A room of institutional building having clear size 7m×3.2m has to be covered with RCC simply supported siab resting on two beams having 230 mm width and 7 m length. Consider live load-4kN/m² and floor finish=1.25kN/m². Use 12 mm diameter bars as main steel, 8 mm diameter as distribution steel. Write the answer in the form of schedule. Do not draw RCC details.

Here is the schedule format for the RCC slab design:

**1. Room Dimensions:**
   - Length (along beam span): 7.0 m
   - Width (slab span): 3.2 m

**2. Slab Details:**
   - Slab Type: Simply Supported One-Way Slab
   - Slab Thickness: To be calculated based on span-to-depth ratio and loading.

**3. Loading Details:**
   - Live Load: 4 kN/m²
   - Floor Finish Load: 1.25 kN/m²
   - Self-Weight of Slab (assuming slab thickness `h`): \( 25 \, \text{kN/m}^3 \times h \, \text{(in meters)} \)

**4. Design Parameters:**
   - Clear Span: 3.2 m (shorter dimension since it's a one-way slab)
   - Effective Depth (d): Slab thickness minus effective cover (typically 20-25 mm)
   - Concrete Grade: Usually M20 (fc = 20 N/mm²)
   - Steel Grade: Fe 415
   - Effective Cover: 20 mm (typical for slab)

**5. Reinforcement Details:**
   - **Main Reinforcement (along the shorter span):**
     - Bar Diameter: 12 mm
     - Spacing: To be calculated based on moment of resistance and slab thickness.
   - **Distribution Reinforcement (along the longer span):**
     - Bar Diameter: 8 mm
     - Spacing: To be calculated based on distribution reinforcement requirements (minimum reinforcement percentage).

**6. Load Calculation:**
   - Total Load (Factored): \( 1.5 \times (\text{Live Load} + \text{Floor Finish Load} + \text{Self-Weight of Slab}) \)

**7. Moment Calculation:**
   - Maximum Bending Moment for One-Way Slab: \( M = \frac{wL^2}{8} \) (Where `w` is the uniformly distributed load, and `L` is the span)
   - Shear Force Calculation: At the supports, check against maximum permissible shear.

**8. Reinforcement Calculation:**
   - Determine required area of steel (Ast) using bending moment and design formulas: \( M_{u} = 0.87 f_y A_{st} d \left(1 - \frac{A_{st} f_y}{f_{ck} bd}\right) \)
   - Spacing of bars: Calculate based on required steel area using the formula \( A_{st} = \frac{\pi \phi^2}{4} \times \frac{1000}{\text{Spacing}} \).

**9. Reinforcement Distribution:**
   - **Main Bars:** 12 mm diameter, Spacing = To be determined.
   - **Distribution Bars:** 8 mm diameter, Spacing = To be determined.

**10. Deflection Check:**
   - Ensure that the slab thickness satisfies the deflection criteria based on the span-to-depth ratio.

**11. Additional Notes:**
   - Minimum Slab Thickness: Generally not less than 100 mm for RCC slab.
   - Provide nominal cover of 20 mm for main reinforcement.
   - Ensure adequate anchorage at supports.

Would you like more details on any specific part of this schedule?

### Follow-Up Questions:
1. How do we calculate the minimum slab thickness based on span-to-depth ratio?
2. What are the detailed steps for finding the spacing of the main steel bars?
3. How do we factor in the self-weight of the slab in the loading calculation?
4. How do we check the slab for deflection criteria?
5. What is the significance of using different diameters for main and distribution steel?

### Tip:
Always check the adequacy of reinforcement spacing to avoid excessive deflection and cracking in the slab.

A room of institutional building having clear size 7m×3.2m has to be covered with RCC simply supported slab resting on two beams having 230 mm width and 7 m length. Consider live load-4kN/m² and floor finish=1.25kN/m². Use 12 mm diameter bars as main steel, 8 mm diameter as distribution steel. Write the answer in the form of schedule. Do not draw RCC details.

Explore a detailed solution for designing an RCC one-way slab with dimensions 7m × 3.2m, considering a live load of 4kN/m² and floor finish of 1.25kN/m². Learn how to calculate reinforcement, bending moments, and deflection for a simply supported slab. Suitable for civil engineering students.

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