Analysis and Design of Slabs — Worked Examples

These examples separate prescriptive thickness screening from explicit analysis and keep every load, depth, and reinforcement calculation consistent with the selected slab geometry.

One-Way Slab Prescriptive Thickness Explorer

Screening rule only — not an analytical deflection or strength verification.

Controls

Applicable span LL4.0 m
Steel yield strength fyf_y420 MPa

Prescriptive calculation

hpresc=L20(0.4+fy700)h_{\text{presc}}=\frac{L}{20}\left(0.4+\frac{f_y}{700}\right)
=400020(1.000)=200.0 mm=\frac{4000}{20}(1.000)=200.0\ \text{mm}

The displayed adopted value rounds the equation result upward to the next 5 mm for a practical teaching selection.

Prescriptive minimum

200 mm

Raw equation value: 200.0 mm. Use only when the slab end rotations are not restrained by continuity in the design direction.

Model scope:nonprestressed solid one-way slab, normal-weight concrete, and the lesson's span-to-thickness rule. A thinner section requires the explicit deflection analysis required by the adopted standard; this tool does not verify flexure, shear, punching, cracking, cover, fire, or vibration.

Example 1: Prescriptive Thickness of a Simply Supported One-Way Slab

Problem: A nonprestressed one-way solid slab has an applicable span L=4.5 mL=4.5\text{ m} and is simply supported. Using the lesson's normal-weight-concrete screening rule with fy=420 MPaf_y=420\text{ MPa}, determine the prescriptive thickness that permits omission of a separate detailed deflection calculation.

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Example 2: One-Way or Two-Way Behavior from Supports and Aspect Ratio

Problem: A 6.0 m×2.5 m6.0\text{ m}\times2.5\text{ m} slab panel is supported by beams on all four sides. Classify its principal slab action using the lesson's aspect-ratio screening rule.

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Example 3: Coherent One-Way Slab Design after the Thickness Check

Problem: Design a representative 1-m1\text{-m} strip of a simply supported one-way slab spanning 4.0 m4.0\text{ m}. The live load is 4.0 kPa4.0\text{ kPa} and superimposed dead load is 2.0 kPa2.0\text{ kPa}. Use fc′=28 MPaf'_c=28\text{ MPa}, fy=420 MPaf_y=420\text{ MPa}, normal-weight concrete at 24 kN/m324\text{ kN/m}^3, 20 mm20\text{ mm} clear cover, and trial 10 mm10\text{ mm} bars. Use the lesson's prescriptive thickness route rather than an independent deflection calculation.

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Example 4: Direct Design Method Total Static Moment

Problem: A qualifying interior panel is being checked for DDM in one direction. The factored uniform load is wu=12 kPaw_u=12\text{ kPa}, the transverse panel dimension is l2=5.0 ml_2=5.0\text{ m}, and the applicable clear span is ln=5.5 ml_n=5.5\text{ m}. Calculate the total static moment M0M_0 before any prescribed positive-negative or strip distribution.

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Example 5: Punching-Shear Critical Perimeter Geometry

Problem: An interior square column is 400 mm×400 mm400\text{ mm}\times400\text{ mm} and the slab effective depth at the connection is d=150 mmd=150\text{ mm}. Determine the simplified critical perimeter bob_o located d/2d/2 from the column faces.

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Example 6: Complete DDM Applicability Screening

Problem: A flat-plate floor has three continuous spans in each direction. Panels are 6.0 m×5.0 m6.0\text{ m}\times5.0\text{ m}, successive spans are equal, columns are on the panel centerlines, loading is gravity-only and uniform, dead load is 4.5 kPa4.5\text{ kPa}, and live load is 3.0 kPa3.0\text{ kPa}. Determine whether the lesson's listed DDM screening conditions are satisfied.

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Example 7: Yield-Line Mechanism Interpretation

Problem: Two admissible yield-line mechanisms for the same idealized slab give collapse-load estimates of 18.0 kPa18.0\text{ kPa} and 16.5 kPa16.5\text{ kPa}. What should be concluded from the upper-bound interpretation of yield-line analysis?

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Example 8: When a More General Slab Analysis Is Required

Problem: A two-way floor has two continuous spans in one direction, a large irregular atrium opening beside a column line, significant nonuniform equipment loads, and the slab participates in the lateral-force-resisting system. Should DDM or a basic gravity-only EFM idealization be treated as sufficient by default?

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