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Stresses and Strains

Kamyar M Davoudi's picture

In classical elasticity, we know that at the interface of two different materials, traction stresses and non-traction strains are cotinuous. Traction stresss are continuous according to Newton's third law, but why non-traction strains are continuous?

Ying Li's picture

Dear Kamyar,

The non-traction strains are continuous because the deformation of the structures could be compatible. That is to say, the dispalcement at the interface should be continuous.

Ying Li Department of Engineering Mechanics Tsinghua University Beijing, 100084, P. R. CHINA

What are non-traction strains?

Kamyar M Davoudi's picture

Dear Ying

 Thank you for your answer. Yes, the displacement should be continuous at the interface, but why some components of its partial derivative are continuous and some of them are discontinuous? Please note that the traction strains (I mean the strains related to traction stresses) are discontious. Suppose we have an antiplane problem and an inclusion (r<R) embeded in an infinite medium. In this case \sigma_{zr}^{(1)}=\sigma_{zr}^{(2)}  on r=R but \epsilon_{zr}^{(1)} != \epsilon_{zr}^{(2)} while \epsilon_{z\theta}^{(1)}=\epsilon_{z\theta}^{(2)} at the interface.

 Could you please explain more?

Wei Hong's picture

Hi Kamyar,

Partial derivatives are nothing but infinitesimal changes of a function along a particular direction.  Since the displacement is continuous across the interface, the change of the function as you move within the plane of interface is also uniquely defined, i.e. continuous across the interface.  However, when you move away from the interface, the change need not to be unique.  Therefore, the partial derivatives on the tangent plane of the interface are continuous, while those on a direction out of the tangent planes need not be continuous.

Wei

Ying Li's picture

Dear Kamyar,

You know, the strain is continuous at interface, which is basically determined by the compatible of the displacement at the interface. Here, you problem is out. For the Cartesian Coordinate (like X and Y), the displacement at the interface should be compatible for u^1=u^2 and v^1=v^2. As the epsilonx=d(u)/d(x) and epsilony=d(v)/d(y), it obviously that the epsilonx^1= epsilonx^2 and epsilony^1= epsilony^2. However, it dose not exist in the polar coordination. As the epsilon r=d(u)/d(r), but epsilon theta = (d(v)/d(theta))/r + u/r, I think the epsilon theta^1= epsilon theta^2. You could get it by yourself.

I hope it could help you.

Ying Li Department of Engineering Mechanics Tsinghua University Beijing, 100084, P. R. CHINA

Nanshu Lu's picture

Dear Kamyar and Ying,

You have brought up a very interesting problem and I would like to join the discussion here.

First let me rephrase Kamyar's problem. You are talking about a cylindrical inclusion with radius R embeded in a matrix whose outer radius goes to infinite. In the following we make use of the conventional cylindrical coordinate to discuss this problem. To distinguish the inclusion and the matrix we use "1" to denote the inclusion and "2" to denote the matrix. By "antiplane problem" you mean we apply shear stresses σzr at the ends of this composite. Maybe it's easier to imagine if the matrix outer radius is finite (set to be Ro) because we also need to apply shear stress at the outer circumference surface of the matrix, pointing to z direction, i.e. the other pair of σzr.

Kamyar's question is that on the inclusion/matrix interface, i.e. at r=R, why εzr1 ≠ εzr2 but εzθ1 = εzθ2?

First, as pointed out by Ying, we all agree that at r=R, u1=u2, v1=v2, w1=w2.

Then let's list all the strain-displacment equations in cylindrical coordinates. We keep consistent with Kamyar's division of strains:

Traction strains (If we make an imaginary cut at the interface, those are the strains that will be exposed. Because the interface normal is in r direction, all strains of this kind should include an r in their subscripts):

εr=∂u/∂r

εzr=[∂w/∂r+∂u/∂z]/2

εrθ=[(1/r)∂u/∂θ+∂v/∂r-v/r]/2

Non-traction strains (all the rest strains)

εθ=(∂v/∂θ+u)/r

εz=∂w/∂z

εθz=[∂v/∂z+(1/r)∂w/∂θ]/2

Looking carefully, we can discover that all traction strains involve a differential term with respect to r while all non-traction terms only include derivatives with respect to z or θ. Why does this matter? Well, remember the difference between strain and displacement is an integral over a length. While the integral limits in z or θ directions are consistent for both the inclusion and the matrix, the starting and ending limits of the r-direction integral is very different. For the cylindrical inclusion, to find the interface displacements we need to integrate the traction strains over [0, R], but [R,Ro] for the matrix. Therefore, because all the displacements pairs are equal at the interface, any strain involving derivatives with respect to r cannot be equal at the interface; but any strain doesn't involve d(displacement)/dr should be equal at r=R.

A final notice is that we need to be careful when we try to generalize the conclusions from this problem. As you know, the strict boundary conditions at interfaces should be tractions equal and displacements equal. Regarding to the strains, we really need to examine the strain-displacment equations along with the problem geometry. Therefore, it is not always true that non-traction strains are cotinuous at the interface.

 

 

 

 

 

Arash_Yavari's picture

Dear Kamyar:

First of all, I don't know what you mean exactly by "Traction stresss" here. Let me assume you mean traction.

I think the best way of looking at problems like this is to think of
balance laws on measure-zero sets (in 3D a surface, for example) and
kinematics of deformation and not to make any general rules.

An interface means a surface on which deformation gradient (or in the
linearized approximation, strain tensor) has a jump discontinuity. But
then the question is weather deformation gradient can have an arbitrary
jump. The answer is no and the jump in deformation gradient has to
satisfy the Hadamard's compatibility equations, i.e. [|F|]= a tensor
product N, where N is unit normal of the undeformed interface at a
given point X on the undeformed interface and "a" is an arbitrary
vector. If you assume that there are no defects like a crack, etc. that
would mean that deformation mapping is single-valued or displacement
field (in the linearized approximation) is continuous. F maps a tangent
vector in the undeformed configuration to a deformed tangent vector
(you could think of these tangent vectors as line elements with
infinitesimal lengths). Given a tangent vector in the undeformed
interface, both F^- and F^+ (deformation gradients on both sides of the
interface) must map it to the same deformed tangent map otherwise the
deformation mapping cannot be single-valued (slip). This would then
give Hadamard's compatibility conditions. In the linearized theory, you
know that displacements are different on both sides of the interface
but displacement field is continuous on the interface (displacement
jump is the zero vector) and hence you can conclude that partial
derivatives of displacements taken in the interface should agree too,
i.e. those components of strain that do not involve partial derivatives
with respect to normal to the interface should be continuous. Other
components are, in general, discontinuous.

Continuity of tractions results from localization of balance of linear
momentum on the interface in the case of stationary interfaces (I don't
like the term "Newton's third law"). Of course, when an interface is
dynamically moving, e.g. a shock wave or a failure front, traction is
not continuous. Instead, jump in traction is proportional to the normal
velocity of the interface.

Arash

In joint material interface between elastic and elastic-plastic material, how can we show the continuity of displacement equation theoretically?It must be continuous and compatible of deformation in practical case.Could anyone help me to understand this answer?

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