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
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.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2f3435e6-2af6-4060-8093-d10f07d52056%2Fdfcc2a1b-37b9-4ff9-a6fc-a5ae4b74510f%2Ffw79n69_processed.jpeg&w=3840&q=75)
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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