Coordination Transformations for Strain & Stress Rates
To keep the presentation as simple as possible, we will look at purely two-dimensional
stress-strain rates. Given an original coordinate system ( x, y ) and a rotated system
( x̂, yˆ ) as shown below:
y
x'
y'
θ
x
Recall that the strain rates in the x-y coordinate system are:
1  ∂u ∂v 
∂v
∂u
ε xx =
ε xy =  + 
ε yy =
2  ∂y ∂x 
∂y
∂x
Or, in index notation:
1  ∂u ∂u j 
ε ij =  i +
2  ∂x j ∂xi 
Also, we note that the unit vectors for the rotated axes are:
K
K
K
iˆ = cos θ i + sin θ j
K
K
K
ĵ = − sin θ i + cos θ j
Thus, the location of a point in ( x̂, yˆ ) is:
 x̂   cos θ
 ŷ  =  − sin θ
 
sin θ 
cos θ 
x
 
 y
Similarly, the velocity components are related by:
 û   cos θ
 v̂  =  − sin θ
 
sin θ   u 
cos θ   v 
For differential changes, we also have
 dx̂   cos θ
 dŷ  =  − sin θ
  
sin θ   dx 
cos θ   dy 
Coordination Transformations for Strain & Stress Rates
Thus, defining T as the rotation matrix, we note that:
∂xˆ 
 ∂xˆ
 ∂x
∂y   cos θ sin θ 
=
T =
∂yˆ   − sin θ cos θ 
 ∂yˆ
 ∂x
∂y 

Inverting this:
∂x 
 ∂x
 ∂xˆ
∂yˆ   cos θ
=
T −1 = 
∂y   − sin θ
 ∂y
 ∂xˆ
∂yˆ 

Thus to find
−1
sin θ 
 cos θ
=

cos θ 
 sin θ
− sin θ 
cos θ 
∂uˆ
in terms of u, v and their derivatives:
∂xˆ
∂uˆ ∂uˆ ∂x ∂uˆ ∂y ∂uˆ
∂uˆ
=
+
=
cos θ + sin θ
∂xˆ ∂x ∂xˆ ∂y ∂xˆ ∂x
∂y
Then, substituting uˆ = u cos θ + v sin θ :
∂uˆ 
∂
∂
= cos θ
+ sin θ  (u cos θ + v sin θ )
∂xˆ 
∂x
∂y 
∂u
∂v
∂u
∂v
= cos2 θ
+ cos θ sin θ
+ cos θ sin θ
+ sin 2 θ
∂x
∂x
∂y
∂y
⇒ ε xxˆ ˆ = cos2 θε xx + 2 cos θ sin θε xy + sin 2 θε yy
∂vˆ 
∂
∂
=  − sin θ
+ cos θ  ( −u sin θ + v cos θ )
∂yˆ 
∂x
∂y 
∂u
∂v
∂u
∂v
= sin 2 θ
− sin θ cos θ
− sin θ cos θ
+ cos2 θ
∂x
∂x
∂y
∂y
⇒
ε yyˆ ˆ = sin 2 θε xx + 2sin θ cos θε xy + cos2 θε yy


1  ∂uˆ ∂vˆ  1  
∂
∂
∂
∂
+  =   − sin θ
+ cos θ  (u cos θ + v sin θ ) + cos θ
+ sin θ  ( − sin θ + v cos θ ) 

∂x
∂y 
∂x
∂y 
2  ∂yˆ ∂xˆ  2  


⇒ ε xyˆˆ = sin θ cos θ (ε yy − ε xx ) + (cos2 θ − sin 2 θ )ε xy
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Coordination Transformations for Strain & Stress Rates
If x − y are the principal strain directions, then ε xy = 0 and ε xˆxˆ , ε xˆyˆ & ε yˆyˆ are
ε xˆxˆ = cos 2 θε xx + sin 2 θε yy
ε yˆyˆ = sin 2 θε xx + cos 2 θε yy
if ε xy = 0
ε xˆyˆ = sin θ cos θ (ε yy − ε xx )
The next step is to relate the stresses in (x, y ) to (xˆ , yˆ ) . Consider a differential surface
with ŷ normal:
τ
τyy
y
τyx
dy
τxx
x'
dx
y'
τyy
θ
θ
x
τxy
τyx
dx
K
The resultant stress is given as the vector τ and the force on the surface is
Decomposing the stress vector into the coordinate axes gives:
K
K
K
τ ds.
K
τ d xˆ = (τ yˆ xˆ i + τ yˆ yˆ j ) d xˆ
K
K
= ( − τ xx d y + τ yx d x ) i + (τ y y d x − τ xy d y ) j
Note that:
dx = cos θ dxˆ
dy = sin θ dxˆ
K
K
K
i = cos θ iˆ − sin θ ˆj
K
K
K
ˆ
j = sin θ i + cos θ ˆj
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Coordination Transformations for Strain & Stress Rates
Thus, the second line becomes:
K
K
( −τ xx sin θ + τ yx cos θ )(cos θ iˆ − sin θ ˆj )dxˆ
K
K
K
K
+(τ yy cos θ − τ xy sin θ )(sin θ iˆ + cos θ ˆj )dxˆ = (τ yxˆ ˆ iˆ + τ yyˆ ˆ ˆj )dxˆ
So collecting all the iˆ & ˆj terms (and enforcing τ xy = τ yx ) gives:
τ yxˆ ˆ = (τ yy − τ xx )sin θ cos θ + τ yx (cos2 θ − sin 2 θ )
τ yyˆ ˆ = τ xx sin 2 θ + τ yy cos2 θ − 2τ yx sin θ cos θ
For the principal strain axes,
τ xx = 2µε xx + λ (ε xx + ε yy )
τ yy = 2 µε yy + λ (ε xx + ε yy )
τ xy = 0
Plugging this into τ yˆxˆ andτ yˆyˆ gives
τ yxˆ ˆ = 2 µ (ε yy − ε xx ) sin θ cos θ
ε xy
ˆˆ
τ yy = 2 µ (ε xx sin 2 θ + ε yy cos2 θ ) + λ (ε xx + ε yy )
ε yy
ˆˆ
ε xx
ˆ ˆ +ε yy
ˆˆ
Thus, we arrive at the known result:
τ yxˆ ˆ = 2 µ ε xyˆˆ
τ yyˆ ˆ = 2 µ ε yyˆ ˆ + λ (ε xxˆ ˆ + ε yyˆ ˆ )
A similar derivation would give:
τ xxˆ ˆ = 2 µε xxˆ ˆ + λ (ε xxˆ ˆ + ε yyˆ ˆ )
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