# Strength of Materials - Mechanical Engineering (MCQ) questions and answers for Q. 11581

Q.  Torque and bending moment of 100 kN.m and 200 kN.m acts on a shaft which has external diameter twice of internal diameter. What is the external diameter of the shaft which is subjected to a maximum shear stress of 90 N/mm2?
- Published on 23 Sep 15

a. 116.5 mm
b. 233.025 mm
c. 587.1 mm
d. 900 mm

#### Discussion

• Sravanthi   -Posted on 25 Nov 15
Given: Torque = 100 kN.m, bending moment = 200 kN.m, external diameter is twice internal diameter (d = 2 d1), maximum shear stress (τ) = 90 N/mm2

Formula: 1) Maximum shear stress (τmax) = (√ M2 + T2 ) / (Zp )

2) Polar sectional modulus (Zp )= π (d4 – di4) / (16 d)

Solution:

Maximum shear stress (τmax) = (√ M2 + T2 ) / (Zp)

90 = [(√ 100 x 106)2 + (200 x 106 )2 ] / (Zp)

(Zp) = 2484519.975 mm3

2) Polar sectional modulus (Zp )= π (d4 – 0.5 d4) / (16 d)

2484519.975 = π (d4 – (0.5 d4) / (16 d)

2484519.975 = π (0.5 d4) / 16 d

2484519.975 = (0.19634 x d4 ) / 16 d

d = 587.1

3) d1 = 0.5 d

= 2 x 58.7.1 = 293.55 mm

Internal diameter = 293.55 mm

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