-22 Two tubes (AB, BC) of the same material arc connected by three pins (pin diameter = d ) just left of B as shown in the figure. Properties and dimensions for each tube are given in the figure. Torque 2ris applied at x = 2L/5 and uniformly distributed torque intensity tQ= 37/L is applied on tube BC. (a)
Find the maximum value of load variable T(N m) based on allowable shear (tx) and bearing(cha ) stresses in the three pins which connect the two tubes at B.
Use the following numerical properties:
L = 1.5m, E = 74GPa, v = 0.33, dp= 18mm, ta=45MPa, =90 MPa, di=85 mm, di = T$ mm, and d3— 60 mm.
(b)
What is the maximum shear stress in the tubes
for the applied torque in part (a)?
(a)
The maximum value of load variable for the given parameters.
Answer to Problem 3.4.22P
The maximum value of load variable for the given parameters is =
Explanation of Solution
Given information:
The total length of the tube is
Figure (1) shows the segment from point (A) to
Figure-(1)
Write the expression for moment about
Here, the reaction force at point (A) is
Substitute
Here,
Substitute
Figure-(2) shows the segment from point
Figure-(2)
Write the expression for moment about
Write the expression for the moment about
Substitute
Substitute
Maximum torsional moment in both
Write the expression for torque.
Here, torque is
Write the expression for force on the pin.
Here, the force on the pin is
Write the expression for bearing stress on the pin due to the torque.
Here, the bearing stress on the pin is
Substitute
Calculation:
Substitute
Substitute
Conclusion:
The maximum value of load variable for the given parameters is =
(b)
The maximum shear stress in the tubes.
Answer to Problem 3.4.22P
The maximum shear stress in the tubes is =
Explanation of Solution
Write the expression for maximum shear stress for the applied torque
Here, the maximum shear stress is
Write the expression for maximum torque.
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
Calculation:
Substitute
Substitute
Substitute
Substitute
Substitute
Conclusion:
The maximum shear stress in the tubes is
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Chapter 3 Solutions
Mechanics of Materials (MindTap Course List)
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- Mechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage Learning