$A bolted joint has four bolts arranged as shown in figure. the cross sectional area of each bolt is 25 m m^{2} . A torque T=200 Nm is acting on the joint. Neglecting friction due to clamping force, maximum shear stress in a bolt is$
40

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Solution

The correct option is A 40

$∴ P_{1} \times r + P_{2} r + P_{3} r + P_{4} r - T = 0$

$P_{1} = P_{2} = P_{3} = P_{4}$ [By symmetry]

$4 P r = T$

$P = \frac{T}{4 r} = \frac{200}{4 \times \frac{50}{1000}} = 1000 N$

$Shear stress = \frac{T}{Area} = \frac{1000}{25} = 40 N / m m^{2}$

= $40 MPa$

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