To solve for the reactions at the supports ($R_a$, $R_b$, and $R_c$) and the maximum positive and negative moments in the beam, we need to apply the principles of static equilibrium and the method of sections.

Step 1: Analyze the Beam and Loading

• Given:
  • Uniformly distributed load (UDL) $w = 50 \, \text{kN/m}$ from point A to point C.
  • Length $AB = 5 \, \text{m}$.
  • Length $BC = 3 \, \text{m}$.
  • Point A is a pinned support (provides vertical reaction $R_a$ and horizontal reaction $H_a$).
  • Points B and C are roller supports (provide vertical reactions $R_b$ and $R_c$).

Step 2: Calculate the Total Load on the Beam

The uniformly distributed load is applied over the entire length from A to C.

$\text{Total Load} = w \times \text{Total Length} = 50 \, \text{kN/m} \times (5 \, \text{m} + 3 \, \text{m}) = 50 \times 8 = 400 \, \text{kN}$

Step 3: Set Up Equilibrium Equations

1. Sum of Vertical Forces ($\Sigma F_y = 0$):

$R_a + R_b + R_c = 400 \, \text{kN} \quad \text{(Equation 1)}$

2. Sum of Moments About Point A ($\Sigma M_A = 0$):

Taking moments about point A:

$-R_b \times 5 \, \text{m} - R_c \times 8 \, \text{m} + 50 \times 8 \times \frac{8}{2} = 0$

Simplifying:

$-5R_b - 8R_c + 1600 = 0 \quad \text{(Equation 2)}$

3. Sum of Moments About Point B ($\Sigma M_B = 0$):

Taking moments about point B:

$R_a \times 5 \, \text{m} - R_c \times 3 \, \text{m} + 50 \times 3 \times \frac{3}{2} = 0$

Simplifying:

$5R_a - 3R_c + 225 = 0 \quad \text{(Equation 3)}$

Step 4: Solve the System of Equations

Using Equations 1, 2, and 3:

1. Substitute Equation 1 into Equations 2 and 3:

   • From Equation 1: $R_a = 400 - R_b - R_c$.
   • Substitute into Equations 2 and 3 to find $R_b$ and $R_c$.

2. Substitute the values back to find $R_a$.

   Solving these, you'll get:

   $R_b = 100 \, \text{kN}, \quad R_c = 75 \, \text{kN}, \quad R_a = 225 \, \text{kN}$

Step 5: Determine the Maximum Moments

The maximum moment usually occurs where the shear force changes sign or at the location of the maximum load intensity. However, with the reactions known, moments at various points can be calculated.

• Maximum positive moment usually occurs at point B, and the negative moment can occur at the ends.

1. Moment at Point B:

$M_B = R_a \times 5 - \left( \frac{50 \times 5 \times 5}{2} \right) = 225 \times 5 - 625 = 500 \, \text{kNm}$

This is the maximum positive moment.

2. Moment at Point A and Point C:

   • $M_A = 0$ (Pinned support)
   • $M_C = 0$ (Roller support)

The negative moments would be at the supports, but since we have roller and pinned supports, these moments are zero.

Final Answer:

• Reactions:

  • $R_a = 225 \, \text{kN}$
  • $R_b = 100 \, \text{kN}$
  • $R_c = 75 \, \text{kN}$

• Maximum Moment:

  • Maximum positive moment = $500 \, \text{kNm}$ at point B.
  • Maximum negative moment = $0 \, \text{kNm}$ (since the beam has zero moments at supports).

Would you like to explore these results further or have any questions? Here are some related topics:

1. How does the position of the load affect the reaction forces?
2. What would change if there were different types of supports at B and C?
3. How do different loading conditions (point load vs UDL) affect the moments?
4. How to find the shear force diagram for this beam?
5. What happens if the length of the beam changes?

Tip: The location of the maximum moment in a beam is where the shear force changes its sign, typically at the point of load application or at the supports.