Audited 29 Jul 2026·Last updated 28 Oct 2026·6 citations·Tier 1·0 uses

Beam Deflection Calculator

Find how far a joist, header or cantilever sags under load, check it against the IRC/IBC L/360, L/240 or L/180 limit, or solve for the moment of inertia needed.

Beam Deflection Calculator

Support and load arrangement
What do you want to find?
Clear distance face to face of supports. For a cantilever this is the projection beyond the support, not the whole member.
ft
Pounds per lineal foot of beam. Used by the two uniform load cases. Enter the LIVE load alone for an IRC-style L/360 check, or the full dead-plus-live load for an IBC D+L check — the page deflects whatever you type.
plf
Used by the two point load cases — a post landing on the beam, a header carrying a girder end, a hot tub corner.
lb
NDS 2018 Supplement Table 4A: Douglas Fir-Larch No.2 1,600,000 · Hem-Fir No.2 1,300,000 · Spruce-Pine-Fir No.1/No.2 1,400,000 · Southern Pine No.2 1,400,000 (Table 4B). Structural steel is 29,000,000. LVL and glulam use the manufacturer's published value.
psi
How is the section defined?
Dressed thickness, not nominal. A 2x is 1.5 in, so a two-ply built-up beam is 3.0 in and a three-ply is 4.5 in.
in
Dressed depth. PS 20 dry sizes: 2x6 = 5.5 · 2x8 = 7.25 · 2x10 = 9.25 · 2x12 = 11.25 · 2x14 = 13.25 in.
in
Used only when the section is entered directly. Take it from the AISC shape tables for steel or the manufacturer's table for I-joists and LVL.
in⁴
Deflection limit
Deflection
0.3277
The largest elastic sag under the load entered, in inches — at midspan on a simply supported beam, at the tip of a cantilever — excluding shear deformation and long-term creep.
Verdict against the code limit
0.3277 in of deflection on a simply supported beam under a uniform load. The limit you selected is L/360, which on this member is 0.4 in — L is the clear span. This member MEETS that limit, at 81.9 % of it, deflecting L/439. Deflection limits come from IRC 2021 Table R301.7 and IBC 2021 Table 1604.3; the edition your jurisdiction has adopted and its local amendments govern, and they differ between jurisdictions. A deflection check is serviceability only — it says nothing about bending, shear or bearing capacity — and a licensed structural engineer or architect must sign off before work proceeds.
Allowable deflection
0.4 in
Deflects L over
439.4531
Percentage of the limit used
81.92
Moment of inertia required
291.6 in⁴
Moment of inertia used
355.957 in⁴
Maximum bending moment
7,200 ft-lb
Maximum shear
2,400 lb

Background.

Take out a wall between a kitchen and a dining room and the joists above it need somewhere new to land: a header or flush beam across the opening. Whether that beam is strong enough is one question. Whether it stays flat enough is another, and it is the one that shows up first — as a crack along the drywall tape, a door that starts to bind, grout lines opening in a tile floor, or a floor that bounces when someone crosses the room.

Codes handle the second question with deflection limits written as a fraction of the span. L/360 means the member may sag no more than its length divided by 360: 0.4 in on a 12 ft beam, 0.6667 in on a 20 ft one. Under IRC 2021 Table R301.7, floors and plaster ceilings get L/360, gypsum-board ceilings and most other members L/240, rafters steeper than 3:12 with nothing attached below L/180, and a lintel carrying brick veneer the strictest value of all, L/600. IBC 2021 Table 1604.3 does the same job for commercial buildings.

Four arrangements cover most residential members: a beam resting on a support at each end with either a uniform load or one load at midspan, and a cantilever with either a uniform load or a load at its tip. For any of them you can ask how far a section you already have will sag, or turn the question round and ask what moment of inertia would land exactly on the limit — the number you take to an LVL table or a steel shape list.

Two exponents explain nearly every result. Span enters to the fourth power, so stretching a beam from 12 to 14 ft under the same load per foot makes it sag 85% more. Depth enters cubed, through the moment of inertia, so adding a second ply of the same lumber halves the sag while going from a pair of 2x8s to a pair of 2x12s cuts it by a factor of 3.7. When a floor feels springy, a deeper member or a shorter span almost always beats more of the same.

What is beam deflection calculator?

Deflection is the distance a loaded beam moves out of line, measured at right angles to its original axis at the point where the movement is largest: midspan for a symmetrically loaded beam on two supports, the free end for a cantilever. It is governed by flexural rigidity, the product E·I. E, the modulus of elasticity, belongs to the material — about 1,600,000 psi for Douglas Fir-Larch No.2, 29,000,000 psi for structural steel. I, the moment of inertia, belongs to the shape; for a solid rectangle it is breadth × depth³ ÷ 12. A deflection limit is a serviceability rule, not a safety one: wood and steel bend visibly long before they break, and the limit exists so that finishes, doors, windows and the people walking on the floor are not troubled by the movement. Because the limit is span ÷ a denominator, it grows with the member. For cantilevers both IRC Table R301.7 footnote b and IBC Table 1604.3 footnote i take L as twice the projection.

How to use this calculator.

  1. Choose the support and load arrangement. A header, joist or girder bearing on a wall or post at each end is simply supported; a balcony joist or bay framing that projects past its support with nothing under the far end is a cantilever. Pick the uniform case for floor, roof or wall loads spread along the member and the point case when a post or girder end lands on it.
  2. Decide what you want back: the deflection of a section you already have, or the moment of inertia that would meet your chosen limit exactly.
  3. Enter the span as the clear distance between the faces of the supports. For a cantilever enter only the projection beyond the support — the doubling the codes require for the limit is applied for you.
  4. Enter the load. For a uniform case, multiply the load per square foot by the width of floor the beam carries: a header picking up joists that span 12 ft on one side carries half of them, 6 ft, so 40 psf of live load becomes 240 plf. Use live load alone for an IRC-style L/360 check and dead plus live for an IBC D+L check.
  5. Enter E and the section. The field hint lists NDS values for common framing lumber. For sawn or built-up lumber give dressed sizes — a 2x12 is 1.5 by 11.25 in, so two plies are 3.0 by 11.25 — or switch to entering I directly for steel shapes, I-joists and LVL.
  6. Select the limit that matches what the member supports, read the verdict, and carry the maximum moment and shear into the bending, shear and bearing checks that deflection does not cover.

The formula.

δ = 5wL⁴ ⁄ (384 · E · I)

Every one of the four cases comes from the same relation between bending moment and curvature, M = E·I·d²y/dx², integrated twice with the support conditions as boundary values. The results all take the form δ = K ÷ (E·I), where K depends only on the load, the span and how the member is held. With the span L converted to inches and a uniform load w converted from pounds per foot to pounds per inch (÷ 12): a simply supported beam under a uniform load has K = 5wL⁴/384; a simply supported beam with a point load P at midspan has K = PL³/48; a cantilever under a uniform load has K = wL⁴/8; a cantilever with P at its tip has K = PL³/3. Because the same K appears in every case, the required moment of inertia is simply K ÷ (E × allowable deflection), and a section with exactly that I deflects exactly to the limit. The allowable deflection is L ÷ denominator, with L doubled for cantilevers. Moment and shear come from statics and do not depend on the section at all: wL²/8 and wL/2, PL/4 and P/2, wL²/2 and wL, PL and P. The comparisons are instructive. The same total load gathered at midspan deflects a beam 1.6 times as much as when it is spread evenly. A cantilever under a uniform load sags 9.6 times as far as a simply supported beam of the same length and load, and a tip load on a cantilever produces 16 times the deflection of the same load at the middle of a simple span. The model assumes small, linear-elastic deflection of a straight member of constant section, braced against sideways buckling, and it ignores shear deformation. That last omission matters for short, deep members and for I-joists with thin webs, where manufacturers add a shear term; it also leaves out the long-term creep of wood under sustained load.

A worked example.

Example

A two-ply 2x12 Douglas Fir-Larch No.2 header spans 12 ft over a widened opening in a wall that carries a floor, with 400 plf on it. Dressed, the section is 3.0 in by 11.25 in, so I = 3.0 × 11.25³ ÷ 12 = 355.96 in⁴, and NDS Table 4A gives E = 1,600,000 psi. Convert to inches: L = 12 × 12 = 144 in and w = 400 ÷ 12 = 33.333 lb/in. The numerator is 5 × 33.333 × 144⁴ = 71,663,616,000, and dividing by 384 gives K = 186,624,000. The flexural rigidity is 1,600,000 × 355.96 = 569,531,250 lb·in². Deflection is 186,624,000 ÷ 569,531,250 = 0.3277 in. At L/360 the allowance is 144 ÷ 360 = 0.4 in, so the header uses 0.3277 ÷ 0.4 = 81.92% of it and passes, deflecting L/439. Solving the other way, 186,624,000 ÷ (1,600,000 × 0.4) = 291.6 in⁴ is the stiffness that would sit exactly on the limit, and the two plies supply about 22% more than that. Statics gives a maximum moment of 400 × 12² ÷ 8 = 7,200 ft-lb and an end reaction of 400 × 12 ÷ 2 = 2,400 lb — the figures to carry into the bending check and into sizing the jack studs and bearing under each end.

depth In11.25
width In3
moment Of Inertia In4355.957
modulus Of Elasticity Psi1,600,000
distributed Load Plf400
section Moderectangular
point Load Lb2,400
deflection Limit Denominator360
span Ft12
solve Fordeflection
load CasesimpleUniform

Frequently asked questions.

How many inches is L/360 on a 16-foot floor joist?
Convert the span to inches and divide: 16 ft is 192 in, and 192 ÷ 360 = 0.5333 in. The same arithmetic gives 0.4 in for a 12 ft member and 0.6667 in for a 20 ft one. A looser L/240 limit on that 16 ft joist would allow 0.8 in. Because the allowance scales with length, a long member is permitted more sag in inches but the same sag relative to its span.
Do floors under ceramic or stone tile need a stricter limit than L/360?
The building code itself stops at L/360 for floors. L/480 is offered as an option because tile and stone setters commonly ask for it — natural stone in particular cracks along grout lines when the substrate flexes — but it comes from industry practice and the tile manufacturer's installation requirements rather than from IRC Table R301.7. If the tile installer or the product's warranty names a limit, design to that.
Should I enter the live load alone or the total load?
That depends on the check. IRC Table R301.7 and the AWC span tables apply L/360 to live load only, on the reasoning that the dead-load sag has already happened when the finishes go on. IBC Table 1604.3 has a separate dead-plus-live column, usually with a looser denominator. The calculator deflects whatever load you type, so match the load to the column you are checking; entering the total load against L/360 errs on the strict side.
Why is a cantilever allowed to sag twice as much?
Both IRC Table R301.7 footnote b and IBC Table 1604.3 footnote i say that for cantilevers L is taken as twice the projection. A cantilever behaves like one half of a simple span of double the length, so doubling keeps the allowance comparable. A 4 ft balcony joist checked at L/240 is therefore allowed 2 × 48 ÷ 240 = 0.4 in, not 0.2 in. The L-over ratio reported for a cantilever uses the same doubled length.
My beam carries a uniform floor load and a post at midspan. How do I check it?
Run the calculator twice — once for the uniform load, once for the post as a midspan point load — and add the deflections, because both peak at the same place and elastic deflections superpose. On the example header, 400 plf gives 0.3277 in and a 2,400 lb post at midspan adds 0.2621 in, a total of 0.5898 in: 147.5% of the 0.4 in allowance, so the combination fails. The moments add the same way, to 14,400 ft-lb.
Which E value should I use for my lumber?
NDS 2018 Supplement Table 4A gives Douglas Fir-Larch 1,900,000 psi for Select Structural, 1,700,000 for No.1 and 1,600,000 for No.2; Hem-Fir 1,600,000, 1,500,000 and 1,300,000 for the same grades; and Spruce-Pine-Fir 1,500,000 for Select Structural and 1,400,000 for No.1/No.2. Table 4B gives Southern Pine 1,800,000, 1,600,000 and 1,400,000. LVL, glulam, PSL and I-joists carry values published by their manufacturer for that specific product.
What moment of inertia does a steel beam need for the same opening?
Use the required-moment-of-inertia option with E = 29,000,000 psi. For the 12 ft span at 400 plf and L/360, the requirement is 186,624,000 ÷ (29,000,000 × 0.4) = 16.09 in⁴, against 291.6 in⁴ for Douglas Fir-Larch No.2 — steel is about 18 times stiffer, so a much shallower section can do the job. Pick a shape whose tabulated I meets or exceeds that figure, then check it for bending and lateral bracing too.
Will a wood beam keep sagging after the house is finished?
Yes, under sustained load. The result here is the immediate elastic deflection. NDS 3.5.2 accounts for creep by multiplying the deflection from long-term load by a factor of 1.5 for seasoned lumber in dry service, or 2.0 for unseasoned lumber or wet service, then adding the short-term deflection. On the example header, if 100 of the 400 plf were dead load, its 0.0819 in would grow to 0.1229 in. Steel does not creep at ordinary temperatures.
What about a beam that runs continuously over a post in the middle?
None of the four cases models continuity. A beam that runs unbroken over a middle support restrains its own rotation there, so under uniform load each span deflects less than a simply supported beam of the same length. Checking each span as simply supported therefore overstates the sag for uniform loads, which errs toward a stiffer member. For a precise answer, or for point loads near the middle support, a frame analysis is needed.
If the beam passes deflection, is it strong enough?
Not necessarily. Deflection is serviceability only. The member must also pass bending (the maximum moment against the section's capacity), horizontal shear, bearing at the supports, and lateral-torsional buckling where the compression edge is not braced. Short, heavily loaded beams are usually governed by bending or shear; long, lightly loaded ones by deflection. Adopted code editions and local amendments vary, and a licensed structural engineer or architect must confirm a structural member before work proceeds.

References& sources.

  1. [1]International Code Council — IRC 2021 Table R301.7, Allowable Deflection of Structural Members, with footnote b: "For cantilever members, L shall be taken as twice the length of the cantilever." Rows used: floors including deck floors L/360, ceilings with gypsum board L/240, rafters steeper than 3:12 with no ceiling attached L/180, all other structural members L/240, lintels supporting masonry veneer L/600.
  2. [2]ICC Building Code Action Committee — code change proposal reproducing Table R301.7 in full with all footnotes, used to confirm every row and the cantilever footnote verbatim.
  3. [3]International Code Council — IBC 2021 Section 1604.3 and Table 1604.3, Deflection Limits, footnote i: "l = Length of the member between supports. For cantilever members, l shall be taken as twice the length of the cantilever."
  4. [4]American Wood Council — NDS 2018 Supplement, Table 4A and Table 4B, Reference Design Values for Visually Graded Dimension Lumber, giving the modulus of elasticity values quoted in the field hint and FAQ.
  5. [5]American Wood Council — Span Tables for Joists and Rafters, 2021 edition, section 3, confirming the PS 20 dressed dry sizes used for the depth hint (2x6 = 1.5 x 5.5, 2x8 = 1.5 x 7.25, 2x10 = 1.5 x 9.25, 2x12 = 1.5 x 11.25) and the L/360 live-load deflection basis.
  6. [6]American Institute of Steel Construction — Steel Construction Manual, Table 3-23, "Shears, Moments and Deflections", the standard tabulation of the four closed-form cases implemented here. Print and member-gated; the four expressions were independently re-derived from the moment-curvature relation.

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Method
δ = 5wL⁴ ⁄ (384 · E · I)
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