1. Assume nominal clear cover (Table 16, cl. 26.4.2) c = __ mm
2. Assume bar diameter = __ mm (20mm, if not given)
3. Effective cover = clear cover + 0.5 × Bar diameter
Effective depth, d = D - effective cover
1. leff = Clear span + d
2. leff = Clear span + support width / 2 + support width / 2
Adopt leff = Minimum of (1) & (2)
Note:
- If only clear span is given or support width not given instead of effective span (leff):
leff = clear span + d
- If only c-c span is given instead of eff. span (leff):
leff = Centre to Centre span
3.1 Dead Load (DL)
Self weight of beam = (b/1000) × (D/1000) × 1 × 25 (Conc. Density, kN/m³) = __ kN/m
+ Other imposed dead load
Total dead load, WDL = __ kN/m
3.2 Live Load (LL)
Total Imposed live load/moving load, WLL = __ kN/m
Factored Design Load
Wu = 1.5 (WDL + WLL) = __ kN/m
4.1 Calculate Ultimate Bending Moment (Mu)
Mu = Wu × leff² / 8 = __ kNm (for UDL)
4.2 Calculate Ultimate Shear Force (Vu)
Vu = Wu × leff / 2
4.3 Calculate Limiting Moment of Resistance of Beam Section (Mu,lim)
Refer Clause G-1-(c), Annexure G, Page 96, IS 456
Mu,lim = 0.36 (xu,max/d) (1 - 0.42 (xu,max/d)) b d² fck
For maximum depth of neutral axis (xu,max) refer Page 70, IS 456
Decision:
- If Mu,lim > Mu: Then beam section is not sufficient to resist the acting bending moment
- If Mu,lim < Mu: Design the beam as Doubly Reinforced section
Generally in singly reinforced beam, the size of the beam section will be increased to make Mu,lim > Mu
5.1 Compressive Steel (Asc)
Refer Clause G-1.2, Page No. 96, IS 456
Mu - Mu,lim = fsc × Asc × (d - d')
After rearranging: Asc = (Mu - Mu,lim) / [fsc × (d - d')]
Where:
- d' = Effective cover to compressive steel
- fsc = Design stress in compression reinforcement corresponding to a strain of Esc (Refer SP 16, clause 1:4, Page No. 6)
- Compute Esc = 0.0035 (xu,max - d') / xu,max (for xu,max refer clause 38.1, page 70, IS 456)
TABLE A: SALIENT POINTS ON THE DESIGN STRESS-STRAIN CURVE FOR COLD-WORKED BARS (SP 16, Clause 1.4)
| STRESS LEVEL |
fy = 415 N/mm² |
fy = 500 N/mm² |
| Strain |
Stress N/mm² |
Strain |
Stress N/mm² |
| 0-80 fyd |
0.00144 |
288.7 |
0.00174 |
347.8 |
| 0-85 fyd |
0.00163 |
306.7 |
0.00195 |
369.6 |
| 0-90 fyd |
0.00192 |
324.8 |
0.00226 |
391.3 |
| 0-95 fyd |
0.00241 |
342.8 |
0.00277 |
413.0 |
| 0-975 fyd |
0.00276 |
351.8 |
0.00312 |
423.9 |
| 1-0 fyd |
0.00380 |
360.9 |
0.00417 |
434.8 |
NOTE: Linear interpolation may be done for intermediate values.
Provide Compressive Steel (Asc)
- Select suitable bar diameter (e.g., 10, 12, 16, 20, 25 & 32mm)
- Calculate area of 1 bar; Abar = __ mm²
- No. of bars required = Asc / Abar = __ no. (Round off to nearest integer)
Note:
- Minimum bars = 2 no.
- Maximum bars = 6 no. in one layer
Calculate Asc,provided = N × Abar
5.2 Tension Steel (Ast)
Refer Clause G-1.2, Page No. 96, IS 456
Ast = Ast1 + Ast2
Where:
- Ast = Area of total tensile reinforcement
- Ast1 = Area of the tensile reinforcement for a singly reinforced section for Mu,lim
- Ast2 = Asc × fsc / (0.87 × fy)
Ast1 = [0.36 × fck × b × xu,max] / (0.87 × fy) = __ mm²
Ast2 = Asc × (fsc / (0.87 × fy)) = __ mm²
Check for Minimum Steel (As,min)
Refer Cl. 26.5.1.1 (a)
Ast,min = 0.85 × b × d / fy = __ mm²
Provide Tension Steel
- Select suitable bar diameter
- Calculate area of one bar; Abar = π × Φ² / 4 = __ mm²
- Calculate number of bars, N = Ast / Abar
Note:
- Minimum bars = 2 no.
- Maximum bar = 6 nos in one layer
Calculate Ast,provided = N × Abar
Provide bars at bottom (tension face of beam)
Calculate Nominal Shear Stress (τv)
Refer Cl. 40.1, IS 456
τv = Vu / (b × d) = __ N/mm²
Get Maximum Shear Stress (τc,max)
Refer Table 20, Page No. 73, IS 456
Table 20: Maximum Shear Stress, τc max, N/mm²
| Concrete Grade |
M 15 |
M 20 |
M 25 |
M 30 |
M 35 |
M 40 and above |
| τc max, N/mm² |
2.5 |
2.8 |
3.1 |
3.5 |
3.7 |
4.0 |
(Clauses 40.2.3, 40.2.3.1, 40.5.1 and 41.3.1)
Check τc,max > τv
Clause 40.2.3, IS 456
Calculate Design Shear Strength of Concrete (τc)
Calculate τc from Table No. 19, Page 73, IS 456
→ % steel provided in Tension side (Pt)
Pt = (100 × Ast,provided) / (b × d)
Table 19: Design Shear Strength of Concrete, τc, N/mm²
| 100As/bd |
Concrete Grade |
| M 15 |
M 20 |
M 25 |
M 30 |
M 35 |
M 40 & above |
| ≤0.15 |
0.28 |
0.28 |
0.29 |
0.29 |
0.29 |
0.30 |
| 0.25 |
0.35 |
0.36 |
0.36 |
0.37 |
0.37 |
0.38 |
| 0.50 |
0.46 |
0.48 |
0.49 |
0.50 |
0.50 |
0.51 |
| 0.75 |
0.54 |
0.56 |
0.57 |
0.59 |
0.59 |
0.60 |
| 1.00 |
0.60 |
0.62 |
0.64 |
0.66 |
0.67 |
0.68 |
| 1.25 |
0.64 |
0.67 |
0.70 |
0.71 |
0.73 |
0.74 |
| 1.50 |
0.68 |
0.72 |
0.74 |
0.76 |
0.78 |
0.79 |
| 1.75 |
0.71 |
0.75 |
0.78 |
0.80 |
0.82 |
0.84 |
| 2.00 |
0.71 |
0.79 |
0.82 |
0.84 |
0.86 |
0.88 |
| 2.25 |
0.71 |
0.81 |
0.85 |
0.88 |
0.90 |
0.92 |
| 2.50 |
0.71 |
0.82 |
0.88 |
0.91 |
0.93 |
0.95 |
| 2.75 |
0.71 |
0.82 |
0.90 |
0.94 |
0.96 |
0.98 |
| 3.00 and above |
0.71 |
0.82 |
0.92 |
0.96 |
0.99 |
1.01 |
(Clauses 40.2.1, 40.2.2, 40.3, 40.4, 40.5.3, 41.3.2, 41.3.3 and 41.4.3)
NOTE: The term As is the area of longitudinal tension reinforcement which continues at least one effective depth beyond the section being considered except at support where the full area of tension reinforcement may be used provided the detailing conforms to 26.2.2 and 26.2.3
Check If τv > τc
If τv > τc, then design shear reinforcement:
- Calculate Vus = Vu - τc × b × d
- Assume legs & diameter of vertical stirrups
- Asv = No. of legs × cross sectional area of stirrup bar
- Calculate spacing of stirrups (Sv) Clause 26.5.1.6, IS 456
Vus = (0.87 × fy × Asv × d) / Sv
Rearrange: Sv = (0.87 × fy × Asv × d) / Vus ①
Calculate minimum shear spacing for stirrup:
Sv,min = (0.87 × fy × Asv) / (0.4 × b) ②
Maximum stirrup spacing:
Sv = 0.75 × d ③ (Clause 26.5.1.5, Page No. 47, IS 456)
Sv = 300 mm ④ (Clause 26.5.1.5, Page No. 47, IS 456)
Provide stirrup spacing minimum of ①, ②, ③ or ④
If τv ≤ τc, then minimum shear reinforcement must be provided
- Assume legs & diameter of vertical stirrups
- Asv = No. of legs × cross sectional area of stirrup bar
- Calculate minimum shear spacing for stirrup:
Sv,min = (0.87 × fy × Asv) / (0.4 × b) ②
Maximum stirrup spacing:
Sv = 0.75 × d ③ (Clause 26.5.1.5, Page No. 47, IS 456)
Sv = 300 mm ④ (Clause 26.5.1.5, Page No. 47, IS 456)
Provide stirrup spacing minimum of ②, ③ or ④
Final stirrup spacing = the smallest value among Sv(min), 0.75d, and 300 mm