Answered: P = 50 kN 600 mm 20 mm P = 50 kN

Engineering

Mechanical Engineering

P = 50 kN 600 mm 20 mm P = 50 kN

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter11: Columns

Section: Chapter Questions

Problem 11.2.15P: Repeat Problem 11.2-14 using L = 12 ft, ß = 0.25 kips/in., ßRl= 1.5ßL2, and ßR2= 2 ßR1.

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Concept explainers

Strain energy

The term strain indicates the variation in an object's length, area, and volume when the object is subjected to an external force. The external force may be of normal, shear, moment, torque, etc. The strain in an object depends on the types of external force and cross-section of the object. Mathematically, the evaluation of strain can be obtained with the help of elastic constants.

Question

A solid circular rod that is 600 mm long and 20 mm in diameter is subjected to an axial force of P = 50 kN. The elongation of the rod is d = 1.40 mm, and its diameter becomes d = 19.9837 mm. Determine the modulus of elasticity and the modulus of rigidity of the material. Assume that the material does not yield.

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