Structural Design Project · ETABS + AutoCAD · Serial No. 2

Building Model Build Spec

Everything you need to build the 10-storey RC frame yourself — grid, stories, sections, loads, the four combinations, BNBC 2020 lateral parameters — plus three hand checks worked all the way to numbers so you can tell whether ETABS is doing what you think it is.

A = 21′ · B = 18′ · C = 11′ 10 storeys + GB BNBC 2020 Live re-calculation

Your serial number sets A, B and C. Change them and every table, diagram and check on this page re-derives. Defaults are S/N 2.

01

What You Are Building

A ten-storey reinforced concrete residential building on a 60′ × 52′ footprint, with a grade beam level, a lift core shear wall, and a stair/lift machine room poking above the roof.

The work splits into three phases, and they have to happen in order. Each one feeds the next, so a mistake in phase 1 shows up as garbage in phase 3.

Phase 1

Model

Grid, stories, sections, loads, combinations. Nothing is designed yet — you are just describing the building to the software.

Phase 2

Analyse & verify

Run it, then check three numbers by hand before you trust anything. This is the step everybody skips and it is the step that catches the errors.

Phase 3

Detail & draw

Pull the design output, choose bars, and draft the column schedule and the Beam AB long section in AutoCAD.

⚠️
One thing this page cannot do for you. The C1/C2/C3 column layout comes from the plan in your assignment PDF, and that pattern is not symmetric — there is a C1 sitting at an interior position on the second row. Any layout invented here would be wrong. Section 04 gives you the grid and a blank schedule to transcribe into; the positions themselves have to come off your drawing.
02

Grid and Story Data

Three bays each way. The X bays are A–B–A; the Y bays are B–16′–B, where that middle 16′ strip is the fixed 7′ + 9′ corridor from the plan and does not scale with your serial number.

X grid — A + B + A
AxisOffsetCumulative
1—0
22121
31839
42160
Y grid — B + 16 + B
AxisOffsetCumulative
A—0
B1818
C1634
D1852

Framing plan — grid, column positions and Beam AB

Sixteen columns at the grid intersections. The highlighted node is the circled column for your schedule; Beam AB is the three-span line along the top.

Story data

Enter these in Edit › Stories and Grid Systems › Modify/Show Story Data. Build them bottom-up. Only the ten typical floors are similar to each other — make Story 1 the master and set Stories 2–10 as similar to it, which is what lets you assign loads once.

LevelHeightElevationMaster / similar toNotes
Base—0—Fixed restraint, all 16 columns
GB6′-0″6′-0″MasterGrade beams only — no slab, no veranda
Story 111′-0″17′-0″MasterBeam AB lives here
Story 2–1011′-0″ eachup to 116′-0″Similar to Story 1Typical floors
Lift / stair top7′-6″123′-6″MasterMachine room over core only
The two things people get wrong here

1. The GB level carries grade beams but no slab — if you draw a floor there, you add roughly 367 kip of dead load that does not exist. 2. The lift/stair top is not a full floor. Draw it over the core footprint only, or your seismic weight and your base shear both come out high.

03

Materials and Sections

Two concrete grades, one steel grade. The trap is that the slabs and stair are 4000 psi while everything else is 4500 — define both materials before you define any section.

Materialf′c / fyE (ksi)Used by
CONC45004500 psi3824All columns, all beams, shear wall
CONC40004000 psi3605Floor slabs, stair slab
Gr6060 ksi29000All reinforcement

\( E_c = 57000\sqrt{f'_c} \) psi — 3824 ksi and 3605 ksi respectively. ETABS fills these in once you type f′c, but check them; a wrong E changes every drift and period you report.

Define › Section Properties › Frame Sections for the line elements, › Slab Sections and › Wall Sections for the shells.

Frame sections — f′c 4500, fy 60
Nameb × h (in)Ag (in²)Type
C112 × 18216Column
C215 × 18270Column
C318 × 18324Column
GB12 × 18216Grade beam
FB12 × 20240Floor beam
SB12 × 12144Secondary beam
Shell sections
Nametf′cModelling type
SLAB55″4000Shell — thin, membrane f11/f22 × 0.25 cracked
STAIR66″4000Shell — thin
SW1010″4500Shell — thin, pier-labelled

Label the core walls as piers (Assign › Shell › Pier Label) or ETABS reports wall forces element by element and you will not be able to design the core.

🔢
Property modifiers matter more than the section sizes. For a dual system the slab in-plane stiffness is what drags load into the core. Use cracked-section modifiers per BNBC 2020 / ACI 318: beams 0.35Ig, columns 0.70Ig, walls 0.35Ig (uncracked 0.70Ig), slabs 0.25Ig. Set them at Assign › Frame › Property Modifiers. Leave them at 1.0 and your period comes out short, your base shear high, and your drift unrealistically small.
04

The Column Layout Problem

Sixteen columns, three section types, and a pattern that is not symmetric. This is the one part of the model you cannot infer — you have to read it off the plan.

The grid gives you sixteen intersections: four X axes × four Y axes. What it does not give you is which of C1, C2 and C3 sits at each one. The plan in your assignment has a C1 at an interior position on the second row, which breaks any rule of thumb like “corners are small, interiors are big.” If you guess, you will get a plausible-looking model that is wrong in a way nobody catches until the column schedule.

Transcribe the layout into this table from your PDF before you draw anything:

GridX (ft)Y (ft)SectionPosition
1-A00from planCorner
2-A210from planEdge
3-A390from planEdge
4-A600from planCorner
1-B018from planEdge
2-B2118C3 — the circled oneInterior
3-B3918from plan — watch for the C1Interior
4-B6018from planEdge
1-C … 4-C—34from planEdge / interior
1-D … 4-D—52from planCorner / edge
Verify the circled column before you commit. Read as the C3 at X = 21′, Y = 18′ — grid 2-B, the first interior intersection. That is the reading this page's Check 1 is built on. If the circle on your PDF is on a different node, the tributary area changes and so does every number in section 08. Confirm it against the original before you draft the schedule.

Columns are drawn once and then replicated up the building. Draw the GB-to-Story-1 lift at the grid intersections, select all of them, and use Edit › Replicate › Story to carry them to Story 10. Orientation matters for C1 and C2 since they are rectangular — the 18″ dimension normally runs parallel to the beam that frames the weaker direction, so check the plan for which way each one is turned.

05

Loads and Where They Go

Four load patterns carry gravity, and the mistake that ruins models is putting a load on the wrong kind of object — area loads onto slabs, line loads onto beams, and never both for the same physical thing.

Define › Load Patterns. Self-weight multiplier is 1 on DEAD only and 0 on everything else. If you leave it at 1 on the superimposed pattern too, you count the whole concrete frame twice.

PatternTypeSelf-wt mult.What it carries
DEADDead1Concrete self-weight, computed by ETABS
SDLSuper Dead0FF 25 psf + PW 30 psf = 55 psf on slabs
WALLSuper Dead0425 lb/ft line load on wall-supporting beams
LIVELive040 psf floors, 100 psf stair
EQX / EQYSeismic0BNBC 2020 auto-lateral
WX / WYWind0BNBC 2020 auto-lateral

Dead load accounting

The 5″ slab weighs 62.5 psf on its own and ETABS adds that itself through DEAD. Your superimposed dead load is only the finish and partition allowance:

ComponentValuePatternApplied to
Slab self-weight62.5 psfDEAD (auto)— do not enter
Floor finish25 psfSDLAll floor slabs
Partition wall allowance30 psfSDLAll floor slabs
Total slab dead load117.5 psf——
Assign › Shell Loads › Uniform  →  SDL, 0.055 ksf, Gravity direction

Wall load — the 425 lb/ft line

This is a 5″ brick wall carried on a beam. It goes on beams, not slabs, and only on the beams that actually have a wall over them: the full perimeter at every floor, plus the interior partition lines shown on your architectural plan.

Assign › Frame Loads › Distributed  →  WALL, 0.425 kip/ft, Gravity
🧩
A quick sanity check on the 425 figure: a 5″ brick wall at about 120 pcf, 10′ clear height, is \(0.417 \times 10 \times 120 \approx 500\) lb/ft gross. Take out the door and window openings and 425 lb/ft is about right for a residential floor. If your plan has a solid party wall with no openings, that line deserves more.
Do not put the wall load on the perimeter at the GB level unless there is a plinth wall there. And do not put it on the roof. Both are easy to catch afterwards because the total will not match: your WALL pattern base reaction should be roughly 1462 kip for ten floors at the assumed wall lengths.

Live load

AreaLive loadNote
All floor slabs40 psfResidential occupancy
Stair slab100 psfEgress — applies to the 6″ stair shell
Roof40 psfUnless your brief says otherwise

Live load reduction: ETABS will not apply it unless you switch it on, and for a ten-storey column it makes a real difference — see Check 1, where it takes the accumulated live load from 133 kip down to 53 kip.

06

The Four Load Combinations

Define exactly these four at Define › Load Combinations. Not ETABS's auto-generated set — these specific four, with these names.

NameDefinitionTypeWhat it is for
UFL1.0 DL + 1.0 LLServiceDeflection, drift, unfactored reactions
FDL1.2 DL + 1.6 LLStrengthGoverning gravity case
FDLEQy0.9 DL + 1.2 LL + 1.32 EQyStrengthGravity + seismic, Y direction
FDLWx0.9 DL + 1.2 LL + 1.2 WxStrengthGravity + wind, X direction
📈
“DL” here means all three dead patterns. When you build these combinations, DL expands to DEAD + SDL + WALL. So FDL is really 1.2(DEAD) + 1.2(SDL) + 1.2(WALL) + 1.6(LIVE) — four rows in the combination dialog, not two. Leaving WALL out of the combinations is the single most common reason a student's beam moments come out low against a hand check.

Two of these are one-directional by design. FDLEQy only covers seismic in Y and FDLWx only covers wind in X, which is what the brief asks for. Be aware that means the combination set is not a complete design envelope — a real design needs EQx and Wy too, plus the negative senses of each. Say so on your sheet rather than quietly implying the four are sufficient.

For the Beam AB envelope in section 11, take the worst of FDL, FDLEQy and FDLWx at each station. Under the lateral combinations the support moments grow and can reverse sign, which is exactly why the bottom steel has to be continued into the supports.

07

Seismic and Wind Parameters

Both are auto-lateral loads in ETABS. Enter the code parameters and let it build the storey force distribution — then check the total against section 10.

Seismic — BNBC 2020

Dhaka, Zone 2

Z = 0.20 · I = 1.0 · Site Class SD
S = 1.5, TB = 0.2 s, TC = 0.8 s, TD = 2.0 s
R = choose for your system, state it
η = 1.0 at 5% damping

Wind — BNBC 2020

Dhaka, urban

V = 147 mph (65.7 m/s), 3-second gust
Exposure A (urban) · I = 1.0
Kd = 0.85, Kzt = 1.0, G = 0.85
Cp = +0.8 windward, −0.5 leeward

Two values to confirm rather than take from this page. Z = 0.20 and the Site Class SD spectral parameters both come from BNBC 2020 Part 6 Chapter 2, and the site class in particular depends on the soil report for your site, not on the city. Look them up and cite the table on your sheet. If your site is on soft Dhaka alluvium, SE is plausible and it changes S from 1.5 to 2.4 — a 60% jump in base shear.

Choosing R

You have a dual system: moment frames plus a lift-core shear wall. The R value depends on the detailing level you commit to, and it is a factor-of-1.3 swing in your design forces, so it is not a throwaway choice:

SystemRResulting VDetailing burden
Dual — intermediate frame + ordinary RC wall5.5620 kModerate; realistic for Zone 2
Dual — special frame + special RC wall7.0487 kFull seismic detailing throughout

Both rows assume the frames can independently resist at least 25% of the base shear, which is the defining condition for a dual system. Check that in your model by running the frames without the wall — if they cannot, it is a wall system and R drops again.

State your assumption

Whichever R you pick, write it on the sheet with a one-line reason. “R = 5.5, dual system with intermediate RC moment frames and ordinary RC shear walls, BNBC 2020 Table 6.2.19” is a defensible sentence. An R with no justification is the first thing a reviewer circles.

08

Check 1 — Column Axial Load

The circled C3 at grid 2-B, at ground floor, where it is carrying everything above it. This is the check that tells you whether your load assignment is right, because it depends on every pattern at once.

Worked check

Factored axial load on C3 at grid 2-B

1
Tributary area. Half a bay each way. X: 10.5 + 9.0 = 19.5 ft. Y: 9.0 + 8.0 = 17.0 ft. \( A_T = \) 331.5 ft² per floor.
2
Slab load into the column. At 117.5 psf dead and 40 psf live: DL = 39.0 kip, LL = 13.3 kip per floor.
3
Beams framing in. The FB 12×20 stem below the slab is 12″ × 15″ = 0.1875 kip/ft. Tributary beam length is 36.5 ft, giving 6.8 kip per floor.
4
Column self-weight. 18×18 over 11 ft = 3.7 kip per storey.
5
Wall load. This is the judgement call. If both grid lines through 2-B carry partition walls, that is 15.5 kip per floor; if only one does, 8.3 kip. Read it off the plan — it swings the answer by 11%.
6
Live load reduction. A column supporting ten floors gets the full reduction. \( L = L_o(0.25 + 15/\sqrt{K_{LL}A_T}) \) with \(K_{LL} = 4\) and \(A_T = \) 3315 ft² gives 16.0 psf, at the \(0.4L_o\) floor.
7
Accumulate and factor. \( P_u = 1.2D + 1.6L \).
Dead load, 10 floors (walls both lines)650 kip
Dead load, 10 floors (wall one line)578 kip
Reduced live load, 10 floors53 kip
Pu — walls both lines865 kip
Pu — wall one line778 kip

Is 18 × 18 enough?

For a tied column the code caps the pure-axial capacity at \( \phi P_{n,max} = 0.80\phi[0.85f'_c(A_g - A_{st}) + f_y A_{st}] \) with \(\phi = 0.65\). At \(A_g\) = 324 in²:

8-#8  (ρ = 1.95%)829 kip
8-#9  (ρ = 2.47%)878 kip
12-#8 (ρ = 2.93%)921 kip
This column is tight, and that is the real finding. Pure axial capacity at 2% steel is about 829 kip against a demand between 778 and 865 kip — and that is before any moment from the lateral combinations, which a corner or interior column always picks up. Expect the ETABS design to want 2.5–3% steel here, or a bigger section in the lower storeys. If ETABS comes back with 1% at ground floor, something is not being applied.
How to read this against ETABS

Pull the axial force at the base of the 2-B column under FDL. It should land in the 780–870 kip band. Over that band means you have doubled a load somewhere — usually self-weight on two patterns, or a slab drawn at the GB level. Well under it means a pattern is not reaching the slab, or live load reduction is set more aggressively than the hand calc.

09

Check 2 — Beam AB Moments

Beam AB runs along the top grid line, Y = 52′, at Story 1. Three continuous spans of 21′ – 18′ – 21′, section FB 12×20.

The ACI approximate moment coefficients are legitimate here, and worth using because they give you an independent number that owes nothing to your model. Check the conditions first — all four hold:

ConditionThis beamOK?
Two or more spansThree✓
Adjacent spans differ by ≤ 20%21 vs 18 → 14.3%✓
Loads uniformly distributedSlab + wall, both uniform✓
Unfactored L/D ≤ 30.19✓
Worked check

Load on the beam, then the moments

1
Slab load onto an edge beam. AB is on the perimeter, so slab arrives from one side only. The panel is 18′ × 21′ and the load pattern is trapezoidal, not a plain half-bay strip. The equivalent uniform load on the long beam is \( w_{eq} = \dfrac{wS}{2}\left(1 - \dfrac{1}{3m^2}\right) \), \( m = L/S \).
2
End span, 21′. \(m = \) 1.167, so \(w_{eq}\) = 0.799 k/ft dead and 0.272 k/ft live from the slab.
3
Add the beam and the wall. Stem 0.1875 k/ft, brick wall 0.425 k/ft. Total D = 1.411 k/ft, L = 0.272 k/ft.
4
Factor. \(w_u = 1.2D + 1.6L = \) 2.128 k/ft on the end spans, 1.965 k/ft on the middle span.
5
Clear spans. With 18″ columns, \(l_n\) = 19.50 ft (end) and 16.50 ft (middle); the interior support uses the average, 18.00 ft.
LocationCoefficientMu (k-ft)As req (in²)Bars
−M exterior supportwuln²/1650.60.702-#6
+M end span (21′)wuln²/1457.80.762-#6
−M first interior supportwuln²/1069.00.912-#7
+M middle span (18′)wuln²/1633.40.702-#6

\(A_s\) from \(R_n = M_u/\phi bd^2\) with \(d\) = 17.5″, then \(\rho = \dfrac{0.85f'_c}{f_y}\left(1-\sqrt{1-\dfrac{2R_n}{0.85f'_c}}\right)\). Minimum steel \(\rho_{min} = 3\sqrt{f'_c}/f_y = 0.00335\) governs three of the four locations — \(A_{s,min}\) = 0.70 in².

The number to check against ETABS: positive moment at the 21′ end span ≈ 58 k-ft under FDL, negative at the interior supports ≈ 69 k-ft. ETABS solves the frame properly rather than with coefficients, so expect it to land within about 10–15% — usually a little lower at midspan and a little higher at the supports.

Shear

\( \phi V_c = 0.75 \times 2\sqrt{f'_c}b_wd = \) 21.1 kip, against a maximum \(V_u\) of 23.9 kip at the face of the first interior support. Stirrups are needed but the demand is small — \(V_s\) required is only 3.6 kip, so the maximum spacing \(d/2\) governs everywhere.

#3 double-leg stirrups @ 8″ c/c throughout, tightened to 4″ over 2h = 40″ from each support face for seismic detailing.
10

Check 3 — Base Shear

The global check. If your seismic weight is wrong, everything lateral is wrong, and this catches it in five minutes.

Worked check

Seismic weight and V = SaW

Floor area is 3120 ft². Build the weight up component by component for one typical floor:

ComponentBasiskip / floor
Slab + finishes117.5 psf over the plan area366.6
Beams0.1875 k/ft stem, all grid lines84.0
Columns16 columns, average 15×1849.5
Brick walls0.425 k/ft, perimeter + partitions146.2
Lift core wall10″ wall, ~26 ft net length35.8
W per floor219 psf682
W total10 floors6,821
1
Period. \(T = C_t h_n^m\) with \(h_n\) = 116 ft = 35.4 m. For a dual system (\(C_t\) = 0.0488, m = 0.75), \(T\) = 0.708 s.
2
Spectrum. That falls between \(T_B\) = 0.2 s and \(T_C\) = 0.8 s, so \(C_s = 2.5S\eta\) = 3.750.
3
Design acceleration. \( S_a = \dfrac{2}{3}\dfrac{ZI}{R}C_s \).
4
Base shear. \(V = S_aW\).
V — R = 5.5, T = 0.71 s620 kip
V — R = 7.0, T = 0.71 s487 kip
V — R = 5.5, longer T = 1.15 s430 kip
Wind, X direction (on the 52′ face)282 kip
Wind, Y direction (on the 60′ face)326 kip
What this tells you

Seismic base shear should land somewhere around 5–9% of W, i.e. roughly 340–620 kip depending on the R and the period you end up with. Wind is smaller but not negligible at 326 kip. Compare both against your ETABS auto-lateral totals under Display › Load Cases › Base Reactions.

🕐
Your period will not match the code formula, and that is normal. \(C_th_n^m\) is a deliberately short empirical estimate; the ETABS modal period from a cracked-section model will typically be 1.3–1.6× longer. The code caps how much of that you may use (\(T \le C_uT_a\)), so the design base shear stays near the empirical value even when the model is softer. Report both periods and say which one drove the design.
11

Beam AB — Detailing and Curtailment

This is what the AutoCAD long section has to show: where the bars are, where they stop, and why they stop there.

Pull the moment envelope from ETABS first. Take the worst of FDL, FDLEQy and FDLWx at every station along the beam, not just the FDL diagram — under the lateral combinations the support moments grow and the end-span support moment can reverse.

Display › Force/Stress Diagrams › Frame/Pier/Spandrel Forces  →  Moment 3-3, envelope, show values

Beam AB long section — bar arrangement and cut-off points

Top steel over the supports, bottom steel in the spans, and the cut-off distances measured from each support face. Dimensions update with your span lengths.

Curtailment rules that set those dimensions

BarRuleThis beam
Top, interior supportExtend ln/4 past the face each side4′-11″
Top, into the 18′ spanln/4 of the shorter span4′-2″
Top, exterior supportln/5, hooked into the column3′-11″
Bottom, into supportAt least ¼ of +As continuous, 6″ min into the support2-#6 through
Development, #7 topld, top-bar factor ψt = 1.34′-3″
Development, #6 bottomld, ψs = 0.81′-9″
Standard hook, #7ldh, 90° hook1′-1″
Check the cut-off against development length, not just against ln/4. The #7 top bar needs 4′-3″ of embedment from the point of maximum stress, and ln/4 here is 4′-11″. Those are close enough that a slightly shorter span or a larger bar flips which one governs. Take the larger of the two and round up to the next 3″.
Also required by ACI at every cut-off

Extend every bar a distance \(d\) or \(12d_b\) beyond the point where it is no longer needed for flexure — here that is 17.5″ and 10.5″ for a #7, so 17.5″ governs. And at least one third of the negative steel must run past the inflection point by \(d\), \(12d_b\), or \(l_n/16\), whichever is greatest. Show both on the drawing; markers are what the checker looks for.

12

The AutoCAD Drawing Set

Four drawings, and each one has a specific job. Draw them at a real scale in model space and plot through a layout — do not draw to fit the paper.

Drawing 1

Framing plan

Grid with dimensions, all sixteen columns with their section marks, beam marks on every line, slab panel callouts. Scale 1/8″ = 1′-0″. This is the drawing that proves your column layout matches the architectural plan.

Drawing 2

Column schedule

The circled column, storey by storey: section size, bar count and size, tie size and spacing, and the splice location. One row per storey group. Include the load it was designed for so the numbers are traceable.

Drawing 3

Beam AB long section

Three spans at 1/4″ = 1′-0″ horizontal. Top and bottom bars with cut-off dimensions, stirrup zones and spacings, support faces marked. Add the moment envelope above it at the same horizontal scale so the steel visibly follows the moment.

Drawing 4

Sections

Cut through the beam at midspan and at a support, 1″ = 1′-0″. Two sections, because the bar arrangement is different at each — that difference is the whole point of the drawing.

Layer discipline

LayerColourLineweightContents
S-GRIDGrey 80.09Grid lines, bubbles
S-CONCCyan 40.35Concrete outlines
S-REBARRed 10.50Main bars — heaviest line on the sheet
S-STIRRUPGreen 30.25Stirrups, ties
S-DIMYellow 20.13Dimensions, leaders
S-TEXTWhite 70.18Notes, bar callouts
📏
Reinforcement is the subject of these drawings, so it gets the heaviest line. The most common presentation failure is a beam section where the concrete outline is bolder than the bars, which reads as a drawing about a rectangle. Set S-REBAR to 0.50 mm and let everything else sit behind it.

Annotate every bar as count – size – length, for example 2-#7 × 12′-6″, and give each a bar mark that matches a bar bending schedule on the same sheet. A drawing with bars but no schedule cannot be fabricated from.

13

When ETABS Disagrees With You

A 15% gap between hand check and model is normal. Beyond that, it is almost always one of these six things, roughly in order of how often it happens.

#SymptomCauseFix
1Everything is ~50% heavySelf-weight multiplier left at 1 on SDL as well as DEADSet SDL and WALL multipliers to 0
2Base reaction high by ~360 kipA slab drawn at the GB levelDelete it — GB carries beams only
3Beam moments low by ~25%WALL pattern missing from the combinationsDL means DEAD + SDL + WALL, three rows each
4Column axial lowSlab not meshed, so load never reaches the beamsAuto-mesh the floor, or check the load path
5Base shear high, drift tinyProperty modifiers left at 1.0Apply cracked-section modifiers
6Wall forces unusableCore walls not pier-labelledAssign pier labels to the whole core
The order to check things in

Always start with the total base reaction under DEAD alone and compare it to 6800 kip ± 10% for a building this size. That one number catches causes 1, 2 and 4 immediately. Only once the gravity total is right does it make sense to look at individual members.

Quick reference

Plan60′ × 52′
Roof elevation116′-0″
Slab dead load117.5 psf
Floor live load40 psf
Wall line load0.425 k/ft
Seismic weight6,821 kip
C3 tributary331.5 ft²
C3 Pu band780–870 kip
Beam AB +M end span57.8 k-ft
Beam AB −M interior69.0 k-ft
Seismic base shear340–620 kip
Wind base shear282–326 kip
✍️
State your assumptions on the sheet. Four of them matter enough to write down: the R value and why; the site class and where it came from; whether the wall load sits on one or both grid lines through the circled column; and the fact that the four required combinations cover only EQy and Wx, so they are not a full design envelope. Assumptions that are written down are engineering judgement. The same assumptions left implicit are just gaps.