2. Introduction
“ The concept of failure is central to the design process, and it is by
thinking in terms of obviating failure that successful designs are achieved.
It has long been practically a truism among practicing engineers and
designers that we learn much more from failures than from success.
Indeed, the history of engineering is full of examples of dramatic failures
that were once considered as confident extrapolations of successful
designs; it was the failures that ultimately revealed the latent flaws in
design logic that were initially masked by large factors of safety and a
design conservatism that relaxed with time.”
Timoshenko History of Strength of Materials, 1953, New York, McGraw Hill.
3. Design Against Failure
Schematic of the design process
for a generic engineering
component.
The failure focus identifies the
type (or mode) of failure the
designer will focus on or
specifically design against,
given the material characteristic
and the operating conditions of
the component.
Engineering
Component
Failure Focus
11. Failure Types
Fracture :
The component breaks. This type of failure is generally signified
by some active part of our component breaking under load.
Yielding :
The component sustains plastic, non-recoverable, deformation.
This type of failure is most commonly localized and often leads
to eventual rupture of the component.
15. Brittle Fracture
Rapid Fractures (speed of sound)
Stresses Below Yield
Little or no Plasticity
Triggered from an Initiation site
Low Temperatures
High Strain Rate
Steels: B.C.C. VS F.C.C.
Phase Transformation (HAZ)
18. Fatigue :
The component experiences a time dependent failure mechanism
known as fatigue. This type of mechanism is still the subject of
intense research, since it is not yet fully understood. Many
engineering components are exposed to time dependent, dynamic,
loading. Under these conditions the material of construction
becomes tired and will fail at a significantly lower stress intensity
than it could sustain in a static manner.
Buckling :
The component experiences a form of elastic instability with large
lateral deflections under compressive loading. This type of failure is
mostly experienced by slender columns under axial loads, and thin
walled pressure vessels subject to external pressure (e.g. vacuum
vessels). A component will buckle under a load significantly lower
than the yield compressive load for the same component. Once
buckled, the component can no longer sustain any significant load.
20. Fatigue of Metals
Crack initiation and propagation.
Cyclic or random loading.
Thermo-mechanical loading.
Fretting fatigue.
Corrosion fatigue.
21. DeHavilland Comet Crash
First production commercial jet
airliner.
Influenced modern aircraft
design by two failures.
Within two years of service two
planes fell apart during ascent.
After 60 design modification it
again resumes flight.
Once again plane crashes after 30
minutes of flight.
Airlines grounded, certificate of
airworthiness revoked and
production line suspended.
24. Comet Crash-I
Fuselage was tested in pressurized water tanks.
Evidence of fatigue cracking was found that originated from the aft lower corner
of the forward escape hatch and also from the right-hand aft corner of the
windows.
These locations feature sharp right hand corners which cause local areas of high
stress-concentration.
25. Creep :
At elevated temperatures there is long-term, relatively slow,
plastic deformation of the component under steady load. The
effect is attributed to the fact that key material properties
such as yield strength and ultimate tensile strength are
generally obtained from tests of the material at ambient
temperatures. Loaded components exposed to higher
temperatures, such as gas turbine blades, or heat exchanger
piping, suffer this type of slow deformation.
Excessive deflection :
The component experiences elastic or plastic deformation
beyond some permitted bound. The technical term for this
type of failure is excessive-deflection.
26. Wear is the progressive damage, involving material loss, which
occurs on the surface of a component as result of its motion
relative to the adjacent working parts.
John Williams
Wear
27. Wear Depends
Geometry of the surface
Applied load
The rolling and sliding velocities
Environmental conditions
Mechanical, Thermal, Chemical and Metallurgical properties
Physical, Thermal and Chemical properties of the lubricant
29. Abrasion Wear
Abrasive wear occurs when a harder material is rubbing against a softer material
Ref.: www.substech.com
Two Body Wear
Three Body Wear
30. Abrasion Wear
Gouging abrasion
Large particles
High compression loads
High stress or grinding abrasion
Smaller particles
High compression load
Low stress or scratching abrasion
No compression load
Scratching abrasion while material is sliding
Polishing abrasion
Ref.: www. mesto.com
31. The impingement of solid particles, or small drops of liquid or gas on the solid
surface cause wear what is known as erosion of materials and components.
Advantages
Cutting, drilling and polishing of brittle material
Ref.: dcu.ie/~stokesjt/Thermal Spraying/Book/Chapter1
Erosion Wear
32. Solid Particle Erosion
Surface wear by impingement of solid particles carried by a gas or fluid.
e.g. Wear of helicopter blade leading edges in dusty environments.
Liquid Drop Erosion
Surface wear by impingement of liquid drops.
e.g. Wear of centrifugal gas compressor blades by condensate droplets.
Cavitation Erosion
Surface wear in a flowing liquid by the generation and implosive collapse of
gas bubbles.
e.g. Fluid-handling machines as marine propellers, dam slipways, gates, and
all other hydraulic turbines.
Type of Erosion Wear
33. Two bodies sliding over or pressed into each other which promote the material
transfer from one to another.
𝑉
𝐿
= 𝐾
𝑃
3σ𝑦
Where
V = wear volume
L = sliding velocity
P = applied load
σy = yield stress of softer material
K = wear coefficient
Ref.: www.substech.com
Friction/Adhesive Wear
34. Galling wear : Severe adhesion actually
leads to material flow up from the
surface.
Galling Wear
35. Fretting wear of splined
shaft– small oscillatory
motion abrades surface –
looks like rust – surface
looks pitted.
Fretting Wear
36. Surface fatigue:
Two surfaces contacting to each other under
pure rolling, or rolling with a small amount of
sliding in contact.
Contact fatigue:
As one element rolls many times
over the other element
Maximum shear stress is higher
than fatigue limit
Ref.:W.A. Glaeser and S.J. Shaffer, Battelle Laboratories
37. Pitting surface fatigue :
large roller thrust bearing
race, compressive stress
developed between roller
bearing and race pitting.
Material actually fatigued
and removed from surface!!
Pitting Wear
38. Brinelling : brinelling of
bearing race due to static
overload. Note brinelling more
of a static failure (indentation)
versus fatigue or wear failure.
Brinelling Wear
39. A wear process where a material loss from the surface by forces
of another surface acting on it in a sliding motion in the form of
thin sheets.
Mechanisms of delamination wear
Plastic deformation of the surface.
Cracks are nucleated below the surface.
Crack propagation from these nucleated
cracks and joining with neighbouring one
After separation from the surface,
laminates form wear debris
Ref.: K Kato, M Bai, N Umehara, Y Miyake
Delamination Wear
40. Material Selection
Durability :
Matching of dominant or primary criteria such as strength,
hardness, elastic behavior, toughness, magnetic, electric &
thermal properties.
Longevity:
Depends on corrosion & heat resistance as well as resistance
to wear, dynamic loading, shock, creep and stress corrosion.
Manufacturability:
Refers to castability, machinability and surface finish
requirements.
42. Principal Stresses
For any state of stress, we can find a set of planes on
which only normal stresses act and the shearing stresses
are zero.
Called Principal Planes and the normal stresses acting on
these planes are Principal Stresses denoted as σ1, σ2 and
σ3.
Convention, σ1> σ2 > σ3.
The principal directions are orthogonal to each other.
43.
44. Ductile vs. Brittle Material
Ductile material : Well defined yield point– Failure on
yielding.
Brittle material : No yield point & sudden failure – Failure on
failure load.
45. Theories of Failure
Max. principal stress theory – Rankine
Max. principal strain theory – St. Venants
Max. strain energy – Beltrami
Distortional energy – von Mises
Max. shear stress theory – Tresca
Octahedral shear stress theory
47. Maximum Principal Stress Theory
Maximum principal stress reaches tensile yield
stress (Y).
Estimate principal stresses σ1, σ2 &σ3.
Apply yield criteria:
48. Maximum Principal Strain Theory
Failure occurs when maximum value of applied strain exceeds
the strain value corresponding to yield point of the material.
If ‘Y’ is the yield stress then under uni-axial loading yield
strain is defined as
εy = Y/E
Maximum strain developed in the design component should
be less than εy.
Principal stresses σ1, σ2 &σ3 corresponds to principal strains
ε1, ε2 &ε3 .
49. Strain Energy Theory
Failure at any point in a body is defined when the energy
density in the body at the applied load equals the energy
density corresponds to the elastic limit of the material.
Uni-axial loading :
i. σ = Eε ( Hooke’s Law)
ii. Strain energy density :
50. Body subjected to principal stresses :
For the onset of yielding :
Yield function
U=1/2E [σ1
2 + σ2
2 + σ3
2 - 2 ν (σ1σ2 + σ2σ3 + σ1σ3)]
Y2/2E=1/2E [σ1
2 + σ2
2 + σ3
2 - 2 ν (σ1σ2 + σ2σ3 + σ1σ3)]
51. von Mises Criteria
(Distortion Energy Criteria)
Failure occurs when equivalent stress (von Mises stress)
reaches the yield stress of the material.
von –Mises yield criteria also suggests a failure or
yielding when the elastic energy of distortion reaches a
critical value. von Mises criteria is also known as
maximum distortion energy criteria.
52. Tresca Theory
(Maximum Shear Stress Theory)
Failure/yielding occurs when the maximum shear
stress at a point equals the maximum shear stress
at yield.
Maximum shear stress less than 0.5 Y (No failure).
Shear stress yield = 0.5 (Tensile stress yield)
53. Tresca Theory (Cont…)
In terms of principal stresses σ1, σ2 & σ3 .
Maximum shear stresses.
Yield function :
54. The Design Problem
(a) Use any necessary assumptions to estimate the maximum
bending moment experienced by the arm when the table is
subjected to your chosen value of the design force, W.
Aeroplane service table
(schematic, not to scale),
dimensions in mm
55. (b) Is a failure predictor (theory of failure) required in order to
predict the failure of these arms by yielding? Why or why not?
(c) List the possible modes of failure for the major elements of the service
table.
(d) On the basis of reasoned argument, decide upon a suitable factor of
safety to be used in the design of the arm to resist yielding in bending.
Tabulate your calculation. Does the resulting factor of safety seem
reasonable?
(e) Design the arm to resist yielding. Ignore details of the end
connections.
56. (f) To ensure the suitability of the table for certain uses (e.g. meals
and writing), it is desirable not only that the table and its
support arms avoid any gross structural damage, but also that
the vertical displacement of the table's near edge under the
design loads be kept to within, say, 10 mm. Indicate the method
by which you would incorporate this second mode of failure
into your design process. There is no need to perform the
calculations for this part of the design.
(g) Suggest reasons why designers might take more care with such
devices than with (say) the design of seats in railway carriages
or street-cars (trams) for structural integrity.
57. Project Title" AIRLINE SERVICE TRAY”
(a) Estimation of design force W & maximum bending moment
Wide range of service loads involving a fair amount of
educated guesswork. Begin by tabulating some likely forces
to get a feel of magnitude of W
Source of load Range Force
Meal tray < 5kg 50 N
books < 1kg 10 N
Writing activity 20 N
Elbow (proportion of torso weight) ~ 100 N
Briefcase/travel bag ~20 kg ~200N
Sitting ~100 kg ~1000 N (excessive)
On the basis of the above list choose W=200/2=100 N
(note: Since we are designing one of the table arms, so the design force W
needs to be force on one arm, hence the division by 2.)
Force estimation
58. Moment Estimation
Arbitrarily/conservatively
choose the line of action of
vertical force W to be 50
mm from the outer tip of
the unfolded table (i.e. 600
mm from the lower hinge)
MP = 100 N x 300mm = 30x10 3Nmm
MQ = 100N x 600mm = 60x10 3Nmm
(Note: moment varies along arm PQ and is maximum at Q)
59. (b) Comments on need for a theory of failure
No, a theory of failure(i.e. a means for combining the various stresses in a multi axial
stress situation into a single quantity to be used for predicting failure) is not necessary
here.
This is because the table arms are subjected only to BENDING stresses, which are
uniaxial.
Laboratory tests (to determine Sy) can be applied directly.
The component will begin to yield when the bending stress equals the yield stress Sy.
Note: The above statement assumes that shear stresses are negligible compared to
bending stresses. An experienced designer will usually make this assumption for long
beams like the table arms. Its validity can be checked latter.
If shear stresses are not ignored, then the stress situation is multi axial, and theory of
failure would be needed to predict the onset of yielding.
Theories of failure such as maximum shear stress (Tresca) and maximum shear strain
energy (Von Mises) can be used.
60. (c) Possible modes of failure for service table
Yielding of table top in bending
Fracture of table top in bending
Yielding of supported arm in bending
Fracture of supported arm in bending
Shearing failure of hinge pin at Q.
Shearing failure of hinge pin at P.
Tearing (shear failure) of slot at “P”
Compression failure of reaction point at “Q”
Excessive deflection of table (vertical).
Excessive deflection of table (lateral sway)
Buckling of lower (compressive) flange of arm.
Excessive friction in hinges.
61. Factors of Safety
Design factor of safety (Fd) is used for handling
systematically the many and varied uncertainties associated
with specific design situation.
Commonly Fd is applied to design in two ways:
i. Reduce the known strength (Sy/Fd)
ii. Increase the predicted load (Fd x W)
The choice of a proper Fd is entirely the result of good
engineering judgment.
62. A General Rule(Fd)
Stationary structures and components (e.g. pressure vessels
and their supporting structures) Fd 2 to 4.
Components where mass and inertia are criteria of operational
performance (I.C. Engines), the constitutive models of
structural performance are generally more precisely
determined. Moreover, substantial care is taken to select and
test material properties. (Fd 1.5 to 2)
Steel ropes made up of many strands of high tensile steel wire
are used in cranes or in structural support applications, where
failure could be life threatening. (Fd 12)
These thumb rules are the result of considerable experience in
the design of engineering structures and components.
63. Estimation of Safety Factor
In the following determination of Fd, the main issue is that
the size of the sub factors l1,l2,l3,s1,s2,s3…s5 depends in each
case upon the level of uncertainty concerning the relevant
quantity, not upon the magnitude of the quantity itself.
Values given on next slide are subjective estimates, lying
within ranges commonly used.
64. Factor Relevance to design Range of
values
Value
Fo Seriousness of failure 1.0-1.4 1.0
Uncertainties associated with load estimates
l1 Magnitude of load 1.0-1.6 1.5
l2 Rate of loading 1.2-3.0 1.3
l3 Load sharing 1.0-1.6 1.5
Material and modeling related uncertainties
s1 Variation in material properties 1.0-1.6 1.1
s2 Manufacturing uncertainties 1.0-1.6 1.1
s3 Environmental, Operational 1.0-1.6 1.1
s4 Effects of stress concentration Can be high 1.5
s5 Reliability of mathematical model 1.0-1.6 1.5
Fd=F0 x l1 x l2 x l3 x s1 x s2 x s3 x s4 x s5 = 6.42 =6.0
Estimation of Fd
65. Bending of Beam
Pure bending results in circular arc deflection.
R is radius of arc.
ϴ is the arc in radians.
c is the distance from n.a. to extreme fiber.
fmax is maximum normal stress in extreme fiber.
M is the bending moment.
I is moment of inertia.
𝑓𝑚𝑎𝑥 =
𝑀𝑐
𝐼
𝑀
𝐼
=
𝑓
𝑐
=
𝐸
𝑅
66. Design Arm Against Yielding
𝐷𝑒𝑠𝑖𝑔𝑛 𝑖𝑛𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦: 𝜎 𝐵,𝑀𝑎𝑥 ≤
𝑆 𝑦
𝐹𝑑
𝜎 𝐵,𝑀𝑎𝑥 = maximum bending stress in the arm.
Max. bending stress occur at the outer fibers of arm.
𝐼𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙, 𝜎 𝐵 =
𝑀𝑦
𝐼 𝑥𝑥
Bending stress will vary along the length of arm,
due to variation in M,y and Ixx.
67. 𝑀 = 60 × 103
N.mm
Design of the arm is to calculate ‘t’.
Express Ixx in terms of ‘t’.
y=15mm
d=30mm
b=20mm
x x
t
71. Cup and Cone
Dimples
Dull Surface
Inclusion at the bottom of the dimple
Ductile
Shiny
Grain Boundary cracking
Brittle Intergranular
Shiny
Cleavage fractures
Flat
Brittle Transgranular
Beachmarks
Striations (SEM)
Initiation sites
Propagation zone
Final fracture zone
Fatigue
Mode of fracture Typical surface characteristics
72. S-N Curve
Some materials, such as steel, show an endurance limit stress below which
the fatigue life is essentially infinite. Other materials may not show such
behavior but an effective endurance limit may be specified at some large
number of cycles.
78. Design Against Fatigue
Modern bicycle wheels are complex structures of a hub, a
light rim and spokes (36).
Each of these spokes are pretension by means of a small
threaded nut, so that net compression of any spoke is
avoided.
Design spokes against fatigue failure
79. Load Distribution in Spokes
Load P is acting between hub & ground.
Spokes directly above hub experience
an increase in tensile force, ΔT = λ1P.
Spokes below experience decrease of
tensile force, ΔT = λ2P.
Horizontal position ΔT = 0.
λ = f (#, geometry, pattern dim spokes).
λ1 = + 0.04 & λ2 = -0.08
80. The Design Problem
a. Determine the values of initial (pretension) force To, which could be
needed to ensure that the tension T in the spokes is always greater
than 0.3To.
Solution:
Design for worst case:
Weight of the rider =95 kg.
Center of mass nearest to the rear wheel (x=0.40L)
𝑆 𝑈 = 880 𝑀𝑃𝑎 𝑆 𝑌 = 880 𝑀𝑃𝑎E = 880 𝑀𝑃𝑎
81. Taking moments about the front axle:
𝑃 = 𝑊
𝐿−𝑥
𝐿
𝑃 = 95 × 9.81
𝐿 − 0.4𝐿
𝐿
P = 559 N
83. 𝑇 𝑚𝑎𝑥 = 𝑇𝑜 + ∆𝑇𝑢𝑝𝑝𝑒𝑟
𝑇 𝑚𝑎𝑥 = (𝑇𝑜+22.4)𝑁
𝑇 𝑚𝑖𝑛 = 𝑇𝑜 + ∆𝑇𝑙𝑜𝑤𝑒𝑟
𝑇 𝑚𝑖𝑛 = (𝑇𝑜−44.7)𝑁
𝑇𝑎𝑚𝑝 =
𝑇 𝑚𝑎𝑥 − 𝑇 𝑚𝑖𝑛
2
= 33.5 𝑁
Tamp is the load amplitude on the spokes.
84. Require 𝑇 ≥ 0.3𝑇𝑜 in order to provide some
safety against compression &buckling .
0.3 To = Tmin
To = 63.9 N
85. RMS Titanic
The Royal Mailing Steamer
Titanic was a British
passenger ship, which sank
on April 15 1912 in North
Atlantic ocean.
86. Titanic Statistics
883 ft long, 93 ft beam wide.
61 ft from waterline to boat deck.
One of the fastest ships.
Built by Harland & Wolff in Belfast.
She was the Largest ship in the world.
Licensed to carry 2603 passengers.
Crew of 944 (3547 in all).
87. Titanic Unsinkable
She was made of medium carbon steel.
She was fitted with 16 transverse separate water-tight
compartments (bulk-heads).
Each compartment can be rendered water-tight quickly
by closing of water tight doors.
Contain and isolate incoming water in case of puncture
(Olympic and Britannic).
92. Why Failed ?
Use of medium carbon steel with low transition temperature.
Brittle Fracture.
Square Hatches.
Welded Vs. Riveted.
Design of Ship (Sagging Vs. Hogging).
Equilibrium.
Sagging
Hogging
93. Why Failed ?
Inadequate lifeboats.
Self-confidence and arrogance
off British Ship Building
Engineers was a major
contributing factor to the
sinking of the Titanic.
94. Lessons Learned
Improved Designs: CAD/CAM
Advances in Materials
Radar can give ships a visual picture of location relative to other ships.
Better weather forecasting
Better navigation systems
Iceberg patrols to warn ships
Advanced satellite communications
96. History of Liberty Ships
Allies Were Losing a Large
Number of Transport Ships due
to Aerial Assaults, Mine Fields
and U-Boat Attacks
Lend-Lease Act: President
Roosevelt Announced a $350
million Emergency Shipbuilding
Program.
Between 1941-1945, 2751 were
built. Cost $127k each.
100. Tacoma Narrows Bridge
Suspension bridge : cables anchored to earth in their ends and
supported by towers.
TNB linked Tacoma and Gig harbor.
Width 39ft, unsupported span 2800ft was necessary due to poor
bottom conditions and swift currents in Narrows.
The stiffness girders were unusually narrow; 8ft in comparison
with their length.
Depth to span ratio of 1/350 over twice that of Golden Gate
bridge. Width to span ratio 1/72, Golden Gate is 1/57.
On November 7, 1940 four months after opening violent
oscillation led to the collapse.
108. What Went Wrong ?
Engineers at Morton-Thiokol knew of potential failure of O-ring and
notified VP Engineering months prior to launch and again on night before
lift off, recommending no launch !!!
109. The Challenger Disaster: Reason
Managers consulted with NASA executives and voted to proceed
responding to business and political pressures.
Engineer responsible for identifying concerns blew whistle in public
investigation; had career ruined and sued for damages.
110. Lessons Learned
The engineer has obligations.
Avoid conflict of interest.
Do not yield to management pressure.
Think! Think! Think of Consequences!
Effect of extreme temp on performance.
Document events/actions/decisions and meetings
for future reference.