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

To find the diameter of the bar, we use the formula for stress: $\sigma = \frac{F}{A}$ where:

• $F = 40 \times 10^3 \, N$
• $\sigma = 20 \times 10^6 \, Pa$

First, we calculate the area: $A = \frac{F}{\sigma} = \frac{40 \times 10^3}{20 \times 10^6} = 2 \times 10^{-3} \, m^2$

The diameter $D$ can be derived from the area: $A = \frac{\pi D^2}{4}$

Rearranging to find $D$ gives: $D = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4(2 \times 10^{-3})}{\pi}} = 0.05046 \, m = 50.46 \, mm$

Strain ($\epsilon$) is calculated as: $\epsilon = \frac{\sigma}{E}$ where:

• $\sigma = 20 \times 10^6 \, Pa$
• $E = 90 \times 10^9 \, Pa$

Substituting in 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 with: $\Delta L = \epsilon L_0$ where:

• $L_0 = 0.3 \, m$
• From part 7.2.2, we found $\epsilon = 0.22 \times 10^{-3}$.

This results in: $\Delta L = (0.22 \times 10^{-3}) \cdot 0.3 = 0.07 \times 10^{-3} \, m = 0.07 \, mm$

Using equilibrium conditions, we consider moments about point B: $\Sigma RHM = \Sigma LHM$

• The equation can be set up as: $(A \cdot 4) + (300 \cdot 6) = (550 \cdot 2) + (80 \cdot 3)$

Solving this gives:

• Left Hand Moments (LHM): $A + 600 = 3300 + 2400 \Rightarrow 8A = 637.5 , N$$

Using similar equilibrium considerations: $\Sigma LHM = \Sigma RHM$ Set up: $(B \cdot 8) = (550 \cdot 2) + (300 \cdot 10)$

• Rearranging gives: $B = 1012.15 \, N$

Thus, we can conclude:

• Reactions at supports A and B are $637.5 \, N$ and $1012.15 \, N$, respectively.