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

The bending moment at points A, B, and C can be calculated as:

1. Bending Moment at A (BM_A):

   $BM_A = RL * 1.5 - 4 N * 1.5 = 6.29 N * 1.5 - 4 N * 1.5 = 9.44 Nm$

2. Bending Moment at B (BM_B):

   $BM_B = RL * 3.5 - 4 N * 1.5 - 5 N * 2 = 6.29 N * 3.5 - 4 N * 1.5 - 5 N * 2 = 14.02 Nm$

3. Bending Moment at C (BM_C):

   $BM_C = RL * 5.5 - 4 N * 1.5 - 5 N * 3.5 - 3 N * (2 m) = 6.29 N * 5.5 - 4 N * 1.5 - 5 N * 3.5 - 3 N * 2 = 8.60 Nm$

To draw the bending-moment diagram, we plot the bending moments calculated at points A, B, and C on a vertical axis against their respective positions along the beam on a horizontal axis. The scale used for the diagram will be consistent, such as 1 mm = 2 Nm. The diagram will depict the values:

• BM_A = 9.44 Nm
• BM_B = 14.02 Nm
• BM_C = 8.60 Nm

The resulting graph should form a triangular shape, illustrating how the bending moments change along the length of the beam.

To calculate the area (A) of the round aluminium bar, use the formula:

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

Where:

• $F = 65 kN = 65 \times 10^3 N$
• $\sigma = 5 MPa = 5 \times 10^6 N/m^2$

Substituting in the values:

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

The diameter (D) of the bar can be calculated using the area:

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

Rearranging gives:

$D = \sqrt{\frac{4A}{\pi}}$

Substituting $A = 13 \times 10^{-3} m^2$:

$D = \sqrt{\frac{4 \times 13 \times 10^{-3}}{\pi}} = 0.1267 m \\ D \approx 128.66 \ mm$

Strain ($\epsilon$) is defined as:

$\epsilon = \frac{\Delta L}{L_0}$

Using Young's modulus to relate stress and strain:

$\sigma = E \cdot \epsilon$

Where:

• $E = 75 \times 10^9$ Pa

Thus:

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

The change in length ($\Delta L$) can be calculated using:

$\Delta L = \epsilon \cdot L_0$

Where:

• $L_0 = 250 mm = 0.25 m$

Thus: $\Delta L = 6.67 \times 10^{-5} \cdot 0.25 = 0.000016675 m = 0.02 mm$