# Boundary condition: null traction on the boundary of an elastic block

• bobinthebox
In summary: The author is trying to impose null traction on the curved boundaries. They do this by imposing a boundary condition where the stress at r1 is zero.
bobinthebox
Hi everyone,

I'm trying to understand the rationale behind the boundary condition for the problem "Finite bending of an incompressible elastic block". (See here from page 180).Here we have as Cauchy Stress tensor (see eq. (5.82)):

##T = - \pi I + \mu (\frac{l_0^2}{4 \bar{\theta}^2 r^2} e_r \otimes e_r + \frac{4 \bar{\theta}^2} r^2{l_0^2} e_{\theta} \otimes e_{\theta}- I)##

At page 183, in order to impose null traction on the curved boundaries, the author writes

##T_r(r_i) = T_{r_i + h}=0##

I'm having some troubles on how to interpret geometrically the first one of the two above conditions. I'll add a sketch hereafter.
My guess is that, since #T_r# is the normal stress (i.e. in #e_r# direction), I'd say that the direction of traction is the same in both conditions, and the only thing that changes is that the evaluation is performed at different radius. In my understanding, what the author wants to impose is that those two blue arrows have 0 length, i.e. the normal component of the stress is identically 0. Is this correct?

Is that right?

No. Null traction means that both the normal stress and the shear stress are zero. Also, the inner blue arrow should be flipped.

So the author doesn't impose explicitely ##T_{r \theta} = 0## because they're already zero, right?Another thing: I agree with you that the inner blue arrow should be flipped. But ##T_r(r_i)= e_r \cdot T(r_i) e_r## and I don't see how it can be flipped, from this computation.

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I've tried to solve by myself the second question. I'd be really happy to have your feedback about this. Let's say that I want to impose null normal stress at ##r=r_i##.

The normal versor there is ##-e_r##, so I should impose ##-e_r \cdot T(r_i)(-e_r)=0## because I want to know the projection in direction ##-e_r## of the stress with outward normal ##-e_r##. Now, by the Cauchy thm. we have ##T(-n)=-T(n)## and hence the condition reads ##-e_r \cdot -(T(r_i)e_r)=0## which is equivalent to ##T_r(r_i)=0## which is indeed what the author asks. Is my argument correct?

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@Chestermiller Did I wrote my last argument too badly to be answered? Let me know if I was not clear enough

bobinthebox said:
So the author doesn't impose explicitely ##T_{r \theta} = 0## because they're already zero, right?Another thing: I agree with you that the inner blue arrow should be flipped. But ##T_r(r_i)= e_r \cdot T(r_i) e_r## and I don't see how it can be flipped, from this computation.
You need to dot it with ##-e_r## to get the traction on the surface. This follows from the Cauchy stress relationship.

bobinthebox
bobinthebox said:
I've tried to solve by myself the second question. I'd be really happy to have your feedback about this. Let's say that I want to impose null normal stress at ##r=r_i##.

The normal versor there is ##-e_r##, so I should impose ##-e_r \cdot T(r_i)(-e_r)=0## because I want to know the projection in direction ##-e_r## of the stress with outward normal ##-e_r##. Now, by the Cauchy thm. we have ##T(-n)=-T(n)## and hence the condition reads ##-e_r \cdot -(T(r_i)e_r)=0## which is equivalent to ##T_r(r_i)=0## which is indeed what the author asks. Is my argument correct?
Yes. Much better.

bobinthebox
The traction vector at ri is $$-T_{rr}e_r-T_{r\theta}e_\theta$$

NEVER MIND. I see from your representation of the stress tensor that the principal axes of stress are aligned with the coordinate directions, so the shear stress components are zero.

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Yes, because you used the fact that ##T(-n) = -T(n)##, where now ##n = e_r##. Since ##T e_r = T_{rr} e_r + T_{r \theta} e_{\theta}## and the normal at ##r= r_i## is ##-e_r## we have

## T(-e_r)=-T(e_r)=-T_{rr} e_r - T_{r \theta} e_{\theta}##, as you wrote. I feel like I got it now.@Chestermiller I have a really last question about the imposition of boundary conditions.

Assume we're using first Piola-Kirchoff stress tensor in order to solve the problem. It is defined at (5.83) in the link I posted above. In order to impose null traction on the curved boundaries the author writes

##S_{r1}(\pm \frac{h_0}{2})=0##

(equation (5.100))

But I can't see the meaning of ##S_{r1}##. I only understand that the evaluation at ##\pm \frac{h_0}{2}## is needed to act on the curved boundaries, but have no idea about why I have to use component ##S_{r1}##.

## 1. What is a boundary condition in elastic blocks?

A boundary condition in elastic blocks refers to the set of conditions that must be satisfied at the boundaries of the block in order for the block to remain in a state of equilibrium. In the case of null traction, this means that there is no external force acting on the boundary of the block.

## 2. Why is null traction important in elastic blocks?

Null traction is important in elastic blocks because it ensures that the block remains in a state of equilibrium and does not experience any external forces that could cause it to deform or move. This is crucial in accurately studying the behavior of elastic blocks and predicting their response to different loads.

## 3. How is null traction applied on the boundary of an elastic block?

Null traction is applied on the boundary of an elastic block by assuming that there are no external forces acting on the boundary. This means that the stresses and strains at the boundary are equal to zero, and the block is in a state of equilibrium.

## 4. What are the consequences of not applying null traction on the boundary of an elastic block?

If null traction is not applied on the boundary of an elastic block, the block may experience external forces that could cause it to deform or move. This could result in inaccurate predictions of the block's behavior and could lead to structural failure in real-life applications.

## 5. Are there any real-life examples of null traction on the boundary of an elastic block?

Yes, there are many real-life examples of null traction on the boundary of an elastic block. One common example is in the design of bridges, where engineers must ensure that the bridge remains in a state of equilibrium and does not experience any external forces that could cause it to deform or collapse.

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