A simple-span beam never asks the most interesting glulam question: what happens when the tension face flips? This example does. A glulam floor beam runs 16 ft between supports inside the building, then cantilevers 6 ft past the exterior wall to carry a balcony — so the bottom fiber is in tension at midspan, but over the exterior support the top fiber goes into tension. For a glulam built from a bending-optimized layup, those two faces have different strengths (Fbx+ = 2,400 psi vs Fbx− = 1,450 psi for the class we pick), and the negative moment ends up governing the whole design.
Every number here was produced by the same calculation engine that runs StructSuite's Wood Beam module, and the agreement is locked by an automated regression test (tests/verification/glulam-balcony-guide-example.test.ts) — the same trust chain used for all StructSuite worked examples.
#The structure
#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 framing decision you read off the plans — nothing is a guess. The balcony live load and the guardrail force are the two most commonly missed: both come from the code, not judgment.
| Item | Value |
|---|---|
| Geometry | 16-ft back span (pin–roller) + 6-ft cantilever; total 22 ft |
| Beam spacing | 6 ft o.c. → tributary width 6 ft |
| Interior floor loads | D = 15 psf, L = 40 psf (residential, ASCE 7-22 Table 4.3-1) |
| Balcony loads | D = 15 psf, L = 60 psf — balconies take 1.5× the live load of the area served, ≤ 100 psf (IBC Table 1607.1) |
| Guardrail | 42-in guard at the cantilever tip; 50 plf or 200 lb top-rail load, any direction (IBC 1607.9) |
| Member | Glulam 5.125 × 13.5, stress class 24F-1.8E, Douglas Fir (Supplement Tables 5A + 1C) |
| Service | Dry (covered balcony), T ≤ 100°F, deflection L/360 live / L/240 total, bearing length 3 in |
#Step 1 — Tributary width: how area loads become line loads
The floor and balcony are framed with joists and sheathing spanning between parallel glulam beams spaced 6 ft on center. Each beam carries the strip of floor halfway to its neighbor on each side — a tributary width of 6 ft (3 ft + 3 ft):
Line load = area load × tributary width:
| Segment | Dead | Live |
|---|---|---|
| Interior (0–16 ft) | 15 psf × 6 ft = 90 plf | 40 psf × 6 ft = 240 plf |
| Balcony (16–22 ft) | 15 psf × 6 ft = 90 plf | 60 psf × 6 ft = 360 plf |
| Beam self-weight | 17.3 plf (69.19 in² at 36 pcf — StructSuite adds this automatically) |
In StructSuite you enter exactly this: a psf value and a tributary width per load, with start/end positions — the app does the multiplication and records it in the report.
#Step 2 — The guardrail end moment
IBC 1607.9 requires guard top rails to resist 50 plf or a 200-lb concentrated load, each applied in any direction, and not concurrently. The beam's tributary length of rail equals its spacing:
- Distributed: 50 plf × 6 ft = 300 lb ← governs
- Concentrated: 200 lb
The horizontal rail force acts 42 in above the beam and comes back to the beam tip as a concentrated moment:
Mrail = 300 lb × 3.5 ft = 1,050 lb-ft, applied at x = 22 ft (a live load — it enters every combination with L).
"Any direction" is a sign choice you must make. Outward push and inward pull produce opposite moment signs at the tip. We ran both through the engine: the sign that adds to the hogging moment governs the flexure check (D/C 0.511 vs 0.495 the other way) — so that is the design case. In StructSuite the rail moment is one "concentrated moment" load entry at 22 ft; flip its sign to check both.
#Step 3 — Section and material: reading Tables 5A and 1C
From NDS Supplement Table 5A, stress class 24F-1.8E (Douglas Fir), the row this example uses (values exactly as tabulated):
| Fbx+ | Fbx− | Fvx | Fc⊥x | Ex |
|---|---|---|---|---|
| 2,400 psi | 1,450 psi | 265 psi | 650 psi | 1.8 × 106 psi |
Fbx+ applies when the bottom laminations are in tension (normal sagging). Fbx− applies when the top is in tension — exactly what happens over our interior support. The manufacturer puts the premium tension-rated laminations on the bottom of a 24F layup; the top face doesn't get them, so it earns only 1,450 psi. (Full layup story: Wood Design with the NDS Supplement — Series 5.)
Geometry from Table 1C (Western species): 5.125 × 13.5 → A = 69.19 in², Sx = 155.70 in³, Ix = 1,051 in4.
#Step 4 — Analysis: reactions, shear, and moment (with a sign reversal)
Unfactored D + L (including self-weight): interior w = 347.3 plf, balcony w = 467.3 plf, Mrail = 1,050 lb-ft at the tip.
Reactions: RA (pin, x = 0) = 2,187 lb; RB (roller at the wall, x = 16 ft) = 6,173 lb. The interior support carries almost 3× the end support — it picks up the whole balcony plus its share of the back span. Statics check: 347.3 × 16 + 467.3 × 6 = 5,557 + 2,804 = 8,361 = 2,187 + 6,173 ✓ (closes exactly).
Governing internal forces (D + L): M⁺ = 6,887 lb-ft at 6.3 ft; M⁻ = 9,461 lb-ft at the interior support; V = 3,370 lb just inside the support on the back-span side. Note the hogging moment is 37% larger than the sagging moment under this arrangement. Load the back span alone, though, and the sagging moment rises to 10,170 lb-ft — larger than the hogging — because the balcony’s counter-moment is removed. It still does not govern, because the tension face it puts in bending is the stronger one (Fbx+); see Why the balcony governs its own deflection.
#Step 5 — Adjustment factors (and CD per combination)
For glulam, Table 5A values are adjusted per NDS Chapter 5. The engine's factor set for this member:
| Factor | Value | Why |
|---|---|---|
| CM, Ct, Ci | 1.0 | dry, ≤ 100°F; glulam is not incised |
| CL | 1.0 | Compression edge taken as continuously braced (NDS 3.3.3.3). Note NDS 4.4.1.2's d/b bracing rules are a sawn lumber provision — glulam is governed by NDS 5.3.5, which refers directly to 3.3.3. Enter an unbraced length in the module to have CL computed from Eq. 3.3-6 instead — the field flags RB > 50 (NDS 3.3.3.7) the moment it is typed, with the longest ℓu the section permits |
| CV (volume factor) | 0.984 | NDS Eq. (5.3-1) with L = 22 ft, d = 13.5 in, b = 5.125 in — glulam's replacement for CF; lesser of CV and CL applies |
| Cfu, Cr | 1.0 | strong-axis bending; not a repetitive sawn member |
| Cvr (shear reduction) | 1.0 | straight prismatic beam — no notches |
| CD | per combination | 0.9 (D alone) · 1.0 (D + L) · 0.9 (0.6D — dead only; no wind entered, NDS 2.3.2.2) |
The engine evaluates every ASD combination with its own CD: the D-alone combination sees capacities × 0.9, D + L sees × 1.0. That is why a report shows different capacities for the same member under different combinations — and why D + L governs here.
One combination is opt-in: in Seismic Design Category D–F, ASCE 7-22 §12.4.4 requires horizontal cantilevers to also be designed for a supplemental net upward force of 0.14D (ASD). The module's Step 2 shows the Horizontal cantilever in Seismic Design Category D–F checkbox whenever the beam has an end cantilever; ticking it adds the §12.4.4 uplift: −0.14D combination, which — despite carrying only a D term — is a seismic effect and runs at CD = 1.6, not the permanent 0.9. This example's site is not asserted to be SDC D–F, so the box stays unticked and the combination list above is complete.
Bracing note for cantilevers. The CL = 1.0 assumption means each compression edge is laterally supported where it is in compression. In the back span that's the top (braced by joists and sheathing). Over the support and along the cantilever, the bottom edge is in compression — provide full-depth blocking at the support and brace the cantilever's bottom flange, or enter the unbraced length so the module computes CL per NDS 3.3.3. Miss this in the field and the calculated capacity doesn't exist.
#Step 6 — Flexure: two checks, two strengths (NDS 3.3.1, Table 5A)
Both faces get the same adjustments; only the reference value differs:
Fb+′ = Fbx+ × CV × CD = 2,400 × 0.984 × 1.0 = 2,361 psi → Mn+ = 2,361 × 155.70 / 12 = 30,633 lb-ft Fb−′ = Fbx− × CV × CD = 1,450 × 0.984 × 1.0 = 1,426 psi → Mn− = 1,426 × 155.70 / 12 = 18,507 lb-ft
| Check | Demand (lb-ft) | Capacity (lb-ft) | D/C |
|---|---|---|---|
| Sagging: M⁺ vs Mn+ | 6,887 | 30,633 | 0.225 |
| Hogging: M⁻ vs Mn− | 9,461 | 18,507 | 0.511 ← governs ✓ PASS |
Read those two rows again — they are the whole lesson. The sagging check uses barely a quarter of the section. The hogging check, with a smaller moment than the section could carry in sagging, consumes 51% — because the top face is 40% weaker. A cantilevered glulam is designed by its weak face.
#Step 7 — Shear (NDS 3.4.2)
Fv′ = 265 × Cvr × CD = 265 psi → Vn = (2/3) Fv′ A = (2/3) × 265 × 69.19 = 12,224 lb
fv = 1.5 V / A = 1.5 × 3,370 / 69.19 = 73 psi. D/C = 3,370 / 12,224 = 0.276 ✓ PASS. (Check it against the reactions above: 2,187 − 347.3 × 16 = 3,370 lb.)
#Step 8 — Deflection (IBC Table 1604.3)
With E′ = 1.8 × 106 psi and I = 1,051 in4. Per NDS 3.2.1 the deflection limit belongs to each span, so the 16-ft back span and the 6-ft cantilever are checked separately; for the cantilever the limit length is twice the projection (2 × 6 = 12 ft) per IBC Table 1604.3. Both cases are now governed by the cantilever, and by an arrangement in which the back span carries no live load at all — see Why the balcony governs its own deflection below. Note the total-load column is not D + L: IBC Table 1604.3 footnote d limits it to "the deflection due to the creep component of long-term dead load deflection plus the short-term live load deflection", and for wood that is dry at installation and used dry the creep component "shall be permitted to be estimated as the immediate dead load deflection resulting from 0.5D". So the case checked is 0.5D + L:
| Case | δ (in) | Governing span | Limit (in) | D/C |
|---|---|---|---|---|
| Live only (L) | 0.291 | cantilever (2 × 6 = 12 ft, footnote i) | L/360 = 144 / 360 = 0.400 | 0.727 ✓ |
| Total: 0.5D + L (footnote d) | 0.277 | cantilever (2 × 6 = 12 ft) | L/240 = 144 / 240 = 0.600 | 0.461 ✓ |
#Why the balcony governs its own deflection
Load the room and the balcony together and the tip barely moves: the back span sags, which rotates the beam over the interior support in the direction that lifts the balcony, and most of the cantilever’s own sag is cancelled. That case gives a tip deflection of just 0.066 in.
It is also not the case that governs. ASCE 7-22 §4.3.3 requires partial loading to be considered where it is worse, and here it is much worse — put people on the balcony while the room behind is empty and the helpful rotation disappears:
| Live load placed on… | Back-span deflection | Cantilever tip |
|---|---|---|
| Both spans | 0.080 in ↓ | 0.066 in ↓ |
| Back span only | 0.187 in ↓ | 0.225 in ↑ (rises) |
| Cantilever only | 0.113 in ↑ | 0.291 in ↓ — governs |
The tip deflection is 4.4× the both-spans figure, and the arrangement that produces it — a loaded balcony off an empty room — is the most ordinary thing in the world. This is the check that sets the member: on a 5.125 × 12 the same beam reaches 0.414 in against a 0.400 in limit and fails, while every strength check still passes comfortably. The published section is 5.125 × 13.5 because of this arrangement and nothing else.
StructSuite generates these arrangements automatically for any member with more than one span; the FAQ works through what it does and does not change.
#Step 9 — Bearing (NDS 3.10.2 check, 3.10.4 Cb)
With a 3-in bearing length, Cb depends on where the bearing sits. NDS 3.10.4 grants the bearing area factor only "for bearings less than 6″ in length and not nearer than 3″ to the end of a member" — reference Fc⊥ already "appl[ies] to bearings of any length at the ends of a member." So the end pin takes Cb = 1.0 → Fc⊥′ = 650 psi, while the interior wall support takes Cb = (3 + 0.375)/3 = 1.125 → Fc⊥′ = 650 × 1.125 = 731 psi (no CD on bearing — Table 2.3.2).
| Support | R (D + L) | fc⊥ = R/(Lb·b) | D/C |
|---|---|---|---|
| End pin (Cb = 1.0) | 2,187 lb | 142 psi | 142 / 650 = 0.219 (0.2188 unrounded) ✓ |
| Interior (wall) (Cb = 1.125) | 6,173 lb | 402 psi | 402 / 731 = 0.549 ✓ |
The wall support works almost three times harder than the end — check the plate and post under it for the same 6,173 lb.
#Two checks engineers skip on cantilevers (don't)
1 — Pattern loading and uplift. Live load doesn't have to occupy the whole floor. Re-run the model with the interior live load removed (balcony live + rail moment only, dead everywhere): the end reaction collapses from 2,187 lb to 267 lb — this beam is 267 lb away from lifting off its end support. A slightly longer cantilever, a heavier balcony finish, or a lighter interior floor and you need a tie-down at the pin. One extra run in StructSuite (remove the interior L load) shows it. And in SDC D–F the code makes uplift mandatory: ASCE 7-22 §12.4.4 adds a supplemental 0.14D net-upward combination for horizontal cantilevers — the Step 2 checkbox in the module.
2 — The stress class is not decoration. Same 5.125 × 13.5 section in class 16F-1.3E (Fbx− = 925 psi): Mn− = 11,806 lb-ft against our 9,461 lb-ft demand — D/C = 0.801 against 0.511 for 24F-1.8E. One class down costs 57% of the margin, all of it on the hogging face, because Fbx− falls from 1,450 to 925 psi while Fbx+ barely moves. The cantilever's negative moment is precisely where cheaper stress classes run out — pick the class for the weak face, not the strong one.
#Summary of results — 5.125 × 13.5 glulam, 24F-1.8E DF
| Check | Demand | Capacity | D/C | Verdict |
|---|---|---|---|---|
| Flexure, hogging (Fbx−) | 9,461 lb-ft | 18,507 lb-ft | 0.511 | governs — ✓ PASS |
| Flexure, sagging (Fbx+) | 6,887 lb-ft | 30,633 lb-ft | 0.225 | ✓ PASS |
| Shear | 3,370 lb | 12,224 lb | 0.276 | ✓ PASS |
| Deflection (0.5D + L, fn d) | 0.277 in | 0.600 in | 0.461 | ✓ PASS |
| Bearing (interior support) | 402 psi | 731 psi | 0.549 | ✓ PASS |
Reproduce it yourself in StructSuite's Wood Beam module: spans 16 + 6 ft (Pinned, Roller, Free); the four uniform loads entered as psf × 6-ft tributary with their start/end positions; a concentrated moment of 1,050 lb-ft at 22 ft (type L); Include member self-weight ticked; material Glulam → Western Species (Douglas Fir is one) → 24F-1.8E → 5.125×13.5; dry service, bearing 3 in, limits L/360 / L/240. Leave the Step 2 Horizontal cantilever in Seismic Design Category D–F checkbox unticked to match this example — tick it only when your structure is assigned to SDC D, E, or F. The detailed calculation panels reproduce every step above — including the two-line flexure check with both Fbx values.
#Frequently Asked Questions
#What is pattern loading, and when does it change my beam?
A continuous member is not worst under its own full load. Take live load off part of it and some effects get bigger — sometimes much bigger. ASCE 7-22 §4.3.3 requires you to look:
The uniform live load applied only to a portion of a structure or member shall be accounted for if it produces a more unfavorable load effect than the uniform live load applied over the full structure or member.
and §4.3.3.1 names the arrangements for roofs — the load "shall be applied to adjacent spans or to alternate spans, whichever produces the greatest unfavorable load effect."
StructSuite does this automatically for any member with more than one span. Each combination that carries L or Lr is run as several cases — all spans loaded, alternate spans, and each adjacent pair — and each appears in the combination list under its own name, so you can see which arrangement produced which number.
What it changes, and what it does not
Two equal spans, uniform load w, span L — the classic case:
| Both spans loaded | One span loaded | Change | |
|---|---|---|---|
| Moment in the span | 0.0703 wL² | 0.0957 wL² | +36% |
| Moment over the support | 0.125 wL² | 0.0625 wL² | −50% |
| Deflection in the loaded span | — | larger | up to ~2× |
Neither arrangement governs everything. The fully loaded case gives the worst support moment — the top steel, or the negative-bending face of a glulam. The patterned case gives the worst span moment and the worst deflection. You need both, which is exactly why the program runs both rather than asking you to choose.
It matters most where you would least expect it
On the balcony beam in this guide the effect is dramatic, and not in the moment:
| Check | Full load only | With pattern loading | |
|---|---|---|---|
| Hogging moment, M− | 9,461 lb-ft | 9,461 lb-ft | unchanged |
| Sagging moment, M+ | 6,887 lb-ft | 10,170 lb-ft | +48% |
| Flexure D/C | 0.511 | 0.511 | unchanged |
| Cantilever tip deflection | 0.066 in | 0.291 in | 4.4× |
The moment result is almost a non-event: the sagging moment jumps 48%, but it lands on the stronger face of the glulam (Fbx+), so the governing flexure ratio does not move at all.
The deflection is the opposite. Load the room and the balcony together and the back span's sag rotates the beam over the support in the direction that lifts the balcony, hiding most of its own sag. Empty the room — the ordinary case, someone standing on the balcony — and that help disappears. The tip deflection is 4.4× the fully-loaded figure, and on a 5.125 × 12 section it fails its L/360 limit while every strength check still passes. That single arrangement is why this guide's beam is a 5.125 × 13.5.
When it cannot matter
- A single-span member. There is one arrangement, so nothing is generated and nothing changes.
- A combination with no L or Lr. D, S, W and E are not patterned by §4.3.3, so a D-only or D + S combination runs exactly once.
- Loads that are genuinely always present. If a "live" load is really permanent, it belongs in D — and then it is not patterned, correctly.
How many arrangements
Three for two spans, four for three, eight for six. Full 2n permutations are not generated: for a uniform load on a prismatic member the arrangements §4.3.3.1 names — alternate spans, and adjacent pairs — are the ones that can govern, and the full set grows to 255 cases on an eight-span beam before being multiplied by every load combination.
If you are reading an older calculation
Calculations produced before this was added checked the fully loaded arrangement only. For a single-span member that is complete and nothing changes. For a continuous or cantilevered member it is not, and the span moments and deflections may be understated — re-run it. The clearest tell is the combination list: if no combination name carries an arrangement like "L on span 1", the patterns were not run.
#Why does a glulam beam have two bending strengths, Fbx+ and Fbx−?
Because a bending-class glulam is built asymmetrically on purpose: the highest-grade tension laminations go on the bottom face, where a normally loaded beam needs them. Fbx+ (2,400 psi here) applies when that bottom face is in tension; Fbx− (1,450 psi) applies when bending is reversed and the ordinary top laminations go into tension — over cantilever supports, at continuous-beam supports, and under uplift. Both values are adjusted by the same factors (CV, CD, …), and each moment region is checked against its own strength. In this example the reversed region governs at D/C = 0.511 even though its moment is smaller than the sagging capacity. Balanced layups (equal faces) exist for continuous members — at a cost in Fbx+.
#How is the tributary width of a beam chosen, and how do psf loads become plf?
The tributary width is the strip of floor that has no closer beam to run to: half the spacing to the neighbor on each side. At 6-ft spacing, that's 3 + 3 = 6 ft, and every area load is multiplied by it — 40 psf of live load becomes 40 × 6 = 240 plf on the beam line. Openings, non-parallel framing, or a beam at the edge of a bay change the arithmetic (an edge beam gets half), which is why StructSuite asks for psf and tributary width separately and prints both in the report — a reviewer can check the framing plan against one multiplication.
#How does a guardrail load become a moment on the supporting beam?
The code rail forces (IBC 1607.9: 50 plf along the top rail or a 200-lb point load, any direction, not concurrent) act at the top of a 42-in guard. When the guard posts mount on the end of a cantilevered beam, a horizontal rail force has a 3.5-ft lever arm to the beam — it arrives as a concentrated moment at the tip: here 50 plf × 6 ft = 300 lb (governing over 200 lb), times 3.5 ft = 1,050 lb-ft. Because "any direction" includes both signs, check the sign that worsens each effect — for this beam, the one that deepens the hogging moment. The moment is visible in the diagram: the moment curve ends at −1,049 lb-ft at the tip instead of zero.