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Stress, Strain  Axial loading (axial forces) 
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#1
Aug1706, 05:34 PM

P: 69

Hello everyone
I study mechanics of materials, curently stress and strain. I don't understand how to figure out those axial forces in specific fields of mechanic construction. Picture 1  Construction 1: HERE Picture 2  Construction 2: HERE2 Picture 3  Construction 3: HERE3 I got those three examples in taskes but for now, I would like to learn a little bit more those axial forces. Construction 1: Here are axial forces from all four field clear to me  I understand them. I cut on four parts the construction from left to right side. N means axial forces but im not sure if the following statement is correct: '' N's are always orientated in the same way as i cut the construction ''. Construction 2: Everything what I learned about axial forces in Construction 1, dropped in the water and thats why I decided to make a post here. I don't understand axial forces in this example. Why is only important first and fourth field, why do I just overlook second and third? As you can see from the both field's picture  here aren't showed A forces which are supposted to be orientated in opposite way as N are, why do I just overlook A axial forces here? In the first field, N is orientated on the left side, which means (remember: first field contains force F!) cutting the construction is from left to right side, which means my statement in Construction 1 was wrong. Also, why do I count third and second field with first one? Because the stress '' pull '' the whole part of construction on the left side? Construction 3: This example was on my latest exam. I tried with setting both axial forces to zero (N1 = 0, N2 = 0) but this was wrong. How am I supposted to cut the construction here then? I would like to please if anyone could tell / learn me something more about axial forces after cutting the consturction (stress, strain) and how they are orientated. I am sorry because Im not really fluent in English language. Thank you 


#2
Aug2006, 08:15 AM

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Are the breaks between blocks actual or simply to show where the crosssection is cut.
What is meant by field? Does it mean 'plane' or 'crosssection'. Does N refer to axial force, which is 'normal' to the plane (field)? 


#3
Aug2006, 01:45 PM

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P: 2,279

What is Fields??
Well anyway the axial deformation or elongation in non prismatic beams or prismatic and/or axially loaded in mid points with or without different materials (must behomogenous, isotropic and linear elastic) can be determined by considering the internal axial forces (N). In the first example, i see 2 cantilever beams, one axially loaded, while the other isn't. One beam is prismatic and not axially loaded, and the other beam is nonprismatic and axially loaded in a mid point. In your first "Field" equation: you have the internal axial load equal to the A load, which is actually the reaction of the beam, and it is 0, because there are not external loads acting on our beam. It's good to notice when there's only one load on one of the extremes, you don't need to consider internal axial forces, because they all will be equal to the applied load. In your 2nd "Field" equation: you seem to be analyzing the nonprismatic beam, before the midpoint load F. Looks correct. In your 3rd and 4th "Field" equation, you have changed the direction of the F load, Why?. Your 2nd example looks correct, but rather confusing... What is this "Fields"? In your 3rd example you have two prismatic nonloaded beams, thereforce both their internal axial loads will be 0. 


#4
Aug2206, 02:25 AM

P: 69

Stress, Strain  Axial loading (axial forces)
When I said ''fields'', i meant ''parts'' or ''cuts'' of construction. For example, if you have construction / material and you would like to see (how many N's) the internal forces in specific location of those construction (no matter how big is it), you need to cut the construction, so you can get to the location where you want to see the internal force. If you would like to know the internal force exsactly in the middle of construction, you can't just check the force in the left or in the right part of construction but you need, but you need to '' come '' to the location of internal force. So when you cut the construction (how many parts of cutting will be there depends of where you want to determinate the internal force), you get different parts of it (it isn't required that those parts are the same size) which means different fields.
Astronuc: Yes, N means axial force. Unfortunately I haven't studied Aerospace (aircraft) engineering, so Im not sure if I can compare this construction with plane. Cyclovenom: When you mentoined first field  the force still moves the left part of construction, even if we cut it, so thats why I put that N1 equals to internal force A. In third and fourth one  I didn't change the direction of F force  it is still the same (same arrow in the construction and third, fourth field) but when I cut the part with force, the other parts that are behind the F, will still have this force of F. Similar task is HERE , where I have to determinate the stress (F / A) in copper part and the stress in steel part of construction. On this task you can see the pole which is ''clinged'' with the ceiling (not sure if this is really a ceiling but this doesn't have the important meaning while solving the task  i meant upper horizontal line). Before force F began to operate, on the downer part appeared the leak  air with nothing there which has 0.05mm of length. So we are searching the stresses in copper and steel fields that appears after the force F began to operate. I think here are reactions in both fields correct. Also deformity equation is correct (I checked the book) but the problem is  when I replaced in formula N (axial force) with the reaction which I got from cutting the fields, I made a mistake. In the similar task which I tried to solve with similar construction, I got correct result and I was solving it on the same way (replacing N with reactions from fields) as I did here now. May I also ask for the reason why I shouldn't replace the axial force in this example? The solution from the book is here: CLICK I don't understand second, third and 5th step. One MPa equals to One N/mm^2. If we try to compate this task with that one  in both, I change axial force (N) with reaction (internal force). If I resume solving out the first one on the way as I did the second one, the solution and result will be correct. So this is why I don't understand why I can't do the same in the second task (correct solution is the one from the book). 


#5
Aug2206, 01:11 PM

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P: 2,279

So that's what's fields. You know when i took Strength of Materials I, we called it sections.
I am confused, which is the problem you are solving? and what is it you want me to compare? on your other post you gave 3 different problems. Be more explicit, provide the problem statement, and i'll gladly help out. 


#6
Aug2206, 02:39 PM

P: 69

I wanted to show on examples too  what do I mean with '' fields ''. You don't need to actually compare anything if you know what do i mean with '' field '' now.
But I pleased you, if you could tell me, whats wrong here ? I also showed the correct solution in my previouns post (solution from the book). Its obvious where have I done the mistake.. I replaced the axis force (N) with the internal force (reaction from field). But I don't understand whats wrong with that? Why is this task so specific that I shouldn't do this if I can in similar taskes? I have been trying to understand the correctly solution but unfortunately without any luck, at least not yet. 


#7
Aug2206, 03:27 PM

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P: 2,279

Hey Andreii, the solution provided by your tutor is correct, except it has a minor mistake in step #2, instead of , it is a +.
Let me make it more clear for you: If you cut from bottom to up the copper part, you will have: [tex] N_{copper} = F + F_{steel} [/tex] and if you cut from the bottom to up the steel part, you will have: [tex] N_{steel} = F_{steel} [/tex] The fun part about this problem is that it is quite obvious you can't solve it throught equilibrium along, but through the constitutive relations and compatibility consideration. If we continue: [tex] \delta_{L} = \sum^{n}_{i=1} \frac{N_{i}L_{i}}{E_{i}A_{i}} [/tex] [tex] \delta_{L} = \frac{F_{steel} L_{steel}}{E_{steel}A_{steel}} + \frac{(F + F_{steel}) L_{copper}}{E_{copper}A_{copper}} [/tex] [tex] \delta_{L} = \frac{N_{steel} L_{steel}}{E_{steel}A_{steel}} + \frac{N_{copper} L_{copper}}{E_{copper}A_{copper}} [/tex] To rewrite it as your teacher did: Recall [itex] \sigma = \frac{F}{A} [/itex] [tex] \delta_{L} = \frac{\sigma_{steel} L_{steel}}{E_{steel}} + \frac{F + \sigma_{steel} A_{steel} L_{copper}}{E_{copper}A_{copper}} [/tex] solving for [itex] \sigma_{steel} [/itex] [tex] \delta_{L}  \frac{FL_{copper}}{A_{copper}E_{copper}} = \frac{\sigma_{steel}L_{steel}}{E_{steel}} + \frac{\sigma_{steel} A_{steel} L_{copper}}{E_{copper}A_{copper}} [/tex] and then it just follows from step 3 to step 5 from your teacher's solution. In regards to your work: You put a force A, which will be the axial reaction from the support in contrast to the force F, this A your teacher calls it the "axial force of copper". In my opinion the drawing it's not too clear without a problem statement. Your solution should be correct anyhow. Maybe check your numbers again. The algebra is correct, and check your teacher's numbers, too. just in case. 


#8
Aug2306, 01:53 PM

P: 69

Hi
Hmm I see, so if I cut from bottom to up, the internal force (reaction) won't be the same for the same axis force (N) as it would be, if I decided to cut from up to bottom. If there wasn't force F (on the picture), both axis forces would be the same. When I come from the field which has bigger width size to the field which has smaller width size (than the previouns one), do I have to put, because of restrict (less width size), minuses instead of pluses in deformity equation? If so, do I put them everywhere, including the part of equation with temperature? I think stress always make bigger size of construction, so I doubt if I change pluses with minuses but anyway.. I rather ask to be sure. And in what cases, do i replace the formula: [itex] \sigma = \frac{F}{A} [/itex] with minus before the fraction? Only if I determinate the stress from opposite way as i used to cut with no matter if in the construction is leak or not? 


#9
Aug2406, 02:08 PM

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P: 2,279

[Note:I say for the loads because for the deformations, the sign convention is different] 


#10
Aug2506, 05:18 AM

P: 69




#11
Aug2506, 12:11 PM

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P: 2,279

I don't see any loads on any of the beams or any problem statement. However, you mentioned temperature. well when the beams or one of them dilates enough to contact the other, yes a compression stress will appear as long as the beam or beam in dilatation gets restricted.



#12
Aug2506, 04:03 PM

P: 69




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