_1}{2(l_1^2+l_2^2)}=\frac{6\times {10}^3\times (40+20)\times 150}{2({20}^2+{150}^2)}=1179.04N$

${\sigma }_t=\frac{1179.04}{\frac{\pi }{4}{d}^2}$

$\tau =\frac{P_P^{\prime }=P_Q^{\prime }}{\frac{\pi }{4}{d}^2}=\frac{1500}{\frac{\pi }{4}{d}^2}$

Maximum shear stress (Guests's theory) in bolts

${\tau }_{max}=\sqrt{{\left(\frac{{\sigma }_t}{2}\right)}^2+{\tau }^2}=\sqrt{{\left(\frac{1179.04}{2\frac{\pi }{4}{d}^2}\right)}^2+{\left(\frac{1500}{\frac{\pi }{4}{d}^2}\right)}^2}$

$\frac{0.5\times 350}{5}=\frac{1}{\frac{\pi }{4}{d}^2}\sqrt{{\left(\frac{1179.04}{2}\right)}^2+(1500{)}^2}$

$d=7.657mm\simeq 8mm$