# A cube of cast iron with sides of length a = 4.0 in. (see figure) is tested in a laboratory under triaxialsire.ss. Gages mounted on the testing machine show that the compressive strains in the material arc a = -225 X l06and,a y = 37.5 X l0_. Determine the following quantities: (a) the norm al stresses i. r,.. and acting on the x, y, and z faces of the cube; (b) the maximum shear stress r in the material; (C) the change ..W in the volume of the cube: (d) the strain energy U stored in the cube; (e) the maximum value of s when the change in volume must be limited to O.O28%; and (f) the required value of when the strain energy must be 38 in.-lb. (Assume £ = 14,000 ksi and v = 0.25.)

### Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347

### Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347
Chapter 7, Problem 7.6.5P
Textbook Problem
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## A cube of cast iron with sides of length a = 4.0 in. (see figure) is tested in a laboratory under triaxialsire.ss. Gages mounted on the testing machine show that the compressive strains in the material arc a= -225 X l06and,ay = 37.5 X l0_. Determine the following quantities: (a) the norm al stresses i. r,.. and acting on the x, y, and z faces of the cube; (b) the maximum shear stress r in the material; (C) the change ..W in the volume of the cube: (d) the strain energy U stored in the cube; (e) the maximum value of s when the change in volume must be limited to O.O28%; and (f) the required value of when the strain energy must be 38 in.-lb. (Assume £ = 14,000 ksi and v = 0.25.)

(a)

To determine

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

### Explanation of Solution

Given information:

A cube of cast iron having side 4 in is tested under triaxial stress. The strain in the x direction is 225 × 10 6 , strain in y direction and z direction is equal which is 37.5 × 10 6 . The modulus of elasticity is 14000 ksi 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 along 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 225 × 10 6 for ε x , 37.5 × 10 6 for ε y , 37.5 × 10 6 for ε z , 14000 ksi for E and 0.25 for ν in Equation (I).

σ x = 14000 ksi ( 1 + 0.25 ) ( 1 ( 2 × 0.25 ) ) [ ( 1 0.25 ) ( 225 × 10 6 ) + 0.25 ( 37.5 × 10 6 37.5 × 10 6 ) ] = 14000 ksi × ( 187.5 × 10 6 ) 0.625 = 4.2 ksi × ( 10 3 psi 1 ksi ) = 4200 psi

Substitute 225 × 10 6 for ε x , 37

(b)

To determine

The maximum shear stress in the material.

(c)

To determine

The change in the volume of the cube.

(d)

To determine

The strain energy stored in the cube.

(e)

To determine

The maximum value of normal stress along the x axis.

(f)

To determine

The required value of strain along the x axis.

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