Failure resulting from static load
Chapter 5
What is Failure?
What is Static load?
M S Dasgupta BITS Pilani
1
Terminologies
1. Failure theory (FT) to use depends on material (ductile or brittle) and type
of loading (static or dynamic).
2. Terminology:
•
Su (or Sut) = ultimate strength in tension
•
Suc = ultimate strength in compression
•
Sy = yield strength in tension
•
Sys = 0.5*Sy = yield strength in shear
•
Sus = 0.75*Su = ultimate strength in shear
•
Se = endurance strength  0.5*Su or get from S-N curve
•
S’e = estimated actual endurance strength = Se(ka) (kb) (kc) (kd) - - -
•
S’se  0.577* S’e = estimated actual endurance strength in shear
2
Ductile materials - extensive plastic deformation and
energy absorption (toughness) before fracture
Brittle materials - little plastic deformation and low energy
absorption before failure
3
Ductility and % Elongation
• Ductility is the degree to which a material
will deform before ultimate fracture.
• Percent elongation is used as a measure
of ductility.
• Ductile Materials have %elong.  5%
• Brittle Materials have %elong. < 5%
• For machine members subject to repeated
or shock or impact loads, materials with
4
%elong > 12% are recommended.
DUCTILE VS BRITTLE FAILURE
(a)
Ductile:
warning before
fracture
(b)
(c)
Brittle:
No
warning
5
Failure Prediction Methods
• Ductile materials are designed based on
yield criteria
– Maximum shear stress (MSS) theory
– Distortion energy (DE) theory
– Ductile Coulomb-Mohr (DCM) theory
• Brittle materials are designed based on
fracture criteria
– Maximum normal stress (MNS) theory
– Brittle Coulomb-Mohr (BCM) theory
– Modified Mohr (MM) theory
6
Maximum-Normal-Stress Theory
• The maximum-normal-stress theory states that
failure occurs whenever one of the three
principal stresses equals or exceeds the
strength.
• For principal stress
• σ1 ≥ σ2 ≥ σ3
σ1 ≥ Sut or σ3 ≤ −Suc
7
Maximum-Shear-Stress Theory
 A   B  0,
1   A, 2   B, 3  0
Failure occurs when
the maximum shear
stress in any element
equals or exceeds the
maximum shear
stress in a tension
test specimen.
0  
 
A
B
1 3 
1 

B
3


n
 S
y
n
,

A
Sy
n
,
 1  0,
 2   A,
3 B
S
Sy
 A  0   B,
 1   A ,  2  0,  3  
,
1 3 
y
n
M S Dasgupta BITS Pilani
Sy
n
,
A

B
B

Sy
n
8
Distortion-Energy (DE) Theory
“Failure occurs when the distortion strain
energy per unit volume reaches or exceeds
the distortion strain energy per unit volume
for yield in simple tension or compression”
For a general state of stress, the Distortion-Energy Theory predicts
yielding when Von Mises stress 
1
2
  1   2 2   2   3 2   3   1 2 
'

S
or

 Sy


y
2


For 2D:-
     A B  
'
2
A

1
2 2
B
9
10
SHEAR YIELD STRENGH:
According to DE (von Mises) criterion, substituting the pure
shear state of stress in the 2-D DE criterion, the two
normal stresses being zero,
3
2
xy
 S y  xy 
Sy
 0.577 S y
3
At yield , S sy  0.577 S y
According to the MSS criterion,
S sy  0.5S y
DE criterion predicts the shear yield strength to be 15 percent more than that
predicted by the MSS criterion. Hence MSS is more conservative.
11
Yield Strength Method
• Uniaxial Static Stress on Ductile Materials
Static
Load
Ductile Material
 In tension:
DESIGN:
 max   d 
S yt
N
 In compression:
DESIGN:
 max   d 
S yc
N
For most ductile materials, Syt = Syc
ANALYSIS:
N
ANALYSIS:
N
S yt
 max
S yt
 max
12
Maximum Shear Stress
• Biaxial Static Stress on Ductile Materials
DESIGN:
max  d 
ANALYSIS:
N
S ys
N

Sy
avg, max
2N
S ys
 max
Ductile materials begin to yield when the maximum shear stress in a load-carrying
component exceeds that in a tensile-test specimen when yielding begins.
13
Distortion Energy
• Static
Biaxial or Triaxial Stress on Ductile Materials
Shear
Diagonal
Sy
2
Best predictor of failure for
ductile materials under static
loads or under completely
reversed normal, shear or
combined stresses.
Sy
Sy
1
'  12   22  1 2
’ = von Mises stress
Sy
Distortion Energy
Failure:
’ > Sy
Design:
’  d = Sy/N
ANALYSIS:
N  Sy/’
14
von Mises Stress
• Alternate Form
'   2x   2y   x  y  3 2xy
For uniaxial stress when y = 0,
'    3
2
x
2
xy
(1 > 2 > 3)
• Triaxial Distortion Energy
( 2   1 )  ( 3   1 )  ( 3   2 )
 '
2
2
2
2
15
Comparison of Static Failure Theories:
Maximum Shear – most conservative
16
TORQUE:
Brittle failure or
ductile failure?
Key: is the
fracture
surface on a
plane of max
shear or max
normal stress.
DUCTILE
BRITTLE
17
From previous Tutorial
Shaft Dia = 1.5 cm
Pulley B = 4 cm
Pulley C = 8 cm
M S Dasgupta BITS Pilani
18
As a Design Problem
Material Selection
Applicable formula selection (based on assumptions)
Computation of safe diameter,
Selection from Manufacturer’s catalogue
M S Dasgupta BITS Pilani
19
Real Shaft
20
Stress Concentration
Changes in cross section causes localized stress concentrations and
severity depends on the geometry of the discontinuity and nature of
the material.
21
Stress concentration factor
Kt = max/ o
– max, maximum stress at
discontinuity and o, nominal
stress.
–
Kt, value depends only on
geometry of the part.
22
23
Design improvement to reduce stress concentration
24
Uni-Bi-Tri axial stress?
Stress State ?
25
Maximum Normal Stress
•Uniaxial Static Loads on Brittle Material:
Brittle Material
Static
Load
–In tension:
DESIGN:
 max
Sut
 K t   d 
N
ANALYSIS:
N
S ut
 max
–In compression:
DESIGN:
 max
Suc
 K t   d 
N
ANALYSIS:
N
Suc
 max
26
Ductile/Brittle Coulomb-Mohr Theory
To be applied when the material has unequal strength in tension
and compression, S yt  S yc
e.g. Gray Cast iron materials, for which
S yc  0 . 5 S yt
DCM is a simplification of the Coulomb- Mohr yield criteria.
OR

where either yield strength or ultimate strength can be used
9/2/2023
27
27
Coulomb-Mohr Theory / Internal Friction Theory
For design equations, incorporating the factor of safety n, divide all strengths
by n.
  1
1
St

3
Sc

n
for plane stress (one of the principal stresses is zero) and assuming that σA ≥ σB:
Case 1: σA ≥ σB ≥ 0. Here, σ1 = σA and σ3 = 0. Equation reduces to  A 
Case 2: σA ≥ 0 ≥ σB . Here, σ1 = σA and σ3 = σB , Equation reduces to
A
St
St
n
B

Case 3: 0 ≥ σA ≥ σB . Here, σ1 = 0 and σ3 = σB, Equation reduces to  B  S c
Sc

1
n
n
28
Modified Mohr Method
• Biaxial Static Stress on Brittle Materials
45° Shear Diagonal
2
Sut
Suc
Sut
1
1, 2
Coulomb
Mohr
Failure when outside of shaded area
Suc
Stress concentrations
applied to stresses before
making the circles
Brittle materials often have a
much larger compressive
strength than tensile strength
29
Summary Static Failure Theories:
• Brittle materials fail on planes of max
normal stress:
– Max Normal Stress Theory
– Modified Mohr Theory
• Ductile materials fail on planes of max
shear stress:
– Max shear stress theory
– Distortion energy theory
30
Brittle failure
For the ASTM Grade 30 cast
Iron, find the maximum force
F with the following failure
models;
a) Maximum Principal Stress
b) Coulomb Mohr
c) Modified Mohr
375
At ‘A’ the stresses are
 = 32M/d3 = 32x350F/ 253=0.228F
 = 16T/ d3 = 16x375F/ 253=0.122F
A = 0.281F,
B = -0.053F
ASTM G30 Sut =214MPa Suc=752MPa
MPS-> F = 214/0.281 = 765N (?)
31
Coulomb Mohr
Slope of load line r = B / A = -0.189
SA = Suc Sut / (Suc – rSut) = 752x214/(752+0.189x214)
= 204MPa
SB = rSA = -38.6 MPa
Now: nA = 204MPa => F = 204 / 0.281 = 726
ASTM G30 Sut =214MPa Suc=752MPa
A = 0.281F
32
• Failure resulting from fluctuating
load
Chapter 6
Fluctuating load?
What is special about it?
M S Dasgupta BITS Pilani
33
Fluctuating / Variable load
• Variable loading results when the applied
load or the induced stress on a component is
not constant but changes with time
• In reality most mechanical components
experience variable loading due to
-Change in the magnitude of applied load
-Change in direction of load application
-Change in point of load application
34
Stress variation: Sinusoidal
 min  minimum stress
 max  maximum stress
 r  range of stress   max   min
 m  midrange or mean stress 
 max   min
 a  amplitude or variabl e stress 
2
 max   min
2
Idealized types of cyclic loading:
Completely Reversed
Sinosoidal: mean stress is zero;
equal reversals on both sides; useful
in conducting experiments
Repeated stress: minimum stress
is zero; mean stress equal to half of
the range stress
Fluctuating stress: maximum,
minimum and mean stress are all
non-zero and arbitrary
Result of Fluctuating stress
Fatigue
• Fatigue is a phenomenon associated with
variable loading or more precisely to cyclic
stressing or straining of a material
• ASTM Definition of fatigue
– The process of progressive localized
permanent structural changes occurring in a
material subjected to conditions that produce
fluctuating stresses at some point or points
and that may result in cracks or complete
fracture after a sufficient number of
38
fluctuations. M S Dasgupta BITS Pilani
Fatigue failure in Metals
Crack initiation, propagation and rupture in a shaft subjected to repeated bending
Final rupture occurs
over a limited area,
characterizing a very
small load required
to cause it
Beach
marks
showing the nature
of crack propagation
Crack initiation at
the outer surface
39
Fatigue Life Prediction
predict the failure in number of cycles N to failure for a specific type of
loading
Low cyclefatigue(LCF): 1  N  103 ; High cyclefatigue(HCF): N  103
•
•
•
Stress life methods
– Based on stress levels only
– Least accurate of the three, particularly for LCF
– It is the most traditional because easiest to implement for a wide range of
applications
– Has ample supporting data
– Represents high cycle fatigue adequately
Strain life methods
– Involves more detailed analysis of plastic deformation at localized regions
– Good for LCF
– Some uncertainties may exist in results because several idealizations get
compounded
– Hence normally not used in regular (special occasions)
Linear elastic fracture mechanics methods (LEFM)
– Assumes that crack is already present and detected
– The crack location is then employed to predict crack growth and sudden rupture with
respect to the stress nature and intensity
40
S-N Diagram
The S-N Diagram for steel (UNS G41300), normalized, Sut=812 MPa.
R. R. Moore highspeed rotating
beam machine.
S’e
Endurance Limit,
Non-Ferrous materials tested up to 5*108 cycles
It is the stress at which the
component can sustain
infinite number of cycles41
Sut – S’e relation
for
S ut  1460 MPa
 0 .5 S ut
S e'  
for
S ut  1460 MPa
 700 MPa
S e'  Endurance limit obtained in reverse bending
S e  Endurance limit in the actual loading conditions
42
Se S’e relation
S e  k a kb k c k d k e S
'
e
k a  surface condition modificati on factor
kb  size modificati on factor
kc  load modificati on factor
k d  temperature modificati on factor
ke  reliability factor
k f  miscellaneous effects modificati on factor
43
Surface cond. Mod. factor (ka)
The surface modification factor depends on the quality of the
finish of the actual part surface and on the tensile strength of
the part material.
b
k a  aSut
Table 6.2
Size modification factor, kb
For rotating circular bars in bending and torsion only :
d / 7.620.107  1.24d 0.107 if
kb  
if
 0.859  0.000837d
2.79  d  51 mm
51  d  254 mm
For axial loading no size effect, kb  1.
What happens when bars are not rotating but
say under bending.
Or non-circular bars like square, or I section?
Concept of Equivalent Diameter de
Kb for non-rotating shapes
Effective dimension “de”
obtained by equating the
volume of material stressed
at and above 95 percent of
the maximum stress to the
same volume in the
rotating-beam specimen
Load modification factor, kc
 1, bending

k c   0 . 85 , axial
 0 . 59 , torsion

Actually the kc is sensitive
to Sut of the material. Tables
6-11 to 6-14 (page no. 333)
in Text Book give the
details. The above values
are representative.
Temperature modifying factor, kd
Brittle fracture is a strong possibility when
operating temp is below RT
At temp. higher than RT, yielding should be
investigated first because the yield strength drops
off rapidly with temperature.
 Creep at elevated temperature
Temperature modifying factor, kd
For carbon and alloy steels experimental result
expressed as a fourth-order polynomial curve fit
to the data underlying
k d  0.975  0.432103 TF  0.115105 TF2  0.104108 TF3  0.5951012 TF4
where
70  TF  1000o F
Or interpolate from a chart / table of
operating temp. vs tensile
Reliability factor, ke
ke  1 0.08 z a
Based on standard
deviation of Endurance
strength data
Miscellaneous effects factor, kf
Accounts for
– Residual stress
– Coating failure
– Frettage corrosion material of mating part.
– Synergic effect of corrosion and temperature
where is Se is function of frequency of loading.
Actual / Fatigue stress concentration factor, Kf
Kf is a reduced value of Kt and it is also called fatigue
strength reduction factor
Kf
maximum stress in notched specimen

stress in notch - free specimen
Kf 1 qKt 1 or Kfs 1 qshearKts 1
q  notch sensitivity value(from Fig. 6 - 20 & 6 - 21)
Kt  Theoretical stress concentration factor (geometricfactor)
Stress-concentration factors for a variety of geometries under
different loading conditions can be found in appendix, Table A–15
52
Notch Sensitivity
53
Estimation of Kf
Kf = 1+q(Kt -1).
•When q=0, the material has no sensitivity to notches, Kf=1.
•When q=1, or when notch radius is large for which q is
almost equal to 1, the material has full notch sensitivity, and
Kf = Kt.
•For all grades of cast iron, use q=0.20.
•Use the different graphs to obtain q for bending/axial and
torsional loading.
•Whenever the graphs do not give values of q for certain
combinations of data, use either Neuber equation or
Heywood equation.
54
Estimation of Kf
Use the Neuber equation when the notch is circular/cylindrical.
1
where
a
r
and
K f  1  q  K t  1
a is Neuber constant and is a material constant
a  f ( S ut ), i.e function of ultimate strength.
r  notch radius
For steel, with Sut in kpsi, the Neuber constant can be
approximated by a third-order polynomial fit of data as
100psi = 0.689MPa
1
q
Bending or axial : a  0.246  3.08(10 3 ) Sut  1.51(10 5 ) Sut2  2.67(10 8 ) Sut3
Torsion :
a  0.19  2.51(10 3 ) Sut  1.35(10 5 ) Sut2  2.67(10 8 ) Sut3
55
Estimation of Kf
Use Heywood equation when the notch is NOT circular/cylindrical but is a
tranverse hole or shoulder or groove.
K
f
Kt

2 K t  1 
1
Kt
a
r
where
a values
are given in the Table 6 - 15; page 335
r= hole/ shoulder/groove size
56
57
Goodman Method
Predictor of failure in ductile materials
experiencing fluctuating stress
a
Sn’ = endurance strength
a = alternating stress
m = mean stress
Sy
Yield Line (Langer line)
Sn’
FATIGUE
FAILURE REGION
Goodman Line
a m

1
Sn S u
NO FATIGUE
FAILURE REGION
-Sy
0
Sy
Su
m
58
Goodman Diagram
Safe Stress Line
a
m
1


S n S u
N
a
Sy
Yield Line
Sn’
Sn’ =endurance strength
a = alternating stress
m = mean stress
FATIGUE
FAILURE REGION
Goodman Line
a m

1
Sn S u
Sn’/N
SAFE ZONE
-Sy
0
Su/N
Sy
Su
Safe Stress Line
m
59
Design under cyclic loading
a
2
 Sm 
    1
Se  Sut 
a
Se
Sa Sm

1
Se S yt

m
S yt

1
nf
Sa Sm

1
Se Sut
2
2
 Sa   Sm 
      1
 
 Se   S yt 
60
Different fatigue failure models
a
Se
a
Se

m

m
S yt
S ut

1
nf
 Soderberg line

1
nf
 Modified Goodman line
2
a
 
1
 n f  m  
Se
nf
 S ut 
2
2
 Gerber line
2
a  m   1 
     ASME Elliptic line
   

  
 S e   S yt   n f 
a m 1


 Langer line (only for checking
S yt S yt n y
for static yielding)
61
Important Intersections in First Quadrant
Modified Goodman and
Langer Failure Criteria
M S Dasgupta BITS Pilani
62
Important Intersections in First Quadrant
Gerber and Langer
Failure Criteria
63
Important Intersections in First Quadrant
ASME-Elliptic and Langer
Failure Criteria
64
Variable loading
Determine SF
1.5 mm Radius
30 mm
DIA
42 mm DIA
Titanium alloy
F varies from 20 to 30.3 kN
FORCE
+
-
MAX = 30.3
30.3  20
 5.15 kN
2
30.3  20
mean 
 25.15 kN
2
alt 
MIN = 20
TIME
65
Example: continued.
• Find the mean stress:
25,150 N
m 
 35.6 MPa

(30 mm )2
4
• Find the alternating stress:
a 
5,150 N

(30 mm )2
4
 7.3 MPa
• Stress concentration from Chart: Table:A-15 Pg. 1028
D 42 mm

 1.4;
d 30 mm
r 1.5 mm

 .05
d 30 mm
 K t  2.3
66
Example: continued.
• Se data not available for titanium so we will guess!
Assume Se = 0.5Su
• TRY Ti-0.2 Pd, Su = 340 MPa, Se = 170 MPa
Table A-24 pg 1047
Kt
a
m
1


S e S u
N
2.3(7.3 MPa)
35.6 MPa 1


 .228
1.(.8)(170 MPa) 340 MPa N
kc Axial
1
Reliability 50%
N
 4.386
kb =1
.248
4.386 is good, need further information on Se for titanium.
67
Find a suitable steel for N = 3 & 90% reliable.
3 mm Radius
50 mm DIA
30 mm
DIA
T varies from 848 N-m to 1272 N-m
TORQUE
+
-
MAX = 1272 N-m
1272  848
 212 N  m
2
1272  848
mean 
 1060 N  m
2
alt 
MIN = 848 N-m
TIME
T = 1060 ± 212 N-m
68
Example: continued.
• Stress concentration from pg. 1028 Fig A-15-8
D 50 mm

 1.667;
d 30 mm
r
3 mm

 .1  K t  1.38
d 30 mm
• Find the mean shear stress:
)
Tm 1060 N  m(1000 mm
m
m 

 200 MPa

Zp
(30 mm )3
16
• Find the alternating shear stress:
Ta 212000 N  mm
a 

 40 MPa
3
Zp
5301 mm
69
Example: continued.
• So,  = 200 ± 40 MPa. Guess a material.
Pg1041 Table A-21
TRY: AISI 1040 Q&T 205°C
Su = 779 MPa, Sy = 593 MPa, %E = 19%
Ductile
• Verify that max  Sys: Pure shear loading
max = 200 + 40 = 240 MPa  Sys  600/2 = 300MPa
So this variety is a possibility
• Find the ultimate shear stress:
Sus = .75Su = .75(779 MPa) = 584 MPa
70
Example: continued.
• Sse  295 MPa
• Assume machined surface
• Find actual endurance strength:
(Fig. 5-8)
S’se = kakbkckdkekfSe
= (0.77)(.86)(.59)(.897) 295MPa = 103.4MPa
ka
kb Size {1.24d-0.107
90% Reliability
Average kc
71
Example: continued.
• Goodman:
a
S sn

m
S su
1

N
(Eqn. 5-28)
1.38(40 MPa) 200 MPa 1


 .876
103.4 MPa
584 MPa N
1
N
 1.14
.
876
No Good!!! We wanted N  3
Need a material with Su about 3 times bigger than this
guess or/and a better surface finish on the part, better
notcg sensitivity etc.
72
Example: continued.
• Guess another material.
TRY: AISI 4340 Q&T 700°F
Su = 1720 MPa, Sy = 1590 MPa, %E = 40%
• Find the ultimate shear stress:
Ductile
Sus = .75Su = .75(1720 MPa) = 1290 MPa
• Find actual endurance strength:
S’se
= kakbkckdkekfSe
-> Above gives High FS, can we chose a different material ?
73
Design Factors, N
(a.k.a. Factor of Safety)
FOR DUCTILE
MATERIALS:
•N = 1.25 to 2.0
Static loading, high level of confidence in all design
data
•N = 2.0 to 2.5
Dynamic loading, average confidence in all design
data
•N = 2.5 to 4.0
Static or dynamic with uncertainty about loads,
material properties, complex stress state, etc…
•N = 4.0 or higher Above + desire to provide extra safety
74
Example: continued.
• Goodman:
K t a
Ssn
m
1


S su N
(Eqn. 5-28)
1.38( 40 MPa) 200 MPa
1

  .378
272 MPa
1140 MPa N
1
N
 2.64
.378
No Good!!! We wanted N  3
Decision Point:
• Accept 2.64 as close enough to 3.0?
• Go to polished surface?
• Change dimensions? Material? (Can’t do much better in
steel since Sn does not improve much for Su > 1500 MPa
75
Failure
Theory:
When Use?
Failure When:
Design Stress:
1. Maximum
Normal Stress
Brittle Material/ Uniaxial
Static Stress
2. Yield Strength
(Basis for MCH T
213)
Ductile Material/
Uniaxial Static Normal
Stress
 max  Syt (for tension)
 max  Syc (for compression)
3. Maximum Shear
Stress (Basis for
MCH T 213)
Ductile Material/ Biaxial Static Stress
 max  Sys where Sys  Sy/2
4. Distortion Energy
(von Mises)
Ductile Material/ Biaxial Static Stress
 '   12   22   1 2  Sy
5. Goodman
Method
Ductile Material/
Fluctuating Normal
Stress (Fatigue Loading)
 max  Kt    Sut (for tension)
 max  Kt    Suc (for compression)
Uniaxial:
Ductile Material/
Fluctuating Combined
Stress (Fatigue Loading)
 d  Syt / N (for tension)
 d  Syc / N (for compression)
Note : Syt  Syc for ductile/wrought material
 d  Sys / N where Sys  Sy/2
where  '  von Mises stress
t a
'
sn
S
see Figure 5 - 13
see Figure 5.15
m
S su
S sn'  0.577 S n' and S su  0.75Su
Bi-axial:
K t ( a ) max ( m ) max

1
S sn'
S su
 'd  Sy / N
K t a  m 1


S n'
Su N
K t a  m

1
S n'
Su
Ductile Material/


Failure
Theories Kfor
 STATIC
1
where Loading
Fluctuating Shear Stress
(Fatigue Loading)
 d  Sut / N (for tension)
 d  Suc / N (for compression)
or
where
S sn'  0.577 S n' and S su  0.75Su
K t a  m
1


'
S sn
S su N
where
S sn'  0.577 S n' and S su  0.75Su
K t ( a ) max ( m ) max

1
S sn'
S su
where
S sn'  0.577 S n' and S su  0.75Su
76
Failure
Theory:
When Use?
Failure When:
Design Stress:
1. Maximum
Normal Stress
Brittle Material/ Uniaxial
Static Stress
2. Yield Strength
(Basis for MCH T
213)
Ductile Material/
Uniaxial Static Normal
Stress
 max  Syt (for tension)
 max  Syc (for compression)
 d  Syt / N (for tension)
 d  Syc / N (for compression)
3. Maximum Shear
Stress (Basis for
MCH T 213)
Ductile Material/ Biaxial Static Stress
 max  Sys where Sys  Sy/2
 d  Sys / N where Sys  Sy/2
4. Distortion Energy
(von Mises)
Ductile Material/ Biaxial Static Stress
 '   12   22   1 2  Sy
5. Goodman
Method
a. Ductile Material/
Fluctuating Normal
Stress (Fatigue Loading)
 max  Kt    Sut (for tension)
 max  Kt    Suc (for compression)
: Syt FATIGUE
Syc for ductile/wrought
material
Failure TheoriesNotefor
Loading
b. Ductile Material/
Fluctuating Shear Stress
(Fatigue Loading)
c. Ductile Material/
Fluctuating Combined
Stress (Fatigue Loading)
where  '  von Mises stress
 'd  Sy / N
see Figure 5 - 13
K t a  m 1


S n'
Su N
K t a  m

1
S n'
Su
K t a  m

1
S sn'
S su
 d  Sut / N (for tension)
 d  Suc / N (for compression)
see Figure 5.15
K t a  m
1


'
S sn
S su N
where
S sn'  0.577 S n' and S su  0.75Su
K t ( a ) max ( m ) max

1
S sn'
S su
where
S sn'  0.577 S n' and S su  0.75Su
where
S sn'  0.577 S n' and S su  0.75Su
K t ( a ) max ( m ) max

1
S sn'
S su
where
S sn'  0.577 S n' and S su  0.75Su
77
Failure
Theory:
When Use?
1. Maximum
Normal Stress
Brittle Material/ Uniaxial
Static Stress
2. Yield Strength
(Basis for MCH T
213)
Ductile Material/
Uniaxial Static Normal
Stress
 max  Syt (for tension)
 max  Syc (for compression)
3. Maximum Shear
Stress (Basis for
MCH T 213)
Ductile Material/ Biaxial Static Stress
 max  Sys where Sys  Sy/2
4. Distortion Energy
(von Mises)
Ductile Material/ Biaxial Static Stress
 '   12   22   1 2  Sy
5. Goodman
Method
a. Ductile Material/
Fluctuating Normal
Stress (Fatigue Loading)
b. Ductile Material/
Fluctuating Shear Stress
(Fatigue Loading)
c. Ductile Material/
Fluctuating Combined
Stress (Fatigue Loading)
Failure When:
Design Stress:
 max  Kt    Sut (for tension)
 max  Kt    Suc (for compression)
 d  Sut / N (for tension)
 d  Suc / N (for compression)
 d  Syt / N (for tension)
 d  Syc / N (for compression)
Note : Syt  Syc for ductile/wrought material
 d  Sys / N where Sys  Sy/2
where  '  von Mises stress
see Figure 5 - 13
K t a  m 1


S n'
Su N
K t a  m

1
S n'
Su
K t a  m

1
S sn'
S su
 'd  Sy / N
see Figure 5.15
K t a  m
1


'
S sn
S su N
where
S sn'  0.577 S n' and S su  0.75Su
K t ( a ) max ( m ) max

1
S sn'
S su
where
S sn'  0.577 S n' and S su  0.75Su
where
S sn'  0.577 S n' and S su  0.75Su
K t ( a ) max ( m ) max

1
S sn'
S su
where
S sn'  0.577 S n' and S su  0.75Su
78
What Failure Theory to Use:
79