Notes On Theories of Failure

8/13/2019 Notes On Theories of Failure

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STRENGTH OF MATERIAL ARSHAD ALI ([email protected])

CHAPTER

4

THE

ORIES

OF

FAIL

URE

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

4

THE

ORIES

OF

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URE

Chapter 4Theories of Failure

2


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CHAPTER

4

THEORIES

OF

FAL

URE

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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CHAPTER

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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


7/50 STRENGTH OF MATERIAL ARSHAD ALI ([email protected])

CHAPTER

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THE

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OF

FAL

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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



CHAPTER

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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



CHAPTER

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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

C



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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

C



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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

C



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12

C



CHAPTER

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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

C



CHAPTER

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ORIES

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URE

Pure Uniaxial Compress ion

y= 0

x= P/A

2

xmax

1= 0

2=

x

C



CHAPTER

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THE

ORIES

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URE

Pure Tors ion

T

T

J

cTxy

xymax

1= xy2= -xy

CHALK

1

C



CHAPTER

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ORIES

OF

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URE

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.

C



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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

C



CHAPTER

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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

C



CHAPTER

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THEORIES

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URE


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

C



CHAPTER

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OF

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URE


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

C



CHAPTER

4

THEORIES

OF

FAIL

URE


00

0

0

xx xy xz


xy yy yz


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



CHAPTER

4

THEORIES

OF

FAIL

URE


0

00

0

xx xy xz


xy yy yz


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




CHAPTER

4

THEORIES

OF

FAIL

URE


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




CHAPTER

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THEORIES

OF

FAIL

URE


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



CHAPTER

4

THEORIES

OF

FAIL

URE

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

C

F i l d t


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THEORIES

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URE

Fai lure due to

Overload / under-design

Elastic yielding

Fatigue

Brittle fracture

Ductile rupture

Creep

Corrosion

Buckling

Wear

Vibration

26

C

S f Th i f f i l


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HAPTER

4

THEORIES

OF

FAILURE

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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HAPTER

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THEORIES

OF

FAILURE

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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HAPTER

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THEORIES

OF

FAILURE

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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HAPTER

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THEORIES

OF

FAILURE

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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HAPTER

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THEORIES

OF

FAILURE

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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HAPTER

4

THEORIES

OF

FAILURE

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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HAPTER

4

THEORIES

OF

FAILURE

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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HAPTER

4

THEORIES

OF

FAILURE

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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HAPTER

4

THEORIES

OF

FAILURE

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


= 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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HAPTER

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THEORIES

OF

FAILURE

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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HAPTER

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THEORIES

OF

FAILURE

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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HAPTER

4

THEO

RIES

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FAILURE

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


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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HAPTER

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THEO

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FAILURE

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



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HAPTER

4

THEO

RIES

OF

FAILURE


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



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HAPTER

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THEO

RIES

OF

FAILURE


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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HAPTER

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THE

ORIES

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FALURE

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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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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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



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APTER

4

THEO

RIES

OF

FAILURE


2 2 2

1 2 2 3 3 1

1

12sU

G


= 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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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


CHA Problem 1: 47


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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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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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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

Thanks

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Notes On Theories of Failure