Calculate the following: 6.1.1 The stress, 6.1.2 The final length of the bar. 6.2.1 The maximum tensile stress, 6.2.2 The minimum stress in the rod, 6.2.3 The total strain.

$$A = w \times t$$ $$A = 0.022 \text{ m} \times 0.012 \text{ m} = 0.000264 \text{ m}^2$$

$$\sigma = \frac{F}{A}$$ $$\sigma = \frac{25 \times 10^3 \text{ N}}{0.000264 \text{ m}^2}$$ $$\sigma \approx 94696969.7 \text{ Pa} \approx 94.70 \text{ MPa}$$

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

6.1.2 The final length of the bar

$$E = \frac{\sigma}{\epsilon} \implies \epsilon = \frac{\sigma}{E}$$ $$\epsilon = \frac{94.69697 \times 10^6 \text{ Pa}}{195 \times 10^9 \text{ Pa}}$$ $$\epsilon \approx 0.0004856$$

$$\epsilon = \frac{\Delta L}{L_0} \implies \Delta L = \epsilon L_0$$ $$\Delta L = 0.0004856 \times 1.8 \text{ m}$$ $$\Delta L \approx 0.000874 \text{ m}$$

$$L_f = L_0 + \Delta L$$ $$L_f = 1.8 \text{ m} + 0.000874 \text{ m}$$ $$L_f = 1.800874 \text{ m}$$

The final length of the bar is $\boxed{\text{1.800874 m}}$.

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

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 $L_2 = 100 \text{ mm} = 0.1 \text{ m}$

6.2.1 The maximum tensile stress

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

$$\sigma_{max} = \frac{F}{A_1}$$ $$\sigma_{max} = \frac{55 \times 10^3 \text{ N}}{0.00049087 \text{ m}^2}$$ $$\sigma_{max} \approx 112040000 \text{ Pa} \approx 112.04 \text{ MPa}$$

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

6.2.2 The minimum stress in the rod

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

$$\sigma_{min} = \frac{F}{A_2}$$ $$\sigma_{min} = \frac{55 \times 10^3 \text{ N}}{0.0019635 \text{ m}^2}$$ $$\sigma_{min} \approx 28011713 \text{ Pa} \approx 28.01 \text{ MPa}$$

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

6.2.3 The total strain

The question asks for "total strain". For a composite rod, this typically refers to the total elongation divided by the total original length.

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

$$\epsilon_2 = \frac{\sigma_{min}}{E} = \frac{28.011713 \times 10^6 \text{ Pa}}{200 \times 10^9 \text{ Pa}} \approx 0.00014006$$ $$\Delta L_2 = \epsilon_2 L_2 = 0.00014006 \times 0.1 \text{ m} \approx 0.000014006 \text{ m}$$

$$\Delta L_{total} = \Delta L_1 + \Delta L_2$$ $$\Delta L_{total} = 0.00005602 \text{ m} + 0.000014006 \text{ m} = 0.000070026 \text{ m}$$

$$\text{Total Strain} = \frac{\Delta L_{total}}{L_{total}}$$ $$\text{Total Strain} = \frac{0.000070026 \text{ m}}{0.2 \text{ m}} \approx 0.00035013$$

The total strain is $\boxed{\text{0.00035013}}$.