Maximum shear stress
• Recall
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
• In principal system
σ XY = l1l2 σ1 + m1m2 σ 2 + n1n2 σ 3
l1l2 + m1m2 + n1n2 = 0 l12 + m12 + n12 = 1
• Assume σ1 ≥ σ 2 ≥ σ 3
• Maximum shear σ XY = l1l2 (σ1 − σ 3 )
stress then
1
τmax =
2
(σ1 − σ 3 )
l22 + m22 + n22 = 1
l1l2 = − n1n2
Example
s=1
1
0
1
1
1
0
1
1
>> [l,sig]=eig(s)
sig =-0.4142
0
0
0
1.0000
0
0
0 2.4142
Maximum shear stress is 1.4142.
Why is this reasonable?
Equilibrium when stresses vary from
point to point
• Differential element
Differential equilibrium equations
• From equilibrium of infinitesimal element
∂ σ xx ∂ σ xy ∂ σ xz
+
+
+ bx = 0
∂x
∂y
∂z
∂ σ xy ∂ σ yy ∂ σ yz
+
+
+ by = 0
∂x
∂y
∂z
∂ σ xz ∂ σ yz ∂ σ zz
+
+
+ bz = 0
∂x
∂y
∂z
bx , by , bz : body force per unit volume in x,y,z directions
• Why do we have derivatives of stresses
but not of body forces?
Equations of equilibrium repair damage
done by kinematic assumptions
• Most commonly used theories start with
assumptions on displacements
• Stresses calculated from displacements often do
not satisfy equilibrium
• Repair via equilibrium equations improves
accuracy
• Stress post-processing in finite element software
is possible and desirable
• Check equations in spherical and cylindrical
coordinate systems in textbook
Beam theory
• Euler-Bernoulli beam assumptions mean no
shear strains
• Equations of equilbrium require shear
stresses!
• Cantilever beam under end load P
My Pxy
=
I
I
∂ σ xx ∂ σ xy
+
=0
∂x
∂y
σ xx =
⇒
∂ σ xy
∂y
=−
Py
I
Deformation of a deformable body
• Deformation will cause a particle P to
move from (x,y,z) to (x*,y*,z*)
Strain of a line element
• Strain measures are typically based on what
happens to an infinitesimal line element or to
angle between to such elements
Displacements
• Components of motion
u = x* − x ⇒ displacement in x-direction
x* = x + u
v = y* − y ⇒ displacement in y-direction
y* = y + v
w = z* − z ⇒ displacement in z-direction
z* = z + w
• What is the difference between
displacements and deformations?
Strain measures
2
2
2
2
• Length of infinitesimal
ds
=
dx
+
dy
+
dz
( ) [ ] ( ) ( )
2
2
2
2
element before and
ds
*
=
dx
*
+
dy
*
+
dz
*
( ) ( ) ( ) ( )
after deformation
ds
*
−
ds
• Engineering strain
εE =
ds
• Maginfication factor of
a line with direction
cosines (l,m,n)
• What does this
remind you of?
⎡⎛ ds * ⎞ 2 ⎤
1 2
ε
εE
M = 0.5 ⎢⎜
1
−
=
+
⎥
E
⎟
2
⎢⎣⎝ ds ⎠
⎥⎦
⎡ε xx ε xy ε xz ⎤ ⎧ l ⎫
⎢
⎥⎪ ⎪
= [l m n ] ⎢ε xy ε yy ε yz ⎥ ⎨m ⎬
⎢ε xz ε yz ε zz ⎥ ⎪⎩ n ⎭⎪
⎣
⎦
Strain components
.
Reading assignment
Sections 2.7-8: Question: Why do we have two kinds of shear
strain?
Source: www.library.veryhelpful.co.uk/ Page11.htm