Plate girders
When a rolled beam is not deep enough, steel plates are welded into an I-shaped plate girder of any depth needed. Plate girders carry heavy loads over long spans in bridges, crane runways, transfer structures in buildings and industrial platforms. Because the designer chooses every dimension, the girder can be made as efficient as the loading allows.
The flanges carry most of the bending moment as a tension-compression couple, so a quick first sizing divides the moment by the distance between flange centroids and by the design strength to get flange area. The web carries most of the shear. A deep, thin web is economical but buckles in shear and under concentrated loads, so it is stiffened. Transverse stiffeners divide the web into panels and raise its shear buckling resistance; bearing stiffeners at supports and under heavy point loads stop the web from crippling; longitudinal stiffeners are added in very deep webs.
After a thin web buckles in shear it can still carry load through tension field action: a diagonal band of tension develops between stiffeners, like the diagonals of a truss, with the stiffeners acting as the verticals. Codes, including IS 800, allow this post-buckling reserve when the panels and end posts are designed for it.
The compression flange must be restrained laterally or checked for lateral-torsional buckling. Flange plates can be curtailed or thinned where the moment is lower. Fatigue matters for bridge and crane girders, so welds of stiffeners and flange splices are detailed with care.
Start from flange area and a web depth of roughly span/10 to span/15, then refine.
Flange area (first estimate): A_f ≈ M / (f_yd × h_f), f_yd = f_y / γ_m0, h_f = distance between flange centroidsWeb shear (first estimate): τ_avg = V / (d t_w)Web slenderness d / t_w governs need for stiffeners
Forgetting bearing stiffeners at supports, where the web carries the full reaction.
Problem. A plate girder must resist a design moment of 3 000 kNm. Steel f_y = 250 MPa, γ_m0 = 1.10, and the distance between flange centroids is 1.5 m. Estimate the flange area.
- f_yd = 250 / 1.10 = 227.3 MPa.
- A_f ≈ 3 000 × 10⁶ / (227.3 × 1 500) = 8 800 mm².
- A 400 mm × 22 mm plate gives 8 800 mm², which is a reasonable first choice.
- Then check the full section (including the web contribution), lateral-torsional buckling and local flange slenderness.
Answer. About 8 800 mm² per flange, for example 400 × 22 mm plates
Quick quiz
1What carries most of the moment in a plate girder?
The flanges, as a tension-compression couple.
2What is tension field action?
Post-buckling shear resistance from a diagonal tension band in the web between stiffeners.
3What do bearing stiffeners prevent?
Web crippling and buckling under concentrated loads and reactions.
- IS 800 (India): General construction in steel: limit state design of steel members and connections.
- EN 1993 (Eurocode 3) (Europe): Design of steel structures.
- AISC 360 (USA): Specification for structural steel buildings.
- IRC:24 (India): Steel road bridges (limit state method).
Clauses and documents to read
- IS 800:2007, cl. 5.4.1 (Table 5): partial safety factor γ_m0
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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