case study group 5
TRANSCRIPT
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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)
$here
d0 F uter diameter of the shaft
k F >atio of inner to outer diameters of the shaft
' k =0 for a solid shaft because inner diameter is
8ero)
α F (olumn-action factor
' α =1.0 for tensile load)
#he term α known as column action factor due to the phenomenon of bucklin! of
lon! slender members which are acted upon by axial compressi"e loads where α is
defined as followsF
3) or L
K 115
α = σ yc
π 2nE ( L K )
2
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$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
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)
(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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