Mechanics of Materials (10th Edition)
Mechanics of Materials (10th Edition)
10th Edition
ISBN: 9780134319650
Author: Russell C. Hibbeler
Publisher: PEARSON
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Chapter 10, Problem 10.1RP

In the case of plane stress, where the in-plane principal strains are given by ε1 and ε2, show that the third principal strain can be obtained from

ε 3 = v ( ε + ε 2 ) ( 1 v )

where v is Poisson’s ratio for the material.

Expert Solution & Answer
Check Mark
To determine

To show that: The third principal strain can be obtained from ε3=ν(ε1+ε2)(1ν).

Answer to Problem 10.1RP

The third principal strain can be obtained from ε3=ν(ε1+ε2)(1ν)_ is proved.

Explanation of Solution

Given information:

The third principal strain ε3=ν(ε1+ε2)(1ν)

Explanation:

For the case of plane stress σ3=0.

Apply the normal strain in x direction as shown below.

ε1=1E(σ1ν(σ2+σ3))

Here, E is the modulus of elasticity, σ1 is the normal stress in x direction, ν is the Poisson’s ratio, σ2 is the normal stress in y direction, and σ3 is the normal stress in z direction.

Substitute 0 for σ3.

ε1=1E(σ1ν(σ2+0))=1E(σ1νσ2)Eε1=σ1νσ2

Multiply both sides of the Equation by ν.

νEε1=(σ1νσ2)ν=νσ1ν2σ2 (1)

Apply the normal strain in y direction as shown below.

ε2=1E(σ2ν(σ1+σ3))

Substitute 0 for σ3.

ε2=1E(σ2ν(σ1+0))=1E(σ2νσ1)Eε2=σ2νσ1 (2)

Apply the normal strain in z direction as shown below.

ε3=1E(σ3ν(σ1+σ2))

Substitute 0 for σ3.

ε3=1E(0ν(σ1+σ2))=1E(ν(σ1+σ2)) (3)

Adding Equation (1) and (2).

νEε1+Eε2=(νσ1ν2σ2)+(σ2νσ1)=νσ1ν2σ2+σ2νσ1=σ2(1ν2)E(νε1+ε2)=σ2(1ν2)

σ2=E(1ν2)(νε1+ε2)

Substitute E(1ν2)(νε1+ε2) for σ2 in Equation (2).

Eε2=E(1ν2)(νε1+ε2)νσ1νσ1=E(1ν2)(νε1+ε2)Eε2

σ1=1ν(E(νε1+ε2)(1ν2)Eε2(1ν2))=Eν(1ν2)(νε1+ε2ε2+ν2ε2)=Eν(1ν2)(νε1+ν2ε2)=E(1ν2)(ε1+νε2)

Substitute E(1ν2)(νε1+ε2) for σ2 and E(1ν2)(ε1+νε2) for σ1 in Equation (3).

ε3=1E(ν(E(1ν2)(ε1+νε2)+E(1ν2)(νε1+ε2)))=νEE(1ν2)(ε1+νε2+νε1+ε2)=ν(1ν)(1+ν)((1+ν)ε1+(1+ν)ε2)=ν(1+ν)(1ν)(1+ν)(ε1+ε2)

ε3=ν(1ν)(ε1+ε2)

Therefore, the third principal strain can be obtained from ε3=ν(ε1+ε2)(1ν)_ is proved.

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