Singly Reinforced Beam Design Procedure

1 Assume Dimensions of Beam

Effective depth: d = Span/10 to Span/8
Assume clear cover for both side spans (Refer to the Table No. 16, Cl. 26.4.2)
Assume diameter of main bars φ = 8, 10, 12 mm
Effective cover = clear cover + φ / 2
Overall Depth: D = d + Eff. Cover

Width: b = D/2 to 2/3 D (where D is the overall depth)

2 Calculate Effective Span (l)

(Assume width of support if not given)

Leff = l + effective depth
Leff = l + support width/2 + support width/2

3 Calculate Self Weight and Design Load

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, W_DL = __ kN/m

3.2 Live Load (LL)

Total Imposed live load/moving load, W_LL = __ kN/m

Factored Design Load

W_u = 1.5 (W_DL + W_LL) = __ kN/m

4 Calculate Factored Design Moment

Calculate factored maximum bending moment (M_u):

M_u = w_u l² / 8

5 Design as Balanced Section/ Check for Depth

The beam section is designed as a balanced section. Equate M_u to M_u,lim and find the effective depth of beam.

Steel Grade	Limiting Moment Formula
Fe 250	M_u,lim = 0.148 f_ck b d²
Fe 415	M_u,lim = 0.138 f_ck b d²
Fe 500	M_u,lim = 0.133 f_ck b d²

Compare depth (d) with the assumed depth:

If calculated depth ≤ assumed depth, proceed with design
If calculated depth > assumed depth, increase beam depth and recalculate

Recalculate self weight and factored moment based on revised depth.

6 Calculate Area of Tension Steel (A_st)

A_st = (0.36 f_ck b x_u,max) / (0.87 f_y)

Steel Grade	x_u,max Value
Fe 250	x_u,max = 0.53 d
Fe 415	x_u,max = 0.48 d
Fe 500	x_u,max = 0.46 d

Select suitable bar diameter and determine number of bars required.

7 Check Steel Area Requirements

Calculate minimum area of steel:

A_s,min = (0.85 b d) / f_y

Calculate maximum area of steel:

A_s,max = 0.04 b d

Verify that provided steel area is between minimum and maximum requirements.

8 Check for Shear and Design Shear Reinforcement

Calculate nominal shear stress:

τ_v = V_u / (b d)

Where:

τ_v = nominal shear stress
V_u = shear force
b = width of section
d = effective depth

If τ_v < τ_c < τ_c,max, the beam is safe in shear. Otherwise, design according to B-5.4.

Permissible Shear Stress in Concrete (τ_c, N/mm²)

Table 19 — Design shear strength of concrete, τ_c (N/mm²)
(Clauses 40.2.1, 40.2.2, 40.3, 40.4, 40.5.3, 41.3.2, 41.3.3 and 41.4.3)
100A_s / bd	M 15 (2)	M 20 (3)	M 25 (4)	M 30 (5)	M 35 (6)	M 40 and above (7)
≤ 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

NOTE — The term A_s is the area of longitudinal tension reinforcement which continues at least one effective depth beyond the section being considered, except at supports where the full area of tension reinforcement may be used provided the detailing conforms to clauses 26.2.2 and 26.2.3.

Table-20 Maximum Shear Stress (τ_c,max, N/mm²)

Concrete Grade	τ_c,max
M15	1.6
M20	1.8
M25	1.9
M30	2.0
M35	2.3
M40+	2.5

9 Check Development Length

Development length L_d is given by:

L_d = (φ σ_s) / (4 τ_bd)

Where:

φ = nominal diameter of bar
σ_s = stress in bar at design load
τ_bd = design bond stress

Design Bond Stress (τ_bd, N/mm²)

Concrete Grade	τ_bd
M20	1.2
M25	1.4
M30	1.5
M35	1.7
M40+	1.9

Notes:

Development length includes anchorage values of hooks in tension reinforcement
For bars of non-circular sections, development length should be sufficient to develop stress by bond

10 Check for Deflection

Follow these steps to check depth from deflection consideration:

Calculate percentage of steel:
p_t = (100 A_s) / (b d)
Calculate f_s:
f_s = 0.58 f_y × [Area of steel required / Area of steel provided]
Determine modification factor for compression reinforcement (K₁) from Fig. 12.1 based on p_t and f_s
- K_t = for Modification factor for tension reinforcement use Fig. 4, IS 456
- K_c = 1 if there no compression reinforcement is provided, if compression rebar is used, use Modification factor as per Fig. 5, IS 456
- K_f = 1 for rectangular sections, For flanged sections, use Modification factor from Fig. 6, IS 456
Calculate:
(l/d)_max = 20K_t × K_c × K_f × f_s
Check
(l/d)_provided < (l/d)_max
Verify calculated ratio

Short-term and long-term deflection (due to shrinkage and creep) can be computed as per Annexure C of IS:456-2000.

11 Face Reinforcement (for deep beams)

If beam depth > 750 mm, provide face reinforcement along both faces:

A_f = 0.1% of web area = (0.1 × bD) / 100

Distribute equally on both faces with spacing ≤ 300 mm or as per IS:456-2000 requirements.

12 Documentation

Document the design steps and prepare a neat sketch showing all reinforcement details(example in below figure).