What a header does and what decides its size
A header is a beam that carries the load a wall would otherwise carry, across the width of an opening, and delivers it to the jack studs at each side. Everything about sizing it comes down to one number — the uniform load in pounds per foot that lands on it — and one span. Get those two right and the rest is arithmetic.
The load arrives from above by tributary area. A roof or floor spans between two supports, so each support picks up half of it. That half-span, measured in feet, is the tributary width, and multiplying it by the load in pounds per square foot gives pounds per foot of header. A house 24 ft deep with trusses bearing on both exterior walls gives a 12 ft tributary width, so a 35 psf roof puts 12 × 35 = 420 lb/ft on the header.
Two independent checks then decide the size. The bending check compares the moment to the section modulus and answers whether the header is strong enough. The deflection check compares the sag to a code limit and answers whether it is stiff enough. Which of the two governs is not a matter of taste, and it is not simply a matter of span — it depends on the depth of the section relative to the span, the deflection limit and the ratio of allowable stress to stiffness. The interpretation section below gives the threshold.
A third check follows the reactions down. Each end delivers half the total load into the jack studs, and those studs crush the plate across the grain long before they buckle — which is why heavily loaded openings need two or three jacks per side.
Working the load through to a size
Step one is the uniform load. w = tributary width × (roof live + roof dead + floors × floor load). Roof live load is the greater of the IRC minimum 20 psf and the local flat-roof snow load. Each floor above adds its own live plus dead load, typically 40 + 10 = 50 psf under IRC Table R301.5.
Step two is the design span. A header does not span the clear opening; it spans from the centre of one bearing to the centre of the other. The calculator uses the clear opening plus 3 in, allowing a full 1.5 in of bearing at each end. On a 6 ft opening that is 6.25 ft — a 4% increase in span, but an 8% increase in moment and a 17% increase in deflection, because of the powers involved.
Step three is the moment and the reactions. For a uniformly loaded simple span, M = wL²/8 and each reaction is wL/2. Both come straight out of statics and neither depends on the material.
Step four is the bending demand. Multiply the moment by twelve to get pound-inches, then divide by the adjusted allowable stress Fb′ = Fb × CD. The load duration factor comes from NDS Table 2.3.2 — 1.15 when snow governs, 1.0 when floor live load governs, 1.6 for wind. The result is the section modulus the header must supply, and the section modulus calculator gives you S for any candidate.
Step five is the stiffness demand. Take the allowable deflection, L divided by 240 or 360, and rearrange the deflection formula for I: I = 5wL⁴ / (384·E·Δ). Everything is in inches and pounds here, so the load in pounds per foot is divided by twelve first. The deflection calculator runs the same equation forwards if you want to check a size you already have.
Step six picks the section. Any member whose S and I both exceed the demands works. Plies multiply both in direct proportion; depth multiplies S by its square and I by its cube.
Worked example: a 6 ft window header carrying a 12 ft tributary of roof
A single-storey house, 24 ft deep, trusses bearing on both exterior walls. Ground snow gives a 20 psf roof live load, the roof assembly weighs 15 psf, and the opening is 6 ft wide. The header is Douglas Fir-Larch No. 2 with Fb = 900 psi and E = 1,600,000 psi, snow governs so CD = 1.15, and the limit is L/240.
- Tributary width. Half of 24 ft = 12 ft.
- Uniform load. w = 12 × (20 + 15) = 12 × 35 = 420 lb/ft.
- Design span. L = 6 ft + 3 in = 6.25 ft = 75 in.
- Design moment. M = 420 × 6.25² ÷ 8 = 420 × 39.0625 ÷ 8 = 2,050.8 lb·ft.
- End reactions. R = 420 × 6.25 ÷ 2 = 1,312.5 lb at each end.
- Adjusted stress. Fb′ = 900 × 1.15 = 1,035 psi.
- Required section modulus. S = 2,050.8 × 12 ÷ 1,035 = 24,609 ÷ 1,035 = 23.78 in³.
- Allowable deflection. Δ = 75 ÷ 240 = 0.3125 in.
- Required moment of inertia. With w = 420 ÷ 12 = 35 lb/in and L = 75 in, L⁴ = 31,640,625, so I = 5 × 35 × 31,640,625 ÷ (384 × 1,600,000 × 0.3125) = 5,537,109,375 ÷ 192,000,000 = 28.84 in⁴.
- Pick the section. A two-ply 2x8 measures 3.0 × 7.25 in, giving S = 3.0 × 7.25² ÷ 6 = 26.28 in³ and I = 3.0 × 7.25³ ÷ 12 = 95.27 in⁴. Both clear the demands, so a two-ply 2x8 is the answer. A two-ply 2x6 gives only 15.13 in³ and fails bending.
- Jack studs. A 1.5 × 3.5 in jack bearing on a DF-L plate at 625 psi carries 625 × 5.25 = 3,281 lb. One jack per end covers the 1,312.5 lb reaction.
Notice which check governed. The two-ply 2x8 supplies 3.3 times the moment of inertia it needs and only 1.1 times the section modulus, so bending decided the size here — as it does for nearly every sawn header checked at L/240.
How to read the result
Compare the two demands against the candidate table and see which one is doing the work. If the smallest section that passes is limited by the section modulus column, your header is strength-controlled and a stronger material — LVL instead of sawn lumber — buys you a size. If it is limited by the moment of inertia column, the header is stiffness-controlled, and material grade helps far less because E varies much less between wood products than Fb does: LVL is roughly three times the bending stress of DF-L No. 2 but only a quarter stiffer.
You can settle the question before you look at the table. For a rectangular member the supplied ratio of I to S is exactly d/2, and the required ratio works out to 1.25 × L × limit × Fb′ / E with L in feet. Setting them equal gives a threshold depth: d = 2.5 × L × limit × Fb′ / E. A section deeper than that is bending-controlled; a shallower one is deflection-controlled. For DF-L No. 2 at L/240 with CD = 1.15 the threshold is 0.39 in of depth per foot of span, so a 6 ft opening only needs 2.4 in of depth before bending takes over — which is why sawn headers checked at L/240 are almost always strength-controlled. For 2.0E LVL at L/240 the threshold is 0.90 in per foot, so a 16 ft LVL header has to reach about 14.6 in deep before bending governs, and below that depth deflection is what stops you.
The jack stud count is a real constraint, not a formality. Each additional jack eats 1.5 in of rough opening at each side, so a garage opening needing three jacks per end takes 9 in more wall than one needing one. Frame the opening before you cut the header, not after.
Treat the candidate list as a starting point rather than an approval. It applies the load duration factor and nothing else. The NDS size factor CF is 1.0 or greater for No. 2 sawn grades up to 2x12, so ignoring it is conservative. The wet service factor CM cuts the other way: above 19% moisture content both Fb and E fall, and these sizes become unconservative.
Finally, check the load path all the way down. The reaction shown here does not stop at the bottom plate: it continues through the rim, the wall below, and into the foundation. On a two-storey opening it is common for the beam to be fine and the post below it, or the footing under that post, to be the real problem.
Section properties of common header build-ups
| Build-up | Actual b × d (in) | S (in³) | I (in⁴) | Typical Fb (psi) |
|---|---|---|---|---|
| 2-ply 2x8 | 3.0 × 7.25 | 26.28 | 95.27 | 900 |
| 3-ply 2x8 | 4.5 × 7.25 | 39.42 | 142.90 | 900 |
| 2-ply 2x10 | 3.0 × 9.25 | 42.78 | 197.86 | 900 |
| 3-ply 2x10 | 4.5 × 9.25 | 64.17 | 296.79 | 900 |
| 2-ply 2x12 | 3.0 × 11.25 | 63.28 | 355.96 | 900 |
| 3-ply 2x12 | 4.5 × 11.25 | 94.92 | 533.94 | 900 |
| 1-ply 11⅞ in LVL | 1.75 × 11.875 | 41.13 | 244.21 | 2,600 |
| 2-ply 11⅞ in LVL | 3.5 × 11.875 | 82.26 | 488.41 | 2,600 |
| 2-ply 14 in LVL | 3.5 × 14 | 114.33 | 800.33 | 2,600 |
| 3-ply 16 in LVL | 5.25 × 16 | 224.00 | 1,792.00 | 2,600 |
Compare the 3-ply 2x12 with the 2-ply 11⅞ in LVL: the LVL supplies less section modulus but pairs it with nearly three times the allowable stress, so it carries substantially more moment in a package 1 in narrower. Confirm Fb and E against the manufacturer's current product report before you rely on them.
IRC minimum design live loads
| Use | Live load (psf) |
|---|---|
| Rooms other than sleeping rooms | 40 |
| Sleeping rooms | 30 |
| Attics with limited storage | 20 |
| Attics without storage | 10 |
| Exterior balconies and decks | 40 |
| Stairs | 40 |
| Roof, tributary area 200 ft² or less, slope under 4:12 | 20 |
Where the ground snow load produces a flat-roof snow load above 20 psf, snow replaces the roof live load and CD becomes 1.15. Local amendments override these values; confirm with your building department.
Mistakes that undersize a header
- Entering the full framing span as the tributary width. The wall picks up half the span, not all of it. This single error doubles the load and is the most common mistake on this page.
- Using the clear opening as the design span. The beam spans bearing centre to bearing centre. Ignoring the bearing understates the moment by a few percent and the deflection by more.
- Forgetting the wall and floor above. A header under a second-storey window carries the roof, the floor, and the wall between them.
- Checking bending and stopping. Whether stiffness or strength governs depends on the depth threshold given above, and for engineered lumber at wide openings it is usually stiffness. A header that passes the stress check by a wide margin can still sag enough to bind the door under it.
- Ignoring bearing and jack stud count. Compression perpendicular to grain on the plates, not the header itself, is what fails first at a heavily loaded opening.
- Treating a header as a substitute for a point load path. A girder truss or a post landing above the opening puts a concentrated load on the header, which this uniform-load calculation does not model.
Prescriptive tables exist and often beat a calculation
The IRC publishes prescriptive header and girder span tables that already embed these calculations for standard load cases. Use those tables when your building fits their assumptions: a defined ground snow load, a stated building width, and the species and grade listed. Use a calculation like this one when your case falls outside them. Anything structural on an occupied building should still be reviewed by a licensed design professional, and the building department has final say.
Where header sizing sits in the framing
A header is one member in a load path that starts at the roof and ends in the soil. Above it, the roof geometry sets the tributary width and the snow load calculation sets the roof live load. Beside it, the wall framing has to accommodate the jack and king studs at each end — the stud count calculator gives you the material list once the opening is fixed. Below it, the reaction runs down through posts to a footing sized for the soil bearing capacity.
Inside the opening, the two checks this page runs are the same ones you would run on any beam. Section properties come first, the deflection check second. Repetitive floor framing beside the opening follows a different route entirely, because joist span tables already fold both checks into a single allowable span. And once the size is settled, a board foot calculation turns it into what the lumberyard will actually charge.
Two situations need an engineer rather than a calculator: any point load landing on or near the header, because the moment then depends on where the load sits, and any header in a shear wall segment, where uplift and shear transfer have to be detailed as well as gravity.
Key terms
- Tributary width
- The strip of roof or floor whose load ends up on this wall, measured in feet. For framing spanning between two supports it is half the span. Multiply it by the area load in psf to get pounds per foot of header.
- Jack stud
- The shortened stud that runs from the bottom plate to the underside of the header and carries the header's reaction. Also called a trimmer. Its capacity is limited by crushing of the plate across the grain.
- Design span
- The length used in the moment and deflection equations. Strictly it runs centre of bearing to centre of bearing; this calculator uses the clear opening plus 3 in, allowing a full 1.5 in of bearing at each end.
- Compression perpendicular to grain (Fc⊥)
- The stress at which wood crushes across the grain, 625 psi for Douglas Fir-Larch and 425 psi for Spruce-Pine-Fir. It governs bearing at plates and never gets the load duration factor.