A Pratt truss supports three forces at joints along the bottom edge, as shown in the figure below. The forces have magnitudes F₁ = 50 kN, F₂ = 150 kN, and F3 = 0 kN. The lengths for all horizontal members is s = 8 m and for all vertical members is h = 8 m. The truss is held in equilibrium by a pin at point A and a roller at point I. Use the method of joints to determine the force in members CD, CE, DE, DF, and EF and indicate whether they are in tension or compression. Note: Express tension forces as positive (+) and compression forces as negative (-). B A त PCD= 50 PCE= 0 PDE= S 0.000678 C ▼ F₁ S E ✓ F₂ S KN kN KN G F3 ?✔100% x 0% X 0% S h

Elements Of Electromagnetics
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## Understanding Truss Analysis Using the Method of Joints

### Problem Statement
In a Pratt truss, three forces act at specific joints along the bottom edge as illustrated in the figure. The magnitude of the forces is \( F_1 = 50 \, \text{kN} \), \( F_2 = 150 \, \text{kN} \), and \( F_3 = 0 \, \text{kN} \). Each horizontal member has a length \( s = 8 \, \text{m} \) and each vertical member is \( h = 8 \, \text{m} \). The truss achieves equilibrium with a pin at point \( A \) and a roller at point \( I \).

### Objective
The goal is to determine the force values for each member (CD, CE, DE, DF, and EF) using the method of joints. Additionally, specify if these members are in tension or compression.

### Method of Joints
1. Assume all unknown forces in the members to be tensile (pulling outward from the joint).
2. Apply equilibrium equations to each joint:
   - \(\Sigma F_x = 0\) (sum of horizontal forces)
   - \(\Sigma F_y = 0\) (sum of vertical forces)
3. Use these equations to solve for the unknown forces.
4. Positive forces denote tension, while negative forces denote compression.

### Note
- Express tension forces as positive (+) and compression forces as negative (-).

### Diagram Description
- The truss is formed by a sequence of triangles connected at joints labeled from A to J.
- All joints are aligned in a horizontal line, with diagonal connections as illustrated.
- The truss diagram gives a clear visualization of how forces impact each member within the structure.

### Calculated Values
- \( P_{CD} = 50 \, \text{kN} \) (100% correct)
- \( P_{CE} = 0 \, \text{kN} \) (0% correct)
- \( P_{DE} = 0.000678 \, \text{kN} \) (0% correct)

These values represent theoretical calculations and their correctness within the specific analysis problem.
Transcribed Image Text:## Understanding Truss Analysis Using the Method of Joints ### Problem Statement In a Pratt truss, three forces act at specific joints along the bottom edge as illustrated in the figure. The magnitude of the forces is \( F_1 = 50 \, \text{kN} \), \( F_2 = 150 \, \text{kN} \), and \( F_3 = 0 \, \text{kN} \). Each horizontal member has a length \( s = 8 \, \text{m} \) and each vertical member is \( h = 8 \, \text{m} \). The truss achieves equilibrium with a pin at point \( A \) and a roller at point \( I \). ### Objective The goal is to determine the force values for each member (CD, CE, DE, DF, and EF) using the method of joints. Additionally, specify if these members are in tension or compression. ### Method of Joints 1. Assume all unknown forces in the members to be tensile (pulling outward from the joint). 2. Apply equilibrium equations to each joint: - \(\Sigma F_x = 0\) (sum of horizontal forces) - \(\Sigma F_y = 0\) (sum of vertical forces) 3. Use these equations to solve for the unknown forces. 4. Positive forces denote tension, while negative forces denote compression. ### Note - Express tension forces as positive (+) and compression forces as negative (-). ### Diagram Description - The truss is formed by a sequence of triangles connected at joints labeled from A to J. - All joints are aligned in a horizontal line, with diagonal connections as illustrated. - The truss diagram gives a clear visualization of how forces impact each member within the structure. ### Calculated Values - \( P_{CD} = 50 \, \text{kN} \) (100% correct) - \( P_{CE} = 0 \, \text{kN} \) (0% correct) - \( P_{DE} = 0.000678 \, \text{kN} \) (0% correct) These values represent theoretical calculations and their correctness within the specific analysis problem.
The image displays a table with force values in kilonewtons (kN) for various elements, identified by subscripts. The table includes the following entries:

- \( P_{CD} = 50 \, \text{kN} \) (Marked with a checkmark and "100%")
- \( P_{CE} = 0 \, \text{kN} \) (Marked with an "x" and "0%")
- \( P_{DE} = 0.000678 \, \text{kN} \) (Marked with an "x" and "0%")
- \( P_{DF} = 0 \, \text{kN} \) (Marked with an "x" and "0%")
- \( P_{EF} = 0 \, \text{kN} \) (Marked with a checkmark and "100%") 

Each entry has a status indicator, showing if the value is deemed correct (checkmark) or incorrect (x), accompanied by a percentage. There are no graphs or diagrams associated with the table.
Transcribed Image Text:The image displays a table with force values in kilonewtons (kN) for various elements, identified by subscripts. The table includes the following entries: - \( P_{CD} = 50 \, \text{kN} \) (Marked with a checkmark and "100%") - \( P_{CE} = 0 \, \text{kN} \) (Marked with an "x" and "0%") - \( P_{DE} = 0.000678 \, \text{kN} \) (Marked with an "x" and "0%") - \( P_{DF} = 0 \, \text{kN} \) (Marked with an "x" and "0%") - \( P_{EF} = 0 \, \text{kN} \) (Marked with a checkmark and "100%") Each entry has a status indicator, showing if the value is deemed correct (checkmark) or incorrect (x), accompanied by a percentage. There are no graphs or diagrams associated with the table.
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