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An irregular shaped hollow tube, with the dimensions shown in figure below, is subjected to a torque of 2.5 kN m. If the tube is fabricated from steel and is 4 m long, determine the maximum value of the average shearing stress and the angle of twist of the tube. Note that the wall thickness of vertical sides is 6 mm and the thickness of the horizontal sides is 5 mm. ● Assume the following values: L₁ = 44 mm • L2 = 50 mm • L3= 88 mm • L4= 25 mm T L4 6 = —L₁- C Tavg= number (rtol-0.01, atol-1e-05) number (rtol-0.01, atol=1e-05) -L3 MPa L2

An irregular shaped hollow tube, with the dimensions shown in figure below, is subjected to a torque of 2.5 kN m. If the tube is fabricated from steel and is 4 m long, determine the maximum value of the average shearing stress and the angle of twist of the tube. Note that the wall thickness of vertical sides is 6 mm and the thickness of the horizontal sides is 5 mm. ● Assume the following values: L₁ = 44 mm • L2 = 50 mm • L3= 88 mm • L4= 25 mm T L4 6 = —L₁- C Tavg= number (rtol-0.01, atol-1e-05) number (rtol-0.01, atol=1e-05) -L3 MPa L2

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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An irregular shaped hollow tube, with the dimensions shown in figure below, is subjected to a torque of 2.5 kN m. If the tube is fabricated from steel and is 4 m long, determine the maximum value of the average shearing stress and the angle of twist of the tube. Note that the wall thickness of vertical sides is 6 mm and the thickness of the horizontal sides is 5 mm. * Assume the following values: • L₁= 44 mm • L₂ = 50 mm • L3= 88 mm • LA = 25 mm Taug ++ S+ O - L₁ number (rtol-0.01, atol-1e-05) number (rtol-0.01, atol-1e-05) L3 MPa L2
Transcribed Image Text:An irregular shaped hollow tube, with the dimensions shown in figure below, is subjected to a torque of 2.5 kN m. If the tube is fabricated from steel and is 4 m long, determine the maximum value of the average shearing stress and the angle of twist of the tube. Note that the wall thickness of vertical sides is 6 mm and the thickness of the horizontal sides is 5 mm. * Assume the following values: • L₁= 44 mm • L₂ = 50 mm • L3= 88 mm • LA = 25 mm Taug ++ S+ O - L₁ number (rtol-0.01, atol-1e-05) number (rtol-0.01, atol-1e-05) L3 MPa L2
Expert Solution
Step 1

Given:

The torque→T=2.5 kN.m

Length of the tube→L=4 m

Wall thickness of the vertical side→t1=6 mm=0.006 m

Thickness of the horizontal side→t2=5 mm=0.005 m = tmin

Because mean line area of the cross section:

→Ao=L4-t2×L1+L3-L1-t1×L2-t2           =25-5×44+88-44-6×50-5           =2590 mm2=0.00259 m2

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