Slab
Slabs are the most fundamental one of structural elements in
construction and give flat surfaces like floors, ceilings or roofs.
Loads are carried by them to beams and columns.
Two groups of slabs are often used: one-way slabs and two-side slabs
One-Way Slabs
A one-way slab is a reinforced concrete slab that mostly bends in one
direction.
This happens when the slab is supported only on two opposing sides, the
other two sides being free-or just spanning between beams.
With a one-way slab, the ratio between length and width (L/B) is 2 or
more.
Characteristics of One-Way Slabs:
- There is only one direction that reinforcement occurs for bending.
- Reinforcement is placed along the shorter route (primary reinforcement) and skew bars are placed perpendicular to it to take up temperature/shrinkage.
- In rectangular rooms, corridors and other rather elongated places versus their powellbroadths it is often encountered.
Two-Way Slabs
A two-way slab bends in both directions and is supported on all four
sides.
This type of slab is used where the ratio of L to B less than 2,
enabling the load to be spread in two perpendicular directions.
Characteristics of Two-Way Slabs:
- Bending occurs in both directions.
- Reinforcement needs to be given in both directions to cope with bending moments.
- This is suitable for square or nearly square spaces like auditoria, warehouses and industrial buildings.
Differences Between One-Way and Two-Way Slabs:
|
Feature |
One-Way Slab |
Two-Way Slab |
|
Load Distribution |
Along one direction only |
Along both directions |
|
Support Conditions |
Two parallel sides |
All four sides |
|
L/B Ratio |
≥ 2 |
< 2 |
|
Reinforcement |
Along shorter span primarily |
In both directions |
|
Application Areas |
Corridors, narrow rooms |
Square or near-square rooms |
Design Example of a One-Way Slab
Problem: Design a
one-way slab for a residential building floor spanning 4 meters (width) and 10
meters (length). The live load is 3 kN/m², and the floor finish load is 1
kN/m². Use M20 grade concrete and Fe415 steel.
Step 1: Determine Effective Span
The effective span for the one-way slab is the
shorter span, i.e., 4 meters.
Step 2: Calculate Total Load
Dead Load = Self-weight of the slab = 25 ×
Thickness (assume 150 mm thickness initially)
Dead Load = 25 × 0.15 = 3.75 kN/m² Total Load =
Dead Load + Live Load + Floor Finish Load = 3.75 + 3 + 1 = 7.75 kN/m²
Step 3: Moment Calculation
Factored Load = 1.5 × 7.75 = 11.625 kN/m² Factored
Bending Moment (M) = wL²/8 = (11.625 × 4²) / 8 = 23.25 kN.m
Step 4: Reinforcement Design
Using M20 concrete and Fe415 steel, calculate the
area of steel (Ast) using the bending moment formula:
Ast = (M × 10⁶) / (0.87 × fy × d)
Assume an effective depth (d) of 125 mm. Ast =
(23.25 × 10⁶) / (0.87 × 415 × 125) = 419.77 mm² Use 12 mm diameter bars spaced
at 150 mm center-to-center.
Design Example of a Two-Way Slab
Problem: Design a
two-way slab for a hall measuring 6 m × 6 m with a live load of 4 kN/m² and
floor finish of 1.5 kN/m². Use M25 grade concrete and Fe500 steel.
Step 1: Check Slab Type
L/B = 6/6 = 1 (less than 2). Therefore, it is a
two-way slab.
Step 2: Calculate Total Load
Dead Load = 25 × Thickness (assume 150 mm
initially) Dead Load = 25 × 0.15 = 3.75 kN/m² Total Load = Dead Load + Live
Load + Floor Finish Load = 3.75 + 4 + 1.5 = 9.25 kN/m²
Step 3: Moment Calculation
Factored Load = 1.5 × 9.25 = 13.875 kN/m² Moments
are calculated using coefficients from IS 456:2000: Mx = αx × w × L², My = αy ×
w × L² Assume αx = 0.052 and αy = 0.045 for simply supported edges.
Mx = 0.052 × 13.875 × 6² = 25.93 kN.m My = 0.045 ×
13.875 × 6² = 22.41 kN.m
Step 4: Reinforcement Design
For Mx and My, calculate the reinforcement using:
Ast = (M × 10⁶) / (0.87 × fy × d) Assume an effective depth (d) of 125 mm for
initial design. Astx = (25.93 × 10⁶) / (0.87 × 500 × 125) = 477.25 mm² Asty =
(22.41 × 10⁶) / (0.87 × 500 × 125) = 412.07 mm²
Use 10 mm bars at 150 mm c/c for Astx and 12 mm bars at 200 mm c/c for Asty.
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