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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Chapter 7, Problem 7.6.6P

Solve the preceding problem if the cube is granite (E = 80 GPa, v = 0.25) with dimensions E = 89 mm and compressive strains E = 690 X l0-6 and = = 255 X 10-6. For part (c) of Problem 7.6-5. find the maximum value of cr when the change in volume must be limited to 0.11%. For part. find the required value of when the strain energy must be 33 J.

  Chapter 7, Problem 7.6.6P, Solve the preceding problem if the cube is granite (E = 80 GPa, v = 0.25) with dimensions E = 89 mm

(a)

Expert Solution
Check Mark
To determine

The normal stresses acting on the x , y and z faces of the cube.

Answer to Problem 7.6.6P

The normal stress acting on the x face is 82.56 MPa .

The normal stress acting on the y face is 54.72 MPa .

The normal stress acting on the z face is 54.72 MPa .

Explanation of Solution

Given information:

A cube of cast iron having side 89 mm is tested under triaxial stress. The strain in the x direction is 690 × 10 6 , strain in y direction and z direction is being equal which is 255 × 10 6 . The modulus of elasticity is 80 GPa and the Poisson’s ratio is 0.25 .

Explanation:

Write the expression for the stress along x axis.

   σ x = E ( 1 + ν ) ( 1 2 ν ) [ ( 1 ν ) ε x + ν ( ε y + ε z ) ] ...... (I)

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

Write the expression for stress along y axis.

   σ y = E ( 1 + ν ) ( 1 2 ν ) [ ( 1 ν ) ε y + ν ( ε x + ε z ) ] ...... (II)

Here, stress along y axis is σ y .

Write the expression for stress along z axis.

   σ z = E ( 1 + ν ) ( 1 2 ν ) [ ( 1 ν ) ε z + ν ( ε x + ε y ) ] ...... (III)

Here, stress along z axis is σ z .

Calculation:

Substitute 690 × 10 6 for ε x , 255 × 10 6 for ε y , 255 × 10 6 for ε z , 80 GPa for E and 0.25 for ν in Equation (I).

   σ x = 80 GPa ( 1 + 0.25 ) ( 1 ( 2 × 0.25 ) ) [ ( 0.75 × 690 ) + ( 0.25 × ( 255 255 ) ) ] × 10 6 = 128 GPa ( 645 × 10 6 ) = 82560 GPa × ( 10 3 MPa 1 GPa ) × 10 6 = 82.56 MPa

Substitute 690 × 10 6 for ε x , 255 × 10 6 for ε y , 255 × 10 6 for ε z , 80 GPa for E and 0.25 for ν in Equation (II).

   σ y = 80 GPa ( 1 + 0.25 ) ( 1 ( 2 × 0.25 ) ) [ ( 0.75 × 255 ) + ( 0.25 × ( 690 255 ) ) ] × 10 6 = 128 GPa ( 427.5 × 10 6 ) = 54720 GPa × ( 10 3 MPa 1 GPa ) × 10 6 = 54.72 MPa

Substitute 690 × 10 6 for ε x , 255 × 10 6 for ε y , 255 × 10 6 for ε z , 80 GPa for E and 0.25 for ν in Equation (III).

   σ z = 80 GPa ( 1 + 0.25 ) ( 1 ( 2 × 0.25 ) ) [ ( 0.75 × 255 ) + ( 0.25 × ( 690 255 ) ) ] × 10 6 = 128 GPa ( 427.5 × 10 6 ) = 54720 GPa × ( 10 3 MPa 1 GPa ) × 10 6 = 54.72 MPa

Conclusion:

The normal stress acting on the x face is 82.56 MPa .

The normal stress acting on the y face is 54.72 MPa .

The normal stress acting on the z face is 54.72 MPa .

(b)

Expert Solution
Check Mark
To determine

The maximum shear stress in the material.

Answer to Problem 7.6.6P

The maximum shear stress in the material is 13.92 MPa .

Explanation of Solution

Write the expression for the maximum shear stress.

   τ max = σ max σ min 2 ...... (IV)

Here, the maximum shear stress is τ max , the maximum stress is σ max and the minimum stress is σ min .

Calculation:

Substitute 54.72 MPa for σ max and 82.56 MPa for σ min in Equation (IV).

   τ max = 54.72 MPa ( 82.56 MPa ) 2 = 27.84 MPa 2 = 13.92 MPa

Conclusion:

The maximum shear stress in the material is 13.92 MPa .

(c)

Expert Solution
Check Mark
To determine

The change in the volume of the cube.

Answer to Problem 7.6.6P

The change in the volume is 846 mm 3 .

Explanation of Solution

Write the expression for the total volumetric strain in the cube.

   ε v = ε x + ε y + ε z ...... (V)

Here, the total volumetric strain in the cube is ε v .

Write the expression for the volume of cube.

   V = a 3 ...... (VI)

Here, the volume of cube is V and side of cube is a .

Write the expression for change in volume.

   Δ V = ( ε v ) V ...... (VII)

Here, the change in the volume is Δ V .

Calculation:

Substitute 690 × 10 6 for ε x , 255 × 10 6 for ε y and 255 × 10 6 for ε z in Equation (V).

   ε v = ( 690 × 10 6 ) + ( 255 × 10 6 ) + ( 255 × 10 6 ) = ( 690 255 255 ) × 10 6 = 1200 × 10 6

Substitute 89 mm for a in Equation (VI).

   V = ( a ) 3 = ( 89 mm ) 3 = 704969 mm 3

Substitute 704969 mm 3 for V and 255 × 10 6 for ε v in Equation (VII).

   Δ V = 704969 mm 3 × ( 1200 × 10 6 ) = 845.962800 mm 3 = 846 mm 3

Conclusion:

The change in the volume is 846 mm 3 .

(d)

Expert Solution
Check Mark
To determine

The strain energy stored in the cube.

Answer to Problem 7.6.6P

The strain energy stored in the cube is 21.43 N m .

Explanation of Solution

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

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

Here, the strain energy is U .

Calculation:

Substitute 704969 mm 3 for V , 82.56 MPa for σ x , 690 × 10 6 for ε x , 54.72 MPa for σ y , 255 × 10 6 for ε y , 54.72 MPa for σ z and 255 × 10 6 for ε z in Equation (VIII).

   U = 704969 mm 3 2 [ ( 82.56 MPa × 690 × 10 6 ) + ( 54.72 MPa × 255 × 10 6 ) + ( 54.72 MPa × 255 × 10 6 ) ] = 352484.5 mm 3 ( 84873.6 × 10 6 MPa ) = ( 21429.22 MPa · mm 3 ) × ( 10 6 N / m 2 1 MPa ) × ( 1 m 3 10 9 mm 3 ) = 21.43 N m

Conclusion:

The strain energy stored in the cube is 21.43 N m .

(e)

Expert Solution
Check Mark
To determine

The maximum value of normal stress along the x axis.

Answer to Problem 7.6.6P

The maximum value of normal stress along the x axis is 72.9 MPa .

Explanation of Solution

Given information:

The change in volume is limited to 0.11 % .

Explanation:

Write the expression for the change in volume.

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

Write the expression for the stress along x axis.

   σ x = E ( 1 + ν ) ( 1 2 ν ) [ ( 1 ν ) ε x + ν ( ε y + ε z ) ] ...... (X)

Calculation:

Substitute 11 × 10 4 for Δ V V , 255 × 10 6 for ε y and 255 × 10 6 for ε z in Equation (IX).

   11 × 10 4 = ε x 255 × 10 6 255 × 10 6 ε x = 11 × 10 4 + 5.1 × 10 4 ε x = 5.9 × 10 4

Substitute 5.9 × 10 4 for ε x , 255 × 10 6 for ε y , 255 × 10 6 for ε z

   80 GPa for E and 0.25 for ν in Equation (X).

   σ x = 80 GPa ( 1 + 0.25 ) ( 1 ( 2 × 0.25 ) ) [ ( 0.75 × 5.9 × 10 4 ) + 0.25 ( 5.1 × 10 4 ) ] = 80 GPa 1.25 × 0.5 × ( 5.7 × 10 4 ) = ( 729.6 × 10 4 GPa ) × ( 10 3 MPa 1 GPa ) = 72.96 MPa

Conclusion:

The maximum value of normal stress along the x axis is 72.9 MPa .

(f)

Expert Solution
Check Mark
To determine

The required value of strain along the x axis.

Answer to Problem 7.6.6P

The required value of the strain along the x axis is 741 × 10 6 .

Explanation of Solution

Given information:

The strain energy of the system is 33 J .

Explanation:

Write the expression for the strain energy.

   U = 1 2 V E ( 1 + ν ) ( 1 2 ν ) [ ( 1 ν ) ( ε x 2 + ε y 2 + ε z 2 ) + 2 ν ( ε x ε y + ε y ε z + ε z ε x ) ] ...... (XI)

Calculation:

Substitute 704969 mm 3 for V , 33 J for U , 0.25 for ν , 80 GPa for E , 255 × 10 6 for ε y , 255 × 10 6 for ε z in Equation (XI).

   ( 33 J × 1 N m 1 J ) = ( 352484.5 mm 3 × 1 m 3 10 9 mm 3 ) × ( 80 GPa × 10 9 N / m 2 1 GPa ) 1.25 × 0.5 [ 0.75 × ( ε x 2 + 0.65025 × 10 7 + 0.65025 × 10 7 ) + 0.5 ( ( 255 × 10 6 ) ε x

+ 0.65025 × 10 7 + ( 255 × 10 6 ) ε x

) ] 7.3142 × 10 7 = [ 0.75 ε x 2 2.55 × 10 4 ε x + 1.3005 × 10 7 ] 0.75 ε x 2 2550 × 10 7 ε x 6.0137 × 10 7 = 0

Now solve the quadratic equation for obtaining the value of ε x .

   ε x = 2550 ± ( 2550 ) 2 + 4 × 0.75 × 6.0137 × 10 7 2 × 0.75 × 10 7 = 2550 ± 13671.31 1.5 × 10 7 = 7414.4208 × 10 7

Conclusion:

The required value of the strain along the x axis is 741 × 10 6 .

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