Designing a reinforced concrete (RCC) staircase is a routine task for structural engineers, but accuracy in calculation and correct interpretation of IS 456:2000 provisions are essential.
This article presents a complete design example of a straight staircase, including geometry, loads, bending moment, reinforcement design, and final detailing.
Parameter from Straight Staircase From Architect Drawing:
| Parameter | Value |
|---|---|
| Floor-to-floor height | 3.150 m |
| Stair clear width | 1.50 m |
| Riser (r) | 150 mm |
| Tread (t) | 250 mm |
| Waist thickness (h) | 150 mm |
| Concrete grade | M25 |
| Steel grade | Fe500 |
| Clear cover | 20 mm |
| Main reinforcement | T12 bars |
| Finish load | 1.0 kN/m² |
| Live load | 3.0 kN/m² |
| Unit weight of concrete | 25 kN/m³ |
Also, read: Staircase: Stair Components | Technical Terms | Staircase Parts
Step 1: Geometrical Design for Straight Staircase
- Number of Riser (r):
\(N_r=\frac Hr=\frac{3150}{150}=21\;risers\)
- Number of Treads:
\(N_t=N_r-1=20-1=20\;treads\)
Stair Run (Horizontal Length)
\(L_{stair}=N_t\times t=20\times.25=5.0m\)
- Slop of Stair (θ)
\(\theta=\tan^{-1}\left(\frac{r}{t}\right)\)=\(\tan^{-1}\left(\frac{0.15}{0.25}\right)\)≈31°
(Satisfies IS 456 recommendation slope of 30°–35° for residential stairs).
Step 2: Load Calculation for Straight Staircase
Unit weight of Reinforced Concrete: 25kN.m3
Waist slab thickness: 150mm=0.15mm
- Dead Load (Self-weight of staircase)
- DL = 0.15 X 1.5 X 25 =5.625kN/m of flight width
- Finishes on staircase =1.0kN/m2 X 1.5 = 1.5kN/m
- Total DL = 5.625 + 1.5 =7.125kN/m
- Load as per IS 875
- Live Load (IS 875 Part 2)= 3.0kN/m2 X 1.5 = 4.5kN/m (For residential building)
- Total Service Load
- w= 7.125 + 4.5 =11.625 ≈11.625 kN/m
- Total Design Load
- w= 1.5 X 11.625 =17.4375 ≈17.44 kN/m
Therefore, the load per meter run of flight is 17.44kN/m.
Step 3: Moment and Shear Calculation
For simply supported flight (assuming cantilever at landing if needed):
Moment:
- Flight span, \(L = \text{Run of flight} = 2.5 m\)
- Maximum bending moment (UDL, simply supported):
\(M_{max}=\left(\frac{wL^2}8\right)=\left(\frac{17.44\times2.5^2}8\right)\)
Maximum moment, \(M_{max}\)=13.63 kN/m
Shear:
\(V_{max}=\left(\frac{wL}2\right)=\left(\frac{17.44\times2.5}2\right)\)
Maximum shear, \(V_{max}\)=21.8 kN
Step 4: Reinforcement Design (Main Bar)
Assume:
- Tension at bottom of waist slab
- Effective depth, \(d=h-cover-\phi/2\)
\(d\)=150-20-12/2 =124 mm≈0.124m
Flexural design formula (IS 456, simplified for singly reinforced):
\(M_u=0.138f_{ck}bd^2\)
Design moment capacity for T12 bars? We’ll compute the required steel area using:
\(M_u=0.138f_yA_{st}d\left(1-\frac{A_{st}f_y}{f_{ck}bd}\right)\)
Iterative, but approximate for preliminary:
\(A_{st}=\frac{M_u}{0.138f_{ck}bd^2}\)
\(A_{st}=\frac{13.625}{0.138\times25\times1500\times\left(124\right)^2}\)
\(A_{st}=\frac{13.625}{79570800}=79.55\)kNm
Moment required 13.625kNm << 79.55kNm.
Number of bars required = \(\frac{A_{st}}{113}\)
Design is safe, so 2-3 T12 bars per width suffice.
Step 5: Shear Check
\(V_c=0.16\sqrt{f_{ck}}bd\)
\(V_c=0.16\sqrt{25\times1500\times124}\)
\(=148.8\)kN
Max Share =21.8kN <<148.8kN, Shear OK, no shear reinforcement required.
Step 6: Bar Detailing
- Main bars (tension at bottom of flight): T12 @ 150 mm c/c (2-3 bars)
- Distribution bars: T8 @ 200 mm c/c top for temperature/shrinkage
- Landing bars: Continuation of main bars with anchorage length 40d ≈ 480 mm
Step 7: IS 456 Checks
- Riser (150 mm) and tread (250 mm) → Within IS 456 recommendations
- Width (1.5 m) → OK
- Waist thickness (150 mm) → Minimum 120 mm → OK
Summary Table of Stair Design:
This is design summery for Straight Staircase as given by the architect in his drawing.
| Parameter | Value |
|---|---|
| Floor-to-floor height | 3.15 m |
| Total risers | 21 |
| Total treads | 20 |
| Landing | After 11th riser, 1.5 m long |
| Stair width | 1.50 m |
| Riser | 150 mm |
| Tread | 250 mm |
| Concrete | M25 (fck 25 MPa) |
| Steel | Fe500 (fy 500 MPa) |
| Waist thickness | 150 mm |
| Main bars | T12 @ 150 mm c/c (2-3 bars) |
| Distribution bars | T8 @ 200 mm c/c |
| Max bending moment | 13.625 kNm |
| Max shear | 21.8 kN |
Also, read: TMT vs TOR Steel: Understanding Differences, Benefits, and Applications
FAQs:
Q: What is the difference between Total Service Load and Total Design Load?
Answer: Total Service Load is the actual load on the structure during normal use (dead load + live load without any factors).
Total Design Load is the factored load used for safety in structural design, obtained by multiplying service loads with partial safety factors (as per IS 456).
Q: Why is the intermediate landing provided?
Answer: The landing breaks the stairs into two manageable flights, improves safety, and reduces the effective span of each flight for structural design.
Q: Why is the staircase treated as a simply supported slab for design?
Answer: Most stair flights rest on beams or landings at both ends, making the simply supported assumption suitable for bending moment and shear calculations.
Q: What is the purpose of providing the main reinforcement at the bottom of the waist slab?
Answer: The bottom of the stairs experiences tension during bending, so the main reinforcement is placed there to resist tensile stresses.
Q: Why is equivalent thickness not used in the design?
Answer: Because the stair is a solid waist slab, the actual waist thickness (150 mm) already represents the structural depth; equivalent thickness is needed only for “stepped” or non-uniform stairs.
References:
- Indian Standard. (2000). Code of Practice for Plain and Reinforced Concrete (IS 456:2000) (4th ed.). Bureau of Indian Standards (BIS).
- Indian Standard. (1987). Code of Practice for Design Loads (Other Than Earthquake for Buildings and Structures) (IS875: 1987) (part-1 Dead Load) (2nd rev.). Bureau of Indian Standards (BIS).
- Indian Standard. (1987). Code of Practice for Design Loads (Other Than Earthquake for Buildings and Structures) (IS875: 1987) (part-1 Imposed Load) (2nd rev.). Bureau of Indian Standards (BIS).
