Mechanics of Materials, 7th Edition
Mechanics of Materials, 7th Edition
7th Edition
ISBN: 9780073398235
Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 11, Problem 123RP

Rod AB is made of a steel for which the yield strength is σY = 450 MPa and E = 200 GPa; rod BC is made of an aluminum alloy for which σY = 280 MPa and E = 73 GPa. Determine the maximum strain energy that can be acquired by the composite rod ABC without causing any permanent deformations.

Chapter 11, Problem 123RP, Rod AB is made of a steel for which the yield strength is Y = 450 MPa and E = 200 GPa; rod BC is

Fig. P11.123

Expert Solution & Answer
Check Mark
To determine

Find the maximum strain energy that can be acquired by the composite rod ABC.

Answer to Problem 123RP

The maximum strain energy of the composite rod ABC is 136.6J_.

Explanation of Solution

Given information:

The diameter of the composite rod AB is dAB=10mm.

The diameter of the composite rod BC is dBC=14mm.

The length of the rod AB is LAB=1.2m.

The length of the rod BC is LBC=1.6m.

The yield strength of the steel rod AB is σY=450MPa.

The modulus of elasticity of the steel rod is E=200GPa

The yield strength of the aluminum alloy BC is σY=280MPa.

The modulus of elasticity of the aluminum alloy is E=73GPa

Calculation:

Calculate the area of the rod (A) as shown below.

A=πd24 (1)

For the steel rod AB.

Substitute 10mm for d in Equation (1).

AAB=π×1024=78.54mm2

For the aluminum alloy BC.

Substitute 14mm for d in Equation (1).

ABC=π×1424=153.94mm2

Calculate the applied load (P) as shown below.

P=σYA (2)

For the steel rod AB.

Substitute 450MPa for σY and 78.54mm2 for A in Equation (2).

P=450MPa×106N/m21MPa×78.54mm2×(1m1,000mm)2=35.343×103N

For the aluminum alloy BC.

Substitute 280MPa for σY and 153.94mm2 for A in Equation (2).

P=280MPa×106N/m21MPa×153.94mm2×(1m1,000mm)2=43.103×103N

Take the smaller value as the applied load, P=35.343×103N.

Calculate the strain energy (U) as shown below.

U=P2LAB2EABAAB+P2LBC2EBCABC

Substitute 35.343×103 N for P, 200GPa for EAB, 1.2m for LAB, 78.54mm2 for AAB, 1.6m for LBC, 73GPa for EBC, and 153.94mm2 for ABC.

U=((35.343×103N)2×1.2m2×200GPa×109N/m21GPa×78.54mm2×(1m1,000mm)2+(35.343×103N)2×1.6m2×73GPa×109N/m21GPa×153.94mm2×(1m1,000mm)2)=47.7+88.9=136.6Nm×1J1Nm=136.6J

Therefore, the maximum strain energy of the composite rod ABC is 136.6J_.

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Chapter 11 Solutions

Mechanics of Materials, 7th Edition

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