RCC Beams (Working Stress Method and Limit State Method)

What to remember

• Two design philosophies: the Working Stress Method (WSM) keeps stresses below permissible limits under working loads; the Limit State Method (LSM, IS 456:2000) uses factored loads, partial safety factors and a stress block at ultimate.
• Under-reinforced beams are preferred because steel yields first, giving a ductile failure with warning. Over-reinforced beams fail suddenly by concrete crushing.
• Key LSM numbers: xu,max/d = 0.53 (Fe 250), 0.48 (Fe 415), 0.46 (Fe 500); partial safety factors 1.5 (concrete) and 1.15 (steel); ultimate concrete strain 0.0035.

1. Basic ideas of reinforced concrete

Concrete is strong in compression and weak in tension. Steel bars carry the tension. A beam bends, so the top fibres (in a simply supported beam) are in compression and the bottom fibres in tension. The neutral axis (N.A.) is the level where bending stress is zero.

Grades are named M15, M20, M25 and so on; the number is the characteristic compressive strength fck in N/mm² (cube strength at 28 days). Reinforcing steel is Fe 250 (mild), Fe 415, Fe 500 and Fe 550; the number is the yield strength fy in N/mm². The modulus of elasticity of steel is Es = 2 × 10⁵ N/mm². For concrete, Ec = 5000 √fck N/mm².

Effective depth d = overall depth D − effective cover (clear cover + half the bar diameter). Example: D = 500 mm, clear cover 30 mm, 20 mm bar gives d = 500 − 30 − 10 = 460 mm.

2. Working Stress Method (WSM)

Assumptions: plane sections remain plane; concrete takes no tension; stress is proportional to strain (elastic); perfect bond between steel and concrete.

• Modular ratio m = 280 / (3 σcbc), where σcbc is the permissible bending compressive stress in concrete. For M20, σcbc = 7 N/mm², so m = 13.33. For M15, σcbc = 5 and m = 18.67.
• Steel is replaced by an equivalent concrete area m·Ast.
• Neutral axis: b x²/2 = m Ast (d − x). The neutral axis factor k = x/d = m σcbc / (m σcbc + σst).
• Lever arm factor j = 1 − k/3. Moment of resistance (concrete side) Mr = ½ σcbc b x (d − x/3) = Q b d², where Q = ½ σcbc k j.
• Moment (steel side) M = σst Ast j d.

Typical design values (M20 concrete):

Steel	σst (N/mm²)	k	j	Q
Fe 250	140	0.400	0.867	1.21
Fe 415	230	0.289	0.904	0.91
Fe 500	275	0.253	0.916	0.81

Balanced section: concrete and steel reach their permissible stresses together. Under-reinforced: steel reaches its limit first (actual N.A. above the balanced N.A.). Over-reinforced: concrete reaches its limit first.

Worked example (WSM): b = 250 mm, d = 450 mm, M20 and Fe 415. Q = 0.91, so balanced Mr = 0.91 × 250 × 450² = 46.1 × 10⁶ N·mm, about 46 kN·m.

3. Limit State Method: stress block and partial factors

Limit states: collapse (flexure, shear, torsion, compression) and serviceability (deflection, cracking).

Partial safety factors: material γc = 1.5 for concrete, γs = 1.15 for steel; load factor 1.5 for dead plus live load (1.2 when wind or earthquake is also considered together with DL + LL). Design strength of steel = fy / 1.15 = 0.87 fy.

The rectangular-parabolic stress diagram of concrete is replaced by an equivalent rectangle:

• Average compressive stress = 0.36 fck over depth xu.
• Resultant compressive force C = 0.36 fck b xu.
• Centroid of C lies 0.42 xu from the extreme compression fibre.
• Strain at the top fibre in flexure = 0.0035.

Limiting neutral axis depth (the section becomes over-reinforced if xu exceeds it):

Steel	xu,max / d	Mu,lim / (fck b d²)
Fe 250	0.53	0.148
Fe 415	0.48	0.138
Fe 500	0.46	0.133

Formulas for a singly reinforced rectangular beam:

• Neutral axis: xu = 0.87 fy Ast / (0.36 fck b).
• Moment of resistance: Mu = 0.87 fy Ast (d − 0.42 xu), valid when xu ≤ xu,max.
• Also Mu = 0.87 fy Ast d (1 − Ast fy / (b d fck)).
• Limiting moment: Mu,lim = 0.36 fck b xu,max (d − 0.42 xu,max).

Worked example: b = 250 mm, d = 450 mm, M20, Fe 415, Ast = 1000 mm². xu = 361.05 × 1000 / (0.36 × 20 × 250) = 361,050/1800 ≈ 200 mm, which is below xu,max = 0.48 × 450 = 216 mm, so the section is under-reinforced. Mu = 361,050 × (450 − 0.42 × 200) ≈ 132 kN·m. Mu,lim = 0.138 × 20 × 250 × 450² ≈ 139.7 kN·m.

4. Doubly reinforced and flanged beams

Doubly reinforced beam: needed when the factored moment exceeds Mu,lim and the section size cannot be increased. Extra tension steel is balanced by compression steel Asc placed near the top. Compression steel also helps reduce long-term deflection.

Flanged (T and L) beams: the slab acts with the beam. Effective flange width bf:

• T-beam (continuous): bf = lo/6 + bw + 6 Df
• L-beam (continuous): bf = lo/12 + bw + 3 Df
• Isolated T-beam: bf = lo / (lo/b + 4) + bw

Here lo is the distance between points of zero moment (for a simply supported beam, the span), bw is the web width and Df the slab (flange) thickness. bf must not exceed the actual spacing of the beams. If xu ≤ Df the N.A. lies in the flange and the section behaves as a rectangle of width bf.

Example: lo = 6000 mm, bw = 300 mm, Df = 100 mm: T-beam bf = 1000 + 300 + 600 = 1900 mm; L-beam bf = 500 + 300 + 300 = 1100 mm.

5. Shear, torsion and bond

Nominal shear stress τv = Vu / (b d). The design shear strength of concrete τc depends on pt = 100 Ast/(b d) and fck. If τv ≤ τc minimum stirrups are provided; if τc < τv ≤ τc,max stirrups are designed; if τv > τc,max the section must be enlarged. Maximum shear stress τc,max: M20 = 2.8, M25 = 3.1, M30 = 3.5 N/mm².

• Strength of vertical stirrups: Vus = 0.87 fy Asv d / sv. Required spacing sv = 0.87 fy Asv d / Vus, with Vus = Vu − τc b d.
• Minimum shear reinforcement: Asv / (b sv) ≥ 0.4 / (0.87 fy).
• Maximum stirrup spacing: 0.75 d or 300 mm, whichever is less.
• Example: 2-legged 8 mm stirrups (Asv = 100.5 mm²), fy = 415, d = 500, sv = 150 mm: Vus = 361.05 × 100.5 × 500 / 150 ≈ 121 kN.

Torsion: design for equivalent shear Ve = Vu + 1.6 Tu / b and equivalent moment Me1 = Mu + Tu (1 + D/b)/1.7. Example: Vu = 100 kN, Tu = 10 kN·m, b = 250 mm gives Ve = 100 + 64 = 164 kN.

Bond and development length: Ld = 0.87 fy φ / (4 τbd). The design bond stress τbd for plain bars in M20 is 1.2 N/mm²; for deformed bars it is increased by 60 percent (1.92 N/mm²). For deformed Fe 415 bars in M20, Ld ≈ 47 φ (a 16 mm bar needs about 752 mm). In compression, τbd is increased by 25 percent.

6. Detailing and serviceability rules

Item	Rule
Minimum tension steel	As / (b d) ≥ 0.85 / fy (Fe 415: about 0.205%)
Maximum tension steel	4% of b D
Maximum compression steel	4% of b D
Clear cover (mild exposure)	20 mm; moderate 30 mm; severe 45 mm; very severe 50 mm; extreme 75 mm
Minimum clear spacing of bars	Greatest of bar diameter, or maximum aggregate size + 5 mm
Side face steel	If web depth exceeds 750 mm: 0.1% of web area, spread on both faces

Deflection control by span/effective depth ratio: basic values are 7 for a cantilever, 20 for a simply supported beam and 26 for a continuous beam (spans up to 10 m). They are multiplied by modification factors for tension steel, compression steel and flanges. Final deflection should not exceed span/250; deflection after partitions and finishes should not exceed span/350 or 20 mm, whichever is less. Crack width limit is about 0.3 mm in normal exposure.

Exam traps

• WSM modular ratio uses σcbc (not fck): m = 280/(3σcbc). Do not confuse with the Es/Ec ratio.
• Xu,max/d changes with the steel grade: 0.53, 0.48, 0.46. The lower the strength, the higher the limit.
• The stress block factors are 0.36 fck and 0.42 xu. Design stress of concrete 0.446 fck is the other form (0.67 fck / 1.5).
• Design steel stress is 0.87 fy, not fy.
• Partial safety factor for concrete is 1.5 and for steel 1.15; do not swap them.
• Minimum steel depends on fy (0.85/fy) but maximum steel is fixed at 4%.
• Maximum stirrup spacing is the lesser of 0.75 d and 300 mm; for inclined stirrups it is d.
• Deformed bars give 60% higher bond stress than plain bars; for compression, 25% more.

One-liners

• 1. Modular ratio m = 280 / (3 σcbc); 13.33 for M20.
• 2. In WSM, k = m σcbc / (m σcbc + σst) and j = 1 − k/3.
• 3. Ultimate concrete strain in flexure is 0.0035.
• 4. Stress block: 0.36 fck × xu, acting 0.42 xu below the top fibre.
• 5. xu,max/d is 0.53, 0.48 and 0.46 for Fe 250, Fe 415 and Fe 500.
• 6. Mu,lim for Fe 415 is 0.138 fck b d².
• 7. Minimum tension steel is 0.85 b d / fy; maximum is 0.04 b D.
• 8. Basic span/depth ratios: 7 (cantilever), 20 (simply supported), 26 (continuous).
• 9. Nominal shear stress is Vu / (b d).
• 10. Vus = 0.87 fy Asv d / sv for vertical stirrups.
• 11. Equivalent shear in torsion is Ve = Vu + 1.6 Tu / b.
• 12. Development length Ld = 0.87 fy φ / (4 τbd).

Practice questions

1. What is the modular ratio m for M20 concrete in the Working Stress Method (σcbc = 7 N/mm²)?

   1. 18.67
   2. 10.00
   3. 9.33
   4. 13.33
   Answer

   D. 13.33

   m = 280/(3σcbc) = 280/21 = 13.33.

2. For M20 concrete and Fe 415 steel (σst = 230 N/mm²) in WSM, the neutral axis factor k is nearly:

   1. 0.333
   2. 0.253
   3. 0.400
   4. 0.289
   Answer

   D. 0.289

   k = mσcbc/(mσcbc + σst) = 93.33/(93.33 + 230) = 0.289.

3. If the neutral axis factor k = 0.30, the lever arm factor j in WSM is:

   1. 0.95
   2. 0.90
   3. 0.70
   4. 0.85
   Answer

   B. 0.90

   j = 1 − k/3 = 1 − 0.10 = 0.90.

4. In a balanced section designed by WSM, which statement is correct?

   1. Steel reaches its permissible stress first
   2. Concrete reaches its permissible stress first
   3. Concrete and steel reach their permissible stresses at the same time
   4. Neither material is stressed to its limit
   Answer

   C. Concrete and steel reach their permissible stresses at the same time

   By definition a balanced section uses both materials fully at once.

5. Why is an under-reinforced beam preferred in design?

   1. It has a smaller neutral axis depth than needed
   2. Concrete crushes first, giving cheaper beams
   3. Steel yields first, giving ductile failure with warning
   4. It needs no stirrups
   Answer

   C. Steel yields first, giving ductile failure with warning

   Yielding of steel gives large deflection and cracks before collapse.

6. The limiting value of xu,max/d for Fe 415 steel in IS 456 is:

   1. 0.55
   2. 0.53
   3. 0.46
   4. 0.48
   Answer

   D. 0.48

   IS 456 gives 0.53 (Fe 250), 0.48 (Fe 415) and 0.46 (Fe 500).

7. The limiting value of xu,max/d for Fe 250 (mild steel) is:

   1. 0.46
   2. 0.53
   3. 0.60
   4. 0.48
   Answer

   B. 0.53

   Lower strength steel allows a deeper neutral axis: 0.53.

8. Mu,lim of a singly reinforced beam with b = 250 mm, d = 450 mm, M20 and Fe 415 is nearly:

   1. 168.0 kN·m
   2. 139.7 kN·m
   3. 100.0 kN·m
   4. 75.0 kN·m
   Answer

   B. 139.7 kN·m

   Mu,lim = 0.138 fck b d² = 0.138 × 20 × 250 × 450² = 139.7 × 10⁶ N·mm.

9. In the LSM stress block, the resultant compressive force acts at a distance from the top fibre of:

   1. 0.50 xu
   2. 0.67 xu
   3. 0.42 xu
   4. 0.36 xu
   Answer

   C. 0.42 xu

   Centroid of the equivalent stress block is 0.42 xu below the extreme compression fibre.

10. The partial safety factors for concrete and steel in IS 456 are respectively:

    1. 1.0 and 1.15
    2. 1.15 and 1.5
    3. 1.5 and 1.5
    4. 1.5 and 1.15
    Answer

    D. 1.5 and 1.15

    γc = 1.5 and γs = 1.15, so design steel stress is 0.87 fy.

11. The design yield stress of Fe 415 steel in LSM is nearly:

    1. 276 N/mm²
    2. 361 N/mm²
    3. 230 N/mm²
    4. 415 N/mm²
    Answer

    B. 361 N/mm²

    0.87 × 415 = 361 N/mm².

12. A beam of b = 250 mm has Ast = 1000 mm², fck = 20 and fy = 415. The neutral axis depth xu (nearest) is:

    1. 150 mm
    2. 250 mm
    3. 200 mm
    4. 300 mm
    Answer

    C. 200 mm

    xu = 0.87 fy Ast/(0.36 fck b) = 361,050/1800 ≈ 200 mm.

13. For the beam above with d = 450 mm and xu ≈ 200 mm, Mu = 0.87 fy Ast (d − 0.42 xu) is nearest to:

    1. 180 kN·m
    2. 110 kN·m
    3. 162 kN·m
    4. 132 kN·m
    Answer

    D. 132 kN·m

    361,050 × (450 − 84) = 132 × 10⁶ N·mm.

14. The minimum area of tension steel for a beam b = 300 mm, d = 500 mm with Fe 415 (As = 0.85 bd/fy) is nearly:

    1. 600 mm²
    2. 307 mm²
    3. 150 mm²
    4. 450 mm²
    Answer

    B. 307 mm²

    0.85 × 300 × 500/415 = 307 mm².

15. The maximum area of tension reinforcement in a beam is limited by IS 456 to:

    1. 4% of bD
    2. 1.5% of bD
    3. 6% of bD
    4. 2% of bD
    Answer

    A. 4% of bD

    Both tension and compression steel are limited to 4% of the gross area.

16. The basic span to effective depth ratio for a continuous beam (span up to 10 m) is:

    1. 7
    2. 15
    3. 26
    4. 20
    Answer

    C. 26

    IS 456 basic values: cantilever 7, simply supported 20, continuous 26.

17. Using the basic ratio only, the minimum effective depth of a simply supported beam of 6 m span is:

    1. 230 mm
    2. 600 mm
    3. 200 mm
    4. 300 mm
    Answer

    D. 300 mm

    d = L/20 = 6000/20 = 300 mm.

18. For a continuous T-beam with lo = 6000 mm, bw = 300 mm and Df = 100 mm, the effective flange width bf is:

    1. 1600 mm
    2. 1300 mm
    3. 2200 mm
    4. 1900 mm
    Answer

    D. 1900 mm

    bf = lo/6 + bw + 6Df = 1000 + 300 + 600 = 1900 mm.

19. For an L-beam with lo = 6000 mm, bw = 300 mm and Df = 100 mm, the effective flange width is:

    1. 1400 mm
    2. 1100 mm
    3. 1900 mm
    4. 800 mm
    Answer

    B. 1100 mm

    bf = lo/12 + bw + 3Df = 500 + 300 + 300 = 1100 mm.

20. A T-beam is analysed as a rectangular beam of width bf when:

    1. The neutral axis lies within the flange (xu ≤ Df)
    2. The neutral axis lies in the web
    3. The web width exceeds the flange depth
    4. The flange is in tension
    Answer

    A. The neutral axis lies within the flange (xu ≤ Df)

    If xu ≤ Df, all compression is in the flange, so the section acts as a rectangle of width bf.

21. A doubly reinforced section becomes necessary when:

    1. Shear stress exceeds τc
    2. The span is more than 10 m
    3. The factored moment exceeds Mu,lim and depth is restricted
    4. The beam is simply supported
    Answer

    C. The factored moment exceeds Mu,lim and depth is restricted

    Extra moment beyond Mu,lim is resisted by compression steel with extra tension steel.

22. A beam of b = 250 mm and d = 500 mm carries Vu = 150 kN. The nominal shear stress τv is:

    1. 0.6 N/mm²
    2. 1.2 N/mm²
    3. 2.4 N/mm²
    4. 3.0 N/mm²
    Answer

    B. 1.2 N/mm²

    τv = Vu/(bd) = 150,000/125,000 = 1.2 N/mm².

23. The maximum shear stress τc,max for M20 concrete is:

    1. 3.1 N/mm²
    2. 3.5 N/mm²
    3. 2.8 N/mm²
    4. 2.0 N/mm²
    Answer

    C. 2.8 N/mm²

    IS 456 gives 2.8 (M20), 3.1 (M25), 3.5 (M30).

24. If τv exceeds τc,max, the designer should:

    1. Provide only minimum stirrups
    2. Reduce the stirrup spacing
    3. Increase the cross-section of the beam
    4. Use plain bars
    Answer

    C. Increase the cross-section of the beam

    Stirrups cannot rescue a section whose shear stress exceeds τc,max.

25. 2-legged 8 mm stirrups (Asv = 100.5 mm²) at 150 mm, fy = 415, d = 500 mm give Vus of nearly:

    1. 242 kN
    2. 150 kN
    3. 60 kN
    4. 121 kN
    Answer

    D. 121 kN

    Vus = 0.87 fy Asv d/sv = 361.05 × 100.5 × 500/150 = 121 kN.

26. The maximum spacing of vertical stirrups in a beam with d = 300 mm is:

    1. 225 mm
    2. 450 mm
    3. 150 mm
    4. 300 mm
    Answer

    A. 225 mm

    Lesser of 0.75d = 225 mm and 300 mm.

27. Minimum shear reinforcement for b = 250 mm, sv = 200 mm, fy = 415 (Asv ≥ 0.4 b sv/0.87 fy) is nearly:

    1. 100 mm²
    2. 55 mm²
    3. 28 mm²
    4. 111 mm²
    Answer

    B. 55 mm²

    0.4 × 250 × 200/361.05 = 55.4 mm².

28. The development length of a 16 mm deformed Fe 415 bar in M20 concrete (τbd = 1.92 N/mm²) is nearly:

    1. 600 mm
    2. 1203 mm
    3. 480 mm
    4. 752 mm
    Answer

    D. 752 mm

    Ld = 0.87 × 415 × 16/(4 × 1.92) = 752 mm (about 47 φ).

29. By how much is design bond stress increased for deformed bars compared with plain bars?

    1. 60%
    2. 100%
    3. 25%
    4. 40%
    Answer

    A. 60%

    IS 456 increases τbd by 60% for deformed bars (and by 25% in compression).

30. The nominal cover for a beam under moderate exposure is:

    1. 30 mm
    2. 20 mm
    3. 45 mm
    4. 50 mm
    Answer

    A. 30 mm

    Moderate exposure: 30 mm; mild: 20 mm; severe: 45 mm.

31. The ultimate strain in concrete at the extreme compression fibre in flexure is:

    1. 0.0015
    2. 0.0045
    3. 0.002
    4. 0.0035
    Answer

    D. 0.0035

    IS 456 limits flexural concrete strain to 0.0035 (0.002 in pure axial compression).

32. In WSM, the area of tension steel is transformed into an equivalent concrete area of:

    1. Ast × σst
    2. m × Ast
    3. Ast/m
    4. Ast × σcbc
    Answer

    B. m × Ast

    Transformed area = m Ast, with m the modular ratio.

33. For M20 and Fe 415 in WSM, the approximate moment of resistance constant Q is 0.91. For b = 250 mm and d = 450 mm, balanced Mr is nearly:

    1. 90 kN·m
    2. 30 kN·m
    3. 46 kN·m
    4. 140 kN·m
    Answer

    C. 46 kN·m

    Mr = Q b d² = 0.91 × 250 × 450² = 46 × 10⁶ N·mm.

34. Which of the following statements are correct? 1. In LSM, a partial safety factor of 1.5 is applied to dead and live loads. 2. LSM design uses permissible stresses of concrete and steel under working loads.

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    A. 1 only

    Statement 1 is true; permissible stresses are the basis of WSM, not LSM.

35. Which of the following statements are correct? 1. Under-reinforced beams fail in a ductile manner. 2. Over-reinforced beams fail by crushing of concrete before steel yields.

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    C. Both 1 and 2

    Both describe the standard failure behaviours.

36. Which of the following statements are correct? 1. Minimum tension steel in a beam depends on the grade of steel. 2. Maximum tension steel is limited to 4% of bD.

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    C. Both 1 and 2

    Minimum is 0.85 bd/fy; maximum is 4% of bD.

37. Stirrups in a reinforced concrete beam are provided mainly to resist:

    1. Bending compression
    2. Torsion only
    3. Temperature stresses
    4. Shear (diagonal tension)
    Answer

    D. Shear (diagonal tension)

    Stirrups carry shear force and hold main bars in position.

38. A beam carries Vu = 100 kN and Tu = 10 kN·m with b = 250 mm. The equivalent shear Ve = Vu + 1.6 Tu/b is:

    1. 116 kN
    2. 164 kN
    3. 260 kN
    4. 100 kN
    Answer

    B. 164 kN

    1.6 × 10 × 10⁶/250 = 64 kN; Ve = 100 + 64 = 164 kN.

39. For 16 mm main bars and 20 mm maximum aggregate, the minimum clear horizontal spacing of bars is:

    1. 16 mm
    2. 25 mm
    3. 20 mm
    4. 40 mm
    Answer

    B. 25 mm

    Greatest of bar diameter (16) and aggregate size + 5 mm (25) = 25 mm.

40. Side face reinforcement is required in beams whose web depth exceeds:

    1. 750 mm
    2. 450 mm
    3. 300 mm
    4. 1000 mm
    Answer

    A. 750 mm

    IS 456 requires 0.1% of web area as side face steel when depth exceeds 750 mm.

41. The final deflection of a beam including shrinkage, creep and temperature should not normally exceed:

    1. Span/500
    2. Span/150
    3. Span/100
    4. Span/250
    Answer

    D. Span/250

    IS 456 limit is span/250; after finishes, span/350 or 20 mm whichever is less.

42. A beam has overall depth 500 mm, clear cover 30 mm and 20 mm main bars. The effective depth is:

    1. 460 mm
    2. 470 mm
    3. 450 mm
    4. 500 mm
    Answer

    A. 460 mm

    d = 500 − 30 − 10 = 460 mm.

43. The lever arm ratio z/d of a balanced LSM section for Fe 415 (z = d − 0.42 xu,max) is:

    1. 0.777
    2. 0.807
    3. 0.900
    4. 0.798
    Answer

    D. 0.798

    1 − 0.42 × 0.48 = 0.798.

44. Which of the following statements are correct? 1. Doubly reinforced beams help reduce long-term deflection. 2. Compression steel in a beam is limited to 4% of bD.

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    C. Both 1 and 2

    Compression steel reduces creep deflection and is limited to 4% of the gross area.

45. In a rectangular beam with no compression steel, the limiting moment of resistance depends on:

    1. Only fy and Ast
    2. fck, b, d and the steel grade
    3. Only the span
    4. Only the stirrup size
    Answer

    B. fck, b, d and the steel grade

    Mu,lim = 0.36 fck b xu,max(d − 0.42 xu,max), and xu,max/d depends on fy.