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- 1. Final reportDG205 Material Behaviour Assignor: Dr. Ir.
F.L.M.Delbressine January March 2012 Eva Palaiologk s112775 Harm
van Hoek s106804 Manual Suarez s118705 David E. Dass s110263
- 2. Table of ContentsIntroduction
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3Objective
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4Tensile Tests
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5 Results
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5 Analysis
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6Bending Test
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7 Process
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7 Results
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9 Error Analysis
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1190degree bending test
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12Studied part
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16 Chapter 5: Flex, sag and wobble stiffness-limited design.
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16 Chapter 10: Keeping it all together: fracture-limited design.
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21Bibliography
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24Appendices..................................................................................................................................................
25 Bending Tests
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25 Tensile Tests
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26 2
- 3. IntroductionThis assignment report comprises the subsequent
and structural steps towards selecting the mostappropriate car body
material.In the beginning of the report one is able to review the
description of the chapters 5 and 10 from thebook Materials
Engineering Science Processing and Design (Michael Ashby, Hugh
Shercliff and DavidCebon; 2007, Cambridge University UK ;), this
was done as part of the research building activity.After the
description of the chapters, two tests, the tensile and the bending
are performed and togetherwith their results it is possible to
understand the behaviour of the materials given. With the
twodifferent graphs that are outcome to the tests, certain
conclusions are drawn about the mechanicalproperties of the
materials given.To conclude this report, an in depth analysis and
selection process of the most suitable car body is givenby each
individual of the group. This is based on the knowledge gained from
the 2 chapters of the book,the tests that were conducted and
analysed. 3
- 4. ObjectiveWithin the first part of the assignment we were
allocated a predefined set of metals namely; IndustrialSteel,
Aluminium 99 Alloy, a precipitate hardened Aluminium, Brass along
with an exotic materialknown as Polycarbonate (PC).These materials
then underwent two tests, a tensile and a bending test,whose
results are illustrated in the appendix of the report.The main goal
of this assignment is to be able as designers to select an
appropriate material based on itsbehaviour (mechanical properties)
to conceptualize our ideas. To practice this aspect, we conducted
thetensile and bending tests of five different materials (St37,
Al99, Al51ST, PC, Brass) and based on theresults as well as the CES
Edupack database, selected an appropriate metal for a car body.With
the knowledge gained through this selection process as well as the
theoretical overviews of theelastic modulus, stress and strain the
group was able to understand basic concepts of elasticdeformation,
plastic deformation and failure of the fore mentioned materials and
their mechanicalproperties i.e., strength, toughness, stiffness and
wear. In the following chapters, the tests and theirrespective
processes, their results and the conclusions are described. 4
- 5. Tensile TestsThe measurement results of the tensile test
performed are displayed in the appendix.ResultsThe graph and brief
explanations of the tensile behaviours are as follows: Tensile Test
( Stress against strain) 16000 14000 12000 10000 Standard Force (N)
Industrial Steel 8000 Aluminium Brass PC 6000 Al 51 st 4000 2000 0
0 10 20 30 40 50 60 70 80 90 Extension (mm) 5
- 6. AnalysisThe aluminium alloy (combination of aluminium,
silicon and magnesium that is precipitated to behardened) has a
standard force of approximately 12000N, this is particularly high
due to the molecularbonding in metallic structure (lattice)(Neuss,
2010)1, and the magnesium and silicon in this case, filled upvacant
spaces in the lattice to therefore have this high yield strength.
Its standard force before fracturetherefore was particularly high
reaching an approximate 13000N.The length however on looking to
thex-axis was much less and at the force of 12336.69N it eventually
cracked at exactly 10.67mm increment.Finally this material is very
tough of having a very high strain order of 13000N.Aluminium 99
Alloy displayed tensile behaviours remarkably different to that of
the aluminium 51 Stalloy. As one can notice on the graph the
standard force is much lower, approximately at 3000N and itstensile
strength considerably lower being about 3250N. This metal had quite
an increment in lengthbefore cracking at a length of
28.611mm.Industrial Steel displayed behaviours quite similar to the
pure aluminium however its standard force wasapproximately 12500N.
Its tensile strength was approximately 12000N and its also showed
quite a largeextension in length of 26.025mm. This is quite a tough
material due to its high strain order of 12000N.This exotic
material had no similarities whatsoever towards the metal
behaviours. It had a standardforce of approximately 2350N and a
tensile stress of just under 2000N. This metal had an extension
of80.167mm before fracture, clearly a brittle material with having
the strain so low.The last material we did was brass, a combination
of copper and zinc. This material had a very strangefinal part, and
showed a zigzag line before fracture on the graph, the reason still
unknown (but possiblydue to the inter-molecular bonding). Brass had
high standard force of approximately 14000N and tensilestrength of
12789N before fracture at a length of 39.9mm, proving to be quite a
tough material.From this tensile test it is possible to draw up
certain conclusions.If there is need for selection for tough
material, it is important to notice its tensile strength i.e.,
thestrain order.1Neuss, Geoffery. Chemistry: IB Diploman Course
Companion. Oxford: Oxford UP, 2010. Print 6
- 7. F Bending Test F l3 l f in m 3 E I f F l3 N h E in 2 3 f I m
b b h3F [N] f [mm] I in m 4 Elasticity limit 12 Inertia moment
Picture: Calculating the E-modulus The measurement results of the
bending test performed are displayed in the appendix. Process The
following are steps that we followed to accomplish the bending
test: The first step of the bending test process was to decide
which weights would be allocated and at what increment would they
be applied to the material We then attached the ruler to a box to
make sure that when reading the scale of the ruler it was constant
for all materials measured (see picture 1) After this we measured
the initial length (L0) of each material sample Next with a G clamp
and a block piece we clamped the edge of the material tested
(picture 1) The clamped area of the material tested was constant
for all the material samples Started to add different weights at
steady increments of all the materials The next step, was to
measure the deflection of the material This was repeated until the
material experienced plastic deformation (in some cases this case
did not occur because of the lack of weights) 7
- 8. Picture 1: a) Illustrating the mounting of the ruler to the
box b) Using a G clamp to hold the material Picture 2: a) Weights
added b) Measurement of deflection 8
- 9. Results A graph illustrating Stress [N/m^2] against Strain
[-] 4E+10 3.5E+10 3E+10 2.5E+10 Stress (N/m^2) St 37 Al 99 2E+10 PC
AL 51 St 1.5E+10 Brass 2mm Brass 1mm 1E+10 5E+09 0 0 0.1 0.2 0.3
0.4 0.5 0.6 Strain (-) 9
- 10. From the graph above one can see that the largest gradient
is that of St 37; this means that it has thelargest E-modulus
compared to the rest of the materials.This can be explained due to
the fact that St 37or Low carbon Steel as it is called has the
highest yield stress and the lowest plastic deformation. This
isthen followed by Brass 2mm and the value is confirmed by the
value in the table below.After this the graphshould illustrate the
E-modulus of Aluminium 51 ST or Al6082-T6 and then Al 99
(weformulated a hypothesis that Al99 is almost pure Aluminium),
which unfortunately cannot be seen in ourgraph,this is due to the
fact that for Al51st the weights that were applied were too large
to start withand the material sample deformed plastically already
at 900 grams. On further discussion, we needed tohave started with
low weights.The most exotic material Polycarbonate displayed the
smallest gradient which stayed true when wecompared this to the
documented E-modulus.Brass 1mm showed a large error. It is assumed
that the thickness of the material does not affect the E-modulus,
in our graph it did. This may have been due to the positioning of
the material when carryingout the bending test and reading off the
deflection. The g-clamp that was holding the material to thetable
was not tightened tight and therefore the deflection was added.
This was a human error as well asother uncertainties (refer to
error analysis) Material Calculated E-modulus (Gpa) Actual
E-modulus(Gpa) [found in CES Edupack 2011]St 37 (Low Carbon Steel)
132 200-215Al 51 St ( Al 6082 T6) 69.9 70 -74Al 99 (Pure Aluminium)
63 69 -72Brass (Wrought Copper: CuZn30) 114 90 -110PC
(Polycarbonate) 1.95 2 - 2.44 A table illustrating calculated
values against documented values 10
- 11. Error AnalysisThe uncertainties that occurred during the
bending test were the following: The ruler The weight piece The
vision of the measurer Picture 3: Vision of the measurer was an
error The perpendicularity of the ruler The positioning of the
weight The positioning of the material when measuring the
deflection The amount of pressure while clamping the material The
value of the gravitational constantIn our opinion these were the
most crucial uncertainties to be mentioned and may have
largelycontributed to the values of shown on the graph; however
there are many more errors that could havecontributed to the
results. 11
- 12. 90degree bending testAs an added part to the bending test,
we tried to bend the material to 90degrees with a force; this wasto
notice which material bent to 90degrees and underwent plastic
deformation and what effects thishad; below are the results and
images. Material ResultsSt 37 Maximum force needed to bend showing
High strength Material does not bend backAl 99 Not enough force to
bend it Very high strength Material does not bend backAl 51 St 5725
g of weight Material bends back to 40degrees when force is removed
Cracks appear in the materialPC Force used for 28 degrees plastic
deformation 5340 gBrass 1mm Force used for 90 degrees plastic
deformation ERROR of how much force was usedBrass 2mm Force used
for 90 degrees plastic deformation 4890 g of weight used No cracks
/ breaks A table illustrating the 90degrees scale testThe results
seen in the table above showed that St37 and Al 99 needed a high
amount of force to causeplastic deformation, however Poly
carbonateunderwent plastic deformation at 28 degrees andnot at 90
degrees, this was due to its lowest elasticmodulus as compared to
the other 6 materials. Picture 4: Tool for carrying 90degree
testing 12
- 13. Picture 5 :PolycarbonatePicture 6: Aluminium 99 13
- 14. Picture 7: Brass 1mmPicture 8: Steel 37 2mm 14
- 15. Picture 9: Aluminium 51 ST Picture 10: Brass 2mm 15
- 16. Studied partChapter 5: Flex, sag and wobble
stiffness-limited design.This chapter starts with the standard
solutions to elastic problems in five different forms.Standard
solutions to elastic problems.Elastic extension or compression.In
this example, a tensile or compressive stress applies to a tie. The
stiffness S can be calculated with S =F/,With F being the load and
the deflection. 2 (M. Ashby, 2007)The stress is uniform over the
whole section A, as can be seen in the graph on the right.Elastic
bending of beams.For the bending of beams, the neutral axis becomes
a curved axis. The lower part of the beam is loadedin compression,
where the top part is loaded in tension. At the neutral axis is the
point wherecompression changes into tension. This result is moment
M 3 (M. Ashby, 2007)2 Ashby, M.; Shercliff, H.; Cebon, D. Materials
Engineering, Science, Processing and Design. University
ofCambridge. 2007. First edition. p. 83. 16
- 17. Torsion of shafts.In the third example the torsion of
shafts is described. Again, there is a neutral axis at which the
stress isneutral. Moving outward from this axis along the radial r,
the shear stress increases, with the oppositedirection of r having
a negative shear stress, working in the opposite direction. 4 (M.
Ashby, 2007)Buckling of columns and plate.There is a critical load
at which a column or plate will fail, which is Fcrit. This critical
load can be calculatedwith the following formula: Fcrit = (n22EI) /
L2With L being the length of the column or plate, El the flexural
rigidity and n the number of halfwavelengths of the object that is
buckled. For instance, for the first situation in the picture
below, n = ,because the shape shows a quarter wavelength.3 Ashby,
M.; Shercliff, H.; Cebon, D. Materials Engineering, Science,
Processing and Design. University of Cambridge.2007. First edition.
p. 83.4 Ashby, M.; Shercliff, H.; Cebon, D. Materials Engineering,
Science, Processing and Design. University ofCambridge. 2007. First
edition. p. 83. 17
- 18. 5 (M. Ashby, 2007)Vibrating beams and plates.When a system
is vibrating in one of its natural frequencies, it can be depicted
as a mass m attachted toa spring with stiffness k. The lowest
natural frequency can be calculated with: f = (1 / 2) (k / m)1/2.In
the following picture you can see three situations with different
end constraints and their lowestnatural frequencies. C2 is a
constant depending on the end constraints. (M. Ashby, 2007)5 Ashby,
M.; Shercliff, H.; Cebon, D. Materials Engineering, Science,
Processing and Design. University ofCambridge. 2007. First edition.
p. 87. 18
- 19. Material indices for elastic design.The next part of this
chapter is about ranking based on objectives, which are criterion
which must eitherbe minimized (like costs or weight) or maximized
(such as energy storage). With this criterion, a list ofobjectives
can be made.As an example, a panel must be constructed with the
objective to be as light as possible, while thedeflection does not
exceed under load F. The thickness h is free. h can be minimized in
order to makethe panel lighter, but this goes at the cost of the
deflection exceeding while under the load F.The weight m of the
panel can be calculated with m=AL=bhLAgain, for the stiffness S can
be calculated with S =F/,but this stiffness must be at least equal
to S* in the formula: S* = (C1EI) / L3C1 is a constant which
depends on the way that the force is distributed over the
cross-section. Theheight of the panel was free, but we can use the
stiffness constraint to eliminate this undefined variable. 6 (M.
Ashby, 2007)Minimizing material cost6 Ashby, M.; Shercliff, H.;
Cebon, D. Materials Engineering, Science, Processing and Design.
University ofCambridge. 2007. First edition. p. 89. 19
- 20. When cost is the factor that must be minimized, C, the
total material cost, is the result of: C = mCm = ALCmwhere Cm is
the material price.However, for a complete product, not only
material costs, but also manufacturing costs for shaping,joining
and finishing the product must be made.Plotting limits and indices
on charts.Screening: attribute limits on charts.Other than
constraints that certain designs imply, there are also constraints
caused by materialsthemselves. To make a selection of effective
materials for a design, these limitations can be plotted onthe axes
of a chart. By applying the requirements of the design, are window
within the chart can beformed, and all the materials within this
window meet the constraints of the design. This is calledScreening.
Then, a more detailed material selection can be made. This is
called Ranking.Ranking: indices on charts.The next step in
selecting the right material is choosing from the screened
materials the one that willoffer maximum performance.Computer-aided
selection. Because of the sheer number of material, selecting them
by hand can be quite unpractical. Differentkinds of software are
available to ease the process of selecting the right material. By
entering limitationsfor the material chose, it is possible to
filter out all the materials that fall outside of these limits.
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- 21. Chapter 10: Keeping it all together: fracture-limited
design.Standard solutions to fracture problemsTensile stress
intensity k1 caused by a crack depends on crack length, component
geometry and the waythe component is loaded. Cracks will not expand
if k1 is kept below the fracture toughness k1c of thematerial of
the structure. 7We can manipulate geometry and points of pressure
in a component design in order to avoid futurefractures. The
non-destructive testing (NDT) to make sure there are no cracks
which have wrong values,this way we can choose materials with
adequate fracture toughness.7 Ashby, M.; Shercliff, H.; Cebon, D.
Materials Engineering, Science, Processing and Design. University
ofCambridge. 2007. First edition. p. 205. 21
- 22. Material indices for fracture-safe designLoad limited
designMaterials with highest values of fracture toughness k1c can
support larger loads. If the fracturetoughness is below M1, it may
fail in a brittle way if the stress exceeds. 8Energy limited
designExamples of designs that are energy limited instead of load
limited are springs and containment systemsfor turbines and
flywheels. 98 Ashby, M.; Shercliff, H.; Cebon, D. Materials
Engineering, Science, Processing and Design. University
ofCambridge. 2007. First edition. p. 207.9 Ashby, M.; Shercliff,
H.; Cebon, D. Materials Engineering, Science, Processing and
Design. University ofCambridge. 2007. First edition. p. 207.
22
- 23. Displacement limited design.Displacement limited designs
must allow enough elastic displacement to allow flexure or
snap-actionwith no failure. Materials with large values of M3 are
the best for displacement limited designs. 10Case studyForensic
fracture mechanics: pressure vessels. 11Pressure vesselss purpose
is to contain a gas under pressure. Their failure can mean a
catastrophe.A filled truck-mounted propane tank exploded when its
driver left it in the sun with the engine running.The tanks
longitudinal welds surface had a crack of 10 mm that was growing
slowly by fatigue everytime it was emptied and refilled. This was
the apparent cause of the failure. According to this, thepressure
needed to generate the explosion was 3,8MPa, while the safety limit
was of 1,5 MPa.Afterfurther tests, heat from the sun and from the
exhaust system of the truck where proved as the cause ofthe high
temperature of the tank. This made the crack propagate by a
pressure higher than 3,8 neededfor it to fail. In normal
circumstances the crack would not have propagated.10 Ashby, M.;
Shercliff, H.; Cebon, D. Materials Engineering, Science, Processing
and Design. University ofCambridge. 2007. First edition. p. 207.11
Ashby, M.; Shercliff, H.; Cebon, D. Materials Engineering, Science,
Processing and Design. University ofCambridge. 2007. First edition.
p. 209. 23
- 24. BibliographyGranta Design Limited. (2011). Edupack 2011.
Cambridge, United Kingdom.Houtzger, Overbeeke, & Vennix.
(1999). Matbase. Retrieved March 11, 2012, from www.matbase.com.M.
Ashby, H. S. (2007). Materials, Engineering, Science, Processing
and Design. London: Elsevier.Neuss, G. (2010). Chemistry: IB
Diploma Course Companion. Oxford: Oxford. 24
- 25. AppendicesBending Tests 25
- 26. Tensile TestsSee C. Meesters Tensile Tests results 26