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Chapter 17 of 17

Plate & Gantry Girders

In the MP Vyapam Sub Engineer Civil syllabus under Steel Structures · 2 parts

Plate Girders · Gantry Girders

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📑 Contents (17 sections)

Part 1 of 2

Plate Girders

Last reviewed 16 Sept 2026 · 5 min read

Why plate girders

When the bending moment or span exceeds what the heaviest rolled beams can carry (roughly beyond 20–25 m spans or large concentrated loads), a plate girder is built up from plates (and sometimes angles) to exactly the required proportions. Uses: railway and road bridges, gantry girders for heavy cranes, transfer girders in buildings, industrial platforms.

Components

Web plate — deep and thin; carries shear and holds the flanges apart.
Flanges — flange plates (welded girders) or flange angles with cover plates (riveted/bolted girders); carry most of the bending moment.
Intermediate transverse stiffeners — prevent shear buckling of the web.
Longitudinal (horizontal) stiffeners — in very deep webs, in the compression zone, to prevent bending buckling.
Bearing (load-carrying) stiffeners — at supports and under concentrated loads.
Connections — flange-to-web welds or rivets/bolts designed for horizontal shear flow; splices in web and flanges.

Proportioning

Economical depth

For a girder of given moment $M$ with web slenderness $k = d / t_{w}$ , minimising the total steel area gives approximately

FormulaEconomical depth

d \approx (\frac{M k}{f _{y} / γ _{m 0}})^{1/3}

Common practice: $k$ = 100–200 for welded girders; depth about 1/10 to 1/12 of span for simply supported girders (deeper for bridges).

At the economical depth, the web area roughly equals the combined flange area (about half the girder's material in the web).

Flanges

Approximately, the flanges carry the moment as a couple:

A_{f} \approx \frac{M}{( f _{y} / γ _{m 0} ) d}

Flange outstand-to-thickness ratios must satisfy the section classification limits (plastic/compact/semi-compact) to avoid local buckling. For lateral stability the compression flange width is commonly about 1/40 to 1/30 of span (or restrained laterally).

Self-weight estimate

A common first estimate for the weight of a plate girder per metre is about $W /300$ of the total superimposed load $W$ (in consistent units) — refined after sizing.

Web slenderness limits (IS 800:2007)

Code ProvisionIS 800 cl. 8.6.1 — web thickness (selected)

Serviceability requirement:

Without transverse stiffeners: $d / t_{w} \leq 200 ε$
With transverse stiffeners only: $d / t_{w} \leq 200 ε$ when $c \geq d$ ; $c / t_{w} \leq 200 ε$ when $0.74 d \leq c < d$ ; $d / t_{w} \leq 270 ε$ when $c < 0.74 d$
With transverse and one longitudinal stiffener (at 0.2 $d$ from the compression flange): $d / t_{w} \leq 250 ε$
With a second longitudinal stiffener at the neutral axis: $d / t_{w} \leq 400 ε$

Compression flange buckling requirement (web must be stiff enough so the compression flange does not buckle into it):

Transverse stiffeners with $c \geq 1.5 d$ : $d / t_{w} \leq 345 ε^{2}$
Transverse stiffeners with $c < 1.5 d$ : $d / t_{w} \leq 345 ε$

Shear buckling of the web

A thin web buckles in diagonal compression before yielding in shear.

Simple post-critical method

V_{cr} = A_{v} τ_{b}

$τ_{b}$ depends on web slenderness $λ_{w} = \frac{f _{y w}}{3 τ _{cr, e}}$ :

$λ_{w}$	$τ_{b}$
≤ 0.8	$f_{y w} / 3$
0.8 to 1.2	$[1 - 0.8 (λ_{w} - 0.8)] f_{y w} / 3$
≥ 1.2	$f_{y w} / (3 λ_{w}^{2})$

Elastic critical shear stress:

τ_{cr, e} = \frac{k _{v} π ^{2} E}{12 ( 1 - μ ^{2} ) ( d / t _{w} ) ^{2}}, k_{v} = ⎩ ⎨ ⎧ 5.35 4 + \frac{5.35}{( c / d ) ^{2}} 5.35 + \frac{4}{( c / d ) ^{2}} no transverse stiffeners c / d < 1 c / d \geq 1

Closer stiffener spacing increases $k_{v}$ and shear strength.

Tension field method

After the web buckles, it continues to carry load through a diagonal tension field anchored by the flanges and stiffeners — the girder behaves like a Pratt truss with the web as tension diagonals and stiffeners as compression verticals. It gives higher shear capacity when stiffeners are provided at $c / d \geq 1$ and the end panels are properly anchored.

Part 2 of 2

Gantry Girders

Last reviewed 16 Sept 2026 · 5 min read

What a gantry girder does

In industrial buildings, an electric overhead travelling (EOT) crane or a hand-operated crane runs along rails. The rails sit on gantry girders spanning between brackets or stepped columns. The crane consists of a crane girder (bridge) spanning across the building, carrying a crab (trolley) with the hook. The gantry girder therefore carries moving wheel loads that are dynamic, repeated and applied eccentrically and laterally — which makes its design different from an ordinary beam.

Loads on a gantry girder

Vertical wheel loads from the crane (self-weight of crane girder, crab and lifted load), with the crab at the nearest approach to the gantry to give maximum wheel reactions.
Impact allowance on vertical loads (sudden lifting, braking, rail irregularities).
Lateral (horizontal surge) force perpendicular to the rails, from acceleration/braking of the crab — applied at rail level.
Longitudinal (drag) force along the rails from acceleration/braking of the whole crane.
Self-weight of the gantry girder and rail.

Code ProvisionIS 875 (Part 2) — impact and crane forces

Crane	Vertical impact	Lateral surge (each side, total)	Longitudinal force
EOT crane	25% of maximum static wheel loads	10% of (weight of crab + lifted load)	5% of static vertical wheel loads
Hand-operated crane	10%	5% of (crab + load)	5%

The lateral and longitudinal forces are not considered to act simultaneously; each is combined with the vertical loads separately.

Maximum wheel load

With crane span $L_{c}$ , minimum hook approach $a$ , crab + lifted load $W_{c}$ and crane girder self-weight $W_{g}$ (assumed shared equally by both end carriages, each with two wheels):

R_{ma x} = \frac{W _{c} ( L _{c} - a )}{L _{c}} + \frac{W _{g}}{2} (on one end carriage), wheel load = \frac{R _{ma x}}{2}

Position of wheels for maximum effects

Maximum bending moment: for two equal wheel loads $W$ at spacing $c$ (wheel base) on a span $L$ (with $c < 0.586 L$ ), place them so that the centre of the span bisects the distance between one wheel and the resultant (Barré's rule, see Influence Lines). If $c \geq 0.586 L$ , a single wheel at mid-span gives the maximum ( $W L /4$ ).
Maximum shear: one wheel just at the support with the other on the span.
Lateral moment: same wheel positions, with the lateral forces at rail level.

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