Transcript Slide 1

Theories of Failure
Failure of a member is defined as one of two conditions.
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.
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.
Several theories are available however, only four important theories are
discussed here.
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.
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.
 max 
 max
2
In case of Biaxial stress state
 max 
 max   min
 max   y
1   2

2

1   2
2
y
2
2
1   2   y
 (3)
 (2)
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.
400 + 200 lb
200 + 500 lb
x 
c
I
Mc
I
d
2
d 4
64
M B  600 6  3600 lb.in
Mc  700 6  4200 lb.in
x 
x 
4200 d
d 4
64
42800
d3
y 0
2
 xy
Tr

J

 xy
4800 d
d 4
 yx
2
32
 xy
x
24,480

d3
 yx
T  (500  200)  16
 300 16  4800 lb.in
OR
 ( 400  200)  24
 200 24  4800 lb.in
 xy
24,480

d3
 xy
42,800
x 
d3
As we know
 1   2  2 max
  x  y 
   xy 2
 1   2  2 
 2 
2
And
y
F .O.S . 
 max
 yield
F .O.S . 
2
( 1   2 )
2
 yield 36000
1   2 

FOS
3
2
2
2
2
 42800  24,480
1   2  2 


3 
3
 2d   d

36,000
 42800  24,480
2 


3 
3
3
 2d   d

2
12,000
 42800  24,480
 


3 
3
2
 2d   d

2
2
 42800  24,480
3610  


3 
3
 2d   d

6
d  1.76' '
2
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 stress,
1   ult
(1)
 2   ult
(2)
Problem 02:The solid circular shaft in Fig. 1 (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 Principal stress theory together with a safety
factor of 3.
400 + 200 lb
200 + 500 lb
Mc
x 
I
d
c
2
I
d 4
64
M B  600 6  3600 lb.in
Mc  700 6  4200 lb.in
x 
4200 d
d 4
64
42800
x 
d3
y 0
2
 xy
Tr

J

 xy
4800 d
d 4
 yx
2
32
 xy
x
24,480

d3
 yx
T  (500  200)  16
 300 16  4800 lb.in
OR
 ( 400  200)  24
 200 24  4800 lb.in
 xy
24,480

d3
 xy
42,800
x 
d3
1 
2 
 x  y
2
 x  y
2
  x  y 
   xy 2
 
 2 
2
  x  y 
   xy 2
 
 2 
2
2
42800  42800  24446.20 
1 
 


3
3 
3
2d
d
 2d  

2
According to maximum normal stress theory .
 ult   1
2
36000 42800  42800  24446.20 

 


3
3 
3
3
2d
d
 2d  

2
2
42800  42800  24446.20 
12000
 


3
3 
3
2d
d
 2d  

2
6
6
6
915
.
92

10
457
.
96

10
597
.
62

10
144106 


3
6
2d
d
d6
9
1
.
514

10
144106 
d6
d 6  10.51
d  1.48"
Example 03
The solid cast-iron shaft shown in Fig. is subjected to a torque of T =
400 Ib . ft. Determine its smallest radius so that it does not fail
according to the maximum-Principal-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
 max
Tc (400 lb. ft )(12in. / ft )r 055.8 lb.in.



4
J
( / 2)r
r3
Mohr's circle for this state of stress (pure shear) is shown in Fig. . Since
R = max, then
 1   2   max
3055 .8lb.in.

r3
The maximum-Principal-stress theory,, requires
|1| ≤ ult
3055 .8lb.in.
2

20
,
000
lb
/
in
r3
Thus, the smallest radius of the shaft is determined from
3055.8lb.in.
2

20
,
000
lb
/
in
r3
r  0.535 in.
Ans.
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/E).
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.
 max   1   y 
 max


  2   y 
 (1)
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.
σx
σy
σz
x= σx / E
x= -μσy / E
y= -μσx / E
y= σy / E
x= -μσ / E
z
y= -μσ / E
z= -μσx / E
z= -μσy / E
z= σ / E
z
z
x
y
z
=
σx / E
-μσy / E
=
σx / E
-μ / E (σy + σz)
=
σy / E
-μσx / E
=
σy / E
-μ / E (σx + σz)
=
σz / E
-μσy / E
=
σz / E
-μ / E (σx + σy)
-μσz / E
-μσz / E
(2)
-μσx / E
Thus
1


 1   ( 2   3 ) 
E E

2 

2 
 ( 1   3 ) 
E E

3 

 3   ( 2   1 ) 
E E

 (3)
Also
1 
1

 2
E
E
Equating 1 and 4
( For Biaxial)  (4)
 yield   1
 yield
E

1
E
and  2 

2
E
 2
E

 1
E
T hen
 y   1   2
 (5)
 y   2   1
 ( 6)
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).
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
Distortion energy per unit volume is given by
1 u
Ud 
2 y2
6E


Strain energy due to distortion per unit volume for biaxial stress system

1 u 2
Ud 
 1   1 2   22
3E

According to distortion energy theory

1 u
1 u 2
2
(2 y ) 
 1   1 2   22
6E
3E
So, equation becomes
 y2  12  1 2   22
