Chapter 7, Problem 7.5.13P

Solve the preceding problem for an aluminum plate with h = 10 in.. i = 0.75 in., E = 10,600 ksi, v = 0.33. P = 96 kips. Pt,. = 24 kips. and V =18 kips.

For part (b) of Problem 7.5-12, assume that the required strain energy stored is 640 in.-lb. In part (c). the change in volume cannot exceed 0.05%.

  

To determine

The change $\Delta V$ in the volume of the plate.

The strain energy $U$ stored in the plate.

Answer to Problem 7.5.13P

The change $\Delta V$ in the volume of the plate is $79.2\times {10}^{-3}{\text{in}}^3$ .

The strain energy $U$ stored in the plate is $495.64\text{in}\cdot \text{lb}$ .

Explanation of Solution

Given information:

The normal force acting on the x-direction is $96\text{kips}$ , the normal force acting along y-direction is $24\text{kips}$ , modulus of elasticity is $10600\text{kips}$ , the width of the plate is $10\text{in}$ , the thickness of the plate is $0.75\text{in}$ , the value of $V$ is $18\text{kips}$ , the strain energy stored is $640\text{in}\cdot \text{lb}$ , the change in the volume is $0.05\%$ , and the Poisson s ratio is $0.35$

  

     Figure (1)

Write the expression for the volumetric strain.

  $\frac{\Delta V}{V_o}={\epsilon }_v$ (I)

Here, the change in the volume of the plate is $\Delta V$ , the volume of plate is $V$ , the volumetric strain is ${\epsilon }_v$ .

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

  $\frac{U}{V_o}=\frac{1}{2E}\left({\sigma }_X{}^2+{\sigma }_Y{}^2-2\nu \left({\sigma }_X\right)\left({\sigma }_Y\right)\right)+\frac{{\tau }_{xy}^2}{2G}$ ....... (II)

Here, the strain energy stored in the plate is $U$ , the stress along x-axis is ${\sigma }_x$ , the stress along y-direction is ${\sigma }_y$ , modulus of the elasticity is $E$ ,the shear stress along x-y direction is ${\tau }_{xy}$ , the shear modulus is $G$ , and the Poisson s ratio is $\nu$ .

Write the expression of the original volume of plate

  $V_o=b\times b\times t$ ....... (III)

Here, the width of the plate is $b$ , and thickness of the plate is $t$ .

Write the expression for the stress along x-direction.

  ${\sigma }_x=\frac{P_x}{\text{area}}$ ....... (IV)

Here the normal force along x-direction is $P_x$ .

Write the expression for the stress along y-direction.

  ${\sigma }_y=\frac{P_y}{\text{area}}$ ....... (V)

Here the normal force along y-direction is $P_y$ .

Write the expression for the area of the plate.

  $\text{area}=b\times t$ ....... (VI)

Write the expression for the shear modulus.

  $G=\frac{E}{2\left(1+\nu \right)}$ (VII)

Write the expression of the volumetric strain.

  $\begin{array}{l}{\epsilon }_{\nu }=\frac{\left(\frac{P_x}{b\times t}\right)+\left(\frac{P_y}{b\times t}\right)+\left(\frac{P_z}{b\times t}\right)}{E}\left(1-2\nu \right)\\ {\epsilon }_{\nu }=\frac{{\sigma }_x+{\sigma }_y+{\sigma }_z}{E}\left(1-2\nu \right)\end{array}$ (VIII)

In the following plate the area of the shear stress for the both side of the plate is $2\times \text{area}$ .

Write the expression of the shear stress.

  ${\tau }_{xy}=\frac{V}{2\times \text{area}}$ (IX)

Calculation:

Substitute $10\text{in}$ for $b$ , and $0.75\text{in}$ for $t$ , in Equation (III).

  $\begin{array}{c}{V}_o=10\text{in}\times 10\text{in}\times 0.75\text{in}\\ \text{=75}{\text{in}}^3\end{array}$

Substitute $10\text{in}$ for $b$ , and $0.75\text{in}$ for $t$ , in Equation (VI).

  $\begin{array}{c}\text{area}=10\text{in}\times 0.75\text{in}\\ \text{=}7.5{\text{in}}^2\end{array}$

Substitute $7.5{\text{in}}^2$ for $\text{area}$ , $96\text{kips}$ for $P_x$ , in Equation (IV).

  $\begin{array}{c}{\sigma }_x=\frac{96\text{kips}}{7.5{\text{in}}^2}\\ =\left(12.8\text{kips}/{\text{in}}^2\right)\\ =12.8\text{ksi}\end{array}$

Substitute $7.5{\text{in}}^2$ for $\text{area}$ , $24\text{kips}$ for $P_y$ , in Equation (V).

  $\begin{array}{c}{\sigma }_y=\frac{24\text{kips}}{7.5{\text{in}}^2}\\ =\frac{3.2\text{kis}\cdot {\text{in}}^{\text{2}}}{1{\text{in}}^2}\\ =3.2\text{kis}\end{array}$

Substitute $12.8\text{ksi}$ for ${\sigma }_x$ , $3.2\text{ksi}$ for ${\sigma }_y$ , $0\text{ksi}$ for ${\sigma }_z$ , $10600\text{ksi}$ for $E$ , and $0.35$ for $\nu$ , in Equation (VIII).

  $\begin{array}{c}{\epsilon }_{\nu }=\frac{\left(12.8\text{ksi}+3.2\text{ksi+0}\text{ksi}\right)}{10600\text{ksi}}\left(1-2\times 0.35\right)\\ =\frac{11.2\text{ksi}}{10600\text{ksi}}\\ =1.056\times {10}^{-3}\end{array}$

Substitute $1.056\times {10}^{-3}$ for ${\epsilon }_{\nu }$ , and $\text{75}{\text{in}}^3$ for $V_o$ , in Equation (I).

  $\begin{array}{c}\Delta V=1.056\times {10}^{-3}\times \text{75}{\text{in}}^3\\ =79.2\times {10}^{-3}{\text{in}}^3\end{array}$

Substitute $10600\text{ksi}$ for $E$ , and $0.35$ for $\nu$ , in Equation (VII).

  $\begin{array}{c}G=\frac{10600\text{ksi}}{2\left(1+0.35\right)}\\ =\frac{10600\text{ksi}}{2.7}\\ =3925.92\text{ksi}\end{array}$

Substitute $\text{75}{\text{in}}^3$ for $\text{area}$ , and $18\text{ksi}$ for $V$ , in Equation (IX).

  $\begin{array}{c}{\tau }_{xy}=\frac{18\text{ksi}}{2\times 75{\text{in}}^3}\\ =\frac{18\text{ksi}}{150}\\ =0.12\text{ksi}\end{array}$

Substitute $12.8\text{ksi}$ for ${\sigma }_x$ , $3.2\text{ksi}$ for ${\sigma }_y$ , $3925.92\text{ksi}$ for $G$ , $10600\text{ksi}$ for $E$ , $0.12\text{ksi}$ for ${\tau }_{xy}$ , $\text{75}{\text{in}}^3$ for $V_o$ , $18\text{ksi}$ for $V$ ,and $0.35$ for $\nu$ , in Equation (II).

  $\begin{array}{c}\frac{U}{75{\text{in}}^3}=\frac{1}{2\times 10600\text{ksi}}\left[{\left(12.8\text{ksi}\right)}^2+{\left(3.2\text{ksi}\right)}^2-\left(2\times 0.35\times 12.8\text{ksi}\times 3.2\text{ksi}\right)\right]\\ +\left(\frac{{\left(0.12\text{ksi}\right)}^2}{2\times 3925.92\text{ksi}}\right)\\ \frac{U}{75{\text{in}}^3}=\frac{1}{21200\text{ksi}}\left[163.84{\text{ksi}}^2+10.24{\text{ksi}}^2-28.672{\text{ksi}}^2\right]\\ +1.83\times {10}^{-3}\text{ksi}\end{array}$

  $\begin{array}{c}\frac{U}{75{\text{in}}^3}=\frac{1}{21200}\left(145.408{\text{ksi}}^2\right)+\left(1.83\times {10}^{-3}\text{ksi}\right)\\ =6.8518\times {10}^{-3}\text{ksi}\left(\frac{1000\text{psi}}{1\text{ksi}}\right)\\ U=6.8518\times {10}^{-3}\text{psi}\times 75{\text{in}}^3\\ =513.88\text{psi}\times {\text{in}}^3\left(\frac{1\text{J}}{0.00512\times 12\times 12\times 12\text{psi}\times {\text{in}}^3}\right)\end{array}$

  $\begin{array}{c}U=58\text{J}\left(\frac{8.85\text{in}\cdot \text{lb}}{1\text{J}}\right)\\ =495.64\text{in}\cdot \text{lb}\end{array}$

Conclusion:

The change $\Delta V$ in the volume of the plate is $79.2\times {10}^{-3}{\text{in}}^3$ .

The strain energy $U$ stored in the plate is $495.64\text{in}\cdot \text{lb}$ .

To determine

The maximum permissible thickness of the plate.

Answer to Problem 7.5.13P

The maximum permissible thickness of the plate is.

Explanation of Solution

Given information:

The strain energy $U$ must be at least $640\text{in}\text{.-lb}$ .

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

  $\begin{array}{c}\frac{U}{V_o}=\frac{1}{2E}\left({\left(\frac{P_x}{b\times t}\right)}^2+{\left(\frac{P_y}{b\times t}\right)}^2-2\nu \left(\frac{P_x}{b\times t}\right)\left(\frac{P_y}{b\times t}\right)\right)\\ +\left({\left(\frac{V}{b\times t}\right)}^2\times \left(\frac{1}{2\times G}\right)\right)\end{array}$ ....... (X)

Calculation:

Substitute $640\text{in}\cdot \text{lb}$ for $U$ , $96\times {10}^3\text{lb}$ for $P_x$ , $24\times {10}^3\text{lb}$ for $P_y$ , $3925.92\text{ksi}$ for $G$ , $10600\text{ksi}$ for $E$ , $10\text{in}\times 10\text{in}\times t\text{in}$ for $V_o$ , $10\text{in}\times t\text{in}$ for $b\times t$ , $18\times {10}^3\text{lb}$ for $V$ ,and $0.35$ for $\nu$ , in Equation (II).

  $\begin{array}{c}\frac{640\text{lb}}{10\text{in}\times 10\text{in}\times t\text{in}}=\left[\begin{array}{l}\frac{1}{\left(2\times 10600\times {10}^3\text{lb}\right)}\left({\left(\frac{96\times {10}^3\text{lb}}{10\text{in}\times t\text{in}}\right)}^2+{\left(\frac{24\times {10}^3\text{lb}}{10\text{in}\times t\text{in}}\right)}^2\right)+\\ \left(\begin{array}{l}-2\times 0.35\times \left(\frac{96\times {10}^3\text{lb}}{10\text{in}\times t\text{in}}\right)\left(\frac{24\times {10}^3\text{lb}}{10\text{in}\times t\text{in}}\right)\\ +{\left(\frac{18\times {10}^3\text{lb}}{10\text{in}\times t\text{in}}\right)}^2\times \left(\frac{1}{2\times 3925.92\text{ksi}}\right)\end{array}\right)\end{array}\right]\\ \frac{6.4\text{lb}}{t{\text{in}}^3}=\frac{1}{2\times 10600\times {10}^3\text{lb}}\left(\left(\frac{92160\times {10}^3{\text{lb}}^2}{t^2{\text{in}}^4}\right)+\left(\frac{5760\times {10}^3{\text{lb}}^{}}{t^2{\text{in}}^4}\right)-\left(\frac{16128\times {10}^3{\text{lb}}^2}{t^2{\text{in}}^4}\right)\right)\\ =+\frac{0.0001273\times {10}^6{\text{lb}}^2}{t^2{\text{in}}^4}\end{array}$$\begin{array}{l}\frac{6.4\text{lb}}{t{\text{in}}^3}=\frac{3.85\text{lb}}{t^4{\text{in}}^2}+\frac{0.1273\text{lb}}{t^4{\text{in}}^2}\\ \frac{6.4\text{lb}}{t{\text{in}}^3}=\frac{3.8512\text{lb}}{t^4{\text{in}}^2}\\ t=0.60175\text{in}\end{array}$

Conclusion:

The maximum permissible thickness of the plate is $0.60175\text{in}$ .

To determine

The minimum width $b$ of the square plate.

Answer to Problem 7.5.13P

The minimum width $b$ of the square plate $9.04\text{in}$ .

Explanation of Solution

Given information:

The change in the volume is $0.05\%$ of the original volume.

Write the expression of the condition of the change in the length.

  $\frac{\Delta V}{V_o}\le \frac{0.05}{100}$ ....... (XI)

Calculation:

Substitute $96\times {10}^3\text{lb}$ for $P_x$ , $24\times {10}^3\text{lb}$ for $P_y$ , $b\times 0.75{\text{in}}^2$ for $b\times t$ , $10600\times {10}^3\text{lb}$ for $E$ , and $0.35$ for $\nu$ in Equation (VIII).

  $\begin{array}{l}{\epsilon }_{\nu }=\frac{\left(\frac{96\times {10}^3\text{lb}}{b\times 0.75{\text{in}}^2}\right)+\left(\frac{24\times {10}^3\text{lb}}{b\times 0.75{\text{in}}^2}\right)+\left(\frac{0\times {10}^3\text{lb}}{b\times 0.75{\text{in}}^2}\right)}{10600\times {10}^3\text{lb}}\left(1-2\times 0.35\right)\\ {\epsilon }_{\nu }=\frac{\frac{128000\text{lb}}{b}+\frac{32000\text{lb}}{b}}{10600\times {10}^3\text{lb}}\left(0.3\right)\\ {\epsilon }_{\nu }=\frac{0.00452}{b}\end{array}$

Substitute $\frac{0.00452}{b}$ for ${\epsilon }_v$ , $b\text{in}\times b\text{in}\times 0.75\text{in}$ for $V_o$ in Equation (I).

  $\begin{array}{l}\Delta V=\frac{0.00452}{b}\times b\text{in}\times b\text{in}\times 0.75\text{in}\\ \Delta V=3.39\times {10}^{-3}\times b{\text{in}}^3\end{array}$

Substitute $3.39\times {10}^{-3}\times b{\text{in}}^3$ for $\Delta V$ , $b\text{in}\times b\text{in}\times 0.75\text{in}$ for $V_o$ in Equation (XI).

  $\begin{array}{l}\frac{3.39\times {10}^{-3}\times b{\text{in}}^3}{b\text{in}\times b\text{in}\times 0.75\text{in}}\le \frac{0.05}{100}\\ \frac{1}{b\text{in}}\le \frac{37.5}{339}\\ b\ge 9.04\text{in}\end{array}$

Conclusion:

The minimum width $b$ of the square plate $9.04\text{in}$ .