Two-Way Slab Design per ACI 318-19: Step-by-Step Worked Example

In reinforced concrete building design, two-way flat plates are among the most economical floor systems for residential and commercial spans up to 7–8 meters. Because there are no interior drop panels or beam projections, storey heights are minimized and formwork costs are cut substantially.

However, flat plates demand rigorous engineering oversight. Because the system lacks beams, flexural moments must be partitioned mathematically into column strips and middle strips to account for slab-column joint stiffness, and punching shear at column interfaces governs in over 80% of structural sizing failures.

This whitepaper walks through a complete, step-by-step design of a $6.0\text{ m} \times 6.0\text{ m}$ interior flat plate panel using the ACI 318-19 Direct Design Method (DDM) (§8.10). Every equation, code limitation, intermediate check, and bar selection is detailed from first principles.

Applicability of Direct Design Method (ACI 318-19 §8.10.2)

Before applying the DDM, all six code limitations under §8.10.2 must be strictly verified:

1. Minimum 3 continuous spans in each direction: Satisfied (interior panel in a multi-bay building).
2. Rectangular panels with aspect ratio $\ell_1 / \ell_2 \leq 2.0$: Here $\ell_1 / \ell_2 = 6.0 / 6.0 = 1.0 \leq 2.0$ — Pass.
3. Successive span lengths in each direction differ by $\leq 33\%$ of longer span: Uniform $6.0\text{ m}$ spans — Pass.
4. Column offset from either axis $\leq 10\%$ of span: Columns are on a regular grid — Pass.
5. Loads are uniformly distributed gravity loads only: Satisfied.
6. Unfactored Live to Dead load ratio $L/D \leq 2.0$: $\frac{L}{D} = \frac{3.0\text{ kPa}}{12.04\text{ kPa}} = 0.25 \ll 2.0 \quad \text{— \textbf{Pass}}.$

Design Parameters & Problem Statement

Step 1 — Minimum Slab Thickness for Deflection Control (§8.3.1.1)

To satisfy serviceability deflection criteria without performing explicit cracked-section time-dependent deflection calculations, ACI 318-19 Table 8.3.1.1 specifies the minimum thickness for interior panels without drop panels:

$h_{min} = \frac{\ell_n \left(0.8 + \dfrac{f_y}{200{,}000}\right)}{33}$

Where $f_y$ is in $\text{psi}$ and $\ell_n$ is clear span:

• $\ell_n = 6.0\text{ m} = 236.22\text{ in}$
• $f_y = 60{,}000\text{ psi}$

$h_{min} = \frac{236.22 \times \left(0.8 + \dfrac{60{,}000}{200{,}000}\right)}{33} = \frac{236.22 \times 1.10}{33} = 7.87\text{ in} \approx 200\text{ mm}$

ACI 318-19 Table 8.3.1.1 also mandates an absolute minimum of $125\text{ mm}$ ($5.0\text{ in}$) for flat plates.

Selected Thickness: We select $h = 220\text{ mm}$ ($8.66\text{ in}$) to provide conservative punching shear margin at interior columns under heavy superimposed dead loads.

Step 2 — Factored Design Gravity Loading (§5.3.1)

Calculate unfactored self-weight using reinforced concrete unit weight $\gamma_c = 24.5\text{ kN/m}^3$: $w_{SW} = 0.220\text{ m} \times 24.5\text{ kN/m}^3 = 5.39\text{ kN/m}^2$

Total unfactored Dead Load: $w_D = w_{SW} + w_{SDL} = 5.39 + 7.00 = 12.39\text{ kN/m}^2$

Factored load combination (Comb 2: $1.2D + 1.6L$): $w_u = 1.2(12.39) + 1.6(3.00) = 14.87 + 4.80 = 19.67\text{ kN/m}^2$

Step 3 — Total Factored Static Design Moment $M_0$ (§8.10.3.2)

For a design strip bounded by panel centerlines ($\ell_2 = 6.40\text{ m}$), the total factored static moment in the direction of span $\ell_1$ is:

$M_0 = \frac{w_u \ell_2 \ell_n^2}{8}$

$M_0 = \frac{19.67\text{ kN/m}^2 \times 6.40\text{ m} \times (6.00\text{ m})^2}{8} = \frac{19.67 \times 6.40 \times 36.00}{8} = 566.50\text{ kN}\cdot\text{m}$

Step 4 — Longitudinal Distribution to Negative and Positive Sections (§8.10.4)

Per ACI 318-19 §8.10.4.2, for an interior span, the total static moment $M_0$ is distributed as:

• Total Negative Factored Moment: $M_u^- = 0.65 \, M_0 = 0.65 \times 566.50 = 368.22\text{ kN}\cdot\text{m}$
• Total Positive Factored Moment: $M_u^+ = 0.35 \, M_0 = 0.35 \times 566.50 = 198.28\text{ kN}\cdot\text{m}$

Step 5 — Transverse Distribution Across Column and Middle Strips (§8.10.5)

Strip geometry:

• Column strip total width $= 2 \times \min\left(\dfrac{\ell_1}{4}, \dfrac{\ell_2}{4}\right) = 2 \times 1.60\text{ m} = 3.20\text{ m}$
• Middle strip width $= \ell_2 - b_{cs} = 6.40\text{ m} - 3.20\text{ m} = 3.20\text{ m}$

Because this is a flat plate with no interior edge beams ($\alpha_{f1} = 0$), Table 8.10.5.1 and Table 8.10.5.5 dictate:

• Negative Column Strip fraction: $75\%$ of $M_u^-$
• Negative Middle Strip fraction: $25\%$ of $M_u^-$
• Positive Column Strip fraction: $60\%$ of $M_u^+$
• Positive Middle Strip fraction: $40\%$ of $M_u^+$

Step 6 — Flexural Reinforcement Design & Verification (§22.2)

Using metric $\phi 12\text{ mm}$ bars ($A_b = 113.1\text{ mm}^2$): $d_{avg} = h - c_c - d_b = 220 - 20 - 12 = 188\text{ mm}$

Flexural resistance equation: $M_n = A_s f_y \left(d - \frac{a}{2}\right), \quad a = \frac{A_s f_y}{0.85 f'_c b}, \quad \phi = 0.90$

Nominal resistance index $R_n = \dfrac{M_u}{\phi b d^2}$: $\rho = \frac{0.85 f'_c}{f_y} \left(1 - \sqrt{1 - \frac{2 R_n}{0.85 f'_c}}\right)$

Minimum Reinforcement per ACI 318-19 §8.6.1.1 & §24.4.3.2: $A_{s,min} = 0.0018 \, b \, h = 0.0018 \times 3{,}200 \times 220 = 1{,}267\text{ mm}^2 \quad (396\text{ mm}^2/\text{m})$

Reinforcement Schedule Table

Step 7 — Two-Way (Punching) Shear Verification at Column (§22.6)

For a square interior column $c_1 = c_2 = 400\text{ mm}$:

1. Critical perimeter $b_0$ at distance $d/2 = 94\text{ mm}$ from face: $b_0 = 4(c + d) = 4(400 + 188) = 2{,}352\text{ mm}$

2. Tributary area for punching shear: $A_{trib} = (\ell_1 \times \ell_2) - (c_1 + d)(c_2 + d) = (6.40 \times 6.40) - (0.588)^2 = 40.96 - 0.35 = 40.61\text{ m}^2$

3. Factored punching shear force: $V_u = w_u \times A_{trib} = 19.67\text{ kN/m}^2 \times 40.61\text{ m}^2 = 798.8\text{ kN}$

4. Nominal concrete shear strength $v_c$ per ACI 318-19 §22.6.5.2:

$v_c = \min \begin{cases} 0.33 \lambda \sqrt{f'_c} = 0.33 \times 1.0 \times \sqrt{28} = 1.746\text{ MPa} \\ 0.17\left(1 + \frac{2}{\beta}\right)\lambda\sqrt{f'_c} = 0.17(3)\sqrt{28} = 2.698\text{ MPa} \\ 0.083\left(2 + \frac{\alpha_s d}{b_0}\right)\lambda\sqrt{f'_c} = 0.083\left(2 + \frac{40 \times 188}{2352}\right)\sqrt{28} = 2.281\text{ MPa} \end{cases}$

Governing concrete shear stress: $v_c = 1.746\text{ MPa}$. With $\phi = 0.75$:

$\phi V_c = \phi v_c b_0 d = 0.75 \times 1.746\text{ N/mm}^2 \times 2{,}352\text{ mm} \times 188\text{ mm} = 579.1\text{ kN}$

Since $V_u = 798.8\text{ kN} > \phi V_c = 579.1\text{ kN}$, shear reinforcement (stud rails per §8.7.7) is required: $\phi v_{n,max} = \phi \cdot 0.50 \sqrt{f'_c} = 0.75 \times 2.645 = 1.984\text{ MPa} \implies \phi V_{n,max} = 877.3\text{ kN} > V_u \quad \text{— Feasible}$

Comparison: Traditional Manual Design vs. CIVORA

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