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.

Here are the solutions to Question 6:

QUESTION 6: STRESS, STRAIN AND YOUNG'S MODULUS

6.1 A tensile force of $25 \text{ kN}$ is applied to a rectangular bar $12 \text{ mm}$ thick and $22 \text{ mm}$ wide. The original length is $1.8 \text{ m}$ and Young's modulus for the steel is $195 \text{ GPa}$.

Given:

• Tensile force, $F = 25 kN = 25 \times 10^3 \text{ N}$
• Thickness, $t = 12 mm = 0.012 \text{ m}$
• Width, $w = 22 mm = 0.022 \text{ m}$
• Original length, $L_0 = 1.8 \text{ m}$
• Young's modulus, $E = 195 GPa = 195 \times 10^9 \text{ Pa}$

6.1.1 The stress

Step 1: Calculate the cross-sectional area ($A$). $A = w \times t$ $A = 0.022 m \times 0.012 m = 0.000264 m^2$

Step 2: Calculate the stress ($\sigma$). $\sigma = \frac{F}{A}$ $\sigma = \frac{25 \times 10^3 N}{0.000264 m^2}$ $\sigma \approx 94696969.7 Pa \approx 94.70 MPa$

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

6.1.2 The final length of the bar

Step 1: Calculate the strain ($\epsilon$). $E = \frac{\sigma}{\epsilon} \implies \epsilon = \frac{\sigma}{E}$ $\epsilon = \frac{94.69697 \times 10^6 Pa}{195 \times 10^9 Pa}$ $\epsilon \approx 0.0004856$

Step 2: Calculate the change in length ($\Delta L$). $\epsilon = \frac{\Delta L}{L_0} \implies \Delta L = \epsilon L_0$ $\Delta L = 0.0004856 \times 1.8 m$ $\Delta L \approx 0.000874 m$

Step 3: Calculate the final length ($L_f$). $L_f = L_0 + \Delta L$ $L_f = 1.8 m + 0.000874 m$ $L_f = 1.800874 m$

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

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

6.2.1 The maximum tensile stress

Step 1: Identify the section with the smallest area (Section 1) and calculate its area ($A_1$). $A_1 = \frac{\pi D_1^2}{4} = \frac{\pi (0.025 m)^2}{4}$ $A_1 = \frac{\pi \times 0.000625 m^2}{4} \approx 0.00049087 m^2$

Step 2: Calculate the maximum tensile stress ($\sigma_{max}$). $\sigma_{max} = \frac{F}{A_1}$ $\sigma_{max} = \frac{55 \times 10^3 N}{0.00049087 m^2}$ $\sigma_{max} \approx 112040000 Pa \approx 112.04 MPa$

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

6.2.2 The minimum stress in the rod

Step 1: Identify the section with the largest area (Section 2) and calculate its area ($A_2$). $A_2 = \frac{\pi D_2^2}{4} = \frac{\pi (0.050 m)^2}{4}$ $A_2 = \frac{\pi \times 0.0025 m^2}{4} \approx 0.0019635 m^2$

Step 2: Calculate the minimum stress ($\sigma_{min}$). $\sigma_{min} = \frac{F}{A_2}$ $\sigma_{min} = \frac{55 \times 10^3 N}{0.0019635 m^2}$ $\sigma_{min} \approx 28011713 Pa \approx 28.01 MPa$

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

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.

Step 1: Calculate the strain and elongation for Section 1. $\epsilon_1 = \frac{\sigma_{max}}{E} = \frac{112.04 \times 10^6 Pa}{200 \times 10^9 Pa} \approx 0.0005602$ $\Delta L_1 = \epsilon_1 L_1 = 0.0005602 \times 0.1 m \approx 0.00005602 m$

Step 2: Calculate the strain and elongation for Section 2. $\epsilon_2 = \frac{\sigma_{min}}{E} = \frac{28.011713 \times 10^6 Pa}{200 \times 10^9 Pa} \approx 0.00014006$ $\Delta L_2 = \epsilon_2 L_2 = 0.00014006 \times 0.1 m \approx 0.000014006 m$

Step 3: Calculate the total elongation ($\Delta L_{total}$). $\Delta L_{total} = \Delta L_1 + \Delta L_2$ $\Delta L_{total} = 0.00005602 m + 0.000014006 m = 0.000070026 m$

Step 4: Calculate the total original length ($L_{total}$). $L_{total} = L_1 + L_2 = 0.1 m + 0.1 m = 0.2 m$

Step 5: Calculate the overall "total strain". $Total Strain = \frac{\Delta L_{total}}{L_{total}}$ $Total Strain = \frac{0.000070026 m}{0.2 m} \approx 0.00035013$

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