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.

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

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

The maximum shear stress in the material.

The change in the volume of the cube.

The strain energy stored in the cube.

The maximum value of normal stress along the x axis.

The required value of strain along the x axis.