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FtuX1.5 3
- Thread starter minny
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It means your supervisor is an idiot...You don't change the allowable for the sake of making your margin positive...that gets people killed.
I think we would probably need to know a bit more about the design to issue a reasonable opinion, but it looks like your supervisor just suggested that you change the mathematical operations of the FAA required 1.5 safety factor on ultimate strength. The allowable is defined in some situations as Ftu / 1.5, not Ftu x 1.5.
I think we would probably need to know a bit more about the design to issue a reasonable opinion, but it looks like your supervisor just suggested that you change the mathematical operations of the FAA required 1.5 safety factor on ultimate strength. The allowable is defined in some situations as Ftu / 1.5, not Ftu x 1.5.
do you know what a negative MS means ?
let's assume you know what a +ve MS means. by extension (and with some thought) you should be able to figure out that a zero MS means that the part is only just strong enough.
i have absolutly no idea why your "supervisor" would suggest increasing the strength of the material by 50%. of course, he may have said "change the material to one that is 50% stronger". ... which probably gives you a +ve MS.
let's assume you know what a +ve MS means. by extension (and with some thought) you should be able to figure out that a zero MS means that the part is only just strong enough.
i have absolutly no idea why your "supervisor" would suggest increasing the strength of the material by 50%. of course, he may have said "change the material to one that is 50% stronger". ... which probably gives you a +ve MS.
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First off, before calling anyone an "idiot", ask your supervisor why. This is what *YOU* need to do. I would recommend in the future you always challenge something that you don't understand or agree with. It helps in the learning process, and quite possibly catch an error in judgment.
Second, sometimes the 1.5Ftu is used as a quick check to account for plastic deformation of the material (in bending). However, there are "more complete" ways of doing this if you are up to the challenge: See Bruhn Chapter A13, Art. A13.10 "Bending Stresses of Homogeneous Beams Stressed above the Elastic Limit Stress Range."
Hope this helps.
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Nert
Second, sometimes the 1.5Ftu is used as a quick check to account for plastic deformation of the material (in bending). However, there are "more complete" ways of doing this if you are up to the challenge: See Bruhn Chapter A13, Art. A13.10 "Bending Stresses of Homogeneous Beams Stressed above the Elastic Limit Stress Range."
Hope this helps.
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Nert
minny
It is not clear from your post what kind of aircraft structure you refer to. Is it a commercial plane or a military use aircraft?
You should follow the specifications for the type of aircraft. For example: MIL-A-8860b - Airplane Strength and Rigidity, General Specification for specifies the following types of airplanes:
6.3.1.6 Model types of airplanes. Airplane model types are defined as follows:
VA - Attack airplane
Vc- Cargo airplane
VE - Electronic airplane
VF - Fighter airplane
VO - Observation airplane
VP - Patrol airplane
VR - Reconnaissance airplane
Vs- Antisubmarine airplane
VT - Trainer airplane
vu - Utility airplane
Vw- Heather airplane
In the same spec limit loads and ultimate loads are defined as follows:
3.1.2 Limit loads. The load factors and load formulas noted in any portion of this specification and the referenced specifications of section 2 represent limit loads, unless otherwise specified
3.1.3 Ultimate loads. Except for loading conditions for which specific ultimate loads are delineated, the ultimate loads are obtained by multiplying the limit loads by the ultimate factor of safety. Failure shall not occur at
the design ultimate load. The ultimate factor of safety to be used for the design of the structure shall be 1.50, except that in certain cases for considerations of added safety, rigidity, quallty assurance, and wear, additional strength or multiplying factors of safety are specified.
One more spec such as MIL-STD-8591 AIRBORNE STORES, SUSPENSION EQUIPMENT AND AIRCRAFT-STORE INTERFACE (CARRIAGE PHASE) gives the following:
5.1.1 Limit loads. The limits loads shall be the maximum and most critical combination of loads which can result from authorized ground and flight use of the air vehicle. This includes maintenance activity, system failures from which recovery is expected, a lifetime usage of the store, and all loads whose frequency of occurrence is greater than or equal to 1x10-7 occurrences per flight. All loads resulting from the requirements of this specification are limit loads unless otherwise specified.
5.1.2 Yield loads. Unless specific yield loads are delineated, yield loads shall be obtained herein by multiplying limit loads by 1.15, which is the yield factor of safety. The yield factor of safety is 1.0 for Army applications. The effects of deformation remaining after application and removal of yield loads shall be not greater than those prohibited in 5.2.
5.1.3 Ultimate loads. Except when specific ultimate loads are delineated, ultimate loads for suspension equipment or airborne stores while in the captive phase (store is within the sphere of influence of the aircraft) shall be obtained by multiplying the limit loads by 1.50, which is the ultimate factor of safety for the captive phase. The airborne store or associated suspension equipment shall not fail during application of ultimate loads. Failure includes unintended separation of the store from the suspension equipment, separation of any part of the store or
suspension equipment at ultimate or lower loads, or a material fracture of the store or suspension
equipment.
So as you can see the 1.5 factor means that under no condition the structure will fracture below 1.5 of the maximum expected load that can apply on the structure. The structure can be deep in the plastic zone but not fail and this is OK. However, at 1.15 of the limit load the deformations (even permanent) should not affect the functionality of the structure.
You should read and study carefully the specifications that are relevant to your structure.
It is not clear from your post what kind of aircraft structure you refer to. Is it a commercial plane or a military use aircraft?
You should follow the specifications for the type of aircraft. For example: MIL-A-8860b - Airplane Strength and Rigidity, General Specification for specifies the following types of airplanes:
6.3.1.6 Model types of airplanes. Airplane model types are defined as follows:
VA - Attack airplane
Vc- Cargo airplane
VE - Electronic airplane
VF - Fighter airplane
VO - Observation airplane
VP - Patrol airplane
VR - Reconnaissance airplane
Vs- Antisubmarine airplane
VT - Trainer airplane
vu - Utility airplane
Vw- Heather airplane
In the same spec limit loads and ultimate loads are defined as follows:
3.1.2 Limit loads. The load factors and load formulas noted in any portion of this specification and the referenced specifications of section 2 represent limit loads, unless otherwise specified
3.1.3 Ultimate loads. Except for loading conditions for which specific ultimate loads are delineated, the ultimate loads are obtained by multiplying the limit loads by the ultimate factor of safety. Failure shall not occur at
the design ultimate load. The ultimate factor of safety to be used for the design of the structure shall be 1.50, except that in certain cases for considerations of added safety, rigidity, quallty assurance, and wear, additional strength or multiplying factors of safety are specified.
One more spec such as MIL-STD-8591 AIRBORNE STORES, SUSPENSION EQUIPMENT AND AIRCRAFT-STORE INTERFACE (CARRIAGE PHASE) gives the following:
5.1.1 Limit loads. The limits loads shall be the maximum and most critical combination of loads which can result from authorized ground and flight use of the air vehicle. This includes maintenance activity, system failures from which recovery is expected, a lifetime usage of the store, and all loads whose frequency of occurrence is greater than or equal to 1x10-7 occurrences per flight. All loads resulting from the requirements of this specification are limit loads unless otherwise specified.
5.1.2 Yield loads. Unless specific yield loads are delineated, yield loads shall be obtained herein by multiplying limit loads by 1.15, which is the yield factor of safety. The yield factor of safety is 1.0 for Army applications. The effects of deformation remaining after application and removal of yield loads shall be not greater than those prohibited in 5.2.
5.1.3 Ultimate loads. Except when specific ultimate loads are delineated, ultimate loads for suspension equipment or airborne stores while in the captive phase (store is within the sphere of influence of the aircraft) shall be obtained by multiplying the limit loads by 1.50, which is the ultimate factor of safety for the captive phase. The airborne store or associated suspension equipment shall not fail during application of ultimate loads. Failure includes unintended separation of the store from the suspension equipment, separation of any part of the store or
suspension equipment at ultimate or lower loads, or a material fracture of the store or suspension
equipment.
So as you can see the 1.5 factor means that under no condition the structure will fracture below 1.5 of the maximum expected load that can apply on the structure. The structure can be deep in the plastic zone but not fail and this is OK. However, at 1.15 of the limit load the deformations (even permanent) should not affect the functionality of the structure.
You should read and study carefully the specifications that are relevant to your structure.
inertia4u is correct... it is simple factor for determining the plastic bending factor for a rectangle... you can find several references for it. Buhn is one of them. And yes, some explaination is needed in the analysis.
The FAA doesn't tell you how to do the analysis. They are concerned with finding compliance with the regulations.
The FAA doesn't tell you how to do the analysis. They are concerned with finding compliance with the regulations.
I've seen the 1.5 factor applied to both Ftu or Fty. It depends on how lofty of a goal you are trying to obtain.
Against Fty, you are calculating moment with yielding of the extreme fiber.
Against Ftu, are you calculating "Ultimate" moment that can be carried by the structure.
You *must* be aware of the implications of using such a factor on Ftu though, and use engineering judgment as to whether or not it is applicable.
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Nert
Against Fty, you are calculating moment with yielding of the extreme fiber.
Against Ftu, are you calculating "Ultimate" moment that can be carried by the structure.
You *must* be aware of the implications of using such a factor on Ftu though, and use engineering judgment as to whether or not it is applicable.
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Nert
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