A floor truss is loaded as shown. Determine the force in members FI, HI, and HJ. 2 kN 4 KN 4 kN 3 kN 2 KN 2 KN 1 KN 0.8 m 0.8 m 0.8 m 0.8 m 0.8 m 0.8 m Во A АО D E G H J L ↓ 0.4 m K↑
A floor truss is loaded as shown. Determine the force in members FI, HI, and HJ. 2 kN 4 KN 4 kN 3 kN 2 KN 2 KN 1 KN 0.8 m 0.8 m 0.8 m 0.8 m 0.8 m 0.8 m Во A АО D E G H J L ↓ 0.4 m K↑
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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Transcribed Image Text:The diagram illustrates a floor truss loaded with various forces at specific intervals. The problem statement is:
"A floor truss is loaded as shown. Determine the force in members FI, HI, and HJ."
**Details of the Diagram:**
- **Structure:** The truss is supported at points A and K.
- **Load Distribution:**
- 2 kN is applied at point D.
- 4 kN at points E and J.
- 3 kN at point G.
- 2 kN at point H.
- 1 kN at point L.
- **Distance Between Load Points:**
- The horizontal distance between each force application point (D, E, G, H, J, L) is 0.8 meters.
- **Vertical Distance:**
- The vertical height at point L to the base K is 0.4 meters.
The objective is to calculate the forces in the truss members FI, HI, and HJ using methods such as static equilibrium equations, considering the given load distribution and geometry.
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