Chapter 7, Problem 7.6.4P

Solve the preceding problem if the element

is steel (E = 200 GPa. p = 0.30) with dimensions

a = 300 mm. h = 150 mm. and c = 150 mm a n d

with the stresses (T( = —62 MPa, r. = -45 MPa,

and = MPa.

For part (e) of Problem 7.6-3, find the maximum value of u. if the change in volume must be limited to —0.028%. For part (0. find the required value of if the strain energy must be 60 J.

  

To determine

The maximum shear stress in the material.

Answer to Problem 7.6.4P

The maximum shear stress on the material is $8.5\text{MPa}$ .

Explanation of Solution

Given information:

The element of length $300\text{mm}$ , width $150\text{mm,}$ and height $150\text{mm}$ is subjected to triaxial stress. The stress in $x$ direction is $-62\text{MPa}$ , in $y$ direction is $-45\text{MPa}$ and in $z$ direction is $-45\text{MPa}$ . The modulus of elasticity is $200\text{GPa}$ and the Poisson’s ratio is $0.30$ .

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, $-45\text{MPa}$ for ${\sigma }_{\text{largest}}$ and $-62\text{MPa}$ for ${\sigma }_{\text{smallest}}$ in Equation (I).

  $\begin{array}{c}{\tau }_{\max }=\frac{-45\text{MPa}+62\text{MPa}}{2}\\ =\frac{17\text{MPa}}{2}\\ =8.5\text{MPa}\end{array}$

Conclusion:

The maximum shear stress on the material is $8.5\text{MPa}$ .

To determine

The changes in the dimensions of the element.

Answer to Problem 7.6.4P

The change in length is $-0.0525\text{mm}$ .

The change in height is $-9.67\times {10}^{-3}\text{mm}$ .

The change in width is $-9.67\times {10}^{-3}\text{mm}$ .

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 $200\text{GPa}$ for $E$ , $-62\text{MPa}$ for ${\sigma }_x$ , $-45\text{MPa}$ for ${\sigma }_y$ , $-45\text{MPa}$ for ${\sigma }_z$ and $0.30$ for $\nu$ in Equation (II).

  $\begin{array}{c}{\epsilon }_x=\frac{1}{200\text{GPa}}\left(\left(-62\text{MPa}\right)-0.30\left(\left(-45\text{MPa}\right)+\left(-45\text{MPa}\right)\right)\right)\\ =\frac{-35\text{MPa}}{200\text{GPa}}\times \left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\times \left(\frac{1\text{GPa}}{{10}^9\text{Pa}}\right)\\ =-0.175\times {10}^{-3}\end{array}$

Substitute $200\text{GPa}$ for $E$ , $-62\text{MPa}$ for ${\sigma }_x$ , $-45\text{MPa}$ for ${\sigma }_y$ , $-45\text{MPa}$ for ${\sigma }_z$ and $0.30$ for $\nu$ in Equation (III).

  $\begin{array}{c}{\epsilon }_y=\frac{1}{200\text{GPa}}\left(\left(-45\text{MPa}\right)-0.30\left(\left(-62\text{MPa}\right)+\left(-45\text{MPa}\right)\right)\right)\\ =\frac{-12.9\text{MPa}}{200\text{GPa}}\times \left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\times \left(\frac{1\text{GPa}}{{10}^9\text{Pa}}\right)\\ =-0.0645\times {10}^{-3}\end{array}$

Substitute $200\text{GPa}$ for $E$ , $-62\text{MPa}$ for ${\sigma }_x$ , $-45\text{MPa}$ for ${\sigma }_y$ , $-45\text{MPa}$ for ${\sigma }_z$ and $0.30$ for $\nu$ in Equation (IV).

  $\begin{array}{c}{\epsilon }_z=\frac{1}{200\text{GPa}}\left(\left(-45\text{MPa}\right)-0.30\left(\left(-62\text{MPa}\right)+\left(-45\text{MPa}\right)\right)\right)\\ =\frac{-12.9\text{MPa}}{200\text{GPa}}\times \left(\frac{1\text{GPa}}{{10}^9\text{Pa}}\right)\times \left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\\ =-0.0645\times {10}^{-3}\end{array}$

Substitute $-0.175\times {10}^{-3}$ for ${\epsilon }_x$ and $300\text{mm}$ for $a$ in Equation (V).

  $\begin{array}{c}\Delta a=300\text{mm}\times \left(-0.175\times {10}^{-3}\right)\\ =-52.5\times {10}^{-3}\text{mm}\end{array}$

Substitute $-0.0645\times {10}^{-3}$ for ${\epsilon }_y$ and $150\text{mm}$ for $b$ in Equation (VI).

  $\begin{array}{c}\Delta b=150\text{mm}\times \left(-0.0645\times {10}^{-3}\right)\\ =-9.675\times {10}^{-3}\text{mm}\end{array}$

Substitute $-0.0645\times {10}^{-3}$ for ${\epsilon }_z$ and $150\text{mm}$ for $c$ in Equation (VII).

  $\begin{array}{c}\Delta c=150\text{mm}\times \left(-0.0645\times {10}^{-3}\right)\\ =-9.67\times {10}^{-3}\text{mm}\end{array}$

Conclusion:

The change in length is $-0.0525\text{mm}$ .

The change in height is $-9.67\times {10}^{-3}\text{mm}$ .

The change in width is $-9.67\times {10}^{-3}\text{mm}$ .

To determine

The change in the volume of the element.

Answer to Problem 7.6.4P

The change in the volume is $-2052{\text{mm}}^3$ .

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 $V_0$ .

Write the expression for the volume.

  $V_0=a\times b\times c$ ...... (IX)

Calculation:

Substitute $300\text{mm}$ for $a$ , $150\text{mm}$ for $b$ and $150\text{mm}$ for $c$ in Equation (IX).

  $\begin{array}{c}{V}_0=300\text{mm}\times 150\text{mm}\times 150\text{mm}\\ =6750000{\text{mm}}^3\end{array}$

Substitute $6750000{\text{mm}}^3$ for $V_0$ , $-0.175\times {10}^{-3}$ for ${\epsilon }_x$ , $-0.0645\times {10}^{-3}$ for ${\epsilon }_y$ and $-0.0645\times {10}^{-3}$ for ${\epsilon }_z$ in Equation (VIII).

  $\begin{array}{c}\Delta V=6750000{\text{mm}}^3\left(\left(-0.175\times {10}^{-3}\right)+\left(-0.0645\times {10}^{-3}\right)+\left(-0.0645\times {10}^{-3}\right)\right)\\ =6750000{\text{mm}}^3\times \left(-0.304\times {10}^{-3}\right)\\ =-2.052\times {10}^3{\text{mm}}^3\end{array}$

Conclusion:

The change in the volume is $-2052{\text{mm}}^3$ .

To determine

The strain energy stored in the element.

Answer to Problem 7.6.4P

The strain energy stored in the element is $56.2\text{N}\cdot \text{m}$ .

Explanation of Solution

Write the expression for the strain energy.

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

Here, the strain energy is $U$ .

Calculation:

Substitute $0.00675{\text{m}}^3$ for $V_0$ , $-62\text{MPa}$ for ${\sigma }_x$ , $-0.175\times {10}^{-3}$ for ${\epsilon }_x$ , $-45\text{MPa}$ for ${\sigma }_y$ , $-0.0645\times {10}^{-3}$ for ${\epsilon }_y$ , $-45\text{MPa}$ for ${\sigma }_z$ and $-0.0645\times {10}^{-3}$ for ${\epsilon }_z$ in Equation (X).

  $\begin{array}{c}U=\frac{0.00675{\text{m}}^3}{2}\left(\begin{array}{l}\left(-62\text{MPa}\times -0.175\times {10}^{-3}\right)+\left(-45\text{MPa}\times -0.0645\times {10}^{-3}\right)\\ +\left(-45\text{MPa}\times -0.0645\times {10}^{-3}\right)\end{array}\right)\\ =\left(3.375\times {10}^{-3}{\text{m}}^3\right)\left(16.655\times {10}^{-3}\text{MPa}\right)\\ =56.2106\times {10}^{-6}\text{MPa}\cdot {\text{m}}^3\times \left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\times \left(\frac{1\text{N}/{\text{m}}^2}{1\text{Pa}}\right)\\ =\text{56}\text{.2106}\text{N}\cdot \text{m}\end{array}$

Conclusion:

The strain energy stored in the element is $56.2\text{N}\cdot \text{m}$ .

To determine

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

Answer to Problem 7.6.4P

The maximum value of normal stress along the $x$ axis is $-50\text{MPa}$ .

Explanation of Solution

Given information:

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

Explanation:

Write the expression for the change in volume.

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

Calculation:

Substitute $-2.8\times {10}^{-4}$ for $\frac{\Delta V}{V_0}$ , $0.30$ for $\nu$ , $200\text{GPa}$ for $E$ , $-45\text{MPa}$ for ${\sigma }_y$ and $-45\text{MPa}$ for ${\sigma }_z$ in Equation (XI).

  $\begin{array}{l}-2.8\times {10}^{-4}=\frac{\left(1-0.6\right)}{200\text{GPa}}\left({\sigma }_x-\left(45\text{MPa}\right)-\left(45\text{MPa}\right)\right)\\ -0.056\text{GPa}\times \left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)=0.4{\sigma }_x-36\text{MPa}\times \left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\\ {\sigma }_x=-50\times {10}^6\text{Pa}\times \left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ {\sigma }_x=-50\text{MPa}\end{array}$

Conclusion:

The maximum value of normal stress along the $x$ axis is $-50\text{MPa}$ .

To determine

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

Answer to Problem 7.6.4P

The required value of the normal stress along the $x$ axis is $-65.1\text{MPa}$ .

Explanation of Solution

Given information:

The strain energy of the system is $60\text{J}$ .

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 $0.00675{\text{m}}^3$ for $V_0$ , $60\text{J}$ for $U$ , $0.30$ for $\nu$ , $200\text{GPa}$ for $E$ , $-45\text{MPa}$ for ${\sigma }_y$ and $-45\text{MPa}$ for ${\sigma }_z$ in Equation (XI).

  $\begin{array}{l}60=\frac{\left(0.00675{\text{m}}^3\times \frac{{10}^9{\text{mm}}^3}{1{\text{m}}^3}\right)}{2\times \left(200\text{GPa}\times \frac{{10}^3\text{MPa}}{1\text{GPa}}\right)}\left(\begin{array}{l}{\sigma }_x^2+{\left(-45\text{MPa}\right)}^2+{\left(-45\text{MPa}\right)}^2\\ -2\times 0.33\left(\begin{array}{l}\left(-45\text{MPa}\right){\sigma }_x+\left(-45\text{MPa}\right)\\ \times \left(-45\text{MPa}\right)+{\sigma }_x\times \left(-45\text{MPa}\right)\end{array}\right)\end{array}\right)\\ 3555.6=\left({\sigma }_x^2+4050{\text{MPa}}^2-\left(\left(-27\text{MPa}\times {\sigma }_x\right)+1215{\text{MPa}}^2+\left(-27\text{MPa}\times {\sigma }_x\right)\right)\right)\\ {\sigma }_x^2+54\text{MPa}{\sigma }_x-720.6{\text{MPa}}^2=0\end{array}$

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

  $\begin{array}{c}{\sigma }_x=\frac{-54\text{MPa}\pm \sqrt{{\left(54\text{MPa}\right)}^2+\left(4\times 720.6{\text{MPa}}^2\right)}}{2}\\ =\frac{-54\text{MPa}\pm 76.15\text{MPa}}{2}\end{array}$

Taking the negative sign.

  $\begin{array}{c}{\sigma }_x=\frac{-54\text{MPa}-76.15\text{MPa}}{2}\\ =-65.07\text{MPa}\end{array}$

Conclusion:

The required value of the normal stress along the $x$ axis is $-65.1\text{MPa}$ .