FIGURE 7.1 below shows a beam subjected to three point loads - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2023 - Paper 1

Using the reactions and the principles of statics to calculate the bending moments:

1. At point A:

   $M_{A} = (R_{L} \times 1.5) - (4 \text{ N} \times 1.5) = 6.29 \times 1.5 - 6 = 9.44 \text{ Nm}$

2. At point B:

   $M_{B} = (R_{L} \times 3.5) - (4 \text{ N} \times 3.5) - (5 \text{ N} \times 1.5) = 6.29(3.5) - 14 - 7.5 = 14.02 \text{ Nm}$

3. At point C:

   $M_{C} = (R_{L} \times 5.5) - (4 \text{ N} \times 5) - (5 \text{ N} \times 3.5) = 6.29(5.5) - 20 - 17.5 = 8.60 \text{ Nm}$

To draw the bending moment diagram, plot the calculated moments at points A, B, and C along a horizontal axis:

1. At A, plot a moment of +9.44 Nm.
2. At B, plot a moment of +14.02 Nm.
3. At C, mark the moment of +8.60 Nm.

Connect these points to form the bending moment diagram, reflecting the appropriate scale.

The area of the bar can be calculated using the formula for stress:

$\sigma = \frac{F}{A} \implies A = \frac{F}{\sigma}$

Substituting the values:

$A = \frac{65 \times 10^3}{5 \times 10^6} = 0.013 \text{ m}^2 = 13 \times 10^{-3} \text{ m}^2$.

Using the area calculated:

$A = \frac{\pi D^2}{4} \implies D = \sqrt{\frac{4A}{\pi}}$

Substitute for A:

$D = \sqrt{\frac{4 \times (13 \times 10^{-3})}{\pi}} = 0.1287 \text{ m} = 128.7 \text{ mm}$.

Strain (( \epsilon )) can be calculated by the relation:

$\epsilon = \frac{\sigma}{E}$

Substituting given values:

$\epsilon = \frac{5 \times 10^6}{75 \times 10^9} = 6.67 \times 10^{-5}$.

To find the change in length (( \Delta L )):

$\Delta L = \epsilon \times L_0$

Where ( L_0 ) is the original length (250 mm):

$\Delta L = 6.67 \times 10^{-5} \times 250 = 0.016675 ext{ m} = 0.02 ext{ m}.$