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Civil Engineering (CE) Exam  >  Civil Engineering (CE) Tests  >  Test: Torsion of Shafts - 3 - Civil Engineering (CE) MCQ

Test: Torsion of Shafts - 3 - Civil Engineering (CE) MCQ


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10 Questions MCQ Test - Test: Torsion of Shafts - 3

Test: Torsion of Shafts - 3 for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Test: Torsion of Shafts - 3 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Torsion of Shafts - 3 MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Torsion of Shafts - 3 below.
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Test: Torsion of Shafts - 3 - Question 1
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Two shaft of same material and same length are connected in series and subjected to a torque T as shown in the figure given below?

If d2/d1 = 2, what is the value of τ1/τ2?
(where τ1 and τ2 are extreme shear stresses in either of the two segments)

Detailed Solution for Test: Torsion of Shafts - 3 - Question 1

Test: Torsion of Shafts - 3 - Question 2
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A shaft PQRS is subjected to torques at P,Q, R, S as shown in the given figure. The maximum torque for the shaft section occurs between

Detailed Solution for Test: Torsion of Shafts - 3 - Question 2

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Test: Torsion of Shafts - 3 - Question 3
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A spring of stiffness 1 kN/m is stretched initially by 100 mm from the underformed position. The work required to stretch it by another 100 mm is

Detailed Solution for Test: Torsion of Shafts - 3 - Question 3

k = 103/103 = 1N/mm
si = 100 mm
Fi = 1 x 100 = 100 N


Additional work required, 

Test: Torsion of Shafts - 3 - Question 4
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A close-coifed helical spring of stiffness 4 N/mm is in series with another spring of stiffness 6 N/mm. What is the stiffness of composite spring?
Detailed Solution for Test: Torsion of Shafts - 3 - Question 4

Let the stiffness of composite spring = k
∴  1/k = 1/k1 + 1/k2
∴ 

Test: Torsion of Shafts - 3 - Question 5
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Stiffness of a close-coiled helical spring of wire diameter d, modulus of rigidity G number of coils n, mean coil radius R, is
Test: Torsion of Shafts - 3 - Question 6
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Wahl’s factor takes into account
Test: Torsion of Shafts - 3 - Question 7
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A spring with 25, active Coils cannot be accommodated within a given space. Hence 5 coils of the spring are cut. What is the stiffness of the new spring.
Detailed Solution for Test: Torsion of Shafts - 3 - Question 7

As we know, k ∝ 1/n
∴ k2/k1 = n1/n2 = 25/20 = 1.25
⇒ k2 = 1.25 k1

Test: Torsion of Shafts - 3 - Question 8
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A close-coiled helical spring has 100 mm mean diameter and is made of 20 turns of 10 mm diameter steel wire. The spring carries an axial load of 100 N. Modulus of rigidity is 84 GPa. The shearing stress developed in the spring in N/mm2 is
Detailed Solution for Test: Torsion of Shafts - 3 - Question 8

Closed coiled helical spring subjected to axial load (W) means that every section is subjected to torsion of WR where R is the radius of spring. From torsion formula,

d= diameter of wire 

d = 10mm
W = 100 N

Shear stress due to load,

Test: Torsion of Shafts - 3 - Question 9
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A carriage spring of span 'l’ consists of n plates each of thickness ‘t' and width b. If the spring carries a load w and modulus of elasticity of the spring is E, then the central deflection of the spring is
Test: Torsion of Shafts - 3 - Question 10
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A solid shaft of diameter d is subjected to twisting moment T, such that the maximum shear stress developed is τ. Now a hole of 0.5 d diameter is drilled throdghout the length of the shaft along the axis. How much torque Tmust be reduced so that shear stress remain the same?
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