Machine design theories of failure

1. 1
Theories of Failure

2. Theories of Failure
1. Fracture of the material of which the member is made. This type
of failure is the characteristic of brittle materials.
2. Initiation of inelastic (Plastic) behavior in the material. This type
of failure is the one generally exhibited by ductile materials.
When an engineer is faced with the problem of design using a specific
material, it becomes important to place an upper limit on the state of stress
that defines the material's failure. If the material is ductile, failure is usually
specified by the initiation of yielding, whereas if the material is brittle it is
specified by fracture.
2

3. These modes of failure are readily defined if the member is subjected to a
uniaxial state of stress, as in the case of simple tension however, if the
member is subjected to biaxial or triaxial stress, the criteria for failure
becomes more difficult to establish.
In this section we will discuss four theories that are often used in
engineering practice to predict the failure of a material subjected to a
multiaxial state of stress.
A failure theory is a criterion that is used in an effort to predict the
failure of a given material when subjected to a complex stress
condition.
3

4. i. Maximum shear stress (Tresca) theory for ductile materials.
ii. Maximum principal stress (Rankine) theory.
iii.Maximum normal strain (Saint Venan’s) theory.
iv. Maximum shear strain (Distortion Energy) theory.
Several theories are available however, only four important
theories are discussed here.
4

5. Maximum shear stress theory for Ductile Materials
The French engineer Tresca proposed this theory. It states that a member
subjected to any state of stress fails (yields) when the maximum shearing
stress (τmax)in the member becomes equal to the yield point stress (τy)in a
simple tension or compression test (Uniaxial test). Since the maximum shear
stress in a material under uniaxial stress condition is one half the value of
normal stress and the maximum normal stress (maximum principal stress) is
max, then from Mohr’s circle.
2
max
max

 
5

6. 6

7. If both of principal stresses are of the same sign
tension compression then
σ1 <σy (4)
σ2 <σy (5)
In case of Biaxial stress state
)
3
(
2
2
)
2
(
2
2
2
1
2
1
max
2
1
min
max
max











y
y
y













7

8. A graph of these equation is given in the figure. Any given state of
stress will be represented in this figure by a point of co-ordinate
1 and 2 where 1 and 2 are two principal stresses. If these point
falls within the area shown, the member is safe and if outside then
member fails as a result of yielding of material. The Hexagon
associated with the initiation of yield in the member is known as
“Tresca Hexagon”. (1814-1885). In the first and third quadrant 1
and 2 have the same signs and
8

9. max is half of the numerically larger value of principal stress 1
or 2. In the second and fourth quad, where 1 and 2 are of
opposite sign, max is half of arithmetical sum of the two 1 and
2. In fourth quadrant, the equation of the boundary or limit
(yield/boundary stress) stress line is
1 - 2 = y
And in the second quadrant the relation is
1 - 2 = -y
9

10. y


 

 2
1
y

 

1
y

 
1
y

 
2
y


 
 2
1
y

 

2
10

11. Problem 01
The solid circular shaft in Fig. (a) is subject to belt pulls at each end
and is simply supported at the two bearings. The material has a yield
point of 36,000 Ib/in2• Determine the required diameter of the shaft
using the maximum shear stress theory together with a safety factor of
3.
11

12. 400 + 200 lb
200 + 500 lb
12

13. 0
42800
64
2
4200
.
4200
6
700
.
3600
6
600
64
2
3
4
4













y
x
x
B
x
d
d
d
in
lb
Mc
in
lb
M
d
I
d
c
I
Mc






13
in
lb
OR
in
lb
T
d
d
d
J
Tr
xy
xy
.
4800
24
200
24
)
200
400
(
.
4800
16
300
16
)
200
500
(
480
,
24
32
2
4800
3
4




















14. 3
480
,
24
d
xy 

3
800
,
42
d
x 

14

15.  2
2
2
1
max
2
1
2
2
2
know
we
As
xy
y
x
















 




3
36000
2
)
(
2
.
.
.
.
.
.
And
2
1
2
1
max






FOS
S
O
F
S
O
F
yield
yield
y








15

16. '
'
76
.
1
480
,
24
2
42800
10
36
480
,
24
2
42800
2
000
,
12
480
,
24
2
42800
2
3
000
,
36
480
,
24
2
42800
2
2
3
2
3
6
2
3
2
3
2
3
2
3
2
3
2
3
2
1



























































d
d
d
d
d
d
d
d
d


16

17. Problem 02
The state of plane stress shown occurs at a critical point of a steel
machine component. As a result of several tensile tests, it has been found
that the tensile yield strength is y = 250 MPa for the grade of steel used.
Determine the factor of safety with respect to yield, using (a) the
maximum-shearing-stress criterion, and (b) the maximum-distortion-
energy criterion.
17

18. 18

19. SOLUTION
Mohr's Circle. We construct Mohr's circle for the given state of stress
and find
MPa
OC y
x
ave 20
)
40
80
(
)
( 2
1
2
1





 


MPa
FX
CF
R
m 65
)
25
(
)
60
(
)
(
)
( 2
2
2
2







Principal Stresses
MPa
CA
OC
a 85
65
20 






MPa
CA
OC
b 45
65
20 






a. Maximum-Shearing-Stress Criterion. Since for the grade of steel used
the tensile strength is ay = 250 MPa, the corresponding shearing stress at
yield is
MPa
MPa
Y
Y 125
)
250
(
2
1
2
1


 
 19

20. MPa
For m 65

 
2
.
19
.
65
125
. 

 S
F
MPa
MPa
S
F
m
Y


20

21. b. Maximum-Distortion-Energy Criterion. Introducing a factor of safety
into Eq. (7.26), we write
2
2
2
.









S
F
Y
b
b
a
a





For a = +85 MPa, b = -45 MPa, and y = 250 MPa, we have
      
2
2
2
.
250
45
45
85
85 









S
F

19
.
2
.
.
250
3
.
114 
 S
F
S
F
21

22. Comment. For a ductile material with y = 250 MPa, we have drawn the
hexagon associated with the maximum-shearing-stress criterion and the
ellipse associated with the maximum-distortion-energy criterion. The
given state of plane stress is represented by point H of coordinates a =
85 MPa and b = -45 MPa. We note that the straight line drawn through
points O and H intersects the hexagon at point T and the ellipse at point
M. For each criterion, the value obtained for F.S. can be verified by
measuring the line segments indicated and computing their ratios:
    19
.
2
.
2
.
19
. 



OH
OM
S
F
b
OH
OT
S
F
a 22

23. 23

24. The solid shaft shown in Fig. 10-41a has a radius of 0.5 in. and is made
of steel having a yield stress of y = 36 ksi. Determine if the loadings
cause the shaft to fail according to the maximum-shear-stress theory
and the maximum-distortion-energy theory.
Example 10-12
Solution
The state of stress in the shaft is caused by both the axial force and the
torque. Since maximum shear stress caused by the torque occurs in the
material at the outer surface, we have
24

25. ksi
in
in
in
kip
J
Tc
ksi
in
kip
A
P
xy
x
55
.
16
2
/
.)
5
.
0
(
.)
5
.
0
.(
.
25
.
3
10
.
19
.)
5
.
0
(
15
4
2










The stress components are shown acting on an element of material at
point A in Fig. 10-41b. Rather than using Mohr's circle, the principal
stresses can also be obtained using the stress-transformation equations,
Eq.9-5.
2
2
2
,
1
2
2
2
,
1
)
55
.
16
(
2
0
10
.
19
2
0
10
.
19
2
2






 













 









 xy
y
x
y
x
25

26. = -9.55 ± 19.11
CTI = 9.56 ksi
CT2 = -28.66 ksi
Maximum-Shear-Stress Theory. Since the principal stresses have
opposite signs, then from Sec. 9.5, the absolute maximum shear stress
will occur in the plane, and therefore, applying the second of Eq. 10-27,
we have
 
36
2
.
38
36
66
.
28
56
.
9
?
2
1





 Y



26

27. Thus, shear failure of the material will occur according to this theory.
Maximum-Distortion-Energy Theory. Applying Eq. 10-30, we have
Using this theory, failure will not occur.
      
 
1296
1187
)
36
(
66
.
28
66
.
28
56
.
9
56
.
9
)
(
2
?
2
2
2
2
2
1
2
1








 y





27

28. Maximum Principal Stress theory or
(Rankine Theory)
According to this theory, it is assumed that when a member is
subjected to any state of stress, fails (fracture of brittle material or
yielding of ductile material) when the principal stress of largest
magnitude. (1) in the member reaches to a limiting value that is equal
to the ultimate normal stress, the material can sustain when subjected to
simple tension or compression.
The equations are shown graphically as
)
1
(
1 ult

  )
2
(
2 ult

 
28

29. ult


These equations are shown graphically if the point obtained by plotting
the values of 1 + 2 falls within the square area the member is safe.
29

30. It can be seen that stress co-ordinate 1 and 2 at a point in the material
falls on the boundary for outside the shaded area, the material is said to
be fractured failed. Experimentally, it has been found to be in close
agreement with the behavior of brittle material that have stress-strain
diagram similar in both tension and compression. It cab be further
noticed that in first and third quad, the boundary is the save as for
maximum shear stress theory.
Problem
A thin-walled cylindrical pressure vessel is subject to an internal
pressure of 600 lb/in2. the mean radius of the cylinder is 15in. If the
material ha a yield point of 39,000lb/in2 and a safety of 3 is employed,
30

31. determine the required wall thickness using (a) the maximum normal
stress theory, and (b) the Huber-von Mises-Hencky theory.
P = 600psi.
r = 15"
y = 39000psi
F.O.S = 3
t = ?
According to

c
 Circumferential stress / girth stress
where
Pr
t
c 

31

32. By comparing 1 = c.
Thus, According to the normal stress theory, maximum Principal stress
should be equal to yield stress/FOS i.e.
|1| ≤ ult |2| ≤ ult
f = maximum (|1|, |2|, |3| - ult = 0)
t
t
t
stress
al
lengitudin
For
t
t
c
l
c
4500
2
"
15
600
2
Pr
9000
"
15
600










32

33. As y = 39000psi
.
69
.
0
9000
13000
9000
3
39000
Ans
t
t
t




Problem 02:-
The solid circular shaft in Fig. 18-12(a) is subject to belt pulls at
each end and is simply supported at the two bearings. The material has
a yield point of 36,000 Ib/in2• Determine the required diameter of the
shaft using the maximum normal stress theory together with a safety
factor of 3. 33

34. 400 + 200 lb
200 + 500 lb
34

35. 35

36. 0
42800
64
2
4200
.
4200
6
700
.
3600
6
600
64
2
3
4
4













y
x
x
B
x
d
d
d
in
lb
Mc
in
lb
M
d
I
d
c
I
Mc






36

37. in
lb
OR
in
lb
T
d
d
d
J
Tr
xy
xy
.
4800
24
200
24
)
200
400
(
.
4800
16
300
16
)
200
500
(
480
,
24
32
2
4800
3
4



















xy

yx

yx

xy

3
480
,
24
d
xy 
 3
800
,
42
d
x 

x

37

38.  
 2
2
2
2
2
1
2
2
2
2
xy
y
x
y
x
xy
y
x
y
x




















 











 



2
3
2
3
3
1
20
.
24446
2
42800
2
42800















d
d
d

According to maximum normal stress theory .
2
3
2
3
3
1
20
.
24446
2
42800
2
42800
3
36000
















d
d
d
ult 

38

39. "
48
.
1
51
.
10
10
514
.
1
10
144
10
62
.
597
10
96
.
457
2
10
92
.
915
10
144
20
.
24446
2
42800
2
42800
12000
6
6
9
6
6
6
6
6
3
6
6
2
3
2
3
3



























d
d
d
d
d
d
d
d
d
39

40. Mohr’s Criterion:-
In some material such as cast iron, have much greater strength
in compression than tension so Mohr proposed that is 1st and third
quadrant of a failure brokes, a maximum principal stress theory was
appropriate based on the ultimate strength of materials in tension or
compression. Therefore in 2nd and 4th quad, where the maximum shear
stress theory should apply.
Pure shear is one in which x and y are equal but of opposite sense.
40

41.  c
ult

Failure or strength envelope
Pure shear
Smaller tension strength
Largest compressive strength
41

42. tension
compression
42

43. Problem 3:-
In a cast-iron component the maximum principal stress is to be limited
to one-third of the tensile strength . Determine the maximum value of
the minimum principal stresses ,using the Mohr theory. What would be
the values of the principal stresses "associated with a maximum shear
stress of 390 MN/m2? The tensile and compressive strengths of the cast
iron are 360 MN/m2 and 1.41 GN/m2 respectively
43

44. stress
Shearing
fourth
and
2
stress
Normal
quadrant
third
and
1
criterion
s
Mohr'
per
As
?
nd
st
2



quadrant

2
)
(
1
/
120
3
360
3
that
indicated
As
m
MN
t
ult





2
6
2
/
10
1414
/
360
m
MN
m
MN
c
ult
t
ult






44

45. By using principal stress theory
2
2
2
1
/
414
/
360
)
(
)
(
m
MN
m
MN
As
b
a
c
ult
t
ult
ult
ult














2
1 /
120 m
MN



Maximum principal stress = 360/3 = 120 MN/m2 (tension). According
to Mohr's theory, in the second and fourth quadrant
From (a) and (b) 45

46. Therefore
2
2
2
/
940
1
1410
360
120
m
MN
and 



 

The Mohr's stress circle construction for the second part of this problem
is shown in Fig. 13.7. If the maximum shear stress is 390 MN/m2, a
circle is drawn of radius 390 units to touch the two envelope lines. The
principal stresses can then be read off as +200 MN/m2 and -580
MN/m2•
1
1
1
2
1
2
1




c
ult
t
ult
c
ult
t
ult
and








46

47. c
ult
 t
ult

360

t
ult

1410

c
ult

390
max 

47

48. Fig 13.7 48

49. Now from second state
2
1
2
1
max
2
1
2
1
2
1
c
ult
2
1
t
ult
2
max
2
2
1414
380
quad
fourth
in
and
theory
stress
shearing
By
/
390






























R
Also
Thus
m
MN
49

50. Example 10-11
The solid cast-iron shaft shown in Fig. 10-40a is subjected to a torque
of T = 400 Ib . ft. Determine its smallest radius so that it does not fail
according to the maximum-normal-stress theory. A specimen of cast
iron, tested in tension, has an ultimate stress of (σult)t = 20 ksi.
Solution
The maximum or critical stress occurs at a point located on the surface
of the shaft. Assuming the shaft to have a radius r, the shear stress is
3
4
max
.
.
8
.
055
)
2
/
(
)
/
.
12
)(
.
400
(
r
in
lb
r
r
ft
in
ft
lb
J
Tc 





50

51. 51

52. 52

53. Mohr's circle for this state of stress (pure shear) is shown in Fig. 10-
40b. Since R = max, then
The maximum-normal-stress theory, Eq. 10-31, requires
|1| ≤ ult
3
max
2
1
.
.
8
.
3055
r
in
lb



 


2
3
/
000
,
20
.
.
8
.
3055
in
lb
r
in
lb

Thus, the smallest radius of the shaft is determined from
.
.
535
.
0
/
000
,
20
.
.
8
.
3055 2
3
Ans
in
r
in
lb
r
in
lb


53

54. The solid shaft shown in Fig. 10-41a has a radius of 0.5 in. and is made
of steel having a yield stress of y = 36 ksi. Determine if the loadings
cause the shaft to fail according to the maximum-shear-stress theory
and the maximum-distortion-energy theory.
Example 10-12
Solution
The state of stress in the shaft is caused by both the axial force and the
torque. Since maximum shear stress caused by the torque occurs in the
material at the outer surface, we have
54

55. ksi
in
in
in
kip
J
Tc
ksi
in
kip
A
P
xy
x
55
.
16
2
/
.)
5
.
0
(
.)
5
.
0
.(
.
25
.
3
10
.
19
.)
5
.
0
(
15
4
2










The stress components are shown acting on an element of material at
point A in Fig. 10-41b. Rather than using Mohr's circle, the principal
stresses can also be obtained using the stress-transformation equations,
Eq.9-5.
2
2
2
,
1
2
2
2
,
1
)
55
.
16
(
2
0
10
.
19
2
0
10
.
19
2
2






 













 









 xy
y
x
y
x
55

56. = -9.55 ± 19.11
CTI = 9.56 ksi
CT2 = -28.66 ksi
Maximum-Shear-Stress Theory. Since the principal stresses have
opposite signs, then from Sec. 9.5, the absolute maximum shear stress
will occur in the plane, and therefore, applying the second of Eq. 10-27,
we have
 
36
2
.
38
36
66
.
28
56
.
9
?
2
1





 Y



56

57. Thus, shear failure of the material will occur according to this theory.
Maximum-Distortion-Energy Theory. Applying Eq. 10-30, we have
Using this theory, failure will not occur.
      
 
1296
1187
)
36
(
66
.
28
66
.
28
56
.
9
56
.
9
)
(
2
?
2
2
2
2
2
1
2
1








 y





57

58. Maximum Normal Strain or Saint Venant’s
Criterion
In this theory, it is assumed that a member subjected to any state of
stress fails (yields) when the maximum normal strain at any point
equals, the yield point strain obtained from a simple tension or
compression test (y = σy/).
Principal strain of largest magnitude |max| could be one of two
principal strain 1 and 2 depending upon the stress conditions acting in
the member . Thus the maximum Principal strain theory may be
represented by the following equation. 58

59. )
1
(
2
max
1
max










y
y






As stress in one direction produces the lateral deformation in the other
two perpendicular directions and using law of superposition, we find
three principal strains of the element.
x=
y=
z=
σx
σx / E
-μσx / E
-μσx / E
σy
μσy / E
σy / E
-μσy / E
σz
–μσz / E
-μσz / E
σz / E
x=
y=
z=
x=
y=
z= 59

60. x = σx / E -μσy / E -μσz / E
= σx / E -μ / E (σy + σz)
y = σy / E -μσx / E -μσz / E
= σy / E -μ / E (σx + σz)
z = σz / E -μσy / E -μσx / E
= σz / E -μ / E (σx + σy)
(2)
Thus
)
3
(
)
(
)
(
)
(
1
2
3
3
3
1
2
2
3
2
1
1




































E
E
E
E
E
E
60

61. Also
)
6
(
)
5
(
Then
and
4
and
1
Equating
)
4
(
)
2
(
1
2
y
2
1
y
1
2
2
2
1
1
2
1
1
































E
E
E
E
E
d
For
E
E
y
y
61

62. The yield surface ABCD is the straight under biaxial tension or biaxial
compression Individual principal stresses greater than σy can occur
without causing yielding.
62

63. Maximum Shear Strain Energy (Distortion
Energy Criterion (Von MISES Criterion)
According to this theory when a member is subjected to any state of
stress fails (yields) when the distortion energy per unit volume at a
point becomes equal to the strain energy of distortion per unit volume at
failure (yielding) in uniaxial tension (or compression).
The distortion strain energy is that energy associated with a change in
the shape of the body.
The total strain energy per unit volume also called strain energy
density is the energy in a body stored internally throughout its volume
due to deformation produced by external loading. If the axial stress 63

64. arising in a tension test is “” and the corresponding axial strain is ε
then the work done or a unit volume of the test specimen is the product
of mean value of the force per unit area (/2) times the displacement in
the direction of the force, or ε. The work is thus
)
1
(
2


U
This work is stored as internal strain energy.
When an elastic element subjected to triaxial loading as show in Fig the
stresses can be resolves in to three principal stresses σ1,σ2 and σ3. where
1,2 and 3 are the principal axes. These three principal stresses will be
accompanied by these principal strain related to the stresses by
equations 3,4and 5 64

65. If it is assumed that loads are applied gradually and simultaneous then
stresses and strain will increase in the same manner. The total strain
energy per unit volume in the sum of energies produced by each of the
stresses (as energy is a scalar quantity )
E
u
E
)
( 3
2
1
1







E
u
E
)
( 3
1
2
2







E
u
E
)
( 2
1
3
3







(2)
(3)
(4)
65

66. Where ε1, ε2, and ε3, are the normal strain in the direction of principal
stresses respectively of strain are expressed in term of stresses than
equation (5) taken the following form
)
5
(
,
2
1
,
,
2
1
,
,
2
1
3
3
2
2 




 


U













































E
E
E
E
E
E
E
E
E
U 2
1
3
3
3
1
2
2
3
2
1
1
2
1
2
1
)
(
2
1 














Which can be reduced to
  )
6
(
)
(
2
2
1
3
2
3
1
2
1
3
3
2
2
2
1 








 





E
U
66

67. As the deformation of a material can be separated in to two parts 
(a) Change in volume, (b) change in shape or distortion.
Similarly the total strain energy can be broken in to two parts. One part
representing the energy needed to cause volume change of the element
with no change in shape (Uv) and the other part representing the energy
needed to distort the element (Ud).
)
7
(
d
v U
U
U 

67

68. 1
3
1
(a)



(b)
1- 
3- 
2- 
(c)
68
The principal stresses σ1,σ2 and σ3 of Fig. (01) can be resolved
in to two states of stresses in Fig. (02) b & c. the state of stress
shown in Fig. b represents a hydrostatic stress condition in
which all these principal stress are equal to the quantity σ.

69. Some materials were subjected to hydrostatic pressure resulting in
appreciable changes in volume but no change in shape and no failure by
yielding. Therefore the remaining portion of the stress (1 - ), (2 - )
and (3 - ) will result in distortion only (No volume change) and the
algebraic sum of the three principal strains produced by the three
principal stresses (1 - ave), (2 - ave) and (3 - ave) must be equal to
zero. i.e
(ε1+ε2+ε3)d=0 (9)
Expressing strains in term of stresses.
69

70.        
 
     
 
     
  0
2
1
3
1
3
2
3
2
1
3
2
1












































 d
E
   
 
   
 
   
 


















2
2
2
2
1
3
1
3
2
3
2
1






























2
2
2
2
1
3
1
3
2
3
2
1

























 6
3
3
3
3
2
2
2
1
1
1 

















 6
3
2
2
2 3
3
2
2
1
1 







70

71. a) One part that causes Volumetric (Uv)
b) Change and one causes distortion an (Ud)
)
4
(
d
v U
U
U 

a) As according to theory of distortion only energy due to
distortion is responsible for failure. Some experimental evidence
supports this assumption some materials were subjected to
hydrostatic pressure result in appreciable changes in volume but
no change in shape and no failure by yielding. The hydrostatic
pressure is the average of three principal stresses 1 2 and 3
known as average stress.
)
5
(
3
3
2
1
max






 71

72. This principal strains ε in the material.
Due to there Principal stresses (1, 2, 3) three principal strains ε1, ε2,
and ε3, are produced. The state of stress in Fig below will result in
distortion only (No volume change) if sum of three normal strains is
zero. That is
       
   
)
6
(
0
2
2
2
1
)
(
2
1
3
3
1
2
3
2
2
1
3
2
1 











































u
E
E
Which eh reduces to
 
)
5
(
3
/
0
3
)
2
1
(
3
2
1
3
2
1

















72

73. The normal strains corresponding to these stresses are found from three
dimension from of Hook’s law given by the following equation.
)
7
(
)
2
1
(
E


 

Strain energy resulting from these stresses or strains can be obtained by
subject through these values is
)
8
(
.
2
3
.
2
1
.
2
1
.
2
1
, 3
2
1
3
2
1


























Uv
73

74. (5)
3
)
(
)
2
1
(
3
)
(
.
2
3 3
2
1
3
2
1 




 





E
u
Uv
2
3
2
1 )
(
6
)
2
1
(


 



E
u
Uv (9)
Strain energy due to distortion
U=Uy + Ud
Ud = U- Uy
 
    












 







2
3
2
1
3
2
3
1
2
1
2
3
2
2
2
1
6
2
1
2
2
1












E
u
u
E
Ud
(10)
  
 
2
3
2
1
3
2
3
1
2
1
2
3
2
2
2
1 2
1
)
(
6
)
(
3
6
1











 








 u
u
E
Ud
74

75. When the third force in the
     
 
2
1
1
3
2
3
2
3
3
2
2
2
2
2
2
1
2
1 2
2
2
6
1











 









E
u
Ud
     
 
2
1
3
2
3
2
2
2
1
6
1





 






E
u
Ud
(12)
(13)
For biaxial stress system, σ3 = 0
 
2
2
2
1
2
1 2
3
1



 



E
u
Ud
(14)
For uniaxial stress system
 
2
1
3
1

E
u
Ud

 (15)
75

76. For Distortion theory the failure occurs when distortion energy of the
member however equal to the strain energy of distortion at failure
(yielding) in uniaxial torsion (or equilibrium). So substituting σ1=σy in
equ (15)
 
2
2
6
1
y
d
E
u
U 

 (16)
And equating it with (13)
     
 
2
1
3
2
3
2
2
2
1
2
6
1
)
2
(
6
1






 







E
u
E
u
y
     2
1
3
2
3
2
2
2
1
2
2 





 





y (17)
76

77. For biaxial stress system
2
2
2
1
2
1
2
2 



 


y (18)
77