Design of Steel Structures (IS 800)

Limit state design of steel tension members, compression members and slenderness, laterally supported beams, plastic analysis with shape factor and collapse loads, and bolted and welded connections.

Steel design in India follows IS 800:2007 (limit state method). GATE asks about member capacities, plastic analysis and connections.

Partial safety factors

γ_m0 = 1.10 (yielding), γ_m1 = 1.25 (ultimate fracture), γ_mb = 1.25 (bolts), γ_mw = 1.25 (shop welds).

Tension members

The design strength is the least of:

  • Gross section yielding: T_dg = A_g f_y/γ_m0.
  • Net section rupture: T_dn = 0.9 A_n f_u/γ_m1 (for plates).
  • Block shear at the end connection.

Compression members

  • Slenderness ratio λ = KL/r. Limits: 180 for members carrying dead and live load, 250 for members carrying only wind/earthquake reversal, 350 for tension members that may reverse.
  • Euler stress f_cc = π²E/λ².
  • IS 800 uses column buckling curves (a, b, c, d by section type) to give the design compressive stress f_cd, and P_d = A_e f_cd.
  • Effective length factor K: 0.65 (both ends fixed), 0.8 (fixed–pinned), 1.0 (pinned–pinned), 2.0 (fixed–free).

Beams

  • Laterally supported beam: M_d = β_b Z_p f_y/γ_m0 (plastic or compact sections, β_b = 1); limited to 1.2 Z_e f_y/γ_m0 for simply supported beams.
  • Shear: V_d = A_v f_y/(√3 γ_m0), with A_v = web area (h t_w).
  • Unsupported beams are checked for lateral torsional buckling.

Plastic analysis

  • Plastic moment M_p = Z_p f_y. Shape factor S = Z_p/Z_e:
SectionShape factor
Rectangle1.5
Solid circle1.7 (16/(3π))
I-section (about major axis)about 1.12–1.15
Diamond2.0
  • Collapse loads (from mechanism method):
  • Simply supported beam, central load: W_c = 4M_p/L.
  • Fixed beam, central load: W_c = 8M_p/L.
  • Fixed beam, UDL: w_c = 16M_p/L².
  • Propped cantilever, UDL: w_c = 11.66 M_p/L².
  • Load factor = collapse load/working load.

Connections

  • Bolts in shear: V_dsb = f_u A_nb/(√3 γ_mb) per shear plane; bearing V_dpb = 2.5 k_b d t f_u/γ_mb.
  • Fillet welds: throat t = 0.7 s; design strength f_wd = f_u/(√3 γ_mw).

Worked example

A rectangular section 100 × 200 mm (bending about the strong axis), f_y = 250 MPa.

  • Z_e = 100 × 200²/6 = 666,667 mm³; Z_p = 100 × 200²/4 = 1,000,000 mm³.
  • Shape factor = 1.5; M_p = 250 kN·m.

Practice

GATE-styleNAT · numerical answer

A fixed beam of span 6 m has M_p = 120 kN·m. What central point load causes collapse in kN?

Show answer and explanationAnswer: 160

W_c = 8M_p/L = 960/6 = 160 kN.

GATE-styleNAT · numerical answer

A flat 200 mm × 10 mm (f_y = 250 MPa) is a tension member. What is its design strength for gross section yielding in kN? (Give 2 decimal places.)

Show answer and explanationAnswer: 454.55

T_dg = A_g f_y/γ_m0 = 2000 × 250/1.10 = 454,545 N = 454.55 kN.

GATE-styleNAT · numerical answer

What is the shape factor of a solid circular section? (Give 3 decimal places.)

Show answer and explanationAnswer: 1.698

Z_p = d³/6 and Z_e = πd³/32, so S = 32/(6π) = 16/(3π) = 1.698.

GATE-styleNAT · numerical answer

A pinned–pinned steel strut 3 m long has a minimum radius of gyration of 30 mm. What is its slenderness ratio?

Show answer and explanationAnswer: 100

λ = KL/r = 1.0 × 3000/30 = 100.