1) For the beam shown below, complete the following: a. Determine the number of degrees of indeterminacy b. Solve for all reactions using reaction B (a roller) as a redundant c. Draw the shear and moment diagrams ww 4 kN/m fftfffttf 8 m - 2 m
1) For the beam shown below, complete the following: a. Determine the number of degrees of indeterminacy b. Solve for all reactions using reaction B (a roller) as a redundant c. Draw the shear and moment diagrams ww 4 kN/m fftfffttf 8 m - 2 m
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
This is the 3rd time I am giving the same questions. Which method is corect ? Method 1 or 2 ?
Frist time I got answers from you as method 1.
The second time your answers were haisy. You gave answers by method 2 as Ra= 12 kn and Rb=28kn but posted SFD and BMD on the basis of method 1 ?
Please clarfy the answers.
- The beam is subjected to a uniformly distributed load (UDL) of \( w = 4 \, \text{kN/m} \).
- Lengths: \( AB = 8 \, \text{m} \), \( BC = 2 \, \text{m} \).
**Step 2:**
**(b)**
*Taking Reaction B as redundant:*
- **Due to UDL:**
- The deflection at B is given by:
\[
(\delta_B)_{\text{udl}} = \frac{wx^2}{24EI} \left( 6L^2 - 4Lx + x^2 \right)
\]
- \( x = 8 \, \text{m} \)
- \( L = 10 \, \text{m} \)
\[
(\delta_B)_{\text{udl}} = \frac{4(8)^2}{24EI} \left( 6(10)^2 - 4(10)(8) + 8^2 \right)
\]
\[
(\delta_B)_{\text{udl}} = \frac{4608}{EI}
\]
- **Due to Reaction Point Load:**
- The deflection at B is given by:
\[
(\delta_B)_R = \frac{R_B x^2}{6EI} (3a - x)
\]
- \( a = x = 8 \)
\[
(\delta_B)_R = \frac{R_B(8)^2}{6EI} (3 \times 8 - 8)
\]
\[
(\delta_B)_R = \frac{170.67R_B}{EI}
\]
- Equating:
\[
(\delta_B)_R = (\delta_B)_{\text{udl}}
\]
\[
\frac{170.67R_B}{EI}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa5c1a5d1-e9a7-475f-89ba-14ff4ec571ed%2Fc9b612d1-5c14-4b9b-baa6-da8b1336d6c6%2Fl5p1da7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
1) For the beam shown below, complete the following:
a. Determine the number of degrees of indeterminacy
b. Solve for all reactions using reaction B (a roller) as a redundant
c. Draw the shear and moment diagrams
- The beam is subjected to a uniformly distributed load (UDL) of \( w = 4 \, \text{kN/m} \).
- Lengths: \( AB = 8 \, \text{m} \), \( BC = 2 \, \text{m} \).
**Step 2:**
**(b)**
*Taking Reaction B as redundant:*
- **Due to UDL:**
- The deflection at B is given by:
\[
(\delta_B)_{\text{udl}} = \frac{wx^2}{24EI} \left( 6L^2 - 4Lx + x^2 \right)
\]
- \( x = 8 \, \text{m} \)
- \( L = 10 \, \text{m} \)
\[
(\delta_B)_{\text{udl}} = \frac{4(8)^2}{24EI} \left( 6(10)^2 - 4(10)(8) + 8^2 \right)
\]
\[
(\delta_B)_{\text{udl}} = \frac{4608}{EI}
\]
- **Due to Reaction Point Load:**
- The deflection at B is given by:
\[
(\delta_B)_R = \frac{R_B x^2}{6EI} (3a - x)
\]
- \( a = x = 8 \)
\[
(\delta_B)_R = \frac{R_B(8)^2}{6EI} (3 \times 8 - 8)
\]
\[
(\delta_B)_R = \frac{170.67R_B}{EI}
\]
- Equating:
\[
(\delta_B)_R = (\delta_B)_{\text{udl}}
\]
\[
\frac{170.67R_B}{EI}
![## Structural Analysis Example
### Problem Description
We have a beam subjected to a uniform distributed load (4 kN/m) with supports at points A, B, and C. The distances are as follows: AB = 3m and BC = 2m.
### Steps
#### a) Degree of Indeterminacy
\[ \text{Degree of Indeterminacy} = R - 2 = 3 - 2 = 1 \]
#### b) Taking \(R_B\) as Redundant Force
**Removing the Redundant Force:**
The redundant force \(R_B\) is temporarily removed to calculate the effect of the load:
- \(\Delta_{B1} = \frac{w \cdot l^4}{8EI} = \frac{4 \cdot 8^3}{8EI} = \frac{2048}{EI}\)
**Removing the External Load:**
Now, consider the beam without the external load:
- \(\Delta_{B2} = \frac{R_B \cdot l^3}{3EI} = \frac{R_B \cdot 8^3}{3EI}\)
Set the deflection at B to zero:
\[
\Delta_{B1} + \Delta_{B2} = 0 \Rightarrow \frac{2048}{EI} + \frac{R_B \cdot 8^3}{3EI} = 0
\]
Solving for \(R_B\):
\[
R_B = 12 \, \text{kN}
\]
Using equilibrium equations:
\[
R_A = 4 \times 8 - R_B = 32 - 12 = 20 \, \text{kN}
\]
### Shear Force and Bending Moment Analysis
**Shear Force Diagram:**
The shear force at various points:
- \(V_A = R_A = 20 \, \text{kN}\)
- \(V_B = V_A - w \cdot x = 20 - 4 \cdot 8 = -12 \, \text{kN}\)
- At \(x = 8\), \(V_C = V_B + R_B = -12 + 12 = 0 \)
Diagram:
A triangle slope from \(0\) at A to \(-12\) kN at B.
**Bending Moment Diagram:**
Calculation](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa5c1a5d1-e9a7-475f-89ba-14ff4ec571ed%2Fc9b612d1-5c14-4b9b-baa6-da8b1336d6c6%2Fcssqepq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Structural Analysis Example
### Problem Description
We have a beam subjected to a uniform distributed load (4 kN/m) with supports at points A, B, and C. The distances are as follows: AB = 3m and BC = 2m.
### Steps
#### a) Degree of Indeterminacy
\[ \text{Degree of Indeterminacy} = R - 2 = 3 - 2 = 1 \]
#### b) Taking \(R_B\) as Redundant Force
**Removing the Redundant Force:**
The redundant force \(R_B\) is temporarily removed to calculate the effect of the load:
- \(\Delta_{B1} = \frac{w \cdot l^4}{8EI} = \frac{4 \cdot 8^3}{8EI} = \frac{2048}{EI}\)
**Removing the External Load:**
Now, consider the beam without the external load:
- \(\Delta_{B2} = \frac{R_B \cdot l^3}{3EI} = \frac{R_B \cdot 8^3}{3EI}\)
Set the deflection at B to zero:
\[
\Delta_{B1} + \Delta_{B2} = 0 \Rightarrow \frac{2048}{EI} + \frac{R_B \cdot 8^3}{3EI} = 0
\]
Solving for \(R_B\):
\[
R_B = 12 \, \text{kN}
\]
Using equilibrium equations:
\[
R_A = 4 \times 8 - R_B = 32 - 12 = 20 \, \text{kN}
\]
### Shear Force and Bending Moment Analysis
**Shear Force Diagram:**
The shear force at various points:
- \(V_A = R_A = 20 \, \text{kN}\)
- \(V_B = V_A - w \cdot x = 20 - 4 \cdot 8 = -12 \, \text{kN}\)
- At \(x = 8\), \(V_C = V_B + R_B = -12 + 12 = 0 \)
Diagram:
A triangle slope from \(0\) at A to \(-12\) kN at B.
**Bending Moment Diagram:**
Calculation
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