IO1070-17/questions and excercises

Rood gemarkeerde vragen zijn vragen die EERDER als belangrijk zijn omcirkeld of aangegeven door Erik Tempelman. UPDATE NA COLLEGE 28/6, markering in geel. Besteed aandacht aan deze vragen! Andere zijn niet van belang voor dit tentamen!�

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Lecture 1.1 / Materials / Voor college.

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Q1.) When did the first polymers begin to appear on the market? And what about glass-
and carbon-fibre reinforced polymers (GFRP and CFRP)?

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       Very first polymers as early as 10000 BC.
       First actual plastics around 1900.

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GFRP around 1930.

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       CFRP around 1990.

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Q2.) How much has the diversity in materials increased since 1800? How can we handle
this diversity in product design?

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   Diversity in materials increased from a hand full to around 160.000 different

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types today.

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   We can handle this through handbooks and digital databases. Computer

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Aided Design/ Computer Aided Selection also helps when it comes to diversity in materials.

Q3.) Materials have properties. These properties are organized into five broad groups.
Which groups are these?

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   Mechanical properties.
   Thermal properties.
   Electrical, Magnetic and Optical properties.
   Chemical properties.
   Environmental properties.

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Q4.) What are the six basic families of materials, according to Materials?

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       Metals
       Elastomers
       Glasses
       Polymers
       Ceramics
       Hybrids

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Q5.) Material families can be organized further in a taxonomy. How is this done in the
book?

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       The book shows:

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The kingdom (Materials)
It’s families (See Q4)
It’s Classes and Subclasses
Members
Attributes

Q6.) What is the full taxonomy for the six reference materials (RM’s) used in this course?
The reference materials in IO1070 are: low carbon steel (AISI 304), stainless steel, age-hardening wrought aluminium alloys, PP with and without 40% glass fibres, and pine wood. You can look them up in CES, level 2 to obtain key data on material properties and further design information, such as typical applications.

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   This way…

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Materials → Metals → Steels → Low Carbon Steel → Attributes
Materials → Metals → Steels → Stainless Steel → Attributes
Materials → Polymers → Polypropylene → Unfilled→ Attributes
Materials → Polymers → Polypropylene → Glass filled → PP with 40% glass fibres → Attributes
Materials → Natural Materials → Woods → Pine Wood → Attributes
Materials → Metals → Al alloys → Age-hardening wrought aluminium alloys → Attributes

Q7.) Material properties can be charted, e.g. with bars or with bubbles. Such charts often
have (double) logarithmic scales. Why is this?
The range can go up to ten million. So it makes sense to use logarithmic scales. When comparing two values with very different ranges in a graph, the graph won’t be readable if the two axes have the same scale. Also some materials have a wider variety of values than others, which can make the graph unnecessarily large. You can separate the different families of materials easier and have a better overview of which materials have matching properties.

Q8.) What is the key difference between original design and redesign? Which one do
you think is most common in industrial product design?
Original design → Starts from a new concept.
Redesign → Starts from an existing product. (Most common, own opinion).

Q9.) When, in a design process, do you typically choose materials by family? When by
class, or subclass? When by member?
You choose your products by:
Family at the concept stage.
Class or subclass at the embodiment stage.
Member at the final stage of detailed design.

Q10.) What are the four essential steps in material selection for design - i.e. what is the
general material selection strategy?
The four steps:
Translation
Screening
Ranking
Documentation
�
Lecture 1.1 / Statics 2D / Voor workshop.

Q1.) In terms of support, what is meant with statically under-determinate, statically
determinate, and statically over-determinate, also known as statically indeterminate? In your answer, refer to the number of unknown reactions (forces plus moments). Consider 2D situations only. Draw examples!
Statically determinate: Structures that can be analyzed using statics

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                   equations only. Equilibrium in all directions.

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   Statically indeterminate:    Can’t be analyzed using statics equations only.
                   They require other material properties, such as
                   deformations, in order to analyze them.

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Statically under-determinate: If a body has less constraint than required.
It is not determinate by the outside forces.

Balk 3 en 4 zijn statically indeterminate (statisch onbepaald), balk 1 en 2 zijn statically determinate (statisch bepaald). Want balk 3 en 4 hebben 4 krachten erop werken en balk 1 en 2 hebben er 3 krachten op werken.

Q2.) How many equations are there to satisfy for static equilibrium? Which equations are
these? Again, consider 2D situations only.
Three equations to satisfy for static equilibrium.
⅀ M0 = 0;
⅀ Fx = 0;
⅀ Fy = 0;

Q3.) In what kind of support is the number of unknown reactions exactly equal to the
number of static equilibrium equations? In what kind is it less, and in what kind is it more?
In a determinate situation the unknown reactions are equal to the equilibrium equation. In an underdetermined situation there are more unknowns than equations. In an overdetermined situation there are more equations than unknowns
→ Statisch onderbepaald: Onbekenden > Vergelijkingen
→ Statisch overbepaald: Onbekenden < Vergelijkingen

Q4.) Do the actions (i.e. external forces and/or moments on the structure) in any way
affect how determinate a structure’s support is: yes or no?
No they don’t. The external forces determine the magnitude of the reaction forces but they do not define how determinate a structure is. The amount of unknowns and equations do.

Q5.) Looking ahead at the next workshop: how many unknown reactions are there for
statically determinate support in 3D situations? How many equilibrium equations are there now, and which are these?
Six equations in 3D situations for equilibrium.
⅀ Mx = 0;
⅀ My = 0;
⅀ Mz = 0;
⅀ Fx = 0;
⅀ Fy = 0;
⅀ Fz = 0;�
Workshop 1.1 / Statics 2D / Tijdens workshop.

E1.) On this page you see nine diagrams. All dimensions are as indicated in the upper-left
diagram. For each situation, determine the reaction forces and use these to draw the free body diagrams. NB: speed matters – this exercise should take you very little time to complete correctly, and certainly less than 15 min in total (!). If you encounter problems, address these during workshop 1.1.

Afbeelding: Freebody diagrams

Lecture 1.2 / Mathematics / Voor college.

For inclusion in our own “map of mathematics for design”, we’ll consider the following ten areas. Your assignment now is (a) to describe what each area is and (b) provide one telling example of how it can be applied to product design. In short: what is it, and what is it good for? Consider:

Q1.) regular differential equations (in Dutch: “reguliere differentiaalvergelijkingen”);
Functies differentiëren met 1 onafhankelijke.
Maximale snelheid van een product berekenen.

Q2.) partial differential equations (“partiële differentiaalvergelijkingen”);
Functies differentiëren met meerdere onafhankelijken.
Gebruiken bij het berekenen van krachten op een meervoudig gebogen oppervlakte.

Q3.) systems of linear equations (“stelsels van lineaire vergelijkingen”);
Uitrekenen van onbekenden.
Verbanden leggen tussen twee verschillende metingen.

Q4.) calculus, i.e. differentiation and integration (“analyse, d.w.z. differentiëren en
integreren”);
Richting en oppervlakte bepalen van functies.
Inhoud van voorwerpen bepalen. Hoek van een gebogen helling bepalen.

Q5.) complex numbers (“complexe getallen”);
Abstracte getallen waarmee gerekend kan worden.
Elektrische circuits.

Q6.) probability and statistics (“kansrekening en statistiek”)
De wetenschap, de methodiek en de techniek van het verzamelen, bewerken, interpreteren en presenteren van gegevens.
Gebruiksonderzoeken analyseren en voorspellingen maken.

Q7.) Fourier analysis (Fourieranalyse);
Uitdrukken van grafieken in goniometrische functies.
Geluid genereren met een elektronisch apparaat.

Q8.) vectors and matrices (“vectoren en matrices”)
De matrix is een middel om samenhangende gegevens en hun bewerkingen op een systematische en overzichtelijke wijze weer te geven.
Krachten op producten berekenen.

Q9.) number theory (“getallenleer”)
Study of integers en bewerken ervan (+,-,x,/,^,√)
Om uit te reken hoeveel producten er zijn

Q10.) crypto- analysis and cryptography (“cryptoanalyse en cryptografie”)
Het verbergen of versleutelen van te verzenden informatie.
Error correcties en beveiliging in computersystemen.

Q11.) Which of these ten areas are needed to find the deflections of a beam in
bending?

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       → Fourierreeks

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Q12.) Which for predicting waiting times for service desks?

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       → Kansrekening

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Q13.) Which for determining reaction forces in structural design problems?

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       → Vectoren

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Q14.) Which for modelling elasticity of fibre-reinforced plastics?
→ Calculus

Q15.) Which for optimizing the aesthetics of all sort of designs, from page lay-out to
building facades?

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       → Statistics�

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Workshop 1.2 / Statics 3D / Tijdens workshop.�
E1.) On the previous page you see a block with dimensions L x B x H. It is supported with
a ball hinge at node 4 (so, providing reactions in x, y and z-directions) and with two rolling supports at nodes 3 and 8 (so, reactions in y-direction only). Is it in statically determinate support? If so, why? If not, add one or more rolling supports to make the support statically determinate.

The block has (like all rigid objects in 3D) six degrees of freedom, but as shown, it has only five (independent) constraints. It therefore cannot be in statically determinate support. And indeed it isn’t: observe that while all three translations and two rotations are constrained, the block can still rotate around the axis through points 2 and 4 (so, this is a rotation in the xz-plane, i.e. rotation around the y-axis). We can make the block’s support statically determinate by constraining this rotation also. This we can do e.g. by adding a rolling support at point 7, oriented along the x-axis.

E2.) First, ensure the block is in statically determinate support. Then, assume the block is
loaded with a certain force F acting on node 5 in the x-direction. Next, write down the system of equations for static equilibrium at/around the origin. Finally, solve this system by hand and draw the FBD.

There are six equations of static equilibrium, one for each degree of freedom: balances of forces and three balances of moments. For the rotations, the smart choice is to draw up the accompanying equations for point 4, as three of the six unknown reactions (= those at point 4) then have no moment arm and are equal to zero. For shorthand, we write all reactions as R4x, R8z and so on, with R4x being the reaction force at point 4 in the x-direction. Important: in the FBD, we should include all reactions (even the ones we might suspect to be zero) and by default, include them as positive. Similarly, the action we write as F5x, for a force at point 5 in the x-direction.

⅀ Fx = 0; R4x + R7x + F5x = 0; (1)
⅀ Fy = 0; R3y + R4y + R8y = 0; (2)
⅀ Fz = 0; R4z = 0; (3)
⅀ Mx = 0; R3y \* B = 0; (4)
⅀ My = 0; - R7x \* B - F5x \* B = 0; (5)
⅀ Mz = 0; R8y \* L + F5x \* H = 0; (6)

```
   L = 0.60 m
   B = H = 0.20 m
   F5x = 100 N

```

Equations (3) and (4) immediately tell us that R4z = 0 and R3y = 0.
Equation (5) tells us that R7x = - F5x = - 100 N.
Together with equation (1), this tells us that R4x = 0.
Equation (6) tells us that R8y = - F5x \* H/L = 100 \* 0.20/0.60 = - 33.3 N.
That, combined with equation (2), then gives the final reaction, R4y = - R8y = 33.3 N.

E3.) Use Maple to solve the same system. Next, do so again, but for a different loading
situation (e.g. for a force F on node 5 in the y-direction, or on node 6 in the z-direction.

E4.) Two pages back you see a tubular structure clamped at the origin, then running
along the z-axis for length L1, running along the x-axis next for length L2 and finally running along the y-axis for length L3. It is loaded with a force F at its free end, oriented in the y-direction. Use Maple to set up and solve the system of equilibrium equations. Bonus question: which part is now in tension, which in bending, and which in torsion? Bonus vraag → L1=torsion, L2=bending, L3=tension

E5.) Three pages back also has a tubular structure clamped at both sides, loaded with a
force F at the joint between both tubes. Draw up the full system of equilibrium equations, with all unknown reactions, respective to the origin. Can you solve this system? Can Maple help you? Does it help if you also draw up this system with respect to the other end of the structure (i.e. the other clamp)?
NB: to determine the reactions in situations of statically indeterminate support, we typically need to take into account how the structure deforms. This makes stiffness the first logical material property to study.

Het antwoord zegt eigenlijk zoiets als: Je hebt hier 12 variabelen (3 krachten 3 momenten in punt A en punt C) en maar 6 formules, je zal die dus theoretisch moeten aanvullen met formules over bending. Hierbij kan Maple niet helpen.

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Lecture 2.1 / Stiffness / Voor college.

Q1-14 refer to Materials chapter 4; Q15 refers to Materials chapter 5. All questions test your base knowledge.

Q1.) What are the units of density? How is it typically measured? What are the densities
of our six reference materials?
Kg/m3 or Mg/m3
Massa met weegschaal meten, volume meten door middel van de verplaatsing van een vloeistof.
Low carbon steel: 7,8e3 - 7,9e3
Stainless steel:7,6e3 - 8,1e3
Age hardening Al. alloy: 2,5e3 - 2,9e3
Pinewood: 440 - 600
PP: 890 - 910
PP + 40% glasvezel (GFRP): 1,18e3 - 1,2e3

Q2.) The term “tie” refers to any structural element loaded primarily in tension. Which
other four structural elements are typically recognized, and what are the accompanying modes of loading?
Column, Beam, Shaft, Shell (tie is the fifth one)
Tie: tension, Column: compression, Beam: bending, Shaft: twisting or torsion, Shell: compression/pressure or bi-axial tension

Q3.) How, in engineering, is tensile stress defined? How is shear stress defined? See
Fig. 4.3 p. 55: what guarantees that we have equilibrium of forces and moments? Similarly, how is tensile strain defined, and how is shear strain defined?
Tensile stress: weerstand van een object dat de neiging heeft om te breken. Dit is de hoogste spanning dat het object kan hebben zonder kapot te gaan. Gemeten in N/m2.
(Trekkracht). Formule → σ = F/A
Er is een evenwicht in krachten omdat er in tegengestelde richting evenveel kracht wordt gezet.
Shear stress: De maximale schuifspanning die door een materiaal of grondsoort kan worden ondergaan.
(Schuifsterkte). Formule → 𝛕 = Fs/A
Op tegengestelde vlakken wordt een tegengestelde kracht uitgeoefent die elkaar opheffen. En er is geen moment omdat de koppels tegengesteld gericht zijn, dus heffen elkaar op.
Tensile strain: Verandering van vorm door tensile stress. Het object rekt uit in de richting waarin getrokken wordt en krimpt in de andere richtingen.
Formule → 𝛆 = (L - L0)/L0
Shear strain: Verandering van vorm door shear stress.
Formule → 𝛄 = w/L0

Q4.) What is Hooke’s law for tension, and what is this law for shear? What, in this
context, are E and G ? What are their typical units, and what are their values for our reference materials (RM’s)?
Hookes Law Tensile: σ = E \* ɛ
Hookes Law Shear: 𝛕 = G \* 𝛄
E = constante van proportionaliteit bij tensile stress [N/m2].

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           Laag koolstof staal: 200 - 215 GPa

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Roestvast staal: 189 - 210 GPa
Aluminium: 68 - 80 GPa
Dennenhout: 0,6 - 0.9 GPa (across grain) / 8,4 - 10,3 GPa (along grain)
Polypropeen: 0,896 - 1,55 GPa
PP + 40% glasvezel (GFRP): 5,72 - 6,01 GPa
G = constante van proportionaliteit bij shear stress [N/m2].

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           Laag koolstof staal: 79 - 84 GPa

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Roestvast staal: 74 - 84 GPa
Aluminium: 25 - 28 GPa
Dennenhout: 0,35 - 0,4 GPa (across grain) / 0,62 - 0,76 GPa (along grain)
Polypropeen: 0,316 - 0,548 GPa
PP + 40% glasvezel (GFRP): 2,08 - 2,19 GPa

Q5.) What are stress-strain curves? How are they determined? Why are they important?
Kromme waarin te zien is hoe de druk zich verhoudt tot de vervorming van het object. Ze worden opgesteld door de hoeveelheid vervorming te meten bij bepaalde tensile of compression stresses.

Q6.) What is the Poisson’s ratio? Again, what are the values for our RM’s? Which
relationship exists between this ratio and E and G ?
G = E / (2(1+𝛎)) is de beschrijving van hoe een materiaal reageert op een trek- of drukbelasting, namelijk welke rek er loodrecht op de trekrichting ontstaat (dwarscontractie). Hierbij is 𝛎 Poisson’s ratio.

Q7.) What is Hooke’s law in three dimensions? Present a numerical example for a shell
under bi-axial stress.
a) The relationship between different stresses in three dimensions on a material.

b) We chose RVS as the material of the cube. This resulted in the following variables and formulas:

Q8.) Besides mechanical stress, are there other stimuli that generate strain as a
response? If so, which ones? Do they apply to all materials, or to a few?

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       Thermal: expansion coefficient.(alles)
       Magnetic: Magnetostrictive constant (ferro-magnetische materials)
       Electric field: piezoelectric constant (piezoelectric materials)

```

Q9.) What is anisotropy? To which one(s) of our RM’s does it apply? What is affected:
stiffness, strength, or both?
Op molecuul basis heeft het invloed hoe de atomen, moleculen en structuren daarin ten opzichte van elkaar zijn georiënteerd en ruimtelijk zijn opgebouwd. PP with 40% glass fibers en dennenhout hebben te maken met anisotropie.

Q10.) What do the terms FCC, BCC and CPH mean? Which applies to which of the three
metals in our RM’s?

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   FCC, BCC en HCP (=CPH, beide gebruikt in het boek) zijn kristalstructuren.

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FCC: Face Centered Cubic Structure (Close-packed structure with more than 2 different layers, ABCABCABC).
BCC: Body Centered Cubic Structure (Non Close-packed structure with multiple layers).
HCP / CPH: Hexagonal Close Packed / Close-Packed Hexagonal Structure. (Two layer structure, ABABAB).

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       Aluminium → CPH en FCC
       Laag koolstof staal → BCC
       Roestvast staal → BCC

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Q11.) Polymers can be thermoplastics (amorphous or semi-crystalline), rubbers, or
thermosets. Give two real-world examples of each (so, eight polymers total), and mention a typical application of each (so, eight applications in total).

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               PET: Water flessen
               HDPE: Melk pakken
           semi-kristallijn:

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PS: Plastic bestek
PP: Plastic bekers

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       Rubbers:
           EPDM: Dakbedekking

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Siliconen: Steunzolen
Butyl rubber: Rubber gloves
Natural rubber: Elastiekjes

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       Thermoharders:
           Bakeliet: oude telefoon behuizingen
           Polyester: Boot

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Q12.) Which two bond types exist in polymers? Which is the stronger (i.e. stiffer) one, and
which is the weaker one? Which is decisive for the stiffness of polymers? How does this bond type compare against the metallic bond?
Covalente bindingen / atoombinding, sterk
Waterstofbruggen, zwak
De stijfheid wordt niet alleen bepaald door de soort binding (covalent of waterstofbrug) maar ook door het aantal crosslinks. het is dus niet te zeggen dat materialen sterker worden door de soort bindingen.

Q13.) For polymers, how does the stiffness vary with temperature? What, in this context,
is the glass transition temperature?
Hoe hoger de temperatuur wordt hoe lager E (young modulus) wordt, dus de stijfheid gaat omlaag. De glass transition temperature is de temperatuur waarop de bindingen gaan smelten, hierdoor kan het makkelijker vervormd worden.

Q14.) Why can we not significantly affect the stiffness of metals? Why, and how, can we
typically do this with plastics? Can we also affect the stiffness of wood, and if so, how?
De molecuulstructuren van metalen zijn (bijna) niet aan te passen, terwijl dit bij plastics wel zo is. Hierdoor kun je de dichtheid en textuur aanpassen.

Q15.) What do the elastic stresses look like over the cross section of a tie in tension, a
beam in bending, and a shaft in torsion? Where in the structural elements are these stresses at their maximum value?

�
Workshop 2.1 / Stiffness / Tijdens workshop.

E1-7 refer to Materials chapter 5. All exercises test your insight. Note that for each equation (in fact, this holds for all equations in the entire course!!), you must be able to:

(a) explain each term (e.g. E , L etc.): what is it, and what are its units?
(b) apply the equation, quickly and correctly, if all but one of the terms are given to find the unknown term.
(c) know when the equation will produce reliable answers, and (eventually) understand why this is so.

E1.) Which equation describes the elastic extension or compression of a tie? Make
sure you go through (a)-(c) as just prescribed – also for subsequent questions, where appropriate.

```
   δ = (L0F)/(AE)
       L = beginlengte
       E = Young’s modulus

```

A = cross-section area
δ = Deflection

E2.) What is the second moment of area I , also known as the area moment of inertia?
What are its units and values for common cross sections such as pipes, solid beams, and I-profiles?

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       Second moment of area I: → Bending of Beams!

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I = ∫((y2b(y)dy)
y = De afstand tot de neutrale as
b = De breedte van het materiaal op hoogte y
Beams Moment I: (bh3)/12 (Onthoudt dat de as waar je over heen buigt tot de macht 3 gaat.
Pipe Moment I: π/4r4
Holle buis: I = (𝞹/4 (rbuitenkant4 - rbinnenkant4)
I Profile: Ix = (a h3 / 12) + (b / 12) (H3 - h3)
Iy = (a3 h / 12) + (b3 / 12) (H - h)

E3.) Rework Example 5.2 (a) for each of the other four RM’s, keeping everything the
same except the material. What are the elastic deflections? Are these less than 10% of the length, which is one of the requirements for the results to be reliable?

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       Het enige wat de materiaal beïnvloed is de E modulus in de formule

```

δ=(FL3)/(C1EIxx). C1 wordt bepaalt door de soort lading.
Met C1 = 3; Ixx = 2.1 \* 10-12; F = 10N; L = 0.25m.
Dus voor low carbon steel δ = (10 \* 0.253 )/(3 \* (207.5 GPa) \* (2.1 \* 10-12 I)= 119 mm
Aluminium = 335 mm
Dennenhout = 33068 mm
Polypropyleen = 24802 mm
De deflections zijn dus niet minder dan 10 % dit komt door het feit dat deze materialen de 10 Newton niet elastisch kunnen dragen, ze vervormen definitief of ze breken.
Om de resultaten dus reliable te maken moet het materiaal de kracht elastisch kunnen dragen.

E4.) What is the polar moment of area J ? What are the units? What is its value for a
solid round shaft 2 cm across? What for a hollow shaft 3 cm across with wall thickness 2 mm? What happens to J if you make a closed cross section open, e.g. going from O-shaped to C-shaped?
Polar moment of area J: → Torsion of Shafts
J=riro2r3 dr
Als r gemeten is vanuit het middelpunt v.d. cirkel.
ro= buitenste straal.
ri = binnenste straal.

E5.) Which equation describes the elastic twisting of a shaft in torsion? Again, make
sure you go through (a)-(c) as just prescribed.

```
       Formule: =TLGJ

```

T = torque
L = lengte
G = shear modulus
J = polar moment → in het materials boek ook wel K
= angle

�
E6.) Which material index should we maximize for a light, stiff tie in tension? Which for a
beam in bending, a shaft in torsion, or a panel in bending?

```
       Tie in tension: Mt=E
       Beam in bending: Mb=(E)12

```

Panel in bending: Mp=E13
Shaft in torsion: Ms=GE

E7.) Make a chart like Figure 5.10 and spot all six RM’s. Rank these five materials with
respect to specific stiffness for beams in bending.
Pinewood
Aluminium
PP+40%GF
Stainless steel
Low carbon steel
PP�
Lecture 2.2 / Strength / Voor college.

Q1-12 refer to Materials chapter 6; Q13-15 refer to Materials chapter 7. All questions test your base knowledge.

Q1.) How, in metals, is the yield strength defined? What about the tensile strength?
And what about the elongation, also known as strain-to-failure?

Yield strength σy for metals: σy is identified with the 0.2% proof stress which is the stress at which the stress-strain curve deviate by a strain of 0.2% form the linear elastic line.

Tensile strength σts in metals: when strained beyond the yield point most metals work harden causing the rising part of the curve until a maximum is reached: the tensile strength. Works for both tension as compression.

Elongation εf: is defined as a percentage that shows how much a specimen can be elongated before it fails: strain-to-failure.

Q2.) Draw accurate stress-strain curves for the metals among our reference materials
(RM’s). Consult CES level 2 for data.

�
Q3.) Draw stress-strain curves for PP (with and without 40% glass fibres) and pine
wood. How is yield strength defined for these materials?

```
       Enter answer here

```

Q4.) Which kinds of hardness measurements are used in engineering? For which
materials are they typically most suited? What are the practical advantages of measuring hardness?

```
   Ceramics: compressive crushing strength which give the elastic limit σel.

```

Hardness test: makes a small indent in the material and measures its size and the force that was needed to create it. Formula: H=F/A and Hv ≈ σy/3. The hardness test makes very small indents in the object and therefore is a better way of measuring the hardness of an object than other destructive tests.

Shore hardness: testing the resistance of a material to indentation by using a durometer, mostly used for rubbers.

Janka hardness: resistance to denting or wear, used for wood

Q5.) What are the definitions of true stress and true strain? Next, superimpose a true
stress-strain curve on a nominal stress-strain curve.
True stress: σt=F/A → takes into account the current dimensions, so in tension for example.
True strain: εt= ln(L/L0) = ln(1 + εn) where εn is the nominal strain and both strains are percentages.

Q6.) What would be the strength of perfect materials, as a function of their stiffness?
Why are actual materials much less strong?
Ideal strength: σideal = Fmax/a02 = S/10a0 = E/10. So the strength is ideal if σideal is a tenth of the modulus E.
No material can be assembled perfectly, there will always be point defects, solute atoms and dislocations. Therefore σideal will never consistently be achieved.

Q7.) What, in metals, are dislocations? What happens to dislocations when the stress
exceeds the yield strength, i.e. when we have plastic flow?

```
   Dislocation: an extra half plane of atoms squeezed into a material.

```

The dislocation starts to move and when it leaves, the crystal has suffered a shear strain γ.

Q8.) What is work hardening, and what does it do to dislocations? How much (Choose
between: + a little i.e. 10-20% or so, ++ a lot i.e. 20-50% or so, and +++ very much i.e. >50%) can it increase yield strength of metals? Which processes are generally used to achieve work hardening?
Work hardening: accumulation of dislocations generated by plastic deformation.
+++ very much i.e. >50% (see tabel 6.1 on p.155)

Q9.) Besides work hardening, mention three other ways to impede dislocation movement
in metals. How much can these ways increase yield strength?
Solution hardening: roughens the slip plane, making it harder for dislocations to move and thereby adding an additional resistance fss opposing dislocation motion.
Dispersion and precipitate strengthening: disperses small yet strong particles in the paths of the dislocations.
Grain boundary hardening: dislocations cannot pass grain boundaries in metals because the slip planes do not line up.

Q10.) Not all heat treatments increase strength of metals! Find two that do (and mention
on which metals they work), and find two heat treatments that actually make metals weaker. Why would anyone want to do that?

```
   Enter answer here

```

Q11.) What, in polymers are crazing, drawing, and shear banding? What is the role of
(the glass transition) temperature in these processes?
Drawing: when pulled in tension the chains of the polymers slide over each other, unraveling, so that they become aligned with the direction of stretch.
Crazing: polymers that do not draw at room temperature craze. Small crack-shaped regions within the polymer appear.
Shear banding: material is in compression and forms shear band (soort uitstulpingen).

Q12.) How can we increase the yield strength of plastics? Specifically, how much
stronger can e.g. PET get if we add (random, short) glass fibres?

```
   By blending, drawing, cross-linking and reinforcement.

```

Q13.) At what force, in [N], will a tie in tension begin to yield? How can you envision that
the external load is applied?

```
   If it exceeds σy it yields.

```

Q14.) At what moment, in [N·m], will a shaft in torsion begin to yield, and where is the
yield stress reached first? (As explained in NSFD - von Mises, yield stress in shear is typically 50-60% of yield stress in tension. FYI, ultimate shear strength is ~80% of ultimate tensile strength).

```
   𝜏max=σy/2, it is reached first at the surface.

```

Q15.) At what force, in [N], will a simply supported beam in bending begin to yield, and
where in the beam’s cross section is the yield stress reached first

```
   When: σmax > σy and it is reached first at the surface.

```

�
Workshop 2.2 / Strength / Tijdens workshop.

E1-2 refer to Materials chapter 7. E3 refers to the equation for buckling given on Materials pp. 103-104. E4 refers to NSFD – von Mises. All exercises test your insight.

E1.) Rework example 7.4 but now for the ball bearing in your bicycle wheel (steel balls,
radius 2.1 mm): determine contact radius, relative displacement, and maximum shear stress. For the force F, use 1 kN (emergency braking, all weight on two ball bearing balls).

E2.) See Figure 7.7 and the accompanying text. Determine the Ksc as a function of c for
one of the six geometries shown in the figure, assuming the loading mode is tension. Set ρsc = 1 mm = 10-3 m for this exercise. Does this Ksc get smaller or larger if instead of tension we would have bending?

E3.) Choose one of the following: a low carbon steel tube (radius R = 11 mm, wall
thickness t = 1 mm), a stainless steel tube (R = 11 mm, t = 1 mm), an aluminium tube (R = 15 mm, t = 2 mm), a PP tube (R = 15 mm, t = 5 mm) or a solid wooden rod (R = 15 mm). Next, determine the critical buckling load Fcrit for this column if L = 1 m, assuming hinged support (n = 1). What is the accompanying compressive stress on the column’s cross section, and does it get close to the (compressive) yield strength or not?

E4.) Make an accurate drawing of the crank of your bike, modelling this part as a solid
beam element with constant, circular cross section. Next, determine the bending moment and torsion moment when, with the crank in the horizontal position, you pedal down as hard as you can. Finally, use the Von Mises yield criterion to determine the equivalent uniaxial stress. As a percentage, how close are you to getting the crank to yield?
�
Lecture 3.1 / Engineering bending theory / Voor college.

Q1-15 refer to NSFD – engineering bending theory. All test your base knowledge.

Q1.) Imagine you are bending a solid beam with a rectangular cross section, L x W x H =
100 x 4 x 2 cm. Estimate how much stiffer it is when bent over the width compared to over the height: 2 times, 4 times, 6 times or 8 times. Also, estimate how much stronger it will be, for the same comparison.

```
   4 keer zo stijf
   8 keer zo sterk

```

Q2.) Now, imagine the same beam is half as long, i.e. L x W x H = 50 x 4 x 2 cm. How
much less will it deflect under the same force as before: 2 times, 4 times, 6 times or 8 times?
8 keer

Q3.) What is meant with a simply supported beam? Draw various examples.

```
   roller aan de ene kant en oplegging aan de andere kant

```

Q4.) Why, in a beam in bending, does the stress over the cross section vary linearly over
the height?

```
       Enter answer here

```

Q5.) Which equation describes the stress in a beam in bending? What does this stress
depend on?

```
       =MyI

```

Q6.) Calculate the area moments of inertia for solid and hollow cross section,
rectangular and round, if given that B = 2, B’ = 1.6, H = 4, H’ = 3.6, R = 2, R’ = 1.6, with all dimensions in cm. Express your answers both in m4 and in mm4 .

```
       Solid rectangular: Ixx= 10,667 m^4 en Iyy=2,667 m^4

```

Hollow rectangular: Ixx=4,45 m^4 en Iyy=1,44 m^4
Solid round: Ixx=Iyy=4pi
Hollow round: Ixx=Iyy=2,36pi

Q7.) Why precisely does there need to be a shear stress if the bending moment M is not
constant?
Er moet een shear stress zijn als het bending moment niet constant is omdat als het buigmoment toeneemt, de spanning toeneemt. Dan moet er een afschuifspanning zijn daarmee nog een tegengestelde spanning. p.55 van het boek

Q8.) Why is the shear stress zero at the upper and lower surface of the beam?
De afschuifspanning wordt veroorzaakt door krachten loodrecht op de balk. de kracht werkt dus niet op de balk maar binnen in de balk (tenzij 2 balken tegen elkaar “schuiven”)
afschuifspanning = V / A
V = de kracht (loodrecht op de balk)
A = oppervlakte van de doorsnede

Q9.) How much more efficient can we make hollow beams in bending, in practice, as
compared to solid beams?

```
       Tot 2x zo efficiënt

```

Q10.) What is the differential equation of flexure and which quantities does it relate to
each other?

```
   M = E * I * d^2y/dx^2

```

curvature of the beam to bending moment M

Q11.) What is a so-called cantilevered beam?

```
   inklemming aan het ene uiteinde, kracht aan het andere uiteinde

```

Q12.) List the six forget-me-not equations for bending of cantilevered beams.

Opmerking: voor afbeelding hierboven geldt; P = F, w = Q

Q13.) What is the wag-effect?
Buigt na de plaats van de kracht nog verder door omdat de balk aan 1 kant zit vastgeklemd. Reken uit door: δtotaal = δ1 + δ2 , met δ1= F·L^3/3·E·I & δ2 = L·F·L^2/2·E·I

Q14.) Can we use the forget-me-nots for cantilevered beams also for simply supported
beams? If so, how? If not, why not?

```
       Jazekers

```

Q15.) Which limitations are there to the used of forget-me-nots for bending (so,
considering deflections only)? Mention at least four different ones.
- Constant young’s modulus
- Length >10x zo groot als breedte en hoogte
- Linear elasticity
- Slender beams
- Overall deformation must be small compared to beams length
- Kracht werkt op de symmetrielijn.�
Workshop 3.1 / Engineering bending theory / Tijdens workshop.

E1-6 refer to NSFD – engineering bending theory. All exercises test your insight.
NB: these exercises all involve hand calculations only. You are recommended to use a calculator or perhaps a spreadsheet program, but nothing more. Workshop 4.1 continues on this topic of engineering bending theory, but then involving Maple as a tool for modelling and analysis.

E1.) Section 2 of NSFD – engineering bending theory closes off with a numerical
example. Rework this example, substituting the (solid) wooden beam for a hollow rectangular aluminium extrusion (B = 4 cm, H = 20 cm, wall thickness t = 0.5 cm). What is the resulting stress, and how does it compare to the yield stress of the material?

```
       Enter answer here

```

E2.) Section 5 of the same document provides a full derivation of the forget-me-nots for a
cantilevered beam loaded with a force F at the free end. For two other load cases (moment M and distributed load Q) it goes straight to the results: see Figure 10. Provide the derivations for these two cases.

```
       Enter answer here

```

E3.) Section 6 talks about the “wag-effect”: your job is to provide a numerical example.
Assume a steel I-profile with dimensions B x H = 3.0 x 6.0 cm, with thickness (for flanges and web) t = 0.3 cm. Assume it has length L = 1 m. Assume it is clamped to the wall on one side – so, a cantilever – and loaded with a bending force F = 2 kN, acting at 0.2, 0.4, 0.6, 0.8 or 1.0 m from the clamped side (NB: with the force at 1.0 m there obviously is no wag-effect). How does the downward deflection of the free end vary with the location of the force?

```
       Enter answer here

```

E4.) Section 6 provides the equation for a simply supported beam loaded with a
distributed force Q over its full length. Reproduce this result.

```
       Enter answer here

```

E5.) Run an internet search for “Michell Structure”. What do you see – and what do these
images remind you of?

```
       Enter answer here

```

E6.) What is the relationship between of the term d2y/dx2 and the radius of curvature R,
in engineering bending theory? So, what then is R for the beam shown in Figure 1, for the region of constant bending moment M? Assumed the dimensions and materials are as indicated in the numerical example at the end of section 2.

```
       Enter answer here�

```

Lecture 4.1 / Mathematics continued / Voor college.

This lecture returns to the Map of Mathematics for Design, as treated in Q&A lecture 1.2. It highlights several common tools that are available to work in various areas of this map, and that you – as a designer – are therefore likely to encounter during your study and your professional career.

One tool that you will use in this course (indeed, that you used already in workshop 1.2) is Maple. As a reminder of how Maple works, watch https://www.youtube.com/watch?v=4jFi7CmxXAk in preparation for this lecture. We recommend you watch this before this Q&A lecture. NB: considering there is a test in week 4 on Monday, prior to this Q&A lecture 4.1, no further preparation is absolutely required, but do watch the video!�
Workshop 4.1 / Engineering bending theory / Tijdens workshop.

E1-3 refer to NSFD – engineering bending theory . All exercises test your insight, as well as your ability to use Maple.

E1.) Use Maple to determine and plot the displacement diagram for the beam shown in
Figure 1. Next, select one of the six reference materials (RM’s) for the beam. Set L = 1.0 m and e = 0.2 m. If both forces F = 50 N, what cross section dimensions (so, B and H ) are required to ensure that the downward deflection of the beam in between both supports is less than 5 mm? (As a rule-of-thumb, observers cannot see deflections if they are less than 0.5% of the beam’s typical length - so, in this case, 5mm or less).

E2.) Use Maple to determine and plot the displacement diagram for the diving board
shown in Figure 13. If the force F = 2.4 kN (= 0.8 kN times a dynamic load factor of 3), and if the downward deflection of the free end should be 0.3 m, what board thickness is required? Assume a solid wooden diving board, 0.30 m wide. Do you think a wooden diving board is feasible?

```
       Enter answer here

```

E3.) Use Maple to determine and plot the displacement diagram for a cantilevered beam
loaded with a distributed force Q [N/m} along its length L = 0.50 m. First, assume a constant cross section B x H = 0.02 x 0.04 m. Next, change the beam to have a variable height H ranging linearly from 0.02 m at the free end to 0.06 m at the clamped end (so, same weight as before). How much do the total deflections differ?

�
Lecture 4.2 / Toughness / Voor college.

Q1 refers to Materials p. 61. Q2 refers to Materials p. 132. Q3-9 refer to Materials chapter 8. Q10 refers to Materials pp. 249-252. All test your base knowledge.

Q1.) Which equation tells us how much elastic energy we can store in a material, per unit
volume?

```
       W=0*d=0*dE=12(max)2E→ Zie vergelijking 4.16 in Materials.

```

Q2.) How much plastic energy can be stored in a material, per unit volume? Determine
this for one of the six RM’s, assuming the material is fully loaded (Assume the stress during plastic deformation to be the average of yield strength and tensile strength).
Wpl=0.2+ts2max of Wpl=0fσdpl of Wpl=pl mag ook gebruikt worden. (eq. 6.2 in materials)
Voor Low carbon steel is dit: Wpl=0fσdpl=[12σf2]=12297,51,262236,16 M Joule/m3

Q3.) What is toughness, as an informal material property? Which common materials
have high toughness, and conversely, which ones are brittle?
The resistance of a material to the propagation of a crack.

```
   → Resistance to crack propagation

```

Low carbon steel
Soda-lime glass

Q4.) How is fracture toughness K1c defined – and is this a formal material property or
not? How is it typically measured? Look up the values for the fracture toughness of the six RM’s.
The critical value of the stress intensity factor. K1c=σYield cY
Measured by loading a sample with a crack, recording the tensile stress at which the crack suddenly propagates.
Low Carbon steel: 41 MPa. m0.5

```
   Stainless Steel: 62 MPa. m0.5
   Age hardening wrought aluminium: 21 MPa. m0.5
   Pinewood: 0.4 MPa. m0.5
   PP: 3 MPa. m0.5
   PP + 40% glass fibres: 4.04 MPa. m0.5

```

Q5.) What is the formal definition of toughness, and how does it relate to other material
properties? Determine the values of Gc for the six RM’s. Compare the outcomes to Q3!
The ability of a material to absorb energy and plastically deform without fracturing.Gc=K1c2E
Low Carbon steel: 4.2

```
   Stainless Steel: 10.2
   Age hardening wrought aluminium: 3.2
   Pinewood: 0.1
   PP: 5.0
   PP + 40% glass fibres: 16.9

```

Q6.) At high strain rates (i.e. during impacts), the stiffness of plastics is higher than at
normal strain rates. If the fracture toughness stays the same, what would happen to the toughness?
Gc=K1c2E→ We verwerken deze situatie in met de formule.
Er gebeurt het volgende: - ==+ waar dus de toughness afneemt als gevolg. (E wordt groter, dus Gc wordt kleiner)

Q7.) Which family of materials is characterized by combining high stiffness with high
fracture toughness? And which is stiff but not tough?
Metals
Ceramics, Glass

Q8.) What is the ductile-to-brittle transition and how is it triggered? Which metals and
which plastics are prone to showing this transition?
It means that some material get brittle when temperature drops.
All polymers, and (some) metals without a FCC structure.

Q9.) How much can – say – aluminium be made stronger without impairing fracture
toughness? And what about PP: which modifications make it stiffer but more brittle, which make it stiffer and tougher? What do you expect is the effect on material price? (The last part of this question will require consulting CES. Advised is to do this at level 3).
About ten times, by alloying
Adding fillers, stiffer but more brittle
Adding glass fibres, stiffer and tougher
Tougher materials have a higher material price

Q10.) In designing for toughness (a.k.a. fracture-resistant design), K1c and Gc are not
the only properties that matter: there is a third one. What is this property, when does it apply, and how do polymers perform regarding this property?
M3 = K1c/E, in displacement limited design

```
   → Displacement limited design: Elongation (=LL) klein houden.

```

Polymers have relative low Young’s moduli so M3 for polymers are quite large and thus good for displacement limited design.

Q11.) Use any remaining time to revisit the questions from the previous weeks.
Zelf uitvoeren!�
Workshop 4.2 / Toughness / Tijdens Workshop.

E1 refers to Materials p. 61. E2 requires high school physics, as well as critical thinking. All exercises test your insight.

E1.) Rank all six RM’s on the maximum elastic energy that they can store, per unit
volume. Here, you should assume that the material gets loaded up to its yield stress. Here we assume that the material can in fact carry its yield stress, which means (for brittle materials) that the sample must be free of cracks. Otherwise, its lack of fracture toughness will likely cause it to crack (well) below yield stress! Which one wins, and by what margin? Next, rank them again, but now on maximum storable elastic energy per unit weight.

```
   Max. elastic energy stored / volume
   W = 12(*)2E
   Low carbon steel:                E = 200 GPa
                           σ0.2 = 250 Mpa
                           1.6 * 105 J/m3

```

```
   Stainless steel:                E = 189 GPa
                           σ0.2 = 170 MPa
                           7.6 * 105 J/m3

```

```
   Age hardening aluminium alloy:         E = 68 GPa
                           σ0.2 = 95 MPa
                           6.6 * 105 J/m3

```

```
   PP:                        E = 0.896 GPa
                           σ0.2 = 20.7 MPa
                           2.4 * 105 J/m3

```

PP + 40% Glass Fibers: E = 7.25 GPa
σ0.2 = 81 MPa
4.4 \* 105 J/m3

Pine wood: E = 0,6 GPa

```
                           σ0.2 = 1,7 GPa
                           0.2 * 105 J/m3

```

Max. storable elastic energy / unit weight

```
   Low carbon steel:            4.8 * 105 J/kg
   Stainless steel:            7,0 * 105 J/kg
   Age hardening Al. alloy:        5.6 * 105 J/kg
   PP:                    5.3 * 105 J/kg
   PP + 40% Glass Fibers:        7.7 * 105 J/kg
   Pinewood:                0.6 * 105 J/kg

```

E2.) To withstand drop impact, we require (among other things: To keep the decelerations
in check, the contact time during drop impact should also be controlled (longer = better). This favours foam-like materials.) that twice the product’s potential energy 2·mgh can be absorbed through elastic deformation only (Since your laptop will bounce up after impact, the casing will in fact need to absorb twice the amount of energy.). Assume you drop your laptop (m = 1.5 kg) from arm’s height (h = 1.5 m). How much potential energy will need to be absorbed? How much material will need to be deformed to do so, in cm3 ? Assume you have a PP laptop casing, dropping onto a perfectly rigid floor (Since the floor is perfectly rigid, it will not absorb any energy.).

```
       2 * m * g * h → 2 *  1.5 * 9.81 * 1.5 = 44.145 J
       W = 12(*)2E= 12(20.7*106)20.896*109= 239112.72 J/m3
       44.145239112.72= 0.000185 m3

```

E3.) Rework example 8.1 for PP with and without glass fibres. Will the ruler fracture
before yield, or yield before fracture?

```
       Enter answer here

```

E4.) (Use any remaining time to discuss your computer test results with your coach: can
you understand why you had the result that you had?)

```
   Zelf uitvoeren!�

```

Lecture 5.1 / Fatigue / Voor college.
We have the following questions about fatigue. Q1-15 refer to Materials chapter 9. All test your base knowledge. They will be treated during Q&A lecture 5.1 on demand.

Q1.) How important is fatigue, for products and structures for which mechanical strength
is required? Estimate the percentage of failures where fatigue plays a role.

```
       Very very important, heel heel belangrijk. → Majority (+++)

```

Q2.) Vibration is a form of fatigue. Which material property helps describe the effects of
vibration? List this property for the reference materials (RM’s).

```
       Mechanical loss coefficient / damping coefficient

```

Low carbon steel: 8,9e-4

```
       Age hardening aluminum: 1e-4
       Stainless steel: 2,9e-4
       PP: 0,0258
       PP + 40% GF: 0,00878
       Pinewood: 0,028

```

Q3.) Fatigue can be low cycle and high cycle. Where is the border between the two, in
number of cycles? (Hint: yes, this is a trick question)
Als hij plastisch vervormd is het low cycle als het elastisch vervormd is het high cycle. Dus onder Yield Strength is het high cycle

Q4.) Which two variables are shown in a so called S-N plot (a.k.a. S-N curve, S-N line)?
What type of scales will turn this plot into two straight lines?
Stress amplitude σ, number of cycles of failure Nf. De as van cycles of failure is in een logaritmische schaal, de stress heeft een normale schaalverdeling.

Q5.) What is Basquin’s law? What is Coffin’s law? What part(s) of the S-N curve do they
linearize?
Basquin’s law - De stress en de high-cycle fatigue life zijn aan elkaar gerelateerd. Zie formule 9.4 op bladzijde 229 → Nfb = C1
Coffin’s law - Deze wordt gebruikt als de peak stress groter is dan de yield stress (In the low cycle). Zie formule 9.6 op bladzijde 230 → pl=C2Nfc

Q6.) Actual measured S-N curves always show large scatter. Why is this? What causes
the largest variation: initiation, crack growth, or final failure?

```
       Enter answer here

```

Q7.) What is the R-value, in the context of fatigue? And what is Goodman’s rule – and
how precisely does the R-value come into this rule?
a) the difference in stress per cycle.
b) m=0(1-mts) it is the R-value with difference in mean stress and stress range.

Q8.) How is the so-called endurance limit (a.k.a. fatigue strength) defined? Consult CES
to list this property for the six RM’s, for R = – 1. Next, use Goodman’s rule to calculate this limit for R = 0.5, again for the six RM’s.
a) Minimum cyclic stress, where the R-value is -1 at 10e7 cycles.
b) Low carbon steel: 203 MPa

```
   Age hardening aluminum: 57 MPa
   Stainless steel: 175 MPa
   PP: 11 MPa
   PP + 40% GF: 42,4 MPa
   Pinewood: 0,96 Mpa

```

Q9.) Instead of constant-amplitude fatigue, we often have to deal with varying-amplitude
fatigue (a.k.a. a load collective, or load spectrum). Miner’s rule then provides the equivalent damage caused by the latter in terms of the former. What is the key assumption behind this important rule?
It assumes that damage always accumulates as N1/Nf1for each level of stress.

�
Q10.) What is the Paris law? What does it aim to model: crack initiation, crack growth, or
final failure?
dcdN=AKm crack initiation

Q11.) When do we already assume that (small) fatigue cracks exist: in safe-life design or
in fail-safe design?
Fail-safe design

Als een systeem failsafe (FS) of faalveilig ontworpen is, dan zal het bij een storing of gebrek niet minder veilig zijn. Het systeem kan wel minder goed of helemaal niet meer werken. Failsafe is een ontwerpprincipe waarbij de eis geldt voorzienbare fouten, gebreken en verstoringen in het ontwerpproces niet mogen leiden tot een minder veilige werking.

In safe-life design products are designed to survive a specific design life with a chosen reserve.

Q12.) Consult Materials pp. 175-176 for examples of stress concentration factors. Next,
calculate the safe nominal stress level (Nominal stress level = stress level in the absence of stress concentrations) for the six RM’s for a beam in bending with a tiny crack in the bottom (= side of tensile bending stress).

```
   Enter answer here

```

Q13.) There are three ways of improving a material’s fatigue strength. Which are these?
What is their effect on the price/kg for your material?
a) Choosing strong materials, stronger materials have higher price/kg
b) Making sure they contain as few as possible defects as possible, higher price/kg
c) Providing a compressive surface stress, does not really affect price/kg

Q14.) Moisture, temperature, and UV radiation all can accelerate fatigue. For each of the
six RM’s, mention what you expect is the main accelerator.
Low carbon steel: temperature
Age hardening aluminum: moisture
Stainless steel: temperature
PP: UV
PP + 40% GF: UV
Pine wood: moisture

Q15.) GLARE, a TU Delft-developed material, has superior fatigue strength compared to
regular aerospace aluminium. Why?
The crack propagation life is significantly longer than aluminum due to the bridges by the intact fibres over the crack
Bron: https://www.sciencedirect.com/science/article/pii/S0142112306000314

Workshop 5.1 / Fatigue / Tijdens Workshop.

We have the following exercises about fatigue. E1-4 refer to Materials chapter 9; E5 requires an online source. All exercises test your insight. They will be treated during workshop 5.1 on demand.

E1.) Rework example 9.1 from Materials, p. 230, on
Basquin’s law. Keeping all other data
the same, what is the allowable stress amplitude if we require that the component should survive 500,000 cycles instead of 200,000 cycles? And what if the fatigue strength exponent b = 0.12 instead of 0.10?

�
E2.) If Basquin’s law is for high cycle fatigue, then Coffin’s law helps us get a grip on low
cycle fatigue. For a typical aluminium alloy, this law takes the form of Δεplastic = 0.2/(N 0.5). Here, Δεplastic is the plastic strain amplitude, N is the number of cycles to failure, 0.2 is a material-dependent parameter and 0.5 is the fatigue strength component, also material-dependent (and clearly larger than the one in Basquin’s law!). What is this N if Δεplastic = 1%?

```
       Enter answer here

```

E3.) Fatigue strength goes down with increasing mean stress, as captured in Goodman’s
rule: see eq. (9.7) on p. 231. Make a plot of the stress range Δσ (σm ) as a function of the mean stress σm . Do so for one of the three metals in our reference materials (RM’s).

```
       Enter answer here

```

E4.) Assume we have a component made of steel AISI 4140 (see CES level 323). It is
cyclically loaded with a stress range σrange = 800 MPa for 1,000 cycles, and a stress range σrange = 400 MPa for 5 million cycles. Use Miner’s rule to predict if this component will fail.

Reken eerst constante C1 uit door gegevens te nemen voor fatigue uit CES (in CES staat bij 10^7 cycles een bepaalde fatigue strength). Met deze deze constante kun je daarna het aantal cycles uitrekenen wat nodig is bij een bepaalde belasting door de formule om te schrijven (vergeet niet b ook om te schrijven, in dit geval tot de macht 1/b omdat het antwoord tot de macht b moet). wanneer je het aantal cycles uit de opdracht deelt door het aantal nodige cycles voor fracture, en je deze twee optelt komt je uit op 0.2695. Dit is lager dan één dus breekt hij niet. Hier is uitgegaan van een b=0.1, zelfde waarde als E1

E5.) See
http://www.grantadesign.com/education/InDepth/html/indepth/materialinfo/metal\_fatiguelifemodel.htm for an explanation of the fatigue strength model used in the CES database. Where does the fatigue strength exponent b come from and what are the model’s limitations?

```
   Enter answer here

```

�
Lecture 5.2 / Temperature / Voor college.
Q1-11 refer to Materials chapter 12. Q12-15 refer to Materials chapter 13. All test your base knowledge. → Grijze vragen hoeven niet (Aldus Erik Tempelman)

Q1.) Which materials have a definite melting temperature, and which have a melting
range? Which have a glass transition temperature?

```
   Enter answer here

```

Q2.) What are the definitions of the minimum and maximum service temperatures, as
used in CES?
Minimum: The temperature at which a material becomes to brittle or otherwise unsafe to be used
Maximum: The temperature at which you can use a material continuously without oxidation, chemical change or excessive distortion becoming a problem.

Q3.) Recall chapter 4, Fig. 4.23. For which type(s) of plastics is the maximum service
temperature essentially defined by the glass transition?
Amorfe elastomeren

Q4.) For plastics, the HDT, or heat deflection temperature, is often used instead of CES’
maximum service temperature. How is this property defined and measured? What is the HDT for PP?
Een balk wordt driepunts gebogen met een bekende constante kracht. De temperatuur in de omgeving stijgt , de temperatuur waarop de balk buigt is de HDT.
PP: 0.45MPa 102 C

```
       1.85MPa 50,1 C

```

Q5.) What is the specific heat, and what are its units? Similarly, what is the linear
thermal expansion coefficient (a.k.a. CLTE)? Finally, what is the thermal conductivity? Rank the six RM’s from low to high specific heat, from low to high CLTE; and from being a poor to good conductor. Which “wins”?
SH-The energy needed to warm up 1Kg of material by 1 degree temperature increase
CLTE-The change in length by an increase in temperature of 1 degree temperature increase
TC- Power per meter kelvin (W/m\*K)

Q6.) Which property of the atomic (or molecular) bond strength is responsible for the
existence of thermal expansion?
The fact that these bonds are nonlinear make it so that materials expand when heated

Q7.) How much does a metal (or any other crystalline solid) expand between 0K and their
melting temperature? Does this vary between metals?

```
   About 2%, this varies slightly due to difference in ‘spring modulus’

```

Q8.) Metals can be strengthened by work hardening, solution hardening and (not all
metals) heat treatments, such as precipitation hardening. How do such hardening mechanisms effect thermal conductivity? What about specific heat, or about CLTE?

Q9.) Plastics can, and commonly are, stiffened and strengthened by addition of glass
fibres. What does this mean for thermal conductivity, specific heat, and CLTE?
Thermal conductivity goes up
Specific heat goes down
CLTE becomes a problem due to the difference in the two (glass and PP), this created thermal stress

Q10.) Which equation describes the stress induced by thermal shock?

Q11.) Which equation can be used to determine the power loss due to heat loss, in
W/m2 ? Estimate this loss for your own room, on a cold winter’s day.

Q12.) Some properties depend linearly on temperature: stiffness is an example. How
much is this particular dependence, between RT and the maximum service temperature? Calculate this stiffness loss for all six RM’s. NB: RT, or “room temperature”, is equal to 293 K.
Low carbon steel:

```
       Pinewood: 0

```

Q13.) Other properties depend non-linearly on temperature: an important case is creep of
polymers, affecting strength and stiffness. Which other variable comes into play here? And how much can e.g. stiffness decrease for PMMA?

```
       Enter answer here

```

Q14.) Which polymers are relatively good at resisting creep, and which are bad? Give four
different options (hint: carefully read pp. 354-355!)
Good:
Phenolics, high Tg
PP + 40% GF, filled with powdered glass, silica or talc
omposites with containing chopped or continuous fibres.
Bad:
PET, not crystalline

Q15.) In creep, elongation (or compression) increases (or decreases) with time and
temperature. What happens in relaxation?
Creep: constante spanning zorgt voor toename deformatie

relaxatie: een constante deformatie zorgt voor afname v/d spanning

�
Workshop 5.2 / Temperature / Tijdens Workshop.

E1.) Some materials feel warm, others feel cold: the difference is captured in the thermal
diffusivity, a material property generally denoted with the letter a (Careful! In Materials, the Greek letter alpha, i.e. 𝛂 is used to denote the thermal expansion of materials. This is also a thermal property, but a very different one than the thermal diffusivity a). Use CES to make a plot of this property for the six reference materials (RM’s). Which is “coldest”, and by what margin as compared to the next material in this plot? NB: thermal diffusivity is not just used to see how a material feels – see also Materials p. 298.

E2.) A bi-metallic strip will bend upwards when heated, as captured in Materials eq.
(12.13). Which combination of metals will produce the largest displacement, at a given ΔT ? Are these metals reasonably affordable or not?

E3.) Use the database Elements from the CES Edupack to plot the stiffness (= Young’s
modulus at 300K) against melting temperature. What trend do you see? Next, plot the stiffness against the thermal expansion. Again, what trend do you see?

E4.) Creep of plastics is complex enough, so let’s start with the facts as measured. On
the next page you see a plot of the strain (blue line) against time for a plastic sample that has been stressed at 2.5 MPa for two hours, i.e. 120 min. Now, answer the following questions:
What was the initial modulus of elasticity of this plastic? Use Hooke’s law at t = 0;
What is, effectively, the modulus at t = 100 min?
Draw the blue curve if the stress were present for double the duration, i.e. from t = 0 to t = 240 min.
Next, draw the blue curve if the stress were doubled i.e. 5.0 MPa during that same period. Assuming linear behaviour;
Now, draw the blue curve if the temperature were raised (make a best guess);
Which plastic do you think was involved in the test: typical amorphous (e.g. PS, ABS, PVC), typical semi-crystalline (e.g. PP, PE), or something else?
Finally, consider the viscous part of the curve: what kind of mathematical function would best describe this part?

```
   Answer on next page….

```

�
�
Lecture 6.1 / Manufacture / Voor college.

Q1-4 refer to Materials chapter 2. Q5 refers to Materials chapter 3. Q6-8 refer to Materials chapter 18. Q9-10 refer to NSFD - NPV analysis. Q11-12 refer to NSFD - Material prices. Q13-14 refer to Materials pp. 586-589. Q15, finally, refers to Materials Figs. 19.2-19.4. All test your base knowledge.

Q1.) Like materials, manufacturing processes can be classified also. Which four broad
classes of processes does Materials recognise?
Primary haping
Secondary processes
Joining
Surface treatment
(verspanen kan zowel primair als secundair gedaan worden)

Q2.) What Materials neglects to mention is that a typical product consist not just of
specifically designed parts for that product (and hence, shaped, treated etc.) but also of standard parts that are bought in. What do you estimate is typically the ratio between the two, in numbers?
80% zijn standaard onderdelen

Q3.) A material’s mechanical properties can be affected by the process used to shape it.
For metals, which are these in particular: stiffness, strength, and/or toughness? In this regard, how does this process-property interaction of forging compare to casting, for metal components?
Stijfheid verandert niet, andere eigenschappen wel.

Q4.) Injection moulding is the #1 shaping process for thermoplastics, but not the only
one. Mention four more. What shapes can these processes make: rods, bars and profiles (“1D”), single-and double-curved shell parts (“2D”), or solid and hollow free-forms (“3D”)–see also Fig. 18.8, p. 523.
Extrusie
Pultrusie
Laser sintering

Q5.) During a typical product design process, is there are definite point, place or moment
for process selection? Tip: see Fig. 3.1, p.35.
Nee, er zijn meerdere material keuze momenten. naarmate het proces ontwikkeld worden de te gebruiken preciezer gespecificeerd.

�
Q6.) What are the steps in process selection? How are these different from materials
selection?
A)
choice of process family
choice of process class
choice of single process
B) Dezelfde stappen zijn aanwezig als bij materiaal selectie.
translation
screening
ranking
supporting documentation
Once past screening, it is a co-selection procedure (with process and materials selection). Materials selection has also shaping, joining and finishing.

Q7.) Part cost is of paramount importance. Which simple equation captures part cost in
its general form? Which three key contributors are listed here?
Cs=Cmaterial+Cdedicatedn+Ccapital+Coverheadn
costs of materials, production and investments.

Q8.) What is the economic batch size of processes? How is it influenced by the
magnitude of all product-specific investments (tools, moulds, jigs, etc.)?
A range of batches for which each process is likely to be the most competitive.
The influence depends on the process speed

Q9.) Investments today can be paid for by future profits. What is the name of the type of
analysis commonly used to see how this works out?
NPV analysis (net present value analysis) is the difference between the present value of cash inflows and the present value of cash outflows over a period of time

Q10.) What is this type of analysis usually used for: analysing single scenarios, or
comparing different scenarios?
comparing different scenarios

Q11.) For high-volume production (i.e. large batches), how much of a product’s sales price
is typically taken up directly by the material(s) price?

```
       Enter answer here

```

Q12.) Which factors influence the material price? Which one(s) of these was already,
implicitly or explicitly, included in the equation from Q7?

```
       Enter answer here

```

Q13.) For metals, stiffness is largely immune to process-property interaction. How is
stiffness affected by processing in plastics?

```
       Enter answer here

```

Q14.) Which plastics have more shrinkage: semi-crystalline ones or amorphous ones?
What does this mean for high-tolerance parts?
Semi-crystalline
Meer shrinkage betekent minder precisie, wanneer je hoge tolerantie hebt maken maten dus minder uit, dus maakt meer shrinkage dus ook minder uit. (logischere vraag zou low tolerance geweest zijn, in dat geval wil je meer precisie en dus minder shrinkage)

Q15.) Nanotechnology makes objects at nm-scale. Which microstructural features of
metals can thus be affected? And which for polymers?

```
   Enter answer here

```

�
Workshop 6.1 / Manufacture / Tijdens Workshop.

E1 refers to NSFD – NPV analysis. E2 stands alone. E3 concerns the integration project. All exercises test your insight.

E1.) Sections 2 and 3 of NSFD – NPV analysis presents worked-out examples for two
specific scenarios. Make a comparable analysis for a third scenario in which the total investment is 100 K€ (instead of 50 or 80 k€ as before) and profits are 2.20 €/product (instead of the 1 or 2 €/product as before). Set idiscount = 6% and iinterest = 8%. What is the ROI for this third scenario?

```
       Enter answer here

```

E2.) When a round metal tube with diameter D is bent over a radius R (as measured from
the tube’s centreline), the strain ε in the outer fibres of the tube is roughly equal to ½ D/R . Conversely, given a certain diameter D and strain-to-failure εmax we can find the minimum bending radius MBR = D/(2·εmax). Use this simple equation to find the MBR for tubes made of low carbon steel, stainless steel, and aluminium (i.e. our three reference metals) if given that D = 22 mm (= the typical diameter for tubular steel as used in furniture).

```
       Enter answer here

```

E3.) Assess which materials and processes have been used in the product you have
selected for the integration project. Use the workshop to get feedback from fellow students and teaching staff.

```
       Enter answer here

```

�
Lecture 6.2 / Spaceframes / Voor college.

Q1-15 refer to NSFD – spaceframes . All test your base knowledge.

Q1.) Which four types of spaceframes can be recognized, for 2D situations?

Q2.) Which type is the strongest and the stiffest? Which type is the safest? Which type
has the least members, at a given number of nodes? Consider 2D spaceframes only.

```
       Strongest and stiffest → Type 2
       Safest → Type 3
       Least members → Type 1 and 4 with both 4 members

```

Q3.) In which frame type(s) is buckling especially relevant?

```
       Type 2 (most relevant) and type 3

```

Q4.) Why, in spaceframes, do we normally require external forces to act on nodes only,
instead of also in between nodes, on members?
Krachten kunnen alleen op de hoekpunten plaatsvinden omdat er anders buiging optreedt.

Q5.) Why do type I spaceframes require at least one rigid node? Why do type IV
spaceframe require at least one rigid node more than a similar type I frame? Again, consider 2D spaceframes only.

```
       Enter answer here

```

Q6.) How many members does a proper 2D spaceframe have, i.e. a type II spaceframe?
Draw examples with 6 and 9 nodes.

```
       # members = 2 * N - 3
       With 6 nodes → 2 * 6 - 3 = 9 members
       With 9 nodes → 2 * 9 - 3 = 15 members

```

Q7.) Now answer Q6 for a proper 3D spaceframe. Also draw the examples asked for.

```
       # members = 3 * N - 6
       With 6 nodes → 3 * 6 - 6 = 12 members
       With 9 nodes → 3 * 9 - 6 = 21 members

```

Q8.) Next, extend your example with 6 nodes into one with 7 nodes. Make sure that
each node can carry forces in any direction.

```
       Enter answer here

```

Q9.) Which two methods are presented for the analysis of type II spaceframe?

```
       Node-method
       Section-method (Snede methode)

```

Q10.) Why, in product design, are type I and type IV spaceframes so common, if they
are the least efficient types? Give three independent reasons.
Gebruiker gedraagt zich niet.
Ziet er mooier uit dan type 2 en 3.
Scharnierende knooppunten zijn niet handig/nodig.

Q11.) Which things are ignored in the quick-and-dirty approach of analysing type I
spaceframes?
tensile and compressive members

Q12.) Which two tricks can we use to analyse type III and type IV spaceframes, i.e. the
statically-indeterminate spaceframes?
exploit symmetry
also look at displacement

Q13.) What is secondary bending, in the context of spaceframe design and engineering?
When an unsymmetrically reinforced node is loaded it deforms and in doing so it bends away from it’s reinforcement.

Q14.) What is meant with the term compliance and how does it relate to stiffness?

```
       stiffness at a structural level, thus not that of a material

```

Q15.) The paper mentions De Saint Venant’s Principle. What is this principle?
Lokale effecten blijven lokaal.�
Workshop 6.2 / Spaceframes / Tijdens Workshop.

E1-2 refer to NSFD – spaceframes. All exercises test your insight.

E1.) Section 4 of NSFD – spaceframes proves a numerical example for a simple type II
spaceframe (2D), shown in Figure 15. Rework this example, but now assuming that an external force F = 5.0 kN is acting on node 3, along the diagonal, away from the frame (so, with equal components in the y+ and z+ directions). Tips: first determine the reactions, next draw the FBD, then determine the member forces, and finally find the required cross sectional areas (for tensile members) and area moments of inertia (for compressive members).

```
       Enter answer here

```

E2.) Section 5 of the same document works through a simple type I spaceframe (again in
2D), shown in Figure 21. Rework this example, but now assuming an external force F = 500 N is acting on node 3, along the diagonal, away from the frame (so, with equal components in the y+ and z+ directions).

```
       Enter answer here

```

E3.) Analyse the product you selected for your integration project in terms of all preceding
theory. Which parts can be modelled well as 1D elements and what is the accompanying loading mode for each part (so, beams in bending, ties in tension, columns in buckling, shafts in torsion). What about all connections: can you model these as hinged nodes or as rigid nodes? And what about static determinacy? As before, use the workshop to get feedback from fellow students and teaching staff.
Enter answer here�
Lecture 7.1 / Sustainability / Voor college.

Q1-11 refer to Materials chapter 20. Q12-14 refers to www.youtube.com/watch?v=6sqnptxlCcw. Q15 requires a quick but critical internet search. All questions test your base knowledge.

Q1.) What is the top-five of materials in terms of annual consumption? Are any of our six
reference materials (RM’s) in this top-five?
1. Concrete
2. Oil and Coal
3. Non-fuel woods
4. Asphalt
5. Cement
Only Pine wood

Q2.) Markets are growing, at least until we create a circular economy. What is the time for
a market to double, if it grows by 3, 5 or 10% per year?
tD=100ln(2)r

```
       3% :    23,1 years

```

5% : 13,9 years
10% : 6,9 years

Q3.) What are critical materials?
(Vrijwel) niet te vervangen materialen en wordt niet in veel diverse landen aangeboden.

Q4.) Which phases are typically distinguished in a so-called life cycle assessment? Can
you add one or more phases between the ones given in Fig. 20.3?

```
       Natural Resources → Material Production
                   Product manufacture

```

Product use
Product disposal

Q5.) What are the embodied energy and CO2 footprint, in terms of units and
measurements? What are these properties for our six RM’s?
A) Embodied energy → Milieu-impact van het reduceren van grondstoffen tot nuttige vormen in termen van energie gemeten in MJ/kg. Embodied energy is een materiaaleigenschap.
B) CO2 footprint → De hoeveelheid vrijgekomen CO2 tijdens het productieproces, gemeten in kgCO2/kgmaterial�
C) embodied energy
Low carbon steel: 25 MJ/kg

```
       Age hardening aluminum: 198 MJ/kg
       Stainless steel: 80,3 MJ/kg
       PP: 75,7 MJ/kg
       PP + 40% GF: 81 MJ/kg
       Pinewood: 8,77 MJ/kg

```

D) CO2 footprint
Low carbon steel: 1,72 kg/kg

```
       Age hardening aluminum: 12,2 kg/kg
       Stainless steel: 4,73 kg/kg
       PP: 2,96 kg/kg
       PP + 40% GF: 4,35 kg/kg
       Pinewood: 0,358 kg/kg

```

Q6.) In terms of ease of recycling, there is a difference between in-house scrap (a.k.a.
production waste) and post-consumer waste?
Yes, in in-house scrap almost 100% of the waste material is recycled as it returns to the primary production loop. The recycling percentages in post-consumer waste are lower as the waste separation is imperfect.

Q7.) What are the current recycling rates for the six RM’s (tip: these data are all in CES)?
How does this affect their embodied energy?
Low carbon steel: 40%

```
       Age hardening aluminum: 33%
       Stainless steel: 35%
       PP: 5,1%
       PP + 40% GF: 0,1%
       Pinewood: 8%

```

Q8.) Name two products for which the materials are the main part of overall life cycle
energy use. Also, name two for which the use phase is dominant.
- Multi-storey car park
- Fibers (carpets)

```
     B)  - Civil aircraft

```

- Family car

Q9.) For each of the four material families of metals, polymers, ceramics and hybrids, list
the ones with the highest and lowest embodied energy. What is the total range spanned for this material property? Which family has the highest range, and what are the “worst” materials in this regard?
high to low embodied energy:

```
           Metals: 10,1 - 10e3 MJ/kg
           Polymers: 10,4 - 10,2e2 MJ/kg
           Ceramics: 1 - 10,1e2 MJ/kg

```

Hybrids: 1,3 - 10, 4e2 MJ/kg

Q10.) What is the definition of sustainable development? What is the triple bottom line?
What are the three capitals?
“Sustainable development is development that meets the needs of the present without compromising the ability of future generations to meet their own needs”

People
Planet
Prosperity

Natural capital (planet)
Human and Social Capital (people)
Manufactured and financial capital (prosperity)

Q11.) What is meant with competing articulations of sustainable development? In this
content, how does the BMW i3 (the world’s first mass produced car with a CFRP body) perform: is it an all-round sustainable option, or does it have its downside(s) – if so, which one(s)?
Different contributions for sustainable development which conflict with each other.

Q12.) The late Swedish professor Hans Rosling made a convincing prediction of how large
the world population will get. What number did he predict?

```
       9 billion

```

Q13.) Similarly, prof. Rosling predicted how energy use will increase. What was this
prediction, and what does this imply for sustainable energy sources?
The energy use will increase with 66% but of which 50% will be sustainable energy. This makes the percentage of unsustainable energy in 2050, 25% less than now.

Q14.) What is “the magic” of the magic washing machine?
Men heeft ineens vrije tijd door de magische wasautomaten, welke het werk nu voor de mens doet.

Q15.) Run a google image search on the price of PV over time. What is the trend you
see?
The price dropped to less than one hundredth of the price of 40 years ago, but it drops inversely exponential so as time progresses, prices drop with a smaller rate.

�
Workshop 7.1 / Sustainability / Tijdens Workshop.

E1-3 refer to Materials chapter 20, and for E3 you might also want to check out the Eco Audit section of the CES database. E4 introduces you to another part of this database. E5 introduces you to a third key resource on materials and sustainability. E6 addresses a current “hot topic” that you are encouraged to debate on. All exercises test your insight, skills, and (especially E6) capacity for critical thinking.

E1.) Assume we need to design a structure that is primarily loaded in bending and that
has to be stiff and lightweight. Consult Materials pp. 107-108 to find the appropriate material index. Assume we also need the structure to be made from materials that have a low gross energy requirement, or GER. Use CES to make a plot of the GER against this index for all materials. What are the top-three, and where are the reference materials (RM’s)?

�
E2.) Redo the plot, but now plot the material’s carbon footprint against this index. Who are
the winners now – or are these the same?

E3.) Now assume the structure is in fact the frame for an electric scooter. Are the
selection results from E1 and E2 still relevant, or is there something else going on now?

```
       Enter answer here

```

E4.) In the CES database (level 3 Sustainability) you can find key information regarding
the world’s legislation and regulations with respect to sustainability. From the EU section, select one of the following: Ecodesign Directive, ELV, RoHS2, or WEEE. What are these directives about, and are they binding or not – i.e. are OEM’s essentially free to ignore the directives or forces to comply?

```
       Enter answer here

```

E5.) Materials provides an excellent introduction to materials and sustainability, and the
CES database puts many data at your fingertips, plus Eco Audit tool. The free (but excellent!) book and website www.withbotheyesopen.com is the next step. Visit this site to get a feel for its contents. Which five (groups of) materials does it cover? And which design options are discussed?

```
       Enter answer here

```

E6.) Run a Google search for the phrase “plastic inzamelen zin of onzin” (= Dutch for
“collection of plastics sense or nonsense”) and scan the first few search results. Next, prepare for a quick debate, to be held during the workshop, in which you either defend the position that separate collection and processing of plastic waste makes sense, or that it makes no sense.

```
   Enter answer here

```