Statics and Mechanics of Materials (5th Edition)
Statics and Mechanics of Materials (5th Edition)
5th Edition
ISBN: 9780134382593
Author: Russell C. Hibbeler
Publisher: PEARSON
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Chapter 8, Problem 1RP

The elastic portion of the tension stress-strain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gage length of 2 in. and a diameter of 0,5 in. When the applied load is 9 kip, the new diameter of the specimen is 0.49935 in. Calculate the shear modulus Gal for the aluminum.

Chapter 8, Problem 1RP, The elastic portion of the tension stress-strain diagram for an aluminum alloy is shown in the

Probs. R8-1/2

Expert Solution & Answer
Check Mark
To determine

Find the shear modulus for an aluminum alloy

Answer to Problem 1RP

The shear modulus for an aluminum alloy is 4.31×103ksi_.

Explanation of Solution

Given information:

Gage length is 2 in..

The diameter of the specimen is 0.5 in..

The axial load acts on the specimen is 9 kips..

The new diameter of the specimen is 0.49935 in.

Calculation:

Calculate the modulus of elasticity for aluminum (Eal) using the relation as shown below:

Eal=σε

Here, the stress is σ and the strain is ε.

Refer the stress-strain diagram.

The value of stress is 70 ksi and the value of strain is 0.00614in.in..

Substitute 70 ksi for σ and 0.00614in.in. for ε.

Eal=700.00614=11400.65ksi

The expression to find the cross-sectional area of the specimen (A) is shown below:

A=π4d2

Here, the diameter of the specimen is d.

Substitute 0.5 in. for d.

A=π4(0.5)2=0.19635in2

Find the value of stress when the specimen is loaded with a 9 kip load using the relation:

σ=PA

Here, the load is P.

Substitute 9 kip for P and 0.19635in2 for A.

σ=90.19635=45.84ksi

The expression to find the strain in the longitudinal or axial direction (εlong) using Hook’s law is shown below:

εlong=σE

Here, the Young’s modulus of the aluminum is E.

Substitute 45.84 ksi for σ and 11,400.65 ksi for E.

εlong=45.8411400.65=0.0040208in.in.

Find the strain in lateral direction (εlat) using the relation as shown below:

εlat=d'dd

Here, the new diameter is d' and the initial diameter is d.

Substitute 0.49935 in. for d' and 0.5 in. for d.

εlat=0.499350.50.5=0.0013in.in.

Find the Poisson’s ratio (ν) using the relation as shown below:

ν=εlatεlong

Substitute 0.0013in.in. for εlat and 0.0040208in.in. for εlong.

ν=0.00130.0040208=0.32332

Calculate the modulus of rigidity for the specimen (Gal) using the relation as shown below:

Gal=Eal2(1+ν)

Substitute 11,400.65 ksi for Eal and 0.32332 for ν.

Gal=11400.652(1+0.32332)=4307.594ksi= 4.31×103ksi

Therefore, the shear modulus for an aluminum alloy is 4.31×103ksi_.

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Chapter 8 Solutions

Statics and Mechanics of Materials (5th Edition)

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