Bruhn - Analysis And Design Of Flight Vehicles Structures Episode 3 part 13 pdf

2 When a significant axial load or bending moment must also be transferred, additional members must de provided or the shear clip must be replaced by a heavier fitting.. The load “bala

HASIG

MATERIAL 7078-76 CLAD

os

9

020 052 040 050 Q63 Ơn O80 O30 IOOO

Fig, D2.23 (Ref 1) Static Strength of Typical Singie

Spot Welds In Tension Using Star Coupons

CENTER TO CENTER SPOT SPACING EN INCE!

Fig D2 24 (Ref 1) Efficiency in Tension for Spot Welding Aluminum Alloys

PROBLEMS:

sg (4) Illustrates a welded plate fittl

tting plate and tube are steel Fry =

O00 What is the maximum desien load P

which the fitting car Se subjected to 1?

a fitting factor of 1.2 is used Fitting

is not subjected to vibration or rotation

9+ nings cin

tape LenS ST ee Tome

Sams as Problem L but tube and fitting ts naat-treated after welding to Fry = 150,060,

By Mise Pe Cravese OTS),

7

WASHERS

_#, (7

P

Fig A

A -.051, 2024-T3 aluminum sheet carries an ultimate tension load of 700 Lbs per inch

It ts spliced by a lap joint involving one row of spot welds spaced at 0.5 inch Is

the spot weld strength satisfactory

A 7075-T6 aluminum sheet * ultimate tensile stress ot Zoo00 ost

sheet is to be soliced Design a scot

welded joint for a leo joint

arries an

‘The

In a wing section involving skin and

stringers, the shear flow in HN

panels to a particular stringer is 400

600 lbs per inch, and acting in the sane

direction Assuming no diagonal tension action due to skin wrinkling, wnat spot spacing ts required to fasten the stringe

to the skin 12 ths skin is 04 thick

204-TS aluminum 2licy material

and

ane

Military Handbook ~ MIL-H)8K-5, August,

1962

ANC=5, Mar:

CHAPTER D2 SOME IMPORTANT DETAILS IN STRUCTURAL DESIGN

BY WILLIAM F McCOMBS (DESIGN SPECIALIST - CHANCE VOUGHT CORP.)

D3 1 Introduction,

In the design and fabrication of an air-

plane the major components receive a thorough

parts, however, are designed at the last minute

and, not receiving so much attention, sometimes

frequently lead to trouble in service and in

tests This chapter represents an attempt to

point out some of the more common details that

seem, somehow, to be overlooked from time to

time This should be of help to those involved

in designing or dealing {n other ways with the

structural components of airplanes or of

similar types of structures With regard to

specific details, many aircraft companies have

standard methods of design The reader should

always consult his company’s data on these, if

available In the event such are not available,

the following suggested practices should be of

practical help

D3,2 Shear Clips

There are nundreds of these in a typical

together both primary structural components

and secondary structural parts such as equip~

ment mounting brackets, etc The function of

the shear clip is to transfer a shear load

from one part to another It is not intended

to transfer axial load or bending moment or

twist, only shear

A typical example is shown in Fig 03.1

Here bracket, or beam, (a) is supported by

beams (b) and (c) The load P is thus "beamed

out" to (b) and (c}, passing as a shear load

through the clips into the webs of (b) and (c)

as 111ustrated

Fig D3 2

When a significant axial load or bending

moment must also be transferred, additional

members must de provided or the shear clip

must be replaced by a heavier fitting This

is illustrated in Fig 03.2 Here a beam, (a),

Fig D3.2

is cantilevered off of a heavy plece of

the shear clip as a shear load from web (a)

plates, "S", are provided for this purpose

They transfer the moment in the form of axial

loads from the flanges of beam (a) to member

{b)

Shear clips are usually seen in two forms;

(a) bent up sheet metal or extruded

angles (the angle being anywhere

from 0° to 180° between the legs)

(>) extruded "tees"

These shear clips are shown in their

The

"ninimum acceptable” form in Fig DS.3

“Minimum” Type Shear Clips Fig D3.3

minimum requirement {s that eacn leg or an

angle type clip must have at least 2 ?

and its leg must also have at least 2 fasteners

each, as snown in (e) and (f}

The load “balance” of an angle type shear

clip is illustrated in Fig D3.4 The corner

edge of the clip should be assumed to carry

D4.1

only the shear being transferred, taken as

fasteners are then as illustrated Once tne

loads are known, the clip and fasteners can

be checked for strength using standard methods

Final Balance

Obtained by Let a=.40", b=1,0"

Adding Above _ 4 aang

Together Then Q= 1000 x~p = 400#

Resuitant Rivet Load:

Rx,/500* +4007 = 6404 Use 2-5/32" Alum, Rivets

Then clip thickness required is t = 032", T5ST Alc,

Fig D3.5 illustrates what will happen if

only one fastener {1s provided in 4 leg of an

(ignoring friction) balance any shear at the

corner In other words, it can receive only

a shear from the web to which it is fastened

This, in turn, puts a twist, P x a, into the

other leg of the clip and, hence, into the

other wed This is unacceptable, of course,

since a much thicker leg would be needed to

carry the torsion, and an undue twist would be

course, Several fasteners, rather than Just

two, may be used when space allows

Fig, D3.4

Clips of type (a), (c) and (e) in Fig

DZ.3 are more efficient than are types (Dd),

(4) and (7) The latter are used when this ts

cases the dimension "a" should be kept as small

3s practical installation will allow

jst For loads on longer leg

in figure, leta=.4", bz1.0",

|) \rwist=10004 42400" 4 This is required to

balance the 1000# which

is out of the plane of

the longer leg This

is unacceptable

1000 14004 SH

400#

An 'Ưnacceptable”

Type Shear Clip

Fig D3.5

Another type of deficiency sometimes

arises when a minimum type shear clip is being

it has been necessary to "joggle” one leg of

the angle clip, say to fit over some locally

thicker part of the member being attached If

\

i ì

i ì

f +

fi 1

Fig D3.6

order of the clip’s thickness or greater, it

can considerably reduce the clip’s rigidity and

cause it to function as 2 "one rivet clips"

with the adverse twist effects mentioned

should be provided on one side of the joggle

tn the joggled leg, as illustrated in Fig

DZ.6(b), to maintain rigidity and proper

be carried by the 2 fasteners above the joggle,

similar to case (b) or (d) in Fig 03.3

Joggles are discussed further in Art D3.4

D3 3 Tension Clips

These are also quite numerous in military airplanes, being used to splice relatively light tension loads from one member to another

The tension clip is a very inefficient type of

particularly, and should be used only when the load 1s small and other design factors prevent the use of the more efficient lap shear splice

It is usually resorted to when some structural member such as a bulkhead web or flange or fitting cannot be efficiently “opened up" to let an axtally loaded member pass

through It is also frequently used to attach cantilevered brackets to bulkheads or ribs or

other structure

on one side of a bulkhead and is to be spliced

to member (b) on the other side There is an axial tension load to be transferred and since the bulkhead cannot be cut, a tension clip arrangement must be used as shown Angle clips

in this case are illustrated

ension Clips

Fig D3.7

AL

ANALYSIS AND DESIGN

S$ the strongest and stiffest for

To obtain maximum strength and stiffness,

bolts should be used for attachment purposes,

Allowadle load data is given in Fig D3.8 for

single angle clip arrangement illustrated

YIELD LOAD FOR SINGLE ANGLES

2 IN AND OVER 30!

Bw

THICKNESS OF ANGLE - INCHES —- YIELD LOAD PER BOLT - LBS

NOTES;

1 In these tests the angles procrudea

LoAD at either end beyond the © of the { boit a distance of 1/2 the bait BOLT HEAD spacing

CLEARANCE 2 For thick angles the bolt may be

eritieal

AN-4 SOLT 1 a Values are for room temperature

AN960 eee Jeet ANGLE use only

Fig D3.8 (Ret, Vought Structures Manual}

ip thick-

1 and begin

ÿ tO smaller solts the thicknesses

ying action

ac the

yoe clip arrangement

P, Segin to "open-up"

sels an increasing tension

gad, <, and the "toss" of the clin bear down

OF FLIGHT VEHICLE STRUCTURES

D3.3

Fig D3.9

moments about ths center of pressure on the

a+d) a)

P Obviously a small anough 201t will yleld

or fail in tension before a thick clip will yield or fail in bending near the washer (Molin = P xe} There is also a prying action in the tse type clip, 2s illustrated

This prying action is the reason why th designer should be cautious in using rivets

even for light tension loads, as is sometimes

done When rivets are used, as in mounting equipment brackets, 1t is best to use steel

types and carefully check the prying load

maintaining an ample margin of safety In

any avent, riveted clips are inferior and no

design data tor them is given here

Another point in using tension clips is frequently overlooked The structure to whicn

the clip is attached must te capable of taking

the loads applied to it These loads consist

of the tension load from the bolt and the load

from the toe action Several examples are

term “unacceptable” means that the allowabis

leads of Fig D3.8 are not applicable

† S Ỷ |

®) (e) dy Ý

|

@ '

Heavy Light Back to Eccentric Criss-

Back-Up Back-Up Back Clip Load — Cross

Structure Structure Clips Path Clips

Accept- Unaccept- Accept- Unaccept- Unaccept-

able able able able able

Fig D3 10 Sases (b), (da) and (¢) require a rearrangement

of, or additional structure in, the vack-up

structure which is receiving the load from the Some aircraft companies have specific

clips in order to achieve the full aliowables strength data and practices for the design of

af The resultant load on the back- Joggled members This should, of course, te

Tension clips, aside from having a low

static strength and stiffness, also exhibit a

of such a nature as to occur many times, say

due to symmetrical flight conditions such as

pull-ups and gusts, there should be a large

due only to some non-recurring type of loading,

such as a crash condition or "jammed" system

load, the large margin of safety would not be

necessary

ther suggested practices

are: involving tension

c11ps

1 Keep the bolt head as

bend radius or fillet

possible

close to the radius as is

repeated loadings are possible, when

dominant

across a joint and part is interrupted,

do not use tension clips to join the

interrusted structure - instead a

heavier, stiffer, machined ?itting is

required

D3.4 Joggled Members

A "Joggzle" is an offset formed in a member

© usually involves one ‘or more flanges of 2

amber or the "open" cross~section type

oggles are quite common in typical metal

irplane structures

‡

They are used most often

is desired to fasten together two

eting members without using an extra

the joint The jJoggle ts a compromise

S$ an extra part but the price pald is 4

IZ tne load in the member at the foint

extra part, instead of or in addition

ation 1s shown in Fig D3.11 where one

an angle member has Deen joggled over

stened to another member lying tn the

lane

t

2

9

n

H «œ

ow

ou H o

Skin or

Floor is

Q Usually

Present, Fig D3 11

Angle Joggied Flange

ratio, others use a 3:1 ratio, or both may be used, Strength or stizfness data for one ratio

of this data indicates that when, in the case

of angle members, the depth of the joggle is

to the order of the thickness of the joggled

leg or more, the loss in strength is about

equivalent to the loss of the leg outboard of

practice is therefore suggested as follows

Assume that the net effect of the Joggle, from

a strength and stiffness standpoint, is equivalent to a slot cut into the joggled leg that extends inward to the bend radius tangent

point This is illustrated in Fig 03.12

A

{c)

OS cà

Member Slot" Effective

Section Through Joggle

Fig D3 12 With this assumption, the flat portion of

the joggled leg will carry no axial load across

the joggled area but will provide support for

the curved element The effective net section,

Fig D3.12(c), can then be checked using

standard methods of analysis for whatever forces are acting on it It is obvious that

the net section shown will have little strength

for carrying bending moment normal to the re- maining leg Thus, care should be taken to

insure that any axial loads are introduced as

near the corner as possible - which in turn means that at least two fasteners should be

used on each side of the jogzle

The above approach, considering the joggle

equivalent to a slot, will give the designer

& much better "feel" of what he is really doing when he specifies 2 joggled member

basic reason ror the loss of strength anc

stiffness can be seen in Fig DS.15

The

Req'd Balancing Loads

at Breaks

in Joggled Flange

Fig D3 13

493

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES The axial loads in the joggled leg being

inclined to each other require a balancing

load Such a balancing load is not available

except a3 shear in the thin leg and this

results in the less of stiffness and strength

If the symmetrical leg of a tee member were

joggled there would be more stiffness than in

the case of an angle, out the same approach,

though more conservative, 1s recommended in

the absence of test data

AS an example of the foregoing discussion

assume that an angle member is supporting

another member locally which is loaded by the

present, as shown, part of the load can be

carried across the joggle by the gusset effect

of the skin This can be approximated by

using the methods of calculating inter-rivet

buckling of skins discussed in another chapter

The rast of the load must be carried across by

the net effective section of the angle in the

jJoggled area

Edge M

Membar, "b'"

Fig D3 14

Thus if the total load at the joint were 2Q

and the load carrying ability of the skin were

R, then the net section of the angle would be

subjected to a load P = 2Q - R and a bending

moment M = 23 x a- Rxb, The stress at the

lower curved edge would be the sum of the

compressive stresses,

Tyet section “Net Section

be a a For ultimate strength, f¢ could be carried up

to Foy, conservatively, in the typical case

In order to realize the maximum strength

and stiffness, the load in the net section must

ve applied in the "corner" This is to prevent

stresses due to bending out of the plane of the

remaining leg This requires that a minimum

of 2 fasteners be provided to receive the load

at the joint The reasoning here {gs the same

as discussed in Art D3.2 concerning minimum

type shear clips, and the fastener loads can

be calculated in the same manner 2s discussed

there

D3.5

section to carry, then an additional member

this are shown in Fig 02.15 Sometimes local requirements are such as

to necessitate both legs of an angle member

being joggled In such cases it should be assumed that the angle has no significant load carrying ability at the Joggle Thus, the existence of any significant load at the Joint would require an additional member and the

angle should be ended just short of the joint

rather than joggled up onto it

Add Member Cut from Tee Extrusion

Add Unjoggied Angle Member

Fig D3 15

The suggested effective net sections of members having other types of cross-sections

are shown in Pig D3.16 where the legs

indicated by dotted lines are joggied In gemeral if the joggle is slight, considerably less than the thickness of the joggled leg, its effect can be ignored, but proper fasteners should still be provided as discussed The

smaller the length to depth ratio used for

joggling, the greater the effect of the Joggle

Joggled members lose stiffness and strength

when subjected to tension loads as well as

when under compression (but any skin present

is, of course, much more effective as a gusset

than when in compression)

fig, D3 18 D3.5 Fillers

As the name implies, fillers are used to

of the structural load path that they need particular attention Fillers also represent

an item that is quite common in typical large

or complicated metal airplane structures

P, are seen to be spliced togetaer by 2 pair

thicker than that of "a", 2 filler is needed

This filler ts part 2f the structural load

"b0, nam

nh

đợt =

t Load

E=—

View A-A

fe Extended € Filler At

Filler

Fig D317

path, ?rom 7c"

Yeslize 2ull st

into memder “a", I

said to bea "“Zloat

later, a floating f

cause a loss in fas

enzth of the fasteners, extended” “and additional

o tle the extended portion this is not done, it 1s

ler, 1f thick enough, will

er strength

t

9

in

11

ten

In the above example let the total load

@ 8000 lbs and assume that 2000 lbs of

from "c™ to "a" by

into filler by

ce taken 4:

fasteners in

the load put

ing pressure can tne

? =

+t

“pinier * “2

204

„€4 + 06

Sufficient fasteners should be out in th

vended part of the filler to transfer this

o los into member "a"

Thus, whenever a thick filler 1s inserted

tetween two members being spliced together in

shear, the 7iller should be assumed to ve a

part of one of them The part of the total

lice load {t will carry can be calculated as

illustrated above The fasteners can then oe

3znsidered as being in two sets One set must

the

rom the Single member to the combination

us member" The ot

Another examcle us

= Hi Shear (Steel) Rivets

32 1

7075-T6 Slum

Alloy Sheet Mtl,

(All Members}

Strengths:

Rivet Shear = 1820+

Bearing In 072 = 16304

Bearing In 081 = 1840#

Fig D3 18

ral with the lower

So be integ

1 Rivets required to splice

"a" to combination 2000 ibs from clus filler:

No Rivets = TE2O = 1,65, or Zz

Tp"):

1500 lbs

No Fasteners = ie = 292 or 1

Total Fasteners required =2+ 123,

Splice is adequate since 3 rivets are present

Had no filler been present, 2 fasteners would

have sufficed

Case Il

P = 5000 lbs

Repeating the same steps as

Ụ require 3000 5 r

2 + Load in Filler = 5000x% 7 _._-Ắ .078 207 + ,072 =

Thus the 3 rivets are required to transfer load, P, from "a" and 2 additional rivets are

needed to unload the filler into member "bd"

The filler should be

additional fasteners dotted lines in Fig

extended over “b" and 2 added as shown by th

D2.18,

edly, the abOVe nracedure is

3 tne effect of the filler when the ft thickness {s less than soout 15%

of the fastener diameter, {ts presence can

te ignored

The effect of the filler {ts to reduce

reason for this can se sesn trom Fig 03.1 wherg tne oresence of the filler causes greater

prying leads and nence mora tension in the

fastener, along with the shear load

the structural Load

be made

any filler in should, o2 course, compatible

path

from a material

in stiffness with that of the

41

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES

—

- a —>

Sa ee Larger Prying

—

“Normal” Eccentricity

with Structural (Extended) Filler Fig D3 19

structure around it That is, one should not

use a soft aluminum filler between high heat-

treated steel parts or a ohenolic or fiberglas

fillers arises not only from design consider-

ations but frequently from manufacturing

between parts sometimes occurs in assembly

To prevent expensive re-work, structural fillers

must be used to make the spliced area adequate

In these cases detailed attention is necessary

in the occasional instances when floating

fillers cannot be avoided, the fasteners should

have quoted allowables well in excess of the

Shear being transferred locally, if the riller

is of Significant thickness It is common

practice also to use a donding agent (glue) in

addition to the fasteners in installing ?iliers

D3.6 Cut-outs in Webs or Skin Panels

The aircraft structure 1s continually

faced with requirements for opening up webs

and panels to provide access or to let other

members such as control rods, hyaraulic lines,

electrical wire bundiss, etc., pass through

The destgner or liaison engineer should be

familiar with some of the various methods of

providing structurally sound cut-outs

There are several ways of providing cut-

outs Three will be mentioned here These

are:

iding suitable framing members around the cut-out

Providing 2 doubler or “bent" where framing as in (a} cannot de done

which have published 2110wab1as

cussed in chapter on beam

T

wy ou

AS an example assume that a deam wed

requires a cut-out as shown in Fig C3.20,

Fig D3 20

Before the cut-out was made the members shown

by solid lines (flanges, stiffeners, weds) are present The members "a" and "b" are added to

frame the cut-out, as shown by the broken

lines

There are 2 ways to determine the loads

in the area framed around the cut-out The first is to assume a shear flow equal and opposite to that present with no cut-out (qd,

in the figure above) and determine the

corresponding balancing loads in the framed area Adding this load system to the original one will give the final leads and, of course,

q = 0 in the cut-out panel The other method

is to use standard procedures assuming the

shear to be carried in reascnable proportions

on each side of the cut-out The first method will ce illustrated here

All shear flows are the edge members (on the

in this discussion

shown as they act on

flanges and stiffeners}

If there were no cut-out there would be 2 constant shear flow, dg, in all of the panels,

equal and opposite to that in the center panel

of Fig (a) 1s applied to the center panel of

Before Cut-Out

| tag IP ime lJ 7218501]

17/1686 {7/44 Ij'118qo- 1

Self- Balancing Internal Loads (Due

to Application of Equai and Opposite

do Assumed in the Cut-Out Panel)

(bì

P

°—y Ị l 9/16 gọ VÌ 1-1728 =——— aot 9/18 a0 + =— ¬=^~—P `

Go} <2 se { °

~—_ IV 9/T82o 0127/86

Pa Final Shear Flow PL

Distribution, (a) + (b)

(c)

Fig D3 21

balancing lead system, no external reactions

outside of the framing areas are required

Tats is an important concept and the reader

These will add or subtract, depending upon their directions, to any loads

present before the cut-out was made

(as in the case of the beam flanges)

areas due to dg in (b) are next determined

To eliminate redundancies, it is usually

assumed that the seme shear flow exists in the

assumed that shear flows are the same in the

panels to the left and right of the cut-out

a}

s)

the shear flow in the panels above

and below the center panel must

statically balance the force due to

do, oF

Since IF, = 0, 40x%7 = 4x (5+3)

q = 7/8 I

the shear flows in the panels to the

left and right of the center panel

must also statically balance the

force due to qq

Since IFy = 0, 49x12 = ax (12+ 12)

q= 1⁄2 qo

the shear flows in the corner panels

must also balance the force due to

the shear flow in the (any) panel

between them Considering the panels

in the right hand bay

BF, = 0; 1/2 dg X7 = 9x (5+3)

1

2 3X”?

4 s“Cg—° Tạ % the final shear flows are gotten by

adding the values in (a) and (b)

(the cut-out) 15 q = 49 ~ dg =O,

as it should be

2 the shear flows above and below and to the left and rignt of the cut-out add, giving a number

greater than the original qo

4 the shear flows in the corner panels are smaller than the original value of do

This is the way the changes always

occur in the area framed about a cut-

out

Finally, and importantly, there are

axial loads developed in all of the

framing members due to the cut-out

The axial loads due to the

can be gotten from Fig (b)

total axial loads in all of the members

These are can be gotten from (c)

illustrated in Fig 03.22

cut-out

P<P, -L2"x1g4o 7 P^P, +12"xïg đọ

—

Pa giên l1-8q l9 l6qg `

Axial Load Distribution in Upper Flange from Fig D3 21¢

fa)

Pa 2X Pxà 2 (1-1/2 9 -

Axial Load Distribution in Framing Member Above Cut-Out Obtained from

D3.21e (Same Result Could be Gotten from Fig D3 21b)

(b)

9/1640 | [1-1/8

PL =8"x(1-1/8qg~ 9/16 a9)

3/46 qu|Ù , „ „z4 SoŸ`1-1/8qo

Axtal Load Distribution in Stiffener Bordering Cut-Out on Left Side, from Fig D3 2ic (Same Result Obtained from D3, 21b}

Fig D3 22 Once the internal loads are known, the members can be checked for strength

Tne cut-out could have been framed without extending the framing members into

case is tllustrated in Fig 03.23

Had the 7" deep cut-out veen required at the 5ottom of bay, the framing could have

preceding cases a1s0) as illustrated in Flg

p3.24 This represents the minimum

adequate framing for any cut-out 7

there must be a minimum of one redis

of nat is, tribution

bay on one side of the cut-out and at least

‘two redistribution bays on the other side,

ane there must be the framing members de- fining the days

Note that in the previous examples in

Case (b) the sum of the loads on all edge members (framing members) is Zero

loads are needed for equilicrium 7

No external nis is

ANALYSIS

Pa ty it Joe it aio 1 | PL in Fig DS.25

ao HR do dt G0, il đọ I 20 — = 1 Ps

2 = II

' \A =) | yuggh © %

fF Teas Pea I 0 | v

Fig D3, 25

P, ° \ = == = PL A8 shown in (a) the doubler whose thickness ts

keep stresses due to curved beam bending

(c} Final Distribution

Fig D3 24

always the case when a set of self-balancing

shear flows are applied to a flat panel

structure or to 4 3 dimensional box structure

with 2 cut-out on any Side The reader should

study the examples closely Although the

method is shown only for a ‘let beam it is

also applicable to any structure with a cut-out,

has actually been illustrated in Solution No &

of that article and the reader should review

it at this time

Sometimes framing 7 ars for a cut-out

are not conveniently available as were the

stiffeners and flanges of the beam used in the

previous examples In such cases they must,

of course, be provided

(0) Framing Cut-Outs with Doublers or Bents

Frequently a cut-out in the web of 4 beam must be so deep that it removes nearly all of

she web In this case the method previously

forces” approach is necessary and a heavy

doubler, or bent, is provided around the

cut-out to carry the shear This is illustrated

flow in the web

The

Strictly speaking, the doubler should be

analyzed as a frame With reasonable symmetry

the loading in (c} can be assumed at the

the total shear, q, x h/2 is resisted in the

top of the frame, one half in the bottom and

a pin joint (no bending moment) exists at the

cut The bending moment axial loads and shears

at any section of the frame follow as a matter

of statics For example,

At A-A,

(there may also be a little relieving motient due to do)

Fy = do X Z The thickness of doubler required to take the

loads can thus be determined using standard methods of stress analysis Tne doubler

should have sufficient out of plane stiffness,

also, to provide simple support for the beam

web, as discussed in Chapter ClO This will

normally be provided oy the thickness required for strength purposes

Sometimes the nature of the cut-out is such that the frame (doubler) can be deeper

In such a case, the

at the top (or bottom)