Interactive Tool

Minimum Slab Thickness Calculator

Equation Used

h = L / 20× (0.4 + f_y / 700)

Required Minimum Thickness

200 mm

* Rounded up to the nearest millimeter. If thickness is less than this value, explicit deflection calculations must be performed.

Solved Problems

Example 1: Minimum Thickness of One-Way Slab

Problem: A one-way simply supported slab spans L=4.5 mL = 4.5 \text{ m}. Determine the minimum thickness hh to avoid deflection calculations. fy=420 MPaf_y = 420 \text{ MPa}.

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Example 2: One-Way Slab Design

Problem: Design a one-way slab to carry a live load of 4 kPa4 \text{ kPa} and superimposed dead load of 2 kPa2 \text{ kPa} on a simply supported span of 4 m4 \text{ m}. fc=28 MPaf'_c = 28 \text{ MPa}, fy=420 MPaf_y = 420 \text{ MPa}. Use h=150 mmh=150 \text{ mm}.

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Example 3: Two-Way Slab Moment Coefficients (Method 2)

Problem: A two-way slab panel is 4 m×5 m4 \text{ m} \times 5 \text{ m} (short span S=4S=4, long span L=5L=5). Supported on all sides, discontinuous edges. Calculate the positive moment in the short direction. wu=12 kPaw_u = 12 \text{ kPa}.

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Example 4: Checking One-Way Shear Capacity in a Slab

Problem: A one-way solid slab with an effective depth d=120 mmd = 120 \text{ mm} supports a uniform factored load wu=15 kPaw_u = 15 \text{ kPa}. The slab spans 4.0 m4.0 \text{ m} between the centerlines of its simple supports. Determine if the slab's concrete shear capacity ϕVc\phi V_c is adequate without stirrups. Use fc=25 MPaf'_c = 25 \text{ MPa}.

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Example 5: Direct Design Method Constraints for Two-Way Slabs

Problem: Determine if the Direct Design Method (DDM) can be used to analyze a two-way flat slab floor system consisting of 3 continuous spans in each direction. The panels are 6 m6 \text{ m} by 5 m5 \text{ m}. The unfactored live load is 3.0 kPa3.0 \text{ kPa} and the unfactored dead load (including self-weight) is 4.5 kPa4.5 \text{ kPa}.

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Case Study 1: Punching Shear Failure at an Interior Column

Problem: An open-plan retail store was built using a flat plate concrete floor system (a two-way slab resting directly on columns without drop panels or column capitals). The floor was originally designed for a live load of 4.8 kPa4.8 \text{ kPa}. Decades later, a tenant placed densely packed heavy library shelving (12 kPa12 \text{ kPa}) around several interior columns. The slab suddenly dropped around one column, leaving the column punching through the floor. Diagnose the structural failure.

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Case Study 2: Flat Slab Progressive Collapse Mechanism

Problem: Following an initial punching shear failure at a single column in an underground parking garage, the entire floor slab progressively collapsed, pulling down the columns next to it, and triggering a chain reaction across the entire structure. Explain why the failure propagated so extensively and how modern codes prevent this.

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