15.3 (B). Show that the strain energy per unit volume of a material under a single direct stress is given by (stress x strain). Hence show that for a material under the action of the principal stresses o1,02 and o3 the strain energy per unit volume becomes 1 2E A thin cylinder 1m diameter and 3 m long is filled with a liquid to a pressure of 2 MN/m². Assuming a yield stress for the material of 240 MN/m2 in simple tension and a safety factor of 4, determine the necessary wall thickness of the cylinder, taking the maximum shear strain energy as the criterion of failure. For the cylinder material, E = 207 GN/m² and v 0.286. [14.4 mm.]

15.3 (B). Show that the strain energy per unit volume of a material under a single direct stress is given by (stress x strain). Hence show that for a material under the action of the principal stresses o1,02 and o3 the strain energy per unit volume becomes 1 2E A thin cylinder 1m diameter and 3 m long is filled with a liquid to a pressure of 2 MN/m². Assuming a yield stress for the material of 240 MN/m2 in simple tension and a safety factor of 4, determine the necessary wall thickness of the cylinder, taking the maximum shear strain energy as the criterion of failure. For the cylinder material, E = 207 GN/m² and v 0.286. [14.4 mm.]

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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Concept explainers

Stress transformation

The term stress indicates the measure of internal forces that develop due to an external force's application on an object. When a specific external force is applied to an object, internal stress generates in the material. The value of stress can obtain by taking the ratio of the externally applied force and cross-sectional area of the object. There are different types of stress depending on the type of external force.

Question

15.3 (B). Show that the strain energy per unit volume of a material under a single direct stress is given by (stress
x strain). Hence show that for a material under the action of the principal stresseso,,02 and o, the strain energy per
unit volume becomes
1
Toi+o +o3 - 2v(0,02+0,03+0203)]
2E
A thin cylinder 1m diameter and 3 m long is filled with a liquid to a pressure of 2 MN/m?. Assuming a yield stress
for the material of 240 MN/m2 in simple tension and a safety factor of 4, determine the necessary wall thickness of
the cylinder, taking the maximum shear strain energy as the criterion of failure.
For the cylinder material, E = 207 GN/m2 and v 0.286.
[14.4 mm.]

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