Designing a one way continuous slab is a crucial aspect of structural engineering, particularly for office buildings and similar structures. This guide will walk you through the process of designing such a slab, including necessary calculations and considerations.
Problem Statement
We are tasked with designing the interior span of a continuous one way slab for an office floor. The slab is continuous over T beams spaced at 3 meters. The parameters provided are:
- Span length: 3 m
- Live load: 4 kN/m²
- Floor finish: 1 kN/m²
- Concrete grade: M20 (f'c = 20 MPa)
- Steel grade: Fe 415 (fy = 415 MPa)
Step 1: Determine Effective and Overall Depth
For a continuous slab, the span-to-depth ratio is typically taken as 26. Given the span length of 3 m (3000 mm), we can calculate the effective depth (d):
- Effective depth (d) = Span length / Span-to-depth ratio = 3000 mm / 26 = 115 mm
- Rounding this value gives us an effective depth of 120 mm.
Next, we consider the effective cover and the diameter of the main reinforcement:
- Effective cover = 15 mm
- Diameter of main reinforcement = 10 mm (which gives a radius of 5 mm)
Thus, the overall depth (D) is calculated as:
- Overall depth (D) = Effective depth + Effective cover + Radius of main reinforcement = 120 mm + 15 mm + 5 mm = 140 mm
Step 2: Load Calculations
To find the dead load (DL), we multiply the overall depth (D) by the unit weight of concrete (25 kN/m³):
- Dead load (DL) = D × Unit weight of concrete = 140 mm × 25 kN/m³ = 3.5 kN/m²
The total dead load is then:
- Total dead load (G) = Dead load + Floor finish = 3.5 kN/m² + 1 kN/m² = 4.5 kN/m²
The live load (Q) is given as 4 kN/m².
Step 3: Calculate Ultimate Moment and Shear
To find the maximum negative bending moment (BM) at the support next to the end support, we refer to the relevant codebook:
- Maximum negative BM = 1/10 × G × L² + 1/9 × Q × L², multiplied by a factor of 1.5 for factored moment.
- After calculations, we find the maximum negative BM to be 12 kNm.
For the positive BM at the center of the span:
- Positive BM = 1/12 × G × L² + 1/10 × Q × L², yielding a value of 10.35 kNm.
Next, we calculate the maximum shear force at the support:
- Maximum shear force = 0.6 × (G + Q) × L, also multiplied by 1.5 for factored shear.
- This results in a maximum shear force of 22.95 kN.
Step 4: Check Effective Depth
Using the calculated values, we apply the check for effective depth:
- Effective depth check = (B × D) / 66, where B is the breadth of the slab.
- The calculated effective depth is greater than the previously determined effective depth of 120 mm, confirming that the section is under-reinforced.
Step 5: Calculate Area of Steel Reinforcement
Using the equation for area of steel (As), we find:
- Area of steel (As) = 292 mm².
We also check for minimum reinforcement:
- Minimum reinforcement = 168 mm², which is less than the calculated As, allowing us to proceed with the design.
Step 6: Determine Reinforcement Spacing
With a diameter of 10 mm for the main reinforcement, we calculate the spacing:
This spacing will be used for both negative and positive BM, as they are approximately equal.
Step 7: Check for Cracking
To ensure safety against cracking, we check:
- Diameter of steel should be less than overall depth / 8. Here, 10 mm < 17.5 mm, confirming safety.
- Spacing of steel should be less than 300 mm, which is satisfied as our spacing is 250 mm.
Step 8: Design Distribution Reinforcement
For distribution reinforcement, we provide the minimum reinforcement of 168 mm² using 8 mm diameter bars:
- Spacing for distribution bars is calculated and rounded to 290 mm, which is less than the maximum allowable spacing of 450 mm, ensuring safety.
Step 9: Check for Shear Stress
Using the shear stress formula, we find:
- Maximum shear force (Vu) = 22.95 kN, and the percentage of steel is calculated as 0.26.
- The value of K is determined based on overall depth, confirming that the design is safe.
Step 10: Check for Deflection
Finally, we check for deflection:
- Using the relevant formula, we find that the deflection is less than the effective depth, confirming safety.
Conclusion
In summary, the design of a one way continuous slab involves careful calculations and checks to ensure safety and structural integrity. The final reinforcement details include:
- Main reinforcement: 10 mm diameter bars at 250 mm spacing for both positive and negative moments.
- Distribution reinforcement: 8 mm diameter bars at 290 mm spacing.
This comprehensive approach ensures that the slab will perform adequately under the specified loads and conditions. Thank you for following this guide on designing a one way continuous slab.
