Chapter 11, Problem 14P

(a)

To determine

The expression for the compressive stress and strain.

Answer to Problem 14P

The expression for the compressive stress and strain is $F/A=Y\left(\alpha \left(\Delta T\right)\right)$.

Explanation of Solution

Given Info:

Expression for the compressive stress and strain is,

$\frac{F}{A}=Y\left(\frac{\Delta L}{L_0}\right)$

• F is the force
• A is the area,
• Y is the Young’s modulus,
• $\Delta L$ is change in length,
• $L_0$ is the original length,

Expression for change in length due to linear expansion is,

$\frac{\Delta L}{L_0}=\alpha \left(\Delta T\right)$

• $\alpha$ is the coefficient of linear expansion,
• $\Delta T$ is the change in temperature,

Use $\alpha \left(\Delta T\right)$ for $\Lambda L/L_0$ in $F/A=Y\left(\Delta L/L_0\right)$ to rewrite.

$\frac{F}{A}=Y\left(\alpha \left(\Delta T\right)\right)$

Conclusion:

Thus, the expression for stress is $F/A=Y\left(\alpha \left(\Delta T\right)\right)$.

(b)

To determine

Obtain symbolic equation for thermal energy transfer.

Answer to Problem 14P

Symbolic equation for thermal energy transfer is $Q=\left(mc/Y\alpha \right)\left(F/A\right)$.

Explanation of Solution

Given Info

Formula for heat transfer is,

$Q=mc\Delta T$

• Q is the heat transfer,
• m is the mass,
• c is the specific heat,
• $\Delta T$ is the change in temperature,

Substitute $\frac{F/A}{Y\alpha }$ for $\Delta T$ (from (a)) to rewrite Q.

$\begin{array}{c}Q=mc\left(\frac{F/A}{Y\alpha }\right)\\ =\frac{mc}{Y\alpha }\left(\frac{F}{A}\right)\end{array}$

Conclusion:

Symbolic equation for thermal energy transfer is $Q=\left(mc/Y\alpha \right)\left(F/A\right)$.

(c)

To determine

The mass of the concrete slab.

Answer to Problem 14P

The mass of the concrete slab is $96.0\text{kg}$.

Explanation of Solution

Given info: Thickness of the slab is 4.00 cm, length of the slab is 1.00 m, and width of the slab is 1.00m and density of the slab is $2.40\times {10}^3\text{kg}\cdot {\text{m}}^{-\text{3}}$.

Formula to calculate the volume of the slab is,

$V=l\cdot b\cdot w$

• V is the volume of the slab,
• l is the length of the slab,
• b is the breadth of the slab,
• w is the thickness of the slab,

Formula to calculate the mass of the concrete slab is,

$m=\rho V$

• m is the mass of the slab,
• $\rho$ is the density of the slab,

Use $l\cdot b\cdot w$ for V in the above expression to rewrite m.

$m=\rho \left(l\cdot b\cdot w\right)$

Substitute $2.40\times {10}^3\text{kg}\cdot {\text{m}}^{-\text{3}}$ for $\rho$, 1.00 m for l, 1.00 m for b, and 4.00 cm for w to find m.

$\begin{array}{c}m=\left(2.40\times {10}^3\text{kg}\cdot {\text{m}}^{-\text{3}}\right)\left[\left(1.00\text{m}\right)\left(1.00\text{m}\right)\left(4.00\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right]\\ =96\text{kg}\end{array}$

Conclusion:

Therefore, the mass of the concrete slab is $96.0\text{kg}$.

(d)

To determine

How much thermal energy must be transferred to the slab to reach the compressive stress.

Answer to Problem 14P

The thermal energy transferred is $6.7\times {10}^6\text{J}$.

Explanation of Solution

Give info: Ultimate compressive strength of concrete is $2.00\times {10}^7\text{Pa}$, specific heat is $880\text{J}/\text{kg}\cdot \text{°C}$, and Young’s modulus is $2.1\times {10}^{10}\text{Pa}$.

Formula to calculate the thermal energy transfer is,

$Q_{\text{MAX}}=\frac{mc}{Y\alpha }{\left(\frac{F}{A}\right)}_{\text{MAX}}$

Substitute 96.0 kg for m, $2.1\times {10}^{10}\text{Pa}$ for Y, $880\text{J}/\text{kg}\cdot \text{°C}$ for c, $2.00\times {10}^7\text{Pa}$ for $F/A$ , $12\times {10}^{-6}{\left(°\text{C}\right)}^{-1}$ for $\alpha$ to find $Q_{\text{MAX}}$.

$\begin{array}{c}{Q}_{\text{MAX}}=\frac{\left(96.0\text{kg}\right)\left(880\text{J}/\text{kg}\cdot \text{°C}\right)}{\left(2.1\times {10}^{10}\text{Pa}\right)\left(12\times {10}^{-6}{\left(°\text{C}\right)}^{-1}\right)}\left(2.00\times {10}^7\text{Pa}\right)\\ =6.7\times {10}^6\text{J}\end{array}$

Conclusion:

The maximum thermal energy the slab can absorb before starting to break is $6.7\times {10}^6\text{J}$.

(e)

To determine

The change in temperature.

Answer to Problem 14P

The change in temperature of the slab is $79°\text{C}$.

Explanation of Solution

Given info: Maximum thermal energy is $6.7\times {10}^6\text{J}$, mass is 96.0 kg, and specific heat of the slab is $880\text{J}/\text{kg}\cdot \text{°C}$.

Formula to calculate change in temperature is,

$\Delta T=\frac{Q}{mc}$

$\Delta T$ is the change in temperature,

Substitute $6.7\times {10}^6\text{J}$ for Q, 96.0 kg for Q, and $880\text{J}/\text{kg}\cdot \text{°C}$ for c to calculate $\Delta T$.

$\begin{array}{c}\Delta T=\frac{6.7\times {10}^6\text{J}}{\left(96.0\text{kg}\right)\left(880\text{J}/\text{kg}\cdot \text{°C}\right)}\\ =79.3°\text{C}\\ \approx \text{79}°\text{C}\end{array}$

Conclusion:

The change in temperature of the slab is $79°\text{C}$.

(f)

To determine

The time required for the slab to reach the point of danger of cracking due to compressive stress.

Answer to Problem 14P

The time required is $\text{3}\text{.7}\text{h}$.

Explanation of Solution

Given info: Power delivered by the sun is $1.00\times {10}^3\text{W}$.

Formula to calculate the time is,

$t=\frac{Q_{\text{MAX}}}{P_{\text{absorb}}}$

• t is the time required for the slab to reach the point of cracking,
• $P_{\text{absorb}}$ is the power absorbed by the slab,

On an average half of the energy delivered by the sun in absorbed $P_{\text{absorb}}=\frac{1}{2}{P}_{\text{solar}}$.

Use $1/2P_{\text{solar}}$ for $P_{\text{absorb}}$ in the above expression to rewrite t.

$t=\frac{Q_{\text{MAX}}}{\left(1/2\right)P_{\text{absorb}}}$

Substitute $1.00\times {10}^3\text{W}$ for $P_{\text{absorb}}$ and $6.7\times {10}^6\text{J}$ for $Q_{\text{MAX}}$ to find t.

$\begin{array}{c}t=\frac{6.7\times {10}^6\text{J}}{\left(1/2\right)\left(1.00\times {10}^3\text{W}\right)}\\ =13400\text{s}\left(\frac{1\text{hr}}{3600\text{s}}\right)\\ =\text{3}\text{.7}\text{h}\end{array}$

Conclusion:

The time required is $\text{3}\text{.7}\text{h}$.