Glulam Beam Design Example — Cantilever Balcony Beam by ASD (NDS 2024)

A complete worked design example for a 5.125×12 24F-1.8E Douglas Fir glulam beam with a 16-ft back span and a 6-ft balcony cantilever: tributary loads, the guardrail end moment, shear and moment diagrams with a sign reversal, the volume factor Cv, positive and negative flexure checks against Fbx+ and Fbx−, shear, deflection, and bearing by Allowable Stress Design per NDS 2024 — verified against StructSuite and locked by a regression test.

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 (F_bx⁺ = 2,400 psi vs F_bx⁻ = 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 (/design/wood-beam), 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

Elevation. Amber = interior floor load, blue = the heavier balcony load, pink = the rail moment applied at the beam end. The guardrail and its horizontal force are drawn as a gray shadow — that force never touches the beam directly; only its moment does.

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 × 12, 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):

Plan. Joists (thin lines) deliver the shaded 6-ft strip to the middle beam — inside the building and out on the balcony.

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	15.4 plf (61.5 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:

M_rail = 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 0.637 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):

F_bx⁺	F_bx⁻	F_vx	F_c⊥x	E_x
2,400 psi	1,450 psi	265 psi	650 psi	1.8 × 10⁶ psi

F_bx⁺ applies when the bottom laminations are in tension (normal sagging). F_bx⁻ 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 × 12 → A = 61.50 in², S_x = 123.0 in³, I_x = 738.0 in⁴.

The cross section is the whole F_bx⁺/F_bx⁻ story: a 24F layup concentrates its premium tension-rated laminations at the bottom (amber). Flip the bending — as this cantilever does over its support — and the ordinary top laminations must carry the tension.

#Step 4 — Analysis: reactions, shear, and moment (with a sign reversal)

Unfactored D + L (including self-weight): interior w = 345.4 plf, balcony w = 465.4 plf, M_rail = 1,050 lb·ft at the tip.

Reactions: R_A (pin, x = 0) = 2,174 lb; R_B (roller at the wall, x = 16 ft) = 6,274 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: 345.4 × 16 + 465.4 × 6 = 5,526 + 2,792 = 8,318 ≈ 2,174 + 6,274 ✓ (small differences are self-weight rounding).

The two things a simple span never shows: the moment changes sign at 12.6 ft, and the diagram does not end at zero — it ends at −1,050 lb·ft, closed out by the applied guardrail moment.

Governing internal forces (D + L): M⁺ = 6,842 lb·ft at 6.3 ft; M⁻ = 9,429 lb·ft at the interior support; V = 3,361 lb just inside the support on the back-span side. Note the hogging moment is 38% larger than the sagging moment — the cantilever runs this design.

#Step 5 — Adjustment factors (and C_D per combination)

For glulam, Table 5A values are adjusted per NDS Chapter 5. The engine's factor set for this member:

Factor	Value	Why
C_M, C_t, C_i	1.0	dry, ≤ 100°F; glulam is not incised
C_L	1.0	d/b = 12/5.125 = 2.3 → NDS 4.4.1.2(b): ends held in position (blocking/hangers) — and the compression edges are braced as discussed below
C_V (volume factor)	0.995	NDS Eq. (5.3-1) with L = 22 ft, d = 12 in, b = 5.125 in — glulam's replacement for C_F; lesser of C_V and C_L applies
C_fu, C_r	1.0	strong-axis bending; not a repetitive sawn member
C_vr (shear reduction)	1.0	straight prismatic beam — no notches
C_D	per combination	0.9 (D alone) · 1.0 (D + L) · 1.6 (0.6D, wind-type)

The engine evaluates every ASD combination with its own C_D: 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.

Bracing note for cantilevers. The C_L = 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 (which 4.4.1.2(b) requires anyway at d/b = 2.3) and brace the cantilever's bottom flange or verify C_L with the unbraced length. 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:

F_b⁺′ = F_bx⁺ × C_V × C_D = 2,400 × 0.995 × 1.0 = 2,389 psi → M_n⁺ = 2,389 × 123.0 / 12 = 24,486 lb·ft F_b⁻′ = F_bx⁻ × C_V × C_D = 1,450 × 0.995 × 1.0 = 1,443 psi → M_n⁻ = 1,443 × 123.0 / 12 = 14,794 lb·ft

Check	Demand (lb·ft)	Capacity (lb·ft)	D/C
Sagging: M⁺ vs M_n⁺	6,842	24,486	0.279
Hogging: M⁻ vs M_n⁻	9,429	14,794	0.637 ← 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 64% — because the top face is 40% weaker. A cantilevered glulam is designed by its weak face.

#Step 7 — Shear (NDS 3.4.2)

F_v′ = 265 × C_vr × C_D = 265 psi → V_n = (2/3) F_v′ A = (2/3) × 265 × 61.50 = 10,865 lb

f_v = 1.5 V / A = 1.5 × 3,361 / 61.50 = 82 psi. D/C = 3,361 / 10,865 = 0.309 ✓ PASS.

#Step 8 — Deflection (IBC Table 1604.3)

With E′ = 1.8 × 10⁶ psi and I = 738 in⁴ (limits taken over the 22-ft total length, per the app's convention):

Case	δ (in)	Limit (in)	D/C
Live only	0.114 (max in the back span at ≈ 7 ft)	L/360 = 0.733	0.155 ✓
Total D + L	0.191	L/240 = 1.100	0.174 ✓

The cantilever tip moves only 0.055 in under D + L — the back span, not the balcony edge, is where this beam deflects. (For balcony comfort, some engineers additionally limit tip deflection to 2·ℓ_cant/240 using twice the cantilever length; at 0.055 in this beam passes any version of that check by a wide margin.)

#Step 9 — Bearing (NDS 4.2.6, 3.10.4)

With a 3-in bearing length and C_b = (3 + 0.375)/3 = 1.125: F_c⊥′ = 650 × 1.125 = 731 psi (no C_D on bearing — Table 2.3.2).

Support	R (D + L)	f_c⊥ = R/(L_b·b)	D/C
End pin	2,174 lb	141 psi	0.193 ✓
Interior (wall)	6,274 lb	408 psi	0.558 ✓

The wall support works almost three times harder than the end — check the plate and post under it for the same 6,274 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 2,174 lb to 254 lb — this beam is 254 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 (delete the interior L item) shows it.

2 — The stress class is not decoration. Same 5.125 × 12 section in class 16F-1.3E (F_bx⁻ = 925 psi): M_n⁻ = 9,437 lb·ft against our 9,429 lb·ft demand — D/C = 0.999. It "passes" by 8 lb·ft. No engineer ships that margin. 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 × 12 glulam, 24F-1.8E DF

Check	Demand	Capacity	D/C	Verdict
Flexure, hogging (F_bx⁻)	9,429 lb·ft	14,794 lb·ft	0.637	governs — ✓ PASS
Flexure, sagging (F_bx⁺)	6,842 lb·ft	24,486 lb·ft	0.279	✓ PASS
Shear	3,361 lb	10,865 lb	0.309	✓ PASS
Deflection (total)	0.191 in	1.100 in	0.174	✓ PASS
Bearing (interior support)	408 psi	731 psi	0.558	✓ PASS

Reproduce it yourself in StructSuite's Wood Beam module (/design/wood-beam): spans 16 + 6 ft (pin, roller, free end); 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); self-weight on; material Glulam → Douglas Fir → 24F-1.8E → 5.125×12; dry service, bearing 3 in, limits L/360 / L/240. The detailed calculation panels reproduce every step above — including the two-line flexure check with both F_bx values.

#Frequently Asked Questions

#Why does a glulam beam have two bending strengths, F_bx⁺ and F_bx⁻?

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. F_bx⁺ (2,400 psi here) applies when that bottom face is in tension; F_bx⁻ (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 (C_V, C_D, …), and each moment region is checked against its own strength. In this example the reversed region governs at D/C = 0.637 even though its moment is smaller than the sagging capacity. Balanced layups (equal faces) exist for continuous members — at a cost in F_bx⁺.

#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 −1,050 lb·ft at the tip instead of zero.

#Where to go next

The loads feeding this beam: Seismic ELF example and Wind Chapter 27 example
Every table this example reads: Wood Design with the NDS Supplement and Engineered Wood Products
Codes, tools, and learning resources: Structural Engineering Software & Resource Directory