Fillet Weld Strength Calculator

A fillet weld fails in shear across its throat, not in tension along its legs, and that single fact drives the whole calculation. This tool gives the nominal, design and allowable capacity of a fillet weld from its leg size, length and electrode classification, reports the capacity per inch that steel detailers work from, checks the base metal at the weld line, and solves backwards for the weld size or length a given load requires. It follows the AISC 360 Chapter J formulation, which AWS D1.1 shares.

Calculator

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Inputs this calculator takes, with typical values
InputWhat to enterExample
Fillet weld leg sizeThe leg dimension measured along the fusion face, which is what a weld gauge reads.0.25 in
Effective length of one weldLength of continuous weld that carries load; exclude craters and unwelded gaps.6 in
Number of weldsCount each separate line of weld, so a lap joint welded both sides is 2.2
Electrode classificationThe classification number is the tensile strength of the deposited metal in ksi.E70XX (70 ksi)
Design methodUse LRFD with factored loads and ASD with service loads; do not mix them.LRFD (φ = 0.75, factored loads)
Applied load on the jointTotal load shared by all the welds, factored for LRFD or unfactored for ASD.60 kips
Base metal tensile strength58 ksi for A36, 65 ksi for A572 Gr 50 and A992; read it from the mill certificate if you have one.58 ksi
Thickness of the connected elementThickness of the thinner part being joined, used for the base metal shear rupture check.0.375 in

It returns

  • Weld capacity (design or allowable) — φRn for LRFD or Rn/Ω for ASD, summed over all the welds.
  • Capacity per inch of weld
  • Nominal capacity Rn
  • Weld size required for the load
  • Base metal capacity at the weld
  • Utilisation of the governing capacity

The formula

Rn=0.60FEXX(0.707w)L
Rn=0.60FutL
wreq=Pφ0.60FEXX0.707L

In plain text: Rn = 0.60 × F_EXX × (0.707 × w) × L

  • RnNominal shear capacity of the weld group (kips)
  • F_EXXElectrode classification strength, e.g. 70 for E70XX (ksi)
  • wFillet leg size (in)
  • 0.707 × wEffective throat of an equal-leg fillet (in)
  • LTotal effective length of weld (in)

Design strength is φRn with φ = 0.75 for LRFD, or Rn/Ω with Ω = 2.00 for ASD. The 0.60 factor is the shear strength of weld metal as a fraction of its tensile classification.

Updated Category Welding & Thermal Cutting Verified against published test cases Reading time 11 min

Why a fillet weld is checked in shear across its throat

A fillet weld is a triangle of deposited metal filling the corner between two parts. Load it any way you like — pull the parts apart, push them along each other, peel them — and testing shows the weld consistently fails by shearing across the narrowest section through the triangle. That section is the effective throat, and for the equal-leg fillet used in almost all fabrication it is the leg size multiplied by the sine of 45°, or 0.707 × w.

Design therefore ignores the leg you can see and works with the throat you cannot. A 1/4 in fillet is checked on a throat of 0.1768 in. This is why doubling the leg size doubles the capacity exactly — throat scales linearly with leg — and why a weld larger than the drawing calls for buys strength at a cost that goes up with the square of the leg, since the cross-sectional area of deposited metal is w²/2. A 3/8 in fillet costs 2.25 times as much filler as a 1/4 in fillet to gain 50 percent more strength.

The nominal shear strength of weld metal is taken as 0.60 times its tensile classification. An E70XX electrode deposits metal with 70 ksi tensile strength, so its nominal shear strength is 42 ksi acting on the throat area. Multiply by throat and by length and you have the joint capacity, before any safety factor.

The formula, the safety factors, and the base metal check

The nominal capacity of a fillet weld group is Rn = 0.60 × FEXX × 0.707w × L, where L is the total effective length. Two reduction conventions convert it to something you can design with. Under LRFD you multiply by φ = 0.75 and compare against factored loads. Under ASD you divide by Ω = 2.00 and compare against service loads. Do not mix them: an ASD comparison against factored loads is roughly a 50 percent error in the unsafe direction.

This produces the number every steel detailer carries in their head. A 1/16 in fillet in E70 metal gives 0.60 × 70 × 0.0442 = 1.856 kips/in nominal, which is 1.392 kips/in factored and 0.928 kips/in allowable. Because throat is linear in leg, you multiply by the number of sixteenths: a 5/16 fillet is 5 × 1.392 = 6.96 kips/in under LRFD. Checking a calculator against 1.392 per sixteenth is the fastest way to confirm you are using the right convention.

The weld is only half the check. Weld metal deposited on a thin or soft plate can be stronger than the plate it is attached to, and the joint then fails by tearing the base metal along the weld line. AISC handles this as base metal shear rupture: Rn = 0.60 × Fu × t × L, with the same φ and Ω. The governing capacity is the smaller of the two. When base metal governs, adding weld size does nothing at all — a point this calculator flags explicitly, because it is the most common way a designer wastes filler.

Two geometry rules constrain the size you may specify. The maximum fillet along the edge of a part 1/4 in thick or thicker is the thickness less 1/16 in, so that the edge remains visible for inspection; below 1/4 in it is the full thickness. Minimum sizes are tabulated by the thickness of the thinner part joined, and exist to control cooling rate rather than strength — a small weld on heavy plate quenches into hard, crack-prone metal.

Worked example: a bracket carrying 100 kips

A gusset plate is welded to a column flange with two 10 in fillets, one each side, using E70XX electrode. The factored load is 100 kips. The gusset is 3/4 in A36 plate.

  1. Effective throat. Try a 5/16 in fillet: 0.707 × 0.3125 = 0.22094 in.
  2. Nominal strength per inch. 0.60 × 70 × 0.22094 = 9.2795 kips/in.
  3. Design strength per inch. 0.75 × 9.2795 = 6.9596 kips/in. Cross-check: 1.392 × 5 sixteenths = 6.960. Agreed.
  4. Total effective length. 2 welds × 10 in = 20 in.
  5. Design capacity. 6.9596 × 20 = 139.19 kips against a 100 kip demand, so utilisation is 72 percent.
  6. Base metal check. 0.60 × 58 × 0.75 × 20 = 522 kips nominal, and 0.75 × 522 = 391.5 kips. The weld governs comfortably.
  7. Solve for the minimum size. w = 100 ÷ (0.75 × 0.60 × 70 × 0.707 × 20) = 100 ÷ 445.4 = 0.2245 in, so a 1/4 in fillet (0.250 in) is sufficient and a 5/16 in fillet has margin.

Check the size rules before committing to 1/4 in. The gusset is 0.750 in thick, so the maximum edge fillet is 0.750 − 0.0625 = 0.6875 in, comfortably above. Minimum size for 3/4 in material joined to a heavier flange is 1/4 in under AISC's table, so 1/4 in is exactly at the floor and 5/16 in gives room for weld profile variation. Specify 5/16 in and the detail passes on strength, on maximum size and on minimum size at once.

Reading the numbers, and where they stop being conservative

Utilisation below about 90 percent is a healthy detail; at 100 percent you have no room for the weld being placed slightly small or slightly short. Effective length is not the length of the drawn weld — craters at each end, gaps and areas of poor fusion do not count, and end returns are treated conservatively.

Three limits move the answer away from the simple formula and are worth knowing.

Short end-loaded welds. Where the effective length of an end-loaded fillet is less than four times the leg size, AISC requires the effective size to be taken as one quarter of the length instead, because such a short weld cannot develop uniform stress. This calculator warns when you cross that threshold rather than silently reporting an unconservative figure.

Long end-loaded welds. Beyond 100 times the leg size, stress is far from uniform along the weld and a reduction factor applies; beyond 300 times, the factor is fixed at 0.60. Long lap-splice welds are the usual case.

Load direction. AISC permits an increase for loads not parallel to the weld axis, the familiar 1.0 + 0.50 sin1.5θ factor, which raises capacity by 50 percent for a weld loaded perpendicular to its axis. This calculator does not apply it, so the figures here are the longitudinal-loading values and are conservative for transverse loading. That is deliberate: the increase requires that you know the direction of load on every weld element, and in weld groups loaded eccentrically the direction varies along the weld.

None of this covers fatigue. Cyclically loaded welds are governed by stress range and detail category, not by static capacity, and a weld with 40 percent static utilisation can still fail in fatigue. If the load cycles, look up the detail category rather than relying on this page.

Fillet weld capacity per inch, E70XX electrode

Values are 0.60 × 70 × 0.707 × leg. The LRFD column is the familiar 1.392 kips per inch per sixteenth of leg; the ASD column is 0.928.
Leg sizeLeg (in)Throat (in)Nominal (kips/in)LRFD φRn (kips/in)ASD Rn/Ω (kips/in)
1/80.12500.088393.71232.78421.8561
3/160.18750.132585.56844.17632.7842
1/40.25000.176787.42465.56843.7123
5/160.31250.220979.28076.96054.6404
3/80.37500.2651711.13698.35275.5685
7/160.43750.3093612.99319.74486.4965
1/20.50000.3535514.849211.13697.4246
5/80.62500.4419418.561413.92119.2807

Multiply by the total effective weld length in inches. For other electrodes, scale by F_EXX / 70 — an E80 weld is 80/70 = 1.143 times these values.

Mistakes that put a welded connection at risk

  • Comparing ASD capacity against factored loads. The two conventions differ by a factor of 1.5, and the error is in the unsafe direction. Pick one method for the whole connection.
  • Forgetting the base metal check. A 1/2 in fillet on 1/4 in plate cannot deliver its capacity, because the plate tears first. When base metal governs, more weld buys nothing.
  • Using drawn length rather than effective length. Craters, stop-starts and unfused areas do not carry load. Effective length is what a weld inspector can certify.
  • Specifying a fillet larger than the edge allows. On material 1/4 in and thicker the maximum edge fillet is thickness less 1/16 in, so the plate edge remains visible for inspection.
  • Ignoring minimum size rules. The minimum fillet on heavy plate exists to slow cooling, not to provide strength. Undersizing it risks HAZ cracking regardless of the load.
  • Applying static capacity to a cyclic load. Fatigue is governed by stress range and detail category. A statically comfortable weld can still fail after a few million cycles.
  • Assuming an intermittent weld equals its total length. Intermittent fillets carry the sum of their segment lengths, but the discontinuities become stress raisers and are barred from many fatigue-sensitive and corrosive applications.

This calculator does not replace a qualified engineer

It implements the AISC 360 Chapter J fillet weld and base metal equations for statically loaded joints in direct shear. Real connections involve eccentricity, weld group geometry, block shear, prying, fatigue and the requirements of AWS D1.1 for procedure and inspection. Use these numbers to size and sanity-check a detail; have the connection designed and sealed by a licensed engineer where the structure demands one.

Fillet welds among the alternatives

Fillet welds dominate fabrication because they need no joint preparation: bring two parts together and weld the corner. Complete-joint-penetration groove welds, by contrast, are checked against the base metal rather than the weld, because a properly executed CJP weld in matching filler develops the full strength of the connected material. That makes CJP welds stronger, but they need edge preparation, backing or back-gouging, and usually ultrasonic or radiographic inspection, so they cost several times as much per inch. Choose a fillet unless the joint genuinely needs to develop the section.

Once the size is fixed, the shop questions follow. Metallurgy in the heat-affected zone is governed by energy per unit length, which the welding heat input calculator quantifies; on hardenable steels the same weld that passes this strength check can crack in the HAZ if it is run too fast and cold. If you are working with unlabelled material and no mill certificate, a hardness reading converted through the hardness conversion calculator gives an approximate tensile strength, which is the Fu this page needs for the base metal check.

Filler consumption follows from the geometry directly. The deposited area of an equal-leg fillet is w²/2 plus reinforcement, so a 5/16 in fillet deposits about 0.049 in² per inch of length. Multiply by length and by the density of steel to estimate electrode weight, the same arithmetic the metal weight calculator performs for stock shapes. That figure drives both cost and, through heat input, distortion.

For members rather than connections, the governing checks move to the section itself — axial stress and elongation from the stress and strain calculator, and bending from a section modulus check. A connection designed to 70 percent utilisation on a member already at 95 percent has not helped anybody.

Frequently asked questions

How strong is a 1/4 inch fillet weld per inch?

With E70XX electrode, 5.57 kips per inch under LRFD or 3.71 kips per inch under ASD. That comes from the standard figure of 1.392 kips per inch per sixteenth of leg size for LRFD, or 0.928 for ASD, multiplied by four sixteenths. The nominal capacity before any factor is 7.42 kips per inch. Always confirm that the base metal at the weld line can carry the same load.

Why is the throat 0.707 times the leg size?

Because the throat is the perpendicular from the root to the hypotenuse of an equal-leg right triangle, which equals the leg times sin 45° = 0.7071. Fillet welds fail by shearing across this shortest section, so it is the dimension design uses. If the fillet is unequal-legged or has a deeply concave face, measure or compute the actual throat rather than applying 0.707.

What size weld do I need for a given load?

Divide the load by the factored capacity per inch and by the total weld length. For 100 kips on two 10 in E70 welds under LRFD: 100 ÷ (0.75 × 0.60 × 70 × 0.707 × 20) = 0.2245 in, so a 1/4 in fillet works. Then check the result against the maximum edge fillet for your plate thickness and the minimum size table, either of which can override the strength answer.

Should I use LRFD or ASD?

Use whichever your project is designed in, and never mix them. LRFD applies φ = 0.75 to the nominal capacity and is compared with factored load combinations; ASD divides by Ω = 2.00 and is compared with service loads. The ratio between the two is exactly 1.5, so quoting an ASD capacity against factored loads overstates the joint by half. Both methods are permitted by AISC 360 and give comparable member sizes when applied consistently.

What is base metal shear rupture and when does it govern?

It is failure of the connected plate along the line of the weld, checked as 0.60 × Fu × t × L with the same φ or Ω. It governs when the plate is thin or its tensile strength is low relative to the weld metal — commonly on thin gussets or when a large fillet has been specified on light material. When it governs, increasing the weld size does not increase joint capacity; you need more weld length, thicker plate or a stronger grade.

Does this include the strength increase for transverse loading?

No, deliberately. AISC permits multiplying capacity by 1.0 + 0.50 sin1.5θ for loads at an angle to the weld axis, giving up to a 50 percent increase for perpendicular loading. Applying it requires knowing the load direction on every weld element, which in eccentrically loaded groups varies along the weld. The figures here are the longitudinal values, so they are conservative for transverse loads.

How does electrode classification change the answer?

Linearly. Capacity is proportional to FEXX, so an E80 weld is 80/70 = 1.14 times an E70 weld of the same size, and E110 is 1.57 times. AWS D1.1 requires filler metal to be matched or undermatched to the base metal according to the material group, so you cannot simply specify a stronger electrode to shrink a weld — check the filler metal requirements for the steel you are joining.

Are intermittent fillet welds treated the same way?

Their capacity is the sum of the effective segment lengths, so ten 2 in segments carry the same static load as one 20 in weld. What changes is everything else: each start and stop is a stress concentration, the gaps admit moisture and cause corrosion, and fatigue performance is significantly worse. Codes restrict intermittent welds in fatigue-sensitive and exposed applications for those reasons rather than on strength grounds.

References

  • ANSI/AISC 360, Specification for Structural Steel Buildings, Chapter J — American Institute of Steel Construction
  • AWS D1.1/D1.1M, Structural Welding Code — Steel — American Welding Society
  • Steel Construction Manual, 16th Edition — American Institute of Steel Construction