Find out the reaction forces at fixed-end A, hinge B and rocker C
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Find out the reaction forces at fixed-end A, hinge B and rocker C
![### Beam Analysis Diagram
This diagram illustrates a beam subjected to different forces and supports, providing an example for studying the principles of statics and beam reactions.
#### Components:
1. **Beam AB**: The beam extends horizontally from point A to point C with a total length of 6 meters.
- **Length Segments**:
- \( \text{Distance from A to B} = 4 \text{ meters (2m + 2m)} \)
- \( \text{Distance from B to C} = 2 \text{ meters} \)
- **Supports**:
- Point A appears to be fixed to a wall or a rigid support.
- Point B is supported by a roller or pin support.
2. **Forces Acting on the Beam**:
- **Point Load**: A point load of 10 kN is applied at 2 meters to the right of point A. The force vector is inclined with a slope ratio of 4:3 (vertical to horizontal).
- **Distributed Load**: From point B to point C, a uniformly distributed load (UDL) of 4 kN/m is acting over a length of 2 meters.
#### Detailed Breakdown of Forces:
- **10 kN Point Load**:
- Located 2 meters from A.
- The force direction is indicated with slope components of 4 (vertical) and 3 (horizontal). This indicates that the load vector can be broken into two components using trigonometry:
- Vertical Component \( = 10 \times \frac{4}{5} = 8 \text{kN} \)
- Horizontal Component \( = 10 \times \frac{3}{5} = 6 \text{kN} \)
- **4 kN/m Distributed Load**:
- Spanning from B (4 meters from A) to C (6 meters from A).
- Total Load = Load Intensity x Length = \( 4 kN/m \times 2 m = 8 \text{kN} \)
- This load acts downwards and its equivalent point load acts at the center of this distribution, i.e., 1 meter from B, or 5 meters from A.
Understanding these forces and their application points is crucial for determining the reactions at the supports (A and B), as well as analyzing bending moments, shear forces, and other](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2de2484e-8c30-4a24-aa43-7ad85b110ab5%2F12a6acd9-e828-45f9-bd13-c19c5020d209%2Fxthuhqd_processed.png&w=3840&q=75)
Transcribed Image Text:### Beam Analysis Diagram
This diagram illustrates a beam subjected to different forces and supports, providing an example for studying the principles of statics and beam reactions.
#### Components:
1. **Beam AB**: The beam extends horizontally from point A to point C with a total length of 6 meters.
- **Length Segments**:
- \( \text{Distance from A to B} = 4 \text{ meters (2m + 2m)} \)
- \( \text{Distance from B to C} = 2 \text{ meters} \)
- **Supports**:
- Point A appears to be fixed to a wall or a rigid support.
- Point B is supported by a roller or pin support.
2. **Forces Acting on the Beam**:
- **Point Load**: A point load of 10 kN is applied at 2 meters to the right of point A. The force vector is inclined with a slope ratio of 4:3 (vertical to horizontal).
- **Distributed Load**: From point B to point C, a uniformly distributed load (UDL) of 4 kN/m is acting over a length of 2 meters.
#### Detailed Breakdown of Forces:
- **10 kN Point Load**:
- Located 2 meters from A.
- The force direction is indicated with slope components of 4 (vertical) and 3 (horizontal). This indicates that the load vector can be broken into two components using trigonometry:
- Vertical Component \( = 10 \times \frac{4}{5} = 8 \text{kN} \)
- Horizontal Component \( = 10 \times \frac{3}{5} = 6 \text{kN} \)
- **4 kN/m Distributed Load**:
- Spanning from B (4 meters from A) to C (6 meters from A).
- Total Load = Load Intensity x Length = \( 4 kN/m \times 2 m = 8 \text{kN} \)
- This load acts downwards and its equivalent point load acts at the center of this distribution, i.e., 1 meter from B, or 5 meters from A.
Understanding these forces and their application points is crucial for determining the reactions at the supports (A and B), as well as analyzing bending moments, shear forces, and other
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