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.
Screening equation
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 and is simply supported. Using the lesson's normal-weight-concrete screening rule with , determine the prescriptive thickness that permits omission of a separate detailed deflection calculation.
Step-by-Step Solution
0 of 2 Steps CompletedExample 2: One-Way or Two-Way Behavior from Supports and Aspect Ratio
Problem: A slab panel is supported by beams on all four sides. Classify its principal slab action using the lesson's aspect-ratio screening rule.
Step-by-Step Solution
0 of 2 Steps CompletedExample 3: Coherent One-Way Slab Design after the Thickness Check
Problem: Design a representative strip of a simply supported one-way slab spanning . The live load is and superimposed dead load is . Use , , normal-weight concrete at , clear cover, and trial bars. Use the lesson's prescriptive thickness route rather than an independent deflection calculation.
Step-by-Step Solution
0 of 6 Steps CompletedExample 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 , the transverse panel dimension is , and the applicable clear span is . Calculate the total static moment before any prescribed positive-negative or strip distribution.
Step-by-Step Solution
0 of 2 Steps CompletedExample 5: Punching-Shear Critical Perimeter Geometry
Problem: An interior square column is and the slab effective depth at the connection is . Determine the simplified critical perimeter located from the column faces.
Step-by-Step Solution
0 of 1 Steps CompletedExample 6: Complete DDM Applicability Screening
Problem: A flat-plate floor has three continuous spans in each direction. Panels are , successive spans are equal, columns are on the panel centerlines, loading is gravity-only and uniform, dead load is , and live load is . Determine whether the lesson's listed DDM screening conditions are satisfied.
Step-by-Step Solution
0 of 3 Steps CompletedExample 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?
Step-by-Step Solution
0 of 2 Steps CompletedExample 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 coefficients?
Step-by-Step Solution
0 of 3 Steps CompletedExample 9: Interior Punching-Shear Demand and Capacity
Problem: An interior column supports a slab with and . The factored column reaction is and factored slab load is . Use the lesson's no-shear-reinforcement ACI 318-14 teaching equations.
Step-by-Step Solution
0 of 4 Steps CompletedExample 10: Edge and Corner Punching with the Same Column Reaction
Problem: Keep the column, , , reaction, and slab load from the preceding example. Compare the lesson's truncated edge and corner critical perimeters.
Step-by-Step Solution
0 of 3 Steps CompletedExample 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.
Step-by-Step Solution
0 of 3 Steps CompletedExample 12: Reinforcement Layout from DDM Design Regions
Problem: A qualifying interior flat-plate panel has . For the beamless interior teaching distribution, lay out the design moments before selecting reinforcement.