A nonprismatic bar ABC with a solid circular cross section is loaded by distributed torques (sec figure). The intensity of the torques, that is, the torque per unit distance, is denoted t(x) and varies linearly from zero at A to a maximum value T0/L at B. Segment BC has linearly distributed torque of intensity r(x) = T0/3L of opposite sign to that applied along AB. Also, the polar moment of inertia of AB is twice that of BC and the shear modulus of elasticity of the material is G.
- Find the reaction torque RA.
(a)
The reaction torque
Answer to Problem 3.4.21P
The reaction torque
Explanation of Solution
Given information:
The intensity of torque at point (A) is
Write the expression for equilibrium torque equation.
Here, the reaction torque is at point (A) is
Write the expression for the reaction torque is at point (A) from Equation (I)
Conclusion:
The reaction torque
(b)
The internal torsional moment in the segment
The internal torsional moment in the segment
Answer to Problem 3.4.21P
The internal torsional moment in the segment
The internal torsional moment in the segment
Explanation of Solution
Write the expression for internal torsional moment in section
Write the expression for internal torsional moment in section
Conclusion:
The internal torsional moment in the segment
The internal torsional moment in the segment
(c)
The twist rotation at point (C).
Answer to Problem 3.4.21P
The twist rotation at point (C) is
Explanation of Solution
Write the expression angle of twist at point (C).
Here, the angle of twist at point (C) is
Write the expression for polar moment of inertia of the bar.
Write the expression for polar moment of inertia in the section
Here, polar moment of inertia is in the section
Integrate Equation (IV).
Write the expression for angle of twist in
Here, angle of twist in section
Substitute
Substitute
Conclusion:
The twist rotation at point (C) is
(d)
The maximum stress and the location of maximum stress in the bar.
Answer to Problem 3.4.21P
The maximum stress is
The maximum stress in the bar is at point (A).
Explanation of Solution
Write the expression for the polar moment of inertia of section
Here, the polar moment of inertia is
Write the expression for the polar moment of inertia of section
Here, the polar moment of inertia is
Write the expression for the shear stress at point (A) from torsional equation.
Here, the shear stress is
Write the expression for the shear stress at point (A) from torsional equation.
Here, the shear stress is
Substitute
Substitute
Substitute,
Substitute
Conclusion:
The maximum stress is=
The maximum stress in the bar is at point (A).
(d)
The torsional moment diagrams.
Explanation of Solution
Write the torsional moment at point (A)
Write the torsional moment at point (B).
The following diagram shows the torsional moment diagram:
Figure-(1)
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Chapter 3 Solutions
Mechanics of Materials (MindTap Course List)
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