notes on theories of failure
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CHAPTER
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Todays Objective
Session 1
Discussion on Assignment
Experimental Demonstration
Break
Session 2
Chapter 4 Start
1Chapter 4: Theories o f Failure
4500 kN = 450 tons approx
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Chapter 4Theories of Failure
2
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3Mohr Circ le of Three Dimens ion
Draw three mohr circles by using the following Matrix of three
dimension on a large graph sheet. Label , 2, 3
Also verify your result of principle stresses by evaluating the roots
of cubic polynomial
3 2
1 2 3
1
2
3
0
xx yy zz
xx xy yy yzxx xz
xy yy yz zzxz zz
xx xy xz
xy yy yz
xz yz zz
I I I
I
I
I
1 2 3
2 4
3 5
1
2
3
4
5
1 2 32 5 6
3 6 8
12053
1
2
0
5
3
xx xy xz
xy yy yz
xz yz zz
a a aa a
a a
for CMS
a
aa
a
a
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4Stress Invariants
3 2
1 2 3
1
2
3
0
xx yy zz
xx xy yy yzxx xz
xy yy yz zzxz zz
xx xy xz
xy yy yz
xz yz zz
I I I
I
I
I
1
2 2 2
2
2 2 2
3 2
yz zx xy
xx yy zz
xy yz zx xx yy yy zz zz xx
xx yy zz xy xz zx xx yy zz
I
I
I
1 1 2 3
2 1 2 2 3 3 1
3 1 2 3
I
I
I
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5
Biaxial Stress State2 2 12
2 11 6
12 6 13
2 2 2 12 11 6
, ,2 11 12 13 6 13
3 2
1 2 3 0I I I
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6Stress
1
2
3
0 0
0 0
0 0
xx xy xz
xy yy yz
xz yz zz
1 2 3
1 2
12
1 3
13
2 3
23
2
2
2
C
C
C
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7Understand ing 3D Stress States
xx xy xz
xy yy yz
xz yz zz
1
2
3
0 0
0 0
0 0
3D Stress System Principal Stresses
2D Views of Principal Stresss
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8Understanding Mohr Circ le Again
The state of plane stress at a point is represented by the
stress element below.
a) Draw the Mohrscircle
b) Determine the principal stresses
c) Determine the maximum shear stresses
d) Draw the corresponding stress elements
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9Draw ing and Labell ing Moh r Circle( yy,xy)=50, 25
(xx
,-xy
)=(-80, -25)
1,= 552,= -85
max= 70
min,= -70
x
y
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102D Princ ipal Stresses to 3D Princ ipal StressesPrincipal Stresses
1= 54.6 Mpa
2= -84.6 MPa
But we have forgotten
about the third principal stress!
Since the element is in plane stress (z= 0), the
third principal stress is zero
1= 54.6 Mpa
2= -84.6 MPa
1= 54.6 Mpa
2= 0 Mpa
3= -84.6 MPa
1 2 3
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11Three Moh r Circles
This means three Mohrs circles can be drawn, each based
on two Principal Stresses
1and 3
1and 2
2and 3
123
3= -84.6 2= 0 1= 54.6
1 2 3
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Pure Uniaxial Tens ion
y= 0
x= P/A 1= x
2
xmax
2= 0
Note when x= Sy,
Sys= Sy/2
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Pure Uniaxial Compress ion
y= 0
x= P/A
2
xmax
1= 0
2=
x
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Pure Tors ion
T
T
J
cTxy
xymax
1= xy2= -xy
CHALK
1
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16Concept of Eigen Vecto r or Eigen Values
Remember the general idea of eigenvalues.
We are looking for values of such that
Ar = r where r is a vector, and A is a matrix
Arr = 0 or (AI) r = 0 where I is the identity matrix
For this equation to be true, either r = 0 or det (AI) = 0.
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17Eigen Value Prob lem1 1
2 2
3 3
1 1
2 2
3 3
0
0
0
1 0 0
0
xx xy xz
xy yy yz
xz yz zz
xx xy xz
xy yy yz
xz yz zz
xx xy xz
xy yy yz
xz yz zz
n n
n n
n n
n n
n n
n n
1
2
3
1
2
3
0
1 0 0
0 0 1 0
0
0
0
xx xy xz
xy yy yz
xz yz zz
n
n
n
n
n
n
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Supplementary Notes for 3D Stress A nalysis 18
0 0
0
1
0 0
0
2
xx xy xz xx xy xz
xy yy yz xy yy yz
xz yz zz xz yz zz
xx xy xz xy xz
xy yy yz yy yz
xz yz zz yz zz
xx xy xz
xy yy yz
xz yz
Expanding by Column
Expanding by Column
0
0 0
0 0
xx xz xy xz
xy yz yy yz
zz xz zz yz zz
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Supplementary Notes for 3D Stress A nalysis 19
0
0 0
0 0
3
0 0
0 0
0 0
xx xy xz xx xz xy xz
xy yy yz xy yz yy yz
xz yz zz xz zz yz zz
xx xy xz xx xy xx xz xy xz
xy yy yz xy yy xy yz yy yz
xz yz zz xz yz xz zz y
Expanding by Column
0
z zz
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Supplementary Notes for 3D Stress A nalysis 20
0
0
xx xy xz
xx xy yy yzxx xz xx yz
xy yy yz
xy yy yz zzxz zz xz zz
xz yz zz
Evaluating
0 0
0 0 0
0 0
xx xy xz xx xy xx xz xy xz
xy yy yz xy yy xy yz yy yz
xz yz zz xz yz xz zz yz zz
0
xx xy xz
xx xy yy yzxx xz
xy yy yz
xy yy yz zzxz zz
xz yz zz
Expanding
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Supplementary Notes for 3D Stress A nalysis 21
00
0
0
xx xy xz
xx xy yy yzxx xz xx yz
xy yy yz
xy yy yz zzxz zz xz zz
xz yz zz
xx xy xz
xx xy yy yzxx xz xx
xy yy yz xy yy yzxz zz xz
xz yz zz
Expanding
2
0
0
0
yy yz
zz yz zz
xx xy xz
xx xy yy yzxx xz
xy yy yz xx yy zz
xy yy yz zzxz zz
xz yz zz
xx xy xz
xx xy xx xz
xy yy yz xx
xy yy xz zz
xz yz zz
2 2 0yy yz
yy zz
yz zz
C
S l t N t f 3D St A l i
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Supplementary Notes for 3D Stress A nalysis 22
0
00
0
xx xy xz
xx xy yy yzxx xz xx yz
xy yy yz
xy yy yz zzxz zz xz zz
xz yz zz
xx xy xz
xx xy yy yzxx xz xx
xy yy yz
xy yy yzxz zz xz
xz yz zz
Expanding
0
0
0
yy yz
zz yz zz
xx xy xz
xx xy yy yzxx xz
xy yy yz xx yy zz
xy yy yz zzxz zz
xz yz zz
Evaluating
C
S l t N t f 3D St A l i
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Supplementary Notes for 3D Stress A nalysis 23
2 2 2
0
xx xy xz
xx xy yy yzxx xz
xy yy yz xx yy zzxy yy yz zzxz zz
xz yz zz
xx xy xz
xx xy yy yzxx xzxy yy yz xx yy zz
xy yy yz zzxz zz
xz yz zz
Evaluating
2 2 2 3
0
0
xx xy xzxx xy yy yzxx xz
xy yy yz xx yy zz
xy yy yz zzxz zz
xz yz zz
Rearrainging
C
S l t N t f 3D St A l i
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Supplementary Notes for 3D Stress A nalysis 24
2 2 2 3
2
0
xx xy xz
xx xy yy yzxx xz
xy yy yz xx yy zz
xy yy yz zzxz zz
xz yz zz
xx xy xz
xx xy yy yzxx xz
xy yy yz xx yy zzxy yy yz zzxz zz
xz yz zz
Rearrainging
3
3 2
3 2
1 2 3
0
0
0
xx xy xz
xx xy yy yzxx xz
xx yy zz xy yy yz
xy yy yz zzxz zz
xz yz zz
I I I
C
THOERIES OF FAILURE
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THOERIES OF FAILURE
Basic idea
Basic idea is that if some combination of the principal stresses
gets too large, the material will fail.
Failure prediction
To predict elastic failure under any condition of applied stressfrom the behaviour of materials in a simple tensile test.
Failure criteria
A number of theoretical criteria exists each seeking to obtainadequate correlation between estimated component life and
that actually achieved under service load conditions for both
brittle and ductile material applications.
25
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Fai lure due to
Overload / under-design
Elastic yielding
Fatigue
Brittle fracture
Ductile rupture
Creep
Corrosion
Buckling
Wear
Vibration
26
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S f Th i f f i l
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Summary of Theor ies of fai lure
Theory Failure Criteria
Maximum principal stress theory
RankineMaximum shear stress theoryTrescaMaximum principal strainSaint VenantTotal strain energy per unit volumeHaighShear strain energy per unit volumevon MisesModified Shear Stress theoryMohr
27
1 yield
1 2 3 yield
1 3 yield
2 2 2 21 2 3 1 2 2 3 3 12 yield
1 2yield
yt yc
2 2 2
1 2 2 3 3 1 2
2 yield
C
1 Maximum princ ipal stress theory
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This theory assumes that when the maximum principal stress in the complex stress
system reaches the elastic limit stress in simple tension, failure occurs. The criterion
of failure is thus
281. Maximum pr inc ipal stress theory RANKINE
1 yield
=y
=y
Since
1 2 3
CH
2 Maximum shear stress theory
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2. Maximum shear stress theory TRESCAThis theory states that failure can be assumed to occur when the maximum shear
stressin the complex stress system becomes equal to that at the yield point in the
simple tensile test.
29
max 1 31
2
3 0
Since the maximum shear stress is half the greatest difference between two
principal stresses therefore criterion of failure becomes
And in simple tensile test all
other stress values are zero
max 1
max 1
max
10
21
2
1
2 yield
= y
= y
1
2
1 31 1
2 2 yield
Equating 1 and 2
1 3 yield
CH3 M i i i l t i th
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3. Maximum pr inc ipal stra in theory SAINT VENANTThis theory assumes that failure occurs when the maximum strain in the complex
stress system equals that at the yield point in the tensile test
30
31 21
E E E
And in simple tensile test all other stress values are zero
We know from chapter of stress analysis
11
1
1
1
0 0
yield
E E E
E
E
1
2
= y
= y31 2 yield
E E E E
1 2 3 yield
Equating 1 and 2
CH4 Total strain energy per unit vo lume theory
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4. Total strain energy per unit vo lume theory HAIGHThe theory assumes that failure occurs when the total strain energy in the complex
stress system is equal to that at the yield point in the tensile test
31
U = The area under the curv e in
the elast ic regio n is cal led the
Elast ic Strain Energ y.
1 1 2 2 3 3
1 1 1
2 2 2
U
For 3D stress state the
strain energy is given as
31 21
32 12
3 2 13
E E E
E E E
E E E
But we know that
CH
Understanding Poissons Ratio
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Understanding Poisson s Ratio
2
11
3
3
22
22
yy E
E
1
2
1 2
21
E
3
2
3 2
23
E
3
2
3
3
2
2
1
1
3 3
33
E
E
1
3
1 3
31
E
2
3
2 3
32
E
1
3
2
1
1 1
11
E
E
2
1
2 1
12
E
3
1
3 1
13
E
CH
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33
Total Strain in y Direction
31 21
31 22
31 2
3
E E E
E E E
E E E
CH
4 Total strain energy per un it vo lume theory
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4. Total strain energy per un it vo lume theory HAIGH34
1 1 2 2 3 3
3 3 31 2 2 1 1 21 2 3
1 1 2 3 2 2 1 3 3 3 1 2
2 2 2
1 1 2 1 3 2 1 2 2 3 3 1 3 2 3
1 1 1
2 2 2
1 1 1
2 2 2
1
2
1
2
U
UE E E E E E E E E
UE
U E
U
2 2 2
1 1 2 1 3 2 1 2 2 3 3 1 3 2 3
2 2 2
1 1 2 1 3 2 2 3 3
2 2 2
1 2 3 1 2 1 3 2 3
2 2 2
1 2 3 1 2 1 3 2 3
1
2
12 2 2
2
1 2 2 22
12
2
E
UE
UE
UE
1
CH
4 Total strain energy per un it vo lume theory
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4. Total strain energy per un it vo lume theory HAIGH 35
2 2 2
1 2 3 1 2 1 3 2 3
2
1 1 1
2
1
2
1 22
10 0 2 0 0 0 0
2
1
21
2 yield
UE
UE
U
E
UE
And in simple tensile test all other stress values are zero
= y
= y2
Equating 1 and 2
2 2 2 21 2 3 1 2 1 3 2 31 122 2
yieldE E
2 2 2 21 2 3 1 2 1 3 2 32 yield
CH5 Shear strain energy per un it volume
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5. Shear strain energy per un it volume VON MISESThis theory states that failure occurs when the maximum shear strain
energy component in the complex stress system is equal to that at the
yield point in the tensile test
36
CH
5 Shear strain energy per un it volume
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37
= +
3
1
2
3
1
2
1 2 3 1 2 3
General State of Stress = Hydrostatic Stress + Deviatoric Stress
Resolving general three-dimensional principal stress state into
hydrostatic and deviatoric components.
5. Shear strain energy per un it volume VON MISES
1 2 33Mean
CH5 Shear strain energy per un it volume
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38
1 2 33
Mean
1 1 2 3 1 2 1 31 1 1
3 3 3
2 1 2 3 2 1 2 31 1 1
3 3 3
3 1 2 3 3 1 3 21 1 1
3 3 3
Writing 1 2 3 in terms of mean stress
5. Shear strain energy per un it volume VON MISES
The mean stress term may be considered as a hydro static or Volum etr ic
stress, equal in all directions..
CH 395 Shear strain energy per un it volume
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5. Shear strain energy per un it volume VON MISESThe strain energy associated with the hydrostatic stress is termed the volumetr ic
strain energy and is found by substituting
1 2 3 1 2 31
3
2 2 21 2 3 1 2 2 3 3 11
22
tUE
in total strain energy equation already discussed in Haighfailure theory
2 2 2
2 2 2 2
12
2
13 2
2
v
v
UE
UE
By replacing each value of stress by mean stress we get volumetric strain
energy
1
CH 405 Shear strain energy per un it volume
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5. Shear strain energy per un it volume VON MISES
2 2
2 2
2
2
1 2 3
13 2 3
2
1 3 62
13 1 2
2
1 13 1 22 3
v
v
v
v
UE
UE
UE
U E
simplifying
t v s
s t v
U U U
U U U
2
1 2 3
1 2
6vU
E
2
Total strain energy = Volumetric Strain Energy + Shear Strain Energy
CH 415 Shear strain energy per un it volume
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5. Shear strain energy per un it volume VON MISES
22 2 2
1 2 3 1 2 2 3 3 1 1 2 3
22 2 2
1 2 3 1 2 2 3 3 1 1 2 3
22 2 2
1 2 3 1 2 2 3 3 1 1 2 3
2 2 2
1 2 3 1 2
1 212
2 6
1 232
6 6
13 2 1 2
6
3 61
6
s t v
s
s
s
s
U U U
UE E
UE E
UE
UE
2
2 3 3 1 1 2 3
2
1 2 3 2
CH 42
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22 2 2
1 2 3 1 2 3 1 2 2 3 3 1
2
1 2 3
2 2 2 2 2 2
1 2 3 1 2 3 1 2 2 3 3 1
2 2 2
1 2 2 3 3 1 1 2 3 1 2 2 3 3 1
2 2 2
1 2 3 1
3 61
6 2
3 3 3 2 2 21
6 2 3 3 3 2 2 2
2 2 2 21
6
s
s
s
U
E
UE
UE
2 2 3 3 1
2 2 2
1 2 2 3 3 1 1 2 3
2 2 2
1 2 3 1 2 2 3 3 1
2 2 2
1 2 3 1 2 2 3 3 1
2 2
2
2 2 2 2 2 21
6 2 2 2 2 2 2sU
E
CH 43
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STRENGTH OF MATERIAL ARSHAD ALI (arshadalibuitems@yahoo.com)
HAPTER
4
THE
ORIES
OF
FALURE
2 2 2
1 2 3 1 2 2 3 3 1
2 2 2
1 2 3 1 2 2 3 3 1
2 2 2
1 2 3 1 2 2 3 3 1
2 2 2
1 2 3 1 2 2 3 3 1
2 2 2 2 2 21 1 2 2 3 3
2 2 2 2 2 21
6 2 2 2 2 2 2
11 2 2 2 2 2 2
6
12 2 2 2 2 2
6
1 26
s
s
s
s
U
E
UE
UE
UE
1 2 2 3 3 1
2 2 2 2 2 2
1 1 2 2 2 3 2 3 1 3 1 3
2 2
12 2 2
6sU
E
CH 44
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STRENGTH OF MATERIAL ARSHAD ALI (arshadalibuitems@yahoo.com)
HAPTER
4
THE
ORIES
OF
FALURE
2 2 2 2 2 2
1 1 2 2 2 3 2 3 1 3 1 3
2 2 2
1 2 2 3 3 1
12 2 2
6
1
6
s
s
U
E
UE
and, since E = 2G (1 + v),
2 2 2
1 2 2 3 3 1
1
12
sU
G
3
CH 455 Shear strain energy per un it volume
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STRENGTH OF MATERIAL ARSHAD ALI (arshadalibuitems@yahoo.com)
APTER
4
THEO
RIES
OF
FAILURE
5. Shear strain energy per un it volume VON MISES
2 2 2
1 2 2 3 3 1
1
12sU
G
And in simple tensile test all other stress values are zero
= y
= y
2 2 2
1 1
2 2
1 1
2
1
21
2
10 0 0 0
12
1
12
12
12
16
1
6
s
s
s
s
s yield
UG
UG
UG
UG
UG
4
CH 465. Shear strain energy per un it volume
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STRENGTH OF MATERIAL ARSHAD ALI (arshadalibuitems@yahoo.com)
APTER
4
THEO
RIES
OF
FAILURE
2 2 2 2
1 2 2 3 3 1
1 1
12 6 yield
G G
2 2 2
1 2 2 3 3 1 2
2 yield
Equating 3 and 4 we get
5. Shear strain energy per un it volume VON MISES
CHA Problem 1: 47
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STRENGTH OF MATERIAL ARSHAD ALI (arshadalibuitems@yahoo.com)
APTER
4
THEO
RIES
OF
FAILU
RE
If the principal stresses at a point in an elastic material are 120 MN/m2 tensile, 180
MN/m2 tensile and 75 MN/m2 compressive, find the stress at the limit of
proportionality expected in a simple tensile test assuming:
(a) Tresca (b) Sain t Venan t (c) Von Mises Assume v = 0.294.
1 3
2
180 75
255 /
yield
yield
yield MN m
2
1
22
2
3
?
180 /
120 /
75 /
yield
MN m
MN m
MN m
(a) Tresca
Data(b) Saint Venant
1 2 3
2
180 0.294 120 0.294 75
166.77 /
yield
yield
yield MN m
2 2 2
1 2 2 3 3 1 2
2 yield
22 2
2
2
180 120 120 75 75 180
2
230.9 /
yield
yield MN m
(c) Von Mises
CHA Problem 2: 48
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STRENGTH OF MATERIAL ARSHAD ALI (arshadalibuitems@yahoo.com)
APTER
4
THEO
RIES
OF
FAILU
RE
A material subjected to a simple tension test shows an elastic limit of 240 MN/m2.
Calculate the factor of safety provided if the principal stresses set up in a complex
two-dimensional stress system are limited to 140 MN/m2 tensile and 45 MN/m2
compressive. The appropriate theories of failure on which your answer should bebased are: (a) Tresca (b) Von Mises.
1 3
240140 45
1.3
yield
n
n
n
2
2
1
2
2
3
240 /
140 /
0
45 /
safety factor =?
yield
MN m
MN m
MN m
n
Now with a factor of safety applied the
design yield point becomes yield/ n
(a) TrescaData
CHA Problem 2 49
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STRENGTH OF MATERIAL ARSHAD ALI (arshadalibuitems@yahoo.com)
APTER
4
THEO
RIES
OF
FAILU
RE
2
2 2 2
1 2 2 3 3 1
22
2 2
2
240140 0 0 45 45 140 2
1.44
yield
n
n
n
Now with a factor of safety applied the design yield point becomes yield/ n
(b) Von Mises
CHA 50
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APTER
4
THEO
RIES
OF
FAILU
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