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Failure and Failure Theories:
Stress-Analysis is performed on a
component to determine:
• The required “size or geometry” (design)
• an allowable load (service)
• cause of failure (forensic)
• For all of these, a limit stress or allowable
stress value for the component material is
required.
• Furthermore, a Failure-Theory is needed to
define the onset of failure.
Failure
Component can no longer function as intended.
Failure Mode :
• yielding: a process of global permanent plastic deformation.
Change in the geometry of the object.
• low stiffness: excessive elastic deflection.
• fracture: a process in which cracks grow to the extent that the
component breaks apart.
•buckling: the loss of stable equilibrium. Compressive loading
can lead to bucking in columns.
•creep: a high-temperature effect. Load carrying capacity drops.
Failure Theories:
1. The Tresca Criterion.
•also known as the Maximum Shear Stress criterion.
•yielding when the shear stress reaches its maximum value.
•In a tensile test, this occurs when the diameter of the largest
Mohr’s circle is equal to the tensile yield strength.
•If the principal stresses are ordered such that s1 > s2 > s3,
then the Tresca criterion is expressed as:
s1  s 3
2
 Sy / 2
2. The maximum principal stress criterion.
• states that a tensile yield (fracture) will occur in a
previously un-cracked isotropic material when the
maximum principal stress reaches a critical value.
• The critical value is usually the yield strength, Sy, or
the ultimate tensile strength, Su.
• This criterion does not characterize fracture in brittle
materials with cracks.
smax = Sy (or Su)
3. Von-Mises Criterion
• Also known as the Maximum Energy of Distortion criterion
• based on a more complex view of the role of the principal stress
differences. In simple terms, the von Mises criterion considers the
diameters of all three Mohr’s circles as contributing to the
characterization of yield onset in isotropic materials.
•When the criterion is applied, its relationship to the the
uniaxial tensile yield strength is:
(s 1  s 2 )  (s 2  s 3 )  (s 3  s 1 )  2S
2
2
2
2
y
Von Mises
• For a state of plane stress (s3=0)
s 1  s 1s 2  s 2  S
2
2
2
y
• It is often convenient to express this as an
equivalent stress, se:


1/ 2
1
(s 1  s 2 ) 2  (s 2  s 3 ) 2  (s 3  s 1 ) 2
2
1
2
2
2
2
2
2 1/ 2
ors e 
(s x  s y )  (s y  s z )  (s x  s z )  6( xy   yz   zx )
2
se 


And the von-Mises failure criterion becomes:
se = Sy
Plane Stress Biaxial Failure Envelopes
Von-Mises
sc= critical value of stress
so= yield stress
sut= ultimate tensile stress
4. Mohr’s failure criterion
• Applies to brittle materials much stronger in compression
than in tension.
• Data from tension and compression tests establish limiting
Mohr’s circle envelope.
• For any given stress state, failure will not occur if the
largest Mohr’s circle lies within
the failure envelope.
S
s
Failure envelope
Sc
S
t
Sn
Try it!
Determine if failiure will occur for the following
Complex stress state, given the material has a
tensile yield strength of 250 MPa and an ultimate
tensile strength of 300 MPa.
 120 50 0


s ij   50 80 0 MPa


0 0
 0
Stress
Material Behaviour in Tension
Ultimate Su =300 MPa
300
Yield Sy =250MPa
200
100
0
0.002
0.010
Strain
Solution:
For the following state of stress, find the principal and critical values.
 120 50 0


s ij   50 80 0 MPa


0 0
 0
Tensor shows that:
sz = 0
and  xz =  yz = 0
y
80 MPa
50 MPa
120 MPa
x
The other 2 faces:
y
80 MPa
x
120 MPa
0 MPa
0 MPa
0 MPa
z
0 MPa
z
3-D Mohr’s Circles
 max = 77 MPa
80
60
H
40
20
0
-20
-40
V
-60
-80
-25
0
25 50 75 100 125 150 175
Normal Stress (MPa)
s1 = 0 MPa; s2 = 45 MPa; s3 = 154 MPa;max = 77 MPa.
1. Tresca Criterion:
Sy / 2 = 250/2 =125 MPa
 max = 77 < 125 MPa, SAFE! FS = 1.62
2. Maximum Principal Stress Criterion
Su = 350MPa
smax= s3 = 154 < 350 MPa, SAFE! FS = 2.27
3. Von-Mises Criterion
se 

1
(s 1  s 2 ) 2  (s 2  s 3 ) 2  (s 3  s 1 ) 2
2
=1/2[(0-45)2 +(45-154)2 +(154-0)2 ] 1/2= 137 MPa
< 250 MPa, SAFE! FS = 250/137=1.82

1/ 2
BUCKLING!
Load, P
 EI
2
Pcr 
L 
2
Le
e
I = 2nd Moment of Area
about weak axis.
E = Young’s Modulus
Deflected shape
The effective length, Le, depends on the Boundary Conditions:
Try It!
Find the Buckling load for a pin-ended aluminum
column 3m high, with a rectangular x-section as
shown:
P
Weak axis:
Iyy = 100 (50)3/12
6 mm4
=
1.04x10
100 mm
 (72000)(104
. x10 )
2
Pcr 
50 mm
6
 3000 2
= 82246 N
P