-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)
The maximum shear stress in the material.
Answer to Problem 7.6.3P
The maximum shear stress on the material is
Explanation of Solution
Given information:
The aluminium element of length
Explanation:
Write the expression for the maximum shear stress.
Here, the maximum shear stress is
Calculation:
Since no shear stresses act on the parallelepiped,
Substitute,
Conclusion:
The maximum shear stress on the material is
(b)
The changes in the dimensions of the element.
Answer to Problem 7.6.3P
The change in length is
The change in height is
The change in width is
Explanation of Solution
Write the expression for the strain along
Here, the strain in the
Write the expression for strain in
Here, the strain in
Write the expression for strain in
Here, the strain in
Write the expression for the change in length.
Here, the length of element is
Write the expression for change in height.
Here, the height of element is
Write the expression for the change in width.
Here, the width of the element is
Calculation:
Substitute
Substitute
Substitute
Substitute
Substitute
Substitute
Conclusion:
The change in length is
The change in height is
The change in width is
(c)
The change in the volume of the element.
Answer to Problem 7.6.3P
The change in the volume is
Explanation of Solution
Write the expression for the change in the volume.
Here, the change in volume is
Calculation:
Substitute
Conclusion:
The change in the volume is
(d)
The strain energy stored in the element.
Answer to Problem 7.6.3P
The strain energy stored in the element is
Explanation of Solution
Write the expression for the strain energy.
Here, the strain energy is
Calculation:
Substitute
Conclusion:
The strain energy stored in the element is
(e)
The maximum value of normal stress along the
Answer to Problem 7.6.3P
The maximum value of normal stress along the
Explanation of Solution
Given information:
The change in volume is limited to
Explanation:
Write the expression for the change in volume.
Calculation:
Substitute
Conclusion:
The maximum value of normal stress along the
(f)
The required value of normal stress along the
Answer to Problem 7.6.3P
The required value of the normal stress along the
Explanation of Solution
Given information:
The strain energy of the system is
Explanation:
Write the expression for the strain energy in terms of stresses using Hooke’s law.
Calculation:
Substitute
Solve the quadratic equation for obtaining the value of
Conclusion:
The required value of the normal stress along the
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Chapter 7 Solutions
Mechanics of Materials (MindTap Course List)
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- Mechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage Learning