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Maximum shear stress
• Recall
• In principal system
• Assume • Maximum shear
stress then
( ) ( ) ( )σ σ σ σ σ σ m σXY 1 2 xx 1 2 yy 1 2 zz 1 2 2 1 yz 1 2 2 1 xz 1 2 2 1 xyl l m m n n m n m n l n l n l m l= + + + + + + + +
σ
10
σ σ σ
11 2 1 2
XY 1 2 1 1 2 2 1 2 32 2 2 2 2
1 22
1 1 1 2 2 2
l
l l
l m m n n
l m n l m nm m n n+ + =
= + +
+ + = + + =
σ σ σ1 2 3≥ ≥
max
σ (σ σ )1 (σ σ )2
XY 1 2 1 3 1 2 1 2
1 3
l l l l n n= − = −
τ = −
Examples = 1 1 0
1 1 10 1 1
>> [l,sig]=eig(s)
sig =-0.4142 0 00 1.0000 00 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
• Why do we have derivatives of stresses but not of body forces?
σσ σ 0 x y zσ σ σ
0x y z
σσ σ 0x y z, , : body force per unit volume in x,y,z directions
xyxx xzx
xy yy yzy
yzxz zzz
x y z
b
b
b
b b b
∂∂ ∂+ + + =
∂ ∂ ∂∂ ∂ ∂
+ + + =∂ ∂ ∂
∂∂ ∂+ + + =
∂ ∂ ∂
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
σ
σ σσ 0 x y y
xx
xy xyxx
My PxyI I
PyI
= =
∂ ∂∂+ = ⇒ = −
∂ ∂ ∂
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
• What is the difference between displacements and deformations?
*
*
*
displacement in x-direction
displacement in y-direction
d
u
isplacement in z-direct
x x
v y y
w nz oz i
=
⇒−
⇒
−
=
⇒
= −
*
*
*
x x u
y y v
z z w
= +
= +
= +
Strain measures
• Length of infinitesimal element before and after deformation
• Engineering strain
• Maginfication factor of a line with direction cosines (l,m,n)
• What does this remind you of?
( ) [ ] ( ) ( )( ) ( ) ( ) ( )
22 2 2
2 2 2 2* * * *
ds dx dy dz
ds dx dy dz
= + +
= + +
[ ]
22
xx xy xz
xy yy yz
xz yz zz
* 10.5 12E E
dsMds
ll m n m
n
ε ε
ε ε εε ε εε ε ε
⎡ ⎤⎛ ⎞= − = +⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦⎡ ⎤ ⎧ ⎫⎢ ⎥ ⎪ ⎪= ⎨ ⎬⎢ ⎥
⎪ ⎪⎢ ⎥ ⎩ ⎭⎣ ⎦
*E
ds dsds
ε −=
Strain components
.
Reading assignmentSections 2.7-8: Question: Why do we have two kinds of shear
strain?
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