21205749 theories of failure 2

8/3/2019 21205749 Theories of Failure 2

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THEORIES OF FAILURE

PRESENTED BY-

PANKAJ SHARMA NAGENDRA PALSINGH100106236 100106218


2/23

Theories of Failure

The material properties are usually determined by simpletension or compression tests.

The mechanical members are subjected to biaxial ortriaxial stresses.

To determine whether a component will fail or not, somefailure theories are proposed which are related to theproperties of materials obtained from uniaxial tension orcompression tests.

Initially we will consider failure of a mechanical membersubjected to biaxial stresses


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The Theories of Failures which are applicable for thissituation are:

Max principal or normal stress theory(Rankines theory)

Maximum shear stress theory(Guests orTrescas theory)

Max. Distortion energy theory(Von Mises &Henckys theory)

Max. strain energy theory

Max. principal strain theory


4/23

Ductile materials usually fail byyielding and hence the limiting strength isthe yield strength of material as determinedfrom simple tension test which is assumed

the same in compression also.

For brittle materials limiting strengthof material is ultimate tensile strength in

tension or compression.


5/23

Max. Principal or Normal stress theory(Rankines Theory):

It is assumed that the failure or yield occursat a point in a member when the max.principal or normal stress in the biaxialstress system reaches the limiting strength ofthe material in a simple tension test.

In this case max. principal stress iscalculated in a biaxial stress case and is

equated to limiting strength of the material.


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Maximum principal stress

2

2

122

xy

yxyx

Minimum principal stress

2

2

222

xy

yxyx

For ductile materials

1 should not exceed in tension,

FOS

Syt

For brittle materials

1 should not exceed in tension

FOS

Sut

FOS=Factor of safety

This theory is basically applicable for brittle materialswhich are relatively stronger in shear and not applicable

to ductile materials which are relatively weak in shear.


7/23

The failure or yielding is assumed to take place at a

point in a member where the max shear stress in abiaxial stress system reaches a value equal to shearstrength of the material obtained from simple tensiontest.

In a biaxial stress case max shear stress developed isgiven by

2.Maximum Shear Stress theory (Guests or Trescastheory):

FOSyt

max

wheremax = FOS2

Sy t

This theory is mostly used for ductile materials.

22

max2

xy

yx


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2

..max

stressdirectMinstressdirectMax

CASE1 (First quadrant )

1 and 2 are +ve

yt

yt

Sei

S

1

1131

max

..

222

0

2

CASE2 (Second quadrant)

1 is -ve and 2 is +ve ,Then

2

2

2

22

)(

1max

11212max

ytS

Then

2max

ytS

CASE3 (Third quadrant)

1 is -ve and 2 is more -ve

,Then

yc

yc

Sei

SThen

. 22

2

0

2

)(

max

223

max

CASE4 (Fourth quadrant)1 is +ve and 2 is -ve ,Then

2

2

2

22

)(

max

2121

max

ytS

Then

Assuming that 1>

2>

3and

3 =0

According to the Maximum shear stress theory,

And also

+1

1=Syt

+2

-1

-2

Syc

Syt

Syc

Syto

1=Syc


9/23

It is assumed that failure or yielding occurs at a pointthe member where the distortion strain energy (alsocalled shear strain energy) per unit volume in a biaxialstress system reaches the limiting distortion energy(distortion energy at yield point) per unit volume asdetermined from a simple tension test.

The maximum distortion energy is the differencebetween the total strain energy and the strain energy dueto uniform stress.

3.Max. Distortion energy theory (Von Mises & Henckystheory):


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3.Max. Distortion energy theory (Von Mises &Henckys theory):

The criteria of failure for the distortionenergy

theory is expressed as

Considering the factor of safety

2132322212

1

FOS

Syt

2

13

2

32

2

212

1 ytS


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3.Max. Distortion energy theory (Von Mises & Henckystheory):

A component subjected to pure shear stresses and thecorresponding Mohrs circle diagram is

Y

X

Element subjected to pure shear stresses

o1-2

Mohrs circle for pure shear stresses


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In the biaxial stress case, principal stress 1, 2 arecalculated based on x ,y & xy which in turn are usedto determine whether the left hand side is more thanright hand side, which indicates failure of the

component.

212221 FOS

Syt

From the figure, 1 = -2 = and 3=0Substituting the values in the equation

We get

Replacing by Ssy, we get

3ytS

yt

yt

sy SS

S 577.03


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+1

+2

-1

-2

Syc

Syt

Syc

Syto

Boundary for distortion energy theory under bi axial stresses

Case 1 (First quadrant)

1 and 2 are +ve and equal

to , then

FOS

SFOS

S

yt

yt

212

2

2

1

Case 4 (Fourth quadrant)

1 is +ve and 2 is -ve and equalto , then

FOS

SFOS

SFOS

SFOS

S

yt

yt

yt

yt

577.0

33 2

21

2

2

2

1

21

2

2

2

1

Case 2 (Second quadrant)

1 is -ve and 2 is +ve and equalto , then

FOS

SFOS

SFOS

SFOS

S

yt

yt

yt

yt

577.0

332

21

2

2

2

1

21

2

2

2

1

Case 3 (Third quadrant)

1 is -ve and 2 is +ve andequal to , then

FOS

SFOS

S

yt

yt

212

2

2

1


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Failure is assumed to take place at a point in a memberwhere strain energy per unit volume in a biaxial stresssystem reaches the limiting strain energy that is strainenergy at yield point per unit volume as determinedfrom a simple tension test.

Strain energy per unit volume in a biaxial system is

The limiting strain energy per unit volume for yieldingas determined from simple tension test is

mEU 21

2

2

2

11

2

2

1

4. Max. Strain energy theory (Heighs Thoery):

2

2 2

1

FOS

S

EU

yt


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Equating the above two equations then we get

In a biaxial case 1

, 2

are calculated based as x

, y

& xy

2

212

2

2

1

2

FOS

S

m

yt

It will be checked whether the Left Hand Side ofEquation is less than Right Hand Side of Equation ornot. This theory is used for ductile materials.


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EFOS

S

mE

E

Ey t21

max

It is assumed that the failure or yielding occurs at a point

in a member where the maximum principal (normal) strainin a biaxial stress exceeds limiting value of strain (strain atyield port) as obtained from simple tension test.

In a biaxial stress case

One can calculate 1 & 2 given x , y & xyand checkwhether the material fails or not, this theory is not usedin general as reliable results could not be detained in

variety of materials.

5.Max. Principal Strain theory (Saint Venants Theory):


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Example :1 The load on a bolt consists of an axial pull of

10kN together with a transverse shear force of5kN. Find the diameter of bolt requiredaccording to

1. Maximum principal stress theory2. Maximum shear stress theory

3. Maximum principal strain theory

4. Maximum strain energy theory

5. Maximum distortion energy theoryPermissible tensile stress at elastic limit =100MPa

and Poissons ratio =0.3


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Solution 1

Cross sectional area of the bolt,

Axial stress,

And transverse shear stress,

227854.0

4ddA

2

221/

73.12

7854.0

10mmkN

ddA

P

2

2/365.6

7854.0

5mmkN

dA

Ps


19/23

According to maximum principal stress theory

Maximum principal stress,

According to maximum principal stresstheory, Syt = 1

22

122

xyxx

2

2

122

xy

yxyx

2

21

2

2

2

221

/15365

365.6

2

73.12

2

73.12

mmNd

ddd

mmd

d

4.1215365

1002


20/23

According to maximum shear stress theory

Maximum shear stress,

According to maximum shear stress,

mmdd

Syt

42.132

1009000

2 2max

2

2

max2

xy

yx

2

2

2

2

2

2

2

2

2

2

max

/9000

/9365.673.12

2

mmNd

mmkNddd

xyx


21/23

According to maximum principal strain theory

The maximum principal stress,

And minimum principal stress,

2

2

1 22 xyyxyx

2

2

222

xy

yxyx

2

2

2

1

15365

22 dxy

xx

2

22

2

2

2

22

22

2

/2635

365.6

2

73.1273.12

22

mmNd

dddxy

xx


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And according to maximum principal strain theory,

mmddd

7.12

1003.0263515365

Sm

E

S

mE

E

22y t2

1

y t21


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According to maximum strain energy theory

According to maximum distortion theory

mmd dddd

Sm

yt

78.12

1003.0263515365

2263515365

2

2

22

2

2

2

2

22122

21

mmd

dddd

Syt

4.13

263515365263515365100

22

2

2

2

2

21

2

2

2

1

21205749 theories of failure 2

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