Chapter 7, Problem 7.6.5P

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.)

  

To determine

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

Answer to Problem 7.6.5P

The normal stress acting on the $x$ face is $-4200\text{psi}$ .

The normal stress acting on the $y$ face is $-2100\text{psi}$ .

The normal stress acting on the $z$ face is $-2100\text{psi}$ .

Explanation of Solution

Given information:

A cube of cast iron having side $4\text{in}$ is tested under triaxial stress. The strain in the $x$ direction is $-225\times {10}^{-6}$ , strain in $y$ direction and $z$ direction is equal which is $-37.5\times {10}^{-6}$ . The modulus of elasticity is $14000\text{ksi}$ and the Poisson’s ratio is $0.25$ .

Explanation:

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

  ${\sigma }_x=\frac{E}{\left(1+\nu \right)\left(1-2\nu \right)}\left[\left(1-\nu \right){\epsilon }_x+\nu \left({\epsilon }_y+{\epsilon }_z\right)\right]$ ...... (I)

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

Write the expression for stress along $y$ axis.

  ${\sigma }_y=\frac{E}{\left(1+\nu \right)\left(1-2\nu \right)}\left[\left(1-\nu \right){\epsilon }_y+\nu \left({\epsilon }_x+{\epsilon }_z\right)\right]$ ...... (II)

Here, stress along $y$ axis is ${\sigma }_y$ .

Write the expression for stress along $z$ axis.

  ${\sigma }_z=\frac{E}{\left(1+\nu \right)\left(1-2\nu \right)}\left[\left(1-\nu \right){\epsilon }_z+\nu \left({\epsilon }_x+{\epsilon }_y\right)\right]$ ...... (III)

Here, stress along $z$ axis is ${\sigma }_z$ .

Calculation:

Substitute $-225\times {10}^{-6}$ for ${\epsilon }_x$ , $-37.5\times {10}^{-6}$ for ${\epsilon }_y$ , $-37.5\times {10}^{-6}$ for ${\epsilon }_z$ , $14000\text{ksi}$ for $E$ and $0.25$ for $\nu$ in Equation (I).

  $\begin{array}{c}{\sigma }_x=\frac{14000\text{ksi}}{\left(1+0.25\right)\left(1-\left(2\times 0.25\right)\right)}\left[\begin{array}{l}\left(1-0.25\right)\left(-225\times {10}^{-6}\right)\\ +0.25\left(-37.5\times {10}^{-6}-37.5\times {10}^{-6}\right)\end{array}\right]\\ =\frac{14000\text{ksi}\times \left(-187.5\times {10}^{-6}\right)}{0.625}\\ =-4.2\text{ksi}\times \left(\frac{{10}^3\text{psi}}{1\text{ksi}}\right)\\ =-4200\text{psi}\end{array}$

Substitute $-225\times {10}^{-6}$ for ${\epsilon }_x$ , $-37.5\times {10}^{-6}$ for ${\epsilon }_y$ , $-37.5\times {10}^{-6}$ for ${\epsilon }_z$ , $14000\text{ksi}$ for $E$ and $0.25$ for $\nu$ in Equation (II).

  $\begin{array}{c}{\sigma }_y=\frac{14000\text{ksi}}{\left(1+0.25\right)\left(1-\left(2\times 0.25\right)\right)}\left[\begin{array}{l}\left(1-0.25\right)\left(-37.5\times {10}^{-6}\right)\\ +0.25\left(-37.5\times {10}^{-6}-225\times {10}^{-6}\right)\end{array}\right]\\ =\frac{14000\text{ksi}\times \left(-93.75\times {10}^{-6}\right)}{0.625}\\ =-2.1\text{ksi}\left(\frac{{10}^3\text{psi}}{1\text{ksi}}\right)\\ =-2100\text{psi}\end{array}$

Substitute $-225\times {10}^{-6}$ for ${\epsilon }_x$ , $-37.5\times {10}^{-6}$ for ${\epsilon }_y$ , $-37.5\times {10}^{-6}$ for ${\epsilon }_z$ , $14000\text{ksi}$ for $E$ and $0.25$ for $\nu$ in Equation (III).

  $\begin{array}{c}{\sigma }_z=\frac{14000\text{ksi}}{\left(1+0.25\right)\left(1-\left(2\times 0.25\right)\right)}\left[\begin{array}{l}\left(1-0.25\right)\left(-37.5\times {10}^{-6}\right)\\ +0.25\left(-225\times {10}^{-6}-37.5\times {10}^{-6}\right)\end{array}\right]\\ =\frac{14000\text{ksi}\times \left(-93.75\times {10}^{-6}\right)}{0.625}\\ =-2.1\text{ksi}\times \left(\frac{{10}^3\text{psi}}{1\text{ksi}}\right)\\ =-2100\text{psi}\end{array}$

Conclusion:

The normal stress acting on the $x$ face is $-4200\text{psi}$ .

The normal stress acting on the $y$ face is $-2100\text{psi}$ .

The normal stress acting on the $z$ face is $-2100\text{psi}$ .

To determine

The maximum shear stress in the material.

Answer to Problem 7.6.5P

The maximum shear stress in the material is $1050\text{psi}$ .

Explanation of Solution

Write the expression for the shear stress in the $xy$ plane.

  ${\tau }_1=\frac{{\sigma }_x-{\sigma }_y}{2}$ ...... (IV)

Here, the shear stress in $xy$ plane is ${\tau }_1$ .

Write the expression for the shear stress in the $yz$ plane.

  ${\tau }_2=\frac{{\sigma }_y-{\sigma }_z}{2}$ ...... (V)

Here, the shear stress in $yz$ plane is ${\tau }_2$ .

Write the expression for the shear stress in $zx$ plane.

  ${\tau }_3=\frac{{\sigma }_z-{\sigma }_x}{2}$ ...... (VI)

Here, the shear stress in the $zx$ plane is ${\tau }_3$ .

Calculation:

Substitute $-4200\text{psi}$ for ${\sigma }_x$ and $-2100\text{psi}$ for ${\sigma }_y$ in Equation (IV).

  $\begin{array}{c}{\tau }_1=\frac{-4200\text{psi}-\left(-2100\text{psi}\right)}{2}\\ =\frac{-4200\text{psi}+2100\text{psi}}{2}\\ =-1050\text{psi}\end{array}$

Substitute $-2100\text{psi}$ for ${\sigma }_y$ and $-2100\text{psi}$ for ${\sigma }_z$ in Equation (V).

  $\begin{array}{c}{\tau }_2=\frac{-2100\text{psi}-\left(-2100\text{psi}\right)}{2}\\ =\frac{-2100\text{psi}+2100\text{psi}}{2}\\ =0\text{psi}\end{array}$

Substitute $-2100\text{psi}$ for ${\sigma }_z$ and $-4200\text{psi}$ for ${\sigma }_x$ in Equation (VI).

  $\begin{array}{c}{\tau }_3=\frac{-2100\text{psi}-\left(-4200\text{psi}\right)}{2}\\ =\frac{-2100\text{psi}+4200\text{psi}}{2}\\ =1050\text{psi}\end{array}$

From the above values of shear stresses, the maximum shear stress be ${\tau }_3$ .

Therefore, ${\tau }_{\max }={\tau }_3=1050\text{psi}$ .

Conclusion:

The maximum shear stress in the material is $1050\text{psi}$ .

To determine

The change in the volume of the cube.

Answer to Problem 7.6.5P

The change in the volume is $-0.0192{\text{in}}^3$ .

Explanation of Solution

Write the expression for the total volumetric strain in the cube.

  ${\epsilon }_v={\epsilon }_x+{\epsilon }_y+{\epsilon }_z$ ...... (VII)

Here, the total volumetric strain in the cube is ${\epsilon }_v$ .

Write the expression for the volume of cube.

  $V=a^3$ ...... (VIII)

Here, the volume of cube is $V$ and side of cube is $a$ .

Write the expression for change in volume.

  $\Delta V=\left({\epsilon }_v\right)V$ ...... (IX)

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

Calculation:

Substitute $-225\times {10}^{-6}$ for ${\epsilon }_x$ , $-37.5\times {10}^{-6}$ for ${\epsilon }_y$ and $-37.5\times {10}^{-6}$ for ${\epsilon }_z$ in Equation (VII).

  $\begin{array}{c}{\epsilon }_v=\left(-225\times {10}^{-6}\right)+\left(-37.5\times {10}^{-6}\right)+\left(-37.5\times {10}^{-6}\right)\\ =\left(-225-37.5-37.5\right)\times {10}^{-6}\\ =-3\times {10}^{-4}\end{array}$

Substitute $4\text{in}$ for $a$ in Equation (VIII)

  $\begin{array}{c}V={\left(a\right)}^3\\ ={\left(4\text{in}\right)}^3\\ =64{\text{in}}^3\end{array}$

Substitute $64{\text{in}}^3$ for $V$ and $-3\times {10}^{-4}$ for ${\epsilon }_v$ in Equation (IX).

  $\begin{array}{c}\Delta V=\left(-3\times {10}^{-4}\right)\times 64{\text{in}}^3\\ =-0.0192{\text{in}}^3\end{array}$

Conclusion:

The change in the volume is $-0.0192{\text{in}}^3$ .

To determine

The strain energy stored in the cube.

Answer to Problem 7.6.5P

The strain energy stored in the cube is $35.3\text{lb}\cdot \text{in}$ .

Explanation of Solution

Write the expression for the strain energy stored in the cube.

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

Here, the strain energy is $U$ .

Calculation:

Substitute $64{\text{in}}^3$ for $V$ , $-4200\text{psi}$ for ${\sigma }_x$ , $-225\times {10}^{-6}$ for ${\epsilon }_x$ , $-2100\text{psi}$ for ${\sigma }_y$ , $-37.5\times {10}^{-6}$ for ${\epsilon }_y$ , $-2100\text{psi}$ for ${\sigma }_z$ and $-37.5\times {10}^{-6}$ for ${\epsilon }_z$ in Equation (X).

  $\begin{array}{c}U=\frac{1}{2}64{\text{in}}^3\left[\begin{array}{l}\left(-4200\text{psi}\times -225\times {10}^{-6}\right)+\left(-2100\text{psi}\times -37.5\times {10}^{-6}\right)\\ +\left(-2100\text{psi}\times -37.5\times {10}^{-6}\right)\end{array}\right]\\ =\frac{1}{2}64{\text{in}}^3\left(0.945\text{psi}+0.07875\text{psi}+0.07875\text{psi}\right)\\ =35.3\text{psi}\cdot {\text{in}}^3\times \left(\frac{1\text{lb}}{1\text{psi}\cdot {\text{in}}^2}\right)\\ =35.3\text{lb}\cdot \text{in}\end{array}$

Conclusion:

The strain energy stored in the cube is $35.3\text{lb}\cdot \text{in}$ .

To determine

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

Answer to Problem 7.6.5P

The maximum value of normal stress along the $x$ axis is $5544\text{psi}$ .

Explanation of Solution

Given information:

The change in volume is limited to $0.028\%$ .

Explanation:

Write the expression for the change in volume.

  $\frac{\Delta V}{V}={\epsilon }_x+{\epsilon }_y+{\epsilon }_z$ ...... (XI)

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

  ${\sigma }_x=\frac{E}{\left(1+\nu \right)\left(1-2\nu \right)}\left[\left(1-\nu \right){\epsilon }_x+\nu \left({\epsilon }_y+{\epsilon }_z\right)\right]$ ...... (XII)

Calculation:

Substitute $2.8\times {10}^{-4}$ for $\frac{\Delta V}{V}$ , $-37.5\times {10}^{-6}$ for ${\epsilon }_y$ and $-37.5\times {10}^{-6}$ for ${\epsilon }_z$ in Equation (XI).

  $\begin{array}{l}2.8\times {10}^{-4}={\epsilon }_x+\left(-37.5\times {10}^{-6}\right)+\left(-37.5\times {10}^{-6}\right)\\ {\epsilon }_x=2.8\times {10}^{-4}+0.75\times {10}^{-4}\\ {\epsilon }_x=3.55\times {10}^{-4}\end{array}$

Substitute $3.55\times {10}^{-4}$ for ${\epsilon }_x$ , $-37.5\times {10}^{-6}$ for ${\epsilon }_y$ , $-37.5\times {10}^{-6}$ for ${\epsilon }_z$ , $14000\text{ksi}$ for $E$ and $0.25$ for $\nu$ in Equation (XII).

  $\begin{array}{c}{\sigma }_x=\frac{14000\text{ksi}}{\left(1+0.25\right)\left(1-\left(2\times 0.25\right)\right)}\left[\begin{array}{l}\left(1-0.25\right)\left(3.55\times {10}^{-4}\right)\\ +0.25\left(-37.5\times {10}^{-6}-37.5\times {10}^{-6}\right)\end{array}\right]\\ =\frac{14000\text{ksi}\times 2.475\times {10}^{-4}}{0.625}\\ =5.544\text{ksi}\times \left(\frac{{10}^3\text{psi}}{1\text{ksi}}\right)\\ =5544\text{psi}\end{array}$

Conclusion:

The maximum value of normal stress along the $x$ axis is $5544\text{psi}$ .

To determine

The required value of strain along the $x$ axis.

Answer to Problem 7.6.5P

The required value of the strain along the $x$ axis is $-234.92\times {10}^{-6}$ .

Explanation of Solution

Given information:

The strain energy of the system is $38\text{lb}\cdot \text{in}$ .

Explanation:

Write the expression for the strain energy.

  $U=\frac{1}{2}V\frac{E}{\left(1+\nu \right)\left(1-2\nu \right)}\left[\left(1-\nu \right)\left({\epsilon }_x^2+{\epsilon }_y^2+{\epsilon }_z^2\right)+2\nu \left({\epsilon }_x{\epsilon }_y+{\epsilon }_y{\epsilon }_z+{\epsilon }_z{\epsilon }_x\right)\right]$ ...... (XIII)

Calculation:

Substitute $64{\text{in}}^3$ for $V$ , $38\text{lb}\cdot \text{in}$ for $U$ , $0.25$ for $\nu$ , $14000\times {10}^3\text{psi}$ for $E$ , $-37.5\times {10}^{-6}$ for ${\epsilon }_y$ , $-37.5\times {10}^{-6}$ for ${\epsilon }_z$ in Equation (XIII).

  $\begin{array}{l}38\text{lb}\cdot \text{in}=\left(0.5\times 64{\text{in}}^3\right)\frac{14000\times {10}^3\text{psi}}{\left(1+0.25\right)\left(1-\left(2\times 0.25\right)\right)}\left[\begin{array}{l}0.75\left(\begin{array}{l}{\epsilon }_x^2+{\left(-37.5\times {10}^{-6}\right)}^2\\ +{\left(-37.5\times {10}^{-6}\right)}^2\end{array}\right)\\ +0.5\left[\begin{array}{l}\left({\epsilon }_x\right)\left(-37.5\times {10}^{-6}\right)\\ +{\left(-37.5\times {10}^{-6}\right)}^2\\ +\left({\epsilon }_x\right)\left(-37.5\times {10}^{-6}\right)\end{array}\right]\end{array}\right]\\ \left(38\text{lb}\cdot \text{in}\times \frac{1\text{psi}\cdot {\text{in}}^2}{1\text{lb}}\right)=\left(537600000\text{psi}\cdot {\text{in}}^3\right){\epsilon }_x^2-\left(26880\text{psi}\cdot {\text{in}}^3\right){\epsilon }_x+\left(2.016\text{psi}\cdot {\text{in}}^3\right)\\ 537600000{\epsilon }_x^2-26880{\epsilon }_x-35.984=0\end{array}$

Now solve the quadratic equation for obtaining the value of ${\epsilon }_x$ .

  $\begin{array}{c}{\epsilon }_x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-\left(-26880\right)\pm \sqrt{{\left(-26880\right)}^2-4\times 5376000}00\times \left(-35.984\right)}{2\times 537600000}\\ =-0.00023492\\ =-234.92\times {10}^{-6}\end{array}$

Conclusion:

The required value of the strain along the $x$ axis is $-234.92\times {10}^{-6}$ .