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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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**Problem 5: Beam Analysis** **Objective:** For the beam loaded as shown, supported by a pin at A and the cord BC: 1. Draw the Free Body Diagram for the beam. 2. Determine the support reactions and tension in the cord necessary for equilibrium. Assume \( F = 50 \, \text{kN} \). Neglect the thickness of the beam. **Diagram Description:** The diagram shows a beam supported at point A and a cord at BC. Forces acting on the beam include a vertical force labeled 26 kN at point F, inclined forces, and the reactions at point A, represented as \( A_x \) and \( A_y \). **Calculations:** - **Horizontal Equilibrium (\( \Sigma F_x = 0 \))**: - \( A_x = \frac{26 \times 5}{13} = 10 \, \text{kN} \) - **Vertical Equilibrium (\( \Sigma F_y = 0 \))**: - \( A_y - 26 \times \frac{12}{13} - F_B \times \frac{3}{5} = 0 \) - \( A_y = 26 \times \frac{12}{13} + 50 \times \frac{3}{5} = 45 \, \text{kN} \, \uparrow \) - **Reaction at B (\( B_y \))**: - \( B_y = -50 \times \frac{3}{5} = -30 \, \text{kN} \) - Corrected to \( B_y = 30 \, \text{kN} \, \downarrow \) - **Horizontal Component at B (\( B_x \))**: - \( B_x = 50 \times \frac{4}{5} = 40 \, \text{kN} \) **Final Reactions:** - \( A_x = 10 \, \text{kN} \, \rightarrow \) - \( A_y = 45 \, \text{kN} \, \uparrow \) - \( B_x = 40 \, \text{kN} \, \rightarrow \) - \( B_y = 30 \, \text{kN} \, \downarrow \) **Note:** A moment equilibrium check is hinted at concluding with
Transcribed Image Text:**Problem 5: Beam Analysis** **Objective:** For the beam loaded as shown, supported by a pin at A and the cord BC: 1. Draw the Free Body Diagram for the beam. 2. Determine the support reactions and tension in the cord necessary for equilibrium. Assume \( F = 50 \, \text{kN} \). Neglect the thickness of the beam. **Diagram Description:** The diagram shows a beam supported at point A and a cord at BC. Forces acting on the beam include a vertical force labeled 26 kN at point F, inclined forces, and the reactions at point A, represented as \( A_x \) and \( A_y \). **Calculations:** - **Horizontal Equilibrium (\( \Sigma F_x = 0 \))**: - \( A_x = \frac{26 \times 5}{13} = 10 \, \text{kN} \) - **Vertical Equilibrium (\( \Sigma F_y = 0 \))**: - \( A_y - 26 \times \frac{12}{13} - F_B \times \frac{3}{5} = 0 \) - \( A_y = 26 \times \frac{12}{13} + 50 \times \frac{3}{5} = 45 \, \text{kN} \, \uparrow \) - **Reaction at B (\( B_y \))**: - \( B_y = -50 \times \frac{3}{5} = -30 \, \text{kN} \) - Corrected to \( B_y = 30 \, \text{kN} \, \downarrow \) - **Horizontal Component at B (\( B_x \))**: - \( B_x = 50 \times \frac{4}{5} = 40 \, \text{kN} \) **Final Reactions:** - \( A_x = 10 \, \text{kN} \, \rightarrow \) - \( A_y = 45 \, \text{kN} \, \uparrow \) - \( B_x = 40 \, \text{kN} \, \rightarrow \) - \( B_y = 30 \, \text{kN} \, \downarrow \) **Note:** A moment equilibrium check is hinted at concluding with
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