Machine Design (EMCH327 Section 001, Fall 2024)
Mid Term Exam 2, 8am, Thursday, Oct. 10, 2024
Name (First/Last): ___________________________________ Score: _______________/100’
1. Static loading design: A cantilevered hollow tube is subject to the combined loadings. The
bending load is 𝐹 = 1.75 kN, the axial tension is 𝑃 = 9 kN, and the torsion is 𝑇 = 72 N ∙ m.
Use Ref. h in Appendix Table 15-19 for the relevant cross-sectional properties of the tube,
determine the von Mises stress for the point A on the top surface of the tube. Also, for the
same point, use the distortion energy theory (DET) to determine the design factor so that the
2014 aluminum alloy with yield strength of 276 MPa can be safely selected as material for
the tube. (30’ in total)
Solutions:
The critical stress element is at point A on the top surface of the tube, where the bending moment is the
largest, and the normal and shear stresses induced by the combined loadings are at their maximum values.
For point A, this is the plane stress problem (specifically, in xy-plane). For
the xyz coordination system established, the 2D plane stress element is
shown on the right, with 𝜎𝑥 = 0.
Using Ref. h in Appendix Table 15-19, we have following relevant cross-
sectional properties provided therein:
Cross-sectional area 𝐴 = 515.2 mm2 = 515.2 × 10−6 m2
Bending section modulus 𝑆 = 4857 mm3 = 4857 × 10−9 m3
Polar section modulus 𝑍𝑝 = 9715 mm3 = 9715 × 10−9 m3
Since the normal stresses induced by both axial tension and bending moment are along the y-axis, they are
additive, giving
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, 𝑃 𝑀 9000 1750 × 0.12
𝜎𝑦 = + = −6
+ = 60.71 MPa
𝐴 𝑆 515.2 × 10 4857 × 10−9
The shear stress component induced by the torque is
𝑇 72
𝜏𝑥𝑦 = = = 7.41 MPa
𝑍𝑝 9715 × 10−9
With these stress components, we can then draw the Mohr’s circle with center position (𝜎𝐶 , 0) =
𝜎𝑥 +𝜎𝑦
( 2
, 0) = (30.355, 0) and reference point position (𝜎𝑦 , 𝜏𝑥𝑦 ) = (60.71, 7.41).
The radius of the Mohr’s circle is 𝑅 = √(60.71 − 30.355)2 + (7.41 − 0)2 = 31.246
Then, the two principal stresses are
𝜎1 = 𝜎𝐶 + 𝑅 = 30.355 + 31.246 = 61.601 MPa
𝜎2 = 0
𝜎3 = 𝜎𝐶 − 𝑅 = 30.355 − 31.25 = −0.891 MPa
The von Mises stress is
(𝜎1 − 𝜎2 )2 + (𝜎2 − 𝜎3 )2 + (𝜎3 − 𝜎1 )2
𝜎𝑒 = √
2
(61.601)2 + (0.891)2 + (61.601 + 0.891)2
=√ = 62.05 MPa
2
A stress is considered safe when von Mises Stress is below the yield strength of the material with design
factor of N:
𝑠𝑦
𝜎𝑒 ≤
𝑁
Given 𝑠𝑦 = 276 MPa, we finally obtain the design factor as
𝑠𝑦 276
𝑁= = = 4.45
𝜎𝑒 62.05
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Mid Term Exam 2, 8am, Thursday, Oct. 10, 2024
Name (First/Last): ___________________________________ Score: _______________/100’
1. Static loading design: A cantilevered hollow tube is subject to the combined loadings. The
bending load is 𝐹 = 1.75 kN, the axial tension is 𝑃 = 9 kN, and the torsion is 𝑇 = 72 N ∙ m.
Use Ref. h in Appendix Table 15-19 for the relevant cross-sectional properties of the tube,
determine the von Mises stress for the point A on the top surface of the tube. Also, for the
same point, use the distortion energy theory (DET) to determine the design factor so that the
2014 aluminum alloy with yield strength of 276 MPa can be safely selected as material for
the tube. (30’ in total)
Solutions:
The critical stress element is at point A on the top surface of the tube, where the bending moment is the
largest, and the normal and shear stresses induced by the combined loadings are at their maximum values.
For point A, this is the plane stress problem (specifically, in xy-plane). For
the xyz coordination system established, the 2D plane stress element is
shown on the right, with 𝜎𝑥 = 0.
Using Ref. h in Appendix Table 15-19, we have following relevant cross-
sectional properties provided therein:
Cross-sectional area 𝐴 = 515.2 mm2 = 515.2 × 10−6 m2
Bending section modulus 𝑆 = 4857 mm3 = 4857 × 10−9 m3
Polar section modulus 𝑍𝑝 = 9715 mm3 = 9715 × 10−9 m3
Since the normal stresses induced by both axial tension and bending moment are along the y-axis, they are
additive, giving
Page 1 of 6
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https://www.coursehero.com/file/244414950/Mid-Term-Exam-2-Solutionspdf/
, 𝑃 𝑀 9000 1750 × 0.12
𝜎𝑦 = + = −6
+ = 60.71 MPa
𝐴 𝑆 515.2 × 10 4857 × 10−9
The shear stress component induced by the torque is
𝑇 72
𝜏𝑥𝑦 = = = 7.41 MPa
𝑍𝑝 9715 × 10−9
With these stress components, we can then draw the Mohr’s circle with center position (𝜎𝐶 , 0) =
𝜎𝑥 +𝜎𝑦
( 2
, 0) = (30.355, 0) and reference point position (𝜎𝑦 , 𝜏𝑥𝑦 ) = (60.71, 7.41).
The radius of the Mohr’s circle is 𝑅 = √(60.71 − 30.355)2 + (7.41 − 0)2 = 31.246
Then, the two principal stresses are
𝜎1 = 𝜎𝐶 + 𝑅 = 30.355 + 31.246 = 61.601 MPa
𝜎2 = 0
𝜎3 = 𝜎𝐶 − 𝑅 = 30.355 − 31.25 = −0.891 MPa
The von Mises stress is
(𝜎1 − 𝜎2 )2 + (𝜎2 − 𝜎3 )2 + (𝜎3 − 𝜎1 )2
𝜎𝑒 = √
2
(61.601)2 + (0.891)2 + (61.601 + 0.891)2
=√ = 62.05 MPa
2
A stress is considered safe when von Mises Stress is below the yield strength of the material with design
factor of N:
𝑠𝑦
𝜎𝑒 ≤
𝑁
Given 𝑠𝑦 = 276 MPa, we finally obtain the design factor as
𝑠𝑦 276
𝑁= = = 4.45
𝜎𝑒 62.05
Page 2 of 6
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