Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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Make correction to this exercice and explain in sentence or two why I made the mistake.
![**Transcription and Explanation for Educational Website**
**Text and Calculations:**
1. **Problem Description:**
- For the beam shown, in the space provided:
1. [3] Draw the free body diagram. (FR = 50.4 KN/m)
2. [2] Resolve the distributed load into 2 point loads and indicate their location as a distance from point O.
2. **Equations and Calculations:**
- \( FR = \sum F = FR_1 + FR_2 + FR_3 \)
- \( FR_A = 6 \text{ kN/m} \)
- \( FR_B = 15 \text{ kN} \)
- \( (+) M_{Ro} = (+) FR_A (5 \text{ m}) - FR_2 (7.5 \text{ m}) \)
- \( M_{Ro} = 6 \text{ kN/m} (5 \text{ m}) - 15 \text{ kN} (7.5 \text{ m}) = 500 \text{ KN/m} \)
- \( HR = 142.5 \text{ KN} \)
- Distance \( d = \frac{HR}{FR} = \frac{142.5 \text{ KN}}{500 \text{ KN/m}} = 0.27 \text{ m} \)
**Diagrams Explanation:**
1. **Beam Diagram:**
- The beam is divided into segments with point loads indicated.
- Multiple forces are acting: a point load of 15 kN at one end and a uniformly distributed load of 6 kN/m over a length of 7.5 m.
- Dimensions along the beam are marked: 7.5 m extends from the left support to the right end, with the distributed load acting evenly across this length.
2. **Free Body Diagram for Bar:**
- A lever arm with a 250 N force pressing downward at a 30-degree angle from vertical.
- The arm is 0.4 m long, and there is a horizontal offset of 0.15 m between two points, denoted as A and B.
**Notes:**
- The calculations show resolution of forces and moments for static equilibrium.
- The aim is to simplify the loading conditions on the beam for structural analysis purposes.](https://content.bartleby.com/qna-images/question/6e7ba281-dd87-4688-be97-9abf26130e05/bab69a6c-6f4a-4a7d-b0a6-456ba04b6231/xvd7uda_processed.jpeg)
Transcribed Image Text:**Transcription and Explanation for Educational Website**
**Text and Calculations:**
1. **Problem Description:**
- For the beam shown, in the space provided:
1. [3] Draw the free body diagram. (FR = 50.4 KN/m)
2. [2] Resolve the distributed load into 2 point loads and indicate their location as a distance from point O.
2. **Equations and Calculations:**
- \( FR = \sum F = FR_1 + FR_2 + FR_3 \)
- \( FR_A = 6 \text{ kN/m} \)
- \( FR_B = 15 \text{ kN} \)
- \( (+) M_{Ro} = (+) FR_A (5 \text{ m}) - FR_2 (7.5 \text{ m}) \)
- \( M_{Ro} = 6 \text{ kN/m} (5 \text{ m}) - 15 \text{ kN} (7.5 \text{ m}) = 500 \text{ KN/m} \)
- \( HR = 142.5 \text{ KN} \)
- Distance \( d = \frac{HR}{FR} = \frac{142.5 \text{ KN}}{500 \text{ KN/m}} = 0.27 \text{ m} \)
**Diagrams Explanation:**
1. **Beam Diagram:**
- The beam is divided into segments with point loads indicated.
- Multiple forces are acting: a point load of 15 kN at one end and a uniformly distributed load of 6 kN/m over a length of 7.5 m.
- Dimensions along the beam are marked: 7.5 m extends from the left support to the right end, with the distributed load acting evenly across this length.
2. **Free Body Diagram for Bar:**
- A lever arm with a 250 N force pressing downward at a 30-degree angle from vertical.
- The arm is 0.4 m long, and there is a horizontal offset of 0.15 m between two points, denoted as A and B.
**Notes:**
- The calculations show resolution of forces and moments for static equilibrium.
- The aim is to simplify the loading conditions on the beam for structural analysis purposes.
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