Two-Way Slab Design — Step-by-Step Worked Example

Design of a Two-Way Slab — Worked Example (All formulas & step-by-step arithmetic)

Written By Mohan Dangi

Table of contents 1. Given data 2. Effective spans & slab type 3. Load calculations 4. Design moments (with formulas) 5. Depth check 6. Flexural design — area of steel 7. Shear check 8. Deflection check 9. Design schedule & detailing 10. Conclusion & notes

1. Given data

Parameter	Value
Clear dimensions (slab)	3.60 m × 6.20 m
Beam width	0.25 m (250 mm)
Concrete unit weight (γc)	25 kN/m³
Live load	2.50 kN/m²
Floor finish/partitions	1.20 kN/m²
Concrete	M25, fck=25 MPa
Steel	Fe500, fy=500 MPa
Clear cover	20 mm
Bar assumed	T10 (10 mm)
Chosen overall slab thickness	D = 140 mm

2. Effective spans & slab type

Short span Lx = 3.60 m
Long span  Ly = 6.20 m
Beam width = 0.25 m

Effective spans:
L_ex = Lx + beam_width = 3.60 + 0.25 = 3.85 m
L_ey = Ly + beam_width = 6.20 + 0.25 = 6.45 m

Aspect ratio = L_ey / L_ex = 6.45 / 3.85 ≈ 1.675 → since < 2, this is a TWO-WAY SLAB.
    

3. Load calculations

Self-weight of slab (w_s):
w_s = γ_c × D = 25 kN/m³ × 0.140 m = 3.50 kN/m²

Service load sum:
w_service = w_s + finish + live = 3.50 + 1.20 + 2.50 = 7.20 kN/m²

Factored (ultimate) load:
w_u = 1.5 × w_service = 1.5 × 7.20 = 10.80 kN/m²
    

4. Design moments (with formulas)

Use representative two-way slab moment coefficients for this aspect ratio (from code tables):

• α_x (short direction negative moment coeff) = 0.075
• α_y (long direction negative moment coeff) = 0.055

Formulas (per 1 m width):
M_x = α_x × w_u × L_ex²
M_y = α_y × w_u × L_ey²

Compute squares:
L_ex² = 3.85 × 3.85 = 14.8225 m²
L_ey² = 6.45 × 6.45 = 41.6025 m²

Now substitute:
M_x = 0.075 × 10.80 × 14.8225
    0.075 × 10.80 = 0.810
    0.810 × 14.8225 = 12.006225
=> M_x = 12.006225 kN·m per metre width  (≈ 12.006 kN·m/m)

M_y = 0.055 × 10.80 × 41.6025
    0.055 × 10.80 = 0.594
    0.594 × 41.6025 = 24.711885
=> M_y = 24.711885 kN·m per metre width (≈ 24.712 kN·m/m)
    

5. Depth check (required effective depth)

Quick working formula used for required effective depth (classroom check):

d_req = sqrt( M_u (N·mm) / (0.138 × f_ck × b) )

Notes:
- M_u in N·mm = kN·m × 10^6
- b = 1000 mm (design per 1 m width)
- f_ck = 25 MPa
    

For M_x = 12.006225 kN·m:
Mu_x = 12.006225 × 10^6 = 12,006,225 N·mm
Denominator = 0.138 × 25 × 1000 = 3,450

d_req_x = sqrt( 12,006,225 / 3,450 ) = sqrt( 3,480.064492753623 ) ≈ 58.992 mm

For M_y = 24.711885 kN·m:
Mu_y = 24.711885 × 10^6 = 24,711,885 N·mm

d_req_y = sqrt( 24,711,885 / 3,450 ) = sqrt( 7,163.2321739 ) ≈ 84.634 mm

Available effective depth:
D = 140 mm ; cover = 20 mm ; bar dia = 10 mm → d = D − cover − (bar_dia/2)
d = 140 − 20 − 5 = 115 mm

Conclusion: available d = 115 mm > d_req_x (≈ 59.0 mm) and > d_req_y (≈ 84.6 mm) → depth adequate for flexure.
    

6. Flexural design — area of steel (worked)

Simplified working relation (limit state approximation):

Mu = 0.87 × f_y × z × A_st
=> A_st = Mu / (0.87 × f_y × z)

Approximate lever arm: z ≈ 0.9 × d (common approximation)
Use d = 115 mm → z = 0.9 × 115 = 103.5 mm
f_y = 500 N/mm²
b = 1000 mm
Mu in N·mm
    

Compute A_st required:

For M_x:
Mu_x = 12,006,225 N·mm
A_st_req_x = 12,006,225 / (0.87 × 500 × 103.5)
            = 12,006,225 / 45,022.5
            ≈ 266.672 mm² per metre width

For M_y:
Mu_y = 24,711,885 N·mm
A_st_req_y = 24,711,885 / 45,022.5
            ≈ 548.879 mm² per metre width
    

Choose practical bars (T10 area = π×10²/4 = 78.5398 mm²):

T10 area ≈ 78.5398 mm²

Provision chosen:
- Short direction (main across short span): T10 @ 175 mm c/c
  bars per metre = 1000 / 175 ≈ 5.7142857 bars/m
  Provided A_st = 5.7142857 × 78.5398 ≈ 448.799 mm²/m

- Long direction (main across long span): T10 @ 125 mm c/c
  bars per metre = 1000 / 125 = 8 bars/m
  Provided A_st = 8 × 78.5398 ≈ 628.3188 mm²/m

Comparison:
- Required (short) ≈ 266.67 mm²/m ; Provided ≈ 448.80 mm²/m → OK
- Required (long)  ≈ 548.88 mm²/m ; Provided ≈ 628.32 mm²/m → OK
    

7. Shear check (clear & consistent)

Use a conservative panel shear (per 1 m width). For uniformly distributed load on strip:

Conservative V_u = w_u × (L_ex / 2)   (kN per metre width)
w_u = 10.80 kN/m² ; L_ex = 3.85 m

V_u = 10.80 × (3.85 / 2) = 10.80 × 1.925 = 20.79 kN/m
    

Convert to shear stress at depth d (in MPa):

T_v = V_u × 1000 / (b × d)   (N / mm² = MPa)
b = 1000 mm ; d = 115 mm

T_v = (20.79 × 1000) / (1000 × 115) = 20.79 / 115 = 0.1807826 MPa ≈ 0.181 MPa
    

Typical tabulated nominal shear resistance (T_c) for slabs with this reinforcement is around 0.35–0.45 MPa (depends on %steel & f_ck). Since:

T_v ≈ 0.181 MPa < typical T_c (≈ 0.39 MPa) → Shear is safe (no shear reinforcement required in panel).
    

Important: This is a simplified panel shear check. For concentrated column loads or punching shear regions, perform detailed punching shear checks per IS 456 (or your national code).

8. Deflection check (practical L/d check)

Actual L/d (short direction) = L_ex / d = 3.85 m / 0.115 m = 3.85 / 0.115 = 33.478 ≈ 33.5

Serviceability limit guidance (practical check):
Permissible L/d is based on K_t × 26, where K_t depends on service stress in steel (tables). For many typical two-way slabs with adequate reinforcement, permissible L/d is well above 30.

Conclusion (practical): Actual L/d ≈ 33.5 is acceptable for continuous two-way slab with provided reinforcement. For formal verification, compute service stress in steel f_s under service load and check IS 456 clause for K_t and exact permissible L/d.
    

9. Design schedule & detailing (final)

Slab	D (mm)	d (mm)	Main short dir	Main long dir
S2	140	115	T10 @ 175 mm c/c (~448.8 mm²/m)	T10 @ 125 mm c/c (~628.3 mm²/m)

• Provide distribution bars perpendicular to main bars (use same or one size smaller spacing as required).
• Provide top/negative reinforcement near supports for continuity (provide similar area at top near supports as required by negative moment). Use continuity slabs detailing per code.
• Maintain minimum and maximum bar spacing per code (for crack control and concrete consolidation). Use chairs to maintain cover = 20 mm.
• Check lap lengths, development lengths and anchorage per IS 456 / project specs.

10. Conclusion & notes

Item	Value
Factored load w_u	10.80 kN/m²
M_x	≈ 12.006 kN·m/m
M_y	≈ 24.712 kN·m/m
Required d (x,y)	≈ 59.0 mm, 84.6 mm
Available d	115 mm
Ast required (x,y)	≈ 266.7 mm²/m, 548.9 mm²/m
Ast provided (x,y)	≈ 448.8 mm²/m (T10@175), 628.3 mm²/m (T10@125)
Shear stress T_v	≈ 0.181 MPa (safe vs typical T_c)
Actual L/d (short)	≈ 33.5 (practical check)

For construction: always perform final checks (punching shear if columns, detailed deflection check per code, detailed negative reinforcement near supports, and have drawings signed by a licensed structural engineer).

Written By Mohan Dangi (Gold Medalist)