stress gradient


by jayanth nivas
Tags: gradient, stress
jayanth nivas
jayanth nivas is offline
#1
Sep1-13, 12:39 AM
P: 15
Hi all,
I have a very basic question regarding the topic of stress.While deriving the equations of equilibrium for a plane element,there is a concept called incremental stress.That is,if stress in the x direction for one face is σx the stress in the opposite face is σx + (δσx/δx)*dx. Where the partial derivative represents the rate of change of stress along the length of element and dx is the length of the element in x direction.

My question is, why is this considered in the first place ?

can anyone please explain this in the case where a bar is under uni axial tension ?

Does stress vary along the length of the bar at different cross sections?.I am having problems in understanding this topic and visualizing this.

I am learning this subject on my own.So I apologise if I have misstated something.

Thanks for going through the post.
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jayanth nivas
jayanth nivas is offline
#2
Sep3-13, 12:11 AM
P: 15
Is my question unclear or am i asking something that doesn't make any sense? .Please correct me if i'm missing something so that i can improve.

Thanks in Advance
timthereaper
timthereaper is offline
#3
Sep3-13, 08:08 AM
P: 341
I think I know what you're talking about. Derivations of these kinds of equilibrium equations are kind of tricky because we're considering differential stresses. Let's say we have a differential volume element of dimensions dx, dy, and dz, and we apply some stress gradient in the x-direction (δσx/δx). We are going to assume that this gradient in a first-order linear function in space, which isn't necessarily a bad assumption since we're considering a differential volume and it becomes more accurate as the volume in the limit approaches zero.

I think you might be getting hung up on the fact that we simply choose a face (in my courses in school, it was always the left face) to be the reference stress σx, which means that the stress on the other face would be σx + (δσx/δx)dx, which is the stress on the left face plus the linear function times the distance across the volume in the stress direction (the δσ the control volume sees from the stress gradient). We had to make a reference point somewhere, and the (left) face was just chosen to illustrate the point that we multiply the stress gradient function by the distance of the volume in that direction. This would work equally well if we had defined σx to be in the middle of the volume, which means on the (left) face we get σx - (δσx/δx)dx/2, and on the right face we get σx + (δσx/δx)dx/2.

For a uniaxial rod element, we only have stress being applied in one direction, so we have this kind of scenario. We can think of a differential volume element in the rod as being under this kind of stress, so this derivation is fitting. The stress in the bar will be different at different points in the cross section of the rod, but it won't matter since we've generalized things by using σx and δσ/δx in our equations to get the final equilibrium equation.

jayanth nivas
jayanth nivas is offline
#4
Sep13-13, 10:33 PM
P: 15

stress gradient


Thanks tim

Sorry for late reply.

I'm still hung up on this topic.(I'm learning calculus and I understand some basic concepts of it).Sorry for repeating the question again.

Still my question is if I take a rod under uni axial tension, will the stress be different for cross section cut at different lengths?

If not why do we consider this in derivation?.And if it has something to do with a differential element only can you please explain?

I actually couldn't follow the generalization part of the stresses.

Does this have anything to do with St.Venant's principal for stress distribution?.That is the stress distribution for elements cut at different proximity from the vicinity of load will be different?

Sorry to repeat the question but I'm still confused and I couldn't carry on with other topics until I understand this.

Thanks for going through the post and apologies if I have stated something wrongly.
jayanth nivas
jayanth nivas is offline
#5
Sep16-13, 01:37 AM
P: 15
Is my Question very elementary or unclear?I asked this question to my professor but the reply he gave was not very convincing.It was the reason I posted the question in this forum.I could understand something from the first reply,But still I'm not completely clear with this.So Please help me in understanding this.Should I attach any reference material or images for this?.Please let me know if such things are required.

Thanks in Advance.
timthereaper
timthereaper is offline
#6
Sep16-13, 08:13 AM
P: 341
Quote Quote by jayanth nivas View Post
I'm still hung up on this topic.(I'm learning calculus and I understand some basic concepts of it).Sorry for repeating the question again.
This is probably why you're getting hung up: you're studying calculus and also trying to derive complicated equations with differential volume elements. It's hard when you're learning 2 things at once. Keep trying to learn, but just expect to struggle more with it because you're trying to ride 2 horses at the same time. As well, this isn't a trivial problem, so there's a lot more to it than you might think.

Quote Quote by jayanth nivas View Post
Still my question is if I take a rod under uni axial tension, will the stress be different for cross section cut at different lengths?
Think about a rod in tension. You've got 2 forces at either end of the rod. If you cut the rod in the middle and did an FBD for one end, what would you get? Newton says you'd still have equal and opposite forces, so it's really the same problem with a shorter length. So, short answer, no.

Quote Quote by jayanth nivas View Post
If not why do we consider this in derivation?.And if it has something to do with a differential element only can you please explain?
You should realize that we're not talking about an actual rod in this derivation. They say "rod" because we're only considering stress in 1 dimension. Rod elements are 1D elements, so we're just saying we don't really care about other stresses for this derivation. It comes into play when we talk about σx + (δσx/δx)dx. We don't care about the y- or z-directions for a 1D element.

Quote Quote by jayanth nivas View Post
I actually couldn't follow the generalization part of the stresses.

Does this have anything to do with St.Venant's principal for stress distribution?.That is the stress distribution for elements cut at different proximity from the vicinity of load will be different?
No. This doesn't have to do with Saint-Venant's principle. All I meant with "generalization" is that this formula is completely independent of where the element is. Inside of an actual rod in uniaxial tension, you can pick *any* volume element and model it this way.

Quote Quote by jayanth nivas View Post
Thanks for going through the post and apologies if I have stated something wrongly.
It's good that you're sticking with this and asking intelligent questions. However, if you don't understand a lot of what I've mentioned in the first post, then you might be trying to run before you can walk.


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