Chapter 7, Problem 7.5.14P

A circle of a diameter d = 200 mm is etched on a brass plate (see figure). The plate has dimensions of 400 x 400 x 20 mm. Forces are applied to the plate, producing uniformly distributed normal stressescr^ =59 MPaander^ = —17 MPa. Calculate the following quantities: (a) the change in length Aac of diameter at: (b) the change in length Abd of diameter bd; (c) the change At in the thickness of the plate; (d) the change AV in the volume of the plate; (e) the strain energy U stored in the plate; (f) the maximum permissible thickness of the plate when strain energy £/must be at least 784 J; and (g) the maximum permissible value of normal stress a_xwhen the change in volume of the plate cannot exceed 0.015% of the original volume. (Assume E = 100 GPa and v = 0.34

To determine

The change in the length $\Delta ac$ of diameter $ac$ .

Answer to Problem 7.5.14P

The change in the length $\Delta ac$ is $0.129\text{mm}$ .

Explanation of Solution

Given information:

The normal stress acting on the x-direction is $59\text{MPa}$ , the normal stress acting along y-direction is $-17\text{MPa}$ , modulus of elasticity is $100\text{GPa}$ , diameter of the circle is $200\text{mm}$ , the plate dimensions is $400\text{mm}\times 400\text{mm}\times 20\text{mm}$ , and the Poisson s ratio is $0.35$

Explanation:

The figure below shows the element.

  

     Figure (1)

Write the expression of the strain along x-direction.

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

Here, the normal stress along x-direction is ${\sigma }_x$ ,the normal stress along y-direction is ${\sigma }_y$ , the strain along x-direction is ${\epsilon }_x$ ,and the modulus of elasticity is $E$ .

Write the expression of the change in the length.

  ${\epsilon }_x=\frac{\Delta ac}{ac}$ (II)

Here, the change in the length is $\Delta ac$ , the length of the diameter is $ac$ , and the strain along x-direction is ${\epsilon }_x$ .

Calculation:

Substitute $59\text{MPa}$ for ${\sigma }_x$ , $-17\text{MPa}$ for ${\sigma }_y$ , $100\times {10}^3\text{MPa}$ for $E$ , and $0.34$ for $\nu$ , in Equation (I).

  $\begin{array}{c}{\epsilon }_x=\frac{1}{100\times {10}^3\text{MPa}}\left(59\text{MPa}-\left(0.34\times -17\text{MPa}\right)\right)\\ =\frac{64.78\text{MPa}}{100\times {10}^3\text{MPa}}\\ =0.64\times {10}^{-3}\end{array}$

Substitute $0.64\times {10}^{-3}$ for ${\epsilon }_x$ , and $200\text{mm}$ for $ac$ , in Equation (II).

  $\begin{array}{l}0.64\times {10}^{-3}=\frac{\Delta ac}{200\text{mm}}\\ \Delta ac=128\times {10}^{-3}\text{mm}\\ \Delta ac=0.129\text{mm}\end{array}$

Conclusion:

The change in the length $\Delta ac$ is $0.129\text{mm}$ .

To determine

The change in the length $\Delta bd$ of the diameter $bd$ .

Answer to Problem 7.5.14P

The change in the length $\Delta bd$ of the diameter $bd$ is $0.07412\text{mm}$ .

Explanation of Solution

Given information:

The length of the $bd$ is $200\text{mm}$ .

Write the expression of the strain along y-axis.

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

Write the expression of the change in the length $\Delta bd$ .

  $\frac{\Delta bd}{bd}={\epsilon }_y$ ....... (IV)

Calculation:

Substitute $59\text{MPa}$ for ${\sigma }_x$ , $-17\text{MPa}$ for ${\sigma }_y$ , $100\times {10}^3\text{MPa}$ for $E$ , and $0.34$ for $\nu$ , in Equation (III).

  $\begin{array}{c}{\epsilon }_Y=\frac{1}{100\times {10}^3\text{MPa}}\left(-17\text{MPa}-\text{0}\text{.34}\times 59\text{MPa}\right)\\ =\frac{37.6\text{MPa}}{100\times {10}^3\text{MPa}}\\ =-0.370\times {10}^{-3}\end{array}$

Substitute $-0.370\times {10}^{-3}$ for ${\epsilon }_x$ , and $200\text{mm}$ for $bd$ , in Equation (IV).

  $\begin{array}{l}\frac{\Delta bd}{200\text{mm}}=-0.370\times {10}^{-3}\\ \Delta bd=-0.370\times {10}^{-3}\times 200\text{mm}\\ \Delta bd=-74.12\times {10}^{-3}\text{mm}\end{array}$

Conclusion:

The change in the length $\Delta bd$ of the diameter $bd$ is $0.07412\text{mm}$ .

To determine

The change in thickness of the plate.

Answer to Problem 7.5.14P

The change in the thickness is $-2.856\times {10}^{-3}\text{mm}$ .

Explanation of Solution

Given information:

Write the expression of the strain along z-direction.

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

Write the expression of the change in thickness.

  $\Delta t=-\frac{\nu }{E}\left({\sigma }_x+{\sigma }_y\right)t$ ....... (VI)

Here, the change in the thickness is $\Delta t$ , and the thickness is $t$ .

Calculation:

Substitute $59\text{MPa}$ for ${\sigma }_x$ , $-17\text{MPa}$ for ${\sigma }_y$ , $100\times {10}^3\text{MPa}$ for $E$ ,and $0.34$ for $\nu$ , in Equation (V).

  $\begin{array}{c}{\epsilon }_z=-\frac{0.34}{100\times {10}^3\text{MPa}}\left(59\text{MPa}+\left(-17\text{MPa}\right)\right)\\ =-\frac{14.28\text{MPa}}{100\times {10}^3\text{MP}}\\ =-0.1428\times {10}^{-3}\end{array}$

Substitute $-0.1428\times {10}^{-3}$ for ${\epsilon }_z$ , and $20\text{mm}$ for $t$ , in Equation (VI).

  $\begin{array}{c}\Delta t=-0.1428\times {10}^{-3}\times 20\text{mm}\\ =-2.856\times {10}^{-3}\text{mm}\end{array}$

Conclusion:

The change in the thickness is $-2.856\times {10}^{-3}\text{mm}$ .

To determine

The change in the volume of the plate.

Answer to Problem 7.5.14P

The change in the volume is $407{\text{mm}}^3$ .

Explanation of Solution

Write the expression for the volumetric strain.

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

Here, the change in the volume is $\Delta V$ , the volume is $V_o$ ,the strain along x-direction is ${\epsilon }_x$ , the strain along y-direction is ${\epsilon }_y$ , and the strain along z-direction is ${\epsilon }_z$ .

Calculation:

Substitute $0.64\times {10}^{-3}$ for ${\epsilon }_x$ , $-0.370\times {10}^{-3}$ for ${\epsilon }_y$ , $-0.1428\times {10}^{-3}$ for ${\epsilon }_z$ , and $400\text{mm}\times 400\text{mm}\times 20\text{mm}$ for $V_o$ , in Equation (VII).

  $\begin{array}{c}\Delta V=400\text{mm}\times 400\text{mm}\times 20\text{mm}\left(0.64\times {10}^{-3}-0.370\times {10}^{-3}-0.1428\times {10}^{-3}\right)\\ =3200000{\text{mm}}^3\left(0.1272\times {10}^{-3}\right)\\ =407{\text{mm}}^3\end{array}$

Conclusion:

The change in the volume is $407{\text{mm}}^3$ .

To determine

The strain energy stored in the plate.

Answer to Problem 7.5.14P

The strain energy stored in the plate is $70.480\text{N}\cdot \text{m}$ .

Explanation of Solution

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

  $U=\frac{V_o}{2}\left({\sigma }_x{\epsilon }_x+{\sigma }_y{\epsilon }_y+{\sigma }_z{\epsilon }_z\right)$ ....... (VII)

Here, the energy in the plate is $U$ .

Calculation:

Substitute $59\text{MPa}$ for ${\sigma }_x$ , $-17\text{MPa}$ for ${\sigma }_y$ , $100\times {10}^3\text{MPa}$ for $E$ , and $0.34$ for $\nu$ , $0.64\times {10}^{-3}$ for ${\epsilon }_x$ , $-0.370\times {10}^{-3}$ for ${\epsilon }_y$ , $-0.1428\times {10}^{-3}$ for ${\epsilon }_z$ , and $400\text{mm}\times 400\text{mm}\times 20\text{mm}$ for $V_o$ , in Equation (VIII).

  $\begin{array}{c}U=\frac{400\text{mm}\times 400\text{mm}\times 20\text{mm}}{2}\left(59\text{MPa}\left(0.64\times {10}^{-3}\right)+17\text{MPa}\times 0.370\times {10}^{-3}+0\text{MPa}\times \left(-0.1428\times {10}^{-3}\right)\right)\\ =1600000{\text{mm}}^3\left(0.04405\text{MPa}\right)\left(\frac{1{\text{m}}^3}{{10}^9{\text{mm}}^3}\right)\left(\frac{{\text{10}}^6\text{N}/{\text{m}}^2}{1\text{MPa}}\right)\\ =70.480\text{N}\cdot \text{m}\end{array}$ The strain energy stored in the plate is $70.480\text{N}\cdot \text{m}$ .

To determine

The maximum permissible thickness of the plate.

Answer to Problem 7.5.14P

The maximum permissible thickness is $22.26\text{mm}$ .

Explanation of Solution

Given information:

The strain energy stored must be at least $78.4\text{J}$ ,the initial strain energy is $70.4\text{J}$ ,and the initial thickness of the plate is $20\text{mm}$ .

Write the expression of the strain energy directly proportional to the thickness.

  $\frac{U_{\text{final}}}{U_{\text{initial}}}=\frac{t_{\text{final}}}{t_{\text{initial}}}$ ....... (IX)

Calculation:

Substitute $78.4\text{J}$ for $U_{\text{final}}$ , $70.4\text{J}$ for $U_{\text{initial}}$ , and $20\text{mm}$ for $t_{\text{initial}}$ in Equation (IX).

  $\begin{array}{c}\frac{78.4\text{J}}{70.7\text{J}}=\frac{t_{\text{final}}}{20\text{mm}}\\ t_{\text{final}}=20\text{mm}\times 1.1136\\ =22.26\text{mm}\end{array}$

Conclusion:

The maximum permissible thickness is $22.26\text{mm}$ .

To determine

The maximum permissible volume of the plate.

Answer to Problem 7.5.14P

The maximum permissible volume is . $63.75\text{MPa}$ .

Explanation of Solution

Write the expression for the volumetric strain.

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

Calculation:

Substitute $59\text{MPa}$ for ${\sigma }_x$ , $-17\text{MPa}$ for ${\sigma }_y$ , $100\times {10}^3\text{MPa}$ for $E$ , and $0.34$ for $\nu$ ,and $0.015\%$ for $\frac{\Delta V}{V}$ in Equation(X).

  $\begin{array}{l}\frac{0.015}{100}=\frac{\left(1-0.68\right)}{100\times {10}^3\text{MPa}}\left({\sigma }_x-17\text{MPa}\right)\\ 1.5\times {10}^{-4}\text{MPa}=3.2\times {10}^{-6}{\sigma }_x-5.4\times {10}^{-5}\text{MPa}\\ {\sigma }_x=63.75\text{MPa}\end{array}$

Conclusion:

The maximum permissible volume is . $63.75\text{MPa}$ .