3. Reactions on support E: Ex, Ey and Me for a 4. Reactions in cable to F: • Equilibrium Equations: 1. Fx = 0: there are FE → = lb; → support; → forces in x direction, hence Ex = lb;

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
icon
Related questions
Question
# Educational Website Example

## Problem Description

The rig shown consists of a 1200-lb horizontal member ABC and a vertical member DBE welded together at B. The rig is being used to raise a 3600-lb crate at a distance \( x = 11.7 \) ft from the vertical member DBE. If the tension in the cable is 3.4 kips, determine the reaction at E, assuming that the cable is anchored at F as shown in the figure.

### Diagram Explanation

- **Components**:
  - A horizontal member labeled as ABC.
  - A vertical member labeled as DBE.
  - The two members are connected at point B.
  - A cable is attached to point D on the rig and anchored at point F.

- **Dimensions**:
  - The horizontal length from A to C is 17.5 ft.
  - The length from D to B is 3.75 ft.
  - The length from B to F is 5 ft.
  - The crate is positioned 6.5 ft (x = 11.7 ft mentioned but related to crate position).

- **Weights and Forces**:
  - The weight of the crate is 3600 lb, depicted hanging from point C.
  - The weight of the member ABC is 1200 lb, acting at the midpoint.
  - The tension in the cable is 3.4 kips (3400 lbs).

## Solution Steps

1. **Free Body Diagram (FBD)**:
   
   Create an FBD for the system by yourself. Take the frame as your FBD and detach it from supports.

2. **Support Reactions**:
   
   - Determine the number of supports and calculate the total forces and reactions on frame ABCDE.

3. **Identify Each Force**:
   
   - Weight of the crate (\( W_{\text{crate}} \)): \( \_\_\_\_ \) lb, direction: \( \_\_\_\_ \).
   - Self-weight of the horizontal member ABC (\( W_{\text{ABC}} \)): \( \_\_\_\_ \) lb, direction: \( \_\_\_\_ \).

4. **Reactions at Support E**:
   
   - \( E_x, E_y \) and moment \( M_E \) about support.

5. **Reactions in Cable to F**:
   
   - Compute the tension resulting from
Transcribed Image Text:# Educational Website Example ## Problem Description The rig shown consists of a 1200-lb horizontal member ABC and a vertical member DBE welded together at B. The rig is being used to raise a 3600-lb crate at a distance \( x = 11.7 \) ft from the vertical member DBE. If the tension in the cable is 3.4 kips, determine the reaction at E, assuming that the cable is anchored at F as shown in the figure. ### Diagram Explanation - **Components**: - A horizontal member labeled as ABC. - A vertical member labeled as DBE. - The two members are connected at point B. - A cable is attached to point D on the rig and anchored at point F. - **Dimensions**: - The horizontal length from A to C is 17.5 ft. - The length from D to B is 3.75 ft. - The length from B to F is 5 ft. - The crate is positioned 6.5 ft (x = 11.7 ft mentioned but related to crate position). - **Weights and Forces**: - The weight of the crate is 3600 lb, depicted hanging from point C. - The weight of the member ABC is 1200 lb, acting at the midpoint. - The tension in the cable is 3.4 kips (3400 lbs). ## Solution Steps 1. **Free Body Diagram (FBD)**: Create an FBD for the system by yourself. Take the frame as your FBD and detach it from supports. 2. **Support Reactions**: - Determine the number of supports and calculate the total forces and reactions on frame ABCDE. 3. **Identify Each Force**: - Weight of the crate (\( W_{\text{crate}} \)): \( \_\_\_\_ \) lb, direction: \( \_\_\_\_ \). - Self-weight of the horizontal member ABC (\( W_{\text{ABC}} \)): \( \_\_\_\_ \) lb, direction: \( \_\_\_\_ \). 4. **Reactions at Support E**: - \( E_x, E_y \) and moment \( M_E \) about support. 5. **Reactions in Cable to F**: - Compute the tension resulting from
**Text Analysis for Educational Purposes**

**Title:** Equilibrium Conditions in Structural Analysis

**Reactions and Equilibrium Equations:**

1. **Reactions on Support E:** 
   - \(E_x\), \(E_y\), and \(M_E\) for a \(\square\) support.

2. **Reactions in Cable to F:** 
   - \(\square =\) \(\square\) lb.

3. **Equilibrium Equations:**

   1. \(\sum F_x = 0\): 
      - There are \(\square\) forces in the x-direction, hence \(E_x =\) \(\square\) lb.

   2. \(\sum F_y = 0\): 
      - There are \(\square\) forces in the y-direction, hence \(E_y =\) \(\square\) lb.

   3. \(\sum M_E = 0\): 
      - \(\square \cdot W_{\text{crate}} + \square \cdot W_{ABC} + \square \cdot E_x + \square \cdot E_y + \square \cdot M_E + \square \cdot T = 0\), hence \(M_E =\) \(\square\) kips-ft.

**Explanation of Symbols:**

- \(E_x\), \(E_y\): Reaction forces at support E in the x and y directions, respectively.
- \(M_E\): Moment at support E.
- \(\sum F_x = 0\): Sum of forces in the x-direction equals zero (horizontal equilibrium).
- \(\sum F_y = 0\): Sum of forces in the y-direction equals zero (vertical equilibrium).
- \(\sum M_E = 0\): Sum of moments about point or axis E equals zero (moment equilibrium).

This content would be pertinent to students or professionals studying civil or structural engineering, focusing on the analysis of forces and moments in static systems.
Transcribed Image Text:**Text Analysis for Educational Purposes** **Title:** Equilibrium Conditions in Structural Analysis **Reactions and Equilibrium Equations:** 1. **Reactions on Support E:** - \(E_x\), \(E_y\), and \(M_E\) for a \(\square\) support. 2. **Reactions in Cable to F:** - \(\square =\) \(\square\) lb. 3. **Equilibrium Equations:** 1. \(\sum F_x = 0\): - There are \(\square\) forces in the x-direction, hence \(E_x =\) \(\square\) lb. 2. \(\sum F_y = 0\): - There are \(\square\) forces in the y-direction, hence \(E_y =\) \(\square\) lb. 3. \(\sum M_E = 0\): - \(\square \cdot W_{\text{crate}} + \square \cdot W_{ABC} + \square \cdot E_x + \square \cdot E_y + \square \cdot M_E + \square \cdot T = 0\), hence \(M_E =\) \(\square\) kips-ft. **Explanation of Symbols:** - \(E_x\), \(E_y\): Reaction forces at support E in the x and y directions, respectively. - \(M_E\): Moment at support E. - \(\sum F_x = 0\): Sum of forces in the x-direction equals zero (horizontal equilibrium). - \(\sum F_y = 0\): Sum of forces in the y-direction equals zero (vertical equilibrium). - \(\sum M_E = 0\): Sum of moments about point or axis E equals zero (moment equilibrium). This content would be pertinent to students or professionals studying civil or structural engineering, focusing on the analysis of forces and moments in static systems.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

Blurred answer
Knowledge Booster
Laws of equilibrium
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Structural Analysis
Structural Analysis
Civil Engineering
ISBN:
9781337630931
Author:
KASSIMALI, Aslam.
Publisher:
Cengage,
Structural Analysis (10th Edition)
Structural Analysis (10th Edition)
Civil Engineering
ISBN:
9780134610672
Author:
Russell C. Hibbeler
Publisher:
PEARSON
Principles of Foundation Engineering (MindTap Cou…
Principles of Foundation Engineering (MindTap Cou…
Civil Engineering
ISBN:
9781337705028
Author:
Braja M. Das, Nagaratnam Sivakugan
Publisher:
Cengage Learning
Fundamentals of Structural Analysis
Fundamentals of Structural Analysis
Civil Engineering
ISBN:
9780073398006
Author:
Kenneth M. Leet Emeritus, Chia-Ming Uang, Joel Lanning
Publisher:
McGraw-Hill Education
Sustainable Energy
Sustainable Energy
Civil Engineering
ISBN:
9781337551663
Author:
DUNLAP, Richard A.
Publisher:
Cengage,
Traffic and Highway Engineering
Traffic and Highway Engineering
Civil Engineering
ISBN:
9781305156241
Author:
Garber, Nicholas J.
Publisher:
Cengage Learning