7.1 Four forces of are acting on the same point, as shown in FIGURE 7.1 below - NSC Mechanical Technology Fitting and Machining - Question 7 - 2018 - Paper 1

To calculate the stress in the bar, use the formula:

$\text{Stress} = \frac{\text{Force}}{\text{Area}}$

1. Calculate the Area (A) of the cross-section:

   • Outer diameter = 62 mm, Inner diameter = 50 mm: $A = \frac{\pi}{4} \times (D^2 - d^2) = \frac{\pi}{4} \times (62^2 - 50^2) \approx 1.06 \times 10^{-2} \text{ m}^2$

2. Calculate the Stress: $\text{Stress} = \frac{80 \times 10^3 \, \text{N}}{1.06 \times 10^{-2} \, \text{m}^2} \approx 75.47 \times 10^{6} \, \text{Pa} = 75.47 \, \text{MPa}$

Strain is defined as:

$\text{Strain} = \frac{\Delta L}{L_0}$

1. Calculate the Change in Length (( \Delta L )) using: $\epsilon = \frac{\text{Stress}}{E}$ where ( E ) is Young's modulus: $\epsilon = \frac{75.47 \times 10^{6}}{90 \times 10^{9}} \approx 0.84 \times 10^{-6}$

2. Substituting into the Strain Formula Maximo:

   • Original length ( L_0 = 100 , ext{mm} ): $\Delta L = \epsilon \times L_0 = 0.84 \times 10^{-6} \times 100 \, ext{mm} = 0.084 \, ext{mm}$

To find the reactions at supports A and B, we will take moments about one support and then use equilibrium conditions:

1. Calculate Moments about B: $\Sigma M_B = 0$ $\frac{(550 \times 8)}{2} + (640 \times 4) + (300 \times 0) - (R_A \times 8) = 0$

   • This gives: $R_A = \frac{(550 \times 4) + (640 \times 2)}{8} \approx 435 \, \text{N}$

2. Calculate Moments about A: $\Sigma M_A = 0$ $\frac{(300 \times 16) + (640 \times 12) + (550 \times 8)}{16} - R_B = 0$ Thus, $R_B = \frac{(550 \times 8) + (640 \times 4) + (300 \times 0)}{16} \approx 1055 \, \text{N}$