A tensile force of 25 kN is applied to a rectangular bar 12 mm thick and 22 mm wide. The original length is 1,8 m and Young's modulus for the steel is 195 GPa. Calculate the following: The stress, The final length of the bar.

$$\sigma = \frac{F}{A}$$ $$\sigma = \frac{25 \times 10^3 \text{ N}}{0.000264 \text{ m}^2}$$ $$\sigma \approx 94.6969 \times 10^6 \text{ Pa}$$ $$\sigma \approx 94.70 \text{ MPa}$$

The stress is $\boxed{\text{94.70 MPa}}$.

6.1.2 The final length of the bar

$$\epsilon = \frac{\sigma}{E}$$ $$\epsilon = \frac{94.6969 \times 10^6 \text{ Pa}}{195 \times 10^9 \text{ Pa}}$$ $$\epsilon \approx 0.0004856$$

$$\Delta L = \epsilon \times L_0$$ $$\Delta L = 0.0004856 \times 1.8 \text{ m}$$ $$\Delta L \approx 0.00087408 \text{ m}$$

$$L_f = L_0 + \Delta L$$ $$L_f = 1.8 \text{ m} + 0.00087408 \text{ m}$$ $$L_f \approx 1.80087 \text{ m}$$

The final length of the bar is $\boxed{\text{1.80087 m}}$.

6.2 FIGURE 3 below shows a round steel rod subjected to a tensile load of 55 kN. Young's modulus for the steel is 200 GPa.

Given:

• Tensile load $F = 55 \text{ kN} = 55 \times 10^3 \text{ N}$
• Young's modulus $E = 200 \text{ GPa} = 200 \times 10^9 \text{ Pa}$
• Section 1: Diameter $D_1 = 25 \text{ mm} = 0.025 \text{ m}$, Length $L_1 = 100 \text{ mm} = 0.1 \text{ m}$
• Section 2: Diameter $D_2 = 50 \text{ mm} = 0.050 \text{ m}$, Length $L_2 = 100 \text{ mm} = 0.1 \text{ m}$

$$A_1 = \frac{\pi D_1^2}{4} = \frac{\pi (0.025 \text{ m})^2}{4} = \frac{\pi \times 0.000625}{4} \approx 0.00049087 \text{ m}^2$$ $$A_2 = \frac{\pi D_2^2}{4} = \frac{\pi (0.050 \text{ m})^2}{4} = \frac{\pi \times 0.0025}{4} \approx 0.0019635 \text{ m}^2$$

6.2.1 The maximum tensile stress

$$\sigma_{max} = \frac{F}{A_1}$$ $$\sigma_{max} = \frac{55 \times 10^3 \text{ N}}{0.00049087 \text{ m}^2}$$ $$\sigma_{max} \approx 112.04 \times 10^6 \text{ Pa}$$ $$\sigma_{max} \approx 112.04 \text{ MPa}$$

The maximum tensile stress is $\boxed{\text{112.04 MPa}}$.

6.2.2 The minimum stress in the rod

$$\sigma_{min} = \frac{F}{A_2}$$ $$\sigma_{min} = \frac{55 \times 10^3 \text{ N}}{0.0019635 \text{ m}^2}$$ $$\sigma_{min} \approx 28.01 \times 10^6 \text{ Pa}$$ $$\sigma_{min} \approx 28.01 \text{ MPa}$$

The minimum stress in the rod is $\boxed{\text{28.01 MPa}}$.

6.2.3 The total strain

The question likely refers to the total elongation (change in length) of the rod, as "total strain" for a composite bar is ambiguous.

$$\epsilon_1 = \frac{\sigma_{max}}{E} = \frac{112.04 \times 10^6 \text{ Pa}}{200 \times 10^9 \text{ Pa}} \approx 0.0005602$$ $$\Delta L_1 = \epsilon_1 \times L_1 = 0.0005602 \times 0.1 \text{ m} = 0.00005602 \text{ m}$$

$$\epsilon_2 = \frac{\sigma_{min}}{E} = \frac{28.01 \times 10^6 \text{ Pa}}{200 \times 10^9 \text{ Pa}} \approx 0.00014005$$ $$\Delta L_2 = \epsilon_2 \times L_2 = 0.00014005 \times 0.1 \text{ m} = 0.000014005 \text{ m}$$

$$\Delta L_{total} = \Delta L_1 + \Delta L_2$$ $$\Delta L_{total} = 0.00005602 \text{ m} + 0.000014005 \text{ m}$$ $$\Delta L_{total} = 0.000070025 \text{ m}$$

The total elongation (change in length) is $\boxed{\text{0.000070025 m}}$.

QUESTION 7: HEAT

7.1 Describe the isothermal process.

An isothermal process is a thermodynamic process in which the temperature of the system remains constant throughout. Any heat transfer into or out of the system occurs slowly enough to maintain thermal equilibrium.

7.2 A right rectangular prism of cast iron has dimensions of $55 \times 45 \times 35 \text{ mm}$ at $25^\circ \text{C}$. The prism is heated to $88^\circ \text{C}$. The coefficient of linear expansion of cast iron is $9 \times 10^{-6} /^\circ \text{C}$.

Given:

• Initial dimensions: $L_0 = 55 \text{ mm}$, $W_0 = 45 \text{ mm}$, $H_0 = 35 \text{ mm}$
• Initial temperature $T_1 = 25^\circ \text{C}$
• Final temperature $T_2 = 88^\circ \text{C}$
• Coefficient of linear expansion $\alpha = 9 \times 10^{-6} /^\circ \text{C}$

$$V_0 = L_0 \times W_0 \times H_0$$ $$V_0 = 55 \text{ mm} \times 45 \text{ mm} \times 35 \text{ mm}$$ $$V_0 = 86625 \text{ mm}^3$$

$$\gamma = 3\alpha = 3 \times (9 \times 10^{-6} /^\circ \text{C}) = 27 \times 10^{-6} /^\circ \text{C}$$

7.2.1 The increase in volume

$$\Delta V = V_0 \gamma \Delta T$$ $$\Delta V = (86625 \text{ mm}^3)(27 \times 10^{-6} /^\circ \text{C})(63^\circ \text{C})$$ $$\Delta V = 86625 \times 0.000027 \times 63 \text{ mm}^3$$ $$\Delta V \approx 147.34 \text{ mm}^3$$

The increase in volume is $\boxed{\text{147.34 mm}^3}$.

7.2.2 The volume at $88^\circ \text{C}$

$$V_f = V_0 + \Delta V$$ $$V_f = 86625 \text{ mm}^3 + 147.34 \text{ mm}^3$$ $$V_f = 86772.34 \text{ mm}^3$$

The volume at $88^\circ \text{C}$ is $\boxed{\text{86772.34 mm}^3}$.