Calculator
Header Span Calculator
Every other calculator on this site multiplies an area or a count by a rate. This one is different: it takes a load, works out the bending moment that load creates across an opening, and compares the header's required capacity against what a standard doubled or tripled dimensional beam can actually resist. That's a real structural calculation, condensed to the one check that's simple enough to run without engineering software — not a coverage estimate with a stricter name.
It's also, deliberately, only one check. This page confirms bending capacity and nothing else. Deflection — how much the header sags under that same load, which is what cracks the drywall or ceiling above it — is not checked here. Shear near the header's ends, and whether the jack studs and wall below can actually bear the reaction load without crushing, are not checked here either. All three matter as much as bending does, and a licensed engineer or your local span table is what actually signs off a header for anything beyond the most modest, lightly loaded opening.
Calculate your quantity
Half of this bears on the wall with the opening.
psf, live plus dead. 40 covers a 30 psf snow zone.
psf, live plus dead.
psi. 875 is typical for No. 2 SPF; LVL runs 2,600.
Required section modulus
34.56 cu in
- Tributary width
- 14 ft
- Roof load on wall
- 560 lb/ft
- Floor load on wall
- 0 lb/ft
- Total uniform load
- 560 lb/ft
- Total load on header
- 3,360 lb
- Maximum moment
- 30,240 in-lb
- Smallest member that reaches it
- 42.78 cu in
- 2 x (2x10) provides 42.78 cu in against 34.56 cu in required in bending alone. Deflection, shear and bearing are separate checks an engineer makes.
Shopping summary
- • 2 x (2x10) spanning 6 ft, plus 2 jack studs each end
This is an estimate — confirm structural work with a professional.
How this calculation works
Load starts from tributary width — half the building's width, on the assumption of a simple rectangular building with a ridge running down its center, so half the roof's weight bears on each of the two long walls. Multiply that tributary width by the roof load (and, for every floor above, by the floor load too) and you have a combined load per linear foot of wall. Multiply that by the opening width and you have the total load the header carries; run it through the standard simply-supported-beam formula, wL²/8, and you have the maximum bending moment at the header's center, converted to inch-pounds.
Divide that moment by your lumber's allowable bending stress and you have the required section modulus — the minimum bending capacity, in cubic inches, the header cross-section needs. This page then checks that figure against a small table of standard doubled and tripled dimensional headers, each with its own section modulus from the standard rectangular-beam formula, and reports the smallest one that clears the requirement. If none of them do, the honest answer is that the opening needs an engineered beam — LVL or steel — not a bigger stack of dimensional lumber.
- This page checks bending only. Deflection, shear, and bearing on the jack studs below are separate checks it does not perform — a header that clears the bending requirement here can still be the wrong header for one of those three reasons.
- Tributary width assumes a simple, centered-ridge rectangular building. A different roof shape, an off-center ridge, or a wall that isn't centered under the ridge changes the real tributary width, and this page's default doesn't detect that for you.
- Allowable bending stress (Fb) depends entirely on your lumber's actual species and grade — the 875 psi default is typical for common No. 2 Spruce-Pine-Fir dimensional lumber, and engineered LVL runs well over triple that.
- The load figures here are combined live-plus-dead assumptions meant as a starting point. Your local snow load, in particular, can run considerably higher than the 40 psf default, and the required section modulus scales directly with it.
The formula
requiredS = (totalPlf × openingWidth² × 12) ÷ 8 ÷ Fb; totalPlf = tributaryWidth × (roofLoad + floorsAbove × floorLoad); tributaryWidth = buildingWidth ÷ 2
- tributaryWidth
- Half the building's width — the share of roof (and any floor above) this wall is assumed to carry, under the simplifying assumption of a rectangular building with a centered ridge.
- totalPlf
- Combined roof and floor load landing on this wall, per linear foot, from tributary width times the stated load-per-square-foot figures — everything above this header pressing down on it, expressed as a rate.
- moment (M)
- The bending force the header must resist at its center, from the standard simply-supported uniform-load beam formula wL²/8, converted to inch-pounds so it can be compared against a section modulus expressed in cubic inches.
- Fb (allowable bending stress)
- The bending stress, in psi, your specific lumber species and grade can safely resist — an input rather than a constant, because it ranges from under 900 psi for common dimensional lumber to well over 2,500 for engineered LVL.
- requiredS (required section modulus)
- The minimum bending capacity, in cubic inches, the header cross-section needs — compared against a table of standard doubled and tripled dimensional members to find the smallest one that clears it, or a verdict that none do.
Moment depends on load times opening width squared, not load times width — which is why widening an opening costs far more structural capacity than the width increase alone suggests. Doubling the opening width, with everything else unchanged, roughly quadruples the required section modulus, not doubles it.
Where these numbers come from
- M = wL² ÷ 8
- Code-derived: the standard structural engineering formula for maximum bending moment at the center of a simply supported beam under uniform load, used universally in header sizing wherever a rough opening spans a uniformly loaded wall.
- S = bd² ÷ 6 per ply
- Code-derived: the standard section modulus formula for a rectangular beam cross-section, applied per ply of dimensional lumber and multiplied by the number of plies for a built-up header.
- 875 psi default allowable bending stress
- Code-derived, representing typical published design values for common No. 2 Spruce-Pine-Fir dimensional lumber. Treat it as a stand-in to replace with your specific lumber's rated Fb from its grade stamp or a design value table — it is not a universal constant.
- 40 psf roof load and 50 psf floor load defaults
- Trade convention representing typical combined live-plus-dead load assumptions, meant as a starting point. Actual design load is set by your local code's snow, live, and dead load requirements for your specific location and structure, and can run considerably higher in a snow-load region.
Worked examples
A 6 ft garage window opening, roof load only
| Opening width | 6 ft |
|---|---|
| Building width | 28 ft |
| Roof load | 40 |
| Floors above | 0 |
| Floor load | 50 |
| Allowable bending stress | 875 |
| Required section modulus | 34.56 cu in |
|---|---|
| Tributary width | 14 ft |
| Roof load on wall | 560 lb/ft |
| Floor load on wall | 0 lb/ft |
| Total uniform load | 560 lb/ft |
| Total load on header | 3360 lb |
| Maximum moment | 30240 in-lb |
| Smallest member that reaches it | 42.78 cu in |
Tributary width is 14 feet — half the 28 foot building — carrying roof load only, since there's no floor above this wall. That gives 560 pounds per linear foot, a maximum moment of 30,240 inch-pounds across the 6 foot opening, and a required section modulus of 34.56 cubic inches. A doubled 2x8 falls short at 26.28; a doubled 2x10, at 42.78, clears it with room to spare.
That doubled 2x10 answer settles bending, and only bending. It says nothing about whether this header will deflect enough under that same load to crack the finish above it, whether it can resist shear near the jack studs, or whether those jack studs and the wall below them can actually take the reaction load without crushing. All three need their own check before this opening is framed — this page's answer is a starting point for that conversation, not the end of it.
A wide 10 ft opening under a second floor and a heavier snow-load roof
| Opening width | 10 ft |
|---|---|
| Building width | 32 ft |
| Roof load | 55 |
| Floors above | 1 |
| Floor load | 50 |
| Allowable bending stress | 875 |
| Required section modulus | 288 cu in |
|---|---|
| Tributary width | 16 ft |
| Roof load on wall | 880 lb/ft |
| Floor load on wall | 800 lb/ft |
| Total uniform load | 1680 lb/ft |
| Total load on header | 16800 lb |
| Maximum moment | 252000 in-lb |
| Smallest member that reaches it | 0 cu in |
Widening the opening from 6 to 10 feet — a 67% increase — combined with one floor above and a higher 55 psf roof load, doesn't scale the required section modulus up proportionally. It goes from 34.56 to 288 cubic inches, more than eightfold, because moment depends on load times opening width squared: the width increase alone accounts for most of that jump before the extra floor and roof load are even added.
No member in this page's table of standard doubled or tripled dimensional headers reaches 288 cubic inches — even a tripled 2x12 tops out around 95. The correct answer here is an engineered beam, LVL or steel, sized by a supplier's own span tables or an engineer, not a fourth or fifth ply of dimensional lumber stacked on top of the others. That's not a page limitation; it's the honest structural answer for an opening this wide carrying this much load.
Common mistakes
- Treating a header that clears this page's bending check as fully designed. Bending is one of at least four checks a real header sizing needs — deflection, shear, and bearing on the jack studs below are separate and this page doesn't perform them.
- Leaving the bending stress at its 875 psi default when the actual lumber is a different species or grade, or is engineered LVL running well over 2,500 psi. The default is a starting point, not a substitute for your lumber's rated value.
- Leaving floors-above at zero on a wall that actually carries a second story. The required section modulus for a floor-bearing wall is substantially higher than the same wall carrying roof load alone.
- Assuming tributary width is always half the building width regardless of roof shape. That's a simplification for a centered-ridge rectangular building; an off-center ridge, a hip roof, or a wall not centered under the ridge changes the real number.
- Ignoring jack stud count and bearing area under a header that passed the bending check. A header sized correctly for bending can still crush undersized jack studs or overload a foundation that isn't ready for the concentrated reaction load.
- Using the 40 psf roof load default in a real snow-load region. Many jurisdictions require considerably more than that, and the required section modulus scales directly and significantly with it.
Shopping summary
This page's output tells you which class of member — dimensional doubled or tripled lumber, or an engineered beam — the bending check calls for. It is not a final cutting list: confirm the result against deflection, shear, and bearing before ordering, and get an engineer's sign-off for anything past a small, lightly loaded opening.
Jack stud count for the header's bearing points comes from the wall stud calculator, which counts a fixed 2 king and 2 jack studs per opening but does not itself check whether that's enough bearing area for the load calculated here — cross-check the two together rather than treating them as independent. If this header sits in a wall carrying a floor above, the floor load figure entered here should come from the actual floor design worked out on the floor joist calculator, not a guess.
FAQ
Why does widening the opening from 6 to 10 feet more than double the required header capacity?
Because bending moment depends on load times the opening width squared, not load times width. A 67% wider opening alone accounts for most of an eightfold jump in required section modulus in the second worked example above — widening an opening is structurally far more expensive than the width number alone suggests.
What does it mean when no standard doubled header reaches the required section modulus?
Practically, this is where cost stops being the deciding factor between dimensional lumber and an engineered beam — past this point dimensional simply isn't an option regardless of price, because no amount of stacking plies closes a gap this size; LVL's allowable bending stress commonly runs 2,600 psi or higher against dimensional lumber's roughly 875, which is most of why one more ply never rescues an opening that misses by a wide margin the way a small shortfall sometimes can.
Why is bending stress a field I have to enter instead of a fixed number the calculator assumes?
Because it depends entirely on your actual lumber's species and grade, which this page has no way to know. Your specific Fb value is printed on the grade stamp on the lumber itself, or available from the supplier's span-table literature for that exact product — not something to eyeball from the species name alone, since two grades of the same species can carry meaningfully different rated stress. Entering the correct value matters most on an opening that's close to the edge of what a standard doubled member can carry, where the difference between a 875 psi guess and your lumber's real 1,000-plus psi rating can be the difference between needing an engineered beam and not.
If a header passes this bending check, does it still need deflection, shear, and bearing checked separately?
Yes, always. This page performs one check out of at least four a complete header design needs. A header that clears bending can still sag enough to crack finishes above it, fail in shear near its ends, or overload the jack studs and wall beneath it — none of which this page evaluates.
My building isn't a simple rectangle with a centered ridge — how do I handle tributary width?
This page's tributary-width assumption (half the building width) only holds for that simple case. A hip roof, an off-center ridge, a wall that doesn't run the full building width, or a roof with valleys all change how much load actually bears on a given wall, and working that out correctly is a structural question worth taking to an engineer rather than guessing at an adjusted width.
Does this page check whether I have enough jack studs under the header?
No. It sizes the header itself for bending; it doesn't check whether the jack stud count and bearing area beneath it can take the resulting reaction load without crushing. The wall stud calculator counts jack studs per opening at a standard convention, but confirming that's adequate for a heavily loaded header is a separate check.
Where to go next
The projects this number is a step of, the guides that explain the method behind it, and the rest of its trade group.