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How do you calculate how much a tube will bend?
smart51 - 14/6/10 at 02:36 PM

I have a steel tube supported at both ends. I am placing a weight in the middle. How do I calculate how much it bends?

SeanStone - 14/6/10 at 02:38 PM

Errm something to do with second moment of area? I can't remember. Deflection of beams was a while back!

austin man - 14/6/10 at 02:46 PM

there will be many factors length diameter, wall thickness, load it will be carrying

http://www.engineersedge.com/beam_calc_menu.shtml

britishtrident - 14/6/10 at 04:56 PM

ISTR for simple supports with point load smack in the centre

(W*L^3) / (48*E*I)

Where
W = Load
L = Distance between supports

E = Youngs Modulus for Material

I = second moment of area of cross section of beam.



[Edited on 14/6/10 by britishtrident]

dinosaurjuice - 14/6/10 at 05:16 PM

your after 'beam deflection' equations.

see HERE

Humbug - 14/6/10 at 05:34 PM

Bend it until it folds, then back a bit

smart51 - 14/6/10 at 06:35 PM

Thank you. My top wishbones will extend inward to push down on inboard dampers. I wanted to know if I needed to reinforce them with a central member in pure compression. They will bend by 73um laden but stationary rising to 3.2mm if the whole vehicle is dropped onto the front wheels at 10g.

Now I need to know how many g they will see in operation. Whilst 0.073mm is nothing, 3mm sounds like a lot.

madteg - 14/6/10 at 07:24 PM

one and a half times the diameter

smart51 - 14/6/10 at 08:44 PM

quote:
Originally posted by madteg
one and a half times the diameter


I don't follow. 1.5 times the diameter of what is what?

Bob C - 14/6/10 at 09:43 PM

I always thought you were supposed to stress for 3G.
I think 10G sounds rather a lot - there will be failures all round the car at that level.

Dingz - 14/6/10 at 10:29 PM

quote:

I don't follow. 1.5 times the diameter of what is what?


Its the minimum radius you should bend tube, 1.5x tube dia, but it depends on other factors too, if it is supported on a mandrel etc.
Unfortunately its not the answer the questioner wanted tho'

86barettaguy - 21/7/10 at 08:27 PM

quote:
Originally posted by Bob C
I always thought you were supposed to stress for 3G.
I think 10G sounds rather a lot - there will be failures all round the car at that level.


I've seen people state everything from 3G to 5G for pothole incidents. Check with the guys over at eng-tips.com, they might have an answer. Or maybe not.

10G and the whole load on the front wheels seems very excessive.
I would start with 5G and consider the vehicle to not be accelerating or decelerating. I would also compare the results to those for a proven design (like that used by mr Champion for the book) to check whether or not they're reasonable.

Rocket_Rabbit - 19/11/10 at 02:21 AM

10G would mean a 1.5 ton corner loading.

Your alloy wheel will never take it, never mind your sus arm lugs!

Unless you are planning a Dukes of Hazzard photoshoot, your car will never see 10G unless it's a crash.

Ian-B - 19/11/10 at 07:20 AM

For info on suspension loads, I've run strain gauges on the race car, the highest loads observed are around 6G vertical, this was only achieved very rarely (in the region 10-20 occurances a season), generally only when the driver had placed the loaded wheel over a very significant kerb. I would expect that for normal road use the pot hole loads would typically be lower, as you are relatively unlikely be hitting a pot hole whilst on limit braking or cornering.
I would be more inerested in the maximium stress rather than deflection for abuse loads. I normally only bother with deflections at normal loads. I abuse load cases extra deflection can be good for reducing shock loads placed on components further up stream, although this is always likely to copromise goodperformance at normal loads.

Regards

Ian

Minicooper - 20/11/10 at 05:07 PM

Another question for you mechanical engineers out there if I have

80 x 40 x 2.64 rectangular box section this way up [ __ ]
30 o/d x 2.64 round tube 2 off side by side OO

Both 1400mm long supported at each end, point load in middle, same material mild steel

Which is stronger in deflection by what margin?

Hopefully people can work out what I'm after

Cheers
David

[Edited on 20/11/10 by Minicooper]

Liam - 20/11/10 at 07:12 PM

quote:
Originally posted by smart51
Thank you. My top wishbones will extend inward to push down on inboard dampers. I wanted to know if I needed to reinforce them with a central member in pure compression. They will bend by 73um laden but stationary rising to 3.2mm if the whole vehicle is dropped onto the front wheels at 10g.

Now I need to know how many g they will see in operation. Whilst 0.073mm is nothing, 3mm sounds like a lot.


What's your design? The inner end welded to the bush tubes sort of mirroring the outer wishbone? If so you should be more worried about stress on the welds than deflection. Also, ultimate load calcs are all well and good, but it's fatigue over time that will cause a failure. I went for an additional member as you describe - see photo archive. Need to think about the pivot point too as there's a huge load on there.

Horizenjob - 1/1/11 at 06:52 AM

quote:

but it's fatigue over time that will cause a failure.


The wonderment of steel is that that is not the practical case. It is the case for aluminum, wheels are counted as having a load cycle every revolution. So a race car's wheels can be quite light because they don't turn very many times. A street car must use a lower design limit because it rotates it's wheels so many times.

Consider the stresses in connecting rods or valve springs. They cycle their load many, many times. Now it is clearly possible to fatigue a piece of steel but I believe it to be relatively unique in being fatigue resistant. I am not a material specialist by any means, but I think if you stay under the yield limit the fatigue life is indefinite? Something to do with the carbon in the steel... I hope someone chimes in...

novacaine - 2/1/11 at 09:37 PM

you are right, so long as you stay under the fatigue limit for steel there will be no failure by fatigue (theoretically)

novacaine - 2/1/11 at 09:50 PM

quote:
Originally posted by Minicooper
Another question for you mechanical engineers out there if I have

80 x 40 x 2.64 rectangular box section this way up [ __ ]
30 o/d x 2.64 round tube 2 off side by side OO

Both 1400mm long supported at each end, point load in middle, same material mild steel

Which is stronger in deflection by what margin?

Hopefully people can work out what I'm after

Cheers
David

[Edited on 20/11/10 by Minicooper]


the round tubes will be stronger in the current loading conditions,

the 2nd moment of area for and individual tube is 1.96x10e-7 compared to the 1.66x10e-7 for the square tube

meaning if you use two round tubes under the same loading conditions it will be approx 2.5 times as strong as the rectangular beam

however

using the rectangular bar "portrait" (as opposed to the landscape orientation it currently is in) it would be significantly better than it currently is giving a 2nd moment of area of 5.0x10e7 meaning it would be stronger than the twin tube setup

the loading conditions dont matter so long as they are the same for both round and rectangular, the only important bit is the 2nd moment of area

hope this is of help to you

courtesy of a 2nd year mechanical engineering student, so no guarantees for its accuracy, even though i am sitting an exam on it a week tomorrow


********edit to correct an error in the calculations********

[Edited on 2/1/11 by novacaine]

indykid - 3/1/11 at 12:52 AM

quote:
Originally posted by novacaine
quote:
Originally posted by Minicooper
Another question for you mechanical engineers out there if I have

80 x 40 x 2.64 rectangular box section this way up [ __ ]
30 o/d x 2.64 round tube 2 off side by side OO

Both 1400mm long supported at each end, point load in middle, same material mild steel

Which is stronger in deflection by what margin?

Hopefully people can work out what I'm after

Cheers
David
[Edited on 20/11/10 by Minicooper]


the round tubes will be stronger in the current loading conditions,

the 2nd moment of area for and individual tube is 1.96x10e-7 compared to the 1.66x10e-7 for the square tube

meaning if you use two round tubes under the same loading conditions it will be approx 2.5 times as strong as the rectangular beam

however

using the rectangular bar "portrait" (as opposed to the landscape orientation it currently is in) it would be significantly better than it currently is giving a 2nd moment of area of 5.0x10e7 meaning it would be stronger than the twin tube setup

the loading conditions dont matter so long as they are the same for both round and rectangular, the only important bit is the 2nd moment of area

hope this is of help to you

courtesy of a 2nd year mechanical engineering student, so no guarantees for its accuracy, even though i am sitting an exam on it a week tomorrow


********edit to correct an error in the calculations********

[Edited on 2/1/11 by novacaine]

probably worth having a look back over your calculations for the round tube there. intuitively, the rectangular tube has far better material distribution for bending loads.

my calcs agree with your answer for the square, but for the round, i only get 2.14x10^-8, so 4.3x10^-8 for both tubes together, neglecting any welds. the rectangular tube wins by a factor of 3.87

what formula did you use? Ixx=pi(do^4-di^4)/64? less material in a less favourable distribution can't possibly yield a higher second moment of area.

courtesy of a 5th year MEng Motorsport student, but still, don't take my word for it

joking aside, good luck with the revision. i've got an FEA and vibration exam on the 21st. i hate exams!