Design of Continuous Beam

Step 1: Depth of Beam

Clear span of beam = ___ mm

Effective depth of beam, d = span/10 to span/15
Eff. depth of beam, d = span/12

Assume clear cover of beam, c = ___ mm

(Refer to table No. 16 for clear cover acc. to diff. exposure condition)

Assume dia. of bar in tension, ø = ___ mm

(generally near 20mm)

Overall depth of beam, D = d + c + ø/2 = ___ mm

Step 2: Effective Span

Refer to clause 22.2 (b)

Clause 22.2(b): Effective Span for Continuous Beam or Slab

In the case of a continuous beam or slab, the effective span depends on the width of the support:

If the width of the support is less than 1/12 of the clear span, the effective span shall be as per Clause 22.2(a).
If the support is wider than 1/12 of the clear span or 600 mm, whichever is less, the effective span shall be taken as follows:

End span with one end fixed and the other continuous, or intermediate spans:
→ Effective span = clear span between supports
End span with one end free and the other continuous:
→ Effective span = clear span + ½ × effective depth of beam or slab
→ or clear span + ½ × width of discontinuous support
→ whichever is less
Spans with roller or rocker bearings:
→ Effective span = distance between centres of bearings

Step 3: Calculation of Load

Dead load (DL)

Self. wt. of the beam = 25 × D/1000 × b/1000 × 1 = ___ kN/m
+ Superimposed dead load (if any)
Total factored dead load, Wu, DL = 1.5 × DL = ___ kN/m

Live load (LL)

Total live load = ___ kN/m
Total factored live load, Wu, LL = 1.5 × LL = ___ kN/m

Step 4: Calculation of Moment & Shear Force

Read Clause 22.5: Moment and Shear Coefficients for Continuous Beams for calculation of Moment and Shear Force in a continuous beam

22.5 Moment and Shear Coefficients for Continuous Beams

22.5.1 Unless more exact estimates are made, for beams of uniform cross-section which support substantially uniformly distributed loads over three or more spans which do not differ by more than 15 percent of the longest, the bending moments and shear forces used in design may be obtained using the coefficients given in Table 12 and Table 13 respectively.

For moments at supports where two unequal spans meet or in case where the spans are not equally loaded, the average of the two values for the negative moment at the support may be taken for design.

Where coefficients given in Table 12 are used for calculation of bending moments, redistribution referred to in 22.7 shall not be permitted.

The Coefficients in Table 12 are applicable to cases of:

Uniformly loaded beams of uniform section with at least three spans
"which do not differ by more than 15% of the longest"

(L₁ − L₂) ≯ 15% L₁
(L₁ − L₃) ≯ 15% L₁

Moment

Mu = Coefficient_DL × (Wu, DL × l²) + Coefficient_LL × (Wu, LL × l²)

Table 12: Bending Moment Coefficients (Clause 22.5.1)

Type of Load	Span Moments		Support Moments
Type of Load	Near Middle of End Span	At Middle of Interior Span	At Support Next to the End Support	At Other Interior Supports
Dead load and imposed load (fixed)	+1/12	+1/16	−1/10	−1/12
Imposed load (not fixed)	+1/10	+1/12	−1/9	−1/9

Note: To obtain the bending moment, multiply the coefficient by the total design load and effective span.

Shear Force

Vu = Coefficient_DL × (Wu, DL × l) + Coefficient_LL × (Wu, LL × l)

Table 13: Shear Force Coefficients (Clauses 22.5.1 and 22.5.2)

Type of Load	At End Support	At Support Next to the End Support (Outer Side)	At Support Next to the End Support (Inner Side)	At All Other Interior Supports
Dead load and imposed load (fixed)	0.4	0.6	0.55	0.5
Imposed load (not fixed)	0.45	0.6	0.6	0.6

Note: For obtaining the shear force, the coefficient shall be multiplied by the total design load.

Step 5: Check for Depth

Max factored Bending moment, Mu = ___ kNm
Limiting Moment of Resistance, Mu, lim = ___ kNm

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.

Step 6: Computation of Steel Required

Ast, required = 0.5 × fck/fy × [1 - √(1-4.6×Mu/(fck×b×d²))] × b×d²

Check for Steel

Ast, min = (0.85 × b × d) / fy

Assume dia. of bars required generally 20mm or more

Area of 1-φ bar = (π × φ²) / 4 = _____ mm²

No. of bars = (Ast, provided) / (Area of 1-φ bar) = ____ no.

Step 7: Check for Shear

Tv = Vu / (b × d) = _____ N/mm²

Check

τv > τc,max

If τc,max is greater than τc, then calculate Design Shear strength

% area of Steel, Pt = (100 × Ast) / (b × d)

for given grade of conc. calculate τc

Tc = ______ N/mm²
for grade of conc. compute τc,max (N/mm²)
Check τc < τc,max
Check τv < τc, if τv > τc then design for shear stress is req.

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

For vertical stirrups (clause 40.4 (a))

Vus = Vu – τc × b × d = ______ N
Asv = 2 (legs) × (π × φ²)/4 [dia. of stirrup]
Sv = (0.87 × fy × Asv × d) / Vus = ______ mm ①

Max. spacing of shear reinforcement (clause 26.5.1.5)

Sv ≠ 0.75d ②
Sv ≠ 300mm ③

Max stirrup spacing permitted = Minimum of 1, 2, 3

Step 8: Write Detailed Summary and Draw a Neat Sketch

Summary:

Grade of Concrete
Grade of Steel
Mention standard code/ guideline followed
development length
Dia. and Spacing/ No. of bars
Cross-sectional details

Ensure all calculations comply with relevant design codes and standards.