The beam is an overhanging beam supported at B (Pin) and D (Roller). We need to determine the bending moment at point C ($M_C$).

Given Data:

• Load at A ($P_A$): $20 \text{ kN}$ (downward)
• Load at C ($P_C$): $40 \text{ kN}$ (downward)
• Distances: $L_{AB} = 2.5 \text{ m}$, $L_{BC} = 3 \text{ m}$, $L_{CD} = 2 \text{ m}$

Method: Sectioning

The bending moment at point C can be found by summing the moments of all forces acting on either the left segment (A to C) or the right segment (C to D). We choose the right segment (CD) as it involves fewer forces, requiring only the calculation of the reaction at D ($R_D$).

Step 1: Calculate Support Reactions ($R_B$ and $R_D$)

We apply the static equilibrium equation by taking the moment about support B ($\sum M_B = 0$, clockwise positive):

$$ (20 \text{ kN} \times 2.5 \text{ m}) - (40 \text{ kN} \times 3 \text{ m}) + (R_D \times 5 \text{ m}) = 0 $$

$$ 50 \text{ kN}\cdot\text{m} - 120 \text{ kN}\cdot\text{m} + 5 R_D = 0 $$

$$ 5 R_D = 70 \text{ kN}\cdot\text{m} $$

$$ R_D = \frac{70}{5} = 14 \text{ kN} $$

($R_D$ acts upwards)

Step 2: Calculate Bending Moment at C ($M_C$)

We analyze the segment to the right of C. The only force is the upward reaction $R_D$ located $2 \text{ m}$ away.

We use the convention where moments causing compression on the top fiber (sagging) are positive. $R_D$ causes positive moment at C.

$$ M_C = R_D \times L_{CD} $$

$$ M_C = 14 \text{ kN} \times 2 \text{ m} $$

$$ M_C = 28 \text{ kN}\cdot\text{m} $$

The value of the bending moment at point C is 28 kN·m.