Chapter 7, Problem 7.5.12P

-12 A square plate of a width h and thickness t is Loaded by normal forces P_xand P and by shear forces V, as shown in the figure. These forces produce uniformly distributed stresses acting on the side faces of the plate.

(a) Calculate the change AV in the volume of the plate and the strain energy U stored in the plate if the dimensions are ft = 600 mm and f = 40 mm; the plate is made of magnesium with E = 41 GPa and v = 0,35; and the forces are P_v= 420 kN, P, = 210 kN, and V = 96 kN. (b) Find the maximum permissible thickness of the plate when the strain energy U must be at least 62 J. [Assume that all other numerical values in part (a) are unchanged.]

(c) Find the minimum width b of the square plate of thickness / = 40 mm when the change in volume of the plate cannot exceed 0.018% of the original volume.

  

To determine

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

The strain energy $U$ stored in the plate.

Answer to Problem 7.5.12P

The change $\Delta V$ in the volume of the plate $2774.88{\text{mm}}^3$ .

The strain energy $U$ stored in the plate is $48\text{J}$ .

Explanation of Solution

Given information:

The normal force acting on the x-direction is $420\text{kN}$ , the normal force acting along y-direction is $210\text{kN}$ , modulus of elasticity is $41\text{GP}$ , the width of the plate is $600\text{mm}$ , the thickness of the plate is $40\text{mm}$ , the value of $V$ is $96\text{KN}$ ,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 we have two surface area at which shear force is applied so, the surface area for the shear stress is $2\times \text{area}$ .

Write the expression of the shear stress.

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

Calculation:

Substitute $600\text{mm}$ for $b$ , and $40\text{mm}$ for $t$ , in Equation (III).

  $\begin{array}{c}{V}_o=600\text{mm}\times 600\text{mm}\times 40\text{mm}\\ \text{=14,400}\times {\text{10}}^3{\text{mm}}^3\end{array}$

Substitute $600\text{mm}$ for $b$ , and $40\text{mm}$ for $t$ , in Equation (VI).

  $\begin{array}{c}\text{area}=600\text{mm}\times 40\text{mm}\\ \text{=24,000}{\text{mm}}^2\end{array}$

Substitute $\text{24,000}{\text{mm}}^2$ for $\text{area}$ , $420\times {10}^3\text{N}$ for $P_x$ , in Equation (IV).

  $\begin{array}{c}{\sigma }_x=\frac{420\times {10}^3\text{N}}{24,000{\text{mm}}^2}\\ =17.5\text{N}/{\text{mm}}^2\left(\frac{1\text{MPa}}{1\text{N}/{\text{mm}}^2}\right)\\ =17.5\text{MPa}\end{array}$

Substitute $\text{24,000}{\text{mm}}^2$ for $\text{area}$ , $210\times {10}^3\text{N}$ for $P_y$ , in Equation (V).

  $\begin{array}{c}{\sigma }_y=\frac{210\times {10}^3\text{N}}{24,000{\text{mm}}^2}\\ =8.75\text{N}/{\text{mm}}^2\left(\frac{1\text{MPa}}{1\text{N}/{\text{mm}}^2}\right)\\ =8.75\text{MPa}\end{array}$

Substitute $17.5\text{MPa}$ for ${\sigma }_x$ , $8.75\text{MPa}$ for ${\sigma }_y$ , $0\text{MPa}$ for ${\sigma }_z$ , $41\times {10}^3\text{MP}$ for $E$ , and $0.35$ for $\nu$ , in Equation (VIII).

  $\begin{array}{c}{\epsilon }_{\nu }=\frac{\left(17.5\text{MPa}+8.75\text{MPa}+0\text{MPa}\right)}{41\times {10}^3\text{MP}}\left(1-2\times 0.35\right)\\ =\frac{26.25\text{MPa}}{41\times {10}^3\text{MP}}\left(0.3\right)\\ =1.9207\times {10}^{-3}\end{array}$

Substitute $1.9207\times {10}^{-3}$ for ${\epsilon }_{\nu }$ , and $\text{14,400}\times {\text{10}}^3{\text{mm}}^3$ for $V_o$ in Equation (I).

  $\begin{array}{c}\Delta V=1.9207\times {10}^{-4}\times \text{14400}\times {\text{10}}^3{\text{mm}}^3\\ =27748.8\times {10}^{-1}{\text{mm}}^3\\ =2774.88{\text{mm}}^3\end{array}$

Substitute $41\times {10}^3\text{MP}$ for $E$ , and $0.35$ for $\nu$ in Equation (VII).

  $\begin{array}{c}G=\frac{41\times {10}^3\text{MP}}{2\left(1+0.35\right)}\\ =15,185.18\text{MPa}\end{array}$

Substitute $\text{24,000}{\text{mm}}^2$ for $\text{area}$ , and $96\times {10}^3\text{N}$ for $V$ in Equation (IX).

  $\begin{array}{c}{\tau }_{xy}=\frac{96\times {10}^3\text{N}}{2\times \text{24000}{\text{mm}}^2}\\ =\left(2\text{N}/{\text{mm}}^{\text{2}}\right)\left(\frac{1\text{Mpa}}{1\text{N}/{\text{mm}}^{\text{2}}}\right)\\ =2\text{MPa}\end{array}$

Substitute $17.5\text{MPa}$ for ${\sigma }_x$ , $8.75\text{MPa}$ for ${\sigma }_y$ , $15,185.18\text{MPa}$ for $G$ , $41\times {10}^3\text{MP}$ for $E$ , $4.375\text{MPa}$ for ${\tau }_{xy}$ , $\text{14,400}\times {\text{10}}^3{\text{mm}}^3$ for $V_o$ , $96\times {10}^3\text{N}$ for $V$ , and $0.35$ for $\nu$ , in Equation (II).

  $\begin{array}{c}\frac{U}{\text{14,400}\times {\text{10}}^3{\text{mm}}^3}=\left[\frac{1}{2\times 41\times {10}^3\text{MP}}\left(\begin{array}{l}{\left(17.5\text{MPa}\right)}^2+{\left(8.75\text{MPa}\right)}^2-\\ 2\times 0.35\times \left(17.5\text{MPa}\right)\left(8.75\text{MPa}\right)\end{array}\right)\right]\\ +\frac{{\left(2\text{MPa}\right)}^2}{\left(2\times 15,185.18\text{MPa}\right)}\\ U=\left[\begin{array}{l}\left(\frac{275.625{\text{MPa}}^2}{82\times {10}^3\text{MP}}+0.0001317\text{MPa}\right)\\ \left(\text{14400}\times {\text{10}}^3{\text{mm}}^3\right)\left(\frac{{10}^6\text{N}/{\text{m}}^2}{1\text{MPa}}\right)\end{array}\right]\\ =\text{48}\text{.39}\text{N}\cdot \text{m}\left(\frac{1\text{J}}{1\text{N}\cdot \text{m}}\right)\\ =48\text{J}\end{array}$

Conclusion:

The change $\Delta V$ in the volume of the plate is $2774.88{\text{mm}}^3$ .

The strain energy $U$ stored in the plate is $48\text{J}$ .

To determine

The maximum permissible thickness of the plate.

Answer to Problem 7.5.12P

The maximum permissible thickness of the plate is $35.72\text{mm}$ .

Explanation of Solution

Given information:

The strain energy $U$ must be at least $62\text{J}$ .

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 $62\text{J}$ for $U$ , $420\times {10}^3\text{N}$ for $P_x$ , $210\times {10}^3\text{N}$ for $P_y$ , $15,185.18\text{MPa}$ for $G$ , $41\times {10}^3\text{MP}$ for $E$ , $600\text{mm}\times 600\text{mm}\times t\text{mm}$ for $V_o$ , $40\text{mm}\times b\text{mm}$ for $b\times t$ , $96\times {10}^3\text{N}$ for $V$ ,and $0.35$ for $\nu$ , in Equation (II).

  $\begin{array}{l}\frac{62\text{J}}{600\text{mm}\times 600\text{mm}\times t\text{mm}}=\left[\begin{array}{l}\frac{1}{\left(2\times 41\times {10}^3\text{MP}\right)}\left(\begin{array}{l}{\left(\frac{420\times {10}^3\text{N}}{600\text{mm}\times t\text{mm}}\right)}^2+\\ {\left(\frac{96\times {10}^3\text{N}}{600\text{mm}\times t\text{mm}}\right)}^2\end{array}\right)+\\ \left(\begin{array}{l}-2\times 0.35\times \left(\frac{420\times {10}^3\text{N}}{600\text{mm}\times t\text{mm}}\right)\left(\frac{96\times {10}^3\text{N}}{600\text{mm}\times t\text{mm}}\right)\\ +{\left(\frac{96\times {10}^3\text{N}}{600\text{mm}\times t\text{mm}}\right)}^2\times \left(\frac{1}{2\times 15,185.18\text{MPa}}\right)\end{array}\right)\end{array}\right]\\ \frac{0.1722}{t{\text{mm}}^3}=\frac{1}{82000}\left(\left(\frac{490000}{t^2{\text{mm}}^4}\right)+\left(\frac{25600}{t^2{\text{mm}}^4}\right)-\left(\frac{11200}{t^2{\text{mm}}^4}\right)\right)+\frac{.00526}{t^2{\text{mm}}^4}\\ \frac{0.1722}{t{\text{mm}}^3}=\frac{6.1512}{t^2{\text{mm}}^3}\\ t=36.12\text{mm}\end{array}$

Conclusion:

The maximum permissible thickness of the plate is $35.72\text{mm}$ .

To determine

The minimum width $b$ of the square plate.

Answer to Problem 7.5.12P

The minimum width $b$ of the square plate $1750\text{mm}$ .

Explanation of Solution

Given information:

The thickness of square plate is $40\text{mm}$ , and the change in the volume is $0.018\%$ of the original volume.

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

  $\frac{\Delta V}{V_o}\le \frac{0.018}{100}$ ....... (XI)

Calculation:

Substitute $420\times {10}^3\text{N}$ for $P_x$ , $96\times {10}^3\text{N}$ for $P_y$ , $b\times 40{\text{mm}}^2$ for $b\times t$ , $41\times {10}^3\text{MP}$ for $E$ , and $0.35$ for $\nu$ in Equation (VIII).

  $\begin{array}{l}{\epsilon }_{\nu }=\frac{\left(\frac{420\times {10}^3\text{N}}{b\times 40{\text{mm}}^2}\right)+\left(\frac{96\times {10}^3\text{N}}{b\times 40{\text{mm}}^2}\right)+\left(\frac{0\text{N}}{b\times 40{\text{mm}}^2}\right)}{41\times {10}^3\text{MPa}}\left(1-2\times 0.35\right)\\ {\epsilon }_{\nu }=\frac{\frac{10,500\text{N}}{b{\text{mm}}^2}+\frac{2400\text{N}}{b{\text{mm}}^2}}{41\times {10}^3\text{MPa}}\left(0.3\right)\left(\frac{1\text{MPa}}{\text{1}\text{N}/{\text{mm}}^2}\right)\\ {\epsilon }_{\nu }=\frac{0.3146}{b}\end{array}$

Substitute $\frac{0.3146}{b}$ for ${\epsilon }_v$ , $40\text{mm}\times b\text{mm}\times b\text{mm}$ for $V_o$ in Equation (I).

  $\begin{array}{l}\Delta V=\frac{0.3146}{b}\times 40\text{mm}\times b\text{mm}\times b\text{mm}\\ \Delta V=\frac{12.58\times b^2{\text{mm}}^3}{b}\\ \Delta V=12.6\times b{\text{mm}}^3\end{array}$

Substitute $12.6\times b{\text{mm}}^3$ for $\Delta V$ , $40\times b\times b{\text{mm}}^3$ for $V_o$ in Equation (XI).

  $\begin{array}{l}\frac{12.6\times b{\text{mm}}^2}{40\times b\times b{\text{mm}}^3}\le \frac{0.018}{100}\\ \frac{1}{b}\le \frac{0.018\times 40}{100\times 12.6\times b\text{mm}}\\ b\ge 1750\text{mm}\end{array}$

Conclusion: The minimum width $b$ of the square plate $1750\text{mm}$ .