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Mechanics of Materials basic exercise

  • #1
Femme_physics
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Homework Statement



A graded steel cylinder AISI 1030 is connected to a wall and loaded with axial forces.

http://img542.imageshack.us/img542/3959/fbd1.jpg [Broken]


Given:

F = 30 [kN]
d1 = 35mm
d2 = 20mm
Elastic Model: E = 2x105 MPa
Safely coeffecient: 2
Yield strength limit: 440 MPa


Required:

1) Describe the course of the internal forces along the beam and the course of the efforts.
2) Check beam for strength
3) Calculate the total distortion of the beam


Homework Equations



Presumably
http://img831.imageshack.us/img831/533/neededformulas.jpg [Broken]

The Attempt at a Solution



Since I don't see a 3D drawing or references should I treat this problem like a 2D problem?


1) I presume they're asking for a free body diagram here:

http://img694.imageshack.us/img694/9629/freebodydiag.jpg [Broken]


2)
If I treat it like 2D:
Before I can use the formula: Sigma = F/A

I need to find A. So,
A = 1 x 0.020 + 2 x 0.035 = 0.09m2

sigma = F/A
sigma = 30+30(2.5)/0.09
sigma = 1166 N/mm2

3) http://img511.imageshack.us/img511/3645/deltal.jpg [Broken]


Is that correct?
 
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Answers and Replies

  • #2
SteamKing
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A = cross sectional area, not the area of the diagram

When you are calculating axial stress, the applied force is spread over the area which is normal (at right angles to) the direction of the force.
 
  • #3
Femme_physics
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A = cross sectional area, not the area of the diagram
Duly noted. Will correct it.

When you are calculating axial stress, the applied force is spread over the area which is normal (at right angles to) the direction of the force.
You're saying it should be like this?:
http://img10.imageshack.us/img10/7915/dongn.jpg [Broken]


I'm not sure how this makes sense.
 
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  • #4
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Hi Fp! :smile:


Since I don't see a 3D drawing or references should I treat this problem like a 2D problem?
It is a 3D problem.
It's a side view of 2 cylinders.


1) I presume they're asking for a free body diagram here:

http://img694.imageshack.us/img694/9629/freebodydiag.jpg [Broken]
Yes, I think so too.
But it puzzles me that they have an external force of F, and what appears to be an internal force of 2.5F.
If it is really an internal force it couldn't be that big, so I guess we'll have to assume it's an external force that compresses the leftmost cylinder.

Can you clarify that?


2)
If I treat it like 2D:
Before I can use the formula: Sigma = F/A

I need to find A. So,
A = 1 x 0.020 + 2 x 0.035 = 0.09m2

sigma = F/A
sigma = 30+30(2.5)/0.09
sigma = 1166 N/mm2
As SteamKing said, you're trying to calculate the tensile stress, which is the force per area of cross section.
We're talking about 2 cylinders here, with each their own cross section and tensile stress.
If a tensile stress is greater than the yield strength limit, the beam would break.

Do you know how to calculate the cross section of a cylinder?

3) http://img511.imageshack.us/img511/3645/deltal.jpg [Broken]

Is that correct?
I think you need to calculate 2 distortions, one for each cylinder.
And then you need to calculate the total distortion.
 
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  • #5
Femme_physics
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Hi ILS. Was really hoping you'd still be here. If legends are true at your help-giving I'm good hands :smile: :wink:

It is a 3D problem.
It's a side view of 2 cylinders.
I see.


Yes, I think so too.
But it puzzles me that they have an external force of F, and what appears to be an internal force of 2.5F.
If it is really an internal force it couldn't be that big, so I guess we'll have to assume it's an external force that compresses the leftmost cylinder.

Can you clarify that?
Hmm. Well, it's saturday and I have no contact to my teacher. What do you think it is likely to be? It sounds like what ur surmising it is.

Is my second Free Body Diagram (based on SteamKing's reply) correct?

As SteamKing said, you're trying to calculate the tensile stress, which is the force per area of cross section.
We're talking about 2 cylinders here, with each their own cross section and tensile stress.
If a tensile stress is greater than the yield strength limit, the beam would break.

Do you know how to calculate the cross section of a cylinder?
Yes, if I have internal diameter. I'm gonna ignore external diameter since it was not given to me.

http://img638.imageshack.us/img638/9237/crosssectarea.jpg [Broken]

So
sigma = F/A
sigma = 30+30(2.5)/0.03239
sigma = 3241.74 N/mm2


I think you need to calculate 2 distortions, one for each cylinder.
And then you need to calculate the total distortion.
Well, you're the master/chief/sensei, I trust what you think.

I think you need to calculate 2 distortions, one for each cylinder.
And then you need to calculate the total distortion.
So I used the correct formula? I wasn't sure.
 
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  • #6
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Hi ILS. Was really hoping you'd still be here. If legends are true at your help-giving I'm good hands :smile: :wink:.
Well I was wondering when or if you'd be back. :tongue:
Say, when the voting starts for special awards (soonish), will you vote for me? :wink:


Hmm. Well, it's saturday and I have no contact to my teacher. What do you think it is likely to be? It sounds like what ur surmising it is..
Since it's a basic exercise, I surmise that the left cylinder is compressed by 2.5F, and the right cylinder is expanded by F.
Otherwise it would really be too complex for a basic exercise in tensile stress.


Is my second Free Body Diagram (based on SteamKing's reply) correct?.
No. Your first FBD was correct.
I'm afraid your 2nd FBD makes no sense to you.
(Did it make sense to you? :rolleyes:)

Note that the forces you drew there extend over the entire cross section of the cylinder.


Yes, if I have internal diameter. I'm gonna ignore external diameter since it was not given to me.

http://img638.imageshack.us/img638/9237/crosssectarea.jpg [Broken]

So
sigma = F/A
sigma = 30+30(2.5)/0.03239
sigma = 3241.74 N/mm2
You have combined both surfaces into one total surface.
I'm afraid you're not supposed to do that.

Btw, there's no such thing as an internal and external diameter.
The cylinders are massive of uniform material.
(It's not a pipe - that's in fluid hydraulics! :wink:)

You need to calculate the tensile stress in the left cylinder, using its compressing force and its cross section.

And separately, you need to calculate the tensile stress in the right cylinder, using the force that expands it and its cross section.



Well, you're the master/chief/sensei, I trust what you think.


So I used the correct formula? I wasn't sure.
The formula for cross sectional area, which is the surface of a circle, is correct.
But you need to calculate and use only 1 surface at a time.
 
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  • #7
Femme_physics
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Well I was wondering when or if you'd be back.
Say, when the voting starts for special awards (soonish), will you vote for me?
Of course! What a question. Just link me in because I don't keep in touch of community activity.

Since it's a basic exercise, I surmise that the left cylinder is compressed by 2.5F, and the right cylinder is expanded by F.
Otherwise it would really be too complex for a basic exercise in tensile stress.
Makes sense. We're just starting, and this is as far as I know non-calculus based.


No. Your first FBD was correct.
I'm afraid your 2nd FBD makes no sense to you.
(Did it make sense to you? )

Note that the forces you drew there extend over the entire cross section of the cylinder.
I thought so! Glad you're here to set the record straight! :smile:

You need to calculate the tensile stress in the left cylinder, using its compressing force and its cross section.

And separately, you need to calculate the tensile stress in the right cylinder, using the force that expands it and its cross section.
http://img195.imageshack.us/img195/4654/penmark.jpg [Broken]

Is that it?

Do I keep the separate results for each one? Or do I combine them eventually?




Btw, there's no such thing as an internal and external diameter.
The cylinders are massive of uniform material.
(It's not a pipe - that's in fluid hydraulics! )
Oh! Thanks for helping me with the separation.

The formula for cross sectional area, which is the surface of a circle, is correct.
But you need to calculate and use only 1 surface at a time.
OK :smile:

http://img848.imageshack.us/img848/4015/penmark2.jpg [Broken]
 
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  • #8
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Of course! What a question. Just link me in because I don't keep in touch of community activity.
I'll get back to you on that! :smile:


Hmm, what happened to the square-symbol in the circle-surface formula?

Did you check your units?
I think there should be a couple more or less zeroes in a couple of places.

And I see you combined the forces for cylinder 2 by adding them, which looks good.
But then for cylinder 1 you should subtract them to get the resulting force.



Do I keep the separate results for each one? Or do I combine them eventually?
No, you do not combine the tensile stresses.
You're supposed to check each against the maximum stress the material can take.



I have a couple of suggestions for improvement.
Let me list them:

1. Check your units. I recommend including them when you substitute the numbers.

2. There should be a couple of parentheses in there.

3. The surfaces-areas need to be updated with the new numbers.

4. Cylinder 1 compresses, so it has a negative delta.

5. The problem asks for the total distortion.
 
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  • #9
Femme_physics
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Hmm, what happened to the square-symbol in the circle-surface formula?

Damn, I keep neglecting that. Even last year.

Did you check your units?
I think there should be a couple more or less zeroes in a couple of places.
Should I keep everything in mm? Since the final result should have the unit "mm". That's at least what the formulas say.

And I see you combined the forces for cylinder 2 by adding them, which looks good.
But then for cylinder 1 you should subtract them to get the resulting force.
Subtract what from what? F1 - F2? Or F2 - F1? Hmm, I fail to see the logic why.

No, you do not combine the tensile stresses.
You're supposed to check each against the maximum stress the material can take.
Ok, duly noted :smile:

I have a couple of suggestions for improvement.
Let me list them:

1. Check your units. I recommend including them when you substitute the numbers.

2. There should be a couple of parentheses in there.

3. The surfaces-areas need to be updated with the new numbers.

4. Cylinder 1 compresses, so it has a negative delta.

5. The problem asks for the total distortion.
1) same question as I asked above

2) Hmm, but the result turns out the same anyway, no?

3) Oh, yea.

4) Ah, will check!

5) Delta l1 + Delta l2?
 
  • #10
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Should I keep everything in mm? Since the final result should have the unit "mm". That's at least what the formulas say.
Yes, that would be fine.



Subtract what from what? F1 - F2? Or F2 - F1? Hmm, I fail to see the logic why.
Can you draw an FBD for cylinder 1?



Ok, duly noted :smile:
Well, does cylinder 1 break? Or cylinder 2?
That's what the problem asks.



1) same question as I asked above
Same answer as above. :wink:


2) Hmm, but the result turns out the same anyway, no?
Dunno, didn't check your calculations.


3) Oh, yea.
Yea, baby! :tongue:


4) Ah, will check!
5) Delta l1 + Delta l2?
Yep! :smile:
 
  • #11
Femme_physics
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Yes, that would be fine.
Got it. Force of habit I guess to convert mm to m, I guess.

http://img411.imageshack.us/img411/9569/fizor.jpg [Broken]



Can you draw an FBD for cylinder 1?
Was thinking to just do that for the FBD isolation:

http://img513.imageshack.us/img513/4356/f123.jpg [Broken]

Whereas F1 and F2 are the forces in the original diagram and F3 is the force applied by cylinder 2.

Well, does cylinder 1 break? Or cylinder 2?
That's what the problem asks.
Only calculations can tell me the answer, I'm still stuck at how to subtract the forces you talked about.

Yea, baby! :tongue:
Did you just Austin Powered me? :wink:
 
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  • #12
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Got it. Force of habit I guess to convert mm to m, I guess.

http://img411.imageshack.us/img411/9569/fizor.jpg [Broken]
Much better!



Was thinking to just do that for the FBD isolation:

http://img513.imageshack.us/img513/4356/f123.jpg [Broken]

Whereas F1 and F2 are the forces in the original diagram and F3 is the force applied by cylinder 2.
Hmm, I can't quite make it out.
What are the magnitudes supposed to be?
You used to be pretty good at drawing FBD's! :smile:

Is F1 the force 2.5F?
Shouldn't it be in the other direction then?

Is F2 the force F?
If so I suggest you draw it where it actually pushes against the cylinder.
And also that you draw it a bit shorter (by about a factor 2.5 ;)).

If F2 is the force from the original diagram, then that is also the force applied by cylinder 2.
So how can F3 be that force?

And what happened to the force that the wall applies?



Did you just Austin Powered me? :wink:
Yeah Baby.
 
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  • #13
Femme_physics
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Much better!
:biggrin:
Hmm, I can't quite make it out.
What are the magnitudes supposed to be?
You used to be pretty good at drawing FBD's!

Is F1 the force 2.5F?
Shouldn't it be in the other direction then?

Is F2 the force F?
If so I suggest you draw it where it actually pushes against the cylinder.
And also that you draw it a bit shorter (by about a factor 2.5 ;)).

If F2 is the force from the original diagram, then that is also the force applied by cylinder 2.
So how can F3 be that force?

And what happened to the force that the wall applies?
That's because I was drawing the other cylinder. Not the walled one.

This is the walled one
http://img442.imageshack.us/img442/6890/fixinginter.jpg [Broken]
 
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  • #14
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:biggrin:
Now that you did your dance of joy, could you switch around A1 and A2 please?
(Just noticed that. :blushing:)



That's because I was drawing the other cylinder. Not the walled one.

This is the walled one
http://img442.imageshack.us/img442/6890/fixinginter.jpg [Broken]
Don't worry about the internal forces - they're not supposed to be in an FBD anyway. :wink:

But what happened to the force that cylinder 2 exerts on cylinder 1?
They are glued together or something, aren't they?
 
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  • #15
Femme_physics
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But what happened to the force that cylinder 2 exerts on cylinder 1?
They are glued together or something, aren't they?
You're right, I don't know why my logic said not to include it. But, perfectly legit and correct t include it.

http://img573.imageshack.us/img573/8638/added.jpg [Broken]
 
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  • #16
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You're right, I don't know why my logic said not to include it. But, perfectly legit and correct t include it.

http://img573.imageshack.us/img573/8638/added.jpg [Broken]
What about the direction of F?
Shouldn't it go the other way?
 
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  • #17
Femme_physics
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Wait, something is quite odd to me. When I isolate an object, aren't I supposed to draw all the forces acting ON THE OBJECT, as opposed to the forces the object is enacting on other objects that are not in the drawing?

That's why I didn't include include a 3rd vector.
 
  • #18
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Yes. That is right.

But I do not understand your conclusion.

In total there are 3 external forces: F, 2.5F, and Nx.
F acts on cylinder 2 and as a reaction, cylinder 1 pulls cylinder 2 back to the left.
The reactive force between cylinder 1 and 2, is that cylinder 2 pulls cylinder 1 to the right.
 
  • #19
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Let's try again. :smile:

Let's look *only* at F.

In the total picture, the wall exerts an equal but opposite power.

In the 2 separate FBD pictures, these forces have to balance out as well.

attachment.php?attachmentid=40742&stc=1&d=1320662843.gif
 

Attachments

  • #20
Femme_physics
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You seemed to have ignored 2.5F, though, am I right?

Sorry it takes me a while to reply sometimes. Appreciate your help. :smile:
 
  • #21
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Yes, I ignored 2.5F. :redface:

Perhaps you can add it?
 
  • #22
Femme_physics
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  • #23
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That's the problem. In the first shot it's easy, I know where to add it. But in the separated FBD, do I add it to both?

http://img407.imageshack.us/img407/3423/sepreal.jpg [Broken]
The first shot is not in equilibrium any more!
Now it will accelerate into the wall! :surprised
Did you know that ƩFx=0?

Add it to both? Uhh, I dunno. :confused: What do you think?
Let's just try that! :cool:
 
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  • #24
Femme_physics
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The first shot is not in equilibrium any more!
Now it will accelerate into the wall!
And how do you know it without calculations?

Add it to both? Uhh, I dunno. What do you think?
Let's just try that!
Well, since it's an external force, it seems to make sense. Still not sure what's wrong with my first FBD, though.

http://img710.imageshack.us/img710/6841/hmmmz.jpg [Broken]
 
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  • #25
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You so serious.
Know a joke?
 

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