How to Size a Hip Rafter — 6:12 Hip Roof (NDS 2024)

A hip rafter is the one roof member that breaks every convenient assumption at once. It is not on the pitch of the roof it carries — it runs diagonally across the corner, so its own slope is flatter than the common rafters framing beside it. It does not carry a uniform load — the jack rafters that die into it get longer and longer, so its load is a triangle. And its two governing load cases are measured on different lengths: roof dead load per square foot of the sloped surface, roof live load per square foot of the horizontal projection.

Get any one of those three wrong and the answer is off by anywhere from 3% to 25% — and the biggest error is not in the moment, it is in the reaction at the top of the hip, which is what sizes the post and the connection nobody drew.

This example works one all the way through: a 19.8-ft (horizontal) hip beam carrying the corner of a clay-tile hip roof, sized as a 6×12 Douglas Fir-Larch No. 2 (Beams & Stringers) by Allowable Stress Design per NDS 2024. Every number below was produced by the same calculation engine that runs StructSuite's Wood Beam module (/design/wood-beam), and the agreement is locked by an automated regression test (tests/verification/hip-rafter-guide-example.test.ts) — the same trust chain used for all StructSuite worked examples.

#The structure

A rectangular 40 ft × 28 ft house with a hip roof pitched 6 : 12 on all four planes. The ridge is 40 − 28 = 12 ft long, and four hips run from the building corners up to the ridge ends. Jack rafters at 24 in. o.c. frame from the top plates into the hips; the ceiling is vaulted, so the gypsum board follows the roof planes and the hip beam is exposed below.

#Why a hip is never on the roof's pitch

This is the single most common hip-rafter mistake, and it is a geometry mistake, not an engineering one.

A common rafter runs perpendicular to the eave: its horizontal run is half the building width, 14 ft, and it rises 14 × 6/12 = 7 ft. The hip runs diagonally across the corner to the same point on the ridge. Its horizontal run is the diagonal of a 14 × 14 square — 14√2 = 19.8 ft — but its rise is the same 7 ft, because it ends at the same ridge height.

Same rise, longer run, so the hip is flatter:

The "17 rule." Carpenters lay out hips on a framing square using 17 in. of run instead of 12 — that is 12√2 = 16.97 in., the plan diagonal of a 12 in. square. It is the same fact as the table above: the hip's rise per foot of its own run is 6/√2 = 4.24 in. Both the engineer's 4.24:12 and the carpenter's "6 and 17" describe one member.

Because the hip is flatter, its sloped-length factor is 1.061, not the roof's 1.118. Using 1.118 for the hip is a 5.4% over-conversion of dead load — small, but it is an error in the conservative direction that costs money, and the same confusion in the other direction (using the hip's 4.24:12 to convert loads on the roof planes) is unconservative.

#Choosing the input parameters — where each one comes from

Every input below is either a code-prescribed value with its section cited on the row, or a geometry fact read off the framing plan. Nothing is a guess.

The dead-load takeoff, all of it per square foot of the sloped roof surface (which is how roofing is sold and installed):

#Step 1 — Hip geometry

Everything downstream comes from four numbers, so derive them once and carry them:

The analysis span is 19.80 ft, the horizontal projection. That is not a simplification — for vertical gravity loads it is statically exact, and it is the convention every span table in the IBC and every manufacturer's roof table is written on. What changes with slope is not the span; it is which loads have to be re-measured onto that span. Step 4 does exactly that, and nothing else.

#Step 2 — Why a hip carries a triangle

Walk along the hip from the building corner toward the ridge. At the corner, the jack rafters dying into it are a few inches long. At the ridge end, they are full-length commons. Each jack delivers half its load to the hip (the other half goes to the wall), so the load the hip picks up grows linearly from zero at the corner to a maximum at the ridge end.

The tributary area, from the geometry. The boundary on each roof plane is the mid-length line of the jacks: it starts at the corner and reaches the hip at the point halfway along the last jack. In plan that is the quadrilateral with corners (0, 0), (14, 7), (14, 14), (7, 14) — shaded green above. Its area is:

That 98 ft² is a plan area. The roof it represents is inclined at 6:12 — the roof plane's pitch, not the hip's — so the actual roof surface hanging on this hip is:

A_surface = 98.0 × 1.118 = 109.6 ft²

#Step 3 — Loads: two intensities, two lengths

Dead load is defined on the roof surface, so start from the surface area and spread the total along the length the hip actually is — 21.0 ft:

Roof live load is defined by ASCE 7-22 §4.8.2 as a load "per ft² of horizontal projection supported by the member" — so it needs no surface correction at all, and it is spread along the 19.8-ft projection:

Self-weight the module adds itself from the section: 63.25 in² ÷ 144 × 35 pcf = 15.4 plf along the hip, and being dead load it converts exactly like D.

Note how close 177.4 and 178.2 plf are — an accident of this roof, and a useful one: it makes the entered dead and live loads look interchangeable while the analysis treats them completely differently. That is the whole point of the next step.

#Step 4 — The slope conversion (horizontal-projection method)

The beam is analysed on its horizontal projection. Loads that are defined along the sloped length have to be re-expressed per horizontal foot, which makes them larger, because one horizontal foot of hip is 1.061 ft of actual member:

Deflection is also computed and limited on L_h — the span-table convention, and the one that matches how IBC Table 1604.3 is written.

#Step 5 — Section and material: reading Table 4D, not Table 4A

6×12 is 5.5 in. × 11.5 in. dressed. That width puts it in the timber size class, so its design values come from NDS 2024 Supplement Table 4D (Beams & Stringers) — not Table 4A, which covers 2 in. to 4 in. dimension lumber. This is not a formality:

The two tables are built from different grading rules and different assumed uses; a 6×12 read off Table 4A is simply the wrong number, in either direction depending on grade. StructSuite ties the two selections together (NDS 4.1.3): the Table 4A/4B/4C/4D choice strictly drives which Table 1B size classes are selectable, the others stay visible but grayed with a note naming the table that enables them, and picking across the 5×5 boundary clears the selection the new table cannot design rather than silently re-reading it from the wrong page.

Section and reference values used here (Table 1B geometry, Table 4D values):

#Step 6 — Analysis: reactions, shear, and moment

With the converted loads on a 19.8-ft simple span, the governing combination is ASCE 7-22 §2.4.1 combination 3a, D + L_r (the module writes it "D + (L_r or 0.7S or R)").

For a triangular load w rising to its peak at the far end, the classical results are R_near = wL/6, R_far = wL/3, and M_max = wL²/(9√3) at x = L/√3. Adding the uniform self-weight:

#Step 7 — Adjustment factors, and C_D per combination

F_b′ = 875 × 1.25 = 1,094 psi for the governing combination. Because C_D is applied per combination, the dead-only case runs against 875 × 0.9 = 788 psi and produces its own D/C of 0.693 — lower, but worth seeing, because on a tile roof the permanent case is closer than engineers expect.

#Step 8 — Flexure (NDS 3.3.1)

#Step 9 — Shear (NDS 3.4.2)

Shear is taken at the support face from statics, and the high end governs because two-thirds of the load lands there.

NDS 3.4.2.2 permits neglecting uniform load within a distance d of the support. StructSuite does not take that reduction — it is a permission the engineer may exercise, not a default the software should apply silently.

#Step 10 — Deflection (IBC 2024 Table 1604.3)

The hip supports a vaulted gypsum ceiling, so the limits are L/240 live and L/180 total, both on the 19.8-ft projection. The total-load case follows IBC Table 1604.3 footnote d: it is the creep component of the long-term dead load plus the short-term live load, estimated as 0.5D + L_r for dry service — not the full D + L_r.

Note what the slope did not do here: dead-load deflection rose with the converted load, but the span used for both the calculation and the limit stayed 19.8 ft. Had the member been modelled on its 21.0-ft sloped length, the limit would have loosened to 1.05 in. while the deflection grew — a double error pointing opposite ways.

#Step 11 — Bearing (NDS 3.10.2, C_b per 3.10.4)

Both ends bear on 3.5-in. seats — the double top plate at the corner, a post cap at the ridge end — over the full 5.5-in. beam width, so A_bearing = 19.25 in².

C_b = 1.0 at both ends: NDS 3.10.4 grants the bearing-area factor only to bearings at least 3 in. from a member end, and both of these are end bearings. C_D never applies to F_c⊥ (NDS Table 2.3.2, footnote 1).

Two practical notes the numbers do not show. First, F_c⊥ is a property of whatever gets crushed — if the plate under the hip is a softer species than the hip itself, name that species in Step 4 so the lower value governs. Second, a hip bears on a sloped cut, so the load is at an angle to the grain and NDS 3.10.3's Hankinson formula would give a higher allowable than F_c⊥; checking against F_c⊥ alone, as here, is the conservative simplification.

#Five mistakes that show up on hip beams

Every alternative below was run through the same engine, so the cost of each mistake is a real number, not a warning.

#Summary of results — 6×12 DF-L No. 2 hip beam, 19.8 ft horizontal

Other combinations, for completeness: 1a. D → D/C 0.693 · 4a. D + 0.75L_r → 0.803 · 7a. 0.6D → 0.416.

Reproduce it yourself in StructSuite's Wood Beam module (/design/wood-beam): one 19.8-ft span, pinned then roller; switch on Sloped member and enter the slope as rise 7 : run 19.8; two linearly varying loads, both starting at w₁ = 0 — type D to w₂ = 177.4 plf and type L_r to w₂ = 178.2 plf; self-weight on; material Sawn Lumber → Douglas Fir-Larch (Timbers) → No. 2 (Beams & Stringers) → 6x12; dry service, bearing 3.5 in., deflection limit "Ceilings with flexible finishes — L/240 (live), L/180 (total)". The pre-filled module further down this page is that exact model, and its detailed calculation panels reproduce every step above.

#Frequently Asked Questions

#What slope do you design a hip rafter for?

The hip's own slope, which on an equal-pitch roof with a square corner is always flatter than the roof planes it carries. Its rise is the same as the common rafter's, but its horizontal run is √2 longer, so its pitch is the roof pitch divided by √2: a 6:12 roof gives a 4.24:12 hip (19.47°), a 4:12 roof gives 2.83:12, a 12:12 roof gives 8.49:12. Carpenters express the same fact as "6 in 17" — the hip rises 6 in. per 17 in. of run instead of per 12. If the two roof planes meeting at the corner have different pitches, or the corner is not 90°, the hip is neither at 45° in plan nor at that ratio, and you have to work the geometry out from the two plane equations — but the principle is unchanged: use the hip's real rise over its real plan run.

#Is the load on a hip rafter triangular or uniform?

Triangular — zero at the eave corner, maximum at the ridge end — because the jack rafters framing into it grow in length as you move up the hip, and each delivers half its load. The total is A_plan × the load, where A_plan = ½ × (half the building width)², and the peak intensity is twice the average. A uniform assumption gets the moment approximately right (2% low in this example) but the reactions badly wrong: it splits the load evenly when in reality two-thirds of it goes to the top support. For a hip on an unequal-pitch or irregular roof the tributary is still bounded by the mid-length line of the jacks — just no longer a clean triangle, in which case a trapezoidal or segmented load is the honest model.

#Do you use the sloped length or the horizontal span for a hip rafter?

The horizontal projection — 19.8 ft here, not 21.0 ft. For vertical gravity loads that model is statically exact, not an approximation: the internal moments and shears in the sloped member about its strong axis are identical to those of a horizontal beam on the projected span carrying the projected loads. It is also the basis on which every prescriptive span table and every deflection limit is written. What the slope changes is the loads: anything defined per foot of sloped length gets multiplied by √(1 + (rise/12)²) before it goes on the projection. Model the member on its 21.0-ft sloped length instead and you inflate the moment by (21.0/19.8)² = 12.5% while simultaneously loosening the deflection limits by 6%.

#Does dead load increase on a sloped roof?

Per horizontal foot, yes. Roofing, sheathing, framing, insulation, and ceiling are all installed and weighed per square foot of the sloped surface, and a steeper roof packs more of that surface above each square foot of floor plan. The multiplier is the sloped-length factor: 1.118 at 6:12, 1.202 at 8:12, 1.414 at 12:12. Live, roof live, and snow loads do not get this treatment — the code defines them on the horizontal projection to begin with (ASCE 7-22 §4.8.2 says so in as many words for L_r, and Chapter 7's sloped-roof snow load p_s is likewise a projection intensity). Wind is a third case: it acts normal to the surface, and its vertical component per sloped foot (× cos θ) spread over the shorter projection (÷ cos θ) comes back to exactly what you entered — factor 1.00.

#What size hip rafter do I need for a 20-foot span?

There is no size that answers that question without the tributary and the loads, and a hip is the worst member to guess at because its triangular tributary makes it lighter than its length suggests. The hip in this example spans 19.8 ft horizontally and carries only 3,950 lb total — an average of 200 plf — because 98 ft² of roof is all that reaches it. A 6×12 DF-L No. 2 handles it at 90% of capacity. Double the tile weight or the tributary and you are into a 6×14, a glulam, or a 5.25-in. PSL very quickly. Run your own numbers: the module below is pre-filled with this example, and changing the two w₂ values and the section re-checks everything.

#How do I run this with snow instead of roof live load?

Enter the sloped-roof snow load p_s from ASCE 7-22 Chapter 7 as a type S load, with the same triangular shape and no slope conversion — p_s is already a horizontal-projection intensity. Two things change downstream. First, C_D drops from 1.25 to 1.15 (NDS Table 2.3.2, two-month snow load), so capacities fall about 8%. Second, ASCE 7-22 writes the ASD combination as D + (L_r or 0.7S or R) — the 0.7 is there because ASCE 7-22 moved to strength-level ground snow loads — so the combination the module builds carries 0.7 × your entered S. Do not also check the unreduced roof live load as if snow were absent unless your jurisdiction requires it; the code takes whichever of L_r, 0.7S, or R produces the greater effect. Drift and unbalanced cases are a separate question, and a hip corner is normally out of the drift zone — but a hip below an adjacent taller roof is not.

#Can I use Table 4A values for a 6×12?

No. NDS Supplement Table 4A covers dimension lumber, 2 in. to 4 in. thick; a 6×12 is a timber and reads Table 4D, Beams & Stringers. The design values genuinely differ — Douglas Fir-Larch No. 1 is 1,000 psi in Table 4A and 1,350 psi in Table 4D, while No. 2 goes the other way, 900 down to 875 psi — and the adjustment factors differ too: Table 4D carries no repetitive-member factor and applies the size factor C_F = (12/d)^1/9 only when the depth exceeds 12 in. There is a further split within Table 4D: Beams & Stringers (nominal width more than 2 in. greater than the thickness, loaded on the narrow face — a 6×12 beam) versus Posts & Timbers (roughly square, loaded axially — a 6×6 post), per NDS 4.1.3.3 and 4.1.3.4. The two carry different tabulated values, so StructSuite filters the grade list to the class the chosen section actually belongs to — pick a 6×12 and only the Beams & Stringers grades remain offered.

#Does the top of a hip rafter need its own support?

Almost always, yes. The hip's high-end reaction — 2,579 lb here, two-thirds of the total — lands where the hip, ridge, and king common rafters converge. A ridge board is a nailing spacer with no bending capacity, so unless a ridge beam, a girder truss, or a post to a bearing wall is present, that load has nowhere to go and the corner sags. Check the reaction, then draw the load path down to the foundation. The same logic applies at the eave corner, where 1,370 lb has to reach the corner post or a properly nailed double top plate corner lap — and if the hip is dropped in a hanger there, note that standard face-mount hangers are level-seat only; a sloped member needs a sloped-seat or slopeable model from the manufacturer's literature. StructSuite makes the hanger step inactive whenever a slope is entered, for exactly that reason.

#Do I need to check the hip for bearing at an angle to the grain?

Only if the simple check fails. A hip seat is cut at the hip's own angle, so the reaction arrives at an angle θ to the grain rather than square to it, and NDS 3.10.3 gives the allowable by the Hankinson formula F_θ = F_cF_c⊥ / (F_c sin²θ + F_c⊥ cos²θ). Because F_c is much larger than F_c⊥, F_θ always exceeds F_c⊥ — so checking against plain F_c⊥, as this example does, is conservative and usually decides the matter (0.214 here). Reach for 3.10.3 when the seat is short, the reaction is large, or the bearing member below is a soft species. The reverse case deserves more care: the supporting member is being crushed perpendicular to its grain with no such bonus.

#What about wind uplift on a hip corner?

Hip corners sit in the highest-suction zones of ASCE 7-22's components-and-cladding maps, so uplift deserves a look even when gravity governs the member. Two checks, and they are different: the member is checked under 0.6D + 0.6W (ASCE 7-22 §2.4.1, combination 7a) for net upward bending, which on a heavy tile roof rarely governs because 0.6D is large; the connections are checked for net uplift regardless, and those are what fail. This example carries no wind load, so the module reports combination 7a as 0.6D alone at D/C = 0.416. Add a type W load — entered at ×1.00, since wind needs no slope conversion — and the same run produces the uplift case. For a full worked wind derivation, the Chapter 27 example linked below takes a gable house from basic wind speed to member forces.