.Calculating the Moment of Turning Force for 25mm Square Tube

[metalworker]

It may sound like a silly question but I really need some help. Not with the maths, I know that's beyond me I'll plug the numbers into an online calculator when I know what it is is I should be doing and that's where I need you.

The set up:

A random length of 25mm square tube with a wall thickness of 1.5mm is held rigid at one end.

The question:

At what point along the piece of square tube will the average person weighing 70kg, be able to stand before the bar bends to the point of plastic deformation?

Do I have to specify a length to find out the moment of turning force in Newtons and then see if this is with the parameters of the material, or can I do a dircect calculation using the 70kg weight?

What stats do I need about the material (plastic modulus, inertia etc) and how do these stats relate to the question?

For your info I've stood on the length of tube and it's around 650mm

 

[xxChrisxx]

Cantilever beam.
You need to know the material data.

You need to know the flexural rigidity. Which is:
E - Youngs Modulus (material property)
I - 2nd moment area (treat a trube as an I beam)

From this you can find an equation for maximum stress on the beam (standard equations).

When max stress = yield stress. You will get plastic deformation.http://www.efunda.com/formulae/solid_mechanics/beams/casestudy_display.cfm?case=cantilever_endload

canilever beam with end load.

 

[metalworker]

Thanks xxChrisxx

I'll give it a try and post my progress if that's okay?

Got to go and help a student make a giant sewing needle, no physics there I can't handle.

All the best

 

[BTown]

I know you said no math, but it might help somebody else looking at this thread:

1. Find the material yield stress (google)
2. Set the material yield stress equal to the equation for bending stress, which is:
σ = (M*y)/I , where:

M = moment = force * distance
y = distance from center to edge of beam (the overall height of the tube divided by two, don't make it harder than it is)
I = 2nd moment of intertia (for a square tube) = [(height)^4 - (inner height)^4] / 12

3. So now you should have:

σ (material yield stress) = (F * x) * y / I

Solve for x, which is the distance at which the material will fail due to the persons weight.

EXAMPLE:
1. Let's just say the material is Aluminum 6061T6, which has a yield of 34,000 psi. This converts to 23.93 kg/mm^2

2. 23.93 kg/mm^2 = (70kg * x) * (25mm/2) / 13031 mm^4

3. Solving for x: x = 356.32 mm

So one of the key factors is what type of material it is made out of. From there you just need to google the yield stress of that material.

I hope this all makes sense... if not let me know.

 

[metalworker]

Okay BTown

a bit scary I wish this language were a familiar to me as to you guys. But I'm not a lost cause, I will try to employ these formula, it will give me a mathmatical pedestal to stand on a survey lesser mortals (currently me)

 

[metalworker]

Okay so here's my first stumbling block.

How can I find the 'moment' if I do not know the distance the the person will be standing at yet. Am I missing something?

 

[xxChrisxx]

Okay so here's my first stumbling block.

How can I find the 'moment' if I do not know the distance the the person will be standing at yet. Am I missing something?

In this case, M is the output of the formula.

σ = (M*y)/I
In this case we are finding a stress for a known moment.


(σ*I)/y = M
In this case, we are finding the moment for a known stress.
We know the stress we want (yield stress).
I is calculated.
Y is the distance from the neutral axis to the edge (half the height of the tube).

 

[BTown]

metalworker,
You're exactly right, you don't know the moment. However remember that moment is equal to force * distance, and you know force, it is the weight of the person.

When I solved for x (which is the max distance the person can walk out before the beam yields), I was breaking moment into its two components, force and distance.

So you'll notice in my final equation for x, which again is the max distance this 70kg person can walk out, there is no 'M' for moment, because it has been broken apart so you could solve for 'x'.

 

[metalworker]

Dear BTown

I can't believe it I really think I understand what you're talking about, you kept the two parts apart and left x unknown in the equation!

Now I need to read your post again and do some more thinking.

Off home now but will post again soon

Slowly a light is glimmering at the end of the tunnel, it could be a train hacking towards me though !

 

[BTown]

metalworker,
you got it man :) let me know if you have any more questions.

 

[metalworker]

I've got my figures together and wondered if you would be so kind as to check this equation, then hopefully help me work through solving it?

250N/mm2 (steel yeild strength) = (70kg * x)*(12.5mm /1.56cm4 (second moment area as stated in tube specs)

Do I have to convert the 1.56cm4 (second moment inertia) into mm?

Also, I'm assuming a couple fo things does * mean multiply and does / mean divide (have to be clear)

 

[xxChrisxx]

It always helps to keep units consistent, you should convert cm^4 to mm^4.
The answer will be a magnitude of 10,000 out if you don't.

Also the 70 kg mass should be Newtons.

 

[metalworker]

A magnitude incorrect of 10,000 is about right for me:

How does this look

250N/mm^2=(686N*x)*(12.5/15.6mm^4)

 

[xxChrisxx]

250N/mm^2=(686N*x)*(12.5/15.6mm^4)

1.56 cm^4 is 15600 mm^4

 

[metalworker]

Man I really thought I was getting somewhere!

 

[xxChrisxx]

You are, it's just a conversion thing, the formula you've written is right.

 

[metalworker]

Okay bear with me, the cogs are whirring

right I've pumped all of these figures into an online calculator and have got the staggering answer of x=1726649842565500

that seems like a pretty big number to me can you help me make more sense of it?

 

[xxChrisxx]

It makes no sense becuase the number is wrong :)

250N/mm^2=(686N*x)*(12.5/15.6mm^4)

250 = (12.5*686x)/15600
250*15600 = 12.5*686x
3900000/(12.5*686) = x
3900000/8575 = x
x= 454mm

 

[metalworker]

You make it look so easy. I wonder why the online calculator didn't work? I was looking for an idiot proof way of solving the equation and got an idiotic answer.

I think my basic maths is missing, I've never had much success with quadratic equations. I won't pester you any more, better I go away and learn some more maths.

Thanks for all you hard work and perserverance.

 

[metalworker]

I'm back

but this time I'm trying to understand more fully the equation.

If below is correct what is the term given to 'half the thickness of the tube? is it youngs modulus?

Yield strength = (moment of turning force) x (half the thickness of the tube ÷second moment of inertia)

 

[BTown]

I'm back

but this time I'm trying to understand more fully the equation.

If below is correct what is the term given to 'half the thickness of the tube? is it youngs modulus?

Yield strength = (moment of turning force) x (half the thickness of the tube ÷second moment of inertia)

Don't over-think it. If the square tube that you have has a total height of 2 inches, then the value you are looking for would be half of that, which is 1 inch.

Even if it wasn't a square tubing, but a rod for example (with a circle cross-section, as opposed to a square), it would still be the radius divided by two.

 

[metalworker]

Okay, thankyou

 

[xxChrisxx]

If you imagine a tube cantilevered on the left, a man standing on the right (so it's bending downwards).

The top of the tube wants to stretch, in tension.
The bottom of the tube wants to squish, in compression.

Convention states that tensile stresses are positive and compressive stresses are negative. When moving from a positive to a negative number, you must cross zero.
So in a tube in bending, the maximum positive stres at the top and maximum negative stress at the bottom, and there is an axis along the tube where there is no stress. This is called the neutral axis.

In that equation for bending stress, stress = My/I, the variable 'y' is 'distance from neutral axis'.
As we know that max stress is at the outer edge, and the neutral axis is in the middle. The value is 'tube height/2'

 

[metalworker]

Hi xxChrisxx

This makes perfect sense. Does it follow that if the tube (beam) is supported at both ends, that the maximum distance would be twice that of the canterlevered calc?

 

[xxChrisxx]

Hi xxChrisxx

This makes perfect sense. Does it follow that if the tube (beam) is supported at both ends, that the maximum distance would be twice that of the canterlevered calc?

The neutral axis is always the same. As determined by the cross section of the beam.

 

[metalworker]

Dear BTown and xxChrisxx

I've been working on a document (with your help) that I've uploaded to a blog called process arts (a project run by the university of the arts London) my employer.

The document is a toe dip for me into creative commons licensing in that I want to encourage readers to refine, alter the document as they feel necessary.

Becuase you were mentioned I thought you might like to see what I produced. This isn't a trfaffic steering post and if innapropriate I aopologise. However, in the spirit of shared knowledge on the internet I hope you get a chance to take a look.

Also, the actual engine powering the blog is not so good so apologies for poor formatting etc. The developers at UAL are working on it.

All the best

Metalworker

 

[metalworker]

oops and here's the link

http://process.arts.ac.uk/content/calculation-point-which-piece-box-section-steel-will-bend-and-not-recover