fracture toughness & fatigue week 3. behaviour of materials in service a material or structure...
TRANSCRIPT
Fracture Toughness & Fatigue
Week 3
Behaviour of Materials in Service
A material or structure is deemed to have failed when it is unable to satisfy the original design function.
Failure may be due to:• Plastic buckling.• Dimensional change with time.• Loss of material through corrosion, erosion or
abrasion. • Fracture - partial or complete.
Failure by Yielding
Metals exhibit both elastic and plastic behaviour.
As the level of stress is increased so the amount of elastic strain increases in direct proportion to the stress applied up to a certain limit – elastic limit.
To minimise the possibility of excessive stress factors of safety are applied.
Failure by Fracture
The terms, tough, ductile, brittle, or fatigue are frequently used to describe the fracture behaviour of a material.
Tough or ductile fracture – failure is preceded by excessive plastic deformation often detectable.
Brittle or non-ductile fracture - involves little or no plastic deformation – often Catastrophic.
Fracture Type
The type of fracture which occurs is largely dependant upon the type and condition of the material.
Other factors include:– the type of stress applied.– the rate of stress application.– temperature and environmental conditions.– component geometry.– size and nature of internal flaws.
Fracture Mechanics This is the study of the relationships
between crack geometry, material strength and toughness and stress systems as they affect the fracture characteristics of a material.
The aim of fracture mechanics is to determine the critical size of a defect necessary for fast fracture to occur.
That is catastrophic crack propagation and failure under service loading.
Fracture Toughness
To improve fracture toughness there is a need to avoid excessive elastic deflections & plastic yielding.
Fast fracture can occur which causes catastrophic failure.
E.g. Welding of ships, oil rigs, bridges, pipelines, pressure vessels.
Fast Fracture
All related to cracks, flaws or defects. Fast fracture caused by growth of these
defects which suddenly become unstable & propagate at the speed of sound.
Why Does This Happen?
There appears to be a critical pressure (stress) related to the size of the internal flaw.
There is therefore a need to determine the critical stress.
Sample with Flaw
a a
Thickness, t
F
F
If the flaw is increased by δa , then:
Work done by loads ≥ Change in elastic energy + energy absorbed at the flaw tip.
i.e.
aGUW cel t
Fig. 1
Definitions
2-
2-6
c
2-c
c
c
10Jm - Glass
Jm10 -Copper
flaw thepropergate toisit harder theGhigher The
Jm - G of Units
RATE RELEASE ENERGY STRAIN CRITICALor
TOUGHNESS called flaw of areaunit makingin absorbedEnergy
:property almathematic a is G
areacrack at
crack of areaunit per absorbedEnergy G
Fast Fracture at Fixed Displacements Consider a plate clamped at both ends
under tension as in fig.1
atGU-
0W
cel
Increase in Load
a
a
E
Consider a small cube of material of unit volume due to load (F)
Cont’d
)2/a)(2/( Uis changeEnergy
energy its all lose &
zero) to tends(relax willarea dotted in theregion The
a'' flaw a Introduce
Eor 1/2 U
energystrain a has thereforecubeunit Each
)(strain & )( stress tosubjected isEach
2el
el
tE
Cont’d
fracturefast at EGaor a/EG
2 offactor aby low toois U of valuetherelaxes, material
hein which t way theregarding sassumption of Because
a/2EG
:givescondition critical The
a/2)aE)(22/(U
a/dadUU
:alengthby spreadscrack theAs
cc
el
c
2el
elel
t
Fast Fracture at Fixed Loads
F
F
Load acts in a more complex way. As the flaw grows material becomes less stiff & relaxes. The applied forces move & do work. W is therefore finite & positive.
Uel is now positive & the final result for fast fracture is found to be the same (some of W goes to increase the strain energy).
Cont’d
c
c
c
G & Eon dependsonly
&property material a be toappears fractureFast
reached is stress
of valuecritical'' some when a''crack a Subject to 2.
a'' size critical a reachescrack a stress a Subject to 1.
:reasons for twooccur can fracturefast Implies
EGa
fracturefast for Condition
EGa Therefore,
:side handRight
:side handLeft
Stress Intensity Factor
cc
c
EGK
KK
: toughnessfracture the torelates &
FACTOR INTENSITY STRESS the
called K'' termsingle a toreducedusually is πaσ termThe
Toughness & fracture toughness values for some metals
Material Toughness
Gc
(kJ m-2)
Fracture toughness
Kc
(MPa m½ or MNm-3/2)
Steels 30 – 150 80 – 170
Cast Irons 0.2 - 3 6 - 20
Aluminium alloys
0.4 - 70 5 - 70
Cont.
No material is free from defects, so it is essential that any crack-like defects are relatively harmless.
Using values of fracture toughness it is possible to calculate the size of defect or magnitude of stress required to cause failure.
Cont.
σy
Applie
d s
tress
Defect size
Failure by
yielding
Failure by
Fast fracture
Critical size
Correction Factors
applied are factors correctionother a, tdimensionsother For
a t , where
a1.12K Therefore,
a dimensionsother & tpracticeIn
dimensions plateother a ti.e.
materials finite-semifor thin only true isresult heStrictly t
Sellotape
MM
t
a
a
Increase the load to the value M that just causes rapid peeling. For this geometry, the quantity Uel is small compared to the work done by M & can be neglected (tape has comparatively little ‘give’).
Cont’d
2-c
2-c
2
c
c
c
490.5JmG
10 x 9.81/2 x 1G
9.81m/sg
1kgM
2cmt
:Example
Mg/tG
atGaMg
atGW
Example
495MPastrength Yield
24.2MNm K
strength. yield at the
fail crack to of size thedetermine Also 0.2mm. of a'' size
crack a containsit if T651-7074alloy aluminiuman for (MPa)
stress fracture thedetermine a,kequation theUsing
3/2-c
c
Solution
fracturefast under fail always willmaterial the0.76mma if I.E.
0.76mmm7.61x10a
/)495/2.24(a
/)/K(a
)/K(a
/Ka Now,
MPa4.965)0.2x10 x (24.2/a/K
aK
4-
2
2fc
2fc
fc
3-cf
fc
Tutorial Exercise 1
Determine the fracture strength for a high strength steel assuming a crack of 0.2mm & kc=55MNm-3/2. Also if the yield strength is 1550 MPa determine the size of crack that will cause failure at this stress.
Solution
mm4.0m4x10a
/)1550/55(a
/)/K(a
)/K(a
/Ka Now,
MPa2.2194
)0.2x10x (/55
a/K
4-
2
2fc
2fc
fc
f
3-f
cf
Ceramics
The strength of ceramics varies considerably from 0.69MPa to about 7x103MPa.
As a class of materials, few have tensile strengths above 172MPa.
There are also large differences in compressive strength, usually between 5 & 10 times higher than the tensile strength.
Many ceramics are hard & have low impact resistance – reflecting their ionic/covalent bonding.
Deformation Mechanisms
Generally lack of plasticity in crystalline ceramics due to bonding.
Bonding directional, dislocations are narrow & do not move easily.
Deformation is mainly due to ionic bond, but this is brittle.
Cracks form at grain boundaries.
Factors Affecting Strength
Mainly occur from structural defects:
Surface cracks.
Voids (porosity).
Inclusions.
Large grains. These are produced during processing.
Cont’d
Once a crack forms - no energy absorbing processes – crack propagates until fracture – general decrease in strength.
Size & volume fraction important. Flaw size can be related to grain size –
finer grain size ceramics have smaller size flaws at their grain boundaries.
Cont’d
Strength is determined by chemical composition, microstructure, surface condition.
Additionally – temperature, environment & type of stress & how it is applied.
Ceramic Abrasive Materials
High hardness makes them useful for grinding, cutting & polishing other lower hardness materials.
Fused alumina & silicon carbide are two commonly used ceramic abrasives.
Products – sheet & wheels manufactured by bonding individual ceramic particles together
Bonding materials include fired ceramics, organic resins & rubbers.
Cont’d
Ceramic particles must be hard with sharp cutting edges.
Must have a certain amount of porosity to provide channels for air or liquid to flow through – act to cool the abrasive & removes debris.
Aluminium oxide grains tougher than silicon carbide ones, but not as hard.
Cont’d
By combining with zirconium oxide improved abrasives have been developed.
Another important ceramic abrasive is cubic boron nitride – borazon- almost as hard as diamond, but better heat stability.
Fracture Toughness of Ceramics
Due to their combination of covalent & ionic bonding, ceramics have inherently low toughness.
Toughness can be improved by processes such as hot pressing with additives & reaction bonding.
Tutorial Exercise 2
The maximum sized internal flaw in a hot-pressed silicon carbide ceramic is 30micron. If this material has a fracture toughness of 3.9MNm-3/2, what is the maximum stress that this material can support?
Solution
MPa1.568
)15x10x (/9.3
a/K
m15x10 /230x10a
2by divided bemust crack theof size the
internal, iscrack thefact that the toDue
f
6-f
cf
6-6-
Fatigue
Effect of Mean Stress – Goodman & Soderberg
Introduction
It been estimated that 75-80% of all failures in engineering components, machines, vehicles, structures & bridges, aircraft, ships,etc..are due to fatigue.
Fatigue is the general failure which occurs after several cycles of loading to a stress level below the ultimate tensile stress & is generally due to localised plastic deformation at the metal surface, culminating in the nucleation of sharp discontinuities, which once formed continue to grow into cracks.
Cont’d
Pre-existing cracks, imperfections & defects, small cracks usually at the surface, weld defects, machining marks, & scratches act as stress concentrators & the failure is characterised as a smooth portion of surface with a ‘clamshell’ or ‘beachmark’ appearance, with a concentric series of lines recording each cyclic advance of the fracture.
Cont’d
This abruptly changes to a granular portion which corresponds to the rapid crack propagation at the point of catastrophic failure.
Fig. 1 shows the conditions for fatigue loading.
Cont’d
Fig. 1 The conditions for fatigue loading.
-
S
TR
ES
S
+
ALTERNATING LOAD
TIME
STRESS RANGE
MEAN STRESS
-
ST
RE
SS
+
PULSATED OR REPEATED LOAD
TIME
MEAN STRESS
STRESS RANGE
-
S
TR
ES
S
+
FLUCTUATING LOAD
TIME
MEAN STRESSSTRESS RANGE
Conditions
Fluctuating load - the mean stress is greater than the stress range.
Pulsating or repeating load – the mean stress is equal to half the stress range.
Alternating load – the mean stress is zero.
S/N Diagrams
Typical S/N diagrams for pieces subject to alternating loads are shown in fig. 2.
Goodman & Soderberg investigated the relationship between stress amplitude, mean stress & fatigue.
Goodman & Soderberg diagrams are shown in fig.3.
S/N Diagrams
Fig. 2 Typical S/N diagrams for pieces subject to alternating loads
Str
ess
am
plit
ud
e (S
)
Failure occurs for stress values above this line
ENDURANCE LIMIT FOR 'N' REVERSALS
Typical S - N diagram for non ferrous alloy
Number of stress cycles (N)
FATIGUE LIMIT
Str
ess
am
plit
ud
e (S
)
Failure occurs for stress values above this line
Typical S - N diagram for steel
Number of stress cycles (N)
S/N Diagrams Fig. 3 Goodman & Soderberg investigated
the relationship between stress amplitude, mean stress & fatigue.
Str
ess
am
plit
ude
(S
)
Typical S - N diagram for non ferrous alloy
MEAN STRESS
Str
ess
am
plit
ude
(S
)
Typical S - N diagram for steel
MEAN STRESS
Yieldstress
Tensilestrength
Material stressed withinthis area should notfail by fatigue
Material stressed withinthis area should notfail by fatigue
Fatigue limit(mean stress = 0)
Fatigue limit(mean stress = 0)
Perfect Alternation
When the mean stress is zero (perfect alternation), the fatigue limit is at a maximum value before failure occurs.
If a steady state of stress is superimposed on the cyclical stress then this must also be taken into account – this steady state stress is the mean stress.
Goodman Diagram
In the Goodman diagram the fatigue limit is zero when the mean stress is equal to the tensile strength of the material, since the material will fail at this value before any cyclical loading can commence.
Cont’d
Therefore, if the point representing the stress amplitude & the mean stress for any given set of conditions lies within the area bounded by the axes & the ‘Goodman Line’( the shaded area), then according to the Goodman relationship the material should not fail by fatigue.
Soderberg Diagram
In the Soderberg diagram the fatigue limit is zero when the mean stress is equal to the yield stress of the material. Again, the point representing the stress amplitude & the mean stress for the material must lie within the shaded area bounded by the axes & the ‘Soderberg Line’, if failure by fatigue is to be avoided.
Note
Since perfect alternation (zero mean stress) rarely occurs in practice, S/N curves should not be used alone without consideration of the mean stress.
Care must also be taken in using even the Goodman & Soderberg diagrams since they tend to give a low value of the fatigue limit for ductile materials & high value of fatigue limit for brittle materials.