   # -3 An element of aluminum in the form of a rectangular parallelepiped (see figure) of dimensions a = 5.5 in., h = 4.5 in, and c = 3.5 in. is subjected to iriaxial stresses = 12,500 psi, o. = —5000 psi, and ci. = —1400 psi acting on the x,i, and z faces, respectively. Determine the following quantities: (a) the maxim um shear stress in the material; (b) the changes ..la, .11. and 1c in the dimensions of the element: (C) the change .IJ’ in the volume: (d) the strain energy U stored in the element: (e) the maximum value of cr1 when the change in volume must be limited to 0.021%; and (f) the required value of o when the strain energy must be 900 in.-lb. (Assume E = 10,400 ksi and v = 0.33.) ### Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347
Chapter 7, Problem 7.6.3P
Textbook Problem
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## -3 An element of aluminum in the form of a rectangular parallelepiped (see figure) of dimensions a = 5.5 in., h = 4.5 in, and c = 3.5 in. is subjected to iriaxial stresses = 12,500 psi, o. = —5000 psi, and ci. = —1400 psi acting on the x,i, and z faces, respectively. Determine the following quantities: (a) the maxim um shear stress in the material; (b) the changes ..la, .11. and 1c in the dimensions of the element: (C) the change .IJ’ in the volume: (d) the strain energy U stored in the element: (e) the maximum value of cr1 when the change in volume must be limited to 0.021%; and (f) the required value of o when the strain energy must be 900 in.-lb. (Assume E = 10,400 ksi and v = 0.33.) (a)

To determine

The maximum shear stress in the material.

### Explanation of Solution

Given information:

The aluminium element of length 5.5 in , width 3.5 in and height 4.5 in is subjected to triaxial stress. The stress in x direction is 12500 psi , in y direction is 5000 psi and in z direction is 1400 psi . The modulus of elasticity is 10400 ksi and the Poisson’s ratio is 0.33 .

Explanation:

Write the expression for the maximum shear stress.

τ max = σ largest σ smallest 2 ...... (I)

Here, the maximum shear stress is τ max , the largest stress acting on the element is σ largest and the smallest stress acting on the element is σ smallest

(b)

To determine

The changes in the dimensions of the element.

(c)

To determine

The change in the volume of the element.

(d)

To determine

The strain energy stored in the element.

(e)

To determine

The maximum value of normal stress along the x axis.

(f)

To determine

The required value of normal stress along the x axis.

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