case study group 5

8/19/2019 Case Study Group 5

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1. FRACTURE ANALYSIS

Machines aren’t supposed to break, and mechanical components such as shafts, fasteners, and

structures aren’t supposed to fail. But when they do fail, they can tell us exactly why.

It may sound a little far-fetched, but experts say that the causes for more than 90 of all plant

failures can be detected with a careful physical examination usin! low power ma!nification

and some basic physical testin!. Inspection of the failure will show the forces in"ol"ed,

whether the load applied cyclically or was sin!le o"erload, the direction of the critical load,

and the influence of outside forces such as residual stresses or corrosion. #hen, accurately

knowin! the physical roots of the failure, you can pursue both the human errors and the latent

causes of these physical roots.

UNDERSTANDING THE BASICS

Before explainin! how to dia!nose a failure, we should review the effects of stress on a

part. $hen a load is put on a part, it distorts. In a sound desi!n the load isn’t excessi"e, the


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stress doesn’t exceed the %yield point%, and the part deforms elastically, i.e., when the load is

released the part returns to its ori!inal shape. #his is shown in Figure 1, a %stress-strain%

dia!ram that shows the relationship between loads and deformation.

In a good design, the part operates in the eastic range! the area "etween the origin and

the #ied strength! the part wi "e per$anent# defor$ed. &"en !reater increases in oad

wi ca%se the part to act%a# "rea&.

Figure 1 illustrates a "ery basic point of desi!n, and applies when the load on a part is

relati"ely constant, such as the load on the frame of a buildin! or the stress in the le!s of your

desk. It is a "ery different case when f%ct%ating oads are appied! such as those in a

hydraulic cylinder or in an automoti"e connectin! rod. These f%ct%ating oads are caedfatig%e oads! and when the fatig%e strength is e'ceeded! a crac& can deveop. #his

fati!ue crack can slowly work its way across a part until a fract%re occ%rs. '(orrosion can

!reatly affect the fati!ue stren!th).

Figure 1

(achine co$ponents can fract%re fro$ either a singe overoad force or fro$ fatig%e

forces. *ookin! at the failure face will tell which of these was in"ol"ed. + singe overoad

can res%t in either a d%ctie fract%re or a "ritte fract%re.

DUCTI)E *+ER)*AD +S. BRITT)E *+ER)*AD ,AI)URES

+ %ductile failure% is one where there is a great dea of distortion of the failed part.

(ommonly, a ductile part fails when it distorts and can no lon!er carry the needed load, like

an o"erloaded steel coat han!er. owe"er, so$e d%ctie parts "rea& into two pieces and

can "e identified "eca%se there is a great dea of distortion aro%nd the fract%re face!

similar to what would happen if you tried to put too much load on a low carbon steel bolt.

#he term %brittle fracture% is used when a part is o"erloaded and "rea&s with no visi"edistortion. #his can happen because the material is "ery brittle, such as !ray cast iron or


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hardened steel, or when a oad is appied e'tre$e# rapid# to a nor$a# d%ctie part. +

se"ere shock load on the most ductile piece can cause it to fract%re i&e gass.

+n important point about failures is that the way the load is applied, i.e., the direction and the

type, can be dia!nosed by lookin! at the failure face. + crack will always !row perpendicular

to the plane of maximum stress. Below we show examples of the difference in appearance

between ductile o"erload and brittle o"erload failures.

Figure 2

rom the examples abo"e in Figure 2, we know we can oo& at an overoad fai%re and

&nowing the t#pe of $ateria! te the direction of the forces that ca%sed the fai%re.

(ommon industrial materials that are ductile include most aluminum and copper alloys, steels

and stainless steels that are not hardened, most non-ferrous metals, and many plastics. Brittle

materials include cast irons, hardened steel parts, hi!h stren!th alloyed non-ferrous metals,

ceramics, and !lass.

ne note of caution is that the type of fracture, ductile or brittle, should be compared with the

nature of the material. #here are some instances where brittle fractures appear in normally

ductile materials. #his indicates that either the load was applied "ery rapidly or some chan!e

has occurred in the material, such as low temperature embrittlement, and the material is no

lon!er ductile. +n example of this was a low carbon steel clip used to hold a conduit in

position in a refri!erated '-/0 ) warehouse. #he clip was made from a "ery ductile material,

yet it failed in a brittle manner. #he in"esti!ation showed it had been hit by a hammer, a blow

that would ha"e deformed it at normal temperatures.


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In a brittle o"erload failure, separation of the two haves isn-t %ite instantaneo%s! "%t

proceeds at a tre$endo%s rate! near# at the speed of so%nd in the $ateria. #he crack

be!ins at the point of maximum stress, then !rows across by clea"a!e of the indi"idual

material !rains. ne of the results of this is that the direction of the fracture path is freuently

indicated by che"ron marks that point toward the ori!in of the failure. Example 1 is a

photo!raph of the input shaft of a reducer where the che"ron marks clearly point toward the

failure ori!in, while Figure 3 is a sketch of the cross section of the wall of a ruptured 10ft.

'2.3 m.) diameter "essel. In both cases, by tracin! the che"ron marks back to their ori!in, we

knew exactly where to take samples to determine if there was a metallur!ical problem.

4otice how theche"ron marks

'hi!h-li!hted)

point toward the

ori!in of the

fracture.

Example 1

Figure 3

,ATIGUE ,AI)URES

5o far we’"e talked about the !ross o"erloads that can result in immediate, almost

instantaneous, catastrophic failures. + "ery important distinction is that fatig%e crac&s ta&e

ti$e to grow across a part. In a fati!ue failure, an incident of a problem can exceed the

material’s fati!ue stren!th and initiate a crac& that wi not res%t in a catastrophic fai%re

for millions of cycles. $e ha"e seen fati!ue failures in 3100 rpm motor shafts that took less

than 31 hours from installation to final fracture, about 670,000 cycles. *n the other hand!

we have aso $onitored crac& growth in sow# rotating process e%ip$ent shafts that

has ta&en $an# $onths and $ore than /0!000!000 c#ces to fai.

Figure 4 shows a simple fati!ue crack with the different !rowth 8ones and the maor physical

features.

#he fatig%e 1one is t#pica# $%ch s$oother than the instantaneo%s 1one! which is

%s%a# "ritte and cr#staine in appearance. 2rogression $ar&s are an indication that the


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growth rate changed as the crack !rew across the shaft and don’t appear on many failure

faces.

Figure 4

#here are some complex mechanisms in"ol"ed in the initiation of a fati!ue crack and once

the crack starts, it is almost impossible to stop because of the stress concentration at the tip.

STRESS C*NCENTRATI*N

+ stress concentration is a physical or metallur!ical condition that increases the local stress in

the part by some factor. + !ood example is the shaft shown in i!ure /. $e see that the stress

in the area of the radius "aries dependin! on the si8e of the radius. + small radius canincrease the stress dramatically.

Figure 5

5tress concentrations, indicated by the symbol :t, can be caused by chan!es in metallur!y,

internal defects, or chan!es in shape. #here is extensi"e data that indicates that the resultant

"alues depends on both the type of stress, i.e., bendin!, torsion, etc., and the !eneral shape of

the part.


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5tress concentrations ha"e a !reat effect on crack initiation because of their effect on

increasin! the local stress. #he crack can start solely as the effect of the operatin! loads or it

can be multiplied by the stress concentration factor.

3HAT T42E *, )*AD 3AS IT5

#he face of a fatig%e fai%re tes %s "oth the t#pe 6"ending! tension! torsion or a

co$"ination7 and the $agnit%de of the oad. #o understand the type of load, look at the

direction of crac& propagation. It is always !oin! to be perpendicular to the plane of

maximum stress. #he four examples in Figure 6 reflects four common fracture paths.

Figure 6

Figure 6 brin!s up the uestion %what type of bendin!;% $as it one-way plane bendin!, like a

leaf sprin! or a di"in! board, or was it rotatin! bendin!, such as a motor shaft with a hea"y

belt load; +s seen in Figure 7 , lookin! at the fracture face a!ain tells us the type of load.

4otice that %rotatin! load% on the ri!ht causes the crack to !row in a non-uniform manner. In

!eneral, when the di"ider of the instantaneous 8one does not point to the ori!in, it shows

there was a rotatin! bendin! in"ol"ed in the failure cause.

Figure 7


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H*3 HEA+I)4 3AS IT )*ADED5

ati!ue failures almost always start on the outside of a shaft at a stress concentration, because

the local stress is increased. owe"er, the instantaneous 8one 'I


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#he small instantaneous 8one indicates the stress at the time when the shaft finally broke was

low, but the multiple ori!ins and the ratchet marks show us there was enou!h stress to cause

crackin! at many points around the perimeter almost simultaneously.

rom this you can conclude that there must ha"e been a si!nificant stress concentration. '#he

calculated stress concentration was in the ran!e of =.0, so the stress in the area of those

ori!ins was four times as much as it should ha"e been.)

$ith this information on the type of load and the ma!nitude of the load, we can start lookin!

at some failures and dia!nosin! where they came from. ollowin! are some examples of

failures and an explanation of their causes.

Figure 10.

+ torsional fati!ue failure resultin! from a

loose hub fit. 4ote the se"ere frettin! 'from

looseness) and the cracked shaft.

Figure 11.

+ rotatin! bendin! fati!ue failure from a

motor shaft. 4otice the small instantaneous

8one that shows the shaft was li!htly loaded

at the time of failure.

Figure 12.

By tracin! the pro!ression marks backward,

Figure13.

Impressi"e brittle fracture of a lar!e uni"ersal


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we can see the failure started at the corner of

the keyway. But, the instantaneous 8one is

tiny. #his indicates the shaft was "ery li!htly

loaded at the time of failure and further

research is needed.

oint. #he che"ron marks point to where the

failure started. #he fact that the surface has

uniform rou!hness tells us that this was an

instantaneous failure.

Figure 14.

+ testimony to an inept repair. #he weld

repair of the shaft should ne"er ha"e been

attempted. #he four !ross weld flaws

initiated fati!ue crackin! of a "ery hea"ily

loaded shaft.

Figure 15.

#ypical rotatin! bendin! failure. Moderate

si8ed instantaneous 8one. >otatin! bendin!

failure ori!ins surround the shaft.

Figure 16.

?!ly plain bendin! failure.

2. MATERIAL SUGGESTION


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High Strength Stainess Stee 6/9:; 2H7

3@-= A is a chromium-nickel !rade of stainless that may be hardened by a sin!le low

temperature precipitation-hardenin! heat treatment. &xcellent mechanical properties at a hi!h

stren!th le"el may be obtained by such treatment. 5calin! and distortion is minimi8ed. 3@-=

A should not be used in the solution treated condition. #he stren!th and corrosion resistance

properties of 3@-= hold up well is ser"ice temperatures up to 600 . abrication techniues

for this steel is similar to those established for the re!ular stainless !rades. 3@-= machines

well, has excellent weldin! characteristics, and for!es easily. #he combination of !ood

mechanical and processin! properties makes this !rade adaptable to a wide ran!e of uses.

ANALYSIS

Carbon(C)Max

Manganese (Mn)

Max

Silicon(Si)Max

Chromiu m (Cr)

Nickel (Ni)

Copper (Cu)

Phosphoru s (P)

Max

Sulfur (S)Max

0.07 1 1 165-17.5 3-5 3-5 0.04 0.3

17-4 stain!ss "!n!#a$ %&n'(s t& ASTM A564 T$)! 630 AMS 5643

(ECHANICA) 2R*2ERTIES

ConditionTensile

Strength(PSI)

Yield Strength

(PSI)

Shear Strength

(PSI)

Elongationin !

"ardness

+ 'annealed) 150*000 110*000 10 40 34

900 'hardened at900 )

200*000 1+5*000 14 50 44

33/0 'hardenedat 33/0 )

145*000 125*000 1, 60 33

T! a&/! /a!s a#! a/!#a"! an (a$ ! %&nsi!#! as #!)#!s!ntati/! &' 17-4 stain!ss

A22)ICATI*NS

3@-= is used where hi!h stren!th and !ood corrosion resistance are reuired, as well as for

applications reuirin! hi!h fati!ue stren!th, !ood resistance to !allin!, sei8in! and stress

corrosion. 5uitable for intricate parts reuirin! extensi"e machinin! and weldin!, andCor

where distortion in con"entional heat treatment is a problem. #he corrosion resistance of 3@-=

is superior to that of hardenable strai!ht chromium !rades such as =32. It approaches thecorrosion resistance of the non-ma!netic chromium-nickel !rades. (orrosion resistin!


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Aower transmission shaft is a "ital element of all rotatin! machinery. Eenerally, a

shaft has a circular cross-section area, and may be hollow or solid. #he shaft is supported on

bearin!s and it rotates a set of !ears or pulleys for the purpose of power transmission. It

!enerally acted upon by bendin! moment, torsion as well as axial force. Desi!n of the shaft

primarily in"ol"es the stresses at critical point in the shaft that is arisin! due to

aforementioned loadin!.

or a desi!n purpose as well as production, a few aspects need to be considered in

order to produce a !ood shaft. *aconically, the desi!n of shaft based on two mechanical

properties which are stren!th and stiffness. Desi!nin! a shaft based on stren!th is carried out

so that stress at any location of the shaft should not exceed the yield stren!th of the material.

+s for the stiffness, it depends on the allowable deflection and twist of the shaft.

irst and foremost, the stren!th of a material plays an important role is desi!nin! a

!ood shaft. Basically, the state of stress to be considered is caused by bendin! due to its

wei!ht or load, axial loadin!, and also torue that bein! transmitted to the shaft. #hese three

basic conditions are !i"en as followsF

5IMA*& B&4DI4E

(ase 3 considers simple bendin!. or a !i"en bendin! moment, M and nominal stress

in bendin!, G b F

σ b= 32 M

π d03(1−k 4)

$here

d0 F uter diameter of the shaft

k F >atio of inner to outer diameters of the shaft

' k =0 for a solid shaft because inner diameter is

8ero)


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+HI+* *+DI4E

(ase 1 considers axial loadin!. or a !i"en axial force acted on the shaft, and

nominal stress in axial force, Ga F

σ a= 4 αF

π d02(1−k 2)

$here




8ero)

α F (olumn-action factor

' α =1.0 for tensile load)

#he term α known as column action factor due to the phenomenon of bucklin! of

lon! slender members which are acted upon by axial compressi"e loads where α is

defined as followsF

3) or L

K 115

α = σ yc

π 2nE ( L K )

2


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$here

: F least radius of !yration

* F shaft len!th

σ yc F yield stress 'compression)

n F 3.0 for hin!ed 1.1/ for fixed point 3.2 for bearin!

A?>& #>J?&

(ase 7 considers pure torue. or a shaft transmittin! power, A0 at a rotational speed,

n the transmitted torue, # can be found fromF

τ xy= 16T

π d03(1−k 4)

$here

τ xy F 5hear stress due to torsion




8ero)

(MBI4+#I4 B&4DI4E, +HI+* *+DI4E, +4D #>5I4

(ase = considers combined bendin!, axial loadin!, as well as the torsion on a circular

shaft. 5ince both bendin! and axial stresses are normal stress, so the net normal stress

is !i"en byF

σ x= 32 M

π d03

(1−k 4

)

+ 4 αF

π d02

(1−k 2

)


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#he net normal stress can be either positi"e or ne!ati"e. 4ormally, shear stress due to

torsion is only considered in a shaft, and shear stress due to load on the shaft is

ne!lected. Desi!n of the shaft mostly uses maximum shear stress theory. It states that

a machine member fails when the maximum shear stress at a point exceeds the

maximum allowable shear stress for the shaft material. #herefore,

τ max=τ allowable=√(σ x

2)2

+τ xy2

5ubstitutin! the "alues ofσ x and Kxy in the abo"e euation will results to,

τ allowable= 16

π d0

3(1−k 4) √( M +αF d

0(1+k 2)

8)2

+T 2

#herefore, the shaft diameter can be calculated in terms of external loads and material

properties. owe"er, the abo"e euation is further standardi8ed for steel shaftin! interms of allowable desi!n stress and load factors in +5M& desi!n code for shaft.

#he shafts are normally acted upon by !radual and sudden loads. ence, the euation

is modified in +5M& code by suitable load factors as followsF

C

(C b M +αF d

0(1+k 2)

8 )

2

+(¿¿ t T )2

τ allowable= 16

π d0

3(1−k 4) √ ¿

$here

C b F bendin! factor

C t F torsion factor


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C b C t

or stationary shaftF- *oad !radually applied

- *oad suddenly applied

3.0

3./ L 1.0

3.0

3./ L 1.0

or rotatin! shaftF

- *oad !radually applied

- *oad suddenly applied 'minor shock)

- *oad suddenly applied 'hea"y shock)

3./

3./ L 1.0

1.0 L 7.0

3.0

3.0 L 3./

3./ L 7.0

+5M& code also su!!ests about the allowable desi!n stress, Kallowable

to be considered for

steel shaftin!,

+5M& (ode for commercial steel shaftin!

Kallowable

// MAa for shaft without keyway

Kallowable

=0 MAa for shaft with keyway

+5M& (ode for steel purchased under definite specifications


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Kallowable

70 of the yield stren!th but not o"er 36 of the ultimate stren!th in

tension for shafts without keyways. #hese "alues are to be reduced by 1/ for the

presence of keyways.

4. ENIRONMENT FACTOR

• a#s !n/i#&n(!nta %&niti&ns !a t& !#&si&n* %#&si&n* !(#itt!(!nt.

• Et#!(! !at %& %an a/! a a(a"in" !''!%t &n 'ati"! 'ai#! an

'ati"! i'!.

Protection Possibilities Checklist

• #&/i! )#&t!%ti&n a"ainst %#&si&n.

• M&nit !t#!(! '#!!nt %an"!s in &' &ain"* t!()!#at#! )#!ss#!.

C&&s! (at!#ias 8it '!8!# (is(at%!s &' t!#(a %&!''i%i!nts &' !)ansi&n

' (atin" )a#ts.

• #&/i! )#&t!%ti&n a"ainst U i"t &t!# a#(' s&#%!s.


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• T! nat#a '#!!n%$ &' t! st#%t#! (st sta$ a8a$ '#&( t! '#!!n%$ &'

its 8in" !n/i#&n(!nt &ain". In%#!as! nat#a '#!!n%$ ' #!%ti&n &'

#!s&nan%! %#&si&n 'ati"!

• Mini(i:! !i(inat! %$%i% st#!ss!s A &a &' ins''i%i!nt (a"nit! t& %as!

'ai#! in a sin"! a))i%ati&n (a$ !a t& 'ai#! i' it is #!(&/! an #!a))i!

#!)!at!$. L&n" a"&* !n"in!!#s is%&/!#! tat i' $& #!)!at!$ a))i! an

t!n #!(&/! a n&(ina &a t& an '#&( a (!ta )a#t ;9n&8n as a


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@ tt)>>888.a))i!t#as&ni%s.%&(>&">'ati"!-'ai#!>)#!/!ntin"-'ati"!-

'ai#!-8it-t#as&ni%-i()a%t-t#!at(!nt
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case study group 5

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