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.
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.
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.
(a)
Expert Solution
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).
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
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
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).
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