/sci/ - Science & Math

[Anonymous No.5286899]

How would you work out the force exerted on a load bearing cylinder?

The cylinder in question is pic related. It has a diameter of 48.3mm and a thickness of 3mm. The cylinder is made of steel.

I need to know what kind of weight I can put on it so it becomes inelastic i.e it bends so much it doens't spring back to shape. The pole is fixed between two other poles, separated by a distance of 0.6m (so the length of the pole = 0.6m)

How do I go about doing that, it's been a while since I did Physics! This isn't homework I legitimately would like to calculate this for a DIY building project i'm doing.

Thanks guys.

[Anonymous No.5286937]

>please respons

[OP No.5286945]

>>5286899
Timoshenko

[Anonymous No.5286962]

First you need to the elastic modulus and moment of inertia. Calculate the bending moment that causes it to yield for a simply supported beam. Blah blah I'm too lazy.

http://en.wikipedia.org/wiki/Bending

[Anonymous No.5286969]

Dude, return all that stolen construction scaffolding right now, and go buy a bench.

[Anonymous No.5286978]

>>5286899
What kind of load?

[OP No.5287007]

This may help you

[Anonymous No.5287013]

This may help you.

[Anonymous No.5287019]

>>5286969

lol, you again

[Anonymous No.5287039]

>>5286962

Holy shit m8, dem equations.

So stress = Fd/y, where Fd is the moment.

This doesn't account for my dimensions or my Youngs modulus of steel.

[Anonymous No.5287046]

tthis looks likelier amirite ?

[Anonymous No.5287050]

>>5286899
>How do I go about doing that, it's been a while since I did Physics! This isn't homework I legitimately would like to calculate this for a DIY building project i'm doing.


Bullllllllllllllllllllllllllllllllllllllshhhhhhhhhhhhhhiiiiiiiiittttttttttttttttttt

[Anonymous No.5287059]

>>5287046
That's the deflection of the beam. You need the maximum bending stress, which you can get with Navier

[Anonymous No.5287056]

>>5287050

Whatever, faggot. IDC what you think. Talk to the hand cos the face ain't listenin.

Yeah you heard that right.

[Anonymous No.5287060]

<span class="quote deadlink">>>5287013[/spoiler]

thank you mate, thanks a lot. let me try and work it out now ....

[Anonymous No.5287066]

>>5286899

Why bother putting so much effort into trying to disguise a homework thread?

You are a horrible person. Seriously. Why not put the same effort into actually solving your problem?

The very least you could do is admit to all of us you're a shit that just didn't bother studying anything then show us your working out and the exact point you're stuck.

FINE

You're stuck at the very very start. OK. The fact you tried to deceive me has upset me so you're not getting any help.

[Anonymous No.5287076]

>>5287066

it's not homework made, look I built this home gym contraption, and I am curious to know how strong the safety bars are (the ones below the barbell).

>please halp

[Anonymous No.5287081]

<span class="quote deadlink">>>5287013[/spoiler]

sorry mate i'm a bit confused.

max stress = Mmax (?)/YM x b/2. What is b and Mmax?

[Anonymous No.5287091]

I found the Young's Modulus of steel, it's 200GPa

so 200GPA = stress/strain.

I remember that much from school.

[Anonymous No.5287095]

>>5287081

First replyer again. I'll try to help a bit now. You need to know the strength of your material.

My = S × σy This is the max moment your tube can take.

Then you calculate your exerted moment on it so it doesn't exceed your moment strength which causes it to yield.

[Anonymous No.5287099]

>>5287095

Mexerted <= Myield. Something like that.

[Anonymous No.5287100]

>>5287076
>Squat Rack
Better than cleaning the bar over your head for each set I suppose

[Anonymous No.5287101]

>>5287099

Forgot the link: http://en.wikipedia.org/wiki/Section_modulus

[Anonymous No.5287111]

>>5287095

Just to be clear. My =Moment= Force x Distance

S is pic related

and thetaY is the yield stress (of steel), which is given?

[Anonymous No.5287117]

>>5287111

so where force is the only unknown in this equation, .. where FD matches S x thetaY is my limit?

[Anonymous No.5287121]

>>5287076
Dude, even if you could figure out the meth, which you obviously can't, that's no guaranty that there aren't structural defects in those scaffolding section or clamps, maybe they let you steal that stuff is because it was going out to the trash anyways.


Buy. A. Bench.

[Anonymous No.5287134]

>>5287133

Assuming you apply a load at the center.

[Anonymous No.5287133]

>>5287117

Yes. You don't want FD/2 to exceed S x SigmaY. D is the length of your tube and F is the point load you apply on it.

[Anonymous No.5287137]

>>5287121
Or use a massive safety factor

[Anonymous No.5287144]

>>5287134

Actually it's FD/4. I'm being dumb :3.

http://www.civilcraftstructures.com/wp-content/uploads/2010/03/ccs-beam-analysis-3.jpg

[Anonymous No.5287145]

>>5287121

>structural defects
>scaffolding

Any defects would be visible, like a dent or a not straight pole. I'm hardly going to use an ultrasound to check for defects. Let's be realistic here.

[Anonymous No.5287154]

>>5287144

so FD/4 = S x SigmaY, why is it 1/4 the moment?

thanks m8 i'm dumb

[Anonymous No.5287175]

alright I got dusted off my old fx-83MS and i'm teh ready to work it out

....

[Anonymous No.5287218]

alright I got F=3956

I changed all the mm to metres, and changed yield stress to N/m^2

also I used Fd/4

so if
F = 3956N
mg = 3956N
m = 403kg

that seems kinda low doesn't it?

am I just a dumbfuck? I was expecting something in the 000s range.

[Anonymous No.5287227]

do I use Yield Strength or Ultimate strength.

[Anonymous No.5287233]

I done goofe'd somewhere
i'll try again

[Anonymous No.5287235]

You use yield strength not tensile strength. I know in the US that the yield strength of steel is typically 50ksi, but that is for structural steel. Your tube could be crap steel for all I know.

[Anonymous No.5287250]

>>5287227
Yield, beyond that plastic deformation will occur. Ultimate strength is when necking occurs, avoid this

[Anonymous No.5288153]

>>5286899

Where is your free body diagram?

[Anonymous No.5288203]

draw a diagram how it'll be used with all the loads, etc. I'll may work out your problem.

[Anonymous No.5288230]

>>5288203
you fucking faggot, hurry up, I'm pumped for some calculation action!!!11!!

[Anonymous No.5288263]

Good morning /sci/

I worked it out once last night and got that ^^^ figure. I'm going to do it again now because pic related

[Anonymous No.5288272]

OK this time I got 1.25 million kg.

[Anonymous No.5288281]

>>5288272
you, faggot, post a diagram how it's going to be loaded and how it is supported!

[Anonymous No.5288286]

OK this time I did it online using this calculator

Entered the units (48mm OD, 3.2mm thickness, L=0.6m), gave me a section modulus 5cm^3.

S = 5 x 10^-6

SigmaY of steel is a little harder to calculate .. I got this http://www.o2scaffold.com/quality.html which gives scaffold steel a min. yield strength of 345MPA
or 345000000N/m^2

So FD/4 now = S x sigma Y
= 1725
FD = 6900
D = 0.6m so F = 11500
F = mg so m = 1173KG

that seems more likely, did I do well bros?

[Anonymous No.5288292]

>>5288286 that seems more likely, did I do well bros?

I HAVE NO IDEA BECAUSE YOU DON'T HAVE A DIAGRAM TO DESCRIBE WHAT THE FUCK YOU'RE DOING

1173kg is certainly more realistic than 1,000,000

But no, I can't imagine pipe that size surviving that much weight.

[Anonymous No.5288295]

>>5288292

here u go m8

[Anonymous No.5288308]

>>5288295
okay, now implying it's made of some crappy steel(which is very likely) whose yield strenght is 185MPa using a safety factor of 1,5 you could apply force of 3600N or 367 kg at the middle point.

[Anonymous No.5288310]

>"The cylinder in question is pic related."
>implying that pic is more helpful than a simple diagram

Muh sides.

[Anonymous No.5288312]

>>5288308

What did you get for the section modulus? I got 5 x 10 ^ -6 metre cubed.

[Anonymous No.5288314]

>>5288308
also I took outer diameter to be 48,3 and inner 42,3

[Anonymous No.5288318]

>>5288312
I got 4,5 * 10^-6 using >>5288314 dimensions

[Anonymous No.5288319]

>>5288310

>why can't I into reading comprehension

[Anonymous No.5288322]

>>5288318

OK that's close enough. So 4.5 x 10 ^ -6 x 185000000, is that right?

So your FD/4 = 832.5?

Where are ou accounting for your safety factor>

[Anonymous No.5288329]

>>5288322
the allowable stress will be 123 MPa, so the moment will be 540 Nm

[Anonymous No.5288328]

Also, if you look at this pic

>>5287076
>>5287076
>>5287076

you can see there are two bars (the ones directly below the barbell). Would it be correct to double the mass, since the load is spread over two bars?

[Anonymous No.5288334]

>>5288329

oh I see, now I see where you got 360kg from.

that is for a static load

how much difference would it be if it were dropped from a metre height, like a barbell coule be, in theory?

[Anonymous No.5288338]

>>5288334
how heavy a barbell?

[Anonymous No.5288340]

>>5288338

Let's say 200kg? or 300kg?

[Anonymous No.5288353]

>>5288340
well, impacts are quite a complex problems. But saying that it weighs 300kg it would have aproximetaly a 3000N force alone, plus the kinetic force from falling one meter. It would have a speed of 4,5 m/s at the point of impact.
Now, impacts have very short times of interaction. Let's say the impact is super long and lasts 10 miliseconds. And let's suppose that the barbell doesn't jump up after the impact, so the reaction force is smaller.
So the average reaction force would be about 135000 N. Pretty rad, dude.

[Anonymous No.5288363]

>>5288353

so you can add the kinetic energy (J) with the F (N)?

sorry m8 but how did you arrive at that figure

[Anonymous No.5288368]

>>5288353

For a 360kg weight at 4.42m/s I got 3516J


My brain isn't processing basic dynamics.

[Anonymous No.5288369]

>>5288363
no, kinetic energy is used to find the speed at the impact.
To find average reaction you need to know the time during which the barbell decelerates and accelerates (if we take that restitution is not zero). We imply that the movements are zero, so know having deceleration and barbell mass we find the inertial forces acting on the pipe.

[Anonymous No.5288374]

>>5288369
of course this is super approximate, because the pipe would deform, if it's material is in poor condition and it's brittle it could shatter. We would need to find the deformation at center point so we could find how much energy without plastic deformations it can absorb. Shit's complicated.

[Anonymous No.5288427]

>>5288369

Ok how do you go about doing that? I too got about 4.5m/s for a 1m high drop. Let's say the barbell weighs 300kg. What equation do you use to find force on bar?

300kg
4.5/ms

[Anonymous No.5288469]

>>5288427
Supposing that the impact lasts 0,01 second, you find the acceleration and using D'Alambert principle you can find what reaction force acts on pipe

[Anonymous No.5288506]

>>5288427
also, fyi bending is the least practical usage of material.

[Anonymous No.5289532]

>>5288506 bending is the least practical usage of material

What do you mean by this?

>>5288295
Where on a weights bench is this?

[Anonymous No.5289587]

Just put enough force on it to overcome it's .2% yield strength and then bam, guaranteed plastic deformation. You probably don't even need to apply as much as that since it's a tube.

[Anonymous No.5290507]

>>5289587

yield strength ia 250MPa

0.2% is 500000Pa

Pressure is Force/Area

lets say area is 2cm^3

so F = 500000 x (2x10)^-3

1000kg. Seems about right, is that correct?

[Anonymous No.5290511]

>>5289532
>>5287076

u can see it here. just the two bars at knee height

[Anonymous No.5290514]

I have two bars .. would I be correct in assuming both can take 2 x 1000kg, since the force is spread across two bars?