FIGURE 7.1 below shows a beam supported by two vertical supports, RL and RR - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2024 - Paper 1

To find the reactions at the supports, we take moments about each support:

1. For RL (taking moments about RR):

   $$ext{RL} imes 10 = (25 ext{ N} imes 8) + (30 imes 3.5) + (15 ext{ N} imes 6.5)$$

   Solving gives:

   $$ext{RL} = 36.5 ext{ N}$$

2. For RR (taking moments about RL):

   $$ext{RR} imes 10 = (15 ext{ N} imes 2) + (30 imes 3) + (25 ext{ N} imes 4)$$

   Resulting in:

   $$ext{RR} = 33.5 ext{ N}$$

Thus, the reactions are RL = 36.5 N and RR = 33.5 N.

To determine the shear forces at points A, B, and C:

1. At A:

   $$S_F = 36.5 ext{ N} ext{ (upward)}$$

2. At B:

   $$S_F = 36.5 ext{ N} - 15 ext{ N} - 30 ext{ N} = -8.5 ext{ N} ext{ (downward)}$$

3. At C:

   $$S_F = 36.5 ext{ N} - 15 ext{ N} - 30 ext{ N} - 25 ext{ N} = -33.5 ext{ N} ext{ (downward)}$$

Thus, the shear forces are:

• At A: 36.5 N (upward)
• At B: -8.5 N (downward)
• At C: -33.5 N (downward).

To draw the shear force diagram:

1. Use a scale where 1 N = 1 mm.
2. Start at point A with a value of 36.5 mm (up).
3. At point B, drop down to -8.5 mm (indicating a downward shear force).
4. Finally, at point C, go down to -33.5 mm.

Ensure to mark the points A, B, and C clearly and use horizontal lines between points for constant shear forces.