Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Question
Chapter 7, Problem 7.6.8P

(a)

To determine

The bulk modulus for the nylon.

(a)

Expert Solution
Check Mark

Answer to Problem 7.6.8P

The bulk modulus of the nylon is 4.95 GPa .

Explanation of Solution

Given information:

The normal stress of the nylon element along the x axis is 3.9 MPa , normal stress along the y axis is 3.2 MPa , stress along the z axis is 1.8 MPa , the strain in the x direction is 640 × 10 6 and the strain in the y direction is 310 × 10 6 .

Explanation:

Write the expression for the bulk modulus.

   K = σ h Δ V V ...... (I)

Here, the bulk modulus is K , the hydrostatic stress is σ h , the change in volume is Δ V and the original volume of the element is V .

Write the expression for the change in volume in terms of strains.

   Δ V V = ε x + ε y + ε z ...... (II)

Here, the strain in the x direction is ε x , strain in the y direction is ε y and the strain in the z direction is ε z .

Write the expression for the hydrostatic stress in terms of stresses.

   σ h = σ x + σ y + σ z 3 ...... (III)

Here,the stress along x axis is σ x , stress along y axis is σ y and stress along z axis is σ z .

Write the expression for strain in the x direction using Hooke’s law.

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

Here, the modulus of elasticity is E and the Poisson’s ratio is ν .

Write the expression for strain in the y direction using Hooke’s law.

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

Write the expression for strain in the z direction using Hooke’s law.

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

Calculation:

Substitute 640 × 10 6 for ε x , 3.9 MPa for σ x , 1.8 MPa for σ z and 3.2 MPa for σ y in Equation (IV).

   640 × 10 6 = 1 E [ 3.9 MPa ν ( 3.2 MPa 1.8 MPa ) ] 640 × 10 6 × E = ( ( 3.9 MPa × 10 6 Pa 1 MPa ) + ( 5 MPa × 10 6 Pa 1 MPa ) × ν ) 640 × 10 6 E 5 Pa × ν × 10 6 = 3.9 Pa × 10 6 ...... (VII)

Substitute 310 × 10 6 for ε y , 3.9 MPa for σ x , 1.8 MPa for σ z and 3.2 MPa for σ y in Equation (V).

   310 × 10 6 = 1 E [ 3.2 MPa × 10 6 Pa 1 MPa ν ( 3.9 MPa × ( 10 6 Pa 1 MPa ) 1.8 MPa × ( 10 6 Pa 1 MPa ) ) ] ( 310 × 10 6 ) E = ( 3.2 Pa + 5.7 Pa × ν ) × 10 6 310 × 10 6 × E 5.7 Pa × ν × 10 6 = 3.2 Pa × 10 6 ...... (VIII)

Solve the Equations (VII) and (VIII) for the value of E and ν .

   2.098 × 10 3 E = 6.23 Pa × 10 6 E = 6.23 Pa × 10 6 2.098 × 10 3 E = 2.9695 × 10 9 Pa

Substitute 2.9695 × 10 9 Pa for E in Equation (VII).

   640 × ( 2.9695 × 10 9 ) Pa × 10 6 5 Pa × ν × 10 6 = ( 3.9 × 10 6 ) Pa 5 Pa × ν × 10 6 = 1999520 Pa ν = 0.3999

Substitute 2.9695 × 10 9 Pa for E , 0.3999 for ν , 3.9 MPa for σ x , 1.8 MPa for σ z and 3.2 MPa for σ y in Equation (VI).

   ε z = 1 2.9695 × 10 9 Pa [ ( 1.8 Pa 0.3999 ( 3.9 Pa 3.2 Pa ) ) × 10 6 ] = 1.03929 × 10 6 Pa 2.9695 × 10 9 Pa = 350 × 10 6

Substitute 3.9 MPa for σ x , 1.8 MPa for σ z and 3.2 MPa for σ y in Equation (III).

   σ h = ( 3.9 MPa ) + ( 3.2 MPa ) + ( 1.8 MPa ) 3 = 8.9 MPa 3 = 2.9667 MPa

Substitute 640 × 10 6 for ε x , 310 × 10 6 for ε y and 350 × 10 6 for ε z in Equation (II).

   Δ V V = ( 640 × 10 6 ) + ( 310 × 10 6 ) + ( 350 × 10 6 ) = ( 640 310 + 350 ) × 10 6 = 600 × 10 6

Substitute 2.9667 MPa for σ h and 600 × 10 6 for Δ V V in Equation (I).

   K = 2.967 × 10 6 Pa 600 × 10 6 = ( 4.95 × 10 9 Pa ) × ( 1 GPa 10 9 Pa ) = 4 .95 GPa

Conclusion:

The bulk modulus of the nylon is 4.95 GPa .

(b)

To determine

The modulus of elasticity.

The Poisson’s ratio.

(b)

Expert Solution
Check Mark

Answer to Problem 7.6.8P

The modulus of elasticity is 1.297 GPa .

The Poisson’s ratio is 0.399 .

Explanation of Solution

Given information:

The bulk modulus of the element is 2162 MPa , the normal stress along the x axis is 3.6 MPa , normal stress along the y axis is 2.1 MPa , normal stress along the z axis is 2.1 MPa and normal strain in the x direction is 1480 × 10 6 .

Explanation:

Write the expression for the bulk modulus.

   σ h K = 1 2 ν E ( σ x + σ y + σ z ) ...... (IX)

Calculation:

Substitute 3.6 MPa for σ x , 2.1 MPa for σ y and 2.1 MPa for σ z in Equation (III).

   σ h = ( 3.6 MPa ) + ( 2.1 MPa ) + ( 2.1 MPa ) 3 = 7.8 MPa 3 = 2.6 MPa

Substitute 3.6 MPa for σ x , 2.1 MPa for σ y , 2.1 MPa for σ z and 2162 MPa for K and 2.6 MPa for σ h in Equation (IX).

   2.6 MPa 2162 MPa = 1 2 ν E ( 3.6 MPa × ( 10 6 Pa 1 MPa ) 2.1 MPa × ( 10 6 Pa 1 MPa ) 2.1 MPa × ( 10 6 Pa 1 MPa ) ) 2.6 2162 = 1 2 ν E ( 3.6 Pa 2.1 Pa 2.1 Pa ) × 10 6 2.6 E = 2162 × ( 1 2 ν ) ( 7.8 Pa × 10 6 ) ...... (X)

Substitute 3.6 MPa for σ x , 2.1 MPa for σ y , 2.1 MPa for σ z and 1480 × 10 6 for ε x in Equation (IV).

   1480 × 10 6 = 1 E ( 3.6 MPa × ( 10 6 Pa 1 MPa ) ν ( 2.1 MPa × ( 10 6 Pa 1 MPa ) 2.1 MPa × ( 10 6 Pa 1 MPa ) ) ) 1480 × 10 6 = 1 E ( 3.6 Pa ν ( 4.2 Pa ) ) × 10 6 1480 × 10 6 × E = ( 3.6 Pa + 4.2 ν Pa ) × 10 6 ...... (XI)

Solve the Equations (X) and (XI) for the value of E and ν .

   2.6 × 10 6 3199760 × 10 6 = 7.8 × 10 6 + 15.6 × 10 6 ν 3.6 + 4.2 ν 38996256 ν = 15598128 ν = 0.399

Substitute 0.399 for ν in Equation (X).

   E = 3.6 Pa × 10 6 + 0.4 × 4.2 Pa × 10 6 1480 × 10 6 = ( 1.297 × 10 9 Pa ) × ( 1 GPa 10 9 Pa ) = 1.297 GPa

Conclusion:

The modulus of elasticity is 1.297 GPa .

The Poisson’s ratio is 0.399 .

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Chapter 7 Solutions

Mechanics of Materials (MindTap Course List)

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