The beam is supported by a pin at A and roller at B and supports a triangular distributed load and a point load. Determine the support reaction at B. Support your answer with an appropriate FBD and equilibrium equation(s). 10 kN

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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### Beam Support Reaction Calculation

#### Problem Statement:
The beam is supported by a pin at A and a roller at B, and it supports a triangular distributed load and a point load. Determine the support reaction at B. Support your answer with an appropriate Free Body Diagram (FBD) and equilibrium equation(s).

#### Diagram:
The beam is detailed as follows:
- A point load of 10 kN is applied at the center of the beam.
- A triangular distributed load with a peak value of 10 kN/m is applied across the length of the beam.
- The length of the beam is 6 meters, with the beam segmented into 3 meters on either side of the central node where the point load is applied.
- Support A is a pin support allowing rotational but not translational movement.
- Support B is a roller support allowing horizontal translational but not vertical movement.

#### Diagram Analysis:
The diagram shows the following:
- A skier-shaped distributed load is represented from point A to point B, peaking at the center.
- The uniformly distributed load is symmetrical about the center of the beam.
- The specific supports are clearly marked - A being a pin support and B being a roller support.

#### Choices:
- 35 kN
- 20 kN
- 15 kN
- 10 kN
- None of the above

Select the correct support reaction at B based on calculations.

#### Free Body Diagram (FBD) and Equilibrium Equations:

1. **Calculate the resultant of the triangular load:**
   - Resultant load (R) of the triangular distributed load = \(\frac{1}{2} \times \text{Base length of distribution} \times \text{Height of triangle} \Rightarrow \frac{1}{2} \times 6 \text{ m} \times 10 \text{ kN/m} = 30 \text{ kN}\)
   - Location of the resultant load from point A = \(\frac{\text{Length}}{3} = \frac{6}{3} = 2 \text{ m from A}\)

2. **Equilibrium Equations:**
   - Sum of vertical forces (\( \Sigma F_y = 0 \)):

     \[
     R_A + R_B - 10 - 30 = 0
     \]
     \[
     R_A + R_B = 40 \text{ k
Transcribed Image Text:### Beam Support Reaction Calculation #### Problem Statement: The beam is supported by a pin at A and a roller at B, and it supports a triangular distributed load and a point load. Determine the support reaction at B. Support your answer with an appropriate Free Body Diagram (FBD) and equilibrium equation(s). #### Diagram: The beam is detailed as follows: - A point load of 10 kN is applied at the center of the beam. - A triangular distributed load with a peak value of 10 kN/m is applied across the length of the beam. - The length of the beam is 6 meters, with the beam segmented into 3 meters on either side of the central node where the point load is applied. - Support A is a pin support allowing rotational but not translational movement. - Support B is a roller support allowing horizontal translational but not vertical movement. #### Diagram Analysis: The diagram shows the following: - A skier-shaped distributed load is represented from point A to point B, peaking at the center. - The uniformly distributed load is symmetrical about the center of the beam. - The specific supports are clearly marked - A being a pin support and B being a roller support. #### Choices: - 35 kN - 20 kN - 15 kN - 10 kN - None of the above Select the correct support reaction at B based on calculations. #### Free Body Diagram (FBD) and Equilibrium Equations: 1. **Calculate the resultant of the triangular load:** - Resultant load (R) of the triangular distributed load = \(\frac{1}{2} \times \text{Base length of distribution} \times \text{Height of triangle} \Rightarrow \frac{1}{2} \times 6 \text{ m} \times 10 \text{ kN/m} = 30 \text{ kN}\) - Location of the resultant load from point A = \(\frac{\text{Length}}{3} = \frac{6}{3} = 2 \text{ m from A}\) 2. **Equilibrium Equations:** - Sum of vertical forces (\( \Sigma F_y = 0 \)): \[ R_A + R_B - 10 - 30 = 0 \] \[ R_A + R_B = 40 \text{ k
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