Factor of Safety

This mechanics of materials tutorial goes over the factor of safety, an important topic in mechanics of materials and structural analysis / design.
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Thanks for watching, I hope it helps!

[@georgemorley1029]

When the o ring seals on the booster rockets for the challenger space shuttle eroded by 30% when under hot gas pressure load, they were deemed by the management (not the engineers) to have a safety factor of 3. The way they figured it, because the o rings had only burned a third of the way through, the o rings were three times stronger or thicker than they needed to be to prevent the blow through of hot gasses into the booster rocket’s fuel system. This is of course absolute horse shit. The o rings had a safety factor of zero, as by eroding up to 30% under normal load, they were reaching ultimate load immediately and were therefore immediately and irreparably damaged beyond their intended operating strength. If they had been able to withstand three times the normal gas pressure of rocket operation and NOT erode at all, then they could have claimed a safety factor of 3 or more. They completely misunderstood the principle. Now, the method of operation on challenger was to throttle up once, get the throttles back as the shuttle and rockets punched through the weather, then rotate and go maximum thrust a second time to accelerate hard once clear of the turbulence the weather caused. That means that the 30% erosion took place over two distinct burn periods where hot gases were pushing at these o-rings for all they were worth. Only problem was, the shuttle had never been operated under such cold conditions as it was being told to by management on the day of the challenger launch. Therefore, when the rockets were operated under abnormally low ambient temperatures on the morning of the challenger launch and had CONTRACTED significantly already due to the cold, the subsequent erosion of the contracted, never before used, completely “safe by a factor of three”(!) o-rings under the hot gas pressure meant that the hot gases eroded the o-rings to such a degree on the first burn, that when they initiated the second max thrust burn, the hot gases escaped past the o-ring seals, contacted the fuel system of the booster rocket and blew the space shuttle up with the loss of everyone onboard. Safety factors! Important stuff!

[@Engineer4Free]

Preach.

[@Samcanplaymusic]

Your videos are legit as hell man

[@Engineer4Free]

Thanks bro, means a lot.

[@progaming-uh8rj]

Thanks alot mate this helped me out with my mechanical principles assignment

[@Engineer4Free]

Glad I could help! If you haven’t already, check out the full series at engineer4free.com/mechanics-of-materials and share it with some classmates 🙂

[@is-ig4zh]

Enjoy your presentation, sir! thank you

[@Engineer4Free]

Glad to hear it! You can find the whole playlist here: engineer4free.com/mechanics-of-materials =)

[@is-ig4zh]

@@Engineer4Free Alright!

[@philipkarlsson1296]

You could mention something about what normal FS value is under some real world problems. Just a thought. Preciate the videos🤜🤛

[@Engineer4Free]

Yeah true. Here's a good excerpt from the wikipedia page on Factor of Safety ( en.wikipedia.org/wiki/Factor_of_safety ) Buildings commonly use a factor of safety of 2.0 for each structural member. The value for buildings is relatively low because the loads are well understood and most structures are redundant. Pressure vessels use 3.5 to 4.0, automobiles use 3.0, and aircraft and spacecraft use 1.2 to 3.0 depending on the application and materials. Ductile, metallic materials tend to use the lower value while brittle materials use the higher values. The field of aerospace engineering uses generally lower design factors because the costs associated with structural weight are high (i.e. an aircraft with an overall safety factor of 5 would probably be too heavy to get off the ground). This low design factor is why aerospace parts and materials are subject to very stringent quality control and strict preventative maintenance schedules to help ensure reliability. A usually applied Safety Factor is 1.5, but for pressurized fuselage it is 2.0, and for main landing gear structures it is often 1.25.[11]" Also, you'll eventually learn about Limit State Design (en.wikipedia.org/wiki/Limit_state_design), which is a bit more complicated than what's going on in this video, but good to know about if you're thinking about the real world design. Thanks for the fist bump, good luck! 🤜🤛

[@kitchen_arm825]

thae echo in the intro got me

[@Engineer4Free]

😂 i forgot about that

[@famiyo]

Thanks for the video, isn't it ultimate stress 'To' is same as Tensile strength of the material, say 250MPa to 350MPa for mild steel?

[@Engineer4Free]

Not necessarily. Shear and Tension are two different things, make sure you are dealing with the appropriate value

[@LT72884]

you probably already caught this, as this video is 6 years old, but its not 44 N/mm^2. When you multiplied by a 1000, you converted it to meters, you just forgot to change the unit is all : )

[@alisadeghi4907]

thanks for saving me

[@Engineer4Free]

Happy to help! Make sure to check out the full Mechanics of Materials playlist here: kzbin.info/aero/PLOAuB8dR35oft2ZLc1sHseypNMAiG_TeJ =)

[@sasankdhara1903]

Why is P(all)=5kN? arent there two loads of 5kN acting on the system?

[@Engineer4Free]

One would be considered the applied load whereas the other would be the reaction. The system is in static equilibrium. So these basic introductory problems are always confusing because they usually don't draw out the entire structure, so you don't realize very easily that there is only on applied load

[@fionaragan9219]

Do you know why a safety factor is not a true margin of spare capacity?

[@Engineer4Free]

There are two different ways to design a structure, one is using the factor of safety method like in this video, and the other is called limit state design, or also load and resistance factor design. Typically, when you are learning this kind of stuff, you will start with factor of safety method, because it is pretty straightforward, but in practice, limit state design is actually what is used, it takes statistics and probability into consideration. If you've never heard of limit state design yet I recommend skimming the wikipedia articles for each method, and then you will have some more vocabulary for further research: en.wikipedia.org/wiki/Factor_of_safety - en.wikipedia.org/wiki/Limit_state_design

[@isabellacorelone3550]

Hi there I need 6 reasons why safety factors are not true margins of spare capacity. One is tensile test represents optimum conditions and one mode of failure (ductile or brittle) can you help me with the other 6 reasons @@Engineer4Free

[@lukes8846]

I have a problem that gives me the ultimate tensile strength. And the saftey factor. It was me to find the largest allowable tension. Is ultimate tensile strength=ultimate stress? Is largest allowable tension=allowable stress? Are they the same thing?

[@Engineer4Free]

Yeah if the units are in kPa (or generally force per unit area) then i believe that your question is referring to ultimate tensile strength as ultimate tensile stress. But If they’re asking for tension, then that is a unit of N (tension in general is a force, not a stress), so you’d just apply the safety factor to find the allowable stress (will be less than the ultimate), and then multiply by cross sectional area to change units of force per area to just just units of force. Without seeing the question,m I think that’s what they’re asking for.

[@lukesample3717]

@@Engineer4Free An 80-m-long wire of 5-mm diameter is made of a steel with E=200 GPa and an ultimate tensile strength of 400MPa. If a factor of safety of 3.2 is desired, determine (a) the largest allowable tension in the wire; (b) the corresponding elongation of the wire.

[@mohammadshidhani5540]

Thanks bru you are the best

[@Engineer4Free]

Thanks Mohammed!! =) =)

[@MM-np4md]

@2:20 Dc-10 cargo door and engine pylon has entered the chat 😅

[@aizawa2222]

hi, Is it possible that factor of safety is yield stress / working stress, not ulitimate tensile stress / working stress?

[@niftywalla]

kzbin.info/www/bejne/jGnOiWqPnseej9U For ductile material and brittle material Fs is different

[@_tim4776]

Hey ,Thanks man

[@Engineer4Free]

Your welcome! :)

[@lightningkhrizz1378]

May I ask why multiplying pi by 6mm turn out to be 113. 10? what I'm missing here?

[@lightningkhrizz1378]

first solve exponential 6mm^2. Thank you

[@Engineer4Free]

Glad you got it sorted =)

[@nijatahmadov3880]

pi * r2=3.14 * (6 mm)2=3,14 * 36 mm2 = 113.10

[@jcnotnot8120]

_how do you do, fellow kids?_ But in all seriousness, thanks for the videos man.