Two-Way Slab Design Calculations: Steps and Formulas

Posted on Jan 5, 2026 in Civil Engineering

Step 1: Design Constants

• For M20 & Fe415:

  • τc max = 1.4 N/mm² (from IS:456 Table 73)

  • σ = 2.76 (used in moment capacity formula)

Step 2: Effective Span Calculation

Slab Type Check

• Cy/Cx = 6/4 = 1.5, which is < 2 → Two-Way Slab Panel

Effective Depth Calculation

Using IS 456 empirical formula:

$$\frac{L}{d} = \text{MF} \times \text{BF}$$

• Basic factor (BF) = 26

• Modification factor (MF) = 1.3

$$d = \frac{L}{\text{MF} \times \text{BF}} = \frac{6000}{1.3 \times 26} = 177.51 \, \text{mm} \approx 180 \, \text{mm}$$

Overall Depth

$$D = 180 + \frac{\phi}{2} + \text{cover} = 180 + 5 + 20 = 206 \, \text{mm}$$

Effective Depth in X-Direction

$$d_x = 180 – \phi = 168 \, \text{mm} \quad (\text{for } \phi = 12 \, \text{mm})$$

Strip Widths Table

Step 3: Load Calculations

Dead Load

$$\text{D.L} = \text{Thickness} \times \text{Density} = 0.206 \times 25 = 5.15 \, \text{kN/m}^2$$

Additional Loads

• Live load = 4 kN/m²

• Floor finish = 1 kN/m²

Total Service Load

$$= 5.15 + 4 + 1 = 10.15 \, \text{kN/m}^2$$

Factored Load

$$= 10.15 \times 1.5 = 15.225 \, \text{kN/m}^2$$

Step 4: Moment Calculations

A. Longer Span (L₂ = 6 m; lₙ = 5.6 m)

Load

$$W_y = 15.225 \times 4 \times 5.6 = 341.05 \, \text{kN}$$

Moment

$$M_0 = \frac{W_y \cdot l_n}{8} = \frac{341.05 \times 5.6}{8} = 238.728 \, \text{kN\cdot m}$$

Distribution (IS 456)

• 65% to negative moment

• 35% to positive moment

Column strip moments:

• Negative moment = 155.17 kN\cdot m

• Positive moment = 83.55 kN\cdot m

Middle strip:

• Negative moment = 25% of column strip

• Positive moment = 40% of column strip

• Values: 39.79 & 33.42 kN\cdot m

B. Shorter Span (L₁ = 4 m; lₙ = 3.6 m)

Load

$$W_x = 15.225 \times 6 \times 3.6 = 328.87 \, \text{kN}$$

Moment

$$M_0 = \frac{W_x \cdot l_n}{8} = \frac{328.87 \times 3.6}{8} = 147.99 \, \text{kN\cdot m}$$

Distribution:

• Column strip: Negative moment = 98.6 kN\cdot m, Positive moment = 51.2 kN\cdot m

• Middle strip: 24.00 & 20.8 kN\cdot m

Step 5: Reinforcement Design

Minimum Steel

$$A_{st,\min} = 0.12\% \times b \times D = 0.0012 \times 1000 \times 206 = 247.2 \, \text{mm}^2/\text{m}$$

Column Strip (Longer Span)

$$M_u = 116.37 \, \text{kN\cdot m}$$

Using IS 456 formula:

$$A_{st} = 0.5 \frac{f_{ck}}{f_y} \left(1 – \sqrt{1-\frac{4.6M_u}{f_{ck} b d_y^2}}\right) b d$$

Design depth checks confirm:

• OK

Column Strip (Shorter Span)

$$M_u = 72.13 \, \text{kN\cdot m}$$

Depth check:

$$d_x = 114.31 \, \text{mm} < 168 \, \text{mm}$$

• Safe