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

Concept and model scope

Serviceability screening rule for the lesson's nonprestressed solid one-way slab basis. It is not an analytical deflection proof or a strength design.

The support sketch uses a fixed physical span scale, so changing span or continuity changes the geometry instead of only changing the denominator. The support selection represents actual structural continuity, not a preference for a smaller thickness. Slider states cover 2.5–8.0 m spans and 280–550 MPa steel; the calculation rejects nonpositive values defensively.

After a trial thickness is selected, the full one-way design laboratory recomputes self-weight, effective depth, flexure, shear, reinforcement, spacing, and serviceability together.

Applicable span LL4.0 m
Steel yield strength fyf_y420 MPa
One-way slab support-condition sketchThe span is drawn with one fixed millimeter-to-pixel scale. Continuous-end states show adjacent slab continuation, while the cantilever shows one restrained end and one free end.applicable span = 4.0 mSimply supported · fixed geometric scale

Screening equation

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}

Adopted screening thickness

200 mm

Use only when end rotation is not restrained by continuity in the design direction. The teaching value rounds the equation upward to the next 5 mm.

Scope: passing this screen does not verify flexure, one-way shear, punching, cracking, long-term deflection, vibration, cover, fire, or constructability. A thinner slab requires the explicit serviceability analysis required by the governing code.

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: 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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Example 8: DDM Applicability Failure and Method Change

Problem: A nominally regular flat plate has only two continuous spans in the east-west direction, four spans north-south, uniform gravity loading, and otherwise regular geometry. Can DDM be used by simply applying the usual M0M_0 coefficients?

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Example 9: Interior Punching-Shear Demand and Capacity

Problem: An interior 400×400 mm400\times400\ \text{mm} column supports a slab with d=170 mmd=170\ \text{mm} and fc′=28 MPaf'_c=28\ \text{MPa}. The factored column reaction is 650 kN650\ \text{kN} and factored slab load is 12 kPa12\ \text{kPa}. Use the lesson's no-shear-reinforcement ACI 318-14 teaching equations.

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Example 10: Edge and Corner Punching with the Same Column Reaction

Problem: Keep the 400×400 mm400\times400\ \text{mm} column, d=170 mmd=170\ \text{mm}, fc′=28 MPaf'_c=28\ \text{MPa}, 650 kN650\ \text{kN} reaction, and 12 kPa12\ \text{kPa} slab load from the preceding example. Compare the lesson's truncated edge and corner critical perimeters.

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Example 11: Opening Near an Interior Column

Problem: For the interior-column case of Example 9, an opening is close enough that the governing geometric rule makes 20% of the original critical perimeter ineffective. Show the effect without pretending the opening area itself is the perimeter reduction.

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Example 12: Reinforcement Layout from DDM Design Regions

Problem: A qualifying interior flat-plate panel has M0=226.9 kN⋅mM_0=226.9\ \text{kN}\cdot\text{m}. For the beamless interior teaching distribution, lay out the design moments before selecting reinforcement.

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