7.1 Four pulling forces of 100 N, 200 N, 300 N and 185 N are acting from the same acting point, as shown in FIGURE 7.1 - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2016 - Paper 1

First, use the relationship between stress, force, and area: $\sigma = \frac{F}{A}$ Where:

• $\sigma = 20 \text{ MPa} = 20 \times 10^6 \text{ Pa}$
• $F = 40 \text{ kN} = 40 \times 10^3 \text{ N}$

We can rearrange for area: $A = \frac{F}{\sigma} = \frac{40 \times 10^3}{20 \times 10^6} = 2 \times 10^{-3} \text{ m}^2$

Using the area of a circle formula: $A = \frac{\pi D^2}{4}$ Rearranging for diameter: $D = 2 \sqrt{\frac{A}{\pi}} = 2 \sqrt{\frac{2 \times 10^{-3}}{\pi}} = 0.05046 \text{ m} \approx 50.46 \text{ mm}$

Strain ($\epsilon$) is calculated using: $\epsilon = \frac{\sigma}{E}$ Where:

• $\sigma = 20 \times 10^6 \text{ Pa}$
• $E = 90 \text{ GPa} = 90 \times 10^9 \text{ Pa}$

Substituting values gives: $\epsilon = \frac{20 \times 10^6}{90 \times 10^9} = 0.22 \times 10^{-3}$

The change in length ($\Delta l$) can be calculated using: $\Delta l = \epsilon \cdot l_0$ Where:

• $\epsilon = 0.22 \times 10^{-3}$
• $l_0 = 0.3 \text{ m}$

Then: $\Delta l = (0.22 \times 10^{-3}) \cdot 0.3 = 0.07 \times 10^{-3} ext{ m} = 0.07 ext{ mm}$

To calculate the reactions at supports A and B, we can start by applying the moment equilibrium equations.

1. Calculate the moments about point B: $\Sigma M_B = 0 = 550 \times 2 + 300 \times 6 - R_A \times 8$ Which rearranges to find $R_A$.

2. Then calculate the upward forces and set them equal to the downward forces: $R_A + R_B = 550 + 300$

3. Solve these equations simultaneously to find the values of $R_A$ and $R_B$.