This image represents a beam subjected to multiple forces and supported by a hinge at point B and a roller at point D. Here's a breakdown of the forces acting on the beam:

Given Data:

• Uniformly distributed load (UDL) at the left end: 400 N/m over 2 m.
• Uniformly distributed load (UDL) on the right section: 200 N/m over 4 m.
• Concentrated load at point C: $P_C = 400 \text{ N}$.
• Weight of the beam: 300 N/m (assumed to be uniformly distributed over the entire length of the beam).

Beam Span:

• $AB = 3 \text{ m}$
• $BD = 4 \text{ m}$
• $DE = 2 \text{ m}$
• Total length of beam $AE = 9 \text{ m}$.

Objective:

We are tasked with finding the reactions at support points B and D. The beam is statically determinate, so we can use equilibrium equations (force and moment balance) to solve this problem.

Step-by-step Plan:

1. Calculate equivalent point loads for the distributed loads.
2. Sum of forces in the vertical direction for equilibrium.
3. Sum of moments about point B (or D) to solve for unknown reactions.

Let me go ahead and compute the reactions based on these principles.The reactions at the supports are as follows:

• Reaction at point D (roller support): $R_D \approx 3191.67 \, \text{N}$
• Reaction at point B (hinge support): $R_B \approx 1908.33 \, \text{N}$

These values were obtained by using equilibrium equations for moments and forces acting on the beam.

Would you like further clarification on any part of this process or details on how these reactions were derived?

Here are five related questions to consider:

1. How does the placement of loads affect the reactions at the supports?
2. How would the reactions change if the uniformly distributed loads were different?
3. What is the significance of the beam being statically determinate?
4. How can shear force and bending moment diagrams be drawn for this beam?
5. What effect does the beam's own weight have on the reactions?

Tip: When calculating reactions, summing moments around a support helps eliminate one of the unknown forces, simplifying the problem.