Doubly Reinforced Beam

Design & Analysis as per IS 456:2000 — Compression Steel, Stress fsc, Moment of Resistance & Solved GATE Examples

Last Updated: March 2026

Key Takeaways 📌

  • A doubly reinforced beam (DRB) has steel in both the tension zone (Ast) and the compression zone (Asc). It is used when the applied moment Mu > Mu,lim — i.e., a singly reinforced section of the given dimensions cannot carry the load.
  • The compression steel Asc supplements the concrete’s compressive capacity, allowing the section to carry the additional moment (Mu − Mu,lim) beyond the singly reinforced limit.
  • The stress in compression steel fsc is NOT simply 0.87fy — it depends on the strain at the compression steel level, which depends on xu and d’. IS 456 provides fsc values from the stress-strain curve.
  • Design approach: Split the total moment Mu into two parts — Mu,lim (carried by the balanced SRB) and (Mu − Mu,lim) (carried by the compression steel couple). Design each part separately, then combine.
  • The neutral axis depth in a DRB is always at the limiting value xu,max when designed for the minimum compression steel.
  • For Fe 415 steel in compression (with d’/d ≤ 0.2): fsc = 353 N/mm² — memorise this for GATE CE.
  • Doubly reinforced beams are used where section depth is architecturally or structurally constrained and cannot be increased.

1. When to Use a Doubly Reinforced Beam

A singly reinforced beam has a maximum moment capacity of Mu,lim = Ru,lim × bd². When the design moment Mu exceeds this limiting value, a singly reinforced section of the given dimensions is insufficient. The engineer then has two options:

  1. Increase the section depth d — this increases Mu,lim proportionally to d². Increasing depth by 20% increases Mu,lim by 44%. This is the most material-efficient solution and is preferred when there is no dimensional constraint.
  2. Use a doubly reinforced beam — add compression steel Asc near the top face to supplement the concrete’s compressive capacity. This is used when the section depth cannot be increased due to architectural constraints (ceiling heights, beam-column proportions), structural constraints (existing construction, openings), or when the beam is continuous and the sagging sections are already deep enough.

Common situations requiring doubly reinforced beams:

  • Beams at the ground floor of a building where headroom is restricted.
  • Transfer beams carrying heavy column loads in a limited depth.
  • Beams in continuous frames where negative (hogging) moments at supports require compression steel in the bottom of the section.
  • Beams subject to reversal of loading (e.g., earthquake loading) — where both top and bottom may need to carry tension at different times, so both are reinforced on both sides.

Additionally, compression steel provides practical benefits beyond pure moment capacity — it reduces long-term deflection (by increasing the compression zone stiffness and reducing creep), improves ductility, helps in positioning stirrups (the compression bars serve as hangers for the stirrups), and provides robustness under accidental loading.

2. Section Geometry and Notation

Notation for Doubly Reinforced Beam

b = breadth of beam [mm]

D = overall depth [mm]

d = effective depth to centroid of tension steel [mm]

d’ = effective cover to centroid of compression steel [mm] (= clear cover + stirrup dia + compression bar radius)

Ast = area of tension steel [mm²]

Asc = area of compression steel [mm²]

xu = neutral axis depth [mm]

fsc = stress in compression steel at ultimate [N/mm²] (NOT simply 0.87fy)

d’/d ratio: This is the critical parameter for finding fsc. Typical values: d’/d = 0.05 to 0.20. For a beam with d’ = 50 mm and d = 500 mm: d’/d = 0.10.

The doubly reinforced beam section has three zones:

  • Top zone (near top face): Compression steel Asc is placed at depth d’ from the compression face. Both concrete and steel contribute to the compressive force here.
  • Middle zone (between compression steel and neutral axis): Pure concrete in compression, stress = 0.36fck (from the stress block).
  • Tension zone (below neutral axis): Tension steel Ast carries all tensile force. Concrete is cracked and carries nothing.

3. Stress in Compression Steel fsc

The stress in the compression steel fsc is the most critical and most misunderstood quantity in doubly reinforced beam design. It is not equal to 0.87fy in general — that value only applies to tension steel that has yielded. Compression steel may or may not have yielded, depending on its depth d’ relative to the neutral axis depth xu.

Strain in Compression Steel at Ultimate

Using the linear strain diagram at the ultimate limit state (εcu = 0.0035 at top fibre):

εsc = εcu × (xu − d’) / xu = 0.0035 × (1 − d’/xu)

Where d’ = depth of compression steel from compression face, xu = neutral axis depth.

For the balanced section (xu = xu,max):

εsc = 0.0035 × (xu,max − d’) / xu,max = 0.0035 × (1 − d’/xu,max)

This strain εsc is then used to find fsc from the steel’s stress-strain curve.

Stress in Compression Steel fsc — IS 456

IS 456 uses a simplified stress-strain curve for steel in compression. The stress fsc is read from IS 456 Table F (for Fe 415) or Table G/H for other grades, based on the strain εsc.

Important simplification for design: In doubly reinforced beam design, xu = xu,max (the neutral axis is at the limiting position for minimum compression steel). This makes εsc determinate — it depends only on d’/d and the steel grade (through xu,max/d).

For Fe 415 with xu,max/d = 0.48:

εsc = 0.0035 × (0.48 − d’/d) / 0.48

For d’/d = 0.10: εsc = 0.0035 × (0.48 − 0.10)/0.48 = 0.0035 × 0.7917 = 0.00277

From the Fe 415 stress-strain curve at ε = 0.00277: fsc ≈ 353 N/mm²

For d’/d = 0.15: εsc = 0.0035 × (0.48 − 0.15)/0.48 = 0.002406 → fsc ≈ 342 N/mm²

For d’/d = 0.20: εsc = 0.0035 × (0.48 − 0.20)/0.48 = 0.002042 → fsc ≈ 329 N/mm²

Why is fsc less than 0.87fy = 361 N/mm²? The compression steel is embedded in concrete that is already being compressed. The strain in the compression steel at the balanced condition is 0.00277 for d’/d = 0.10 — this is close to the yield strain of Fe 415 (0.00380) but slightly below it. The steel has not fully yielded, so it cannot develop its full design stress. For very shallow d’ (d’/d close to 0), the compression steel strain approaches 0.0035 and the stress approaches the yield stress. For deeper compression steel (larger d’/d), the strain is lower and fsc is lower.

4. fsc Values — IS 456 Reference ⭐ GATE

d’/dFe 250 (xu,max/d = 0.53)Fe 415 (xu,max/d = 0.48)Fe 500 (xu,max/d = 0.46)
0.05217 N/mm²355 N/mm²424 N/mm²
0.10217 N/mm²353 N/mm²412 N/mm²
0.15217 N/mm²342 N/mm²395 N/mm²
0.20217 N/mm²329 N/mm²370 N/mm²

Notes:

  • For Fe 250, fsc = 217 N/mm² = permissible stress, constant for all d’/d — mild steel yields at very low strain, so it always reaches yield in the compression zone.
  • For Fe 415, the most important value to memorise: fsc = 353 N/mm² for d’/d = 0.10 (the most common assumption in GATE CE problems).
  • These values include a deduction for the displaced concrete — the compression steel occupies space that would otherwise be concrete. IS 456 accounts for this by using (fsc − 0.36fck) as the net additional stress from the steel over and above the concrete it displaces. Some formulations explicitly use (fsc − 0.36fck) while others separate the concrete and steel contributions — both give the same result.

5. Force Equilibrium — Neutral Axis

At the ultimate limit state, the total compressive force equals the total tensile force. In a doubly reinforced beam, the total compressive force has two components: compression in the concrete (Cc) and compression in the steel (Cs).

Force Equilibrium — DRB

Total compression C = Cc + Cs

Cc = 0.36 fck × b × xu (from concrete stress block)

Cs = (fsc − 0.36 fck) × Asc

The (fsc − 0.36 fck) term accounts for the fact that the compression steel displaces concrete — we first add back the concrete stress block contribution over the full b×xu area (which incorrectly includes the area occupied by the steel), then subtract it for the steel area. The net contribution of the compression steel is therefore the steel stress minus the concrete stress at that level: (fsc − 0.36fck) × Asc.

Total tension T = 0.87 fy × Ast

Force equilibrium: Cc + Cs = T

0.36 fck b xu + (fsc − 0.36 fck) Asc = 0.87 fy Ast

For the design case (xu = xu,max):

0.36 fck b xu,max + (fsc − 0.36 fck) Asc = 0.87 fy Ast

6. Moment of Resistance Mu

Moment of Resistance — DRB (Taking Moments About Tension Steel)

Mu = Cc × (d − 0.42xu) + Cs × (d − d’)

Mu = 0.36 fck b xu (d − 0.42xu) + (fsc − 0.36fck) Asc (d − d’)

Where:

First term = Moment from concrete compression = Mu,lim (when xu = xu,max)

Second term = Additional moment from compression steel = (Mu − Mu,lim)

This decomposition is the key insight for design: Mu = Mu,lim + ΔM

Where ΔM = Mu − Mu,lim = moment to be carried by the compression steel couple.

Physical Interpretation — Two-Couple Model

The doubly reinforced beam can be conceptualised as two beams acting together:

Beam 1 (Singly reinforced, balanced):

Concrete compression + Tension steel Ast1 → carries Mu,lim

Ast1 = 0.36 fck b xu,max / (0.87 fy) = Ast,bal

Beam 2 (Steel couple):

Compression steel Asc + Additional tension steel Ast2 → carries ΔM = Mu − Mu,lim

Lever arm for couple = d − d’

Asc = ΔM / [(fsc − 0.36fck)(d − d’)]

Ast2 = ΔM / [0.87fy(d − d’)] = (fsc − 0.36fck) × Asc / (0.87fy)

Total tension steel: Ast = Ast1 + Ast2

7. Design Procedure — Two-Moment Method

Given: Design moment Mu, section dimensions b and d (fixed), material grades fck and fy, compression steel cover d’.

  1. Step 1 — Check if DRB is needed:

    Compute Mu,lim = Ru,lim × bd²

    If Mu ≤ Mu,lim: design as singly reinforced.

    If Mu > Mu,lim: doubly reinforced beam is needed.

  2. Step 2 — Find ΔM:

    ΔM = Mu − Mu,lim

  3. Step 3 — Find fsc:

    Compute d’/d ratio. Read fsc from IS 456 table or use standard values:

    Fe 415: fsc = 353 N/mm² for d’/d = 0.10 (most common assumption).

  4. Step 4 — Find Asc:

    Asc = ΔM / [(fsc − 0.36fck) × (d − d’)]

  5. Step 5 — Find Ast,bal (steel for the balanced SRB part):

    Ast1 = 0.36 fck × b × xu,max / (0.87 fy)

  6. Step 6 — Find Ast2 (additional steel for the compression couple):

    Ast2 = (fsc − 0.36fck) × Asc / (0.87 fy)

    Or equivalently: Ast2 = ΔM / [0.87fy × (d − d’)]

  7. Step 7 — Total tension steel:

    Ast = Ast1 + Ast2

  8. Step 8 — Select bars:

    Choose bar diameters and numbers for both Asc (top) and Ast (bottom). Check minimum/maximum steel limits.

8. Analysis Procedure — Finding Mu

Given: Section dimensions, Ast, Asc, d’, material grades. Find Mu.

  1. Assume xu = xu,max as an initial estimate.
  2. Find fsc from d’/d ratio and IS 456 table.
  3. Check force equilibrium:

    LHS = 0.36fckbxu + (fsc − 0.36fck)Asc

    RHS = 0.87fyAst

    If LHS = RHS: xu assumed is correct → proceed.

    If LHS ≠ RHS: find xu iteratively or solve from equilibrium equation.

  4. Find xu from equilibrium (if needed):

    0.36fckbxu = 0.87fyAst − (fsc − 0.36fck)Asc

    xu = [0.87fyAst − (fsc − 0.36fck)Asc] / (0.36fckb)

  5. Check xu ≤ xu,max:

    If xu < xu,max: under-reinforced → proceed with calculated xu.

    If xu > xu,max: over-reinforced → use xu = xu,max and re-evaluate fsc.

  6. Compute Mu:

    Mu = 0.36fckbxu(d − 0.42xu) + (fsc − 0.36fck)Asc(d − d’)

9. Worked Example 1 — Design of Doubly Reinforced Beam

Problem: Design a doubly reinforced rectangular beam for Mu = 280 kN·m. Section: b = 250 mm, d = 500 mm. Cover to compression steel d’ = 50 mm. fck = 20 N/mm², fy = 415 N/mm².

Step 1 — Check if DRB is Needed

Mu,lim = Ru,lim × bd² = 2.76 × 250 × 500²

= 2.76 × 250 × 250,000 = 172,500,000 N·mm = 172.5 kN·m

Mu = 280 kN·m > Mu,lim = 172.5 kN·m

→ Doubly reinforced beam required.

Step 2 — Excess Moment ΔM

ΔM = Mu − Mu,lim = 280 − 172.5 = 107.5 kN·m = 107.5 × 10⁶ N·mm

Step 3 — Stress in Compression Steel fsc

d’/d = 50/500 = 0.10

For Fe 415, d’/d = 0.10: fsc = 353 N/mm²

fsc − 0.36fck = 353 − 0.36×20 = 353 − 7.2 = 345.8 N/mm²

Step 4 — Compression Steel Asc

Asc = ΔM / [(fsc − 0.36fck) × (d − d’)]

= 107.5 × 10⁶ / [345.8 × (500 − 50)]

= 107,500,000 / (345.8 × 450)

= 107,500,000 / 155,610

Asc = 691.0 mm²

Provide 3 bars of 18 mm dia: Asc = 3 × 254.47 = 763.4 mm² (adopt)

(Rounding up is standard practice — always provide area ≥ required)

Step 5 — Tension Steel Ast

Part 1 — Steel for balanced SRB (Ast1):

xu,max = 0.48 × 500 = 240 mm

Ast1 = 0.36 × 20 × 250 × 240 / (0.87 × 415)

= 0.36 × 20 × 250 × 240 / 361.05

= 432,000 / 361.05 = 1196.4 mm²

Part 2 — Additional steel for compression couple (Ast2):

Ast2 = (fsc − 0.36fck) × Asc / (0.87fy)

= 345.8 × 691.0 / 361.05

= 238,927.8 / 361.05 = 661.8 mm²

Or using: Ast2 = ΔM / [0.87fy(d−d’)] = 107.5×10⁶ / (361.05×450) = 107,500,000/162,472.5 = 661.7 mm² ✓

Total Ast = Ast1 + Ast2 = 1196.4 + 661.8 = 1858.2 mm²

Provide 4 bars of 25 mm dia: Ast = 4 × 490.87 = 1963.5 mm² (adopt)

Step 6 — Final Design Summary

b = 250 mm | D = 550 mm (d=500+cover50) | d = 500 mm | d’ = 50 mm

Compression steel: 3–18 mm dia (Asc = 763.4 mm²) at top

Tension steel: 4–25 mm dia (Ast = 1963.5 mm²) at bottom

Verification check — force equilibrium with provided steel:

Cc = 0.36×20×250×240 = 432,000 N (xu = xu,max = 240 mm assumed)

Cs = 345.8 × 763.4 = 264,003 N

Total C = 432,000 + 264,003 = 696,003 N

T = 0.87 × 415 × 1963.5 = 709,178 N

Close but not equal → actual xu is slightly > xu,max due to providing more steel than minimum.

For provided steel, re-check xu: [0.87×415×1963.5 − 345.8×763.4] / (0.36×20×250) = [709,178 − 264,003]/1800 = 445,175/1800 = 247.3 mm

xu = 247.3 mm > xu,max = 240 mm → slightly over-reinforced with provided steel

Use xu = xu,max = 240 mm for Mu calculation:

Mu = 432,000×(500−0.42×240) + 264,003×(500−50)

= 432,000×399.2 + 264,003×450

= 172,454,400 + 118,801,350 = 291,255,750 N·mm = 291.3 kN·m > 280 kN·m ✓

10. Worked Example 2 — Analysis of Doubly Reinforced Beam

Problem: A doubly reinforced beam has b = 300 mm, d = 550 mm, d’ = 55 mm. Ast = 2945 mm² (6–25 mm dia), Asc = 982 mm² (2–25 mm dia). fck = 25 N/mm², fy = 415 N/mm². Find Mu.

Step 1 — Find fsc

d’/d = 55/550 = 0.10 → fsc = 353 N/mm² (Fe 415, d’/d = 0.10)

fsc − 0.36fck = 353 − 0.36×25 = 353 − 9 = 344 N/mm²

Step 2 — Find xu from Force Equilibrium

0.36fckbxu + (fsc−0.36fck)Asc = 0.87fyAst

0.36×25×300×xu + 344×982 = 0.87×415×2945

2700xu + 337,808 = 1,062,743

2700xu = 1,062,743 − 337,808 = 724,935

xu = 724,935 / 2700 = 268.5 mm

xu,max = 0.48 × 550 = 264 mm

xu = 268.5 mm > xu,max = 264 mm → slightly over-reinforced

→ Use xu = xu,max = 264 mm for Mu calculation (IS 456 limits)

Re-check fsc at xu = 264 mm: εsc = 0.0035×(264−55)/264 = 0.0035×0.7917 = 0.00277 → fsc = 353 N/mm² (same as before — d’/d = 0.10 unchanged) ✓

Step 3 — Moment of Resistance Mu

Cc = 0.36×25×300×264 = 712,800 N

Cs = 344×982 = 337,808 N

Mu = Cc×(d−0.42xu) + Cs×(d−d’)

= 712,800×(550−0.42×264) + 337,808×(550−55)

= 712,800×(550−110.88) + 337,808×495

= 712,800×439.12 + 337,808×495

= 313,016,736 + 167,214,960

= 480,231,696 N·mm

Mu = 480.2 kN·m

Verification using two-couple model:

Mu,lim for M25, Fe 415 = 3.45 × 300 × 550² = 3.45 × 300 × 302,500 = 313,087,500 N·mm = 313.1 kN·m

ΔM = Cs×(d−d’) = 337,808×495 = 167,214,960 N·mm = 167.2 kN·m

Mu = 313.1 + 167.2 = 480.3 kN·m ✓

11. Worked Example 3 — GATE-Style Problem

Problem (GATE-style): A doubly reinforced beam (b = 250 mm, d = 450 mm, d’ = 45 mm) is made with M20 concrete and Fe 415 steel. The compression steel Asc = 400 mm². What is the additional moment capacity provided by the compression steel over the singly reinforced limiting moment? (Use fsc = 353 N/mm² for d’/d = 0.10)

Solution

d’/d = 45/450 = 0.10 → fsc = 353 N/mm² ✓

fsc − 0.36fck = 353 − 0.36×20 = 353 − 7.2 = 345.8 N/mm²

Lever arm for compression couple = d − d’ = 450 − 45 = 405 mm

Additional moment = Cs × (d − d’) = (fsc − 0.36fck) × Asc × (d − d’)

= 345.8 × 400 × 405

= 345.8 × 162,000

= 56,019,600 N·mm

ΔM = 56.02 kN·m

This type of problem is the most common GATE DRB question — the examiner gives Asc and asks for the additional moment. The answer is always: ΔM = (fsc − 0.36fck) × Asc × (d − d’).

12. Practical Design Notes

Minimum Area of Compression Steel — IS 456 Cl. 26.5.1.2

IS 456 does not specify a mandatory minimum for compression steel in beams (it is provided as needed by design). However, where compression steel is provided:

It must be enclosed by ties (lateral ties similar to column ties) to prevent buckling of the compression bars.

Minimum tie diameter = φsc/4 or 6 mm, whichever is greater (same rule as column ties).

Maximum tie spacing = lesser of 12 × compression bar diameter, least compression dimension, or 300 mm.

When fsc = 0.87fy is Approximately Valid

If d’/d is very small (d’/d ≤ 0.05 to 0.07), the compression steel strain is close to or exceeds the yield strain, and fsc approaches 0.87fy. In such cases, using fsc = 0.87fy slightly overestimates the compression steel contribution — it is slightly unconservative but acceptable for quick calculations.

For Fe 415 with xu,max/d = 0.48:

εsc = 0.0035 × (0.48 − d’/d)/0.48

For fsc = 0.87fy = 361 N/mm², need εsc ≥ 0.00380:

0.0035 × (0.48 − d’/d)/0.48 ≥ 0.00380 → d’/d ≤ 0

This is never satisfied for realistic d’/d values. So for Fe 415, fsc is always < 0.87fy — always use the tabulated fsc values.

AspectSingly ReinforcedDoubly Reinforced
Steel locationTension zone onlyBoth tension and compression zones
Max moment capacityMu,lim = Ru,lim × bd²Mu,lim + (fsc−0.36fck) Asc(d−d’)
Neutral axisxu varies with Astxu = xu,max (for design with min Asc)
DeflectionMore deflection (less stiff)Less deflection (compression steel reduces creep)
DuctilityGood (under-reinforced)Better (compression steel improves ductility)
CostLower (less steel)Higher (more steel, more complex)
Use whenMu ≤ Mu,limMu > Mu,lim and depth is constrained

13. Common Mistakes Students Make

  • Using fsc = 0.87fy for compression steel: This is the single most common error in doubly reinforced beam problems. The compression steel stress fsc is less than 0.87fy for realistic d’/d values with Fe 415 and Fe 500 steel. For Fe 415 with d’/d = 0.10, fsc = 353 N/mm², not 361 N/mm². Always determine fsc from the d’/d ratio and IS 456 table, or use 353 N/mm² as the standard GATE CE assumption.
  • Forgetting to subtract the displaced concrete in Cs: The compression force from compression steel is (fsc − 0.36fck) × Asc, not fsc × Asc. The 0.36fck term subtracts the concrete that the steel bars displaced — without this deduction, the compression of that concrete area is double-counted. This error gives a slightly over-optimistic moment capacity.
  • Using (d − 0.42xu) as the lever arm for the compression steel couple: The lever arm for the compression steel couple is (d − d’) — from the tension steel to the compression steel. The lever arm (d − 0.42xu) applies to the concrete compressive force only. Using the wrong lever arm for the compression steel term significantly over- or underestimates ΔM.
  • Not checking whether the section is over-reinforced after providing actual bars: When you round up bar areas (as required in practice), the provided Ast and Asc may push xu slightly beyond xu,max. Always verify with the provided areas. If over-reinforced, cap xu at xu,max — the moment capacity is Mu,lim + Cs(d − d’) evaluated at xu,max.
  • Designing Ast2 separately without verifying against Asc: The additional tension steel Ast2 and compression steel Asc form a couple. Their forces must be equal: (fsc − 0.36fck) × Asc = 0.87fy × Ast2. If you use provided (rounded) Asc but calculated (unreduced) Ast2, there is an imbalance. Always recalculate Ast2 based on the actual provided Asc.

14. Frequently Asked Questions

Why is the neutral axis always at xu,max in a doubly reinforced beam design?

In the design problem, we choose the minimum amount of compression steel needed to carry the excess moment (Mu − Mu,lim). The singly reinforced part of the design already pushes the neutral axis to xu,max — that is the definition of Mu,lim. Adding compression steel increases the total compression force, which would push the neutral axis deeper if Ast were kept the same. To maintain the balanced condition (xu = xu,max), we also add extra tension steel Ast2 to match the extra compression Cs. So both Asc and Ast2 are added together, and the neutral axis stays at xu,max. If the provided steel areas differ from the minimum design values (due to bar size rounding), xu will shift slightly from xu,max, which must then be checked in the analysis direction.

Can the compression steel in a DRB ever yield?

For Fe 415 and Fe 500, the compression steel strain at the balanced condition is less than the yield strain for typical d’/d values (0.05 to 0.20). So technically, the compression steel does not fully yield in these cases. For Fe 250 (mild steel), the yield strain is lower, and the compression steel does reach yield — which is why fsc = 217 N/mm² (the permissible stress, close to full yield) for Fe 250 regardless of d’/d. In practice, engineers often conservatively assume fsc = 0.87fy for Fe 250 since it is so close to yield.

What is the difference between a doubly reinforced beam and a T-beam?

Both have steel in the compression zone, but for very different reasons. A doubly reinforced beam is a rectangular section where compression steel Asc is deliberately added to supplement the concrete’s compressive capacity when the moment exceeds Mu,lim. A T-beam is a beam that is physically part of a floor slab system — the slab acts as a wide compression flange, dramatically increasing the effective compression area. In a T-beam, no compression reinforcement is typically needed because the large flange area provides far more than enough compression capacity. The distinction is: DRB = rectangular section with added compression steel; T-beam = non-rectangular section (T-shaped) taking advantage of the slab flange.

How does compression steel help reduce long-term deflection?

Long-term deflection in RCC beams is dominated by creep — the gradual increase in strain in the compression zone under sustained loading. Concrete creeps significantly under compression; steel does not. Compression steel shares the compression load with the concrete, reducing the sustained compressive stress in the concrete and therefore reducing creep strain. IS 456 Cl. 23.2.1 acknowledges this by allowing higher span-to-depth ratios (less deflection concern) when compression steel is present — the modification factor Mc in Fig. 5 of IS 456 increases the allowable span/d ratio by up to 1.5 times depending on the compression steel percentage. For heavily loaded continuous beams where long-term deflection is the governing design criterion, compression steel is often added even when the strength is adequate, purely to control deflection.

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