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Plate Girders

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Why Structural Steel?

Built-up I-sections, tension-field action, stiffener design.

⏱ 120 min★ Advanced✎ 3 objectives
§01Section 01

Engineering story

Engineering story
Chapter 10 · Plate Girders

Built-up I-sections, tension-field action, stiffener design.

A real project narrative for this chapter will be authored as this chapter migrates to the v3.0 structured schema.

§02Section 02

Learning objectives

After this chapter you will be able to
  • Design a built-up I-section
  • Evaluate tension-field action
  • Detail transverse stiffeners
§03Section 03

Engineering motivation

§04Section 04

Failure mechanisms

Failure mechanisms & lessons learned

Photographs and lessons-learned case studies for this topic will be added during chapter migration.

AISC Reference Box
  • AISC 360-22 — Specification chapter governing this topic
  • AISC Manual 16th Ed. — Design tables and worked examples

Why This Chapter Matters

When a rolled W-shape is not deep enough, a built-up I-girder is welded from plates. Plate girders open up tension-field action, stiffener design, and flange-web proportioning issues that don't arise in rolled shapes.

Learning Objectives

  • Proportion a plate girder to minimize weight for a given moment demand.
  • Design transverse and bearing stiffeners per AISC G2.3 and J10.
  • Compute tension-field capacity Vn per AISC G2.2.
  • Check flange local buckling, web local buckling, and web sidesway buckling.
  • Detail full-penetration welds joining flanges to web.

Where This Chapter Is Used

Long-span bridge girders, industrial crane runways, and heavy-transfer girders in commercial buildings.

ANSI / AISC 360-22Specification for Structural Steel Buildings16.1-70 to 16.1-100
Chapter
F
AISC 360-22

Chapter F. Design of Members for Flexure — §F5 & F13 (Built-Up Girders)

Use this reference to flip directly to the correct page of the AISC 360-22 Specification while solving problems in this course chapter.

§Section titlePage
F5Doubly Symmetric I-Shapes with Slender Webs (Plate Girders)16.1-70
F13Proportions of Beams and Girders (Flange/Web Proportions)16.1-90
G2.2Tension-Field Action (Post-Buckling Shear)16.1-103
G2.3Transverse Stiffeners16.1-104
J10Bearing Stiffener Design16.1-146

Companion reference: AISC Design Guide 25 — Plate Girders

Lecture Notes

Chapter 10 — Plate Girders (AISC 360-22 §F5, §G2, §J10)

Chapter focus. When a rolled W-shape cannot span far enough or deep enough — long-span industrial floors, crane runways, transfer girders, bridge stringers — engineers weld up a plate girder: a deep, thin web plate with two thick flange plates. Making the web thin is exactly what makes the girder efficient, and exactly what forces you out of §F2/§F3 and into the slender-web provisions of §F5, the tension-field shear provisions of §G2.2, and the stiffener requirements of §G2.3 and §J10. This chapter is where beam design and stability theory meet fabrication reality.

1. Why Fabricate a Girder at All?

The AISC Manual's deepest rolled shape is roughly 44 in. A plate girder has no depth limit — 70 in., 96 in., 120 in. webs are routine. Since flexural efficiency scales roughly with depth squared while material scales with depth, going deeper is the cheapest way to buy moment capacity. Fabricate when:

  • Span exceeds ~60–70 ft with meaningful load, so no rolled shape satisfies both strength and deflection.
  • Depth is constrained but load is huge — a transfer girder under a column drop can need a 4 in. thick flange that simply is not rolled.
  • The moment diagram varies strongly — you can taper the web depth or step the flange thickness to follow it, saving 15–25% of steel.
  • Special geometry — haunched girders, curved girders, unequal flanges for composite action.
Economics caution. Fabricated girders cost 2–4× rolled shapes per pound because of cutting, fit-up, submerged-arc welding, stiffener welding, camber and NDT. Never fabricate a girder that a W40×397 could do. The break-even is usually around a span/depth ratio where no rolled shape works at all.
Anatomy of a welded plate girder — every part has a job bearing stiffener bearing stiffener intermediate intermediate intermediate SLENDER WEB — carries shear, contributes little to moment h/t_w typically 150–260; buckles diagonally well before yielding COMPRESSION FLANGE — carries ~90% of the moment; braced against LTB TENSION FLANGE — carries the other half of the couple; check net section at splices h a (end panel — tighter) a (interior — wider) Shear governs near the supports (close stiffener spacing); moment governs at midspan (spacing can open up or vanish).
Figure 10.1 — The three components of a plate girder and their distinct roles. Flanges carry the moment couple; the web carries the shear; the stiffeners make the thin web behave.

2. Web Slenderness Decides Everything

The single classifying parameter is h/tw, compared against the Table B4.1b limits for webs in flexure:

Table B4.1b Case 15λpw = 3.76√(E/Fy) λrw = 5.70√(E/Fy)
FyλpwλrwAbsolute max h/tw (§F13.2, no long. stiffener)
36 ksi107162322
50 ksi90.6137260
65 ksi79.4120228
70 ksi76.5116220
Which chapter governs? A web-slenderness decision tree Compute h/t_w and λ_pw, λ_rw from Table B4.1b h/t_w ≤ λ_pw COMPACT WEB §F2 / §F3 M_p attainable λ_pw < h/t_w ≤ λ_rw NONCOMPACT WEB §F4 R_pc between M_y and M_p h/t_w > λ_rw SLENDER WEB §F5 — this chapter R_pg < 1; M_y is the ceiling Shear: §G2.1 (C_v1) or §G2.2 (TFA) Stiffeners: §G2.3 and §J10 λ_pw = 3.76√(E/F_y) = 90.6 · λ_rw = 5.70√(E/F_y) = 137 (F_y = 50 ksi) Every rolled W-shape in the AISC Manual has a compact web. Only fabricated girders reach §F5.
Figure 10.2 — The decision tree. Crossing λrw moves you from §F4 to §F5 and changes both the flexure and the shear provisions.

The §F13.2 upper limit exists because an extremely thin web has essentially no out-of-plane stiffness: it will "oil-can" under its own welding distortion, and web fatigue cracking at the stiffener terminations becomes a real durability issue. If you need to exceed it, add a longitudinal stiffener at h/5 from the compression flange, which raises the limit to 0.40E/Fy.

3. Flexural Strength — §F5 in Four Limit States

A slender-web member can never reach Mp — the compression portion of the web buckles out of plane and sheds its stress to the compression flange long before the section plastifies. §F5 handles this by capping the strength at first yield (SxcFy) and then applying a further reduction Rpg.

3.1 The bend-buckling reduction factor Rpg

Spec. Eq. F5-6Rpg = 1 − [aw / (1200 + 300aw)]·(hc/tw − 5.70√(E/Fy)) ≤ 1.0
Web-to-flange area ratioaw = hc·tw / (bfc·tfc) ≤ 10

Read Rpg physically: it is the fraction of first-yield moment that survives after the buckled compression web sheds load. It depends on how slender the web is (hc/tw beyond λrw) and on how much web there is relative to the flange (aw). A big, stiff flange (small aw) can absorb the shed stress, so Rpg stays near 1.0. A thin flange over a big web is penalized hard. Typical values run 0.90–0.98.

3.2 The four §F5 limit states

Limit stateSectionNominal strengthWhat sets it
Compression-flange yieldingF5.2, Eq. F5-1Mn = RpgFySxcMaterial; the ceiling for every other check
Lateral-torsional bucklingF5.2, Eqs. F5-2/3/4Mn = RpgFcrSxcLb vs Lp, Lr; usually governs
Compression-flange local bucklingF5.3Mn = RpgFcrSxcbf/2tf vs λpf, λrf
Tension-flange yieldingF5.4, Eq. F5-10Mn = FySxtOnly matters when Sxt < Sxc (unequal flanges)

Note the two things §F5 does not have: no Rpc (that is §F4's noncompact-web factor), and no plastic moment anywhere. φb = 0.90, Ωb = 1.67 throughout.

3.3 LTB with the correct radius of gyration

Spec. Eq. F4-11rt = bfc / √[12(1 + ⅙aw)]
Eq. F4-7 / F5-5Lp = 1.1rt√(E/Fy) Lr = π·rt√(E / 0.7Fy)
Eq. F5-3 (Zone 2)Fcr = Cb[Fy − 0.3Fy(Lb − Lp)/(Lr − Lp)] ≤ Fy
Eq. F5-4 (Zone 3)Fcr = Cbπ²E / (Lb/rt)² ≤ Fy

Do not use ry of the full section. rt models the compression flange plus one-third of the compressed portion of the web acting as an isolated strut — which is the actual buckling mechanism. Because the web is thin, rt is close to bfc/√12, so flange width is your primary LTB control variable. Widening the compression flange from 16 in. to 20 in. buys 25% more Lp at almost no weight penalty relative to the whole girder.

4. Shear — the Web's Real Job (§G2)

A slender web buckles in shear at a small fraction of its yield strength. But unlike a column, a buckled shear panel does not fail: it re-organizes. The buckled sheet forms diagonal tension bands anchored by the flanges above and below and by the transverse stiffeners on each side, exactly like the diagonals of a Pratt truss. This post-buckling reserve is called tension-field action (TFA), and it typically triples or quadruples the panel shear strength.

4.1 Web plate buckling coefficient

Spec. Eq. G2-5kv = 5 + 5/(a/h)² (stiffened panels) ; kv = 5.34 when a/h > 3.0 or a/h > [260/(h/tw)]²

Tighter stiffener spacing → smaller a/h → bigger kv → higher shear buckling stress. This is the whole design lever.

4.2 Without tension-field action (§G2.1)

Spec. Eq. G2-1Vn = 0.6·Fy·Aw·Cv1 (Aw = d·tw)
Eq. G2-3 / G2-4Cv1 = 1.0 if h/tw ≤ 1.10√(kvE/Fy); otherwise Cv1 = 1.10√(kvE/Fy)/(h/tw)

4.3 With tension-field action (§G2.2)

Spec. Eq. G2-7Vn = 0.6FyAw[ Cv2 + (1 − Cv2) / (1.15√(1 + (a/h)²)) ]
Eq. G2-9 / G2-10 / G2-11Cv2 = 1.0, or 1.10√(kvE/Fy)/(h/tw), or 1.51kvE/[(h/tw)²Fy] depending on the slenderness range

TFA is permitted only when all of these hold:

  • The panel is not an end panel (there must be an adjacent panel to anchor the tension field).
  • 2Aw/(Afc + Aft) ≤ 2.5 — the flanges must be substantial enough to anchor the field.
  • h/bfc ≤ 6.0 and h/bft ≤ 6.0.
  • a/h ≤ 3.0.
  • The panel does not contain a large web opening.
End panels never get TFA. There is no adjacent panel beyond the support to resist the horizontal component of the diagonal tension, so the field cannot anchor. The standard cure is to space the first stiffener close to the support, which raises kv and Cv1 enough to carry the (maximum) end shear on §G2.1 alone.

Resistance factors: φv = 0.90, Ωv = 1.67 for plate girders. (The φv = 1.00 value applies only to rolled I-shapes with h/tw ≤ 2.24√(E/Fy).)

5. Transverse Stiffeners (§G2.3)

Slendernessbst/tst ≤ 0.56√(E/Fyst)
Spec. Eq. G2-13Ist ≥ b·tw³·j, j = 2.5/(a/h)² − 2 ≥ 0.5, b = min(a, h)

Where tension-field action is used, §G2.3 adds a second requirement (Eq. G2-14) proportioning Ist to the actual shear demand and the web/stiffener yield strengths. In practice both are satisfied by a very modest plate — the stiffness requirement almost never governs the plate size; slenderness and fabricator preference do.

Detailing that matters more than the arithmetic

  • Intermediate stiffeners may be stopped short of the tension flange by 4tw to 6tw. Do this — welding hard to the tension flange creates a severe fatigue detail (Category C′ or worse).
  • Where a stiffener also anchors a brace or a cross-frame, it must be attached to the compression flange to carry 1% of the flange force.
  • Single-sided stiffeners are cheaper and fine for shear; use pairs at bracing points and at concentrated loads.
  • Fillet welds to the web can be intermittent for intermediate stiffeners; bearing stiffeners need continuous welds sized to deliver the full reaction.

Additional Design Aids & Stratified Equations

Plate girder — transverse stiffener layout a (end panel — no TFA) a (interior — TFA OK) a a (end panel — no TFA) h (web depth)
End panels (adjacent to supports) may NOT use tension-field action; interior panels may. Tighten the end-panel spacing to compensate.
Panel aspect ratio limita/h ≤ (260/(h/t_w))² and ≤ 3.0
Stiffener stiffnessI_st ≥ b · t_w³ · j, j = 2.5/(a/h)² − 2 ≥ 0.5
Stiffener slendernessb_st/t_st ≤ 0.56 √(E/F_yst)

6. Concentrated Forces and Bearing Stiffeners (§J10)

A ⁵⁄₁₆ in. web is extremely weak in local bearing. At every support and every large concentrated load, check:

Limit stateEquationφNote
Web local yieldingJ10-2 (interior), J10-3 (end)1.00Rn = Fywtw(5k + lb) or (2.5k + lb)
Web local cripplingJ10-4, J10-5a/b0.75Usually the binding check on a thin web
Web sidesway bucklingJ10-6/70.85Only if the compression flange is unrestrained
Web compression bucklingJ10-80.90Concentrated forces on both flanges

When any of these fails, provide a bearing stiffener (§J10.8). It is designed as a short compression member consisting of the stiffener plates plus an effective strip of web — 12tw at the member end, 25tw at an interior location — with Lc = 0.75h. Fit it to bear (or weld it to develop the force) and clip the corners to clear the web-to-flange weld.

7. Design Workflow

  1. Set the depth. Start near span/12 for floors, span/10 for crane runways, span/20 for depth-limited transfers.
  2. Size the web. Pick tw so that h/tw ≤ 260 (or 322 for Fy = 36). Thinner is cheaper — the web is a big chunk of the weight.
  3. Size the flanges from the moment couple: Af ≈ Mu/(φbFy·h) as a first pass, then refine with Rpg.
  4. Check flange local buckling: keep bf/2tf ≤ 0.38√(E/Fy) to stay compact.
  5. Compute aw, Rpg, and all four §F5 limit states at the governing Lb. Verify φbMn ≥ Mu.
  6. Lay out stiffeners: tight near supports (shear), open or omitted at midspan (moment). Check each panel with its own a/h.
  7. Check the end panel on §G2.1 only. Adjust the first spacing until it works.
  8. Size intermediate stiffeners for slenderness and Ist.
  9. Design bearing stiffeners at every reaction and concentrated load per §J10.
  10. Serviceability: live-load deflection, camber for dead load, and vibration for long floor spans.
  11. Fatigue (Appendix 3) for crane runways and bridges — stiffener terminations and web-to-flange welds are the critical details.

8. Practical and Fabrication Notes

  • Standardize plate thicknesses. Mills stock certain thicknesses; ⁵⁄₁₆, ⅜, ½, ⅝, ¾, 1, 1¼ are easy. Odd thicknesses invite substitutions.
  • Keep one web thickness for the whole girder even if the shear diagram says otherwise. Splicing the web is more expensive than the steel you save.
  • Step the flange, not the web. A flange thickness change at a shop splice is cheap and follows the moment diagram nicely.
  • Camber for dead load. A 70 in. deep, 80 ft girder will deflect visibly under wet concrete. Specify camber in the shop.
  • Watch shipping. Roughly 12 ft height and 100 ft length are practical highway limits; beyond that you need field splices, which need their own §J design.

⚠ Common mistakes

  • Using Zx and Mp for a slender-web girder — §F5 is written in Sxc and can never reach Mp.
  • Taking Rpg = 1.0 without evaluating Eq. F5-6, or missing the aw ≤ 10 cap.
  • Using ry instead of rt in the LTB equations.
  • Applying tension-field action to an end panel — never permitted.
  • Confusing Cv1 (§G2.1) with Cv2 (§G2.2, inside the TFA bracket).
  • Using φv = 1.00 on a slender web. That value is only for stocky rolled I-shapes.
  • Checking shear only at the support instead of panel by panel.
  • Forgetting bearing stiffeners under concentrated loads — web crippling on a thin web is brutally low.
  • Using 25tw of effective web at an end bearing stiffener (it is 12tw).
  • Ignoring fatigue at stiffener terminations on a crane runway girder — that is where these girders actually crack.

🎞 Watch the Phenomenon — Loading Stages

Slender web plate under axial / shear compression σ σ
Slender web plate under compression / shear — flat, wave, buckle, post-buckling.
StageWhat is happening in the animation
1 — Flat / elastic (σ < Fcr)Plate carries uniform in-plane stress. Any minor out-of-flatness is negligible.
2 — Onset of waveVery small out-of-plane wave becomes visible near the critical stress. The plate is behaving like a wide, thin column that hasn't quite buckled.
3 — Elastic plate bucklingσcr = kπ²E / [12(1−ν²)(b/t)²]. Plates buckle into waves, not a single half-sine, because the edges are supported.
4 — Post-buckling resistanceUnlike columns, plates can carry more load after buckling — the edges keep supporting the flow of stress. Effective-width method (AISC E7) uses this.
5 — Yielding of edge strips / crushingUltimate failure is when the edge strips finally yield. This is why HSS wall slenderness and web h/tw get so much attention.

📖 Using the AISC Manual — Plate Girders

Companion reference: Chapter F13 / G — no dedicated selection tables (built-up sections). In practice, engineers rarely compute every quantity from first principles — the AISC Steel Construction Manual (16th Ed.) tabulates φRn (or Rn/Ω) for every rolled shape so you can pick a member in seconds. Formulas remain essential for understanding, verifying, and for anything the tables do not cover.

Because plate girders are custom-fabricated, the Manual does not tabulate φMn or φVn — you must compute from Chapter F13 (flexure with slender web) and G2.2 (tension-field action) directly. However, two aids help:

  • Manual Part 1 lists standard plate thicknesses and widths available from mills — always specify a stocked size.
  • Design Guide 25 (Frame Design Using Web-Tapered Members) gives complete worked examples for the girder + stiffener check sequence.
Table vs. formula — which to use? Use tables to select a shape quickly. Use formulas to verify odd geometry, non-standard grades (Fy≠50 ksi), unusual K-factors or Lb, and to answer exam problems that hand you a section not in the current Manual.
Welded plate girder (stiffeners shown)
Welded plate girder with intermediate transverse stiffeners and tension-field action (AISC Ch. F + App. 6).

Formula Sheet

NameEquationAISC Ref
Design strengthφ Rn ≥ RuAISC 360-22 B3.1

Worked Example

Common data for Examples 10.1–10.4. A welded plate girder spans 80 ft simply supported. Web PL 70 × ⁵⁄₁₆; flanges PL 18 × 1 (both). All plate is ASTM A572 Grade 50 (Fy = 50 ksi, Fu = 65 ksi, E = 29,000 ksi). Compression flange braced at the ⅕ points (Lb = 20 ft). Intermediate stiffeners at a = 200 in. in the interior; end panel a = 100 in. End reaction Ru = 250 kips over a 10 in. bearing length.

Worked Example 10.1 — Slender-Web Flexural Strength (§F5)

Problem statement. Determine the available flexural strength of the plate girder for compression-flange yielding and lateral-torsional buckling, and verify that the section is slender-web.

Welded plate girder — PL 18×1 flanges, PL 70×⁵⁄₁₆ web, ASTM A572 Gr. 50 b_f = 18 in. t_f = 1 in. h = 70 in. d = 72 in. t_w = ⁵⁄₁₆ in. → h/t_w = 224 λ_rw = 5.70√(E/F_y) = 137 224 > 137 → SLENDER WEB (§F5) A_w = 21.9 in² I_x = 54,300 in⁴ S_x = 1,509 in³ a_w = 1.215 Simple span 80 ft; L_b = 20 ft (braced at ⅕ points); C_b = 1.0 assumed conservatively
Figure 10.1a — Girder cross-section with all dimensions and computed section properties
Step 1 — Classify the web (§B4.1, Table B4.1b).
h/tw = 70 / 0.3125 = 224; λrw = 5.70√(E/Fy) = 5.70√(29,000/50) = 5.70(24.1) = 137
224 > 137 → the web is slender, so §F5 governs the flexural design. Also confirm the §F13.2 limit for girders without longitudinal stiffeners: h/tw = 224 ≤ 260 o.k.
Step 2 — Section properties.
ComponentA (in²)d from N.A. (in.)A·d² (in⁴)Iown (in⁴)
Top flange 18 × 118.035.522,6851.5
Web 70 × ⁵⁄₁₆21.9008,932
Bottom flange 18 × 118.035.522,6851.5
Total57.9——Ix = 54,300
Sxc = Sxt = Ix/(d/2) = 54,300/36.0 = 1,509 in³
Step 3 — Bend-buckling reduction factor Rpg.
Spec. Eq. F5-6aw = hctw / (bfctfc) ≤ 10
aw = 70(0.3125)/[18(1.0)] = 21.9/18.0 = 1.215 ≤ 10 ✓
Spec. Eq. F5-6Rpg = 1 − [aw/(1200 + 300aw)]·(hc/tw − 5.70√(E/Fy)) ≤ 1.0
Rpg = 1 − [1.215/(1200 + 300(1.215))](224 − 137)
= 1 − [1.215/1565](86.7) = 1 − (0.000777)(86.7) = 1 − 0.0673 = 0.933
Step 4 — Compression-flange yielding (§F5.2).
Spec. Eq. F5-1Mn = Rpg·Fy·Sxc
Mn = 0.933(50)(1,509) = 70,370 kip-in = 5,860 kip-ft
LRFD: φbMn = 0.90(5,860) = 5,280 kip-ft; ASD: Mn/Ωb = 5,860/1.67 = 3,510 kip-ft
Step 5 — Compression-flange local buckling (§F5.3). λ = bfc/(2tfc) = 18/(2×1.0) = 9.00; λpf = 0.38√(E/Fy) = 0.38(24.1) = 9.15
9.00 < 9.15 → the flange is compact, so CFLB does not reduce Mn. (A ⅞ in. flange would have λ = 10.3 and would be non-compact — a good reminder that flange thickness is a stability decision, not just a strength one.)
Step 6 — Lateral-torsional buckling (§F5.2) at Lb = 20 ft.
Spec. Eq. F4-11rt = bfc / √[12(1 + ⅙aw)] = 18/√[12(1 + 0.203)] = 18/3.80 = 4.74 in.
Spec. Eq. F4-7Lp = 1.1·rt·√(E/Fy) = 1.1(4.74)(24.1) = 126 in. = 10.5 ft
Spec. Eq. F5-5Lr = π·rt·√(E/(0.7Fy)) = π(4.74)√(29,000/35) = 14.9(28.8) = 428 in. = 35.7 ft
Lp = 10.5 ft < Lb = 20 ft < Lr = 35.7 ft → Zone 2, inelastic LTB
Spec. Eq. F5-3Fcr = Cb[Fy − 0.3Fy·(Lb − Lp)/(Lr − Lp)] ≤ Fy
With Cb = 1.0 (conservative): Fcr = 50 − 15(240 − 126)/(428 − 126) = 50 − 15(0.378) = 44.3 ksi
Spec. Eq. F5-2Mn = Rpg·Fcr·Sxc = 0.933(44.3)(1,509) = 62,350 kip-in = 5,200 kip-ft
LRFD: φbMn = 0.90(5,200) = 4,680 kip-ft
Limit stateEquationMn (kip-ft)φbMn (kip-ft)
Compression-flange yieldingF5-15,8605,280
Lateral-torsional bucklingF5-2/F5-35,2004,680
Compression-flange local bucklingF5-3 (compact)n/an/a
Tension-flange yielding (Sxt = Sxc)F5-106,2905,660
Result: φbMn = 4,680 kip-ft, governed by inelastic lateral-torsional buckling at Lb = 20 ft.

Engineering interpretation: the slender web costs only 6.7% of capacity (Rpg = 0.933) — the web is not carrying much moment anyway, so bend-buckling of the web is a modest penalty. Bracing costs far more: going from Lb = 10.5 ft to 20 ft dropped the capacity 11%. That is the classic plate-girder trade: a deep, thin web is very efficient in bending, but the compression flange is a slender strut sitting on a flexible web, so bracing discipline matters more than in a rolled shape.

Common mistakes:
  • Using Zx and Mp. A slender-web girder can never reach the plastic moment; §F5 is written entirely in terms of Sxc.
  • Taking Rpg = 1.0 without evaluating Eq. F5-6.
  • Using ry of the whole section in the LTB equations instead of rt (Eq. F4-11), which represents the compression flange plus ⅓ of the compression web.
  • Missing the aw ≤ 10 cap, which matters for very thin flanges.
  • Forgetting to check the tension flange when Sxt < Sxc (unequal flanges).

Worked Example 10.2 — Shear Strength with Tension-Field Action (§G2.2)

Problem statement. Determine the available shear strength of an interior panel (a = 200 in.) and of the end panel (a = 100 in.) of the same girder.

Tension-field action: the stiffened panel behaves as a Pratt truss after the web buckles stiffener stiffener stiffener diagonal tension field (post-buckling) a = 200 in., a/h = 2.86 END PANEL no TFA (§G2.1) V
Figure 10.2a — After the web buckles diagonally, the panel acts as a truss: the web carries a diagonal tension field and the stiffeners act as compression posts. End panels have no anchoring panel beyond them, so TFA is not permitted there.
Step 1 — Confirm tension-field action is permitted (§G2.2).
RequirementComputedLimitStatus
2Aw/(Afc + Aft)2(21.9)/(18+18) = 1.22≤ 2.5o.k.
h/bfc70/18 = 3.89≤ 6.0o.k.
h/bft70/18 = 3.89≤ 6.0o.k.
Panel locationinteriornot an end panelo.k.
Step 2 — Web plate buckling coefficient (§G2.1). a/h = 200/70 = 2.86 (≤ 3.0, so the panel is stiffened)
Spec. Eq. G2-5kv = 5 + 5/(a/h)² = 5 + 5/(2.86)² = 5 + 0.61 = 5.61
Step 3 — Web shear buckling coefficient Cv2. 1.37√(kvE/Fy) = 1.37√[5.61(29,000)/50] = 1.37(57.0) = 78.1
h/tw = 224 > 78.1 → the web buckles elastically:
Spec. Eq. G2-11Cv2 = 1.51kvE / [(h/tw)²Fy] = 1.51(5.61)(29,000)/[(224)²(50)] = 245,700/2,508,800 = 0.0979
Only 9.8% of the shear-yield strength survives elastic buckling. Without TFA, this girder would be almost useless in shear.
Step 4 — Interior panel shear strength with TFA.
Spec. Eq. G2-7Vn = 0.6FyAw·[ Cv2 + (1 − Cv2) / (1.15√(1 + (a/h)²)) ]
0.6FyAw = 0.6(50)(21.9) = 656 kips
√(1 + 2.86²) = √9.16 = 3.03; 1.15(3.03) = 3.48
Bracket = 0.0979 + (1 − 0.0979)/3.48 = 0.0979 + 0.259 = 0.357
Vn = 656(0.357) = 234 kips → LRFD: φvVn = 0.90(234) = 211 kips; ASD: 234/1.67 = 140 kips
Step 5 — End panel, no tension-field action (§G2.1). a/h = 100/70 = 1.43 → kv = 5 + 5/(1.43)² = 7.45
1.10√(kvE/Fy) = 1.10√[7.45(29,000)/50] = 1.10(65.7) = 72.3 < 224
Spec. Eq. G2-4Cv1 = 1.10√(kvE/Fy) / (h/tw) = 72.3/224 = 0.323
Spec. Eq. G2-1Vn = 0.6FyAwCv1 = 656(0.323) = 212 kips
LRFD: φvVn = 0.90(212) = 191 kips ≥ Ru = 250 kips? NO — 191 < 250
Result: The interior panel is good for φvVn = 211 kips, but the end panel fails at 191 kips against the 250-kip reaction. Fix by tightening the end-panel spacing: at a = 55 in. (a/h = 0.786), kv = 5 + 5/0.617 = 13.1, Cv1 = 1.10(87.2)/224 = 0.428, and φvVn = 0.90(656)(0.428) = 253 kips ≥ 250 kips o.k.

Engineering interpretation: tension-field action nearly quadruples the panel shear strength (C = 0.098 → 0.357). That is not a code trick — it is real post-buckling behaviour, verified in hundreds of tests, in which the buckled web sheet acts as a set of diagonal tension bands anchored by the flanges and stiffeners. But the anchorage has to exist. At the end of the girder there is no adjacent panel to pull against, so the code forbids TFA there. The standard detailing response is exactly what we did: bunch the stiffeners near the supports where shear is highest, and spread them out at midspan where moment governs.

Common mistakes:
  • Applying Eq. G2-7 to the end panel. Never permitted (§G2.2).
  • Confusing Cv1 (§G2.1, shear buckling) with Cv2 (§G2.2, used inside the TFA equation).
  • Using kv = 5.34 (unstiffened web) in a stiffened panel, or using the stiffened kv when a/h > 3.0.
  • Using φv = 1.00 for a slender web. That value applies only to rolled I-shapes with h/tw ≤ 2.24√(E/Fy); here φv = 0.90.
  • Checking shear only at the support. Every panel needs its own check with its own a/h and its own Vu.

Worked Example 10.3 — Intermediate Transverse Stiffener Design (§G2.3)

Problem statement. Design intermediate transverse stiffeners for the interior panels (a = 200 in.) of the girder, using tension-field action.

Step 1 — Stiffener slenderness limit (§G2.3). Try 2 PL 4 × ⅜ in., one each side of the web.
§G2.3b/t ≤ 0.56√(E/Fyst)
b/t = 4.0/0.375 = 10.7; 0.56√(29,000/50) = 0.56(24.1) = 13.5
10.7 ≤ 13.5 o.k.
Step 2 — Required moment of inertia (§G2.3).
Spec. Eq. G2-13Ist ≥ b·tw³·j, where j = 2.5/(a/h)² − 2 ≥ 0.5
j = 2.5/(2.86)² − 2 = 0.306 − 2 = −1.69 → use j = 0.5
b = the smaller of a and h = min(200, 70) = 70 in.
Ist1 = 70(0.3125)³(0.5) = 70(0.0305)(0.5) = 1.07 in⁴
Step 3 — Provided moment of inertia about the web centerline.
Ist,prov = tst(2bst + tw)³ / 12 = 0.375(2×4 + 0.3125)³/12
= 0.375(8.31)³/12 = 0.375(574)/12 = 17.9 in⁴ ≫ 1.07 in⁴ o.k.
Step 4 — Detailing requirements.
  • Intermediate stiffeners need not bear on the tension flange; they may be stopped short of it by 4tw to 6tw = 1¼ in. to 1⅞ in.
  • Stop the weld 4tw to 6tw short of the web-to-flange weld to avoid a triaxially restrained intersection and to reduce the fatigue-detail category.
  • Where the stiffener also serves as a connection point for a brace or a beam, it must be welded to the compression flange to transmit 1% of the flange force (§G2.3).
Result: Use 2 PL 4 × ⅜ in. intermediate stiffeners in pairs at 200 in. on centre, stopped 1½ in. short of the tension flange, with intermittent fillet welds to the web.

Engineering interpretation: notice how small the required Ist is (1.07 in⁴) compared with what any practical plate provides (17.9 in⁴). The stiffness requirement almost never governs; the b/t slenderness limit and the fabricator's minimum practical plate size do. Do not over-think stiffener sizing — but do think hard about stiffener spacing, which directly sets kv and therefore the shear capacity, and about the welding, which sets the fatigue category for a crane-runway or bridge girder.

Common mistakes:
  • Forgetting the j ≥ 0.5 floor and getting a negative required Ist.
  • Using b = a instead of b = min(a, h) in Eq. G2-13.
  • Computing Ist about the stiffener's own centroid rather than about the web centerline (for a pair) or the web face (for a single-sided stiffener).
  • Welding intermediate stiffeners hard against the tension flange, creating a severe fatigue detail.
  • Treating an intermediate stiffener as a bearing stiffener. Different section (§J10.8), different requirements.

Worked Example 10.4 — Bearing Stiffener at the End Reaction (§J10)

Problem statement. The end reaction is Ru = 250 kips applied over a bearing length lb = 10 in. Check web local yielding and web local crippling, then design a bearing stiffener if required.

Bearing stiffener at the end reaction — cruciform column of stiffener plates + effective web girder web t_w = ⁵⁄₁₆ in. R_u = 250 kips bearing length l_b = 10 in. Section A-A (plan of the stiffener pair) 2 PL 8 × ⅝ in. 8 in. 12t_w = 3.75 in. of effective web
Figure 10.4a — The bearing stiffener pair plus an effective strip of web acts as a short column carrying the reaction directly into the bearing
Step 1 — Web local yielding at the girder end (§J10.2). Take k = tf + weld leg ≈ 1.0 + 0.3125 = 1.31 in. Because the reaction is at the end (distance from end < d):
Spec. Eq. J10-3Rn = Fywtw(2.5k + lb)
Rn = 50(0.3125)[2.5(1.31) + 10] = 15.6(13.3) = 208 kips; φ = 1.00 → φRn = 208 kips
208 kips < 250 kips → web local yielding is exceeded; stiffeners required.
Step 2 — Web local crippling at the girder end (§J10.3). lb/d = 10/72 = 0.139 ≤ 0.2, so:
Spec. Eq. J10-5aRn = 0.40tw²[1 + 3(lb/d)(tw/tf)1.5]√(EFywtf/tw)
(tw/tf)1.5 = (0.3125)1.5 = 0.175 → bracket = 1 + 3(0.139)(0.175) = 1.073
√(29,000 × 50 × 1.0/0.3125) = √4,640,000 = 2,154
Rn = 0.40(0.0977)(1.073)(2,154) = 90.2 kips; φ = 0.75 → φRn = 67.6 kips ≪ 250 kips
Step 3 — Size the bearing stiffener as a compression member (§J10.8, §E3). Try 2 PL 8 × ⅝ in., full depth, fitted to bear on the bottom flange.
Slenderness: b/t = 8/0.625 = 12.8 ≤ 0.56√(E/Fy) = 13.5 o.k.

Effective column section = stiffener pair + 12tw of web (end of member):
A = 2(8)(0.625) + 12(0.3125)(0.3125) = 10.0 + 1.17 = 11.2 in²
I = tst(2bst + tw)³/12 = 0.625(16.31)³/12 = 0.625(4,341)/12 = 226 in⁴
r = √(I/A) = √(226/11.2) = 4.50 in.
§J10.8Lc = 0.75h = 0.75(70) = 52.5 in. → Lc/r = 52.5/4.50 = 11.7
Very stocky → Fe = π²E/(Lc/r)² = π²(29,000)/137 = 2,090 ksi; Fy/Fe = 0.024 → Fcr = 0.6580.024(50) = 49.5 ksi
Spec. Eq. E3-1φcPn = 0.90(49.5)(11.2) = 499 kips ≥ 250 kips o.k.
Step 4 — Bearing on the contact area (§J7). Clip ¾ in. off each stiffener corner to clear the web-to-flange weld: Apb = 2(8 − 0.75)(0.625) = 9.06 in²
Spec. Eq. J7-1Rn = 1.8FyApb = 1.8(50)(9.06) = 815 kips; φ = 0.75 → 612 kips ≥ 250 kips o.k.
CheckReferenceAvailableRequiredVerdict
Web local yielding (unstiffened)Eq. J10-3208 k250 kFAILS → stiffener needed
Web local crippling (unstiffened)Eq. J10-5a67.6 k250 kFAILS → stiffener needed
Bearing stiffener as a column§J10.8, E3-1499 k250 ko.k. (D/C 0.50)
Bearing on the fitted contact areaEq. J7-1612 k250 ko.k. (D/C 0.41)
Result: Provide 2 PL 8 × ⅝ in. full-depth bearing stiffeners, fitted to bear on the bottom flange and clipped ¾ in. at the web fillets, welded to the web to develop the 250-kip reaction.

Engineering interpretation: a ⁵⁄₁₆ in. web is a sheet of paper in local bearing — web crippling gives only 68 kips. The bearing stiffener does not "strengthen" the web; it bypasses it, carrying the reaction as a short column directly from the bearing to the top flange. That is why it must be fitted to bear (or welded to develop the force) and why it is checked as a compression member with Lc = 0.75h — the flanges provide partial rotational restraint at both ends.

Common mistakes:
  • Using 25tw of effective web at an end bearing. The end value is 12tw; 25tw applies to interior stiffeners.
  • Using Lc = h instead of 0.75h (§J10.8).
  • Forgetting the corner clip in Apb — the stiffener cannot bear where the flange fillet weld is.
  • Sizing the stiffener for strength but not checking the b/t slenderness limit.
  • Using Eq. J10-2 (interior) instead of J10-3 (end) for web local yielding, which is unconservative by roughly 2.5k.
Chapter 10 takeaway. A plate girder is a design in three parts that must be reconciled: the flanges carry the moment (§F5, with Rpg and LTB), the web carries the shear (§G2, with or without tension-field action), and the stiffeners make the web work (§G2.3 for shear panels, §J10 for concentrated forces). Shear governs near the supports, moment governs at midspan, and the stiffener layout is the design variable that lets you satisfy both with one web plate.

FE-Style Worked Examples(6)

Each example mirrors the NCEES FE Civil Reference Handbook style: brief givens, a labeled figure, AISC section reference, step-by-step numeric solution, and a single boxed answer.

Given
Span 80 ft, gravity wu=4 k/ft, no rolled W deep enough/economical.
AISC Reference
AISC Ch. F + App. F13
Step-by-step solution
  1. Decision
    Built-up I with h/tw > 260 → plate girder territory.
Answer Use welded plate girder, design per §F13.
Bend the rules: when to use a plate girder
Problem statement image
Welded plate girder (stiffeners shown)
DIMDimensions from the problem statement
wu = 4k
Plate girder — flange + web dimensions
  • Web h × t_w (slender if h/t_w > 5.7√(E/F_y))
  • Flange b_f × t_f (compact / non-compact / slender per B4.1)
  • Transverse stiffeners at spacing a control web shear

Graded Chapter Quiz(13 FE-style questions · AISC Manual required)

These questions reference AISC Steel Construction Manual (16th ed.) — sections, equations, and tables are cited explicitly. Use a calculator. Each question offers a clue you may reveal before answering. Submissions are recorded to your account once signed in.

C10-01AISC 360-22 Table B4.1b Case 15
1. Plate girder web is classified 'slender' when h/tw exceeds:
Welded plate girder h/tw stiff. web tw flange PL
C10-02AISC 360-22 §F5
2. §F5 gives Mn for slender-web I-shape as controlled by:
Welded plate girder h/tw stiff. web tw flange PL
C10-03AISC 360-22 Eq. F5-6
3. Bending strength reduction Rpg for slender web:
C10-04AISC 360-22 §G2.2
4. Tension-field action (TFA) uses:
C10-05AISC 360-22 §G2.2(b)
5. TFA not permitted in:
C10-06AISC 360-22 §G2.2
6. Intermediate stiffener spacing recommended a/h:
Welded plate girder h/tw stiff. web tw flange PL
C10-07AISC 360-22 Eq. G2-13
7. Stiffener area required (Ast) per Eq. G2-13:
C10-08AISC 360-22 §J10
8. Bearing stiffeners are required at:
C10-09AISC 360-22 §F5
9. Flange local buckling (FLB) for slender-web girder governs when:
C10-10AISC 360-22 §F5
10. aw (web-to-flange ratio) =
C10-11AISC 360-22 §J1
11. Girder web splice designed for:
C10-12AISC 360-22 §F5 & Comm.
12. Hybrid girder (flange Fy > web Fy) requires:
C10-13AISC 360-22 Eq. G2-6
13. For a girder h/tw=200, Fy=50, a/h=2, kv from AISC:
Welded plate girder h/tw stiff. web tw flange PL
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§11Section 11

Chapter summary

Formula sheet
  • Design strength
    φ Rn ≥ Ru
    AISC 360-22 B3.1
Engineering checklist
  • Module 10: Plate Girders
  • Key limit states and AISC references are listed in the reference box.
  • Use φRn ≥ Ru for every check.
  • Verify section properties with the official AISC Manual.
Professional tips
  • • Mixing ASD and LRFD load combinations in the same problem.
  • • Using nominal strength Rn instead of design strength φRn.
  • • Forgetting to check every limit state listed in the AISC chapter.
§13Section 13

FE exam preparation

FE exam preparation
Concept review
Concept summary coming soon.
Calculator tips

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Common exam traps

Traps coming soon.

Time management

Aim for ~3 minutes per FE problem; skip and return to any item that takes longer than 5 minutes.