2.4.1: Transformation of Stress
• Direction cosines
z
X
x
y
z
X
l1
m1
n1
Y
l2
m2
n2
Z
l3
m3
n3
l12 + l22 + l32 = 1
l12 + m12 + n12 = 1
m12 + m22 + m32 = 1
l22 + m22 + n22 = 1
n12 + n22 + n32 = 1
l32 + m32 + n32 = 1
• Is there a less redundant
description?
θxZ
θ xX
y
x
• Orthogonality properties
and unit length
Z
θ xY
Y
l1l2 + m1m2 + n1n2 = 0
l1m1 + l2 m2 + l3 m3 = 0
l1l3 + m1m3 + n1n3 = 0
l1n1 + l2 n2 + l3 n3 = 0
l2l3 + m2 m3 + n2 n3 = 0
m1n1 + m2 n2 + m3 n3 = 0
Stresses on an inclined plane
uur
N = l iˆ + m ˆj + n kˆ
• Normal vector to plane
y
−σ
N
z
AOBC
A
A
= l; OAC = m; OBA = n
AABC
AABC
AABC
x
P
A
σP
−σ
x
O
−σ
C
Stress vector on plane P:
uur
uur
uur
uuur
σP = l σx + m σ y + n σz
uur
σ P = σ Px iˆ + σ Py ˆj + σ Pz kˆ
y
z
uur
Components of the stress vector σ P :
σ Px = l σ xx + m σ yx + n σ zx
σ Py = l σ xy + m σ yy + n σ zy
σ Pz = l σ xz + m σ yz + n σ zz
Normal and shear stresses
• Normal and shear components
σ PN
JG JJG
= N • σP
( )
( )
σ PN = l 2 σ xx + m 2 σ yx + n 2 σ zx + 2m n σ yz + 2n l ( σ xz ) + 2l m σ xy
σ PS = σ P2 − σ P2N = σ P2x + σ P2y + σ P2z − σ P2N
• We often need the maximal and minimal
normal stresses and maximum shear
stress. Why?
Stress transformation
• Using the equation for normal stresses
σ XX = l12 σ xx + m12 σ yy + n12 σ zz + 2m1n1 σ yz + 2n1l1 σ xz + 2l1m1 σ xy
σYY = l22 σ xx + m22 σ yy + n22 σ zz + 2m2 n2 σ yz + 2n2l2 σ xz + 2l2 m2 σ xy
σ ZZ = l32 σ xx + m32 σ yy + n32 σ zz + 2m3 n3 σ yz + 2n3l3 σ xz + 2l3 m3 σ xy
• Similarly, obtain
σ XY = l1l2 σ xx + m1m2 σ yy + n1n2 σ zz + ( m1n2 + m2 n1 ) σ yz + ( l1n2 + l2 n1 ) σ xz + ( l1m2 + l2 m1 ) σ xy
σ XZ = l1l3 σ xx + m1m3 σ yy + n1n3 σ zz + ( m1n3 + m3 n1 ) σ yz + ( l1n3 + l3 n1 ) σ xz + ( l1m3 + l3 m1 ) σ xy
σYZ = l2l3 σ xx + m2 m3 σ yy + n2 n3 σ zz + ( m2 n3 + m3 n2 ) σ yz + ( l2 n3 + l3 n2 ) σ xz + ( l2 m3 + l3 m 2 ) σ xy
• How do we reduce these to a single short line?
More compact notation
• Stress tensor transformation
Tnew
⎡ l1 m1 n1 ⎤ ⎡σ xx
⎢
⎢
⎥
= ⎢l2 m 2 n 2 ⎥ ⎢ σ xy
⎢⎣ l3 m3 n 3 ⎥⎦ ⎢ σ xz
⎣
σ xy
σ yy
σ yz
σ xz ⎤ ⎡ l1 l2 l3 ⎤
⎥⎢
σ yz ⎥ ⎢ m1 m 2 m3 ⎥⎥
⎥
σzz ⎦ ⎢⎣ n1 n 2 n 3 ⎥⎦
• Matrix notation
⎡ l1 l2 l3 ⎤
r = ⎢⎢ m1 m 2 m3 ⎥⎥ = rotation matrix:measured from old system
⎢⎣ n1 n 2 n 3 ⎥⎦
Tnew = r T Told r
Principal stresses
• Can obtain principal stresses by posing
two seemingly unrelated questions
– What plane will give us maximum or minimal
normal stresses?
– What plane will give us zero shear stresses?
– Both approaches give the same answer: A
coordinate system with zero shear stresses
and extreme normal stresses
• Textbook derives equations based on zero
shear stresses
Derivation
• Stress vector normal to plane, hence
σ Px = l σ xx + m σ yx + n σ zx = l σ
σ Py = l σ xy + m σ yy + n σ zy = m σ
σ Pz = l σ xz + m σ yz + n σ zz = n σ
• That is, [l,m,n] is eigenvector of stress matrix
⎡σ xx − σ
σ xy
σ xz ⎤ ⎧ l ⎫ ⎧0 ⎫
σ xx − σ
σ xy
σ xz
⎢
⎥⎪ ⎪ ⎪ ⎪
σ yy − σ
σ yz ⎥ ⎨m ⎬= ⎨0 ⎬ → σ xy
σ yy − σ
σ yz = 0
⎢ σ xy
⎢
⎥ ⎪ m ⎪ ⎪0 ⎪
σ
σ
σ
−
σ
σ xz
σ yz
σ zz − σ
yz
yy
⎣ xz
⎦⎩ ⎭ ⎩ ⎭
• Principal stresses are the eigenvalues.
• What properties eigenvalues of symmetric matrix?
Stress invariants
• Setting determinant to zero gives
σ3 − I1 σ 2 − I 2 σ − I3 = 0
I1 = σ xx + σ yy + σ zz
I2 = −
σ xx
σ xy
σ xy
σ yy
σ xx
−
σ xz
σ xz σ yy
−
σ zz σ yz
σ yz
σ zz
= σ 2xy + σ 2xz + σ 2yz − σ xx σ yy − σ xx σ zz − σ yy σ zz
σ xx
σ xy
σ xz
I3 = σ xy
σ yy
σ yz = σ xx σ yy σ zz + 2 σ xy σ yz σ xz − σ xx σ 2yz − σ yy σ 2xz − σ zz σ 2xy
σ xz
σ yz
σ zz
• Why are the I’s invariant to coordinate system?
Example
s=[1,1,1;1,1,1;1,1,1]
s=1 1 1
1 1 1
1 1 1
>> [l,sig]=eig(s)
l =0.4082 0.7071 0.5774
0.4082 -0.7071 0.5774
-0.8165
0 0.5774
sig = 0
0
0
0
0
0
0
0
3
What are easy checks that the answers
obtained from Matlab are correct?
Another example
s=1
1
0
1
1
1
0
1
1
If we swapped the solutions
of the two examples. How
could we tell that something is
wrong?
>> [l,sig]=eig(s)
l =0.5000 -0.7071
-0.7071 0.0000
0.5000 0.7071
0.5000
0.7071
0.5000
sig =-0.4142
0
0
0
1.0000
0
0
0 2.4142
Mean and deviator stresses
• Mean normal stress
• We divide stress tensor
as
σ xx + σ yy + σ zz
1
σ m = I1 =
3
3
T = Tm + Td
⎡σ xx − σ m
0⎤
σ xy
σ xz ⎤
⎡σ m 0
⎢
⎥
σ yy − σ m σ yz ⎥
Tm = ⎢⎢ 0 σ m 0 ⎥⎥ Td = ⎢ σ xy
⎢ σ xz
⎢⎣ 0
0 σ m ⎥⎦
σ yz
σ m ⎥⎦
⎣
• Mean stress responsible
for volume change
• Deviator for yielding
Plane stress
• 2-D state of stress
σ zz = σ xz = σ yz
⎡ σ xx
⎢
= 0 → T = ⎢ σ yx
⎢ 0
⎣
σ xy
σ yy
0
0⎤
⎥ ⎡σ xx
0⎥ = ⎢
⎢⎣ σ yx
⎥
0⎦
σ xy ⎤
⎥
σ yy ⎥⎦
• Direction cosines
x
y
z
y
Y
X
θ
π −θ
2
X
π +θ
2
l1 = cos θ m1 = sin θ n1=0
Y
l2 = − sin θ
m 2 = cos θ
n2=0
Z
l3=0
m3=0
n3=1
θ
x
Stress transformation
• Matrix version
• Or
⎡σXX
T=⎢
⎣σYX
σXY ⎤ ⎡ cos θ sin θ ⎤ ⎡σ xx
=
⎢
σYY ⎥⎦ ⎢⎣ − sin θ cos θ ⎥⎦ ⎢⎣ σ yx
σ xy ⎤ ⎡cos θ − sin θ ⎤
⎥
σ yy ⎥⎦ ⎢⎣ sin θ cos θ ⎥⎦
σ XX = σ xx cos 2 θ + σ yy sin 2 θ + 2 σ xy cos θ sin θ
σYY = σ xx sin 2 θ + σ yy cos 2 θ − 2 σ xy cos θ sin θ
(
σ XY = − σ xx − σ yy
)
(
sin θ cos θ + σ xy cos 2 θ − sin 2 θ
)
( σ xx + σ yy ) + 12 ( σ xx − σ yy ) cos 2θ + σ xy sin 2θ
σYY = 12 ( σ xx + σ yy ) − 12 ( σ xx − σ yy ) cos 2θ − σ xy sin 2θ
σ XY = − 12 ( σ xx − σ yy ) sin 2θ + σ xy cos 2θ
σ XX =
1
2
• Review Mohr’s circle in textbook
Reading assignment
Sections 2.5-6: Question: Where do we use the first of
equations 2.45 in elementary beam theory?
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