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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Textbook Question
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Chapter 7, Problem 7.5.10P

A brass cube of 48 mm on each edge is comp ressed in two perpendicular directions by forces P = 160 kN (see figure).

(a) Calculate the change ...IV in the volume of the cube and the strain energy U stored in the cube. assuming E = 100 GPa and i’ = 0.34.

(b) Repeat part (a) if the cube is made of an alumim um alloy with E = 73 GPa and v = 0.33.

  Chapter 7, Problem 7.5.10P, A brass cube of 48 mm on each edge is comp ressed in two perpendicular directions by forces P = 160 , example  1Chapter 7, Problem 7.5.10P, A brass cube of 48 mm on each edge is comp ressed in two perpendicular directions by forces P = 160 , example  2

(a)

Expert Solution
Check Mark
To determine

The change Δ V in the volume of the cube and strain energy U stored in the cube.

Answer to Problem 7.5.10P

The change Δ V in the volume of the cube is 49.15 mm 3 .

The strain energy U stored in the cube is 3.5195 J .

Explanation of Solution

Given information:

Forces acting on the cube in the perpendicular directions is 160 KN , modulus of elasticity is 100 GP , the length of edges is 48 mm , and the Poisson’s ratio is 0.34 .

  Mechanics of Materials (MindTap Course List), Chapter 7, Problem 7.5.10P

     Figure (1)

Write the expression for the volumetric strain.

   Δ V V = ε x + ε y + ε z .....(I)

Here, the change in the volume of the cube is Δ V , the volume of cube is V , the strain along x-axis is ε x , the strain along y-axis is ε y , and the strain along z-axis is ε z .

Write the expression for the strain energy stored in the cube.

   U = 1 2 V ο ( σ X ε x + σ Y ε y + σ Z ε z ) .....(II)

Here, the strain energy stored in the cube is U .

Write the expression for the stresses.

   σ x = Force Area .....(III)

Write the expression for the stresses along x-direction and y-direction.

   σ y = Force Area .....(IV)

Here, the stress along x-axis is σ x , the stress along y-direction is σ y .

Write the expression for the strain along x-direction.

   ε x = 1 E ( σ x ν σ y ) .....(V)

Here the Poisson’s ratio is ν .

Write the expression for the strain along y-direction.

   ε y = 1 E ( σ y ν σ x ) .....(VI)

Write the expression for the strain along z-direction.

   ε z = ν E ( σ x + σ y ) .....(VII)

Calculation:

Substitute 160 × 10 3 KN for Force , 0.048 m × 0.048 m for Area in Equation (III).

   σ x = 160 × 10 3 KN 0.048 m × 0.048 m = 69.44 × 10 6 kN / m 2 ( 1 MPa 10 6 kN / m 2 ) = 69.44 MPa

Substitute 160 × 10 3 KN for Force , 0.048 m × 0.048 m for Area in Equation (IV).

   σ y = 160 × 10 3 KN 0.048 m × 0.048 m = 69.44 × 10 6 kN / m 2 ( 1 MPa 10 6 kN / m 2 ) = 69.44 MPa

Substitute 100 × 10 9 Pa for E , 0.34 for ν , 69.44 MPa for σ x , and 69.44 MPa for σ y , in Equation (V).

   ε x = 1 100 × 10 6 MPa ( 69.44 MPa 0.34 × ( 69.44 MPa ) ) = 45.83 MPa 100 × 10 3 MPa = 485.3 × 10 6

Substitute 100 × 10 9 Pa for E , 0.34 for ν , 69.44 MPa for σ x , and 69.44 MPa for σ y , in Equation (VI).

   ε y = 1 100 × 10 6 MPa ( 69.44 MPa 0.34 × ( 69.44 MPa ) ) = 45.83 MPa 100 × 10 3 MPa = 485.3 × 10 6

Substitute 100 × 10 9 Pa for E , 0.34 for ν , 69.44 MPa for σ x , and 69.44 MPa for σ y in Equation (VII).

   ε z = 0.34 100 × 10 3 MPa ( 69.44 MPa + ( 69.44 MPa ) ) = 47.219 MPa 100 × 10 3 MPa = 472.19 × 10 6

Substitute 485.3 × 10 6 for ε x , 485.3 × 10 6 for ε y , 472.19 × 10 6 for ε z , and 110 , 592 mm 3 for V ο , in Equation (I).

   Δ V = 110592 mm 3 × ( 458.304 458.304 + 472.192 ) × 10 6 = 110592 mm 3 × ( 444.416 ) × 10 6 = 49.15 mm 3

Substitute 69.44 MPa for σ x , and 69.44 MPa for σ y

   485.3 × 10 6

   0 MPa for σ Z , for ε x , 485.3 × 10 6 for ε y , 472.19 × 10 6 for ε z , and 110 , 592 mm 3 for V ο , in Equation (II).

   U = 1 2 ( 110 , 592 mm 3 ) ( 2 × 69.44 MPa × 485.3 × 10 6 ) = 3726 mm 3 MPa ( 1 m 3 10 9 mm 3 ) ( 1 kN / m 2 1 MPa ) = 3.7 N m ( 1 J 1 N m ) = 3.7 J

Conclusion:

The change Δ V in the volume of the cube is 49.15 mm 3 .

The strain energy U stored in the cube is 7.4 J .

(b)

Expert Solution
Check Mark
To determine

The change Δ V in the volume of the cube and the strain energy U stored in the cube.

Answer to Problem 7.5.10P

The change Δ V in the volume of the cube is 71.54 mm 3 .

The strain energy U stored in the cube is 4.8943 J .

Explanation of Solution

Given information:

Modulus of elasticity is 73 GP , the length of edges is 48 mm , and the Poisson’s ratio is 0.33 .

Calculation:

Substitute 160 × 10 3 KN for Force , 0.048 m × 0.048 m for Area in Equation (III).

   σ x = 160 × 10 3 KN 0.048 m × 0.048 m = 69.44 × 10 6 kN / m 2 ( 1 MPa 10 6 kN / m 2 ) = 69.44 MPa

Substitute 160 × 10 3 KN for Force , 0.048 m × 0.048 m for Area in Equation (IV).

   σ y = 160 × 10 3 KN 0.048 m × 0.048 m = 69.44 × 10 6 kN / m 2 ( 1 MPa 10 6 kN / m 2 ) = 69.44 MPa

Substitute 73 × 10 9 Pa for E , 0.33 for ν , 69.44 MPa for σ x , and 69.44 MPa for σ y , in Equation (V).

   ε x = 1 73 × 10 6 MPa ( 69.44 MPa 0.33 × ( 69.44 MPa ) ) = 46.523 MPa 73 × 10 3 MPa = 637.326 × 10 6

Substitute 73 × 10 9 Pa for E , 0.33 for ν , 69.44 MPa for σ x , and 69.44 MPa for σ y , in Equation (VI).

   ε x = 1 73 × 10 6 MPa ( 69.44 MPa 0.33 × ( 69.44 MPa ) ) = 46.523 MPa 73 × 10 3 MPa = 637.326 × 10 6

Substitute 73 × 10 9 Pa for E , 0.33 for ν , 69.44 MPa for σ x , and 69.44 MPa for σ y in Equation (VII).

   ε z = 0.33 73 × 10 3 MPa ( 69.44 MPa + ( 69.44 MPa ) ) = 44.8304 MPa 73 × 10 3 MPa = 627.814 × 10 6

Substitute 637.326 × 10 6 for ε x , 637.326 × 10 6 for ε y , 627.814 × 10 6 for ε z , and 110 , 592 mm 3 for V ο , in Equation (I).

   Δ V = 110 , 592 mm 3 × ( 637.326 × 10 6 637.326 × 10 6 + 627.814 × 10 6 ) × 10 6 = 110592 mm 3 × 646 × 10 6 = 71.54 mm 3

Substitute 69.44 MPa for σ x , and 69.44 MPa for σ y

   485.3 × 10 6 , 0 MPa for σ Z , for 637.326 × 10 6 for ε x , 637.326 × 10 6 for ε y , 627.814 × 10 6 for ε z , and 110 , 592 mm 3 for V ο in Equation (II).

   U = 1 2 ( 110592 mm 3 ) ( 2 × 69.44 MPa × 637.326 × 10 6 ) = 4894 mm 3 MPa ( 1 m 3 10 9 mm 3 ) ( 1 kN / m 2 1 MPa ) = 4.8943 N m ( 1 J 1 N m ) = 4.8943 J

Conclusion:

The change Δ V in the volume of the cube is 71.54 mm 3 .

The strain energy U stored in the cube is 4.8943 J .

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Chapter 7 Solutions

Mechanics of Materials (MindTap Course List)

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