PHY454H1S Continuum Mechanics. Lecture 3. Strain tensor review. Stress tensor. Taught by Prof. K. Das. Originally appeared at: http://sites.google.com/site/peeterjoot2/math2012/continuumL3.pdf Peeter Joot — [email protected] Jan 18, 2012

continuumL3.tex

Contents 1

Review. Strain.

1

2

Stress tensor. 2.1 Examples of the stress tensor . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Example 1. stretch in two opposing directions. . . . . . . . . . . . 2.1.2 Example 2. stretch in a pair of mutually perpendicular directions 2.1.3 Example 3. radial stretch . . . . . . . . . . . . . . . . . . . . . . . .

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3 4 4 5 5

1. Review. Strain. Strain is the measure of stretching. This is illustrated pictorially in figure (1)

Figure 1: Stretched line elements. 2

ds0 − ds2 = 2eik dxi dxk , where eik is the strain tensor. We found   1 ∂ei ∂ek ∂el ∂el eik = + + 2 ∂xk ∂xi ∂xi ∂xk Why do we have a factor two? Observe that if the deformation is small we can write

1

(1)

(2)

2

ds0 − ds2 = (ds0 − ds)(ds0 + ds)

≈ (ds0 − ds)2ds so that we find 2

ds0 − ds ds0 − ds2 ≈ ds2 ds Suppose for example, that we have a diagonalized strain tensor, then we find 02



2

ds − ds = 2eii

dxi ds

(3)

2 (4)

so that 2

ds0 − ds2 = 2eii dxi2 ds2 Observe that here again we see this factor of two. If we have a diagonalized strain tensor, the tensor is of the form   e11 0 0  0 e22 0  0 0 e33

(5)

(6)

we have 2

dxi0 − dxi2 = 2eii dxi2

(7)

ds0 = (1 + 2e11 )dx12 + (1 + 2e22 )dx22 + (1 + 2e33 )dx32

(8)

ds2 = dx12 + dx22 + dx32

(9)

2

so dx10 =

p

1 + 2e11 dx1 ∼ (1 + e11 )dx1

(10)

dx20 =

p

1 + 2e22 dx2 ∼ (1 + e22 )dx2

(11)

dx30 =

p

1 + 2e33 dx3 ∼ (1 + e33 )dx3

(12)

Observe that the change in the volume element becomes the trace dV 0 = dx10 dx20 dx30 = dV (1 + eii )

(13)

How do we use this? Suppose that you are given a strain tensor. This should allow you to compute the stretch in any given direction. FIXME: find problem and try this.

2

2. Stress tensor. Reading for this section is §2 from the text associated with the prepared notes [1]. We’d like to consider a macroscopic model that contains the net effects of all the internal forces in the object as depicted in figure (2)

Figure 2: Internal forces. We will consider a volume big enough that we won’t have to consider the individual atomic interactions, only the average effects of those interactions. Will will look at the force per unit volume on a differential volume element The total force on the body is ZZZ

(14)

FdV,

where F is the force per unit volume. We will evaluate this by utilizing the divergence theorem. Recall that this was ZZZ

(∇ · A)dV =

ZZ

A · ds

(15)

We have a small problem, since we have a non-divergence expression of the force here, and it is not immediately obvious that we can apply the divergence theorem. We can deal with this by assuming that we can find a vector valued tensor, so that if we take the divergence of this tensor, we end up with the force. We introduce the quantity ∂σik , ∂xk and require this to be a vector. We can then apply the divergence theorem F=

ZZZ

ZZZ

ZZ

∂σik 3 dx ∂xk where dsk is a surface element. We identify this tensor FdV =

σik =

Force Unit Area

σik dsk ,

(16)

(17)

(18)

and f i = σik dsk , 3

(19)

as the force on the surface element dsk . In two dimensions this is illustrated in the following figures (3)

Figure 3: 2D strain tensor. Observe that we use the index i above as the direction of the force, and index k as the direction normal to the surface. Note that the strain tensor has the matrix form   σ11 σ12 σ13 σ21 σ22 σ23  (20) σ31 σ32 σ33 We will show later that this tensor is in fact symmetric. FIXME: given some 3D forces, compute the stress tensor that is associated with it. 2.1. Examples of the stress tensor 2.1.1

Example 1. stretch in two opposing directions.

Figure 4: Opposing stresses in one direction. Here, as illustrated in figure (4), the associated (2D) stress tensor takes the simple form   σ11 0 0 0

4

(21)

2.1.2

Example 2. stretch in a pair of mutually perpendicular directions

For a pair of perpendicular forces applied in two dimensions, as illustrated in figure (5)

Figure 5: Mutually perpendicular forces our stress tensor now just takes the form 

σ11 0 0 σ22

 (22)

It’s easy to imagine now how to get some more general stress tensors, should we make a change of basis that rotates our frame. 2.1.3

Example 3. radial stretch

Suppose we have a fire fighter’s safety net, used to catch somebody jumping from a burning building (do they ever do that outside of movies?), as in figure (6). Each of the firefighters contributes to the stretch.

Figure 6: Radial forces. FIXME: what form would the tensor take for this? Would we have to use a radial form of the tensor? What would that be? References [1] L.D. Landau, EM Lifshitz, JB Sykes, WH Reid, and E.H. Dill. Theory of elasticity: Vol. 7 of course of theoretical physics. Physics Today, 13:44, 1960. 2 5

PHY454H1S Continuum Mechanics. Lecture 3 ... - Peeter Joot's Blog

Jan 18, 2012 - (dxi ds. )2. (4) so that ds. 2 − ds2 ds2. = 2eiidx2 i. (5). Observe that here again we see this factor of two. If we have a diagonalized strain tensor, the tensor is of the form ... We will consider a volume big enough that we won't have to consider the individual atomic interactions, only the average effects of those ...

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