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Size of web and flange plate in plate girder

Short answer

The optimum web depth is d = (M k / f)^(1/3) with k = d/t_w around 100–200; web thickness t_w = d/k checked for shear and slenderness. Flange area A_f ≈ M γ_m0/(f_y d), width b_f ≈ d/3 to d/5, and thickness t_f = A_f/b_f, ensuring compactness.

Describe the process to determine (estimate) the optimum size of web plate and flange plate of a plate girder.

A plate girder is a built-up I-beam (a web plate with flange plates welded or bolted to it) used when rolled sections cannot carry the moment or shear. Its size is fixed by economical (minimum-weight) proportioning, followed by checks.

Process (step by step)

  1. Loads and forces. Find the factored maximum bending moment MM and shear force VV from the span and loads. Include an assumed self-weight, which is revised once the section is known.
  2. Economical depth of web. Take d≈L/10d \approx L/10 to L/12L/12 for a start, then use the minimum-weight condition. If the flanges carry all the moment, the flange area is Af=M/(f d)A_f = M/(f\,d) and the web area is Aw=d tw=d2/kA_w = d\,t_w = d^2/k, where k=d/twk = d/t_w. Total area A=2M/(fd)+d2/kA = 2M/(f d) + d^2/k. For minimum AA, dA/dd=0dA/dd = 0:
d=(M kf)1/3,f=fyγm0d = \left(\frac{M\,k}{f}\right)^{1/3}, \qquad f = \frac{f_y}{\gamma_{m0}}

Here kk is chosen as about 100 to 200 for a web with stiffeners (a deeper girder with a thinner web is lighter but needs more stiffeners). 3. Web thickness. tw=d/kt_w = d/k. Then check the limits of IS 800:2007 cl. 8.6.1: d/twd/t_w not more than about 200ε200\varepsilon to 270ε270\varepsilon, depending on the stiffener spacing (serviceability), and d/tw≤345εf2d/t_w \le 345\varepsilon_f^2 to prevent the compression flange buckling into the web. Also check shear: V≤Vd=d twfyw/(3γm0)V \le V_d = d\,t_w f_{yw}/(\sqrt3\gamma_{m0}) (with post-buckling strength if stiffened). The thickness is rounded up (not less than 8 mm; 6 mm in sheltered work). 4. Flange area. The flanges are assumed to take the whole moment (the web is conservatively ignored, or a part Aw/6A_w/6 is allowed for):

Af=Mγm0fy (d+tf)−Aw6 (approx.)A_f = \frac{M\gamma_{m0}}{f_y\,(d + t_f)} - \frac{A_w}{6} \ (\text{approx.})
  1. Flange plate size. Choose the flange width bf≈d/3b_f \approx d/3 to d/5d/5 (or L/40L/40 to L/45L/45 for lateral stability), then tf=Af/bft_f = A_f/b_f (rounded up to a standard plate thickness). Check the outstand: b/tf≤8.4εb/t_f \le 8.4\varepsilon (plastic) or 9.4ε9.4\varepsilon (compact), and up to 13.6ε13.6\varepsilon for semi-compact welded flanges.
  2. Check the trial section. Compute IzzI_{zz} and ZzzZ_{zz}, the moment capacity (with lateral-torsional buckling if the flange is unsupported), the shear capacity and the deflection; then design the stiffeners, flange-to-web welds and splices.

Sketch of the girder section

   |<-- b_f -->|
   +-----------+  ---  t_f
        | |
        | |  d  (web, t_w)
        | |
   +-----------+  ---  t_f

The flanges give the moment resistance and the web gives the shear resistance. Increasing the depth, rather than the thickness, is the most effective way to gain strength, so the web is made as deep and as thin as the code limits allow.

✓ Answer checked by Claude Opus 5.5 (high effort)

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