The 10-ft long beam shown below is subjected to a uniformly-distributed load of w = 5 kip/ft. W ……….…..…... B a) What is the total magnitude of the force applied to the beam? b) If the distributed load were replaced with a single point force, where on the beam would it need to act in order for the beam supports to experience the same effect?

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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### Image Transcription for Educational Use

#### Description:
The image illustrates a 10-foot long beam subjected to a uniformly-distributed load. The specified load is \( w = 5 \, \text{kip/ft} \).

#### Diagram:
- The beam is horizontally positioned and supported at two ends, labeled as points \( A \) and \( B \).
- Arrows pointing downward across the beam represent the uniformly-distributed load, denoted by \( w \).
- The length of the beam is marked as \( L = 10 \, \text{ft} \).

#### Questions:
a) What is the total magnitude of the force applied to the beam?

b) If the distributed load were replaced with a single point force, where on the beam would it need to act in order for the beam supports to experience the same effect?

---

### Explanation:
- **Graph/Diagram Details:** The diagram visually explains the setup of the beam with supporting symbols at the ends. The distributed load is represented by evenly spaced arrows, indicating the uniform application of force across the beam's length.
- **Key Concepts:**
  - **Uniformly-Distributed Load:** A type of load spread out over a length rather than applied at a single point.
  - **Calculation of Force:** For part (a), the total force from the distributed load can be calculated by multiplying the load per unit length by the total length of the beam (\( F = w \times L \)).
  - **Single Point Force:** For part (b), the equivalent point force would act at the centroid of the load distribution, often at the midpoint of the beam for uniform loads. 

These principles are foundational in structural analysis and mechanics.
Transcribed Image Text:### Image Transcription for Educational Use #### Description: The image illustrates a 10-foot long beam subjected to a uniformly-distributed load. The specified load is \( w = 5 \, \text{kip/ft} \). #### Diagram: - The beam is horizontally positioned and supported at two ends, labeled as points \( A \) and \( B \). - Arrows pointing downward across the beam represent the uniformly-distributed load, denoted by \( w \). - The length of the beam is marked as \( L = 10 \, \text{ft} \). #### Questions: a) What is the total magnitude of the force applied to the beam? b) If the distributed load were replaced with a single point force, where on the beam would it need to act in order for the beam supports to experience the same effect? --- ### Explanation: - **Graph/Diagram Details:** The diagram visually explains the setup of the beam with supporting symbols at the ends. The distributed load is represented by evenly spaced arrows, indicating the uniform application of force across the beam's length. - **Key Concepts:** - **Uniformly-Distributed Load:** A type of load spread out over a length rather than applied at a single point. - **Calculation of Force:** For part (a), the total force from the distributed load can be calculated by multiplying the load per unit length by the total length of the beam (\( F = w \times L \)). - **Single Point Force:** For part (b), the equivalent point force would act at the centroid of the load distribution, often at the midpoint of the beam for uniform loads. These principles are foundational in structural analysis and mechanics.
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