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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;

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
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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# 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.
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