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CHAPTER3 FAILURE THEORIES AND MATERIAL STRENGTHOn the base of engineering mechanics, this chapter will develop more specific

understanding in failure theories and material strength for machinery design. The

failure usually reflect one of the most important perspectives ([pE5spektiv]) in

assessing the safety of mechanical system and components.

3.1 THEORIES OF FATIGUE

The maximum-normal-stress theory (the first strength theory)

The maximum-normal-strain theory (the second strength theory)

The maximum-shear-stress theory (the third strength theory)

The distortion-energy theory (the fourth strength theory)

3.2 BULK STRENGTHS OF THE MACHINE COMPONENTWS

3.2.1 Load and Stress

static load static stress

Load stress

varying load varying (fluctuating) stress

Load: Loads are the external action between two bodies. In terms of the behaviour,

load can be presented in a fashion of force, bending moment and torque.

Static load: The magnitudes or directions of the load remain unchanged or change a

little and slowly within a given time.

Varying load: The magnitudes or direction of the load is continuously time-varying.

Stress: Stresses are the external resistances or forces, which are set up in thematerial when a load acts on a component.

Static stress: static stresses are the stresses whose magnitudes and directions remain

unchanged or change a little and slowly within a given time.

Varying (fluctuating) stress: A varying stress varies its magnitude and directions all

the time. Static or varying load both can cause varying stress.

o t

=constant


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Fig.3.1 Stress-time relationships for some typical stress

(a) Static stress;(b) and (c) Non-symmetrical (or nonsinusoidal) fluctuating stresses;

(d) Sinusoidal fluctuating stress; (e) Repeated alternating (pulsant) stress; (f)

Completely reversed (symmetrical) stress

Stress

Working stress:Using the formula in mechanics of materials calculate the stress on

section plane of element

Calculated stress:Followed the theory of strength, calculate the stress that is equal

to simple tension

Ultimate stress:Some utmost of mechanical characters of material

Strength utmost: off set limit, limit of fatigue,

Allowable stress: Calculate the allowable maximum of stress

Safety factor :

][

lim

=S

Calculated value of safety factor


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1.Cycle Performance

The parameters used to describing the stress condition

min=minimum stress

max=maximum stress

a =alternating stress (or stress amplitude)m=mean stress

r=stress range

s=steady ,or static stress

From fig.3.1, the mean stress and alternating stress can be calculated by

max min

max min

2

2

m

a

+=

=

Cycle Performance :

-1, completely reversed stress

min

max

r

= = 0, repeated alternating stress

1, static stress

2. -N fatigue curve (material fatigue curve )

The -N fatigue curve is shown as follow. It taken the limit of fatigue of material

by the parameter, based on the experiment we can get the curve from.

Repeated alternatingstress

r

=0

Sinusoidalfluctuating

stress

Completelyreversed

stress

r=-1 max

m

T

max

min

a

a

m

max

min

a

a

o t

o t

a

a

min

r

=+1

o

t


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In the origin, the number of stress cycles is N=1/4. It means the material will be

snapped when it is loaded to the maximum, so the value is the strength limit B .


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AB N< 310 , max has little changed, so it can be approximated to static stress

strength.

BC As the N augments, max declines, at the point C,410N ,so this kind of

destroy is called low-cycle fatigue or strain fatigue.

CD: As the N augments, max declines rapidly, this phase is called high cycle

fatigue or finite lifetime. Most fatigue of machine element happen in this phase.

Describe as follow:m

rN C D( )N C N N N =

Explanation: machinery design mainly discuss the high cycle fatigue. Sometimes it

is necessary to do the strength check of static stress for the element which have

prodigious peak value but little action.

Since point D: Curve goes to be horizontal, it means as the N augments no longer

declines, it is called infinite life. 6 710 25 10N :

The equation is: rN r D )N N = >

Because of the value of DN is large, when doing the fatigue experiment we always

ordain a cycle index ON (cycle radix). We use ON and its fatigue limit r

approximatively stand for ,D rN ,m m

rN r 0 N N C = =

max

N

r

N0107

C

DrN

N

BA

N=1/4 104

C

B

103


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interval CD: the relationship between N and fatigue limit rN

0mrN r

N

N

=

m

r

0

rN

N N

=

Among the equation above, the value of is definited by material testing.

As the testing result shown in interval CD, after relevant numbers of varying

stresses the check bar will happen endurance failure. But since the point D, if the

maximum stress less than the stress of the point D, In that way no matter how many

times of the circles, the material will never be destroyed.

High circle fatigue:

CD-----fatigue for finite life

After D----fatigue for infinite life

0mrN r

N

N =

0mn

NK

N= life factor

Eg: In the strength calculation of gear and worm, use the following equation to

calculate the allowable stress.

[ ]r

nKS

=

3. Constant Life Fatigue Curve (Material Fatigue Curve )

Material fatigue curve can also be shown by using the relationship among limit

stress amplitudes in the given N, it is called constant life fatigue curve.

Two reduced method in practical application


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The explanation of stress limit

(1)Every point on the curve stands for limiting fatigue stress of material under the

different cycle performance.

(2)Point A correspond the cycle of r=-1, 1 max0,m a = = =

Point D correspond the cycle of r=0, min 0 max0, / 2 / 2m a = = = =

Line OC correspond the static stress r=1, 0a = ,choose the coordinate values of

point C be equal to the off set limit s of material.

(3) A straight line is drawn from point C with an angle of 45to the mean-stress axis

and stops at point G. Every point on line CG stands for the varying static condition

that max m a s = + = .

(4)The area inside the broken line ' ' ' ' A D G C is yielding and fatigue safety zone.

(5)For simplifying the calculation, use the equation but not the curve.

2.The stress limit of element

Because of the geometrical shape, size dimension,Surface texture and intensifying

factor, the fatigue value of material is above and beyond the Fatigue value of element.

fatigue value of material fatigue limit

fatigue value of element influence coefficient K

Kbending fatigue limit influential factorsKshear fatigue limit influential factors

eg

1

1e

K

=

-1esymmetric cycle bending fatigue limit of element

-1 symmetric cycle bending fatigue limit of material

Amending the stress limit of material diagram AG:

The coordinate value changed:

A(0,-1)A0 or0 1

D(0

2

,

0

2

)D(

0

2

,

0

2K

)

The equation of AD

1 ' 'ae meK = +

Where is a constant with regarding to the material and can be determined by

fatigue tests of following equation:

=


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1 0

0

2

=

----material characteristic

---- material bending characteristic

--- material shearing characteristic



1 0

0

2

=

The equation of line CG( considered by static stress )

' 'ae me s + =

In the condition of shearing stress , the similar form linear equation can get

1 ' '

' '

ae me

ae me s

K

= +

+ =

3.3.2.3 The Strength Calculation of Machine Elements Under the Static Stresses

The characteristic of static stress: max min, holds the value on minimum life section

Or max min, holds the value on minimum life section. Pure static varying stress: one of

the normal or tangential stress Doubleaction static varying stress: both normal and

tangential stress.

The design method: theoretical design

Use for calculation: safety factor checking

lim

caS S

=

Machine elements fatigue value of pure static varying stress when calculating the

endurance bending strength of machine element, first calculate the maximum

(minimum) stress on minimum life section, second calculate the, and then definite the

operating point M on the limit stress curve of element . Which point is the value of the

limit stress depends on the varying discipline of the element working stress.

1.In the case ofr=C(a constant),the cycle performance of the varying stress remains

the same .It is the stress condition of the majority of the rotating shaft.


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a

m

max min

max min

=

+1

1

r

r

=

+'C=

The limit stress of the point M

' ' '

max ae me = +

1( )

m a

a mK

+

=+

1 max

a mK

=+

The calculation of safety factor and stress condition:

lim

caS

=

'

max

max

= 1

a m

SK

= +

Static stress condition:

lim

max

s s

ca

a m

S S

= = =

+

2. In the case of m C = , the mean stress of the varying stress remains the same.

It is the stress condition of the vibrating loaded spring.

The limit stress of point M:

' 1

max 1

( )(1 ) m

e m

K

K K

+ = + =


'

max 1lim

max

( )

( )

m

ca

a m

KS S

K

+

= = = +

Static stress condition:


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lim

max

s s

ca

a m

S S

= = =

+

3.In the case of min C = theminimum stress of the varying stress remains the same.

It is the stress condition of the bolt joining under the axial varying load.

min m aC = =

The limit stress of point M:

' ' ' 1 min

max

2 ( )me ae

K

K

+

= + =+


'

max 1 minlim

max min

2 ( )

( )( )ca

a

KS S

K

+

= = = + +

illustrationwhen the law of how the stress varies is dubious,we always choose the

calculation method of r=C.

1

ca

a m

SK

=+

working stress amplitude stress amplitude influence coefficientS=

the fati gue l i mi t of symmetri c cycl e

Also , for the shear stress, the only thing we have to do is using the instead of in

the equation above.

Machine elements fatigue value of pure unstable varying stress

Eg: The stress condition of automobile leaf-spring is affected by the load, vehiclespeed, pavement, tyre and road sense, etc.


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Fig3-9 regular unstable varying stress diagram

Fig3-10 unstable varying stress on the -N curve

Adopt: fatigue damage accumulation hypothesis

1 2 3

1 2 3

1n n n

N N N

+ + =

Regular unstable varying stress diagram

Unstable varying stress on the N curve

If every circle of stresses make the equicohesive destruct on the material, the

material damage ratio of stress 1 is 11/ N in every circle, so in 1n circles the material


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damage ratio of 1 is 1 1/n N and in 2n circles the material damage ratio of 2 is

2 2/n N .

If the stress is lower than 1 , we can believe that the stress is not damaging.

When the damage ratio is up to 100%,the material happens to endurance failure. So

we obtain:

31 2

1 1 2 3

...... 1z

i

i i

n nn n

N N N N =

= + + + = 10m

i

i

N N

=

where

As the experiment shown

1)When the sequence of stress operation is increased and then decreased, we obtain:

1

1z

i

i i

nN

=

In general case:1

0.7 ~ 2.2z

i

i i

n

N=

=

In limit case: 11 1 2 2

0 1 0 1

1( ... ) 1

z

m

i i

m m m i

z zm m

n

n n nN N

=

+ + + = =

If under these stress the material isnt destroyed,we can obtain:

0 1

1

z

m m

i i

i

n N

=


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of safety factor and strength condition is :

2 2ca

S SS S

S S

= +

Among this:

S only under the normal stress , the calculation value of safety factor

1

a m

SK

=+

Sonly under the shearing stress , the calculation value of safety factor

1

a m

S

K

=+

3.3.3 the Measure for Increasing the Fatigue Value of Machine Element

In the design phase the design measure should be taken to increasing the fatigue

value of machine element :

. Lower the stress concentration as much as possible

. Choose the material of high fatigue value;

.Choose the heat treatment method and reinforced technics to increasing the

fatigue value of machine element;. Increase the surface texture of machine element;

.Decrease or eliminate the original crack size of the element surface.

For the case of constant ratio r, When the working points are on fatigue safety

zone:

1 [ ]( )

N

a a

D a m

kS S

k

= +



1 0

0

2

=

When the working points are on plastic safety zone:

[ ]s

a s

S S

=

+

Illumination :

Two other cases (m =constant and min =constant) are not given in detail here forthey are not very common.


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MAIN CONTENTS OF THE CHAPTER

Knowing the significance and use of the fatigue curve and limit stress diagram, can

draw the simplified limit stress diagram by knowing some basic mechanical

characters ( 1 0, , ,B S )and the geometrical property.

Grasp the calculation method of strength under the pure varying stress, and know

the concept of equivalent stress.

Know the significance and application mode of fatigue damage accumulation

hypothesis (Miner law).

Grasp the checking method of doubleaction varying stress.

Can use the line graph and numerical tables in the appendix.

-1 s

7

0

5 6 8

3.1 The fatigue limit of symmetric 275 ,off set limit 355MPa,the critical number of cycles is N 10 ,index of life is m=9.Estimate: when the number

of cycles Nis 10 ,5 10 ,or 10 ,the l

Example Mpa = ==

N -1N

ife factor K andfatigue limit .

0 1 1

Example 3.4 The material of shaft is 40Cr, thermal refining, its mechanical character is

0.2, 355MPa, 205MPa, 0.1. The diameter of the its minimum

life section is d=40mm, the bending moment

= = = =

M=300N m, the torque T=800N m, fatigue

limit influential factors K 2.5, 1.5.K

= =

g g

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