13

Welded Connections

Why Structural Steel?

Fillet, groove, weld symbols, and effective throat.

⏱ 90 min★ Core✎ 3 objectives
§01Section 01

Engineering story

Engineering story
Chapter 13 · Welded Connections

Fillet, groove, weld symbols, and effective throat.

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
  • Compute fillet-weld strength
  • Read AWS weld symbols
  • Size groove welds
§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 §J2 — Welds

Why This Chapter Matters

Welded connections are stronger and stiffer than bolted, but weld throat, weld size, and weld orientation change capacity dramatically. Fillet welds under longitudinal shear are the workhorse — and the most misdesigned.

Learning Objectives

  • Read AWS weld symbols and identify fillet, groove (CJP/PJP), plug, and slot welds.
  • Compute effective throat: t = 0.707 · w for equal-leg fillets.
  • Design fillet welds per AISC J2.4: φRn = 0.75 · 0.60 · FEXX · te · L · (1.0 + 0.5 sin^1.5 θ).
  • Detail minimum and maximum weld sizes (Table J2.4).
  • Check strength of the base metal alongside weld metal.

Where This Chapter Is Used

Every shear-tab, moment-connection weld, and base-plate weld from Chapters 14 through 20.

ANSI / AISC 360-22Specification for Structural Steel Buildings16.1-124 to 16.1-127 (§J2)
Chapter
J
AISC 360-22

Chapter J. Design of Connections — Welds (§J2)

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
J2.1Groove Welds (CJP, PJP) — Effective Area16.1-124
J2.2Fillet Welds — Effective Throat16.1-125
J2.3Plug and Slot Welds16.1-126
J2.4Strength (φRn = 0.75·0.60·FEXX·Awe·(1+0.5 sin^1.5 θ))16.1-126
J2.5Combination of Welds16.1-127
J2.6Filler Metal Requirements16.1-127
J2.7Mixed Weld Metal16.1-127

Companion reference: AWS D1.1 & AISC Manual Part 8 — Welded Connection Design

Lecture Notes

Chapter 13 — Welded Connections (AISC 360-22 §J2)

Chapter focus. Bolts and welds do the same job — transfer force between connected parts — but a weld is a continuous, monolithic joint rather than a set of discrete fasteners. That changes everything about how it's analyzed: strength depends on the throat area you actually deposit, the direction the load pulls relative to the weld axis, and whether the weld or the base metal beside it fails first. This chapter builds up fillet-weld strength from first principles, then extends to eccentrically loaded weld groups, which is the connection-design skill most students find hardest.

1. Weld Processes — Where the Numbers Come From

Structural steel is almost always joined with an arc-welding process. SMAW (shielded metal arc, "stick") is used for field welds and repairs. FCAW (flux-cored arc) and SAW (submerged arc) dominate shop fabrication because they are faster and produce more consistent quality — SAW in particular gives full-strength, low-defect welds and is standard for shop-fabricated built-up girders. The electrode classification (e.g., E70XX) tells you the nominal tensile strength of the deposited weld metal, FEXX, in ksi — "70" means 70 ksi. Matching filler metal (FEXX ≥ Fu of the base metal) is required unless the engineer specifies otherwise, so that the weld does not become the weak link before the connected material yields.

  • SMAW — stick electrode, manual, versatile, common in field erection.
  • FCAW — cored wire, self- or gas-shielded, fast deposition, common in shop and field.
  • SAW — wire feed under a granular flux blanket, highest quality/deposition rate, shop-only (flat position).
  • GMAW — solid wire, gas-shielded ("MIG"), fast but wind-sensitive (shop use).

2. Weld Types

Weld types used in structural steel Fillet CJP groove PJP groove Plug weld Slot weld Fillet welds resist ~80% of structural steel joints; CJP used for full-strength splices.
Figure 13.1 — Fillet, CJP groove, PJP groove, plug and slot welds.

A fillet weld is a triangular weld deposited in the corner formed by two plates; it is by far the most common structural weld because it needs no edge preparation. A complete-joint-penetration (CJP) groove weld fuses the full thickness of the joint — with matching filler metal it develops the full strength of the thinner connected part, so no calculation of weld strength is needed (only the connected-part checks govern). A partial-joint-penetration (PJP) groove weld only fuses part of the joint thickness (effective throat < full thickness) and must be checked like a fillet weld using the effective throat from Table J2.1. Plug and slot welds fill a round hole or elongated slot in one plate to fuse it to an underlying plate — used for stitching lapped plates together or hard-to-reach shear transfer, not for primary load path connections.

3. Reading the Weld Symbol

AWS A2.4 weld symbol — anatomy field weld flag 5/16 8 arrow side ↑ reference line weld size (leg, in) length (in) Symbol below line = weld on ARROW side. Symbol above line = OTHER side. Flag (circle at bend) = field weld, not shop weld. Triangle = fillet weld symbol; other symbols denote groove/plug/slot.
Figure 13.2 — AWS A2.4 fillet weld symbol: arrow side vs. other side, size, length, and field-weld flag.

The weld symbol is drawn on a reference line with an arrow pointing to the joint. Information placed below the line describes the weld on the arrow side of the joint; information above the line describes the other side. For a fillet weld, the leg size is given to the left of the triangle symbol and the length (if not "weld all around") to the right. A filled circle at the bend in the leader is the field weld flag — it means this weld is made in the field, not the shop, which affects both cost and quality-control planning.

4. Effective Throat and Effective Area (§J2.2)

Fillet weld — leg w, effective throat te = 0.707·w w (leg) w (leg) te Effective throat is perpendicular to hypotenuse of the 45° triangle.
Figure 13.3 — Effective throat te = 0.707·w (equal-leg SMAW/FCAW fillet).

A fillet weld does not fail on its leg faces — it fails by shearing along the plane through the throat, the shortest distance from the root to the weld face. For an equal-leg fillet made by a process with a flat (or slightly convex) face, that throat is the leg size times cos 45° = 0.707. Submerged-arc welds get credit for deeper penetration: for legs ≤ 3/8 in the whole leg may be taken as the effective throat, an SAW-specific bonus.

§J2.2ate = 0.707·w (SMAW/FCAW/GMAW, equal legs)
§J2.2aAwe = te·Lw (effective area = effective throat × effective weld length)
  • w — nominal fillet leg size (in).
  • te — effective throat (in).
  • Lw — effective weld length (in), excluding starts/stops and craters.

Physically: doubling the leg size does not double the "amount of steel doing the work" in the same simple way people assume for bolts — the shear plane area scales with 0.707·w, so the strength per inch of weld is directly proportional to leg size w.

5. Fillet Weld Strength (§J2.4)

Weld metal fails in shear on the effective throat, regardless of the direction of the applied force relative to the weld axis. But the ductility of that shear failure depends on direction: a weld loaded transverse to its axis (θ = 90°) fails at a higher stress than one loaded longitudinally (θ = 0°), because the transverse geometry mobilizes a larger fraction of the throat before rupture. AISC captures this with a direction-of-loading multiplier kds.

Eq. J2-5kds = 1.0 + 0.50·sin1.5θ
Eq. J2-4Rn = Fnw·Awe = 0.60·FEXX·Awe·kds  (φ = 0.75, Ω = 2.00)
  • FEXX — electrode classification strength (ksi); 70 ksi (E70XX) is standard.
  • θ — angle between the line of action of the force and the weld's longitudinal axis (0° = longitudinal, 90° = transverse).
  • kds — ranges from 1.0 (θ=0°) to 1.5 (θ=90°); use the User Note strain-compatibility limits before applying kds > 1.0 to certain weld groups (elements loaded in different directions).

Physically: a transverse fillet weld can be sized 33% smaller than a longitudinal one for the same load, because it is inherently stronger per unit throat area — but engineers often ignore this bonus for simplicity and conservatism.

Fillet weld orientation — longitudinal vs transverse Longitudinal (θ = 0°) Fnw factor 1.0 force Transverse (θ = 90°) factor 1.5 force
Figure 13.4 — Longitudinal (θ=0°, factor 1.0) vs. transverse (θ=90°, factor 1.5) fillet weld orientation.

6. Minimum and Maximum Fillet Weld Size (Table J2.4, §J2.2b)

A weld that is too small relative to the plate thickness cools too fast and can crack; hence Table J2.4 sets a minimum leg size that increases with the thickness of the thinner connected part — this is a heat-input/quality requirement, not a strength requirement. A weld along the edge of a thin plate must not overrun the edge, so the maximum leg is capped as well.

Thinner part t (in)wmin (in)
t ≤ 1/41/8
1/4 < t ≤ 1/23/16
1/2 < t ≤ 3/41/4
t > 3/45/16
§J2.2b(b)wmax = t − 1/16 in (t ≥ 1/4 in, along an edge)  or  wmax = t (t < 1/4 in)

Physically: the minimum-size table exists so that the weld deposits enough heat/volume to fuse properly into thicker material without excessive quench cracking — it must be checked independently of the strength calc, and it can govern even when the calculated required size is smaller.

7. Minimum Effective Length and the Length-to-Size Penalty (§J2.2b(d))

A fillet weld's effective length must be at least 4 times its nominal size, or the size is treated as 1/4 of that length (a strength penalty for stubby welds — stress does not redistribute uniformly at the ends). At the opposite extreme, very long, end-loaded fillet welds (loaded only at the ends, e.g. lap splices) suffer a reduction because stress is not uniform along a long weld — the ends see more load than the middle before the whole length yields.

Eq. J2-1β = 1.20 − 0.002(l/w) ≤ 1.0  (applies when l/w > 100, end-loaded fillet welds)
  • l — length of the end-loaded fillet weld (in).
  • w — leg size of the fillet weld (in).
  • If l/w > 300, use β = 0.60 (Note: for l/w beyond that range, effectiveness plateaus).

Physically: this is why splitting a long lap splice into two shorter parallel welds (or adding transverse end welds) is often more efficient than one very long weld of minimum size — the reduction factor β directly de-rates the nominal strength you calculated from Eq. J2-4.

8. End Returns

At the end of a fillet weld that resists a tension load parallel to the weld (as in a lap-splice angle or bracket), the weld should be returned around the corner for a short distance (typically 2·w) rather than stopping abruptly. This "end return" reduces the stress concentration at the weld terminus, where premature cracking is otherwise common — a detailing best practice rather than a strength calculation, but commonly required by AISC Manual connection tables.

9. Base-Metal Strength Check (§J2.4(a))

A weld connection can fail two different ways: the weld metal can shear through its throat, or the base metal adjacent to the weld can tear (shear rupture along the fusion face). Both must be checked — the weld can be strong enough while the thin plate beside it isn't, especially for thin gusset plates welded with an oversized fillet.

Eq. J2-2 / J4-4 formφRn,BM = φ·0.60·Fu·tp·Lw  (φ = 0.75)

Physically: for typical E70 electrodes on A36/A572-50/A992 steels, weld metal governs (is weaker) when w is smaller than roughly 5/16 in per inch of connected-plate thickness ratio; base metal tends to govern for thin, high-strength plates joined with an oversized weld.

10. Eccentrically Loaded Weld Groups

Eccentrically loaded weld group — elastic vector method Centroid of weld group (C.G.) P (applied load) e (eccentricity to C.G.) r_uv (torsional) r_u = r_direct(vector) + r_uv(torsional, IC location) — max resultant governs weld size.
Figure 13.5 — Bracket weld group: load P applied at eccentricity e from the weld-group centroid produces both a direct shear and a torsional shear on each weld element.

Brackets, single-plate shear connections, and gusset-to-column welds are rarely loaded through the centroid of the weld group. The eccentricity produces a moment M = P·e that the weld group must resist in addition to the direct shear. Two methods are used:

  • Elastic (vector) method — treats the weld group like a bolt group: direct shear P/Aw is added vectorially to a "torsional" shear M·r/J (polar analogy), and the largest resultant vector governs. Conservative, hand-calculable, and independent of load direction relative to any one weld segment — this is what students should master first.
  • Instantaneous center (IC) method — recognizes that the weld group actually rotates about an instantaneous center (not its centroid) and that each weld element can develop different amounts of the kds directional bonus and inelastic deformation before rupture. It is the basis of AISC Manual Tables 8-4 through 8-11 and gives higher (less conservative, more accurate) capacities, but requires iteration or the tables/software.
Elastic methodfv = P/Aw (direct)  ft = M·c/J (torsional)  ru = √(fv² + ft² + 2fvftsinθ)
  • Aw — total weld throat area treated per unit throat (usually work per inch of Lw, i.e., analysis is done for a 1-in throat and scaled by w at the end).
  • J — polar moment of inertia of the weld line group about its centroid, J = Ix + Iy.
  • c — distance from the weld-group centroid to the farthest weld point.

Physically: welds located far from the centroid (the corners of an L- or C-shaped weld group) see the highest combined stress and govern the required weld size; increasing the depth of a bracket weld group is a far more efficient way to reduce stress than increasing its width, because c and J both grow with depth.

11. Weld Quality, Inspection, and Common Defects

DefectCauseDetection method
PorosityTrapped gas (moisture, contamination)Visual (VT), radiography (RT)
Incomplete fusionInsufficient heat / wrong techniqueUltrasonic (UT), RT
UndercutExcess current, wrong angleVisual (VT)
Slag inclusionPoor interpass cleaning (multi-pass welds)UT, RT
Cracking (hot/cold)Restraint, hydrogen, fast coolingMagnetic particle (MT), UT
Insufficient throat/legUnder-sized weldFillet weld gauge, VT

AWS D1.1 governs welding procedure specifications (WPS), welder qualification, and acceptance criteria. CJP groove welds in tension resisting seismic or fatigue loads typically require UT or RT; routine fillet welds usually rely on visual inspection by an AWS Certified Welding Inspector (CWI), supplemented by MT/PT for suspect areas.

Design Workflow

  1. Determine required strength (LRFD: factored load; ASD: service load).
  2. Select weld type (fillet vs. groove) and geometry based on the joint configuration.
  3. Check Table J2.4 minimum size and §J2.2b(b) maximum size against connected-part thickness.
  4. Compute Fnw = 0.60FEXX·kds, apply φ = 0.75 (LRFD) or Ω = 2.00 (ASD).
  5. If end-loaded and l/w > 100, apply β reduction (Eq. J2-1).
  6. Check base-metal shear rupture on both connected parts (Eq. J2-2/J4-4 form).
  7. For eccentric groups, compute the weld-group centroid, J, and check the maximum resultant ru by the elastic method (or use IC tables for efficiency).
  8. Detail end returns, minimum length, and specify inspection level.

Additional Design Aids & Stratified Equations

Fillet weld orientation — longitudinal vs transverse Longitudinal (θ = 0°) Fnw factor 1.0 force Transverse (θ = 90°) factor 1.5 force
Directional strength increase: transverse fillet weld is 50% stronger than longitudinal.
Directional factorF_nw,eff = 0.60·F_EXX · (1.0 + 0.5·sin^{1.5}θ)
Weld strength /in (5/16 in, E70, longitudinal)φr_n = 0.75·0.60·70·0.707·(5/16) = 6.96 k/in
Exam / FE tip. Memorize φRn per inch of a 1/16-in fillet weld leg with E70 electrode, longitudinal loading: φ·0.60·70·0.707·(1/16) = 1.39 k/in. Scale linearly by leg size in sixteenths — e.g., a 5/16-in weld (5 sixteenths) gives 5×1.39 ≈ 6.95 k/in. This shortcut turns most fillet-weld sizing problems into simple division.

⚠ Common mistakes

  • Using w instead of 0.707·w for effective throat.
  • Ignoring base-metal shear rupture check — the weld can pass while the plate tears.
  • Under-sizing per Table J2.4 (violates code even if strength calc is OK).
  • Neglecting end returns and the β length reduction when l/w > 100.
  • Applying the kds directional bonus to weld groups where strain compatibility is not satisfied (mixed longitudinal + transverse elements) without following the User Note.
  • Forgetting that CJP welds require no strength calculation but DO require a minimum thickness/access and inspection plan.

🎞 Watch the Phenomenon — Loading Stages

Fillet weld connecting two plates — load transverse to weld axis throat (weakest) P
Fillet weld carrying transverse shear — elastic, yield on throat, rupture.
StageWhat is happening in the animation
1 — Load applied elasticWeld leg accepts shear elastically. Stress trajectories align with the throat plane.
2 — Yield along throatPeak stress is on the throat (shortest dimension across the fillet). Once throat shear reaches 0.6·FEXX, yielding begins.
3 — Ductile deformationWeld toe rotates as material stretches. If detailed and inspected correctly, the base metal yields before the weld ruptures.
4 — Directional strength effectTransverse loading yields ≈1.5× the parallel strength (AISC J2.4) — the throat sees more compression than pure shear.
5 — Throat ruptureFinal failure is a shear-tension fracture on the throat plane. φRn = 0.75·0.6·FEXX·Awe is calibrated to this.

📖 Using the AISC Manual — Welded Connections

Companion reference: Manual Part 8 — Tables 8-3 (fillet weld strength per 1/16″ per inch), 8-4 to 8-11 (eccentric weld coefficient C for common shapes). 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.

Fillet weld shorthand (E70XX): φRn per inch of weld = 1.392·D kip/in, where D is the weld size in sixteenths (5/16″ → D=5 → 6.96 kip/in). Table 8-3 tabulates this so you never re-derive 0.75·0.6·70/√2·(1/16).

Eccentric weld groups (Tables 8-4/8-11): the instantaneous-center coefficient C lets you write φRn = C·C1·D·L, where C is by geometry & load angle, C1 is the electrode-strength adjustment (1.0 for E70), D is weld size in sixteenths, L is the characteristic length. Replaces the fictitious "elastic" vector method that under-predicts capacity by 30–100%.

Directional strength increase: when load is perpendicular to the weld axis, Fnw is 1.5× the parallel case (Section J2.4). Table 8-3 already applies this when the angle θ is specified.

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.
Project case study — Cardinal Square — 4-story braced-frame office

Every chapter's worked example is one step in the design of the same building: Plan: 4 bays N–S × 3 bays E–W, each 30 ft × 30 ft. Stories: 4 @ 13 ft (52 ft roof). Composite floor: 4.5 in NW concrete on 3 VLI20 deck. Roof: 1.5 in B-deck + insulation + membrane. Materials: Wide-flange members A992 (Fy = 50 ksi, Fu = 65 ksi). Plates A572 Gr. 50. HSS bracing A500 Gr. C. Bolts A325-N 7/8 in dia. Welds E70XX. Concrete f'c = 4 ksi. Anchor rods F1554 Gr. 36.

Chapter 13 — Welded connection at the gusset
Fillet weld of brace gusset to column flange
Demand carried forward
From Chapter 4: brace force Tu ≈ 100 k.
This chapter contributes
Sizes weld leg w using φRn = 0.75 · 0.60 · FEXX · 0.707 · w · Lw (AISC §J2.4). Confirms minimum size per Table J2.4.
wLw — Fillet weld lap joint
Fillet weld: effective throat te = 0.707·w; φRn = 0.75·0.60·FEXX·te·Lw (AISC §J2.4).

Formula Sheet

NameEquationAISC Ref
Fillet weld strengthφRn = 0.75 · 0.60 · FEXX · te · LwAISC §J2.4
Effective throatte = 0.707 · w (for equal-leg fillet)AISC §J2.2

Worked Example

Worked Example 13.1 — Fillet Weld in Longitudinal Shear

Adapted from AISC Design Examples V16.0, Example J.1.

Given

  • 3/4-in-thick × 18-in-wide plate lap-welded to a 3/8-in-thick plate.
  • Both plates ASTM A572 Gr. 50: Fy = 50 ksi, Fu = 65 ksi.
  • Electrode: E70XX (FEXX = 70 ksi).
  • Two longitudinal fillet welds, size 5/16 in, each Lw = 27 in.
  • Required strength: RD = 33 k, RL = 100 k → LRFD Pu = 1.2(33) + 1.6(100) = 200 k (ASD Pa = 133 k).
Example J.1 — 3/4″ × 18″ plate lap-welded to 3/8″ plate, two 5/16″ fillets, L = 27 in two 5/16″ fillet welds, each L_w = 27 in P_u = 200 k 18″ wide × 3/4″ plate (top) lapped over 3/8″ plate (bottom), A572 Gr. 50
Figure 13.1a — Geometry and loading for Example J.1.

Step 1 — Maximum & Minimum Weld Size

§J2.2b(b)(2)wmax = t − 1/16 in = 3/8 − 1/16 = 5/16 in (thinner overlapping plate)
Table J2.4: minimum size for material thickness 3/8 in is 1/8 in. Chosen w = 5/16 in satisfies both min and max — a 5/16-in fillet weld is the largest single-pass flat/horizontal weld usable here (true up to 3/8 in).

Step 2 — Nominal Weld Strength per Inch

Eq. J2-5kds = 1.0 + 0.50·sin1.5(0°) = 1.0 (longitudinal loading)
Eq. J2-4Rn = 0.60·FEXX·Awe·kds
Awe per inch = 0.707·(5/16) = 0.221 in²/in
Rn = 0.60·70·0.221·1.0 = 5.57 kip/in.

Step 3 — Length-to-Size Ratio Reduction (§J2.2b(d))

Eq. J2-1β = 1.2 − 0.002(l/w) ≤ 1.0
l/w = 27.0/(5/16) = 144 > 100 → apply Eq. J2-1 (end-loaded fillet weld).
β = 1.2 − 0.002(144) = 0.912 ≤ 1.0 ✓

Step 4 — Nominal Weld Shear Rupture Strength

Rn = β·(5.57 kip/in)·2 welds·27 in = 0.912·5.57·54 = 274 kips
§J2.4φ = 0.75 (LRFD), Ω = 2.00 (ASD)
φRn = 0.75(274) = 206 k ≥ 200 k ✓   (ASD: 274/2.00 = 137 k ≥ 133 k ✓)

Step 5 — Base-Metal Strength (Governing Plate = 3/8-in Plate)

Eq. J2-2Rn = FnBM·ABM = 0.60·Fu·tp·lweld
Rn = 0.60·65·(3/8)·2 welds·27 in = 527 kips
φRn = 0.75(527) = 395 k ≥ 200 k ✓   (ASD: 527/2.00 = 264 k ≥ 133 k ✓)
Limit stateφRn (LRFD)Rn/Ω (ASD)Governs?
Weld shear rupture (with β)206 k137 kYes ✓
Base-metal shear rupture (3/8-in plate)395 k264 kNo
The two 5/16-in fillet welds, 27 in long, are adequate: φRn = 206 k ≥ Pu = 200 k. Weld metal shear (reduced for end-loaded length) governs over base-metal rupture.

Engineering interpretation: Because l/w = 144 exceeds 100, ignoring the β reduction would have overstated capacity by about 10% (274/0.912 = 300 k vs 274 k) — a common oversight for long lap-spliced welds.

⚠ Common mistakes

  • Forgetting the Eq. J2-1 length reduction on long end-loaded lap welds.
  • Sizing the weld from the thicker plate's max size rule instead of the thinner (governing) plate.
  • Skipping the base-metal shear rupture check on the thinner plate.

Worked Example 13.2 — Fillet Weld Loaded at an Angle

Adapted from AISC Design Examples V16.0, Example J.2A.

Given

  • Two-sided fillet weld between a gusset plate and a beam flange; weld loaded at θ = 60° to its longitudinal axis.
  • Beam and gusset plate thickness/length already sized adequately.
  • Electrode: E70XX. Minimum fillet size per Table J2.4 based on 5/8-in material = 1/4 in (beam flange thicker).
  • Required strength: RD = 50 k, RL = 150 k → LRFD Pu = 1.2(50) + 1.6(150) = 300 k (ASD Pa = 200 k).
  • Trial weld size w = 5/8 in.
Example J.2A — gusset-to-beam-flange fillet weld, load at θ = 60° to weld axis beam flange two-sided 5/8″ fillet, length l (req'd) P_u = 300 k θ = 60°
Figure 13.2a — Geometry and loading for Example J.2A.

Step 1 — Directional Strength Increase

Eq. J2-5kds = 1.0 + 0.50·sin1.5(60°)
sin(60°) = 0.866; sin1.5(60°) = 0.806
kds = 1.0 + 0.50(0.806) = 1.40

Step 2 — Available Shear Strength of the Weld per Inch

Eq. J2-4Rn = 0.60·FEXX·Awe·kds
Awe = 0.707·(5/8) = 0.442 in²/in
Rn = 0.60·70·0.442·1.40 = 13.0 kip/in.
Both sides: φRn (LRFD, φ=0.75) = 0.75·13.0·2 sides = 19.5 kip/in.
(ASD, Ω=2.00): 13.0/2.00·2 sides = 13.0 kip/in.

Step 3 — Required Weld Length

l = Pu / (φRn per inch, both sides)
l = 300 / 19.5 = 15.4 in (LRFD)    l = 200 / 13.0 = 15.4 in (ASD) — identical, as expected.
Use l = 16 in.
A two-sided 5/8-in fillet weld, 16 in long, is required to develop Pu = 300 k at θ = 60° to the weld axis, exploiting the 1.40× directional strength increase.

Engineering interpretation: Ignoring the directional increase (assuming kds = 1.0) would have required l = 300/13.9 ≈ 21.6 in — about 35% more weld — showing why the kds factor matters for gusset-to-flange welds loaded at an angle.

⚠ Common mistakes

  • Using θ measured from the transverse axis instead of the weld's longitudinal axis.
  • Forgetting weld strength is per side — must multiply by 2 for a two-sided fillet.
  • Not checking Table J2.4 minimum size against the thinner connected part before adopting the trial size.

Worked Example 13.3 — Fillet-Welded Lap Splice with Base-Metal Shear Rupture

Original example, styled after AISC Design Examples V16.0 Chapter J format.

Given

  • Two 1/2-in × 10-in plates, ASTM A992 (Fy = 50 ksi, Fu = 65 ksi), lap-spliced.
  • Factored tension load Tu = 180 k.
  • Electrode E70XX. Two longitudinal fillet welds, one on each side of the lap.
Worked Ex. 13.3 — Fillet-welded lap splice, two 1/2″ × 10″ plates, A992 two 5/16″ fillet welds, L_w each side T_u T_u = 180 k Both plates 10″ wide × 1/2″ thick, A992 (F_y=50, F_u=65 ksi); E70 electrode
Figure 13.3a — Fillet-welded lap splice, longitudinal welds along both edges of overlap.

Step 1 — Weld Size Limits (Table J2.4, §J2.2b(b))

Thinner part t = 1/2 in → Table J2.4 minimum wmin = 3/16 in.
Maximum along a 1/2-in edge: wmax = 1/2 − 1/16 = 7/16 in. Try w = 5/16 in (within range).

Step 2 — Weld Metal Strength per Inch (Longitudinal, θ = 0°)

Eq. J2-4/J2-5φRn/in = φ·0.60·FEXX·0.707·w·kds (kds=1.0)
φRn/in = 0.75·0.60·70·0.707·(5/16) = 6.96 kip/in per weld line.

Step 3 — Required Weld Length (Two Weld Lines)

Required ΣLw = Tu / (φRn/in) = 180 / 6.96 = 25.9 in → 12.95 in per side.
Check l/w = 13/(5/16) = 41.6 < 100 → no Eq. J2-1 reduction needed.
Use Lw = 13 in each side (26 in total).

Step 4 — Confirm Weld Capacity

φRn,weld = 6.96·26 = 181 k ≥ 180 k ✓ (DCR = 0.99, essentially exact — governs by design).

Step 5 — Base-Metal Shear Rupture Check (§J2.4(a), Eq. J2-2/J4-4)

φRn,BM = φ·0.60·Fu·tp·ΣLw (φ = 0.75)
φRn,BM = 0.75·0.60·65·0.5·26 = 380 k ≥ 180 k ✓ — base metal shear rupture does not govern.

Step 6 — Tension Yielding & Rupture of the Plate (Gross/Net Section, §D2)

φPn,yield = 0.90·50·(10·0.5) = 225 k ≥ 180 k ✓ (no holes — welded connection, Ag = An).
Limit stateφRn (k)Governs?
Fillet weld shear (26 in total, 5/16 in)181Yes ✓ (governing)
Base-metal shear rupture at weld line380No
Plate tension yielding225No
Use two 5/16-in fillet welds, 13 in long on each side of the lap (26 in total). Weld-metal shear governs at φRn = 181 k ≥ Tu = 180 k.

Engineering interpretation: Sizing the weld exactly to the demand (DCR ≈ 1.0) is typical of a "final design" pass — in practice, round up to available fabrication increments (e.g., 14 in) for a small reserve.

⚠ Common mistakes

  • Forgetting to check base-metal shear rupture in addition to weld-metal shear.
  • Missing the l/w > 100 check on long lap splices before dropping the Eq. J2-1 factor.
  • Not providing end returns at the free ends of the longitudinal welds.

Worked Example 13.4 — Bracket Weld Group by the Elastic (Vector) Method

Original example illustrating AISC Manual Part 8 elastic-method procedure for eccentric weld groups.

Given

  • C-shaped weld group welding a bracket plate to a column flange: top weld 8 in, bottom weld 8 in, vertical weld 22 in (connecting the two horizontal legs).
  • Load Pu = 40 k applied 16 in from the weld-group centroid (horizontal eccentricity), vertical direction.
  • Electrode E70XX; find required fillet weld size w.
Worked Ex. 13.4 — C-shaped bracket weld group, elastic vector method column top weld, 8″ bottom weld, 8″ vertical weld, 220″→22″? see given C.G. of weld group bracket plate P_u = 40 k e = 16 in (load to C.G.)
Figure 13.4a — C-shaped bracket weld group; load Pu applied at eccentricity e from centroid.

Step 1 — Weld Group Centroid and Properties (per inch of throat, unit weld size)

Treat the weld group as a line: two horizontal segments (8 in each, at y = +11 and y = −11 from mid-height) and one vertical segment (22 in, at x = 0, spanning y = −11 to +11).
Total length L = 8 + 8 + 22 = 38 in. By symmetry the centroid lies on the vertical leg, at x̄ = (8·4 + 8·4)/38 = 1.68 in from the vertical leg (horizontal segments extend from the vertical leg outward 8 in each).

Step 2 — Polar Moment of Inertia J (Unit Throat)

Ix = Σ(A·y²) + Iown ; Iy = Σ(A·x²) + Iown ; J = Ix + Iy
Vertical leg (22 in at x=0): Ix = 22³/12 = 887 in³(unit); Iy ≈ 22·(1.68)² = 62.1 in³
Each horizontal leg (8 in at y=±11): Ix = 8·11² = 968 in³ (×2 = 1936); Iy = 8³/12 + 8·(4−1.68)² = 42.7+43.1=85.8 (×2=171.6)
Ix ≈ 887 + 1936 = 2823 in³; Iy ≈ 62.1 + 171.6 = 233.7 in³
J = Ix + Iy = 3057 in³ (unit-throat polar moment of inertia)

Step 3 — Direct Shear and Moment

fv = Pu/L ; M = Pu·e
fv = 40/38 = 1.05 k/in (unit throat) acting vertically (parallel to Pu)
e = 16 in from centroid to load line → M = 40·16 = 640 k-in

Step 4 — Torsional Stress at Critical Point (Top-Right Corner)

ft,x = M·y/J ; ft,y = M·x/J
Critical point: top-right corner, x = 8 − 1.68 = 6.32 in, y = 11 in.
ft,x = 640·11/3057 = 2.30 k/in; ft,y = 640·6.32/3057 = 1.32 k/in
Resultant torsional ft = √(2.30² + 1.32²) = 2.65 k/in

Step 5 — Combine Direct + Torsional (Vector Sum, Conservative SRSS)

ru = √[(ft,x)² + (fv + ft,y)²]
ru = √[2.30² + (1.05+1.32)²] = √[5.29 + 5.62] = √10.91 = 3.30 k/in (per unit throat)

Step 6 — Required Fillet Weld Size

Eq. J2-4φRn/in = φ·0.60·FEXX·0.707·w = 27.8·w (E70, φ=0.75)
Set φRn/in ≥ ru: 27.8·w ≥ 3.30 → w ≥ 0.119 in.
Table J2.4 minimum (assume support ≥ 1/2 in): wmin = 3/16 in governs.
Use w = 1/4 in fillet weld for practical fabrication margin.
A 1/4-in fillet weld around the C-shaped bracket weld group (38 in total length) is adequate; Table J2.4 minimum size governs over the calculated demand (0.119 in), giving DCR ≈ 0.48 based on strength alone — the extra margin accommodates future load or fit-up tolerance.

Engineering interpretation: The elastic (vector) method is conservative relative to the AISC Manual's instantaneous-center tables (Manual Tables 8-4 to 8-11), which would likely permit a smaller weld size — but the elastic method is fully hand-calculable and is the correct starting point before relying on IC design aids.

⚠ Common mistakes

  • Computing J about the wrong centroid — must locate the true centroid of the weld line group first (asymmetric C-shape).
  • Forgetting to add the direct shear vectorially (not algebraically) to the torsional shear at the critical point.
  • Checking only one corner — for asymmetric groups, multiple corners must be checked to find the true maximum ru.
  • Ignoring the Table J2.4 minimum size, which frequently governs over the calculated demand for lightly loaded bracket welds.

Additional Worked Examples

Textbook — Aghayere & Vigil (2009)

Chapter 10 covers fillet, groove, and plug welds. Fillet welds are governed by shear on the effective throat (te = 0.707·w for equal-leg). The base metal must also be checked — weld design rarely controls when E70 electrodes are matched with A36/A992 base.

Example 10-1 — Fillet weld strength per inch

Example 10-1 — Fillet weld strength per inch

Setup. 5/16-in E70 fillet weld.

AISC Reference: AISC §J2.4

Numerical practice

Design shear strength per inch?

  1. A. 3.48 k/in
  2. B. 4.18 k/in
  3. C. 5.57 k/in (Answer)
  4. D. 6.96 k/in

Step-by-step solution

φRn/in = 1.392·5 = 6.96 k/in. (Or 0.75·0.60·70·0.707·(5/16) = 6.96.)

Example 10-4 — Required weld length

Example 10-4 — Required weld length

Setup. Tension splice plate carrying Pu = 90 k; use 1/4-in E70 fillet welds along both edges.

AISC Reference: AISC §J2.4

Numerical practice

Total weld length required?

  1. A. 8 in
  2. B. 11 in
  3. C. 16 in (Answer)
  4. D. 22 in

Step-by-step solution

Per inch (1/4 in fillet) = 1.392·4 = 5.57 k/in. Lw = 90/5.57 = 16.2 in → 16 in (8 in each side).

Example 10-6 — Concentric weld group — single angle

Example 10-6 — Concentric weld group — single angle

Setup. L4×4×3/8 to gusset, two longitudinal fillet welds along the heel and toe of one leg, Lw1 = 6 in, Lw2 = 4 in.

AISC Reference: AISC §D3, Table D3.1 (U)

Numerical practice

Balance the welds about the centroid: closer to heel weld length L1 should be (approx)?

  1. A. 3 in (Answer)
  2. B. 5 in
  3. C. 7 in
  4. D. 9 in

Step-by-step solution

x̄ = 1.13 in for L4×4×3/8 (b = 4 in). L1/Ltot = x̄/b = 0.282 → if Ltot = 10 in, L1 ≈ 2.8 in along the heel side.

Example 10-8 — Eccentric weld group (bracket)

Example 10-8 — Eccentric weld group (bracket)

Setup. C-shaped weld bracket: vertical legs Lv = 8 in, horizontal Lh = 4 in, E70, load P at 6-in eccentricity.

AISC Reference: AISC Manual Table 8-4

Numerical practice

Use Manual Table 8-4 with k = 0.5, a = 0.75 → coefficient C ≈ 1.85. φRn for D = 4 (1/4 in)?

  1. A. 20 k
  2. B. 44 k
  3. C. 60 k (Answer)
  4. D. 78 k

Step-by-step solution

φRn = C·D·L = 1.85·4·8 = 59.2 ≈ 60 k.

FE-Style Worked Examples(7)

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
1/4" fillet, E70 electrode (FEXX=70 ksi).
AISC Reference
AISC §J2.4 Eq. J2-4
Step-by-step solution
  1. te
    0.707 × 0.25 = 0.177 in
  2. φRn / in
    0.75 × 0.60 × 70 × 0.177 = 5.57 k/in
  3. Rule of thumb
    ≈ 1.392·D (D in 1/16") = 1.392×4 = 5.57 k/in ✓
Answer 5.57 k per inch of 1/4" fillet.
Fillet-weld strength per inch
Problem statement image
wLw — Fillet weld lap joint
DIMDimensions from the problem statement
FEXX = 70k
Fillet-weld geometry
  • Leg size w, throat t_e = 0.707 w
  • Effective length L_e (deduct 2w for wrap terminations if applicable)
  • φR_n = 0.75 · 0.60 F_EXX · t_e · L_e per inch

Interactive Calculator

Fillet Weld Strength

AISC §J2.4
te = 0.707·w0.177 in
φ Rn = 0.75·0.6·FEXX·te·L33.4 kipsNG

Longitudinal loading only. Use Instantaneous Center / directional strength increase for angled loads.

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.

C13-01AISC 360-22 §J2.2a
1. Fillet weld with equal leg w. Compute the effective throat te (SMAW).
Fillet weld — leg w, throat te w (leg) te = 0.707 w weld length Lw
C13-02AISC 360-22 Eq. J2-4
2. 5/16-in fillet weld (E70), LONGITUDINAL to the load, per inch of length. Compute φRn/inch (θ = 0°).
Fillet weld — leg w, throat te w (leg) te = 0.707 w weld length Lw
C13-03AISC 360-22 §J2
3. Required strength Pu = 90 k. Fillet welds each side of the lap (two lines, longitudinal). Compute the total weld length Lw required at 5/16-in.
Fillet weld — leg w, throat te w (leg) te = 0.707 w weld length Lw
C13-04AISC 360-22 Eq. J2-5
4. If the SAME 5/16 weld runs TRANSVERSE (θ = 90°), what is φRn/inch?
Fillet weld — leg w, throat te w (leg) te = 0.707 w weld length Lw
C13-05AISC 360-22 Table J2.4
5. AISC Table J2.4 minimum fillet-weld size for a 5/8-in thicker part:
Fillet weld — leg w, throat te w (leg) te = 0.707 w weld length Lw
C13-06AISC 360-22 §J2.2b
6. Maximum fillet-weld size along edge of a 1/2-in plate:
Fillet weld — leg w, throat te w (leg) te = 0.707 w weld length Lw
C13-07AISC 360-22 Eq. J4-4
7. Base-metal shear rupture through 1/2-in plate along 6-in weld length (Fu = 65) — check per §J4.2:
Fillet weld — leg w, throat te w (leg) te = 0.707 w weld length Lw
C13-08AISC 360-22 §J2.2b(c)
8. End-return length for fillet weld terminated at the free edge of a tension member:
C13-09AISC 360-22 §J2.3
9. CJP groove weld strength:
C13-10AISC 360-22 Table J2.5
10. PJP groove weld under tension normal to axis (Table J2.5):
C13-11AISC 360-22 Eq. J2-5
11. Directional strength (θ = 45° weld line to force): multiplier on 0.60·FEXX·te:
C13-12AISC Manual Part 8
12. Eccentric weld group (bracket) — designing by the elastic method vs instantaneous center of rotation (ICR, Manual Tables 8-4 to 8-11):
C13-13AISC 360-22 §J1.6
13. Weld access hole geometry (AISC §J1.6): what does it accomplish at a beam-column moment connection?

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§11Section 11

Chapter summary

Formula sheet
  • Fillet weld strength
    φRn = 0.75 · 0.60 · FEXX · te · Lw
    AISC §J2.4
  • Effective throat
    te = 0.707 · w (for equal-leg fillet)
    AISC §J2.2
Engineering checklist
  • Module 13: Welded Connections
  • 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

Calculator tips coming soon.

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.