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Tension members

Intermediate Steel Design Reviewed 24 Oct 2026 2 min read

Tension members are the simplest steel members because tension straightens a member instead of buckling it. Ties in roof trusses, bracing rods, hangers in suspension bridges and bottom chords of trusses all work in tension. The design question is which of several possible failures comes first.

There are three limit states. Yielding of the gross section causes excessive elongation; its strength is the gross area times fy divided by γm0. Rupture of the net section happens at the bolt holes, where area is reduced; its strength uses the net area and fu with the larger factor γm1, and a factor that reflects stress concentration. Block shear is a tearing failure where a chunk of plate tears out along a shear plane and a tension plane around a bolt group. The design strength is the least of the three.

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Diagram: Indiacon original, not to scale.

Net area depends on the hole pattern. For a chain of holes across the plate, subtract each hole. For staggered holes, a zig-zag path may be critical, and each inclined segment adds back s²/(4g), where s is the stagger and g the gauge. Angles connected by one leg suffer shear lag, because the outstanding leg is not fully effective near the connection, so codes reduce their rupture strength. Slenderness limits prevent sagging and vibration of long tension members, and members that can reverse into compression under wind must also be checked as struts.

Key formulas
  • Gross yield: Tdg = Ag fy / γm0
  • Net rupture (plate): Tdn = 0.9 An fu / γm1
  • Staggered holes: An = (b − n d_h + Σ s²/(4g)) t
  • Block shear: least of two combinations of shear and tension areas
Solved example

Problem. A 150 × 10 mm E250 plate (fu = 410 MPa) is connected by one row of three 20 mm bolts across the width, in 22 mm holes. Find its tension capacity (yield and rupture).

  1. Tdg = 1 500 × 250 / 1.10 = 341 kN.
  2. An = (150 − 3 × 22) × 10 = 840 mm².
  3. Tdn = 0.9 × 840 × 410 / 1.25 = 248 kN.
  4. Block shear should also be checked for the actual bolt layout; here rupture governs at about 248 kN.

Answer. About 248 kN, governed by net section rupture

Quick quiz

1Name the three tension limit states.

Gross yield, net section rupture and block shear.

2Why is γm1 larger than γm0?

Rupture is a brittle, less forgiving failure than yield.

3What is shear lag?

Uneven stress in a member connected through only part of its section, such as one leg of an angle.

Code note
  • IS 800 (India): General construction in steel: limit state design of steel members and connections.

Clauses and documents to read

  • IS 800:2007 cl. 6.2 (yielding of gross section)
  • IS 800:2007 cl. 6.3 (rupture of critical section)
  • IS 800:2007 cl. 6.4 (block shear)
  • IS 800:2007 cl. 3.8 (maximum slenderness ratio)

Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.

References and further reading

Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.

  • N. Subramanian, Design of Steel Structures, Oxford University Press, latest edition.
  • C. G. Salmon, J. E. Johnson, F. A. Malhas, Steel Structures: Design and Behavior, Pearson, 5th edition.
  • W. T. Segui, Steel Design, Cengage, latest edition.

Free and official sources

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