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W, complete following: a. Determine the number of degrees of indeterminaey b. Solve for all reactions using reaction B (a roller) as a redundant e. Draw the shear and moment diagrams ww 4 kN/m 2 m

W, complete following: a. Determine the number of degrees of indeterminaey b. Solve for all reactions using reaction B (a roller) as a redundant e. Draw the shear and moment diagrams ww 4 kN/m 2 m

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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### Structural Analysis of a Beam #### Notations and Assumptions - **M<sub>A</sub>**: Moment at support A - **R<sub>B</sub>**: Reaction at support B - **w**: Uniformly distributed load (UDL) of 4 kN/m - Span sections: AB = 3m, BC = 2m #### (a) Degree of Indeterminacy - Total reactions, R = 2 - Number of equilibrium equations, e = 3 - Degree of indeterminacy = R - e = 3 - 2 = 1 #### (b) Analysis Taking R<sub>B</sub> as Redundant Force 1. **Removing the Redundant Force:** - Consider the beam without R<sub>B</sub>. - Calculation for deflection due to UDL at B: - \(\Delta_{D1} = \frac{wL^4}{8EI} = \frac{4 \times 8^4}{8EI} = \frac{2048}{EI}\) 2. **Removing the External Load:** - Consider the beam without the external load. - Calculation for deflection due to R<sub>B</sub>: - \(\Delta_{B2} = \frac{R_B L^3}{3EI} + \frac{R_B \times 8^3}{3EI} = \frac{R_B \times 8^3}{3EI}\) 3. **Deflection at B is Zero:** - Set the total deflection to zero: \(\Delta_{B1} + \Delta_{B2} = 0\) - Solving: \(\frac{2048}{EI} + \frac{R_B \times 8^3}{3EI} = 0\) - Find \(R_B\): \(R_B = 12 \text{ kN}\) 4. **Reactions:** - \(R_A + R_B = \text{Total Load}\) - \(R_A = 40 - 12 = 28 \text{ kN}\) #### Shear Force and Bending Moment Analysis 1. **Shear Force Diagram:** - VA = R_A = 28 kN - V_AB =
Transcribed Image Text:### Structural Analysis of a Beam #### Notations and Assumptions - **M<sub>A</sub>**: Moment at support A - **R<sub>B</sub>**: Reaction at support B - **w**: Uniformly distributed load (UDL) of 4 kN/m - Span sections: AB = 3m, BC = 2m #### (a) Degree of Indeterminacy - Total reactions, R = 2 - Number of equilibrium equations, e = 3 - Degree of indeterminacy = R - e = 3 - 2 = 1 #### (b) Analysis Taking R<sub>B</sub> as Redundant Force 1. **Removing the Redundant Force:** - Consider the beam without R<sub>B</sub>. - Calculation for deflection due to UDL at B: - \(\Delta_{D1} = \frac{wL^4}{8EI} = \frac{4 \times 8^4}{8EI} = \frac{2048}{EI}\) 2. **Removing the External Load:** - Consider the beam without the external load. - Calculation for deflection due to R<sub>B</sub>: - \(\Delta_{B2} = \frac{R_B L^3}{3EI} + \frac{R_B \times 8^3}{3EI} = \frac{R_B \times 8^3}{3EI}\) 3. **Deflection at B is Zero:** - Set the total deflection to zero: \(\Delta_{B1} + \Delta_{B2} = 0\) - Solving: \(\frac{2048}{EI} + \frac{R_B \times 8^3}{3EI} = 0\) - Find \(R_B\): \(R_B = 12 \text{ kN}\) 4. **Reactions:** - \(R_A + R_B = \text{Total Load}\) - \(R_A = 40 - 12 = 28 \text{ kN}\) #### Shear Force and Bending Moment Analysis 1. **Shear Force Diagram:** - VA = R_A = 28 kN - V_AB =
# Beam Analysis and Indeterminacy ### Problem Statement For the beam shown below, complete the following: 1. Determine the number of degrees of indeterminacy. 2. Solve for all reactions using reaction B (a roller) as a redundant. 3. Draw the shear and moment diagrams. ![Beam Diagram](beam-diagram.jpg) - **Uniformly Distributed Load (UDL)**: \( w = 4 \, \text{kN/m} \) - **Beam Span:** \( 8 \, \text{m} \) to point A, \( 2 \, \text{m} \) to point C ### Analytical Steps #### Step 2: Analysis with Redundant Reaction **Part (b)** - *Taking Reaction B as redundant:* - **UDL Contribution:** \[ (\delta B)_{\text{udl}} = \frac{wx^2}{24EI}(6L^2 - 4Lx + x^2) \] - \( x = 8 \, \text{m} \) - \( L = 10 \, \text{m} \) \[ (\delta B)_{\text{udl}} = \frac{468 \times 8^2}{24 \times 10^4} \left( (6 \times 10^2) - 4 \times (10) \times (8) + 8^2 \right) \] \[ (\delta B)_{\text{udl}} = \frac{468 \times 64}{240} \times \left( 3600 - 320 + 64 \right) \] - **Point Load Contribution:** \[ (\delta B)_R = \frac{-Ka^2}{6EI}(3a - x) \] - \( a = x = 8 \, \text{m} \) \[ (\delta B)_R = \frac{-R \times 64}{6 \times 10} \times (3 \times 8 - 8) \] \[ (\delta B)_R = \frac{-R\times 64}{60} \times 16 \] - **Balance Equations:** \[ (\delta B)_R
Transcribed Image Text:# Beam Analysis and Indeterminacy ### Problem Statement For the beam shown below, complete the following: 1. Determine the number of degrees of indeterminacy. 2. Solve for all reactions using reaction B (a roller) as a redundant. 3. Draw the shear and moment diagrams. ![Beam Diagram](beam-diagram.jpg) - **Uniformly Distributed Load (UDL)**: \( w = 4 \, \text{kN/m} \) - **Beam Span:** \( 8 \, \text{m} \) to point A, \( 2 \, \text{m} \) to point C ### Analytical Steps #### Step 2: Analysis with Redundant Reaction **Part (b)** - *Taking Reaction B as redundant:* - **UDL Contribution:** \[ (\delta B)_{\text{udl}} = \frac{wx^2}{24EI}(6L^2 - 4Lx + x^2) \] - \( x = 8 \, \text{m} \) - \( L = 10 \, \text{m} \) \[ (\delta B)_{\text{udl}} = \frac{468 \times 8^2}{24 \times 10^4} \left( (6 \times 10^2) - 4 \times (10) \times (8) + 8^2 \right) \] \[ (\delta B)_{\text{udl}} = \frac{468 \times 64}{240} \times \left( 3600 - 320 + 64 \right) \] - **Point Load Contribution:** \[ (\delta B)_R = \frac{-Ka^2}{6EI}(3a - x) \] - \( a = x = 8 \, \text{m} \) \[ (\delta B)_R = \frac{-R \times 64}{6 \times 10} \times (3 \times 8 - 8) \] \[ (\delta B)_R = \frac{-R\times 64}{60} \times 16 \] - **Balance Equations:** \[ (\delta B)_R
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