Chapter 7, Problem 7.6.6P

Solve the preceding problem if the cube is granite (E = 80 GPa, v = 0.25) with dimensions E = 89 mm and compressive strains E = 690 X l0-6 and = = 255 X 10-6. For part (c) of Problem 7.6-5. find the maximum value of cr when the change in volume must be limited to 0.11%. For part. find the required value of when the strain energy must be 33 J.

  

To determine

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

Answer to Problem 7.6.6P

The normal stress acting on the $x$ face is $-82.56\text{MPa}$ .

The normal stress acting on the $y$ face is $-54.72\text{MPa}$ .

The normal stress acting on the $z$ face is $-54.72\text{MPa}$ .

Explanation of Solution

Given information:

A cube of cast iron having side $89\text{mm}$ is tested under triaxial stress. The strain in the $x$ direction is $-690\times {10}^{-6}$ , strain in $y$ direction and $z$ direction is being equal which is $-255\times {10}^{-6}$ . The modulus of elasticity is $80\text{GPa}$ 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 alon $x$ 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 $-690\times {10}^{-6}$ for ${\epsilon }_x$ , $-255\times {10}^{-6}$ for ${\epsilon }_y$ , $-255\times {10}^{-6}$ for ${\epsilon }_z$ , $80\text{GPa}$ for $E$ and $0.25$ for $\nu$ in Equation (I).

  $\begin{array}{c}{\sigma }_x=\frac{80\text{GPa}}{\left(1+0.25\right)\left(1-\left(2\times 0.25\right)\right)}\left[\left(0.75\times -690\right)+\left(0.25\times \left(-255-255\right)\right)\right]\times {10}^{-6}\\ =128\text{GPa}\left(-645\times {10}^{-6}\right)\\ =-82560\text{GPa}\times \left(\frac{{10}^3\text{MPa}}{1\text{GPa}}\right)\times {10}^{-6}\\ =-82.56\text{MPa}\end{array}$

Substitute $-690\times {10}^{-6}$ for ${\epsilon }_x$ , $-255\times {10}^{-6}$ for ${\epsilon }_y$ , $-255\times {10}^{-6}$ for ${\epsilon }_z$ , $80\text{GPa}$ for $E$ and $0.25$ for $\nu$ in Equation (II).

  $\begin{array}{c}{\sigma }_y=\frac{80\text{GPa}}{\left(1+0.25\right)\left(1-\left(2\times 0.25\right)\right)}\left[\left(0.75\times -255\right)+\left(0.25\times \left(-690-255\right)\right)\right]\times {10}^{-6}\\ =128\text{GPa}\left(-427.5\times {10}^{-6}\right)\\ =-54720\text{GPa}\times \left(\frac{{10}^3\text{MPa}}{1\text{GPa}}\right)\times {10}^{-6}\\ =-54.72\text{MPa}\end{array}$

Substitute $-690\times {10}^{-6}$ for ${\epsilon }_x$ , $-255\times {10}^{-6}$ for ${\epsilon }_y$ , $-255\times {10}^{-6}$ for ${\epsilon }_z$ , $80\text{GPa}$ for $E$ and $0.25$ for $\nu$ in Equation (III).

  $\begin{array}{c}{\sigma }_z=\frac{80\text{GPa}}{\left(1+0.25\right)\left(1-\left(2\times 0.25\right)\right)}\left[\left(0.75\times -255\right)+\left(0.25\times \left(-690-255\right)\right)\right]\times {10}^{-6}\\ =128\text{GPa}\left(-427.5\times {10}^{-6}\right)\\ =-54720\text{GPa}\times \left(\frac{{10}^3\text{MPa}}{1\text{GPa}}\right)\times {10}^{-6}\\ =-54.72\text{MPa}\end{array}$

Conclusion:

The normal stress acting on the $x$ face is $-82.56\text{MPa}$ .

The normal stress acting on the $y$ face is $-54.72\text{MPa}$ .

The normal stress acting on the $z$ face is $-54.72\text{MPa}$ .

To determine

The maximum shear stress in the material.

Answer to Problem 7.6.6P

The maximum shear stress in the material is $13.92\text{MPa}$ .

Explanation of Solution

Write the expression for the maximum shear stress.

  ${\tau }_{\max }=\frac{{\sigma }_{\max }-{\sigma }_{\min }}{2}$ ...... (IV)

Here, the maximum shear stress is ${\tau }_{\max }$ , the maximum stress is ${\sigma }_{\max }$ and the minimum stress is ${\sigma }_{\min }$ .

Calculation:

Substitute $-54.72\text{MPa}$ for ${\sigma }_{\max }$ and $-82.56\text{MPa}$ for ${\sigma }_{\min }$ in Equation (IV).

  $\begin{array}{c}{\tau }_{\max }=\frac{-54.72\text{MPa}-\left(-82.56\text{MPa}\right)}{2}\\ =\frac{27.84\text{MPa}}{2}\\ =13.92\text{MPa}\end{array}$

Conclusion:

The maximum shear stress in the material is $13.92\text{MPa}$ .

To determine

The change in the volume of the cube.

Answer to Problem 7.6.6P

The change in the volume is $-846{\text{mm}}^3$ .

Explanation of Solution

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

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

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

Write the expression for the volume of cube.

  $V=a^3$ ...... (VI)

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

Write the expression for change in volume.

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

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

Calculation:

Substitute $-690\times {10}^{-6}$ for ${\epsilon }_x$ , $-255\times {10}^{-6}$ for ${\epsilon }_y$ and $-255\times {10}^{-6}$ for ${\epsilon }_z$ in Equation (V).

  $\begin{array}{c}{\epsilon }_v=\left(-690\times {10}^{-6}\right)+\left(-255\times {10}^{-6}\right)+\left(-255\times {10}^{-6}\right)\\ =\left(-690-255-255\right)\times {10}^{-6}\\ =-1200\times {10}^{-6}\end{array}$

Substitute $89\text{mm}$ for $a$ in Equation (VI).

  $\begin{array}{c}V={\left(a\right)}^3\\ ={\left(89\text{mm}\right)}^3\\ =704969{\text{mm}}^3\end{array}$

Substitute $704969{\text{mm}}^3$ for $V$ and $-255\times {10}^{-6}$ for ${\epsilon }_v$ in Equation (VII).

  $\begin{array}{c}\Delta V=704969{\text{mm}}^3\times \left(-1200\times {10}^{-6}\right)\\ =-845.962800{\text{mm}}^3\\ =-846{\text{mm}}^3\end{array}$

Conclusion:

The change in the volume is $-846{\text{mm}}^3$ .

To determine

The strain energy stored in the cube.

Answer to Problem 7.6.6P

The strain energy stored in the cube is $21.43\text{N}\cdot \text{m}$ .

Explanation of Solution

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

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

Here, the strain energy is $U$ .

Calculation:

Substitute $704969{\text{mm}}^3$ for $V$ , $-82.56\text{MPa}$ for ${\sigma }_x$ , $-690\times {10}^{-6}$ for ${\epsilon }_x$ , $-54.72\text{MPa}$ for ${\sigma }_y$ , $-255\times {10}^{-6}$ for ${\epsilon }_y$ , $-54.72\text{MPa}$ for ${\sigma }_z$ and $-255\times {10}^{-6}$ for ${\epsilon }_z$ in Equation (VIII).

  $\begin{array}{c}U=\frac{704969{\text{mm}}^3}{2}\left[\begin{array}{l}\left(-82.56\text{MPa}\times -690\times {10}^{-6}\right)+\left(-54.72\text{MPa}\times -255\times {10}^{-6}\right)\\ +\left(-54.72\text{MPa}\times -255\times {10}^{-6}\right)\end{array}\right]\\ =352484.5{\text{mm}}^3\left(84873.6\times {10}^{-6}\text{MPa}\right)\\ =\left(21429.22\text{MPa}·{\text{mm}}^3\right)\times \left(\frac{{10}^6\text{N}/{\text{m}}^2}{1\text{MPa}}\right)\times \left(\frac{1{\text{m}}^3}{{10}^9{\text{mm}}^3}\right)\\ =21.43\text{N}\cdot \text{m}\end{array}$

Conclusion:

The strain energy stored in the cube is $21.43\text{N}\cdot \text{m}$ .

To determine

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

Answer to Problem 7.6.6P

The maximum value of normal stress along the $x$ axis is $-72.9\text{MPa}$ .

Explanation of Solution

Given information:

The change in volume is limited to $0.11\%$ .

Explanation:

Write the expression for the change in volume.

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

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

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

Calculation:

Substitute $-11\times {10}^{-4}$ for $\frac{\Delta V}{V}$ , $-255\times {10}^{-6}$ for ${\epsilon }_y$ and $-255\times {10}^{-6}$ for ${\epsilon }_z$ in Equation (IX).

  $\begin{array}{l}-11\times {10}^{-4}={\epsilon }_x-255\times {10}^{-6}-255\times {10}^{-6}\\ {\epsilon }_x=-11\times {10}^{-4}+5.1\times {10}^{-4}\\ {\epsilon }_x=-5.9\times {10}^{-4}\end{array}$

Substitute $-5.9\times {10}^{-4}$ for ${\epsilon }_x$ , $-255\times {10}^{-6}$ for ${\epsilon }_y$ , $-255\times {10}^{-6}$ for ${\epsilon }_z$

  $80\text{GPa}$ for $E$ and $0.25$ for $\nu$ in Equation (X).

  $\begin{array}{c}{\sigma }_x=\frac{80\text{GPa}}{\left(1+0.25\right)\left(1-\left(2\times 0.25\right)\right)}\left[\left(0.75\times -5.9\times {10}^{-4}\right)+0.25\left(-5.1\times {10}^{-4}\right)\right]\\ =\frac{80\text{GPa}}{1.25\times 0.5}\times \left(-5.7\times {10}^{-4}\right)\\ =\left(-729.6\times {10}^{-4}\text{GPa}\right)\times \left(\frac{{10}^3\text{MPa}}{1\text{GPa}}\right)\\ =-72.96\text{MPa}\end{array}$

Conclusion:

The maximum value of normal stress along the $x$ axis is $-72.9\text{MPa}$ .

To determine

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

Answer to Problem 7.6.6P

The required value of the strain along the $x$ axis is $741\times {10}^{-6}$ .

Explanation of Solution

Given information:

The strain energy of the system is $33\text{J}$ .

Explanation:

Write the expression for the strain energy.

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

Calculation:

Substitute $704969{\text{mm}}^3$ for $V$ , $33\text{J}$ for $U$ , $0.25$ for $\nu$ , $80\text{GPa}$ for $E$ , $-255\times {10}^{-6}$ for ${\epsilon }_y$ , $-255\times {10}^{-6}$ for ${\epsilon }_z$ in Equation (XI).

  $\begin{array}{l}\left(33\text{J}\times \frac{1\text{N}\cdot \text{m}}{1\text{J}}\right)=\left(352484.5{\text{mm}}^3\times \frac{1{\text{m}}^3}{{10}^9{\text{mm}}^3}\right)\times \frac{\left(80\text{GPa}\times \frac{{10}^9\text{N}/{\text{m}}^2}{1\text{GPa}}\right)}{1.25\times 0.5}[\begin{array}{l}0.75\times \left(\begin{array}{l}{\epsilon }_x^2+0.65025\times {10}^{-7}\\ +0.65025\times {10}^{-7}\end{array}\right)\\ +0.5(\begin{array}{l}\left(-255\times {10}^{-6}\right){\epsilon }_x\end{array}\end{array}\end{array}$

+ 0.65025 × 10 − 7 + ( − 255 × 10 − 6 ) ε x

) ] 7.3142 × 10 − 7 = [ 0.75 ε x 2 − 2.55 × 10 − 4 ε x + 1.3005 × 10 − 7 ] 0.75 ε x 2 − 2550 × 10 − 7 ε x − 6.0137 × 10 − 7 = 0

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

  $\begin{array}{c}{\epsilon }_x=\frac{-2550\pm \sqrt{{\left(-2550\right)}^2+4\times 0.75\times 6.0137\times {10}^7}}{2\times 0.75\times {10}^7}\\ =\frac{-2550\pm 13671.31}{1.5\times {10}^7}\\ =-7414.4208\times {10}^{-7}\end{array}$

Conclusion:

The required value of the strain along the $x$ axis is $-741\times {10}^{-6}$ .