T-Beam & L-Beam

Effective Flange Width, Neutral Axis in Flange vs Web, Moment of Resistance & Solved GATE CE Examples — IS 456:2000

Last Updated: March 2026

Key Takeaways 📌

  • In a typical floor system, beams and slabs are cast monolithically — the slab acts as a compression flange for the beam, making it a T-beam (intermediate beams) or L-beam (edge beams).
  • The slab flange dramatically increases the effective compression area, making the section far more efficient than a rectangular beam of the same web width.
  • The effective flange width bf is determined from IS 456 Cl. 23.1 — it is the width over which the slab is considered to act compositely with the beam web.
  • The critical design decision: Is the neutral axis within the flange (xu ≤ Df) or in the web (xu > Df)?
  • If NA is within the flange: design as a rectangular beam of width bf and depth d — simple and most common for sagging sections.
  • If NA is in the web: a modified formula is needed that accounts for the flange and web separately — the stress block is no longer uniform over the full compression width.
  • For GATE CE: most T-beam problems have NA in the flange — always check this first. The NA-in-web case appears occasionally as a more challenging problem.

1. What is a T-Beam? — Physical Concept

In modern building construction, floor beams and slabs are almost always cast as a single monolithic unit — the concrete is poured continuously from the beam web up through the slab. As a result, when the beam is loaded and develops a sagging bending moment (concave upward), the top portion — which is in compression — includes not just the beam web but also a width of the slab on either side.

This composite action between the beam web and the adjacent slab creates an effective T-shaped cross-section in the compression zone. The wide slab portion at the top is called the flange and the narrower beam stem below is called the web. The resulting section resembles the letter T in cross-section — hence the name T-beam.

The structural advantage of the T-section is significant. The large flange area provides a massive compression zone, which means the neutral axis is very shallow — often within the flange depth itself. A shallow neutral axis means a large lever arm between the compression resultant and the tension steel, which gives a very high moment capacity for the amount of tension steel used. T-beams are therefore among the most material-efficient flexural members in structural engineering.

Consider a 250 mm wide × 550 mm deep rectangular beam. Its limiting moment capacity is approximately 2.76 × 250 × 500² ≈ 172 kN·m (Fe 415, M20). The same web dimensions with a 1500 mm wide, 120 mm deep slab flange acting compositely might have a moment capacity of 400–500 kN·m — three times more capacity for the same amount of web concrete.

However, only a portion of the slab on each side of the beam actually participates in carrying the compression — the stress in the slab decreases as you move farther from the beam web (shear lag effect). IS 456 accounts for this by defining an effective flange width bf, which is the width of slab that can be assumed to act uniformly at the full compressive stress. The actual slab width beyond bf is ignored in design.

2. T-Beam vs L-Beam vs Isolated Beam

Beam TypeLocation in FloorFlange ConfigurationIS 456 Reference
T-beamIntermediate beam — slab on both sidesSymmetric flange on both sides of webCl. 23.1(a)
L-beamEdge/spandrel beam — slab on one side onlyFlange on one side only — asymmetricCl. 23.1(b)
Isolated beamNot connected to slab, or slab not consideredNo flange — rectangular section onlyCl. 23.1(c)

In practice, the same physical beam may behave as a T-beam for positive (sagging) moments and as a rectangular beam (of web width bw) for negative (hogging) moments at supports — because in hogging, the slab is in tension (below the neutral axis) and the web is in compression. The tension in the slab concrete is ignored (cracked), so the effective compression section is just the web.

3. Effective Flange Width — IS 456 Cl. 23.1

The effective flange width bf is the width of the slab assumed to act as the compression flange of the T-beam or L-beam. IS 456 gives separate formulae for T-beams and L-beams, and applies a further restriction that bf cannot exceed the actual slab width available on each side.

Effective Flange Width — T-Beam (IS 456 Cl. 23.1(a))

bf = lo/6 + bw + 6Df

Subject to: bf ≤ actual width of slab available (centre-to-centre distance between beams)

Where:

lo = distance between points of zero moment (contraflexure) in the beam [mm]

For simply supported beams: lo = effective span L

For continuous beams: lo ≈ 0.7 × effective span (approximation for internal spans)

bw = breadth of web [mm]

Df = thickness of flange (slab depth) [mm]

The three terms represent: (lo/6) = contribution from slab span, (bw) = web itself, (6Df) = contribution from slab thickness, totalling the effective width.

Effective Flange Width — L-Beam (IS 456 Cl. 23.1(b))

bf = lo/12 + bw + 3Df

Subject to: bf ≤ bw + half the clear distance to the next beam

The coefficients are halved compared to the T-beam formula because the L-beam has a flange on one side only.

Isolated T-Beam (IS 456 Cl. 23.1(c))

For an isolated beam (not connected to slab, or used as an independent member):

bf = bw + lo/5

Subject to: bf ≤ (bw + 4 × Df) and bf ≤ actual flange width available

Practical note: In GATE CE problems, the effective flange width bf is usually given directly. The effective flange width calculation from IS 456 Cl. 23.1 appears occasionally as a separate question or as part of a multi-step problem.

4. Case 1 — Neutral Axis in Flange (xu ≤ Df)

When the neutral axis lies within the flange (xu ≤ Df), the entire compression zone is within the uniform flange. The web below the flange is entirely in tension (cracked and ignored). In this case, the T-beam behaves exactly like a rectangular beam of width bf and effective depth d. All the rectangular beam formulae apply directly with b replaced by bf.

NA in Flange — All Formulae

Neutral axis depth:

xu = 0.87 fy Ast / (0.36 fck × bf)

Check: If xu ≤ Df → assumption confirmed ✓

Moment of resistance:

Mu = 0.87 fy Ast (d − 0.42 xu)

OR: Mu = 0.36 fck bf xu (d − 0.42 xu)

Limiting moment (balanced section):

Mu,lim = 0.36 fck bf xu,max (d − 0.42 xu,max) = Ru,lim × bf × d²

These are identical to the singly reinforced rectangular beam formulas — just with bf instead of b.

Why is this case most common? In practice, the flange is wide and the slab is thick enough that the neutral axis almost always falls within the flange depth for sagging moments. A typical floor slab is 100–200 mm thick, and the flange width can be 1000–2000 mm. The neutral axis depth for typical tension steel areas in a floor beam is often only 50–100 mm — well within the flange. This is the principal reason T-beams are so efficient: the deep neutral axis of a rectangular beam is avoided, keeping the NA shallow and the lever arm large.

5. Case 2 — Neutral Axis in Web (xu > Df)

When the neutral axis lies below the flange (xu > Df), the compression zone extends into the web. Now the compression area is no longer uniform — the flange provides uniform compression over width bf up to depth Df, and the web provides compression over width bw from Df to xu. The total compressive force is the sum of contributions from both regions.

IS 456 Simplified Formula — NA in Web

IS 456 Annex G provides a simplified approach. The total moment of resistance is split into two parts:

Part 1 — Web acting as rectangular beam of width bw:

Mu1 = 0.36 fck bw xu (d − 0.42 xu)

Part 2 — Flange overhang contribution (bf − bw) acting as added compression:

Mu2 = 0.45 fck (bf − bw) yf (d − yf/2)

Where yf = effective depth of flange in compression (IS 456 approximation):

yf = 0.15 xu + 0.65 Df (IS 456 Annex G, Eq. G-2)

But yf ≤ Df (flange cannot exceed its actual thickness)

Total Mu = Mu1 + Mu2

Total compressive force = Tensile force (for force equilibrium):

0.36 fck bw xu + 0.45 fck (bf − bw) yf = 0.87 fy Ast

This equation is used to find xu iteratively (since yf also depends on xu).

Alternative — Direct Integration Approach

Some textbooks use direct force equilibrium without the yf approximation:

Total compression = Flange contribution + Web contribution

Cflange = 0.36 fck × bf × Df (if xu > Df, the whole flange is compressed)

Cweb = 0.36 fck × bw × (xu − Df) (web compression below flange)

Total C = Cflange + Cweb = 0.87 fy Ast = T

This direct approach is used for analysis when xu is given or when the IS 456 approximation is not required by the problem.

Moment about T: Mu = Cflange(d − Df/2) + Cweb(d − Df − (xu−Df)/2 ×0.42… )

More carefully: Cflange acts at Df/2 from top. Cweb acts at Df + 0.42(xu−Df) from top (centroid of rectangular stress block in web).

Mu = Cflange × (d − Df/2) + Cweb × (d − Df − 0.42(xu−Df))

This approach is conceptually clean and is preferred for GATE CE problems asking for exact analysis.

6. How to Determine NA Location — Decision Flowchart

For any T-beam or L-beam problem, follow this decision process:

Step-by-Step NA Location Check

Step 1: Assume NA is in the flange (xu ≤ Df).

Step 2: Calculate xu assuming full flange width bf:

xu = 0.87 fy Ast / (0.36 fck × bf)

Step 3: Compare xu with Df:

If xu ≤ Df → NA is in flange ✓ — use rectangular beam formula with width bf.

If xu > Df → NA is in web — must use the more complex flanged beam formula.

For design problems (finding Ast): First check if the required xu for the given Mu falls within the flange. For most practical floor beams with wide flanges, the NA will be in the flange and the design reduces to a rectangular beam problem with b = bf.

Quick check: If the moment capacity of the flange alone (assuming xu = Df, rectangular beam of width bf and depth Df) is greater than Mu, then the NA is definitely in the flange. Compute Mflange = 0.36 fck × bf × Df × (d − 0.42 Df). If Mu ≤ Mflange, NA is in flange.

7. Design Procedure — T-Beam

Given: Design moment Mu, span, bf, bw, Df, d, fck, fy.

  1. Check if NA is likely in flange: Calculate Mflange = 0.36fck × bf × Df × (d − 0.42Df). If Mu ≤ Mflange, NA is in flange — proceed as rectangular beam of width bf.
  2. If NA in flange — design as SRB with b = bf:
    • Find xu from Mu = 0.36fckbfxu(d − 0.42xu)
    • Check xu ≤ Df (confirm assumption)
    • Check xu ≤ xu,max (confirm under-reinforced)
    • Find Ast = 0.36fckbfxu / (0.87fy)
  3. If NA in web — use IS 456 flanged beam formula:
    • Set up force equilibrium with yf = 0.15xu + 0.65Df
    • Solve iteratively for xu
    • Find Ast from equilibrium
  4. Check minimum steel: Ast,min = 0.85bwd/fy (based on web width, not flange width)
  5. Select bars and check spacing

8. Analysis Procedure — Finding Mu

  1. Assume NA in flange. Compute xu = 0.87fyAst / (0.36fckbf)
  2. If xu ≤ Df: NA in flange confirmed. Mu = 0.87fyAst(d − 0.42xu). Done.
  3. If xu > Df: NA is in web. Use direct force equilibrium:
    • Cflange = 0.36fck(bf − bw)Df (overhang only)
    • Cweb = 0.36fckbwxu
    • Total C = Cflange + Cweb = T = 0.87fyAst
    • Solve for xu: xu = [0.87fyAst − 0.36fck(bf−bw)Df] / (0.36fckbw)
    • Check xu ≤ xu,max
    • Mu = Cflange(d − Df/2) + Cweb(d − 0.42xu)

9. Worked Example 1 — NA in Flange (Analysis)

Problem: A T-beam has bf = 1200 mm, bw = 300 mm, Df = 120 mm, d = 550 mm. Ast = 1884 mm² (6 bars of 20 mm dia). fck = 20 N/mm², fy = 415 N/mm². Find Mu.

Step 1 — Assume NA in Flange, Find xu

xu = 0.87 × 415 × 1884 / (0.36 × 20 × 1200)

= 679,869.6 / 8640

xu = 78.7 mm

Step 2 — Check NA Location

xu = 78.7 mm < Df = 120 mm → NA is in flange ✓

xu,max = 0.48 × 550 = 264 mm → xu = 78.7 mm << xu,max → Heavily under-reinforced

(This is typical for T-beams — large bf keeps xu very small)

Step 3 — Moment of Resistance Mu

Mu = 0.87 × 415 × 1884 × (550 − 0.42 × 78.7)

= 679,869.6 × (550 − 33.05)

= 679,869.6 × 516.95

= 351,638,000 N·mm

Mu = 351.6 kN·m

Compare with rectangular beam of same web (b = 300 mm):

xu = 0.87×415×1884/(0.36×20×300) = 679,870/2160 = 314.8 mm > xu,max = 264 mm → over-reinforced!

Mu for rectangular 300mm wide beam = Mu,lim = 2.76×300×550² = 249.7 kN·m

The T-beam provides 351.6 kN·m vs 249.7 kN·m for the same steel — 41% more capacity.

10. Worked Example 2 — NA in Web (Analysis)

Problem: A T-beam has bf = 800 mm, bw = 250 mm, Df = 100 mm, d = 500 mm. Ast = 3927 mm² (5 bars of 32 mm dia). fck = 25 N/mm², fy = 415 N/mm². Find Mu.

Step 1 — Check if NA is in Flange

xu assuming NA in flange (use bf):

xu = 0.87 × 415 × 3927 / (0.36 × 25 × 800)

= 1,417,099 / 7200 = 196.8 mm

xu = 196.8 mm > Df = 100 mm → NA is NOT in flange — NA is in web.

Step 2 — Find Actual xu (NA in Web)

Using direct force equilibrium:

Cflange = 0.36 × 25 × (800 − 250) × 100 = 9 × 550 × 100 = 495,000 N

Cweb = 0.36 × 25 × 250 × xu = 2250 xu

T = 0.87 × 415 × 3927 = 1,417,099 N

Force equilibrium: Cflange + Cweb = T

495,000 + 2250 xu = 1,417,099

2250 xu = 922,099

xu = 409.8 mm

Wait — xu,max = 0.48 × 500 = 240 mm. xu = 409.8 mm > xu,max → Over-reinforced!

Use xu = xu,max = 240 mm for moment calculation.

Verify: Cflange = 495,000 N (unchanged), Cweb = 2250 × 240 = 540,000 N

Total C = 495,000 + 540,000 = 1,035,000 N ≠ T = 1,417,099 N → confirms over-reinforced (the section cannot fully utilise the tension steel)

Step 3 — Moment of Resistance (at xu = xu,max = 240 mm)

Cflange acts at Df/2 = 50 mm from top → lever arm = d − 50 = 450 mm

Cweb acts at Df + 0.42(xu − Df) from top

= 100 + 0.42 × (240 − 100) = 100 + 58.8 = 158.8 mm from top

Lever arm for Cweb = d − 158.8 = 500 − 158.8 = 341.2 mm

Mu = Cflange × 450 + Cweb × 341.2

= 495,000 × 450 + 540,000 × 341.2

= 222,750,000 + 184,248,000

= 406,998,000 N·mm

Mu = 407.0 kN·m

11. Worked Example 3 — T-Beam Design

Problem: Design a T-beam to carry Mu = 320 kN·m. Given: bf = 1000 mm, bw = 250 mm, Df = 110 mm, d = 520 mm. fck = 20 N/mm², fy = 415 N/mm².

Step 1 — Check NA Location

Capacity with NA at Df = 110 mm:

Mflange = 0.36 × 20 × 1000 × 110 × (520 − 0.42 × 110)

= 792,000 × (520 − 46.2) = 792,000 × 473.8

= 375,249,600 N·mm = 375.2 kN·m

Mu = 320 kN·m < Mflange = 375.2 kN·m

→ NA is within the flange. Design as rectangular beam with b = bf = 1000 mm.

Step 2 — Find xu

Mu = 0.36 fck bf xu (d − 0.42 xu)

320 × 10⁶ = 0.36 × 20 × 1000 × xu × (520 − 0.42 xu)

320 × 10⁶ = 7200 xu (520 − 0.42 xu)

320 × 10⁶ = 3,744,000 xu − 3024 xu²

3024 xu² − 3,744,000 xu + 320,000,000 = 0

xu² − 1238.1 xu + 105,820.1 = 0

xu = [1238.1 − √(1238.1² − 4 × 105,820.1)] / 2

= [1238.1 − √(1,532,991.6 − 423,280.4)] / 2

= [1238.1 − √1,109,711.2] / 2

= [1238.1 − 1053.4] / 2 = 184.7/2 = 92.35 mm

Check: xu = 92.35 mm < Df = 110 mm ✓ (NA in flange confirmed)

xu,max = 0.48 × 520 = 249.6 mm → xu = 92.35 mm ≪ xu,max ✓ (under-reinforced)

Step 3 — Find Ast

Ast = 0.36 fck bf xu / (0.87 fy)

= (0.36 × 20 × 1000 × 92.35) / (0.87 × 415)

= 664,920 / 361.05

Ast = 1840.3 mm²

Provide 4 bars of 25 mm dia: Ast = 4 × 490.87 = 1963.5 mm² ✓

Step 4 — Check Minimum Steel

Ast,min = 0.85 × bw × d / fy = 0.85 × 250 × 520 / 415 = 265.8 mm²

Ast,provided = 1963.5 mm² ≫ 265.8 mm² ✓

Final: 4 bars of 25 mm dia in the tension zone of the T-beam.

12. Worked Example 4 — Effective Flange Width Calculation

Problem: A simply supported T-beam has a clear span of 6 m, effective span 6.3 m. Web width bw = 300 mm. Slab thickness Df = 120 mm. Centre-to-centre spacing between beams = 2.4 m. Find the effective flange width bf.

Solution — IS 456 Cl. 23.1(a)

For a simply supported beam: lo = effective span = 6300 mm

bf = lo/6 + bw + 6Df

= 6300/6 + 300 + 6 × 120

= 1050 + 300 + 720

= 2070 mm

Check against actual slab width:

C/C spacing between beams = 2400 mm → actual slab width available = 2400 mm

bf = min(2070, 2400) = 2070 mm (governs)

Effective overhang on each side = (bf − bw)/2 = (2070 − 300)/2 = 885 mm

Actual overhang available = (2400 − 300)/2 = 1050 mm > 885 mm ✓

13. L-Beam Design

An L-beam (spandrel or edge beam) has the slab on one side only. The effective flange width for an L-beam is smaller than for a T-beam of the same span, because only one side of the flange is available.

L-Beam Effective Flange Width (IS 456 Cl. 23.1(b))

bf = lo/12 + bw + 3Df

Subject to: bf ≤ bw + (half the clear distance to the next beam)

The design procedure for an L-beam is identical to a T-beam, with the appropriate (smaller) bf.

For the same section dimensions, an L-beam will have a smaller effective compression area than a T-beam, so the NA will be slightly deeper and the moment capacity slightly lower for the same steel area.

NA Check for L-Beam — Same Logic

The same two-case approach applies:

Case 1 (NA in flange): xu = 0.87fyAst / (0.36fckbf)

If xu ≤ Df → use bf as the effective width, treat as rectangular beam.

Case 2 (NA in web): Use the same flanged beam formula with the L-beam bf.

The only difference from T-beam analysis is the value of bf — everything else is identical.

14. Common Mistakes Students Make

  • Using bw instead of bf for the NA calculation: The most common error. When checking whether NA is in the flange, you must use the full effective flange width bf to calculate xu. Using the web width bw gives a much larger xu — it makes the NA appear to be in the web when it is actually well within the flange. Always start the NA check with bf.
  • Treating the T-beam as a rectangular beam of web width bw even when NA is in the flange: When the neutral axis is within the flange, the T-beam behaves as a rectangular beam of width bf, not bw. The web is entirely in tension and plays no role in the compression calculation. Using bw in this case severely underestimates the moment capacity and leads to unnecessary over-design.
  • Using the full slab width as the effective flange width: The effective flange width bf from IS 456 Cl. 23.1 is always ≤ the actual centre-to-centre distance between beams. Do not use the full slab span as bf without checking against the IS 456 formula — the actual effective width may be significantly less than the full slab width, especially for short-span beams or thick slabs.
  • Wrong lever arm for the NA-in-web case: When NA is in the web, the flange force Cflange acts at Df/2 from the top, and the web force Cweb acts at a different location. Students often use (d − 0.42xu) as the lever arm for both forces — this is only valid for the web compression block. The flange force lever arm is (d − Df/2), which is different and must be used separately.
  • Not checking minimum steel based on bw: IS 456 specifies minimum tension steel as Ast,min = 0.85bwd/fy — using the web width bw, not the flange width bf. Students sometimes use bf for minimum steel calculation, which gives a much larger (and unnecessarily conservative) minimum steel requirement.

15. Frequently Asked Questions

Why is the T-beam so much more efficient than a rectangular beam?

The efficiency of a T-beam comes from the large compression area provided by the flange. In a rectangular beam, the compression zone deepens as more tension steel is added — the neutral axis moves down, reducing the lever arm and limiting the moment capacity. In a T-beam, the wide flange provides a very large compression area, keeping the neutral axis very shallow even for large amounts of tension steel. A shallow neutral axis means a large lever arm (d − 0.42xu approaches d), so the same amount of tension steel produces a much larger moment. In essence, the T-beam uses the slab’s compressive area (which exists anyway as part of the floor system) to enhance the beam’s efficiency — no additional concrete is consumed.

What happens to a T-beam at a continuous support (hogging moment)?

At an interior support of a continuous T-beam, the moment is hogging (concave downward). The top of the beam is now in tension and the bottom is in compression. The slab, which is at the top, is in tension — cracked concrete does not carry tension, so the slab flange is ignored. The effective compression section is just the rectangular web (bw × depth in compression). At supports, the beam is effectively designed as a rectangular beam of width bw, with tension steel at the top (in the slab region) and compression zone in the bottom of the web. This is why continuous T-beams require top steel at supports — the structural situation is completely different from the midspan sagging region.

What is the significance of the IS 456 formula yf = 0.15xu + 0.65Df?

The formula yf = 0.15xu + 0.65Df (IS 456 Annex G) is an empirical approximation for the equivalent depth of the flange in compression when the neutral axis is below the flange. In reality, the stress in the flange is not uniform — the stress block extends uniformly up to 0.42xu from the compression face, but the flange thickness Df may be less than or greater than 0.42xu depending on xu. The yf formula accounts for the parabolic shape of the stress block at the flange-web junction by interpolating between the flange depth Df and 0.42xu. For most practical cases where the difference between yf and Df is small, the simpler direct force equilibrium method gives nearly identical results and is preferred for hand calculation.

Can a T-beam be doubly reinforced?

Yes, though it is uncommon. A doubly reinforced T-beam would have compression steel in the flange (at the top of the web, near the underside of the slab) in addition to the tension steel. This would be needed if the T-beam’s moment capacity (even with the full flange acting) is insufficient for the applied moment — a rare situation given the large compression capacity of the flange. More commonly, if a T-beam section is insufficient, the web dimensions are increased or additional tension steel is added while remaining within the under-reinforced limit. The analysis of a doubly reinforced T-beam combines the T-beam flanged section analysis with the compression steel treatment from the doubly reinforced beam theory.

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