
I1 = σxx + σyy + σzz
I2 = σxx σyy + σyy σzz + σzz σxx - (σxy)2 - (σyz)2 - (σzx)2
I3 = σxxσyyσzz + 2σxyσyzσzx - σxx(σyz)2 - σyy(σzx)2 - σzz(σxy)2
An external force is that applied to the body due to the
action of another body or field
∆𝐹
∆𝐹𝑛
∆𝐹𝑡
𝑛
𝜎 = = lim
𝜎𝑛𝑛 = lim
𝜎𝑛𝑡 = lim
𝑇 ∆𝐴→0 ∆𝐴
∆𝐴→0 ∆𝐴
∆𝐴→0 ∆𝐴
𝐸𝑖𝑔𝑒𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛: 𝜎 3 − 𝐼1 𝜎 2 + 𝐼2 𝜎 − 𝐼3 = 0
Stresses in any 2D direction (on any plane n-t)
𝜎𝑥𝑥 + 𝜎𝑦𝑦 𝜎𝑥𝑥 − 𝜎𝑦𝑦
𝜎𝑛𝑛 =
+
𝑐𝑜𝑠2𝜃 + 𝜏𝑥𝑦 𝑠𝑖𝑛2𝜃
2
2
𝜎𝑦𝑦 − 𝜎𝑥𝑥
𝜎𝑡𝑛 =
𝑠𝑖𝑛2𝜃 + 𝜎𝑥𝑦 𝑐𝑜𝑠2𝜃
2
1
𝑁𝑜𝑡𝑒: 𝑐𝑜𝑠 2 𝜃 = (1 + 𝑐𝑜𝑠2𝜃)
2
1
2
𝑠𝑖𝑛 𝜃 = (1 − 𝑐𝑜𝑠2𝜃)& 𝑠𝑖𝑛2𝜃 = 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
2
Stresses with coordinate axis rotation
𝜎𝑥 ′ 𝑥 ′ = 𝜎𝑥𝑥 𝑐𝑜𝑠 2 𝜃 + 𝜎𝑦𝑦 𝑠𝑖𝑛2 𝜃 + 2𝜎𝑥𝑦 𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃
𝜎𝑦 ′ 𝑦 ′ = 𝜎𝑥𝑥 𝑠𝑖𝑛2 𝜃 + 𝜎𝑦𝑦 𝑐𝑜𝑠 2 𝜃 − 2𝜎𝑥𝑦 𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃
1
𝜎𝑥 ′ 𝑦 ′ = (𝜎𝑦𝑦 − 𝜎𝑥𝑥)𝑠𝑖𝑛2𝜃 + 𝜎𝑥𝑦 𝑐𝑜𝑠2𝜃
2
𝜎𝑥 ′ 𝑥 ′ + 𝜎𝑦 ′ 𝑦 ′ = 𝜎𝑥𝑥 + 𝜎𝑦𝑦
Principal stresses and maximum shear stress
𝜎𝑥𝑥 + 𝜎𝑦𝑦
𝜎𝑥𝑥 − 𝜎𝑦𝑦 2
± √[
] + 𝜏𝑥𝑦 2
2
2
2𝜎𝑥𝑦
𝑡𝑎𝑛2𝜃 =
𝜎𝑥𝑥 − 𝜎𝑦𝑦
 We call the maximum σnn the principal stresses, their
corresponding directions the principal directions, and the
planes on which they act the principal planes
 Physical definition of principal stresses is where the shear
stresses on the plane are zero
 Max and min shear stresses are in the directions of 45° and
135° with respect to the principal direction 1
1
𝜏 𝑚𝑎𝑥
} = ± (𝜎1 − 𝜎3)
𝜏 𝑚𝑖𝑛
2
 Draw up the Mohr’s circle on a σ-τ axis. Where τ = 0, σ1 is at
the furthest right and σ3 is at the furthest left. Locate the
points (σxx , -σxy) and (σyy , σxy). The angle made from the
σxx line to σ1 and σ3 taken in an anticlockwise direction
corresponds to 2θ1 and 2θ3 respectively. This determines the
orientation of the infinitesimal element.
𝜎1,3 =
Cylindrical pressure vessels
𝐴𝑥𝑖𝑎𝑙 (𝑙𝑜𝑛𝑔)𝑠𝑡𝑟𝑒𝑠𝑠: 𝜎𝑥𝑥 =
𝑝𝑟
𝑝𝑟
& 𝐻𝑜𝑜𝑝 𝑠𝑡𝑟𝑒𝑠𝑠: 𝜎𝑦𝑦 =
2𝑡
𝑡
Principal stresses as Eigenvalues of the stress matrix
 The principal stresses are the Eigenvalues of the stress
tensor, while the corresponding eigenvector is the direction
cosine
𝜎𝑥𝑥 𝜎𝑥𝑦 𝜎𝑥𝑧 𝑙
𝜎𝑛𝑥
{𝜎𝑛𝑦} = [𝜎𝑦𝑥 𝜎𝑦𝑦 𝜎𝑦𝑧] {𝑚}
𝜎𝑧𝑥 𝜎𝑧𝑦 𝜎𝑧𝑧 𝑛
𝜎𝑛𝑧
3D stresses with coordinate rotation
𝜎(𝑛𝑒𝑤) = 𝑅𝜎(𝑜𝑙𝑑)𝑅𝑇
Where R is the transformation matrix:
𝑙
𝑚
𝑛
𝑅 = [ 𝑙′ 𝑚′ 𝑛′ ]
𝑙′′ 𝑚′′ 𝑛′′
Invariants
First invariant:
I1 = σ1 + σ2 + σ3
Second invariant:
I2 = σ1 σ2 + σ2 σ3 + σ3 σ1
Third invariant:
I3 = σ1 σ2 σ3
Principal direction cosine
Linear equations:
𝜎𝑥𝑥 − 𝜎
𝜎𝑥𝑦
𝜎𝑥𝑧
𝑙𝑝
𝜎𝑦𝑦 − 𝜎
𝜎𝑦𝑧 ] {𝑚𝑝}
[ 𝜎𝑦𝑥
𝜎𝑧𝑥
𝜎𝑧𝑦
𝜎𝑧𝑧 − 𝜎 𝑛𝑝
Geometric equation:
(𝑙 2 + 𝑚2 + 𝑛2 ) = 1


Pick any two linear equations and combine with the geometric
equation to solve for l, m and n simultaneously
ℓ = cos θ and m = cos φ on the x-y plane, where φ = other nonright angle other than θ
MATLAB
Solve for Eigen equation:
a = [1 -60 -9100 4300]
x = roots(a)
Determine Eigenvalues and vectors:
A = [the stress tensor]
[v,d] = eig(A)
Equations of motion
Static equilibrium (acceleration = zero)
𝜕𝜎𝑥𝑥 𝜕𝜎𝑥𝑦 𝜕𝜎𝑥𝑧
+
+
+ 𝑏𝑥 = 0
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝜎𝑦𝑥 𝜕𝜎𝑦𝑦 𝜕𝜎𝑦𝑧
+
+
+ 𝑏𝑦 = 0
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝜎𝑧𝑥 𝜕𝜎𝑧𝑦 𝜕𝜎𝑧𝑧
+
+
+ 𝑏𝑧 = 0
𝜕𝑥
𝜕𝑦
𝜕𝑧
Remove bx, by and bz if body forces are negligible.
Summary
1. Write up stress tensor
2. Reduce to 2D problem. Don’t forget third principal stress!
3. Calculate principal stress σ1 and σ3, then rank all three.
4. Determine principal directions. Check which direction
corresponds to which principal stress using:
𝜎𝑥 ′ 𝑥 ′ = 𝜎𝑥𝑥 𝑐𝑜𝑠 2 𝜃 + 𝜎𝑦𝑦 𝑠𝑖𝑛2 𝜃 + 2𝜎𝑥𝑦 𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃
5. Find principal direction cosines l, m and n for all three principal
directions (corresponds to angles from the x, y and z axes
respectively).
Displacement and strain
 Occurs when a solid is subject to external loading and the
positions of materials points in the solid will change
 Rigid body displacement- relative position between points
remain unchanged
 Deformation- relative position changed
 Displacements are non uniform and should be some
functions of coordinates
 Direct strain- change relative distance of two ends of a
segment ie ε = (ℓ-ℓo)/ ℓo
 Shear strain- change the included angle of two segments ie γ
= ½ tan ψ ≈ ½ ψ. If shear strain causes angle reduction, it is
positive. If angle increase, then negative
Strain displacement relation
𝜕𝑢
𝜕𝑣
𝜕𝑤
𝜀𝑥𝑥 =
𝜀𝑦𝑦 =
𝜀𝑧𝑧 =
𝜕𝑥
𝜕𝑦
𝜕𝑧
1 𝜕𝑣 𝜕𝑢
1 𝜕𝑤 𝜕𝑣
1 𝜕𝑢 𝜕𝑤
𝜀𝑥𝑦 = ( + ) 𝜀𝑦𝑧 = (
+ ) 𝜀𝑧𝑥 = ( +
)
2 𝜕𝑧 𝜕𝑦
2 𝜕𝑦 𝜕𝑧
2 𝜕𝑧 𝜕𝑥
Where u, v and w are changes in the x, y and z directions.
Volume strain
Can be derived using direct strain equation and V0 = lx ly lz
Dilation ΔV = (V - V0)/V0 = εxx + εyy + εzz
Strains can be found using the same equations as stresses,
where θ is the orientation of the strain gauges from the axis:
𝜀𝜃 = 𝜀𝑥𝑥 𝑐𝑜𝑠 2 𝜃 + 𝜀𝑦𝑦 𝑠𝑖𝑛2 𝜃 + 2𝜀𝑥𝑦 𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃
Principal strains can be found using the EXACT SAME METHOD
AND EQUATIONS as principal stresses.
Compatibility equation
𝜕 2 𝜀𝑥𝑥 𝜕 2 𝜀𝑦𝑦
𝜕 2 𝜀𝑥𝑦
+
=
2
𝜕𝑦 2
𝜕𝑥 2
𝜕𝑥𝜕𝑦
Hooke’s Law for 1D
Poisson’s ration
Strain in lateral direction
𝐸
{ 𝜀𝑧𝑧 = −𝑣𝜀𝑥𝑥 = −
Bulk modulus indicates the ability of resisting the volume change
of a material under hydrostatic stress K = E/3(1-2v)
1
𝐸
𝜎 = (𝜎1 + 𝜎2 + 𝜎3) =
𝐼 𝜀 = 𝐾𝐼1 𝜀
3
3(1 − 2𝑣) 1
 Large G → small shear deformation (impossible to shear)
 Small volume strain → Large K (incompressible e.g. rubber)
0
0]
𝜎
𝑣𝜎𝑥𝑥
𝐸
𝑣𝜎𝑥𝑥
𝐸
Lamé’s constants
𝜆 = 𝐶12 𝜇 =
G = E/2(1 + v)
εxy = σxy(1+v)/E and σxy = 2Gεxy
In pure shear (left) and hydrostatic (right):
0
𝜎𝑥𝑦 0
𝜎 0
[𝜎𝑦𝑥
0
0] 𝑎𝑛𝑑 [ 0 𝜎
0 0
0
0
0
𝜎𝑥𝑥
𝜀𝑦𝑦 = −𝑣𝜀𝑥𝑥 = −
Shear modulus
Shear Hooke’s rule
In a uniaxial tensile test the stress and strain states are:
𝜎1 0 0
0
0
1 𝜎1
[𝜎] = [ 0 0 0] 𝑎𝑛𝑑 [𝜀] = [ 0 −𝑣𝜎1
0 ]
𝐸
0 0 0
0
0
−𝑣𝜎1
σ=Eε
v = - ε lateral/ε axial
𝜀𝑥𝑥 =
Relationship between material properties
𝜇(3𝜆 + 2𝜇)
𝜆
𝐸=
𝑣=
𝜆+𝜇
2(𝜆 + 𝜇)
𝐸
𝐸𝑣
𝜇=
𝜆=
(1 + 𝑣)(1 − 2𝑣)
2(1 + 𝑣)
Note: These have been derived with uniaxial stress conditions
Thermal effect does not change Hooke’s Law (from stress to
strain condition), just add the α(ΔT) component to εxx etc
Don’t forget the VQ/It stuff from last year!
1
(𝐶11 − 𝐶12)
2
𝜎1 = 2𝜇𝜀1 + 𝜆𝐼1𝜀
Principal planes {𝜎2 = 2𝜇𝜀2 + 𝜆𝐼1𝜀
𝜎3 = 2𝜇𝜀3 + 𝜆𝐼1𝜀
Where I1E is the first STRAIN invariant εxx + εyy + εzz
𝜎𝑥𝑥 = 2𝜇𝜀𝑥𝑥 + 𝜆𝐼1𝜀 𝜎𝑥𝑦 = 2𝜇𝜀𝑥𝑦
General planes {𝜎𝑦𝑦 = 2𝜇𝜀𝑦𝑦 + 𝜆𝐼1𝜀 { 𝜎𝑦𝑧 = 2𝜇𝜀𝑦𝑧
𝜎𝑧𝑧 = 2𝜇𝜀𝑧𝑧 + 𝜆𝐼1𝜀 𝜎𝑧𝑥 = 2𝜇𝜀𝑧𝑥
Hooke’s Law (from strain to stress)
𝐸
[(1 − 𝑣)𝜀𝑥𝑥 + 𝑣(𝜀𝑦𝑦 + 𝜀𝑧𝑧)]
𝜎𝑥𝑥 =
(1 + 𝑣)(1 − 2𝑣)
𝐸
𝜎𝑥𝑦 =
𝜀𝑥𝑦
1+𝑣
(Replace xx with yy, zz, xy, yz and zx as needed)
𝑣
(𝜀𝑥𝑥 + 𝜀𝑦𝑦)
𝜀𝑧𝑧 = −
1−𝑣
The σxx equation and the εxx equation (directly above) can help
us arrive at the following equation:
𝐸
[𝜀𝑥𝑥 + 𝑣𝜀𝑦𝑦] − ONLY FOR PLANE STRESS
𝜎𝑥𝑥 =
1 − 𝑣2
Derived by: Establish σzz equation, equate to 0, make εzz the
subject. Then establish σxx, sub in εzz and derive σxx in terms of
εxx and εyy.
Hooke’s Law (from stress to strain)
1
1+𝑣
1
𝜀𝑥𝑥 = [𝜎𝑥𝑥 − 𝑣(𝜎𝑦𝑦 + 𝜎𝑧𝑧)] 𝜀𝑥𝑦 =
𝜎𝑥𝑦 =
𝜎𝑥𝑦
𝐸
𝐸
2𝐺
(Replace xx with yy etc as needed)
From plane stress to plane strain:
𝑅𝑒𝑝𝑙𝑎𝑐𝑒 𝐸 𝑎𝑛𝑑 𝑣 𝑤𝑖𝑡ℎ
𝐸
𝑣
𝑎𝑛𝑑
1 − 𝑣2
1−𝑣
From plane strain to plane stress:
𝐸(1 + 2𝑣)
𝑣
𝑅𝑒𝑝𝑙𝑎𝑐𝑒 𝐸 𝑎𝑛𝑑 𝑣 𝑤𝑖𝑡ℎ
𝑎𝑛𝑑
(1 + 𝑣)2
1+𝑣
MODELLING AND SOLUTION
After defining BC, solve for displacements, strains or stresses.
Displacement method, derive: (usually get rid of ρ, equals 0)
𝜕𝐼
(𝜆 + 𝜇)
+ 𝜇∇2 𝑢 + 𝜌𝑓𝑥 = 𝜌𝑎𝑥
𝜕𝑥
𝜕𝐼
(𝜆 + 𝜇)
+ 𝜇∇2 𝑣 + 𝜌𝑓𝑦 = 𝜌𝑎𝑦
𝜕𝑦
𝜕𝐼
2
(𝜆
{ + 𝜇) 𝜕𝑧 + 𝜇∇ 𝑤 + 𝜌𝑓𝑧 = 𝜌𝑎𝑧
Where:
𝜕2
𝜕2
𝜕2
∇2 = 2 + 2 + 2
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑢 𝜕𝑣 𝜕𝑤
𝑎𝑛𝑑 𝐼 𝜀 =
+
+
𝜕𝑥 𝜕𝑦 𝜕𝑧
After obtaining u,v and w, one can calculate strain using straindisplacement method and then stress using Hooke’s Law.
P
2 R
  zz ( rd )( dr ) 

0 0
2 R
   zz rddr
0 0
M yy  M 
2 R
  r zz ( rd )( dr ) cos
0 0
Stress method: solve for stress, then strain & displacements
Strain method: solve for strain, then stress & displacements
M xx 
Remember: plain strain deformation is independent of z and θ,
applicable to very long structures.
Q 
2 R
  r zz ( rd )( dr ) sin   0
0 0
2 R
   z r
2
ddr
0 0
 zr  0
At r=R  L  z  0 , 0    2 (cylindrical surface):
 rr   r   rz  0
3D POLAR SYSTEM
Equilibrium equations:
𝜕𝜎𝑟𝑟 1 𝜕𝜎𝑟𝜃 𝜕𝜎𝑟𝑧 𝜎𝑟𝑟 − 𝜎𝜃𝜃
+
+
+
+ 𝜌𝑓𝑟 = 𝜌𝑎𝑟
𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑧
𝑟
𝜕𝜎𝑟𝜃 1 𝜕𝜎𝜃𝜃 𝜕𝜎𝜃𝑧 2𝜎𝑟𝜃
+
+
+
+ 𝜌𝑓𝜃 = 𝜌𝑎𝜃
𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑧
𝑟
𝜕𝜎𝑟𝑧 1 𝜕𝜎𝜃𝑧 𝜕𝜎𝑧𝑧 𝜎𝑟𝑧
+
+
+
+ 𝜌𝑓𝑧 = 𝜌𝑎𝑧
𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑧
𝑟
NOTE: For pure bending, there is no shear force, shear strain,
shear stress and transverse normal stress.
𝜀(𝑙𝑎𝑡𝑒𝑟𝑎𝑙) = 𝑐𝑦
𝜀(𝑎𝑥𝑖𝑎𝑙) = −𝑣𝑐𝑦
ALSO: To eliminate rigid body motion, assume to fix one end of
the beam/ structure.
Strain-displacement relations:
1 𝜕𝑣 1 𝜕𝑢 𝑣
𝜀𝑟𝜃 = ( +
− )
2 𝜕𝑟 𝑟 𝜕𝜃 𝑟
1 1 𝜕𝑤 𝜕𝑣
𝜀𝜃𝑧 = (
+ )
2 𝑟 𝜕𝜃 𝜕𝑧
1 𝜕𝑢 𝜕𝑤
{ 𝜀𝑧𝑟 = 2 ( 𝜕𝑧 + 𝜕𝑥 )
𝜕𝑢
𝜀𝑟𝑟 =
𝜕𝑟
1 𝜕𝑣 𝑢
𝜀𝜃𝜃 =
+
𝑟 𝜕𝜃 𝑟
𝜕𝑤
𝜀𝑧𝑧 =
{
𝜕𝑧















Hooke’s Law for Polar Coordinates remains the same
Can convert between a plane-stress and plain-strain
problem by replacing the respective E and v
σrr and σθθ are independent of material properties (for
hollow cylinder). Displacement & strain aren’t.
For very long, plain strain cylinder:
 rr

  
u

Solution of a thermal static deformation problem without body
forces and surface stresses but with a steady field of temp
change ΔT is equivalent to a solution of a statics problem
subjected to a set of body forces and surface stresses.
𝐸𝛼 𝜕(∆𝑇)
1 − 2𝑣 𝜕𝑥
𝐸𝛼 𝜕(∆𝑇)
𝐵𝑜𝑑𝑦 𝑓𝑜𝑟𝑐𝑒𝑠 𝑓𝑦 = −
1 − 2𝑣 𝜕𝑦
𝐸𝛼 𝜕(∆𝑇)
{ 𝑓𝑧 = − 1 − 2𝑣 𝜕𝑧
𝑓𝑥 = −
𝐸𝛼
(∆𝑇)𝑙 = 𝑙𝜎𝑥𝑥 + 𝑚𝜎𝑥𝑦 + 𝑛𝜎𝑥𝑧
1 − 2𝑣
𝐸𝛼
𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝜎 𝜎"𝑛𝑦 =
(∆𝑇)𝑚 = 𝑙𝜎𝑦𝑥 + 𝑚𝜎𝑦𝑦 + 𝑛𝜎𝑦𝑧
1 − 2𝑣
𝐸𝛼
{ 𝜎"𝑛𝑧 = 1 − 2𝑣 (∆𝑇)𝑛 = 𝑙𝜎𝑧𝑥 + 𝑚𝜎𝑧𝑦 + 𝑛𝜎𝑧𝑧
𝜎"𝑛𝑥 =
p R 2  p0 R02  pi  p0 Ri2 R02
 i i2

,
R0  Ri2
R02  Ri2 r 2
pi Ri2  p0 R02
R02  Ri2
THERMAL ANALYSIS

 pi  p0 Ri2 R02 ,
R02  Ri2  r 2
(1   )(1  2 )  pi Ri2  po Ro2  (1   )   pi  po Ri2 Ro2  1
 R 2  R 2 r  E 
r,
E
Ro2  Ri2
o
i




v  0,
Plane stress to plane strain:
𝐸, 𝑣 & 𝛼 𝑎𝑟𝑒
Plain stresses = plain strain
RECAP ST VENANT PRINCIPLE (POLAR)
x
Q
M
M
P
P
Q
Plane strain to plane stress:
x
z
dr
d rd
𝐸
𝑣
,
𝑎𝑛𝑑 (1 + 𝑣)𝛼
2
1−𝑣 1−𝑣
𝐸, 𝑣 & 𝛼 𝑎𝑟𝑒
y
(1 + 𝑣)𝛼
𝐸(1 + 2𝑣)
𝑣
,
𝑎𝑛𝑑
2
(1 + 𝑣)
1+𝑣
1 + 2𝑣
Some retarded polar equilibrium equation:
y
At z=0, z= L 0  r  R , 0    2 (flat end faces):
 dA  ( rd )( dr )
𝜕 2 1 𝜕𝑢 𝑢
𝜕(∆𝑇)
+
− 2 = −𝑓𝑥 = (1 + 𝑣)𝛼
2
𝜕𝑟
𝑟 𝜕𝑟 𝑟
𝜕𝑟
THERMAL FIT OF A HOLLOW DISK ONTO A SHAFT
BC:
At r = Ri and Ro, σrr = σrθ = 0
At r = Ri, ΔT = ΔTi
At r = Ro, ΔT = ΔTo
𝐻𝑒𝑎𝑡 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛:
1 𝑑 𝑑∆𝑇
[𝑟
]=0
𝑟 𝑑𝑟
𝑑𝑟
9 ( R1  R2 )  P 
 


16 R1R2  E * 
1
1
𝜀𝑟𝑟 = (1 + 𝑣)𝛼[∆𝑇 − 2 ∫(∆𝑇)𝑟𝑑𝑟] + 𝐴 − 𝐵 2
𝑟
𝑟
𝑅𝑖
q  q0 1 
(1 + 𝑣)(𝛼)
1
𝜀𝜃𝜃 =
∫(∆𝑇)𝑟𝑑𝑟 + 𝐴 + 𝐵 2
2
𝑟
𝑟
{
3
Where q0
𝐸𝛼
𝐸𝛼
𝐸𝛼 𝐵
𝜎𝑟𝑟 = − 2 ∫(∆𝑇)𝑟𝑑𝑟 +
𝐴−
𝑟
1−𝑣
1 + 𝑣 𝑟2
𝑅𝑖
𝑟
{
𝐸𝛼
𝐸𝛼
𝐸𝛼 𝐵
𝜎𝜃𝜃 = 2 ∫(∆𝑇)𝑟𝑑𝑟 − 𝐸𝛼(∆𝑇) +
𝐴−
𝑟
1−𝑣
1 + 𝑣 𝑟2
𝐴=
𝑅𝑖
𝑅𝑜
𝐵=
𝑅𝑖
b2 
(1 + 𝑣)𝑅𝑖 2
∫ (∆𝑇)𝑟𝑑𝑟
𝑅𝑜 2 − 𝑅𝑖 2
𝑅𝑖
3 
𝑟
1
𝐷
𝑢(𝑟) = (1 + 𝑣)𝛼 ∫(∆𝑇)𝑟𝑑𝑟 + 𝐶𝑟 +
𝑟
𝑟
𝑅𝑖
𝑅𝑜
𝑅𝑖
b2 
1 − 𝑣12 1 − 𝑣22
+
)
𝐸
𝐸
3 R1R2
P
4 ( R1  R2 ) E *
q  q0 1 
r
2

2a
9 1  P 
 


16 R1  E * 
4

R1
p
E*
b
b
E2,2
q  q0 1 
𝑞𝑜 2 =
x2
b2
Where q0 
2
E*p
R1
𝑥2 2
𝑞 = 𝑞𝑜 (1 − 2 )
𝑏
1 𝑅2 − 𝑅1
(
) 𝑝𝐸 ∗
𝜋 𝑅𝑖𝑅2
E1,1
P
AIRY STRESS FUNCTION
R2
𝜕2∅
𝜕𝑦 2
𝜕2∅
𝜎𝑦𝑦 = 2
𝜕𝑥
𝜕2∅
𝜎𝑥𝑦 =
{
𝜕𝑥𝜕𝑦
𝜎𝑥𝑥 =
P
E2,2
(Distribution of contact stress)
3
R2
1  R1  R2 

E * p
  R1R2 
R1
6  R  R2 
 E *2 P
 3  1
  R1R2 
3
P
a  R1
4 E*
q
1
E1,1
P
A sphere in contact with flat half space
( R2   )
3
q0
R1
4 𝑅1𝑅2 𝑝
𝑏2 =
𝜋 𝑅2 − 𝑅1 𝐸 ∗
2
3
Where q0
E1,1
b2
−1
2
a2
p
Cylinder with large concave cylindrical surface
Two balls in contact
9 ( R1  R2 )  P 


16 R1R2  E * 
x
2
Cylinders with flat half-space
When the size of the contact area is much smaller than the
characteristic dimension of a component, the component
can be considered to be semi-infinite bounded by a surface.
3 
R1R2
p
 ( R1  R2 ) E *
4
2
2 p 1  12  2 R1
 1  2  2 R2

ln

0
.
407

 0.407 


 ln

  E1  b
E2 
b


q02 
CONTACT PROBLEMS
a3 
a2
Where
𝑅𝑖
𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑌𝑜𝑢𝑛𝑔′ 𝑠 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝐸 ∗ = (
r2
6  R  R2 
 E *2 P
 3  1
  R1R2 
q  q0 1 
𝑅𝑜
Required temp: Ri + u(r=Ri) ≥ R shaft


R1
l
𝛼(1 + 𝑣)𝑅𝑖 2
𝐷=
∫ (∆𝑇)𝑟𝑑𝑟
𝑅𝑜 2 − 𝑅𝑖 2
𝛼(1 − 𝑣)
𝐶=
∫ (∆𝑇)𝑟𝑑𝑟
𝑅𝑜 2 − 𝑅𝑖 2
E1,1
P
Two Parallel Cylinders
𝑅𝑜
1−𝑣
∫ (∆𝑇)𝑟𝑑𝑟
𝑅𝑜 2 − 𝑅𝑖 2
2
2
𝑅𝑖
𝑟
3
Where q0
3
𝑟
𝑟
2
6  1
 3   E *2 P
q  q0 1  2
a
  R1 
A sphere in contact with concave half space ( R2   R2 )
3 R1R2
P
a3 
4 ( R1  R2 ) E *
R2 “ – ”
r2
∇2 ∇2 ∅ = 0 ∇2 =

2
R1
R2
1 𝜕∅ 1 𝜕 2 ∅
+
𝑟 𝜕𝑟 𝑟 2 𝜕𝜃 2
𝜕2∅
𝜎𝜃𝜃 = 2
𝜕𝑟
𝜕 1 𝜕∅
{ 𝜎𝑟𝜃 = − 𝜕𝑟 (𝑟 𝜕𝜃 )
𝜎𝑟𝑟 =
𝜕2
𝜕2
𝜕2 1 𝜕2
1 𝜕2
+ 2 𝑂𝑅
+
+ 2 2
2
2
2
𝜕𝑥
𝜕𝑦
𝜕𝑟
𝑟 𝜕𝑟
𝑟 𝜕𝜃
Define BC, select stress function, apply BC to determine
unknowns in stress function, differentiate stress function to find
stresses.
Stress concentration around circular hole
b
T
a

1
 xx  xx   yy  yy   zz zz  2( xy  xy   yz  yz   zx zx )
2
U
T


 
2
2
𝜎𝑟𝑟 = 𝜎𝑥𝑥𝑐𝑜𝑠 𝜃 + 𝜎𝑦𝑦𝑠𝑖𝑛 𝜃 + 2𝜎𝑥𝑦𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
1
= 𝑇(1 + 𝑐𝑜𝑠2𝜃)
2
𝜎𝑟𝜃 = (𝜎𝑦𝑦 − 𝜎𝑥𝑥)𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃 + 2𝜎𝑥𝑦(cos 2 𝜃 − sin2 𝜃)
1
= − 𝑇𝑠𝑖𝑛2𝜃
2

T  a 2  T  3a 4 4a 2 
 rr  1  2   1  4  2  cos 2
2
r  2
r
r 


T  a 2  T  3a 4 

   1  2   1  4  cos 2
2
r  2
r 


4
2
   T 1  3a  2a  sin 2
 r
2 
r4
r 2 

When cos 2  1 ,   reaches the maximum
  r  a  
Thus
and stress concentration factor k:
max
 
 k nominal
(k=3 in this small hole case)
PLASTICITY AND FAILURE

Considering the principal directions as the coordinate axes, a
special oriented plane in which a normal vector makes equal
angles with each of the principal axes, i.e. having direction
cosines equal to l = m = n = |1/√3|, is called an octahedral
plane.
Octahedral normal stress
1
3
 o   nn  ( 1   2   3 )
𝑃𝑜𝑤𝑒𝑟 𝑙𝑎𝑤: 𝜎 = 𝐴𝜀 𝑛
𝑅𝑎𝑚𝑏𝑒𝑟𝑦 − 𝑂𝑠𝑔𝑜𝑜𝑑 𝑀𝑜𝑑𝑒𝑙: 𝜎 =



Distortion energy: External work = Internal strain energy

𝜎
𝜎 𝑛
+𝑘( )
𝐸
𝐸
Tresca yield criterion: σ1 − σ3 = Y
Von Mises criterion: (σ1 − σ2)2 + (σ2 − σ3)2 + (σ3 − σ1)2 = 2Y2
Failure theories (ductile)
 Max shear stress theory: | σ1 − σ3 | ≤ Y/α (Tresca)
 Max distortion energy theory (Von Mises):
2Y 2
( 1   2 ) 2  ( 2   3 ) 2  ( 3   1 ) 2 

2
2
( xx   yy ) 2  ( yy   zz ) 2  ( zz   xx ) 2  6( xy
  2yz   zx
)  2Y 2
Failure theories (brittle- no yield)
 Max normal stress theory: | σ1 | ≤ σf/α
 Max normal strain theory: ε1 ≤ εf/α
 σf and εf are failure normal stress and strain
 Safety factor is α
 1   ( 2   3 ) 
It is the mean principal stress or hydrostatic stress. This value is
the same in all eight octahedral planes.
Octahedral shear stress
The octahedral shear stress = resultant shear stress on an
octahedral plane. Total stress on the octahedral plane:
1
(𝜎12 + 𝜎22 + 𝜎32 )
3
The octahedral shear stress can be calculated from the total
stress and normal stress as:
1
𝜏𝑜 = √(𝜎1 − 𝜎2)2 + (𝜎2 − 𝜎3)2 + (𝜎3 − 𝜎1)2
3

Uv represents strain energy density due to volume change
caused by hydrostatic stress and Ud corresponds to the shape
change due to shear stress, called distortion energy density
T  a 2  T  3a 4 
1

1  4 cos    3T
2  a 2  2 
a 
max
 
 3T

1
( 12   22   32 )  2 ( 1 2   2 3   3 1 )
2E
2 1 
1
 K I1 
( 1   2 ) 2  ( 2   3 ) 2  ( 3   1 ) 2
2
6E
 Uv  Ud
U

f

FEM





Geometrical discretization error
Piecewise approximation error
Elemental stiffness matrices, global elemental SM
(expanded), global SM, global equilibrium equation, apply
boundary conditions & solve for disp. and forces
Be sure to swap nodes where needed, ie u1 then u4
DOF = no. of directions in which the nodes can move
 k1
 k1
 k k  k
 1 1 2
 k2
 0
0   u1  
f1(1)   F1 

  
 
 k 2  u2    f 2(1)  f1( 2 )    F2 

k 2  u3   f 2( 2 )   F3 
Bar elements
N1 ( x )  1 
For stuff like 2EA and 3EA, carry the 2 and 3 into the SM so only
EA remains outside. Treat bar elements same way, different SM
Method 1
Disp. (u)
u ( x)  (1  x / L)ui  ( x / L)u j



  E 
u j  ui
L


L
Force or
W=U
E
L
E
 EA 
F  A 
A
  k
L
 L 
Equil. eqn.
ku  f
Stiffness
matrix
k
k
 k
 k  EA  1  1

k  L  1 1 
Method 2
Disp. (u)
u ( )  Ni ( )ui  N j ( )u j  (1   )ui  u j
du d
d 
 Nu   N u  Bu
dx dx
 dx 
  E  EBu



Force or
W=U

1 
1
U uT   BT EB dV u  uTf
2 V
2

Equil. eqn.
ku  f
Stiffness
matrix
k


N 3 ( x) 
3
2
3
2
x2 
L
x2 
L
x3,
L
2
3
x3,
L
N 2 ( x)  x 
N 4 ( x)  
Where l and m are direction cosines from the x and y axis
respectively (ie l = cosθx and m = sinθy)
'
l
 lm   ui   f i ' 
 '   
 lm  m 2   vi   0 
 '  
l2
lm  u j   f j' 
 '
lm
m 2  v j   0 
Beam elements
Shape functions: (where x is a local coordinate)
L
x2 
L
2
x3
L
1
2
x3
L
 N1 ( x )0  N 2 ( x )0  N 3 ( x )( 1)  N 4 ( x )(1)
2
1
 3
  1

  2 x 2  3 x 3     x 2  2 x 3 
L
L
L
  L

 ( 3 x 2  2 x 3  x 2  x 3 )
 3x 3  4 x 2
Find v(x) for all elements involved and plot e.g.:
v( x )   x 3  x 2  x  1
2D Elements
Constant strain triangle (CST or T3) - Linear triangular, 3 nodes
numbered counter clockwise. Linear disp. constant strain. Use
where strain gradient is small, use in mesh transition areas, avoid
using in stress conc. or crucial areas (holes, corners)
v  b4  b5 x  b6 y
(2dof x 3 nodes =
6dof)
Shape function
1
Ni 
(i  i x   i y )
2A
(i=1,2,3)
1 x1 y1
1
A  1 x2 y2
2
1 x3 y3
Strain
Constant strain
u
 xx 
 b2
x
v
 yy 
 b6
y
 u v 
2 xy    
 y x 
 b3  b5
1
( x2 y3  x3 y2 )  ( y2  y3 ) x  ( x3  x2 ) y 
2A
1
( x3 y1  x1 y3 )  ( y3  y1 ) x  ( x1  x3 ) y 
N2 
2A
1
( x1 y2  x2 y1 )  ( y1  y2 ) x  ( x2  x1 ) y 
N3 
2A
N1 
𝑣𝑖
𝐹𝑖
6𝐿 −12 6𝐿
𝑀𝑖
4𝐿2 −6𝐿 2𝐿2 𝜃𝑖
] { } = { 𝐹𝑗 }
−6𝐿 12 −6𝐿 𝑣𝑗
𝑀𝑗
2𝐿2 −6𝐿 4𝐿2 𝜃𝑗
1
x2 
 N1 ( x )v1  N 2 ( x )1  N 3 ( x )v2  N 4 ( x ) 2
Displacement
Linear:
u  b1  b2 x  b3 y
2
1
2
 vi 
 
 i
v ( x )  Nu  N1 ( x ) N 2 ( x ) N 3 ( x ) N 4 ( x ) 
v j 
 j 
 N1 ( x )vi  N 2 ( x ) i  N 3 ( x )v j  N 4 ( x ) j
v ( x )  3x 3  4 x 2
2D Bar Element
12
𝐸𝐼 6𝐿
[
𝐿3 −12
6𝐿
3
Calculating deflection for an element where v2 = -1 and θ2 = 1,
assuming that L=1, P=1, M=1, EI=1/24:
EA  1  1
L  1 1 
 l2
lm

m2
EA  lm
L   l 2  lm

2
 lm  m
2
Linear strain triangle (LST or T6) - Quadratic triangular, quadratic
displacement & fully linear strain
Quadratic:
u  b1  b2 x  b3 y 
b4 x 2  b5 xy  b6 y 2
v  b7  b8 x  b9 y 
b10x 2  b11xy  b12 y 2
(2dof x 6 nodes =
12dof)
N1   ( 2  1)
Fully-linear
 xx  b2  2b4 x  b5 y
N 2   ( 2  1)
N 3  (1     )
 yy  b9  b11x  2b12 y
2(1     )  1
N 4  4
N 5  4 (1     )
N 6  4(1     )
2 xy  b3  b5  
b3  2b10 x  2b6  2b11y
3D Elements





Tetrahedron (tet)- linear 4 nodes (tri pyramid), quadratic 10
Hexahedron (brick)- linear 8 nodes (rect prism), quadratic 20
Penta- linear 6 nodes (tri prism), quadratic 15 nodes
Each node has 3 dof
4-node tet element is a constant strain element, do not use
for high stress/strain gradients
Linear
Hexahedron
Bi-linear quadratic element (Q4)
Jacobian matrix is the transitional conversion between different
matrices used to obtain strain results through the chain rule.
(Def: relation of derivatives with respect to different coordinate
systems) Also, something about Gaussian integration points
Bilinear
Half-linear
1
N1  (1   )(1   ) ,
4
1
N 2  (1   )(1   ) ,
4
1
N 3  (1   )(1   ) ,
4
1
N 4  (1   )(1   )
4
u  b1  b2 x 
b3 y  b4 xy
v  b5  b6 x 
b7 y  b8 xy
(8 d.o.f.)
u
 b2  b4 y
x
v
 yy 
 b7  b8 x
y
v u
2 xy 

x y
 b6  b8 y  b3  b4 x
 xx 
Quadratic quadrilateral element (Q8)
Quadratic
b12 x 2  b13xy  b14 y 2
1
N1   (1   )(  1)(    1)
4
1
N 2   (1   )(  1)(    1)
4
1
N 3   (1   )(  1)(1     )
4
1
N 4   (1   )(  1)(1     )
4
 b15x 2 y  b16 xy 2
N5
(16 d.o.f.)
N6
u  b1  b2 x  b3 y 
b4 x  b5 xy  b6 y 
2
2
b7 x 2 y  b8 xy 2
v  b9  b10 x  b11 y 
N7
N8



1
 (1   2 )(1   )
2
1
 (1   )(1   2 )
2
1
 (1   2 )(1   )
2
1
 (1   )(1   2 )
2
Half-quadratic
 xx  b2 x  2b4 x  b5 y 
2b7 xy  b8 y 2
 yy  b11 y  b13 y  2b14 y
 b15x 2  2b16 xy
 xy  b3  b5 x  2b6 y 
b7 x 2  2b8 xy 
b10 x  2b12 x  b13 y
 2b15xy  b16 y 2
Q4 and T3 used together in mesh with linear elements
Q8 and T6 applied in mesh with quadratic elements
Quadratic elements preferred for stress analysis due to high
accuracy and flexibility in modeling complex geometry
Gaussian points:
m,n
11
22
33
𝐴
, 
0.0
± √3/3 = ± 0.5774
0
±√0.6 = ± 0.7746
𝐴
𝐴
Wi
1
0.8888
0.5555
𝐴
𝑥 = ∑ 𝑁𝑖𝑥𝑖 𝑦 = ∑ 𝑁𝑖𝑦𝑖 𝑢 = ∑ 𝑁𝑖𝑢𝑖 𝑣 = ∑ 𝑁𝑖𝑣𝑖
𝑖=1
𝑖=1
𝑖=1
𝑖=1
Where x, y are natural co-ord. systems and u, v are disp. fields
and A = 6 for T6, 4 for Q4 and 8 for Q8.
Isoparamatric Elements: The same shape functions are used as
for the displacement field, but with z and w added and A = 8
1
N1 ( , ,  )  (1   )(1   )(1   ),
8
1
N3 ( , ,  )  (1   )(1   )(1   ),
8
1
N5 ( , ,  )  (1   )(1   )(1   ),
8
1
N 7 ( , ,  )  (1   )(1   )(1   ),
8
1
N 2 ( , ,  )  (1   )(1   )(1   ),
8
1
N 4 ( , ,  )  (1   )(1   )(1   ),
8
1
N 6 ( , ,  )  (1   )(1   )(1   ),
8
1
N8 ( , ,  )  (1   )(1   )(1   ),
8
Jacobian Matrix
 x
 

x
J
 
 x

 
y

y

y

8
z   Ni xi
   i 1 
z   8 Ni
 
xi
   i 1 
z   8 N
i

    xi
 i 1
N
 i yi
i 1
8
N
 i yi
i 1
8
8
N
  i yi
i 1
8
N

 i zi 

Ni 
  zi 

i 1
8
N 
  i zi 
i 1

i 1
8
Axisymmetric Problems

Involves both rotational structures AND axisymmetric
loading
 Cylindrical coordinate system (x,y,z) → (r,θ,z)
 Disp. field: u(r,z) w(r,z) (2dof each node due to axisymmetry)
v = εrθ =εθz = 0
 Strain displacement relation:
u (r , z )
u (r , z )
w(r , z )
w(r , z ) u (r , z )
 rr 
,  
,  zz 
,  rz 

r
r
z
r
z
Stress-strain relation:


0  
1 
 rr 
 
  rr 
 
1



0
E
  

  



  


1


0
 zz  (1   )(1  2 ) 
1  2   zz 
0
0
 rz 
 
 0
2   rz 

EXAMPLES
(D) Torsion of circular shaft: u=v=w=σrr=σrθ=σrz=σzz=σzr=0
2 R
(A) In the CST element below, use shape function to determine
the displacements uA, uB, uD at the mid points of each edge if
nodal displacements are found as; u1=1, u2=3, u3=2. Go on to
determine uc at the centroid of the triangular element.
u  N1u1  N 2u2  N 3u3
v  N1v1  N 2v2  N 3v3
Using N1= 1/2A shape functions where A = 1/2 x 1 x 1 = 0.5
Point A (0.5,0):
1
(1  1  0  0)  (0  1)  0.5  (0  1)  0  0.5
2  0.5
1
(0  0  0  1)  (1  0)  0.5  (0  0)  0  0.5
N 2 (0.5,0) 
2  0.5
1
(0  0  1  0)  (0  0)  0.5  (1  0)  0  0
N 3 (0.5,0) 
2  0.5
 r
(B) The 8-node iso-paramatric brick element as shown in
Cartesian coordinate system (x,y,z) below is mapped to the
natural coordinate (,,). Use shape function to determine (1)
Cartesian coordinate of centroid (2) displacement u, v, w at the
centroid if the nodal displacement is:
d T  0 0 d3
d4
Non-zero nodal displacements:
0 0 d7
d8 124
d3  1,2,1 ,
d4   2,0,2, d7  2,2,2 , d8   1,1,2
1
1
1
1
 (1  0)(1  0)(1  0)(0)  (1  0)(1  0)(1  0)( 2)  (1  0)(1  0)(1  0)( 2)  (1  0)(1  0)(1  0)(0)
8
8
8
8
1
1
1
1
 (1  0)(1  0)(1  0)(0)  (1  0)(1  0)(1  0)(3)  (1  0)(1  0)(1  0)(3)  (1  0)(1  0)(1  0)(0)
8
8
8
8
2 2
3 3
10
x0  0    0  0    0 
 1.25
8 8
8 8
8
The Cartesian coordinate for centroid: x0=1.25 y0=1.375 z0= 1.75
Apply similar method to find displacement.
(C)
L
, w0

1  v 1 u v  1  z
z 
 r  2  r  r   r   2  L  0  L   0






1  1 w v  
 
r
 z  
2  r  z  2 L


1  u w 
000
 zr   
2  z x 

G

r
 z  2G z 
L

 rr   r   zz      zr  0
2 R
 r
 z ddr 
2
0 0
2 R
 r
0 0
2
2  G

 G 
r ddr  

  d
 L 
 L 0

2 LQ
GR
4


LQ
G R / 2
4


R
r
3
0
 G  1 4
dr  
2 R  Q
 L  4
LQ
GJ
(E) Inner radius, outer member Q: Use σθθ equation for innear &
outer pressure, set outer pressure = 0 and find p for required
value of σθθ. For temperature: uT=R2-R3
R  R3
uT
  r  R2 
 2
 T
R2
R2
T 
u  0,0,0, 0,0,0, 1,2,1,  2,0,2, 0,0,0, 0,0,0, 2,2,2,  1,1,2T
x0 (  0,  0,   0)  N1x1  N 2 x2  N 3 x3  N 4 x4  N5 x5  N 6 x6  N 7 x7  N8 x8
zr
u

 rr  r  0

1 v u

 0
  
r  r

w

 zz    0
x  0,0,0, 2,0,0, 2,1,0, 0,3,0, 0,0,4, 3,0,3, 3,3,3, 0,4,4T
Using N1 = 1/8 shape functions,
u  0, v 
0 0
N1 (0.5,0) 
Similar for centroid (x=1/3 and y =1/3)
 z ddr  Q
2
u2
0.007424

 12.69 o C
R2 50.002424  11.7  10 6
(F) Strain in CST element: ε = Nu for both elements, then sketch.
Remember u = (u1, v1, u2, v2 etc)
(G) Shaft τ = (T/J)ρ where ρ = radius, J = polar moment of inertia
Beams σ = (-My)/I where M = bending moment & I = 2nd moment
of inertia. Also, shear stress τ = (VQ)/It
Combined loads σ = F/A - My/I & shear stress τ = VQ/It ± Tρ/J
Deformation: Bars δ = FL/AE and shaft φ = TL/GJ (angle of twist)
Beams: v = ∫∫ M(x)/EI dxdx + Cx + D where C and D need to be
determined using boundary conditions.
Slope θ = dv/dx = ∫ M(x)/EI dx + C