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Home The objective About the Author Contacts Aircraft Structural Design

MAY 25, 2012

Aircraft requirements -The CS25.305 structural design requirement, what is the structural failure of an aircraft .

When you are developing an aircraft

structure you need to perform the

stress analysis regarding the loads vs

the static resistance of the material

used. You need to check if the

material and geometry used in the design as the adequate strength for

the job. But to develop the analysis

you need to know when the structure

as fail or what is considered a failure

of the aircraft structure. A ircraft

structure failure is specified in the

airworthiness requirements, check in the CS25 or FAR 25 for large aircraft's the CS25.305

or FAR25.305 requirement. What do we find in this requirement? The

CS25.305/FAR25.305 specify the criteria used to define the failure for aircraft structural

components. But, what is failure? When can we say that my aircraft structure as fa il?

The requirement CS25.305/FAR25.305 shows us several different types of fa ilures that

can occurs in the aircraft structure.

The first is of course, the simple one!

If the aircraft structure breaks, this is

failure. This is the obvious mode of

failure, the loss of structural efficiency

by the rupture or crack of my

structure. The structure will be no

longer capable to maintain the

applied loads causing the aircraft

crash. This type of failure is not

permitted up to Ultimate Load. What

is Ultimate Load? Ultimate Load is

the Umit Load multiplied by the factor

of safety. According to airworthiness

code CS25/FAR25 factor of safety

for large aircraft's is 1.5, in some specific structures or details additional factors of safety

are added to the general1 .5 value. Limit Load is the maximum load applied to the aircraft

during service life. Therefore, if we have maximum load applied to the Wing structure = 1000 lbs, Ultimate Load will be = 1.5 x 1000 lbs = 1500 lbs. The material/geometry used

must be capable to resist failure up to 1500 lbs applied on the wing as a static strength

point of view. According to the requirements the margin of safety must be higher than

CHAPTERS

Aircraft damage (3)

Aircraft design (5)

Aircraft hardware (6)

Aircraft materials (3)

Aircraft protection and finish (2)

Aircraft requirements (3)

Aircraft stress (1)

Aircraft structures (3)

Aircraft weight (1)

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2 of 7 7/27/2012 11:44 AM

zero, because the structure must be capable to resist u~imate load application for 3

seconds. If the authority as doubts about it, they can request a structural test to prove the

1.5 factor of safety and the 3 seconds rule. All aircraft wings do this test for certification

purposes, see picture of one of this test.

cs 25.305 Strength and deformat ion

(a) 'I'M ~~ muSl ~ ~bl~ to suppon wwt load; \\~!bout ~lrimenul ~ ~formation.

At :my load up to lumt l01.ds. the ~formation m:t)'

not IDt~~ wnb ~ c:peraucn

(b) 'I'M ~IU:-e mwl ~ .tbl~ to suppo11 ultunat~ loatb w•tbout liulu..--e !cr at lta~t > $~ODds.

The second type of failure, where in

the aircraft structural requirements is

considered also fa ilure, is that no

permanent deformation is permitted

up to Limit Load. Remember Limit

Loads? Maximum loads applied to

aircraft during service life. The

aircraft structures must always work

in the elastic domain of the material. Applying load, the Wing will deform, removing load

the wing will return to its initial state and form. No permanent deformation of the wing is

permitted. This means that the yield strength of the material used in a effective area must

always be higher than the Limit Loads applied to the aircraft structure. For the 1500 lbs

Ultimate Load wing, the wing will not present permanent deformation of the structure up to

1000 lbs applied maximum load= Limit Load. The material yield strength resists the 1000

lbs load application. The word "detrimental" in the airworthiness requirements means that

permanent deformation must occur in a effective area of the material. The stress

concentration factor and punctual yield deformation are not accounted for this type of

failure. Permanent deformation in concentrated points is permitted.

The third type of failure, no excessive

elastic deformation is permitted to

aircraft structure where the applied

deformation changes significantly the

purpose or function of the

structure.Yes, the deformation of the

structure inside the elastic region of

the material can be considered

failure? When? When the function of

the structure changes. For example, if

the elastic deformation of the wing

change significantly the aerodynamic

characteristics of the wing, this will be considered failure of the structure. Because the wing

will be unable to give to the aircraft the required lift for safe operation. In this particular

case the wing as not suffer rupture, no permanent plastic deformation, but the maximum

linear elastic deformation obtained is considered failure of the wing structure. This type of

failure occurs in the buckling of the upper panel of the wing. The buckling failure mode will

cause an elastic deformation of the upper panel of the wing, changing the airfo il geometry,

and lift and drag characteristics of the wing. This cannot occur, is not permitted in the

aircraft structure. We are given the wing as example but the same apply to fuselage skin

panel buckling failure, beam web buckling fa ilure, etc ... lmagine the aircraft passengers

looking by the window and see wing panels o r fuselage panels wrinkled and buckling in the

elastic domain of the material, fear, panic, ...

This are 3 failure criterion specified in

the airworthiness code C$25.305 or

FAR25.305 for static strength

analysis of aircraft structures, you

must always prove and demonstrate

this 3 requirements for the structure.

Remember: No rupture up to U~imate

Load, No permanent deformation up

to Limit Load, no elastic deformation

changing the performance and

characteristics of the structure.

Design always your structure for this

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MAY 16, 2012

Aircraft hardware- The hi-lok fastener system, the HL 18, HL 19, HL20, HL21 pins and the HL70 collar.

The hi-lok fasteners are the second most used

fasteners in structural assembly of aircraft's

after the solid rivets MS20470 and MS20426. Why? What are the characteristics that makes

this fastener the second choice after the solid

rivets?

Hi-lok fasteners are defined by a pin and a

co llar. The pin is similar to a screw and the

co llar to a nut. The assembly of the pin, and the

co llar makes the hi-lock fastener system . The

pin is like a screw but with additional advantages regarding a simple screw. The collar is not

a simple nut, but a special nut with unique properties. What are the reasons to use this type

of fasteners? The reasons are simple and can be summarize to the following properties:

1. Higher tension and shear strength, the material used and the geometry of the

fastener gives him the best properties regarding strength.

2. Lightweight fasteners, the hi-lok system as the best relation between weight and

strength regarding the assembly between two elements. Better that the use of screws

and nuts.

3. Easy and quick assembly, with the specific tools the assembly time is must lower

that the rivet or screws/nut assembly time. The assembly is very easy with no

complex steps and the final assembly is always perfect with no defects.

4. Good stress concentration factor, the specific geometry of the pin element gives to

the hole an increase in fatigue life, much better that if we use the screw/nut system.

5. Fastener cost, accessible and cost effective element in the installation on the

aircraft.

6. Correct torque application, the co llar property gives us the guaranty that the correct

torque is applied to the assembly, removing the problem of vibration, strength and

fatigue of the pin itself .

..._, .. ~"""" ....... ... , • • CG-I~rt,$1'-op· H."-~~~T..-.cn tU1.~ ... ~

..... HI~Ph ...... t-U)f('"Pt'l MlloG,"'*'4..0~ J¥ HA:s . .. ~1'n

These are the mains advantages of the hi-lok fasteners. These fasteners are property of

the hi-shear corporation, the company that develops this fastener. For that reason the

standard designation of these fasteners is defined by the HL letters. There exist NAS or

specific manufacturer specification that are equivalent to the HL designation. The hi-lok is

the same but the NAS system or the manufacturer company internal systems adopts

different numbering system to identify the hi-lok part number. Generally the P/N is defined

by the Hi-lok manufacturer company Part number.

The P/N is specified by the sequencing letters and numbers: HL 18PB-5-16. The HL letters

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4 of 7 7/27/2012 11:44 AM

define that we are specifying an Hi-lok

fastener, the number 18 specify the

geometry and material of the fastener,

the letter PB defines the finish

protection of the fastener, the -5

number defines the diameter of the

fastener per multiples of 32inchs (

diameter =5/32 =0, 156 in) and the -16

defines the grip length of the fastener

per multiples of 16 inchs (length =

16/16= 1 in).

This P/N is only to the pin, the collar

as a similar P/N. The most used co llar

is the HL70-5, the letter HL defines

the hi-lok system, the number 70

defines the material and geometry of

the collar and the -5 defines the

diameter in mu~iples of 32inch

(diameter=5/32=0, 156 in). If you need

to specify the hi-lok system in your assembly you must request the pin and the co llar P/N

and not only the pin P/N. Therefore HL 18 is pin and HL70 is a collar. geometry and

material is defined by the numbers 18 and 70. The quantity of fasteners is so great that we

can find numbers from 1 to 1000 in the material/geometry designation.

During the installation of this hi-lok fastener

system we need access to both sides of the

structure, we insert the pin in the hole. with or

without interference depending of the target of

the hi-lok ( shear or tension application), we apply

the collar to the pin and apply torque with the

required tool. We apply the torque to a specific

part of the collar, this part of the co llar will break

at the correct torque value. The permanent part

of the collar will continue with the pin in the

assembly. This part of the collar that breaks

permits that the correct torque be applied to the installation, not requiring additional

supervision to confirm the correct torque value.

There are several type of pins and several type

of collars, depending of the application, you must

select the correct one. The most used hi-loks are

the HL 18, HL 19, HL20 and HL21. We can divide

the hi-loks pins for tension and shear application.

The tension pins have greater heads than the

shear pins. After dividing the pins for tension and

shear they can be also divided in protruding head

or countersunk head pins. You must select the

tension pins when the main load applied is

tension and shear pins when the shear is the

dominant load. The HL 18 is the shear protruding pin, the HL20 is the tension protruding pin,

the HL 19 is the shear countersunk pin and the HL 21 is the countersunk pin.

Each normal hi-lok fasteners system diameter as two oversize fasteners that can replace

the first diameter in case of a repair or damage to the hole. That means that if you do

repairs to aircraft's and you need to replace a HL 18PB-6 pin because the hole as cracks or

is oversize. You do not need to select the next hi-lok pin diameter the HL 18PB-7. You must

use the 1st oversize of the HL 18PB-6, that presents a small increment in diameter just

used for hi-lok repairs. They are called the 1st and 2nd oversize hi-lok pins, defined by a

specific number after the HL letters.

In shear application select the diameter of the pin according to the standard procedure: the

upper higher shear strength of the bearing load failure of the plate. That means that in your


5 of 7 7/27/2012 11:44 AM

joint, the failure criterion used must be

the plate bearing failure. The hi-lok

must present a higher shear strength

than the bearing of the plate used. For

that use the closer diameter of the

fastener that guaranties the fai lure of

the assembly by bearing and not by

fastener shear.

In tension application use the hi-lok

with adequate tension ultimate

strength for the assembly.

http://www.hi-shear.com

• • A 8

lfastener_hl_stds.htm

http://www.gen-aircraft-hardware.com/template.asp?pagename=hiloks

http://www.gen-aircraft-hardware.com/images/pdf/hilok.pdf



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MAY 6, 2012

Aircraft materials - Selecting metallic material allowable for structural stress analysis with approved design data

When you are developing a structural

analysis, you have the requirement to

know the allowable of the metallic

material that you are using in the

design of the structural part. Certainly

you can't go to the Internet and find

the 2024T3 material allowable

properties and use this values to

design your aircraft structure. You

need a reliable and approved source

of metallic material properties. Then,

where can I find this reliable material

properties source? This information is

defined in the aircraft airworthiness certification requirements. Go to the CS25 or FAR25 in

the CS25.613 or FAR25.613 requirement paragraph and you can find the answer. The

material strength properties must be based on enough tests of material meeting approved

specifications to establish design values on a statistical basis. This is the requirement, you

need to make coupon testing regarding the material that you what to use based on

authority approved specification. But I want to use aluminium alloys, this is a material well

know by the aircraft industry, must I make tests? If you are from a small aircraft company is

difficult to make test just to use aluminium alloys in the aircraft. For that reason you must

use authority approved material data. According to CS25.613 or FAR25.613 approved

metallic material data is given by the MMPDS handbook or the ESDU00932. This two

references for material properties are the reliable and approved source for metall ic

material properties in aircraft application. If you work in a small company and don't have a

material properties handbook approved by the authority you must use MMPDS or

ESDU00932 handbooks. I generally use the MMPDS handbook. And What type of

properties can we find in the MMPDS handbook? The required mechanical properties to


6 of 7 7/27/2012 11:44 AM

do your analysis.

Material properties are presented in a table format. Each table are referring to the material

designation name. As example I will use the 2024 material table figure. In order to correctly

select the material properties you must select first the Manufacturing Specification, Form ,

Temper, thickness and data Statistic basis. Only after this selection you obtain the correct

allowable for the material that you are using in the aircraft. The Specification, Form ,

Temper and Thickness of the material is defined in the manufacturing drawing that you

produce as design engineer.

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As stress engineer you must find the manufactured part properties from the material. Now,

what type of statistical basis I must use? The table present the A basis and B basis. The A

basis defines that 99% of the samples fail at superior strength values with 95% of

confidence. The B basis defines that 95% of the samples fa il at superior strength values

with 95% of confidence. If you are developing a one loading path structural part (landing

gear) the A basis must be used. If you are developing a redundant structure (two or more

loading paths, stringers as example) the B basis can be used. If weight is not a problem ,

use the conservative value of the table: the lower value.

t sr dor~ctaon Other parameter that we can f ind in

the table is the applicable

LT d .roctt.on

grain direction. The L, LT and ST

designation define the properties of

the part according to the

manufacturing grain direction of the

material. If you check, you can find

that the grain direction can influence significantly the material properties. For yield stress,

the influence can go up to 5 Ksi of stress resistance. For that reason if you are not

controlling the grain direction in the installation process, be conservative and use the lower

value of all directions.

As presented is the table the direction

of the material grain will present

higher mechanical properties the L

direction. The perpendicular direction

of the grain as lower mechanical

properties, the LT direction. The

ST direction is the vertical axis of the

grain direction and for sheets and

plates this propertie are not required.

Remember selecting material allowable:

1. Go to MMPDS

LTd~reclloni

2. Select material designation chapter (2024)

·-LT direction

3. Select table for the material specification, form and temper

(AMS-00-A-250/4, Sheet, T3)

4. Select the thickness of the used (0.063 in)

5. Select the A orB column according to the airworthiness

STdirection -t ST d irection

E.lrtruslon Angle


7 of 7 7/27/2012 11:44 AM

requirements (use conservative, A basis)

6. Select the proper grain direction properties (use conservative,

lower of the two)

7. Obtain the mechanical properties of the material.

Hope this help to obtain approved metallic material data to your aircraft structural design. If

you work on a manufacturer design company just go to your company design manual and

select the materials allowable from there. The same way that from MM PDS handbook. Use

always approved design data for your analysis, otherwise EASA or FAA will reprove the

modification or repair design.



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