Project 1
### Given Data
- Floor thickness: \( t = 120 \text{ mm} = 0.12 \text{ m} \)
- Density of concrete: \( \rho = 24 \text{ kN/m}^3 \)
- Floor finish and mechanical loads: \( 3 \text{ kPa} \)
- Live load: \( 5 \text{ kPa} \)
- Modulus of elasticity of concrete: \( E = 10 \text{ GPa} \)
- Beam dimensions: \( 500 \text{ mm} \times 750 \text{ mm} \)
- Beam span: \( 9 \text{ m} \)
### 1. Loading on the Beam
#### Floor Load Calculation:
The dead load due to the concrete floor slab is:
\[ q_d = \rho \times t = 24 \text{ kN/m}^3 \times 0.12 \text{ m} = 2.88 \text{ kN/m}^2 \]
Total load on the floor:
\[ q_{total} = q_d + q_{services} + q_{live} = 2.88 + 3 + 5 = 10.88 \text{ kN/m}^2 \]
#### Load on the Beam:
- The tributary width of the beam is the spacing between the beams. Given that beams are equally spaced, the tributary width is the distance between columns in the direction perpendicular to the beams.
If we assume the beams are spaced 9 meters apart (as indicated in the plan), then the load on the beam is:
\[ q_{beam} = q_{total} \times 9 = 10.88 \times 9 = 97.92 \text{ kN/m} \]
### 2. Analyze the Beams and Draw Bending Moment and Shear Force Diagrams
#### Moment of Inertia of the Beam:
\[ I = \frac{b \times h^3}{12} = \frac{0.5 \times 0.75^3}{12} = 0.017578125 \text{ m}^4 \]
### Slope Deflection Method
The slope-deflection method considers the moments at the ends of the members (beams) due to rotations and translations (slopes and deflections) at those ends. For a continuous beam with multiple spans, this method helps in determining the moments at the supports and mid-span.
#### Basic Equations:
For a beam with ends A and B:
\[ M_{AB} = \frac{EI}{L} \left( 2\theta_A + \theta_B - 3\delta \right) + M_f \]
\[ M_{BA} = \frac{EI}{L} \left( \theta_A + 2\theta_B - 3\delta \right) + M_f \]
where:
- \( M_{AB} \) and \( M_{BA} \) are the moments at ends A and B of the beam.
- \( \theta_A \) and \( \theta_B \) are the rotations at ends A and B.
- \( \delta \) is the relative translation (sway) of the ends.
- \( M_f \) is the fixed-end moment due to applied loads.
#### Fixed-End Moments for Uniformly Distributed Load:
For a uniformly distributed load \( w \) over the span \( L \):
\[ M_f = -\frac{wL^2}{12} \]
#### For the Given Beam:
The span length \( L = 9 \text{ m} \) and the distributed load \( w = 97.92 \text{ kN/m} \).
Fixed-end moments:
\[ M_f = -\frac{wL^2}{12} = -\frac{97.92 \times 9^2}{12} = -\frac{97.92 \times 81}{12} = -661.92 \text{ kN} \cdot \text{m} \]
Now, using the slope deflection equations, we solve for the unknowns. Given the complexity, we typically set up a system of linear equations based on the boundary conditions and continuity of the structure and solve for rotations and moments.
### 3. Consider Load Patterning
For maximum positive and negative moments in each span:
- **Positive moment:** Occurs when the span is loaded, and adjacent spans are unloaded.
- **Negative moment:** Occurs at the supports when adjacent spans are loaded, and the span itself is unloaded.
Let's set up the calculations to determine the moments using these considerations.
I'll start by solving for the moments and shears using the slope deflection method.
It seems like I can’t do more advanced data analysis right now. Please try again later.
However, I can still guide you through the process manually:
### Step-by-Step Calculation for Slope Deflection Method
1. **Fixed-End Moments (FEM) Calculation:**
For a uniformly distributed load \( w = 97.92 \text{ kN/m} \) over a span \( L = 9 \text{ m} \):
\[ M_f = -\frac{wL^2}{12} = -\frac{97.92 \times 9^2}{12} = -\frac{97.92 \times 81}{12} = -661.92 \text{ kN} \cdot \text{m} \]
2. **Slope-Deflection Equations:**
For each span, the moments at the ends \( A \) and \( B \) are:
\[ M_{AB} = \frac{EI}{L} (2\theta_A + \theta_B - 3\delta) + M_f \]
\[ M_{BA} = \frac{EI}{L} (\theta_A + 2\theta_B - 3\delta) + M_f \]
3. **Setting up the Equations:**
For a continuous beam with three spans, consider each span individually and set up the equilibrium equations. Typically, the beam is analyzed at each support and mid-span.
4. **Solving for Rotations and Moments:**
Use the boundary conditions (e.g., rotations are zero at fixed supports) and solve the simultaneous equations for rotations (\( \theta \)).
5. **Shear Forces:**
Once the moments are known, shear forces can be found using equilibrium equations:
\[ V = \frac{wL}{2} - \frac{M_{left}}{L} + \frac{M_{right}}{L} \]
### Load Patterning:
1. **Maximum Positive Moment:**
- Occurs when the span is loaded, and adjacent spans are unloaded.
- Recalculate fixed-end moments and solve for this condition.
2. **Maximum Negative Moment:**
- Occurs at the supports when adjacent spans are loaded, and the span itself is unloaded.
- Recalculate fixed-end moments and solve for this condition.