Chapter 7, Problem 7.6.3P

-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.)

  

To determine

The maximum shear stress in the material.

Answer to Problem 7.6.3P

The maximum shear stress on the material is $8750\text{psi}$ .

Explanation of Solution

Given information:

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

Explanation:

Write the expression for the maximum shear stress.

  ${\tau }_{\max }=\frac{{\sigma }_{\text{largest}}-{\sigma }_{\text{smallest}}}{2}$ ...... (I)

Here, the maximum shear stress is ${\tau }_{\max }$ , the largest stress acting on the element is ${\sigma }_{\text{largest}}$ and the smallest stress acting on the element is ${\sigma }_{\text{smallest}}$ .

Calculation:

Since no shear stresses act on the parallelepiped, $x$ , $y$ , $z$ directions coincide with the principal stress directions.

Substitute, $12500\text{psi}$ for ${\sigma }_{\text{largest}}$ and $-5000\text{psi}$ for ${\sigma }_{\text{smallest}}$ in Equation (I).

  $\begin{array}{c}{\tau }_{\max }=\frac{12500\text{psi+5000}\text{psi}}{2}\\ =\frac{17500\text{psi}}{2}\\ =8750\text{psi}\end{array}$

Conclusion:

The maximum shear stress on the material is $8750\text{psi}$ .

To determine

The changes in the dimensions of the element.

Answer to Problem 7.6.3P

The change in length is $7.73\times {10}^{-3}\text{in}$ .

The change in height is $3.75\times {10}^{-3}\text{in}$ .

The change in width is $-1.3\times {10}^{-3}\text{in}$ .

Explanation of Solution

Write the expression for the strain along $x$ axis.

  ${\epsilon }_x=\frac{1}{E}\left({\sigma }_x-\nu \left({\sigma }_y+{\sigma }_z\right)\right)$ ...... (II)

Here, the strain in the $x$ axis is ${\epsilon }_x$ , the modulus of elasticity is $E$ , the stress along $x$ axis is ${\sigma }_x$ , stress along $y$ axis is ${\sigma }_y$ , stress along $z$ axis is ${\sigma }_z$ and the Poisson’s ratio is $\nu$ .

Write the expression for strain in $y$ direction.

  ${\epsilon }_y=\frac{1}{E}\left({\sigma }_y-\nu \left({\sigma }_x+{\sigma }_z\right)\right)$ ...... (III)

Here, the strain in $y$ direction is ${\epsilon }_y$ .

Write the expression for strain in $z$ direction.

  ${\epsilon }_z=\frac{1}{E}\left({\sigma }_z-\nu \left({\sigma }_x+{\sigma }_y\right)\right)$ ...... (IV)

Here, the strain in $z$ direction is ${\epsilon }_z$ .

Write the expression for the change in length.

  $\Delta a=a\times {\epsilon }_x$ ...... (V)

Here, the length of element is $a$ .

Write the expression for change in height.

  $\Delta b=b\times {\epsilon }_y$ ...... (VI)

Here, the height of element is $b$ .

Write the expression for the change in width.

  $\Delta c=c\times {\epsilon }_z$ ...... (VII)

Here, the width of the element is $c$ .

Calculation:

Substitute $10400\times {10}^3\text{psi}$ for $E$ , $12500\text{psi}$ for ${\sigma }_x$ , $-5000\text{psi}$ for ${\sigma }_y$ , $-1400\text{psi}$ for ${\sigma }_z$ and $0.33$ for $\nu$ in Equation (II).

  $\begin{array}{c}{\epsilon }_x=\frac{1}{10400\times {10}^3\text{psi}}\left(12500\text{psi}-0.33\left(-5000\text{psi}-1400\text{psi}\right)\right)\\ =\frac{14612\times {10}^{-3}\text{psi}}{10400\text{psi}}\\ =1.405\times {10}^{-3}\end{array}$

Substitute $10400\times {10}^3\text{psi}$ for $E$ , $12500\text{psi}$ for ${\sigma }_x$ , $-5000\text{psi}$ for ${\sigma }_y$ , $-1400\text{psi}$ for ${\sigma }_z$ and $0.33$ for $\nu$ in Equation (III).

  $\begin{array}{c}{\epsilon }_y=\frac{1}{10400\times {10}^3\text{psi}}\left(-5000\text{psi}-0.33\left(12500\text{psi}-1400\text{psi}\right)\right)\\ =\frac{-8663\times {10}^{-3}\text{psi}}{10400\text{psi}}\\ =-0.83298\times {10}^{-3}\end{array}$

Substitute $10400\times {10}^3\text{psi}$ for $E$ , $12500\text{psi}$ for ${\sigma }_x$ , $-5000\text{psi}$ for ${\sigma }_y$ , $-1400\text{psi}$ for ${\sigma }_z$ and $0.33$ for $\nu$ in Equation (IV).

  $\begin{array}{c}{\epsilon }_z=\frac{1}{10400\times {10}^3\text{psi}}\left(-1400\text{psi}-0.33\left(12500\text{psi}-5000\text{psi}\right)\right)\\ =\frac{-3875\times {10}^{-3}\text{psi}}{10400\text{psi}}\\ =-0.3726\times {10}^{-3}\end{array}$

Substitute $1.405\times {10}^{-3}$ for ${\epsilon }_x$ and $5.5\text{in}$ for $a$ in Equation (V).

  $\begin{array}{c}\Delta a=5.5\text{in}\times 1.405\times {10}^{-3}\\ =7.73\times {10}^{-3}\text{in}\end{array}$

Substitute $-0.83298\times {10}^{-3}$ for ${\epsilon }_y$ and $4.5\text{in}$ for $b$ in Equation (VI).

  $\begin{array}{c}\Delta b=4.5\text{in}\times \left(-0.83298\times {10}^{-3}\right)\\ =-3.7484\times {10}^{-3}\text{in}\end{array}$

Substitute $-0.3726\times {10}^{-3}$ for ${\epsilon }_z$ and $3.5\text{in}$ for $c$ in Equation (VII).

  $\begin{array}{c}\Delta c=3.5\text{in}\times \left(-0.3726\times {10}^{-3}\right)\\ =-1.3041\times {10}^{-3}\text{in}\end{array}$

Conclusion:

The change in length is $7.73\times {10}^{-3}\text{in}$ .

The change in height is $3.75\times {10}^{-3}\text{in}$ .

The change in width is $-1.3\times {10}^{-3}\text{in}$ .

To determine

The change in the volume of the element.

Answer to Problem 7.6.3P

The change in the volume is $0.0173{\text{in}}^3$ .

Explanation of Solution

Write the expression for the change in the volume.

  $\Delta V=V_0\left({\epsilon }_x+{\epsilon }_y+{\epsilon }_z\right)$ ...... (VIII)

Here, the change in volume is $\Delta V$ and the original volume is $V_0$ .

Calculation:

Substitute $\left(5.5\text{in}\times 4.5\text{in}\times 3.5\text{in}\right)$ for $V_0$ , $1.405\times {10}^{-3}$ for ${\epsilon }_x$ , $-0.83298\times {10}^{-3}$ for ${\epsilon }_y$ and $-0.3726\times {10}^{-3}$ for ${\epsilon }_z$ in Equation (VIII).

  $\begin{array}{c}\Delta V=\left(5.5\text{in}\times 4.5\text{in}\times 3.5\text{in}\right)\left(1.405\times {10}^{-3}-0.83298\times {10}^{-3}-0.3726\times {10}^{-3}\right)\\ =86.625{\text{in}}^3\times 0.19942\times {10}^{-3}\\ =0.0173{\text{in}}^3\end{array}$

Conclusion:

The change in the volume is $0.0173{\text{in}}^3$ .

To determine

The strain energy stored in the element.

Answer to Problem 7.6.3P

The strain energy stored in the element is $964\text{lb}\cdot \text{in}$ .

Explanation of Solution

Write the expression for the strain energy.

  $U=\frac{1}{2}{V}_0\left({\sigma }_x{\epsilon }_x+{\sigma }_y{\epsilon }_y+{\sigma }_z{\epsilon }_z\right)$ ...... (IX)

Here, the strain energy is $U$ .

Calculation:

Substitute $86.625{\text{in}}^3$ for $V_0$ , $12500\text{psi}$ for ${\sigma }_x$ , $1.405\times {10}^{-3}$ for ${\epsilon }_x$ , $-5000\text{psi}$ for ${\sigma }_y$ , $-0.83298\times {10}^{-3}$ for ${\epsilon }_y$ , $-1400\text{psi}$ for ${\sigma }_z$ and $-0.3726\times {10}^{-3}$ for ${\epsilon }_z$ in Equation (IX).

  $\begin{array}{c}U=\left[\frac{86.625{\text{in}}^3}{2}\left(\begin{array}{l}\left(12500\text{psi}\times 1.405\times {10}^{-3}\right)\\ +\left(-5000\text{psi}\times -0.83298\times {10}^{-3}\right)\\ +\left(-1400\text{psi}\times -0.3726\times {10}^{-3}\right)\end{array}\right)\right]\\ =\left(963661.545\times {10}^{-3}{\text{in}}^3\cdot \text{psi}\right)\left(\frac{1\text{lb}/{\text{in}}^{\text{2}}}{1\text{psi}}\right)\\ =963.66\text{lb}\cdot \text{in}\\ \approx \text{964}\text{lb}\cdot \text{in}\end{array}$

Conclusion:

The strain energy stored in the element is $964\text{lb}\cdot \text{in}$ .

To determine

The maximum value of normal stress along the $x$ axis.

Answer to Problem 7.6.3P

The maximum value of normal stress along the $x$ axis is $12824\text{psi}$ .

Explanation of Solution

Given information:

The change in volume is limited to $0.021\raisebox{1ex}{$0$}\!\left/ \!\raisebox{-1ex}{$0$}\right.$ .

Explanation:

Write the expression for the change in volume.

  $\frac{\Delta V}{V_0}=\frac{\left(1-2\nu \right)}{E}\left({\sigma }_x+{\sigma }_y+{\sigma }_z\right)$ ...... (X)

Calculation:

Substitute $2.1\times {10}^{-4}$ for $\frac{\Delta V}{V_0}$ , $0.33$ for $\nu$ , $10400\times {10}^3\text{psi}$ for $E$ , $-5000\text{psi}$ for ${\sigma }_y$ and $-1400\text{psi}$ for ${\sigma }_z$ in Equation (X).

  $\begin{array}{l}2.1\times {10}^{-4}=\frac{0.34}{10400\times {10}^3\text{psi}}\left({\sigma }_x-5000\text{psi}-1400\text{psi}\right)\\ 0.34{\sigma }_x=4360\text{psi}\\ {\sigma }_x=12823.53\text{psi}\end{array}$

Conclusion:

The maximum value of normal stress along the $x$ axis is $12824\text{psi}$ .

To determine

The required value of normal stress along the $x$ axis.

Answer to Problem 7.6.3P

The required value of the normal stress along the $x$ axis is $11967\text{psi}$ .

Explanation of Solution

Given information:

The strain energy of the system is $900\text{lb}\cdot \text{in}$ .

Explanation:

Write the expression for the strain energy in terms of stresses using Hooke’s law.

  $U=\frac{V_0}{2E}\left({\sigma }_x^2+{\sigma }_y^2+{\sigma }_z^2-2\nu \left({\sigma }_x{\sigma }_y+{\sigma }_y{\sigma }_x+{\sigma }_x{\sigma }_z\right)\right)$ ...... (XI)

Calculation:

Substitute $86.625{\text{in}}^3$ for $V_0$ , $900\text{lb}\cdot \text{in}$ for $U$ , $0.33$ for $\nu$ , $10400\times {10}^3\text{psi}$ for $E$ , $-5000\text{psi}$ for ${\sigma }_y$ and $-1400\text{psi}$ for ${\sigma }_z$ .

  $\begin{array}{l}900=\left[\frac{86.625}{2\times 10400\times {10}^3}\left(\begin{array}{l}{\sigma }_x^2+{5000}^2+{1400}^2\\ -2\times 0.33\left(-5000{\sigma }_x+5000\times 1400+{\sigma }_x\times -1400\right)\end{array}\right)\right]\\ \frac{1800\times 10400\times {10}^3}{86.625}={\sigma }_x^2+25\times {10}^6+1.96\times {10}^6+3300{\sigma }_x-4.76\times {10}^6+952{\sigma }_x\\ {\sigma }_x^2+4224{\sigma }_x-193762900=0\end{array}$

Solve the quadratic equation for obtaining the value of ${\sigma }_x$ .

  $\begin{array}{c}{\sigma }_x=\frac{-4224\text{psi}\pm \sqrt{{4224}^2{\text{psi}}^{\text{2}}+4\times 193762900{\text{psi}}^{\text{2}}}}{2}\\ =\frac{23934}{2}\text{psi}\\ =11967\text{psi}\end{array}$

Conclusion:

The required value of the normal stress along the $x$ axis is $11967\text{psi}$ .