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6 Problem Set #2

Problem 4 For the stress tensor ⎞⎛

σij = ⎜⎝

1 1 01 1 0 ⎟⎠

0 0 2

a) Find the principal stresses and directions. (Use a right­handed coordinate system)

b) Compare the three Invariants for the original stress tensor and the prin­cipal stress tensor.

σdevc) Find the deviatoric stress tensor ij

d) Find the principal stresses and directions of the deviatoric stress tensor. Find the Invariants of the deviatoric stress tensor.

e) What are the relations of the principal stresses, directions, shears, etc.= for these two tensors (the complete and deviatoric tensors investigated above)? Can you say anything in general about these relations for an arbitrary stress tensor and its deviator?

f) Construct Mohr’s circle for σij.

g) Construct Mohr’s circle for σdev .ij

h) What is the relation between the maximum shear for the two tensors?

Solution a) Finding the principal stresses and directions is done in the same way

as diagonalizing a matrix or finding the principal coordinate system (aka the eigenbasis) — in fact these are all the same problem. In essence, we want to find the right­handed coordinate system, expressed in terms of the given coordinate system, in which only normal stresses act along each of the basis vectors. There is a lot of linear algebra hidden in here, but I will just state that given any stress tensor, you can find a pricipal coordinate system. Mathematically, we can express the above statement like this:

σij · nj = λˆˆ nj (1)

which just says that the traction on the nj plane (remember we define our planes by use of the vector normal to it) is along the normal vector (i.e. the stress is normal to the plane). We rewrite equation 1 as

ˆ(σij − λ δij) nj = 0 (2) · ·

�

� � � �� � � � � � �� � �

� �

�

�

7 Problem Set #2

and the λ’s and nj’s for which this is true are the eigenvalues (principal stresses) and the eigenvectors (principal coordinate systems), respectively. There will be three eigenvalues and associated eigenvecors which will solve equation 2, not neccisarily unique. Two further constraints are placed on the nj’s, they must be normalized (unit length) and they must for a right­handed coordinate system. We can express these two conditions as

|nj| nj · nj = n1 + n2 + n2 = 1normalized: ˆ = ˆ ˆ 2 23

right­handedness: ˆ ˆ nkni = nj × ˆ

We proceed by finding the eigenvalues (principal stresses), which are the λ’s which solve equation 2, we denote them in a matrix as: ⎞⎛ ⎜⎝

σ1 0 0 0 σ2 0 0 0 σ3

⎟⎠

and are found by solving for the roots of the characteristic polynomial

det (σij − λδij) = 0

which can be simplified by use of the Invariants or solved directly. I will solve this directly, and you can refer to your notes for how the characteristic polynomial simplifies with the Invariants.

(1 − λ) 1 0 1 0

(1 − λ) 0 0

= (2 − λ)· (λ− 1)2 = (2 − λ)(λ− 2)λ = 0− 1 (2 − λ)

The solutions to this characteristic polynomial are λ = 2, 2 and 0, these are the principal stresses. Following the convention σ1 > σ2 > σ3, σ1 = σ2 = 2 and σ3 = 0.

We then find the associated eigenvectors (principal directions), by solving for a nj for each λ in equation 2. For simplicity, I am going to call the

⎛⎞⎛components of the eigenvectors xj. 1­0 1 0 x1 0

⎞⎛⎞

For λ = 0,⎜⎝ 1 1­0 0 ⎟⎠ ·⎜⎝ x2 ⎟⎠ =⎜⎝ 0 ⎟⎠

0 0 2­0 x3 0

x1 + x2 = 0 x1 = s

⇒x3 = 0

⇒ x2 = −s x3 = 0

where s is an undetermined variable. We now enforce the normalization constraint

2s2 + (−s)2 + 02 = √

2s = √

2 s = 1

�

� � �� � � � ��

�� � � � ��

8 Problem Set #2

so s =2

(note since this equation involves squaring, the sign is ambiguous,

so we just choose one, in this case positive), and the eigenvector associated

1√

with σ3 = 0 is ⎞⎛ 1

(3) j =

1√2

⎜⎝ ⎟⎠−1

0n .

⎞⎛We then find the eigenvectors associated with σ1 and σ2 in the same way: 0

⎞ 11 − 2

1

⎛⎞⎛ x1 0 ⎜⎝

⎟⎠ ⎜⎝

⎟⎠ ⎜⎝

⎟⎠For λ = 2, 01− 2 0

0x2 = · 0 02 − 2 x3

x1 = s x1 − x2 = 0

0 x2 = s⇒ ⇒

0= · x3 x3 = t

where s and t are undetermined variables. This gives us two vectors:⎞⎛⎞⎛ 0 s

(1) j =⎜⎝ 0 ⎟⎠ and n

(2) j =⎜⎝ s ⎟⎠ .n

t 0

We now enforce the normalization constraint on n(2) j

√s2 + s2 + 02 =

√2s2 =

√2 s = 1

1√2

(again note eigenvector associated with σ2 = 2 is

the sign ambiguity, we choose positive), and theso s =

⎞⎛ 1

(2) j =

1√2

⎜⎝ ⎟⎠1n .

0

(1) jNow, the normalization constraint on n is trivial, and it tells us that

t = ±1, so to resolve the ambiguity we now utalize the right­handedness constraint:

ı j k 1 1 01 −1 0

= ı·0−j·

⎞⎛ 0

1 0+ˆ 1

k·(−1−1)· ⎜⎝ 0−2/

√2

⎟⎠(1) i = (2)

j n× ˆ(3) kn n √

2 √

2 = =

so we find, subject to the normalization constraint above, ⎞⎛ 0

(1) j =⎜⎝

⎟⎠ .n 0−1

� �

9 Problem Set #2

We can then put this all together as ⎞⎛⎞⎛ 2 0 0 0 1 1

1 σprincipal

nn =⎜⎝ ⎟⎠

⎜⎝ ⎟⎠ ,0 1 −1

−√

2 0 0 2 0 and nj(n) = √

2 ·

0 0 0 0

where the first matrix is the principal stress tensor, and the columns of the second matrix define the principal frame, given in terms of the original coordinate system. Note that the final solution to this problem is non­unique in terms of the signs of the eigenvectors, which is to say it is non­unique in terms of π rotations about the axes.

2

b) For the origninal stress tensor: ⎞⎛

1 1 0 σij =⎜⎝ 1 1 0 ⎟⎠

0 0 2

I1 = σii = 1 + 1 + 2 = 4

1 1 2 11 + 2σ2

22 + σ2σ2 12 + σ2

33 − (σ11 + σ22 + σ33)I2 =2·(σijσij − σiiσjj) =

2· = −4

I3 = = det (σij) = 2(1 − 1) = 0|σij|

For the principal stress tensor: ⎞⎛

σij = ⎜⎝

2 0 00 2 0 ⎟⎠

0 0 0

I1 = σii = 2 + 2 + 0 = 4

1 I2 = · (σijσij − σiiσjj) = −σ11σ22 = −4

2

I3 = |σij| = det (σij) = 0

The Invariants of the original stress tensor and the principal stress tensor are the same — they are invariant under coordinate system rotation.

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10 Problem Set #2

c) The deviatoric stress tensor is found by σdev = σij − σii δij, and is ij 3

· ⎞⎛⎞⎛⎞⎛ 1 1 0 4 0 0 −1/3 1 0

1 1

σdev ij =⎜⎝

⎟⎠ ⎜⎝

⎟⎠ =⎜⎝ ⎟⎠1 1 0 0 4 0 0−1/3

0−

3 ·

0 0 2 0 0 4 0 2/3

d) Following the proceedure in part a, we find the eigenvalues (principal stresses) and the eigenvectors (principal coordinates) as ⎞⎛⎞⎛

0 1 12/3 0 0 0 2/3 0 0 0 −4/3

1 σprincipal

nn =⎜⎝ ⎟⎠

⎜⎝ ⎟⎠ ,0 1 −1

−√

2 0 and nj(n)√

2 ·

0

The Invariants are Idev = 01

4 Idev = 2 3

Idev −16 = 3 27

is σdeve) The definition of the deviatoric stress tensor ij = σij − σkk δij,3 ·

which simply says that the deviatoric stress tensor is the stress tensor minus the average of the normal stresses. So we decrease the principal stresses by the same amount. Since the shear stresses stay the same, while the normal stresses all decrease by the same amount, the principal coordinates stay the same.

The relationship between the Invariants of the stress tensor and its devi­ator can be found by (keep track of the indecies):

Idev = σdev = σii − σkk

= σii − σkk

3 = 01 ii δii3 3

1 Idev = 3 σdev

ij − σdev σdev ij σ2

ii jj 2 �� � � � � ��1 1 1 1 1

= σij − σkkδij σij − σkkδij σii − σkkδii σjj − σkkδjj 2 3 3

− 3 3

note that δii = 3, δijδij = 3, and σijδij = σii, and expanding the multiplied terms

1 2 3 = σijσij − σkkσjj + σkkσkk − σiiσjj + 2σkkσjj − σkkσjj

2 3 9

1 6 2−1 +σijσij − σiiσjj + σkkσjj =

3 −

2 3 3

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11 Problem Set #2

1 1 2 = (σijσij − σiiσjj) + σkkσjj

2 2 3

1 Idev I2 ⇒ = I2 +

3 12

For Idev I will derive this for the principal coordinate system, which no3

loss of generality, setting p = −31σkk = −1I1:3

Idev = det σdev = σdev σdev σdev = (σ11 + p) (σ22 + p) (σ33 + p)3 ij 11 22 33

σ11σ22 + σ11σ33 + σ22σ33 + p (σ11 + σ22 + σ33) + p 2 = σ11σ22σ33 + p

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= I3 + p −I2 − 3p 2 + p 2 = I2 −−1

I1 (−I2)− 2 −1

I13 3

1 2 Idev = I3 + I3 3⇒

3 I1I2 +

27 1

f) Mohr’s circle is given by

σn = σc + σr cos 2θ

τ = σr sin 2θ

σ1−σ2where σr = 2

, σr = σ1+σ2 and σ1,2 are the principal stresses (σ1 > σ2).2

For σij

σ2 = 0, σ1 = 2

which gives σr = σc = 1.

The plot is given below. g) For σdev

ij

2 σ2 =

−4 , σ1 =

3 3

which gives −1

σr = 1 and σc = . 3

The plot of the Mohr’s circle for the full stress tensor (σij) and its deviatoric component (σdev , labelled with a superscript d in the plot) is below. ij

12 Problem Set #2

sn

t

0 2sc

sr

2q

tmax

sij

-1/3 2/3scd

srd

2qd

tmax

sijd

h) The maximum shear stress occurs when sin 2θ = 0, which gives 2θ = . (This is also apparent from examining the plot of Mohr’s circle.) For 2θ = , τ = σr is the maximum that the shear stress can be. Since σr = 1 for the two tensors (σij and σdev ) it follows that the maximum shear stress is the ij

same for both of the tensors.

π 2π 2