Steel Structures

What to remember

• IS 800:2007 uses the Limit State Method with partial safety factors γm0 = 1.10 (yielding) and γm1 = 1.25 (ultimate rupture). Structural steel E250 has fy = 250 N/mm² and fu = 410 N/mm²; E = 2 × 10⁵ N/mm².
• Tension members are checked for yielding of gross section, rupture of net section and block shear; compression members are governed by slenderness ratio KL/r and buckling curves.
• Connections: bolts (bearing and HSFG) and welds. A fillet weld has throat thickness 0.7 × size; bolt holes are 2 mm larger than the bolt for diameters up to 24 mm.

1. Material and design basics

Mild steel is ductile, equally strong in tension and compression, and has a unit weight of 78.5 kN/m³. Common structural shapes are the I-sections (ISMB, ISLB, ISHB, ISWB), channels (ISMC), angles (ISA), tees and hollow tubes.

Design strengths use partial factors: γm0 = 1.10 for yielding and buckling, γm1 = 1.25 for ultimate stress (net section rupture), γmb = 1.25 for bolts, γmw = 1.25 for shop welds and 1.50 for field welds.

The yield stress ratio ε = √(250/fy). For E250, ε = 1; for E350, ε = 0.845.

Local buckling classes (cross-section): plastic (can form a hinge and has rotation capacity), compact (reaches plastic moment but limited rotation), semi-compact (reaches yield in extreme fibre only) and slender (local buckling before yield).

2. Tension members

Examples: ties in trusses, hangers, bracings. The design tensile strength is the least of three values.

• 1. Yielding of gross section: Tdg = Ag fy / γm0.
• 2. Rupture of net section: Tdn = 0.9 An fu / γm1.
• 3. Block shear: tearing along a shear plane combined with tension on the perpendicular plane.

Example: Ag = 1000 mm², fy = 250: Tdg = 250,000/1.1 ≈ 227 kN. With An = 800 mm², fu = 410: Tdn = 0.9 × 800 × 410/1.25 ≈ 236 kN. The lower governs.

Net area: An = (b − n d0) t for holes in a line, where d0 is the hole diameter. Standard hole diameter d0 = bolt diameter + 2 mm for bolts up to 24 mm (+3 mm above 24 mm). Example: plate 200 × 10 mm, two holes for 20 mm bolts in a line: An = (200 − 2 × 22) × 10 = 1560 mm².

Staggered holes: net width = b − n d0 + Σ p²/(4g), where p is the stagger pitch and g the gauge. Example: b = 200 mm, three holes of 22 mm, two staggers with p = 40 mm and g = 50 mm: width = 200 − 66 + 2 × (1600/200) = 150 mm.

Single angle tension members: connected by one leg only, so shear lag reduces the effective area. Lug angles or longer connections reduce shear lag and the number of bolts in a connection.

3. Bolted connections

Bolt grade 4.6 means fub = 400 N/mm² and fyb = 0.6 × 400 = 240 N/mm². Grades 8.8 and 10.9 are high strength. Common bolt types: black bolts (bearing type) and high strength friction grip (HSFG) bolts, which transfer load by friction, not bearing.

Failure/limit	Formula
Bolt shear	Vnsb = fub/√3 × (nn Anb + ns Asb) / γmb
Bolt bearing	Vnpb = 2.5 kb d t fu / γmb
Bearing factor	kb = least of e/(3d0), p/(3d0) − 0.25, fub/fu, 1.0

Here Anb is the net tensile area of the bolt (about 0.78 of nominal area), nn and ns are the numbers of shear planes through threaded and shank portions. Example: M20, grade 4.6, one threaded shear plane (Anb = 245 mm²): Vnsb = 400/1.732 × 245/1.25 ≈ 45.3 kN.

Spacing rules (d0 = hole diameter, t = thinner plate):

• Minimum pitch: 2.5 × nominal bolt diameter.
• Maximum pitch: 32 t or 300 mm, whichever is less (general); in tension members 16 t or 200 mm; in compression members 12 t or 200 mm.
• Minimum edge distance: 1.7 d0 for sheared or hand flame cut edges; 1.5 d0 for rolled, machine flame cut or planed edges.
• Maximum edge distance: 12 t ε.

Efficiency of a joint = strength of the joint / strength of the solid plate. It is below 100% because holes remove area.

Shear lag, prying action, block shear are secondary checks for connections.

4. Welded connections

Types: fillet welds (most common in lap and tee joints), butt (groove) welds, plug and slot welds. Shop welds use γmw = 1.25; field welds use 1.5.

• Throat thickness tt = K × size (K = 0.7 when the angle between fusion faces is 60° to 90°). A 6 mm fillet has throat 4.2 mm.
• Design strength per unit length = fu / (√3 γmw) × tt. Example: fu = 410, shop weld, 6 mm fillet: 410/(1.732 × 1.25) × 4.2 ≈ 796 N/mm.
• Minimum effective length: 4 × weld size.
• Maximum fillet size along a square plate edge: thickness − 1.5 mm. Minimum size depends on the thicker part (3 mm for plates up to 10 mm).
• Butt weld strength equals the parent metal (full penetration).

5. Compression members (columns, struts)

The slenderness ratio λ = KL/r, where KL is effective length and r = √(I/A) is the radius of gyration. Limiting λ for members carrying dead and live loads is 180.

Effective length factors (IS 800 recommended design values):

End conditions	KL
Both ends fixed	0.65 L
One end fixed, other hinged	0.80 L
Both ends hinged	1.0 L
One end fixed, other free	2.0 L

Euler's critical load: Pe = π² E I / (KL)². Example: E = 2 × 10⁵, I = 10⁷ mm⁴, hinged, L = 4 m: Pe = 9.8696 × 2 × 10⁵ × 10⁷ / (4000)² ≈ 1234 kN. Euler load does not depend on yield strength and falls as the square of the length.

IS 800 column design: uses buckling curves a, b, c, d with imperfection factors α = 0.21, 0.34, 0.49 and 0.76. The design compressive stress is fcd = fy/γm0 / (φ + √(φ² − λ²)) with φ = 0.5[1 + α(λ − 0.2) + λ²] and non-dimensional slenderness λ = √(fy/fcc). Pd = Ae fcd.

Built-up columns: two or more rolled sections joined by lacing (single or double diagonal bars at 40° to 70° to the axis) or battens (plates). Battens are spaced so that the slenderness of a single component between battens is not more than 50 or 0.7 times the slenderness of the whole column, whichever is less. Lacing gives better shear resistance; battens are cheaper but less stiff.

Column bases: slab base (for light loads, plate designed in bending), gusseted base (heavy loads), grillage foundation for very heavy loads.

6. Beams and plate girders

Elastic and plastic moments: Me = fy Ze; Mp = fy Zp. Shape factor = Zp/Ze: rectangle 1.5, circle about 1.7, I-sections about 1.12 to 1.15. Example: rectangle 50 × 100 mm, fy = 250: Zp = bd²/4 = 125,000 mm³, Mp = 31.25 kN·m; Ze = bd²/6, Me = 20.8 kN·m, ratio 1.5.

Design bending strength (laterally supported): Md = βb Zp fy / γm0, limited to 1.2 Ze fy/γm0 for simply supported and 1.5 Ze fy/γm0 for cantilever beams (βb = 1 for plastic and compact sections).

Shear strength: Vd = Av fy / (√3 γm0), where Av is the shear area (web area). Example: Av = 3000 mm²: Vd = 3000 × 250 / (1.732 × 1.1) ≈ 394 kN.

Lateral torsional buckling: the compression flange of an unrestrained beam buckles sideways. It is prevented by lateral restraint (floor slab, bracing) or by reducing the bending strength.

Plate girder: a built-up I-girder with a thin web and large flanges, used for long spans and heavy loads. Flanges resist bending (flange area method); the web resists shear. Stiffeners: bearing stiffeners at supports and concentrated loads; intermediate transverse stiffeners to improve web shear buckling resistance; longitudinal stiffeners for deep webs.

7. Roof trusses and gusset plates

Trusses carry roof loads over large spans. Common types: Howe, Pratt, Fink, fan, north-light (for daylight in factories). Purlins rest on the rafters at panel points so the rafter carries axial load only. Loads: dead load, live load, wind load (IS 875). Joint design uses gusset plates, which connect members meeting at a joint.

Exam traps

• Net section rupture uses 0.9 An fu/γm1 (γm1 = 1.25), while gross yielding uses fy/γm0 (γm0 = 1.10).
• Bolt hole is 2 mm larger than the bolt (up to 24 mm), not equal to bolt diameter.
• Bolt grade 4.6: first number × 100 gives fub (400), and 0.6 of it gives fyb.
• HSFG bolts transfer load by friction; ordinary black bolts by bearing.
• Fillet weld throat is 0.7 × size, not size.
• Euler load depends on E and I, not on fy.
• Effective length factor fixed-free is 2.0 and fixed-fixed is 0.65.
• Shape factor of a rectangle is 1.5; circle 1.7.

One-liners

• 1. E for steel is 2 × 10⁵ N/mm².
• 2. E250 steel has fy = 250 and fu = 410 N/mm².
• 3. γm0 = 1.10; γm1 = 1.25; γmw (field) = 1.5.
• 4. Standard hole diameter = bolt diameter + 2 mm (up to 24 mm bolts).
• 5. Minimum bolt pitch is 2.5 d; maximum edge distance is 12 t ε.
• 6. Bolt bearing strength contains the factor 2.5.
• 7. Minimum effective fillet weld length is 4 × size.
• 8. Maximum slenderness for a compression member under dead and live load is 180.
• 9. Imperfection factors for curves a, b, c, d are 0.21, 0.34, 0.49, 0.76.
• 10. Lacing angle is 40° to 70° with the member axis.
• 11. Plastic moment Mp = fy Zp; shape factor Zp/Ze.
• 12. Shear strength Vd = Av fy / (√3 γm0).

Practice questions

1. The modulus of elasticity of structural steel adopted in IS 800 is:

   1. 5 × 10⁵ N/mm²
   2. 2 × 10⁵ N/mm²
   3. 2 × 10⁴ N/mm²
   4. 2 × 10⁶ N/mm²
   Answer

   B. 2 × 10⁵ N/mm²

   Es = 2 × 10⁵ N/mm² (200 GPa).

2. The ultimate tensile strength fu of E250 structural steel is:

   1. 410 N/mm²
   2. 540 N/mm²
   3. 340 N/mm²
   4. 250 N/mm²
   Answer

   A. 410 N/mm²

   E250 has fy = 250 and fu = 410 N/mm².

3. The partial safety factor for material strength against yielding (γm0) in IS 800:2007 is:

   1. 1.50
   2. 1.10
   3. 1.00
   4. 1.25
   Answer

   B. 1.10

   γm0 = 1.10 for yielding and buckling; γm1 = 1.25 for ultimate stress.

4. A tension member has Ag = 1000 mm² and fy = 250 N/mm². Its design strength due to yielding of gross section is nearly:

   1. 200 kN
   2. 284 kN
   3. 250 kN
   4. 227 kN
   Answer

   D. 227 kN

   Tdg = Ag fy/γm0 = 250,000/1.1 = 227 kN.

5. A tension member with An = 800 mm² and fu = 410 N/mm² has rupture strength Tdn = 0.9 An fu/γm1 of nearly:

   1. 295 kN
   2. 236 kN
   3. 328 kN
   4. 262 kN
   Answer

   B. 236 kN

   0.9 × 800 × 410/1.25 = 236 kN.

6. The diameter of the hole for a 20 mm bolt is taken as:

   1. 22 mm
   2. 20 mm
   3. 23 mm
   4. 21 mm
   Answer

   A. 22 mm

   Standard clearance for bolts up to 24 mm is 2 mm.

7. A 200 mm × 10 mm plate has two 20 mm bolt holes in a line. The net area is:

   1. 1600 mm²
   2. 1520 mm²
   3. 1560 mm²
   4. 1560 cm²
   Answer

   C. 1560 mm²

   An = (200 − 2 × 22) × 10 = 1560 mm².

8. A plate 200 mm wide has three holes of 22 mm diameter in a zig-zag with two staggers (p = 40 mm, g = 50 mm). The net width is:

   1. 134 mm
   2. 150 mm
   3. 166 mm
   4. 142 mm
   Answer

   B. 150 mm

   200 − 3 × 22 + 2 × (40²/(4 × 50)) = 200 − 66 + 16 = 150 mm.

9. In the bolt grade designation 4.6, the ultimate tensile strength of the bolt is:

   1. 460 N/mm²
   2. 600 N/mm²
   3. 400 N/mm²
   4. 240 N/mm²
   Answer

   C. 400 N/mm²

   First number × 100 = 400 N/mm²; yield = 0.6 × 400 = 240 N/mm².

10. A single M20 bolt of grade 4.6 has Anb = 245 mm² and γmb = 1.25. In single shear through threads, the design shear strength is nearly:

    1. 36.2 kN
    2. 90.5 kN
    3. 56.6 kN
    4. 45.3 kN
    Answer

    D. 45.3 kN

    Vnsb = (400/√3) × 245/1.25 = 45.3 kN.

11. The minimum pitch of bolts as per IS 800 is:

    1. 2.5 times the nominal bolt diameter
    2. 1.5 times the bolt diameter
    3. 3 times the hole diameter
    4. 1.0 times the bolt diameter
    Answer

    A. 2.5 times the nominal bolt diameter

    Minimum pitch = 2.5 d; maximum 32t or 300 mm.

12. The minimum edge distance for a 22 mm hole in a sheared or hand flame cut edge (1.7 d0) is nearly:

    1. 44 mm
    2. 55 mm
    3. 33 mm
    4. 37.4 mm
    Answer

    D. 37.4 mm

    1.7 × 22 = 37.4 mm.

13. The constant factor used in the bearing capacity formula Vnpb = k d t fu/γmb is:

    1. 3.0 kb
    2. 1.5 kb
    3. 2.5 kb
    4. 0.9 kb
    Answer

    C. 2.5 kb

    Vnpb = 2.5 kb d t fu/γmb.

14. The efficiency of a riveted or bolted joint is defined as:

    1. Strength of solid plate divided by strength of the joint
    2. Strength of the joint divided by strength of the solid plate
    3. Net area divided by gross area times 100
    4. Strength of bolts divided by weight
    Answer

    B. Strength of the joint divided by strength of the solid plate

    Holes weaken the plate so efficiency is below 100%.

15. The throat thickness of a 6 mm fillet weld with a 90° fusion angle is:

    1. 4.2 mm
    2. 8.6 mm
    3. 3.0 mm
    4. 6.0 mm
    Answer

    A. 4.2 mm

    tt = 0.7 × 6 = 4.2 mm.

16. The design strength of a 6 mm shop fillet weld (fu = 410, γmw = 1.25) per mm length is nearly:

    1. 1146 N/mm
    2. 796 N/mm
    3. 400 N/mm
    4. 600 N/mm
    Answer

    B. 796 N/mm

    fu/(√3 γmw) × 4.2 = 189.4 × 4.2 = 796 N/mm.

17. The partial safety factor for field welds as per IS 800 is:

    1. 1.25
    2. 1.10
    3. 1.50
    4. 1.00
    Answer

    C. 1.50

    Shop welds 1.25; field welds 1.5.

18. The minimum effective length of an 8 mm fillet weld is:

    1. 40 mm
    2. 64 mm
    3. 16 mm
    4. 32 mm
    Answer

    D. 32 mm

    Minimum effective length = 4 × size = 32 mm.

19. The maximum size of a fillet weld along the square edge of a 10 mm plate is:

    1. 8.5 mm
    2. 7 mm
    3. 10 mm
    4. 5 mm
    Answer

    A. 8.5 mm

    Maximum size = thickness − 1.5 mm = 8.5 mm.

20. The maximum permitted slenderness ratio for a compression member carrying dead and imposed loads is:

    1. 180
    2. 350
    3. 250
    4. 400
    Answer

    A. 180

    IS 800 limits such members to KL/r = 180.

21. A column of 3 m length fixed at the base and free at the top has an effective length of:

    1. 2 m
    2. 4.5 m
    3. 3 m
    4. 6 m
    Answer

    D. 6 m

    Fixed-free: KL = 2.0 L = 6 m.

22. The Euler critical load of a hinged column with E = 2 × 10⁵ N/mm², I = 10⁷ mm⁴ and L = 4 m is nearly:

    1. 2467 kN
    2. 1234 kN
    3. 308 kN
    4. 617 kN
    Answer

    B. 1234 kN

    Pe = π² E I/L² = 9.87 × 2 × 10⁵ × 10⁷/(4000)² = 1.234 × 10⁶ N.

23. A column has effective length 3000 mm and radius of gyration 50 mm. The slenderness ratio is:

    1. 150
    2. 6
    3. 60
    4. 30
    Answer

    C. 60

    λ = 3000/50 = 60.

24. The imperfection factor for buckling curve c in IS 800 is:

    1. 0.49
    2. 0.34
    3. 0.21
    4. 0.76
    Answer

    A. 0.49

    Curves a, b, c, d use α = 0.21, 0.34, 0.49, 0.76.

25. Which section class can form a plastic hinge with the rotation capacity needed for plastic analysis?

    1. Slender
    2. Compact
    3. Plastic
    4. Semi-compact
    Answer

    C. Plastic

    Plastic sections reach Mp and rotate enough to redistribute moment.

26. The shape factor of a rectangular section is:

    1. 1.7
    2. 1.5
    3. 1.12
    4. 2.0
    Answer

    B. 1.5

    Zp/Ze = (bd²/4)/(bd²/6) = 1.5.

27. The shape factor of a solid circular section is nearly:

    1. 1.5
    2. 1.15
    3. 2.0
    4. 1.7
    Answer

    D. 1.7

    Zp/Ze for a circle = 16/(3π) ≈ 1.70.

28. A rectangular steel section 50 mm wide and 100 mm deep has fy = 250 N/mm². Its plastic moment is:

    1. 31.25 kN·m
    2. 20.8 kN·m
    3. 15.6 kN·m
    4. 62.5 kN·m
    Answer

    A. 31.25 kN·m

    Zp = 50 × 100²/4 = 125,000 mm³; Mp = 250 × 125,000 = 31.25 × 10⁶ N·mm.

29. For a laterally supported simply supported beam, Md is limited to:

    1. 2.0 Ze fy/γm0
    2. 1.0 Zp fy
    3. 1.2 Ze fy/γm0
    4. 1.5 Ze fy/γm0
    Answer

    C. 1.2 Ze fy/γm0

    The limit is 1.2 for simply supported and 1.5 for cantilever beams.

30. A beam web has shear area Av = 3000 mm² and fy = 250 N/mm². The design shear strength Vd is nearly:

    1. 433 kN
    2. 273 kN
    3. 750 kN
    4. 394 kN
    Answer

    D. 394 kN

    Vd = 3000 × 250/(√3 × 1.1) = 394 kN.

31. Lateral torsional buckling of a beam is prevented mainly by:

    1. Lateral restraint to the compression flange
    2. Increasing the web thickness only
    3. Using plain bars
    4. Reducing the span of tension flange
    Answer

    A. Lateral restraint to the compression flange

    Restraining the compression flange against sideways movement prevents LTB.

32. Bearing stiffeners in a plate girder are provided:

    1. At mid span only
    2. In the tension flange only
    3. At supports and points of concentrated loads
    4. Along the neutral axis only
    Answer

    C. At supports and points of concentrated loads

    They prevent web crippling under concentrated reactions or loads.

33. A gusset plate in a truss is used to:

    1. Carry the roof sheet
    2. Connect the members meeting at a joint
    3. Increase the depth of the rafter
    4. Reduce wind load
    Answer

    B. Connect the members meeting at a joint

    Members are bolted or welded to the gusset at a joint.

34. A truss type widely used in factories to admit north light for the shop floor is the:

    1. Howe truss only
    2. Warren bridge truss
    3. Parker truss
    4. North-light truss
    Answer

    D. North-light truss

    Its vertical glazed face lets in daylight with no direct sun.

35. Purlins are placed at the panel points of a truss so that the rafters:

    1. Carry no load
    2. Carry mainly axial force without bending
    3. Carry large bending
    4. Are free of compression
    Answer

    B. Carry mainly axial force without bending

    Panel-point loading keeps truss members in axial tension or compression.

36. In IS 800, lacing bars in a built-up column are inclined to the column axis at an angle of:

    1. 0° to 20°
    2. 75° to 90°
    3. 10° to 30°
    4. 40° to 70°
    Answer

    D. 40° to 70°

    The recommended inclination is 40° to 70° for single or double lacing.

37. Battens in a battened column are spaced so that the slenderness ratio of a single component between battens does not exceed 50 or what fraction of the slenderness of the whole column?

    1. 0.7
    2. 0.3
    3. 0.5
    4. 1.0
    Answer

    A. 0.7

    The limit is 50 or 0.7 times the column slenderness, whichever is less.

38. Block shear failure of a tension member connection involves:

    1. Only compression of bolt holes
    2. Only buckling of the web
    3. Bending of the bolts alone
    4. Shear along one plane and tension along a perpendicular plane
    Answer

    D. Shear along one plane and tension along a perpendicular plane

    A block of material tears out along bolt lines.

39. Lug angles are provided in channel and angle tension members mainly to:

    1. Increase slenderness
    2. Increase the radius of gyration
    3. Reduce shear lag and the length of the connection
    4. Reduce corrosion
    Answer

    C. Reduce shear lag and the length of the connection

    Lug angles let a connection engage the outstanding leg and shorten the joint.

40. Which of the following statements are correct? 1. For a fillet weld with a 90° fusion angle, throat thickness is 0.7 times the size. 2. HSFG bolts transfer load mainly by bearing against the hole.

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

    A. 1 only

    HSFG bolts rely on friction between plates, not bearing.

41. Which of the following statements are correct? 1. The yield stress ratio ε = √(250/fy). 2. For E350 steel, ε is greater than 1.

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

    A. 1 only

    For E350, ε = √(250/350) = 0.845, which is below 1.

42. Which of the following statements are correct? 1. Euler's critical load does not depend on the yield stress of steel. 2. Euler's load increases when effective length increases.

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

    A. 1 only

    Pe = π²EI/(KL)² falls as length rises.

43. Which of the following gives the correct column buckling load order, from highest to lowest, for the same section and length?

    1. Fixed-free > hinged-hinged > fixed-hinged > fixed-fixed
    2. Fixed-fixed > fixed-hinged > hinged-hinged > fixed-free
    3. Hinged-hinged > fixed-fixed > fixed-free > fixed-hinged
    4. Fixed-hinged > fixed-fixed > fixed-free > hinged-hinged
    Answer

    B. Fixed-fixed > fixed-hinged > hinged-hinged > fixed-free

    Effective lengths 0.65L, 0.8L, 1.0L, 2.0L give the loads in that order.

44. A slab base for a steel column is suitable when:

    1. No base plate is needed
    2. The column is a cantilever
    3. The column load is relatively light
    4. The load is very heavy
    Answer

    C. The column load is relatively light

    A gusseted base is used for heavier loads.

45. The unit weight of structural steel is taken as:

    1. 25 kN/m³
    2. 78.5 kN/m³
    3. 98.1 kN/m³
    4. 7.85 kN/m³
    Answer

    B. 78.5 kN/m³

    Steel density is 7850 kg/m³, i.e. 78.5 kN/m³.