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Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347

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BuyFindarrow_forward

Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347
Chapter 7, Problem 7.6.2P
Textbook Problem
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An element of aluminum is subjected to tri- axial stresses. Calculate the strains in the element in Chapter 7, Problem 7.6.2P, An element of aluminum is subjected to tri- axial stresses. Calculate the strains in the element in , example  1Chapter 7, Problem 7.6.2P, An element of aluminum is subjected to tri- axial stresses. Calculate the strains in the element in , example  2and Chapter 7, Problem 7.6.2P, An element of aluminum is subjected to tri- axial stresses. Calculate the strains in the element in , example  3directions if the stresses are -20 MPa, 28 MPa, and -18 MPa. respectively. Assume

E = 70 GPa and v = 0.33. Also, find the strain energy density of the element.

  Chapter 7, Problem 7.6.2P, An element of aluminum is subjected to tri- axial stresses. Calculate the strains in the element in , example  4

To determine

The strains in the element in x , y and z directions, strain in x, y direction and strain in energy density.

Explanation of Solution

Given information:

The normal stress acting in x direction is 20 MPa , the normal stress acting in y direction is 28 MPa ,the normal stress acting in z direction is 18 MPa ,the modulus of elasticity is 70 GPa and the Poisson’s ratio is 0.33 .

Explanation:

Write the expression for the normal strain in the x direction.

   ε x = 1 E ( σ x ν ( σ y + σ z ) ) .....(I)

Here, the Poisson’s ratio is ν , stress in x direction is σ x , stress in y direction is σ y , the stress in z direction is σ z , the modulus of elasticity is E and the strain in the x direction is ε x .

Write the expression for the normal strain in y direction.

   ε y = 1 E ( σ y ν ( σ x + σ z ) ) .....(II)

Here, the strain in y direction is ε y .

Write the expression for the normal strain in z direction.

   ε z = 1 E ( σ z ν ( σ x + σ y ) ) .....(III)

Here, the strain in z direction is ε z .

Write the expression for the strain energy density.

   U = 1 2 ( σ x ε x + σ y ε y + σ z ε z ) .....(IV)

Here, the strain energy is U .

Calculation:

Substitute 20 MPa for σ x , 28 MPa for σ y , 18 MPa for σ z , 70 GPa for E and 0.33 for ν in Equation (I).

   ε x = 1 70 × 10 9 Pa ( 20 × 10 6 Pa 0.33 ( 28 × 10 6 Pa 18 × 10 6 Pa ) ) = 23.3 × 10 6 Pa 70 × 10 9 Pa = 3.33 × 10 4

Substitute 20 MPa for σ x , 28 MPa for σ y , 18 MPa for σ z , 70 GPa for E and 0.33 for ν in Equation (II).

   ε y = 1 70 × 10 9 Pa ( 28 × 10 6 Pa 0

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Chapter 7 Solutions

Mechanics of Materials (MindTap Course List)
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