21. An aluminum tube with the cross section shown has a constant wallthickness of 2 mm. A torque T = 400 N·m is applied to the tube. Use G= 28 GPa20 mm21. What is the area of the tube?a. 957.08mm2Ob. 968.09mm2Oc.966.08mm2O d. 977.08mm240 mm40 mm10 mm 22.What is the total length of the median line? *Oa. 0.1313mb. 0.1314mOC. 0.1315mO d. 0.1316m23. What will be the result shear flow of the object?O a. 207,967.94 N/mOb. 208,968.98 N/mOc. 209,978.96 N/mO d. 209,979.00 N/m24. The shearing stress value will be *Oa. 104.5 MPaOb. 105.6 MPaOc. 104.6 MPaOd. 105.7 MPa25. The angle of twist result is *a. 14.68 Deg/mOb. 12.66 Deg/mOc. 14.68 Deg/md. None of the above.

Answered: Image /qna-images/answer/37335794-51e4-4819-b343-817f7c993271.jpg

21. An aluminum tube with the cross section shown has a constant Wall thickness of 2 mm. A torque T = 400 N·m is applied to the tube. Use G= 28 GPa LLILU OC 21. What is the area of the tube? a. 957.08mm2 Ob. 968.09mm2 Oc.966.08mm2 O d. 977.08mm2 40 mm 40 mm 10 mm

21. An aluminum tube with the cross section shown has a constant Wall thickness of 2 mm. A torque T = 400 N·m is applied to the tube. Use G= 28 GPa LLILU OC 21. What is the area of the tube? a. 957.08mm2 Ob. 968.09mm2 Oc.966.08mm2 O d. 977.08mm2 40 mm 40 mm 10 mm

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

British Gravitational Bg System Of Units

Measuring is the comparison of a particular physical quantity with its reference value. A unit is a fixed physical quantity. The length is a physical quantity. A physical quantity is the product of a unit and the numerical value we need to compare with. 

Question

21. An aluminum tube with the cross section shown has a constant wall
thickness of 2 mm. A torque T = 400 N·m is applied to the tube. Use G= 28 GPa
20 mm
21. What is the area of the tube?
a. 957.08mm2
Ob. 968.09mm2
Oc.966.08mm2
O d. 977.08mm2
40 mm
40 mm
10 mm

22.What is the total length of the median line? *
O
a. 0.1313m
b. 0.1314m
OC. 0.1315m
O d. 0.1316m
23. What will be the result shear flow of the object?
O a. 207,967.94 N/m
Ob. 208,968.98 N/m
Oc. 209,978.96 N/m
O d. 209,979.00 N/m
24. The shearing stress value will be *
Oa. 104.5 MPa
Ob. 105.6 MPa
Oc. 104.6 MPa
Od. 105.7 MPa
25. The angle of twist result is *
a. 14.68 Deg/m
Ob. 12.66 Deg/m
Oc. 14.68 Deg/m
d. None of the above.

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