Project 1

Step 1: Understanding the Geometry and Beam Layout

From the image and data:

• Floor dimensions:

  • East-West direction: 5 bays × 9 m = 45 m total

  • North-South direction: 3 bays × 9 m = 27 m total

• Beam of interest: East-West spanning beam (between two North-South girders)

• Beam spacing (North-South): From the drawing, we can count 4 bays, giving 3 spans of 9 m each

Step 2: Calculate Loading on Beam

1. Dead Load (excluding self-weight)

• Floor thickness = 120 mm = 0.12 m

• Unit weight of normal concrete ≈ 24 kN/m³

• Dead load from slab = $0.12 \times 24 = 2.88 \, \text{kPa}$

2. Superimposed Dead Load (finish + mechanical)

• Given = 3 kPa

3. Live Load

• Given = 5 kPa

4. Tributary Width

• Beams are equally spaced, each beam takes half the span on either side

• Spacing = 3 m (assuming evenly spaced in the 9 m direction divided into 3 bays)

• So tributary width = 3 m

5. Total Uniform Load on Beam (per m length):

• Dead Load:

  $$(2.88 + 3) \times 3 = 17.64 \, \text{kN/m}$$

• Live Load:

  $$5 \times 3 = 15 \, \text{kN/m}$$

• Total:

  • Dead: 17.64 kN/m

  • Live: 15 kN/m

  • Combined: 32.64 kN/m

Step 3: Beam Properties

• Cross section: 500 mm × 750 mm

• $I = \frac{1}{12} b h^3 = \frac{1}{12} \times 0.5 \times 0.75^3 = 0.0176 \, \text{m}^4 = 17.6 \times 10^6 \, \text{mm}^4$

• $E = 10 \, \text{GPa} = 10^4 \, \text{MPa}$

Step 4: Slope Deflection Method Setup

For a continuous beam with 3 spans (each 9 m) and uniform loading:

Let:

• $I$ be constant

• Nodes: A–B–C–D (3 spans)

• Moments: $M_{AB}, M_{BA}, M_{BC}, M_{CB}, M_{CD}, M_{DC}$

Slope Deflection Equation (for ends of each span):

$$M_{AB} = \frac{2EI}{L} (2\theta_A + \theta_B) + M_{F,AB}$$ $$M_{BA} = \frac{2EI}{L} (2\theta_B + \theta_A) + M_{F,BA}$$

Where $M_{F,AB}$ = fixed-end moment due to UDL on span AB:

$$M_{F,AB} = -wL^2 / 12, \quad M_{F,BA} = +wL^2 / 12$$

Since all spans and loads are equal, we can write similar equations for all spans and solve the system of equations for rotations ($\theta_A, \theta_B, \theta_C$).

Step 5: Fixed-End Moments (FEMs)

For UDL $w = 32.64 \, \text{kN/m}$, $L = 9 \, \text{m}$:

$$M_F = \frac{wL^2}{12} = \frac{32.64 \times 81}{12} = 220.32 \, \text{kNm}$$

So:

• $M_{AB} = -220.32$

• $M_{BA} = +220.32$

• Similarly for BC and CD

Now insert into slope-deflection equations and solve for moments using support conditions (θ at ends A and D = 0 since those are end supports).

Step 6: Draw SFD and BMD

After solving for all moments, use these to construct the shear force diagram and bending moment diagram across the 3 spans.

Step 7: Load Patterning (Live Load Case)

For maximum positive moment at midspan:

• Apply live load only to that span.

For maximum negative moment at supports:

• Apply live load on adjacent spans to induce hogging.

So:

• Run two cases:

  1. Live load on all spans → base case

  2. Live load on two adjacent spans for negative moment at support

  3. Live load only on one span for max positive midspan moment

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