The magnitude of the force in member DE is: O 13.3 kN O 10.0 kN 20.0 KN O Zero O 15.0 kN DE O None of the above
The magnitude of the force in member DE is: O 13.3 kN O 10.0 kN 20.0 KN O Zero O 15.0 kN DE O None of the above
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:### Problem Statement
The magnitude of the force in member DE is:
- ○ 13.3 kN
- ○ 10.0 kN
- ○ 20.0 kN
- ○ Zero
- ○ 15.0 kN
- ○ None of the above

Transcribed Image Text:### Structural Analysis of a Truss
#### Description:
The image depicts a truss structure commonly used in engineering and architecture.
#### Components and Dimensions:
- **Points & Connections:**
- **A, B, C, D, E, F, G:**
- These points denote the joints in the truss where different members (bars) are connected.
- **Members:**
- The truss consists of multiple members forming triangular and rectangular elements.
- **Loads:**
- There are two downward forces, each with a magnitude of **12 kN**, acting at points G and F.
#### Dimensions:
- Horizontal distances between points are:
- From **A to G**: 3 meters
- From **G to F**: 3 meters
- From **F to E**: 3 meters
- The entire span from **A to E** is 9 meters (3 + 3 + 3 meters).
- Vertical distance:
- The height from point **D to E** is **4 meters**.
#### Setup:
- **Support at A:** Represented by a triangular base, indicating a pin support which allows rotation but not translation.
- **Support at E:** Represented by a roller support, which allows for horizontal movement but restricts vertical movement, indicating it is a hinge support.
#### Equilibrium Conditions:
For the truss to be in equilibrium:
- The sum of all vertical forces must be zero.
- The sum of all horizontal forces must be zero.
- The sum of moments about any point must be zero.
This image is fundamental for understanding how to analyze a truss structure by breaking it down into manageable sections and applying static equilibrium equations to determine the forces in each member.
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