Def & Fracture | Exam 1 Content

0.0(0)
Studied by 0 people
call kaiCall Kai
Locked
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
GameKnowt Play
Card Sorting

1/47

encourage image

There's no tags or description

Looks like no tags are added yet.

Last updated 3:26 PM on 9/8/26
Name
Mastery
Learn
Test
Matching
Spaced
Call with Kai
Chat

No analytics yet

Send a link to your students to track their progress

48 Terms

1
New cards

Engineering Strain

How much a material’s length changes compared with its original length. Equation to left.

<p>How much a material’s length changes compared with its original length. Equation to left. </p>
2
New cards

True Strain

Measures deformation by considering the continuous change in length as the material deforms, rather than comparing everything to the original length.



<p>Measures deformation by considering the continuous change in length as the material deforms, rather than comparing everything to the original length. </p><p></p><p></p>
3
New cards

Why are engineering and true strain different?

They use different references! See left!

<p>They use different references! See left!</p>
4
New cards

Poisson Ratio

Poisson Ratio = - (transverse strain = fat) / (axial strain = longitudinal strain)

<p>Poisson Ratio = - (transverse strain = fat) / (axial strain = longitudinal strain)</p>
5
New cards

Relationship Between E and G

E is the numerator; G is the result; 2 and (1 + ν) live underneath.

Emily’s number, (G)irlfriend is the result, and (after they get married?) 1(1+ poisson’s ratio) lives underneath.

<p>E is the numerator; G is the result; 2 and (1 + ν) live underneath.</p><p>Emily’s number, (G)irlfriend is the result, and (after they get married?) 1(1+ poisson’s ratio) lives underneath. </p>
6
New cards

Poisson Ratio = 0

Since Poisson Ratio = - (transverse strain = fat) / (axial strain = longitudinal strain), when Poisson Ratio = 0, assuming there is a nonzero longitudinal strain, then transverse strain (= fatness) = 0.

Physically, if you pull the material in one direction, it stretches in the loading direction (okay => since there is nonzero longitudinal strain [aka a relation to length]). It does not contract or expand in the transverse direction (= fatness) (okay! makes sense since transverse strain is 0). So, for example, the material gets longer but doesn’t get narrower.

=> also material is brittle/might break? => Chat: when Poisson ratio = 0, material could theoretically still be ductile or brittle.

But, materials like ceramics and glass are brittle (but that does not mean Poisson ratio = 0) (according to Chat).

7
New cards

33 direction

x1 = diagonal direction toward the bottom-left

x2 = horizontal direction toward the right

x3 = vertical direction up

So: 3-direction=x3=vertical direction

The first number tells you which strain direction you're talking about, and the second number tells you which direction the deformation is measured in.

For normal strain: ε33 means strain in the 3-direction caused by deformation along the 3-direction. In this example, you're pulling the block upward:

↑ x₃ (3-direction)

┌────────┐
│ │
│ block │
│ │
└────────┘

Therefore:

ε33=vertical/axial strain

And the sideways strains are:

ε11=strain in the x1 direction

ε22=strain in the x2 direction

So for the Poisson ratio on your slide:

ν=−ε11ε33

you're basically saying:

Poisson's ratio = sideways deformation ÷ vertical deformation (vertical deformation = axial strain => strain = (change in length / original length)

where 33 is the direction you're pulling in this particular diagram.

<p><span>x1</span> = diagonal direction toward the <strong>bottom-left</strong></p><p><span>x2</span> = horizontal direction toward the <strong>right</strong></p><p><span>x3</span> = vertical direction <strong>up</strong></p><p>So: 3-direction=x3=vertical&nbsp;direction</p><p><span style="color: oklch(0.159066 0 none);">The first number tells you </span><strong>which strain direction you're talking about</strong><span style="color: oklch(0.159066 0 none);">, and the second number tells you </span><strong>which direction the deformation is measured in</strong><span style="color: oklch(0.159066 0 none);">.</span></p><p>For normal strain: <span>ε33 </span>means <strong>strain in the 3-direction caused by deformation along the 3-direction</strong>. In this example, you're pulling the block <strong>upward</strong>:</p><p><span style="background-color: color(srgb 0.976409 0.976496 0.976504); color: oklch(0.159066 0 none);">                  ↑  x₃ (3-direction)<br>                  │<br>       ┌────────┐<br>       │                     │<br>       │  block           │<br>       │                     │<br>       └────────┘</span></p><p>Therefore:</p><p><span>ε33=vertical/axial&nbsp;strain</span></p><p>And the sideways strains are:</p><p><span>ε11=strain&nbsp;in&nbsp;the&nbsp;x1&nbsp;direction</span></p><p><span>ε22=strain&nbsp;in&nbsp;the&nbsp;x2&nbsp;direction</span></p><p>So for the Poisson ratio on your slide:</p><p><span>ν=−ε11ε33</span></p><p>you're basically saying:</p><p>Poisson's ratio = sideways deformation ÷ vertical deformation (vertical deformation = axial strain =&gt; strain = (change in length / original length)</p><p><span style="color: oklch(0.159066 0 none);">where 33 is the </span><strong>direction you're pulling</strong><span style="color: oklch(0.159066 0 none);"> in this particular diagram.</span></p>
8
New cards

Fatigue

Failure of a material caused by repeated or cyclic loading over time

9
New cards

Yield Point Phenomena

Chat: Yield point phenomenon is when a material begins to deform PLASTICALLY at a distinct upper and lower yield point, instead of smoothly transitioning from elastic to plastic deformation. So if you see a stress–strain graph with a sudden drop/plateau at yielding, think yield point phenomenon. On a stress–strain curve. 1.) Elastic region → material returns to original shape. 2.) Upper yield point → stress reaches a maximum. 3.) Stress suddenly drops → to the lower yield point. 4.) Plastic deformation continues at roughly constant stress for a while.

Memo 3: “Also, it is interesting to note that hot rolled steel exhibits a peak and plateau region. This region is caused by the yield point phenomenon [19]. In low carbon steels, the foreign atoms like carbon accumulate around dislocations, forming Cottrell atmospheres [19]. These atmospheres pin the dislocations, restricting their motion [19]. Only when enough stress is applied, the dislocations break free from these pinning points, resulting in the upper yield point followed by a drop in stress [19]. As deformation continues, dislocations are repeatedly pinned and unpinned by successive Cottrell atmospheres, producing the zig-zag plateau region seen in the stress-strain curve [19]. This behavior is observed between strains of 0.011 to 0.037 in Figure 5.” Chat: What is yield point phenomena? In some steels, when you pull the material, the stress–strain curve does something unusual: stress rises → reaches a peak → drops → stays around a plateau → continues deforming. That peak + plateau is the yield point phenomenon. Why does this happen in low-carbon steel? Low-carbon steel contains small amounts of carbon atoms. The carbon atoms move toward dislocations. Think of a dislocation as a place where the crystal structure is imperfect. The carbon atoms gather around these dislocations: Carbon atoms + dislocation => Cottrell atmosphere. A Cottrell atmosphere is basically a group of carbon atoms surrounding a dislocation. The carbon atoms essentially hold the DISLOCATION in place. So: Cottrell atmosphere => pins dislocation. A pinned dislocation can't move easily. And remember: Dislocation movement = plastic deformation. So if the dislocations can't move, the material resists deformation. Then you apply more stress. Eventually, you apply enough stress to break the dislocations free. Pinned dislocation =enough stress​> dislocation breaks free. This requires a relatively high stress → upper yield point. Then the DISLOCATIONS suddenly start moving. Because deformation becomes easier, the required stress drops. That's the drop after the upper yield point. Why is there a plateau/zig-zag? As the material keeps deforming, more dislocations encounter carbon atoms and become pinned/unpinned. So you get repeated little changes in stress: pin => break free => pin => break free. This creates the zig-zag/plateau region on the stress–strain curve. Super simple chain: Carbon → Cottrell atmosphere → pins dislocations → need more stress → dislocations break free → stress drops → plateau.

<p>Chat: Yield point phenomenon is when a material begins to deform <strong>PLASTICALLY </strong>at a distinct upper and lower yield point, instead of smoothly transitioning from elastic to plastic deformation. So if you see a stress–strain graph with a sudden drop/plateau at yielding, think yield point phenomenon. On a stress–strain curve. 1.)<strong> Elastic region → material returns to original shape</strong>. 2.) Upper yield point → stress reaches a maximum. 3.) Stress suddenly drops → to the lower yield point. 4.) Plastic deformation continues at roughly constant stress for a while.</p><p>Memo 3: “<span style="font-family: Aptos, sans-serif; line-height: 115%;">Also, it is interesting to note that hot rolled steel exhibits a peak and plateau region. This region is caused by the yield point phenomenon [19]. <strong>In low carbon steels, the foreign atoms like carbon accumulate around dislocations, forming Cottrell atmospheres </strong>[19]. <strong>These atmospheres pin the dislocations, restricting their motion [19]. Only when enough stress is applied, the dislocations break free from these pinning points, resulting in the upper yield point followed by a drop in stress </strong>[19].<strong> As deformation continues, dislocations are repeatedly pinned and unpinned by successive Cottrell atmospheres, producing the zig-zag plateau region seen in the stress-strain curve [19]. This behavior is observed between strains of</strong></span><span style="font-family: &quot;Cambria Math&quot;, serif; line-height: 115%;"><strong> </strong></span><span style="font-family: Aptos, sans-serif; line-height: 115%;"><strong>0.011 to 0.037 in Figure 5.” </strong>Chat: What is yield point phenomena? </span>In some steels, when you pull the material, the stress–strain curve does something unusual: stress rises → reaches a peak → drops → stays around a plateau → continues deforming. That peak + plateau is the yield point phenomenon. Why does this happen in low-carbon steel?<strong> Low-carbon steel contains small amounts of carbon atoms</strong>. <strong>The carbon atoms move toward</strong> <strong>dislocations</strong>.<strong> Think of a dislocation as a place where the crystal structure is imperfect.</strong> <strong>The carbon atoms gather around these dislocations: Carbon atoms + dislocation =&gt; Cottrell atmosphere.</strong> <strong>A Cottrell atmosphere is basically a group of carbon atoms surrounding a dislocation. The carbon atoms essentially hold the DISLOCATION in place</strong>. <strong>So: Cottrell atmosphere =&gt; pins dislocation.</strong> A pinned dislocation can't move easily. And remember: Dislocation movement = plastic deformation. So if the dislocations can't move, the material resists deformation. Then you apply more stress. Eventually, you apply<strong> </strong>enough stress to break the dislocations free. Pinned&nbsp;dislocation =enough&nbsp;stress​&gt; dislocation&nbsp;breaks&nbsp;free. This requires a relatively high stress → <strong>upper yield point</strong>. Then the DISLOCATIONS suddenly start moving. <strong>Because deformation becomes easier, the required stress drops</strong>. That's the <strong>drop after the upper yield point</strong>. Why is there a plateau/zig-zag?  As the material keeps deforming,<strong> more dislocations encounter carbon atoms and become pinned/unpinned</strong>. So you get repeated little changes in stress: pin =&gt; break free  =&gt; pin =&gt; break free. This creates the <strong>zig-zag/plateau region</strong> on the stress–strain curve. Super simple chain: Carbon → Cottrell atmosphere → pins dislocations → need more stress → dislocations break free → stress drops → plateau. </p>
10
New cards

Stress in 33 direction

For a NORMAL stress, first and second numbers are the same. Stress in 3-3 direction is simply normal stress in the 3-direction.

Same numbers => normal stress.

<p>For a NORMAL stress, first and second numbers are the same. Stress in 3-3 direction is simply normal stress in the 3-direction. </p><p>Same numbers =&gt; normal stress. </p>
11
New cards

Stress in 12 direction, etc.

Different numbers => shear stress.

Stress in 12 direction is shear stress on the 1-face, in the 2-direction.

<p>Different numbers =&gt; <strong>shear stress</strong>.</p><p>Stress in 12 direction is <strong>shear stress on the 1-face, in the 2-direction.</strong></p>
12
New cards

Tensile Response of Materials: 3-Dimensional Stress State

knowt flashcard image
13
New cards

Memorizing x1,x2,x3

Look left (BL = 1), right (x2), then up (x3).

<p>Look left (BL = 1), right (x2), then up (x3). </p>
14
New cards

Plane STRESS

Plane STRESS = a 2D stress state where there is no stress acting in the 3rd direction! So => sigma_33 = 0!!

The material can STILL have strain in the 3rd direction because of Poisson’s effect (contrast to homework, 2.2, where we said plane STRAIN, so one of the strains were = 0). Def of plane stress = a 2D stress state where there is no stress acting in the 3rd direction.

Look to the left too!

<p>Plane STRESS = a 2D stress state where there is no stress acting in the 3rd direction! So =&gt; sigma_33 = 0!!</p><p>The material can STILL have strain in the 3rd direction because of Poisson’s effect (contrast to homework, 2.2, where we said plane STRAIN, so one of the strains were = 0). Def of plane stress = a 2D stress state where there is no stress acting in the 3rd direction. </p><p>Look to the left too! </p>
15
New cards

Slip planes

Slip planes = where DEFORMATION happens preferentially. Slip planes are the specific planes inside a crystal where atoms can slide past one another most easily, so PLASTIC deformation happens there preferentially. Think of it like a deck of cards. Imagine a stack of cards. Each card = an atomic plane. Push the stack sideways → the cards slide past each other. Some planes allow this sliding more easily than others. Those easier-to-slide planes are the slip planes!! Another wording: Slip planes are planes in the crystal where dislocations move most easily, so plastic deformation tends to occur there.



16
New cards

Stress & Strain as Tensors

Yes — stress and strain are tensors! The easiest way to think about it is: 1.) Stress = tensor. Stress tells you how forces act on different planes and in different directions at a point!!

Look to left!

So, stress isn’t just one number. It depends on the plane AND direction you’re considering.

<p>Yes — stress and strain are tensors! The easiest way to think about it is: 1.) Stress = tensor. Stress tells you how forces act on different planes and in different directions at a point!!</p><p>Look to left!</p><p>So, stress isn’t just one number. It depends on the plane AND direction you’re considering.</p>
17
New cards

Stress & Strain as Tensors Cont.

Strain tells you how much the material deforms in different directions.

Look to left!

<p>Strain tells you how much the material deforms in different directions.</p><p>Look to left!</p>
18
New cards

Stress & Strain as Tensors Cont.

Honestly: more of just an interesting thing but!

Look to left!

Memory Trick:

Stress tensor = what forces are doing. Strain tensor = what the material is doing in response.

<p>Honestly: more of just an interesting thing but!</p><p>Look to left!</p><p>Memory Trick:</p><p>Stress tensor = what forces are doing. Strain tensor = what the material is doing in response. </p>
19
New cards

Elastic Anisotropy (cont.)

Chat: Absolutely. The slide is explaining elastic anisotropy (= it is the property of a material where its elastic behavior [stiffness, deformation, or wave velocity] depends on the direction in which it is measured) using the generalized Hooke’s law relationship between stress and strain. 1.) The main equation. At the top: σij=Cijklϵkl. Think of this as: stress=stiffness×strain. σ = stress, ϵ = strain, C = elastic stiffness tensor (= describes how a material's stress changes when it is subjected to strain. Interesting??). Because each quantity has directions, the stiffness Cijkl tells you how a STRAIN in one direction produces a stress in another direction 2. Why is C so complicated? The slide writes the relationship as a matrix: (σ11σ22σ33σ23σ13σ12)=(C1111C1122C1133⋯C2211C2222C2233⋯⋮)(ϵ11ϵ22ϵ33ϵ23ϵ13ϵ12). There are 6 independent stress/strain components: 11,22,33: normal components 23,13,12: shear components So the C tensor becomes a 6 × 6 stiffness matrix!! 3.) What does "anisotropic" mean? The key idea is: The stiffness depends on direction. Imagine pulling a crystal. If you pull it along direction 1, it might be very stiff. If you pull it along direction 2, it might be softer. So: E1≠E2≠E3 and similarly, the response to shear can depend on which plane/direction you're shearing. That's elastic anisotropy.

* Also!! Note the green highlighted! Need to memorize for the exam!


<p>Chat: Absolutely. The slide is explaining <strong>elastic anisotropy</strong> (= it is the property of a material where its elastic behavior [stiffness, deformation, or wave velocity] depends on the direction in which it is measured) using the generalized Hooke’s law relationship between stress and strain. 1.) The main equation. At the top: σij=Cijklϵkl. Think of this as: s<strong>tress=stiffness×strain</strong>. σ = stress, ϵ = strain, C = <strong>elastic stiffness tensor </strong>(= <span style="color: oklch(0.159066 0 none);">describes how a material's </span><strong>stress</strong><span style="color: oklch(0.159066 0 none);"> changes when it is subjected to </span><strong>strain</strong><span style="color: oklch(0.159066 0 none);">. Interesting??). </span>Because each quantity has directions, the stiffness Cijkl tells you how a STRAIN in one direction produces a stress in another direction 2. Why is C so complicated? The slide writes the relationship as a matrix: (σ11σ22σ33σ23σ13σ12)=(C1111C1122C1133⋯C2211C2222C2233⋯⋮)(ϵ11ϵ22ϵ33ϵ23ϵ13ϵ12). There are 6 independent stress/strain components: 11,22,33: normal components 23,13,12: shear components So the C tensor becomes a <strong>6 × 6 stiffness matrix</strong>!! 3.) What does "anisotropic" mean? The key idea is: The stiffness depends on direction. Imagine pulling a crystal. If you pull it along direction 1, it might be very stiff. If you pull it along direction 2, it might be softer. So: E1≠E2≠E3 and similarly, <strong>the response to shear can depend on which plane/direction you're shearing</strong>. That's <strong>elastic anisotropy</strong>.</p><p>* Also!! Note the green highlighted! Need to memorize for the exam!</p><p></p>
20
New cards

Elastic Stiffness Tensor

The elastic stiffness tensor describes how a material's stress changes when it is subjected to strain!! In linear elasticity, the fundamental relationship is: σi j= Cijkl * ϵkl where: σij = stress tensor, ϵkl = strain tensor, Cijkl = elastic stiffness tensor, i,j,k,l=1,2,3. The stiffness tensor therefore has 4 indices!!!!, so in 3D it initially contains 34=81 components.

21
New cards

C and S

C = how resistant the material is to deformation => stiffness. Elastic STIFFNESS tensor! Memory = Can’t Deform; C = Stiffness; A material with a high C is stiff! => it resists deformation. Stress = C * strain.

S = how much DEFORMATION occurs for a given stress => compliance. SuCks. S = Squishy => S = Compliance. A material with a high! S! is compliant => it DEFORMS easily! Strain = S * stress. Think!!: S tells you how much strain you get from a given stress.

Reminder => stress = stiffness [think: C] * strain.

22
New cards

Four indices

Why Cijkl have four indices? Each index corresponds to a direction. Look to left!

So => the four indices tell you which strain component causes which stress component? [C = ELASTIC STIFFNESS tensor!!! => stress = C * strain]. C = Can’t deform = stiffness. High C = high stiffness.

S = Squishy = Compliance => A material with a high S is compliant => It deforms easily! Strain = S * Stress. Think: S tells you how much strain you get from a given!! stress!!!

<p>Why Cijkl have four indices? Each index corresponds to a direction. Look to left!</p><p>So =&gt; the four indices tell you which strain component causes which stress component? [C = ELASTIC STIFFNESS tensor!!! =&gt; stress = C * strain]. C = Can’t deform = stiffness. High C = high stiffness. </p><p>S = Squishy = Compliance =&gt; A material with a high S is compliant =&gt; It deforms easily! Strain = S * Stress. Think: S tells you how much strain you get from a given!! stress!!!</p>
23
New cards

Elastic Anisotropy (cont.)

6. The important part about symmetry: The material's crystal symmetry can make many of those Cijkl values equal to each other—or force some to be zero. For example, a highly symmetric material doesn't need dozens of independent stiffness constants. This is what your note: "simple cubic FCC" is getting at. For a cubic crystal, the elastic response has very high symmetry!! Instead of needing all the possible Cijkl, only three independent elastic stiffness constants are needed: C11,C12,C44 The full cubic stiffness matrix becomes: (Look to left!).

<p>6. The important part about symmetry: The material's crystal symmetry can make many of those <span>Cijkl</span> values equal to each other<strong>—or force some to be zero</strong>. For example,<strong> a highly symmetric material doesn't need dozens of independent stiffness constants. </strong>This is what your note: "simple cubic FCC" is getting at. For a <strong>cubic crystal</strong>, the elastic response has very high symmetry!! Instead of needing all the possible <span>Cijkl</span>, only <strong>three independent elastic stiffness constants</strong> are needed: <span>C11,C12,C44 </span>The full cubic stiffness matrix becomes: (Look to left!). </p>
24
New cards

Elastic Anisotropy (cont.)

  1. What your red note about C11,C12,C13 is getting at. Look to left!


<ol start="7"><li><p>What your red note about C11,C12,C13 is getting at. Look to left!</p></li></ol><p></p>
25
New cards

Elastic Anisotropy (cont.)

8. "Shear forces in any of these directions are similar". This handwritten statement is essentially saying: For a cubic crystal, the symmetry makes the shear RESPONSE equivalent along the symmetry-equivalent directions: C44=C55=C66. So shearing on equivalent planes produces the same stiffness.

Reminder: Strain = S * Stress (Think: given!! stress!!). S = how much deformation occurs for a given stress. C = how resistant the material is to deformation => stiffness!

The entire slide can be reduced to this: Stress=stiffness×strain. For an anisotropic material, stiffness depends on direction. But crystal symmetry reduces the number of independent stiffness constants. For a cubic crystal: 3 independent constants: C11,C12,C44. And if you rotate the coordinate system, you must transform the tensor components, even though the underlying physical material hasn't changed (Chat: The key idea is the physical object stays the same, but its numerical description depends on the coordinate system you use. In one coordinate system, you describe the stress using components like σxx,σxy,σyy. Now rotate your x- and y-axes by 45°. You haven't rotated or changed the rubber block at all—you've only changed how you're looking at it. Therefore, the tensor components must change so that they continue to describe the same physical stress.). (The key conceptual distinction: changing coordinates can change the numbers Cijkl you write down; changing the material's physical symmetry does not.).

26
New cards

Zener Ratio

In crystal/material anisotropy, the Zener anisotropy ratio measures how different a cubic crystal’s elastic response is in different crystallographic directions!!

If Zener Ratio = A = 1, the material is [elastically] isotropic!!

If Zener Ratio = A NOT = 1, the material is [elastically anisotropic].

The farther Zener Ratio = A is from 1, the stronger the elastic anisotropy.

Chat: Zener = shear stiffness / normal-stiffness difference. [Factor of 2 is also part of the definition].

Chat: “Elastically isotropic” => a material has the same ELASTIC response regardless of the direction you apply the force.

<p>In crystal/material anisotropy, the Zener anisotropy ratio measures how different a cubic crystal’s elastic response is in different crystallographic directions!!</p><p>If Zener Ratio = A = 1, the material is [elastically] isotropic!!</p><p>If Zener Ratio = A NOT = 1, the material is [elastically anisotropic]. </p><p>The farther Zener Ratio = A is from 1, the stronger the elastic anisotropy.</p><p>Chat: Zener = shear stiffness / normal-stiffness difference. [Factor of 2 is also part of the definition]. </p><p>Chat: “Elastically isotropic” =&gt; a material has the same ELASTIC response regardless of the direction you apply the force. </p>
27
New cards

Zener Ratio Cont.

Zener = 2-4 over 11-12.

A = ( 2 (C44) ) / ( C11 - C12 ). Look to left!

<p>Zener = 2-4 over 11-12.</p><p>A = ( 2 (C44) ) / ( C11 - C12 ). Look to left!</p>
28
New cards

Elastic Anisotropy - Cubic Systems

For Isotropic Case. Look to Left! 2 independent constants. C_44 = (C_11 - C_12) / 2.

<p>For Isotropic Case. Look to Left! 2 independent constants. C_44 = (C_11 - C_12) / 2. </p>
29
New cards

Determination of Elastic Moduli Along Different Directions in Cubic Materials

knowt flashcard image
30
New cards

Elastic Properties of Polycrystals

Voigt Average and Reuss Average.

Voigt and Reuss averages are ways to estimate the effective elastic properties of a polycrystalline material when individual crystals are anisotropic!!

1. Voigt average — uniform strain. Voigt assumes: same strain in every grain. So all grains deform by the same amount, but they may experience different stresses!! Memory: Voigt = same strain This gives an upper-bound estimate of stiffness. Memory: Think of the Vietnamese name VU! Memory:

2. Reuss average — uniform stress. Reuss assumes: same stress in every grain. So every grain experiences the same stress, but different grains may undergo different strains. Memory: Reuss = same stress. 2 ss in a row in Reuss, 2 ss in stress. Same stress in every grain! This gives a lower-bound estimate of stiffness. Memory: RoLL! => Reuss = lower bound estimate for [effective] stiffness!

Put together: Reuss <= actual polycrystal <= Voigt. [Think VU = upper bound!].

31
New cards

How Young’s Modulus (Elastic Stiffness!) Changes Periodically with Atomic Number, especially for the Transition Metals

How does the strength of metallic bonding change as we move through the periodic table?

Periodic pattern => Look to left!

Young’s Modulus vs. Atomic Number!

Graph has a similar rise => maximum => fall pattern in each series (3d series, 4d series, 5d series) (as you increase the atomic number aka go from left to right of the periodic table within each series).

<p>How does the strength of metallic bonding change as we move through the periodic table?</p><p>Periodic pattern =&gt; Look to left!</p><p>Young’s Modulus vs. Atomic Number!</p><p>Graph has a similar rise =&gt; maximum =&gt; fall pattern in each series (3d series, 4d series, 5d series) (as you increase the atomic number aka go from left to right of the periodic table within each series). </p>
32
New cards

How Young’s Modulus (Elastic Stiffness!) Changes Periodically with Atomic Number, especially for the Transition Metals Cont.

3. The key is electronic structure.

These d-electrons participate significantly in metallic bonding. More effective d-electron bonding → stronger bonds → higher Young’s modulus. This is why the modulus rises as we move from the early transition metals toward the middle/late-middle of the transition series.

4. Why does it reach a maximum? Look to left.

<p> 3. The key is electronic structure. </p><p><span style="color: oklch(0.159066 0 none);">These d-electrons participate significantly in </span>metallic bonding<span style="color: oklch(0.159066 0 none);">. </span>More effective <span>d</span>-electron bonding → stronger bonds → higher Young’s modulus. This is why the modulus rises as we move from the early <strong>transition metals </strong>toward the middle/late-middle of the transition series.</p><p>4. Why does it reach a maximum? Look to left.</p>
33
New cards

How Young’s Modulus (Elastic Stiffness!) Changes Periodically with Atomic Number, especially for the Transition Metals Cont.

knowt flashcard image
34
New cards

How Young’s Modulus (Elastic Stiffness!) Changes Periodically with Atomic Number, especially for the Transition Metals Cont.

Atomic number→electron configuration→d-band filling→bond strength→Young’s modulus.

Summary: The Young’s modulus of transition metals varies periodically because the d-electron population changes across each transition series, causing the strength of metallic bonding—and therefore the stiffness—to rise to a maximum and then fall.

<p>Atomic&nbsp;number→electron&nbsp;configuration→d-band&nbsp;filling→bond&nbsp;strength→Young’s&nbsp;modulus. </p><p>Summary: The Young’s modulus of transition metals varies periodically because the <span>d</span>-electron population changes across each transition series, causing the strength of metallic bonding—and therefore the stiffness—to rise to a maximum and then fall.</p>
35
New cards

Effect of Structure on Elastic Behavior Slide

As Young’s modulus increases, tensile strength also increases. => Chat: Actually not a fundamental rule. A material can have: High E + low tensile strength → very stiff but breaks relatively easily. Low E + high tensile strength → less stiff but can withstand high stress before breaking. For example, ceramics generally have high Young's modulus but can be relatively brittle, while some polymers have low Young's modulus but can have substantial tensile strength!

36
New cards

Linear Strain

e = deformation / L0 = (change in length) / L0

<p>e = deformation / L0 = (change in length) / L0</p>
37
New cards

Direct (pure) shear stress

tress produced when forces act parallel to a surface and try to slide one part of a material past another.

38
New cards

Torsional (or bending) force

Yes.

39
New cards

Engineering (nominal) Stress vs True Stress

Engineering (nominal) stress is based on A0 = original area while true stress is based on A = instantaneous area.

40
New cards

Relationship between Engineering Stress and True Stress

sigma_t = stress_e (1 + strain_e).

sigma_t = true stress.

stress_e = engineering stress.

strain_e = engineering strain

t = e (1 + e). Strain to see the second one (inside parentheses). (Engineering = more complicated expression).

(end of slide 9).

<p>sigma_t = stress_e (1 + strain_e). </p><p>sigma_t = true stress. </p><p>stress_e = engineering stress. </p><p>strain_e = engineering strain</p><p>t = e (1 + e). Strain to see the second one (inside parentheses). (Engineering = more complicated expression). </p><p>(end of slide 9). </p>
41
New cards

Shear Stress and Shear Strain Formulas

knowt flashcard image
42
New cards

Polycrystalline aggregates (without texturing), E is isotropic or the same in all directions!

43
New cards

Poisson’s Ratio

Allowable (Theoretical) Range of Poisson Ratios:

v (actual greek? letter) = -1 <= v <= 0.5!

44
New cards

End of Elastic Limit = Yield Stress

Ok.

45
New cards

Resilience of a Material

Area under curve (under elastic loading conditions) = energy PER unit volume.

Kinda similar to toughness. Stress * strain = units of stress. Kind of: See next as well. Resilience units = J / m³ = Pa.

<p>Area under curve (under elastic loading conditions)  = energy PER unit volume. </p><p>Kinda similar to toughness. <s>Stress * strain = units of stress. </s> Kind of: See next as well. Resilience units = <strong>J / m³ = Pa</strong>.</p>
46
New cards

Resilience of a Material Units

knowt flashcard image
47
New cards

Toughness of a Material Equation

knowt flashcard image
48
New cards

3-dim Stress State

A 3-dimensional (3D) stress state means that stresses can act in all three spatial directions: x, y, and z.

(end of slide 25).