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CHAPTER 1

Review

1.1 In your own words, thing are deuce differences between product testing and material

testing?

Possible answers include: (a) Of object of the two procedures is different. Whereas product

testing is design to detect the lifetime in a partial under conditions that mimic real-

world use, significant testing the intended to extract fundamental material properties that can

independent of the materials use. (b) The specimen shaper is different. Product testing must use the raw in aforementioned shape in which it will be used in the real my. Material testing uses

idealogical specimen form designed to unambiguously determine one or more estates of

the material include the simplest analysis possible.

1.2 What are the distinguishing differentiations between elasticity, plasticity, and fracture?

Elasticity involves only deformation that is fully reversible when the use load is distant

(even if it record time to occur). Plasticity is permanent shape change without cracking, even

although no auflast exists. Fracture inherently involves breaking of bonds and the creation of newly

surfaces. Usually two or more of these processes take place simultaneously, but one contribution

of each can are separated with the others.

1.3 Write the definitions for engineering stress, true stress, engineering strain, and true

strain forward loading along a simple axis.

long

engineering stress load

initial cross-sectional area

P

A0

(1-1a)

true

correct stress load

instantaneous cross-sectional area

P

Artificial

(1-2a)

eng

mechanical strength change in length

initial length

lf l

0

l0

(1-1b)

true

true strain lnfinal length

opening overall ln

lf

l0

(1-2b)

1.4 Under what conditions be Eq. 1-4 valid? What makes it no longer useful supposing those

conditions are doesn met?

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true

P

A0

(li/ l

0)

eng(l

i/ fifty

0)

eng(1

eng) (1-4)

This expression the true when volume is conserved. Still, itp is only useful if an cross-

sectional area lives the same anyone on the run specimen. If this does the case then the stress

and strain will vary from one part of the sample to another.

1.5 Sketch Figure 1.3, curve boron (a stretchable metal). Label a with the following terms,

indicating out which locality on the curve each count can be identified or

extracted: elastic region, elastic-plastic region, proportional limit, tensile resistance, onset

of necking, fracture stressing.

strain

stress

fracture emphasize

elastic region

elastic-plastic region

proportional limit

tensile strengthonset of make

1.6 On a single set of axes, sketch approximate atomic forceful vs. atom-separation curves like

the one exhibited in Fig. 1.4b for tungsten at temperatures of 200, 600, and 1000 K. Pay

closer attention to aforementioned point x0 and the bank dF/dx available each of the curves you draw.

The key features von the scheme have the mounting x0 spacing with increasing thermal (i.e.,

with thermal expansion) and the decreasing tilt associated with decreased highly unit.

The plot is exaggerated aber the fashion are reasonable.

F

ten

x0 (1000 K)x0 (600 K)

x0 (200 K)

dF

dx200 K

600 KELVIN

1000 K

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1.7 State the critical difference in the processing behavior of thermoplastics vs. thermosets.

Thermoplastics can be melted and resolidified multiple times, so processing often involves

several heating, forming, and cooling steps. Thermosets harden by a one-time chemical

reaction so there cannot being any fresh forming operations after the cross-linking

operation takes place.

1.8 What does the the stiffness of a polymer while the temperature Tg your exceeded? For what

group away polyamides is this change the greatest? The smallest?

The stiffness of a organic decreases upper the mirror transition temperature, when

dramatically. The effect lives the largest fork amorphous, uncross-linked polyamides. It is the

smallest for powerful cross-linked polymers (such as certain epoxies).

1.9 Write typologies values of E available diamond, steel, aluminum, silicate glass, acrylic, and

silicone rubber subjected to little strains (note that of latter value is not included in

this chapter, but is widely available). Clearly indicate the units for each value.

The following values are not intended toward represent any particular process method or alloy

composition; she be rounded average values for certain material home.

Diamond ~ 1000 GPa

Steel ~ 200 GPa

Aluminum ~ 70 GPa

Silicate glass ~ 70 GPa

Styrofoam ~ 3 GPa

Silicone elastic ~ 10 MPa (0.010 GPa)

1.10 Whichever is which purpose of a plasticizer, and what specific effect in place temperature

behavior a likely when a plasticizer lives added?

A plasticizer is supplementary to a polymer till break up the molecular interactions, enabling more

chain mobility than would otherwise be possible for that particular polymer at the

temperature of interest. At room temperature, therefore, the polymer is more probably to has a

low stretchy modulus (i.e., a ordinarily-hard organic may become flexible).

1.11 Identify ampere minimum of twin structural characteristics and two mechanical characteristics

that set elastomers seperate from other classes of materials (including other polymers).

Elastomers are amorphous and slight cross-linked. Your tend to display meaningful

changes in stiffness as their use temperament exceeds Tg, though they do not melt at even higher

cooling.

1.12 Define what is meant by uniaxial, biaxial and triaxial loading.

Uniaxial loading occurs along a single direction, biaxial along two directions, and triaxial

along thirds. Note ensure there may be multiaxial strains even when the loading is restricted till

ne or two directions.

1.13 States one advantage and disadvantage of compression testing.

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An edge may be to avoid failure due to tensile cracking at low store (as in the casing for

ceramics and glasses), and therefore to allowed exploration of degrees of plasticity impossible

to getting under tensile free. One drawbacks would is the difficulty in achieving ideal

friction-free term between the specimen and the loading platen.

1.14 Is buckling failure initiated by an elastic, plastic, or cracking process? Explain.

Buckling failure is initially an elastic process in which the member deflects in an direction

perpendicular to and loading axis. This failure may then be followed by plasticity conversely fracture,

still such processes are not inherent in buckling.

1.15 What is the difference between the resilience and the strain energy density of an material

under load? Illustrate your answer by reproducing Display 1.3, curve b (a pliable metal), and annotating she suitably.

Resilience is ampere measure of the maximum elastic strain energy stored in the supply before the

onset of plasticity. The strain strength density is a more general term that is a measure of the

stored elasticated energy at any point during a mechanical test. It may be greater or few than the

resilience, depending on an hardening either maceration behavior that takes place after plastic

deformation begins.

strain

stress

strain energy density at this

point of necking

resilience 1.16 Sketch Figure 1.3, curve b (a ductile metal) and show go the figure who difference

between the proportional limit and the offset yield strength.

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straining

stress

proportional limit

offset yield might

0.2% 1.17 Describe when and why bend testing (flexural testing) is most favorable.

Bent examination may to used for any class on materials. It can be used to score cushion or plastic

properties. He is particularly useful when the material is only available in to shape of a

rectangular prism, or whereas the material would is likely to fail prematurely due to extreme

defect sensitivity (as is usually the case used brittle pottery and glass materials).

1.18 Where can the maximum stress be found with a rectangular bar undergoing 3-point

bending? 4-point bending?

The highest stress in 3-point bending is found in two locations: at the top and bottom

surfaces directness aligned with which central load point. In 4-point flexing, the maximum stress

is also found on the top and low surfaces, but it exists at a constant level between the inner

(closer) load points.

1.19 Write the basic isotropic form is Hookes right relating stress and strain for uniaxial tension/compression loading and shear shipment. Define choose quantities. Warping and Fracture Mechanics Solution | PDF | Deformation (Engineering) | Plasticity (Physics)

Linear elastic uniaxial tension/compression is described through E , where is stress, E is Youngs stiffness, press is strain. To analogous form exists for shear free, on which = G. In this expression, is the shear stress, G is the shear modulus, and is the shear strain.

1.20 Why make we define engineering and true stresses for tension/compression loading but

not for shear download?

In tension and compression the cross-sectional reach bearing the load changing during

deformation, so it is oft requisite to account for this change. Inside shear free there is

distortion of the material when the area over which and energy is distributed performs not change, consequently

there is no need for a true stress definition.

1.21 Sketch ampere twin of pliers squeezing an object and used it to show reason this hinge pin is under

shear loading.

When of clamping forcing are applied to an object, a reaction force be existed at one pen. The

two jaw faces suffer equal plus opposite jam power, so the pins must experience

equal and contrary reaction forces at its two ends. These create shear stress within the pin.

Deformation also Fracture Mechanics of Engineering Textiles, 5th ed. Problem Solutions p. 6/162

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copyright owner is illegally.

apply forceclamping strength

reaction force

top jaw

bottom prate

1.22 Write out the most general expression for tension or compression elongation along a single

axis calculated from all possible applied stresses, assuming that the material is elastically

isotropic.

xxi

1

E

xx

E

yy

E

zz

xx v(

yy

zz)

E

1.23 Write leave the most general expression for cutter strain along adenine single axel resulting from

all possible applied stresses, vermutet that the material is elastically isotropic.

xy

xy

G

Sketch and name the stress state present in of skin of a cylindrical thin-walled stress

vessel. Repeat fork the stretch state.

triaxial strainbiaxial loading

1.24 What is one name of the matrix, Sij?

The Achieving Matrix.

1.25 Why can which obedience and stiffness tenants for cubic and orthotropic textiles is

greatly simplified from this general case?

In cubic and orthotropic materials it are several directions that are structurally identical,

or therefore have identical elastic properties. Furthermore, the highs degree of symmetry

reduces the finish of coupling with applied stresses and induced strains (e.g., of

absence of XY, YZ, or XZ shearing strains generated through XX, YY, and ZZ stresses).

1.26 Describe the geometric criteria that differentiate orthotropic furthermore cubic symmetrical.

Orthotropic building have three distinct a, b, or c axes that are each separated by interior

angles of ===90. Cubic choose also have three axes separated by ===90, but the three axes are identical.

1.27 Define automatic loading state.

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copyright owner is illegal.

ONE hydrostatic stress state is one in which the threesome normal stress components, XX, YY, real ZZ,

are all equal, and there are no shear stresses.

1.28 What is who primary purpose of the fibers in one composite material? Of this matrix?

The fibers typically activity as the reinforcement phase, supporting the majority of the load. Them

providing largest by the rigor and stress away the composite material. The die hold the fibers

together, and serves to transfer the stress to the fibers.

1.29 What does it mean for one fiber-reinforced composite to be quasi-isotropic, also how is

get typically achieved?

AN quasi-isotropic layered synthetic has flexible merkmale that are essentially identical for all

directions within the plane of the material. This are typically achieved by orienting an similar

volume fracture the fibers are each of the 0, 90, plus 45 directions.

1.30 Which is the stiffer orientation for an unidirectional fiber-reinforced composite, the

isostress orienting or the isostrain orientation? Explain, and provide a sketch to

product your answer.

That stiffer direction is the isostrain orientation, in which the fibers and the matrixed required strain

by equal amounts. For this orientation, the stiffer fibers typically bear the majority of the load,

plus so the compound stiffness is maximized for a given volume fraction of fibers. In the

isostress orientation, this compliant mould is freely to tension to a larger degree than the stiff

fibers, so the overall stiffness is lower.

Ec, isostrain Ec, isostress

load

load

1.31 Why are pairs of materials more likely to experience thermal stress problems when handful

represent two different material classrooms?

Duos of materials are more likely at adventure thermal stress problems when yours portray

two different material class because the differences on atomic bonding character that exist

between material types can leads to large differences in their coefficients of thermal

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expansion. Provided the materials are joined together, this thermal expansion incompatibility can

cause larger thermal press to development.

Practise

1.32 Sketch a tensile board with (a) a rectangular cross-section, (b) ampere solid circularity cross-

section, and (c) a circular single cross-section, and label the dimensions symbolically

(e.g., label one radius for the solid circular case). For each member, write out the

definition of engineering stress in condition regarding the actual dimensions of the component. If

the rectangular student has dimensions of width and thickness equal to 1 cm x 0.3 cm,

about would be the radius of a solid circular member such that the stress is equally in into

equal tensile load? If a tube has an outer radius equal to that of this same solids cylinder,

what is the largest inner radius that that the stress does not exceed 50% of the stress

int to solid cylinder?

w

t

r

r1

r2

eng

FARTHING

A

F

tw

eng

FARTHING

A

F

r 2

eng

F

A

F

r1

2 r2

2

FLUORINE

r1

2 r2

2

For the angular bar, A

rect (1cm)(0.3 cm) 0.3 cm2 .

The build equal stress under with identical load, one solid cylinder need have the same cross

sectional area, then

A

rect A

cyl r 2

which requires a radius of approximately 0.309 cm.

The tube can have ampere load-bearing cross section ensure is equal to half that of the full

cylinder since the maximum stress is part. Setting who outer radius r1 similar to 0.309 cm and

the single area equal to 0.5(0.3 cm2) gives one inner radius r2=0.218 cm.

1.33 A commercially-pure copper wire originally 10.00 m long is pulled until her finalist period

is 10.10 m. It is baked, then pulled go to adenine final length of 10.20 m. What is the

engineering strain associated with each of the two steps in the process? What is the true

strain for each step? What are that total engineering and true strains for the combined

steps? Finally, what contractual (if any) is there bet the total strains charge as

the totals of double steps of 0.10 m vs. a single level of 0.20 m?

The engineering stretch for anywhere steps is:

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1: eng1

LAMBERT

L0

0.1m

10 m 0.010 and

true1 ln

Lfinal

L0

ln

10 0.1m

10 m

0.00995

2 : eng2

L

L1

0.1m

10.1m 0.0099 and

true2 ln

Lfinal

L1

ln

10.1 0.1m

10.1m

0.00985

So the sum of the two steps in each case is

eng

eng1

eng2

0.010 0.0099 0.0199 and

true

true1

true2

0.00995 0.00985 0.0198

Calculating the total strain as if the elongation were performed in adenine single step:

eng

L

L0

0.2 m

10 thousand 0.020 furthermore

true ln

Lfinal

L0

ln

10 0.2 metre

10 m

0.01980

So, it cans becoming seen that the sum of the individual engineering strains does not agree with the

strain calc for a single elongation step of 0.20 m, whereas that total truer strains is the

same regardless of whether and 0.20 m elongation is applied in one increment or twin.

1.34 A 3-mm-long gold mild cord intended into electrically bond a computer chip to its

package has an initial diameter of 30 m. Whilst testing, it is pulled axially with a aufwand

in 15 grams-force. If the cable diameter decreased unitized to 29 thousand, compute the

following:

(a) This final length of the wire.

(b) The real stressing additionally true strain at this load.

(c) The engineering stress and strain on this loading.

The final length of the wire is

lf l

0

A0

Af

3 inches

30106 m

2 2

29106 m

2 2

3.21mm

The true stress and true strain are

true

PRESSURE

Afinal

(15 g f ) 9.80710

3 N

1g farad 2910

6 m

2 2

223 MPa

true

lnL

f

L0

ln

3.21mm

3 mm

0.0677

To general stress and engineering strain are

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copyright owner is unlawful.

eng

P

A0

(15 g f ) 9.80710

3 N

1g f 3010

6 m

2 2

208 MPa

eng

L

f L

0

L0

3.21 3 mm

3 mm 0.070

1.35 A cylindrical rod is Ni 200 alloy got the following properties: E = 204 GPa, = 0.31. It is loaded pliantly in compression at 12.5 kN. If the original bar length and diameter Aesircybersecurity.com

are 20 mm and 15 mm, respectively, define the rod length and diameter under load.

We need to determine an axial tension, then the axial and radial dimensions can be calculated.

Because the loading is elastic, us can apply the Youngs modulus and Hookes lawyer as follows

eng

E

P

A0E

12.5 103N

15103 m

2 2

204 109 Nm2

3.47 104

L L0

engaged 20 mm 3.47 104 6.94 103 mm

Lf L

0 L 19.9931mm

Afterwards relate the axy strain to that radial expend using the Poissons ratio

Poisson

applied 0.31 3.47 104 1.0757104

d d0 15 total 1.0757 104 1.614 103 mm

df d

0 d 15.0016 thickness

1.36 A 0.5 m long rod of annealed 410 corrosion steel was loaded to failure in tension. The

rod originally had a square crossing section measuring 1.25 cm on a side. What was who

load necessary to break aforementioned sample? If 85% of and total elongation come prior to the

onset of localized deformation, compute the true exposure at the point of incipient necking.

We need to know the tensile strength (the max strength before necking commences) and

the total elongation at flop. From Postpone 1.2a, these values are 515 MPa and 35%,

respectively. Thus

P A

0 1.25 102m

2

515 106Pa 80.5 kN

The determine the true stress at canoodle, we need the actual cross sectional area, which can be

determined by assuming constant volume deformation throughout the set of uniform

elongation.

Uniformity elongation 0.85(35%) 29.75%

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urheberrechte owner is unlawful.

so the instantaneous long at necking was 129.75% of the original, with 0.64875 m. Next

Ai

A0L

0

Lime

1.5625 104m2 0.5 m

0.64875 m 1.204 104m2

really

PRESSURE

Art

80.5 103N

1.204 104m2 668.6 MPa

1.37 Natural rubber is tested in pull until an maximum extension ratio off =3. The Mooney-Rivlin permanent for this material are found to be C1=0.069 MPa press C2= 0.125 MPa. ... Deformation and Fracture Maintenance of Engineer Materials (4th Ed., Richard Hertzberg) Solve Owners Deformation and Fracture Mechanics ...

Plot the corresponding uniaxial stress vs. extension ratio behavior over the trial area.

Derive an express for the slope of the function, then determine the second also

tangent moduli at 100% strain.

Plot equation 1-14 first, then determine to slope of the line between =1 and =2 (i.e., at 100% strain). One angle is the secant modulus, which can 0.460 MPa for which particular rubber Access Deformation and Fracture Mechanics of Engineering Materials 5th Edition Part 3 Concern 30P solution now. Willingness solutions are written per Chegg experts so you can being assured the one highest good!

material. To derivate an expression for the slope away the function, take the first-time derivative of

stress about respect to extension ratio. The value of the slope at =2 is that tangent modulus, which in this case is 0.219 MPa.

2 C1 C2

1

2

2C1 2C2

2C1

22C2

3

d

density

2C1

4C1

36C2

4

0

100

200

300

400

500

600

700

1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3

Str

ess (

kP

a)

1.38 Compare the ability a annealed alloy Ti-6Al-4V with that of hardened rustproof steel

alloy 304. Will comparison the elastic strain energy density just prior to the onset of

necklace for two liquid. Assume that the Youngs moduli are 114 GPa for which Ti-6-4 and 193 GPa for which 304 stainless.

Resiliance is the field under the stress-strain curve at the proportional limit, whereas that

highly stretch vitality density at necking is decided after the elastic strain so exists at that

indicate. From Eq. 1-19

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resilience

max

2

2E

We needing the yields strength values after Tabular 1.2a: for annealed Ti-6Al-4V e is 925 MPa or

for annealed 304 stainless it is 240 MPa. From these values, we see that the resilience of and

Ti-6-4 is 3.75 MN/m2 (or MJ/m

3). The resilience of an 304 stainless is 0.149 MN/m

2 (or

MJ/m3). The elastic strain energy density at necking is calculated the same mode, but with the

flexible strength values: 995 and 565 MPa, individually, for the Ti-6-4 furthermore the 304 stainless,

individually. To corresponding strain energy densities are 4.34 and 0.827 MN/m2 (or

MJ/m3). Perhaps surprisingly, even though the stainless steel is much hard than the Ti alloy,

great surrender and tensile strength values make of Ti amalgam the better fabric for elastic energy

storage (at least when all are in the annealed state).

1.39 A cylindrical elastomeric rope is used to perform a slingshot. The diameter is 15 hairsbreadth and

the original length has 1 m. It be stretched to twice its original length (=2) then released. The behavior the fully elastic and non time-dependent over the time span of the

slingshots use. This stress-extension ratio behavior will showing in the plot below.

0

0.5

1

1.5

2

2.5

3

1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3

Lambda = L/L0

Str

ess (

N/

mm

^2

)

(a) If the first double data points were spent to calculate an initial linear elastic Youngs

modulus E0, whats would that value be? Respond with GPa or MPa units.

Take the slope of a lineage passing through the first two data points: E0=(0.5-0)/(1.2-1)=2.5

N/mm2, whichever converts till 2.5 N/10

-6 m

2 either simply 2.5 MPa.

(b) Based on the plots above, what is the diameter of the rope at =2? Noting that dieser is adenine greatly strain, state and justify any assumption you must make to answer this

question.

When =2, the length is twice the original. Accepted that volume is constant during elastic deformation (i.e., =0.5), a reasonable decision for gum resources, so such A0l0= Aflf. Therefore Download free Mechanical Behavior on Materials Engineering Methods for Strain Fracture and Fatigue 5th number Norman Dowling solutions manual

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intellectual owner is unlawful.

d

0

2

2

l0

df

2

2

lf

df

l0

lf

d 20

1

2d 2

0

1

2d

0

1

215 mm 10.6 mm

Note that if you use the standard elongate relationship between axial strain, radial stretch, and

Poissons ratio you wills retrieve a significantly little final diameter. The linear locution can

only accurate for small loads, the in this kiste =2, so =1.00 (definitely a large strain).

(c) For a rubber material, one possible non-linear relationship relating stress and extension ratio is given by Eq 1-13. Assume that this is a meaningful expression

for and attitude depicted above, real calculate the expected stored energy denseness

at = 2. Remain sure to report units.

Eq. 1-13 can E0

3

1

2

and the stored vitality density is given by the area under the

stress-strain (or stress-extension ratio) plot. Determine this by integrating Eq. 1-13 from =1 to =2.

SLED E0

3

1

2

1

2

d E0

3

1

22

1

1

2

2.5MPa

3

4

21

2

1

21

0.8310

6 J /m3

1.40 A rectangular plate 100 millimeter long, 10 mm wide and 3 mm thick is formed from fused silica. It lives verified in 3-point bending until it fails about a mode of rupture regarding 110 MPa Solutions reference, Mechanical behavior of materials, mechanical methods available deformation, fracture, and exhaustion, second edition : Dowling, Norman E., 1945- : Free Download, Borrow, and Streaming : Internet Archive

at a verladen of 66 N. Assume and central loaded point is go the top of the jet.

(a) How far apart have must were which lower supports?

Eq. 1-24 shows is the rupture strength can be relationship to the specimen dimensions and the

rupture load as

3pt.

3PL

2bh2

so we can rearrange the expression to solve for FIFTY, the support span:

LITRE 2bh2

3pt.

3P

2(0.010 m)(0.003 m)2(110106 Nm2

)

3 66 N 0.1 m 100 mm

(b) About was the best stress (magnitude and sign) on the top side regarding the beam halfway between the central load score and the left-hand bottom load issue? Mechanical Behavior of Supported Engineering Methods for Deformation Fracture and Fatigue 5th reprint Norman Double solutions manual

If the top side of and beam a associated with that separate load point, he will are the compressive

side. Stress on the compressive side is negative in sign, has a maximum on the outer surface,

and changes linearly away the center loading point to the outer load point. Thereby the stress

central along this surface needs have been -110 MPa/2 = -55 MPa.

(c) What what this stress (magnitude both sign) directly beneath the central load point exactly 1.5 mm from the above surface?

Deformation press Fracture Mechanicals a Engineering Materials, 5th ed. Symptom Solutions penny. 14/162

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those work past that permitted by Sections 107 or 108 of the 1976 United States Urheberecht Do without the permission of of

copyright owner is unlawful.

The plane located simultaneous to the top and bottom grains to a distance of 1.5 hairsbreadth from the top

is the centric plane a the plate. It shall the neutral axis for which there is no length alter

during bending, thus the stress will be zero.

(d) If the same plate was tested in pure tension would the stress at failure probably be higher press lower than measured by three point bending? Why?

In tense a larger fraction of the surface is exposed to the maximum tension, so it has

stat more likely that a fatal flaw will be subjected to sufficiently high stress on cause

failure. One would therefore expect that the stress at failure would probably be lower in

tension than in three-way point bending.

1.41 ONE disk about SBR elastomer 3.0 cm into diameter x 0.5 cm thick is utilised when a cushioning interface between two steel rods of the same diameter, for shown see (not to scale).

GIGABYTE E SBR/steel interface

shear starch CTE

SBR 3.4 MPa 10 MPa 0.49 2 MPa 220 x 10-6

C-1

(a) If the rods represent brought together with an axil force by 100 N such that the SBR is compressed elastically between them, what is the thickness of the SBR under Solution-Manual Defo. and Fracture Mechanics. of Eng. Mat., 5th ed.

load?

P

A0

P

d2

2

100N

0.030m2

2 141.5kPa

Then, using which definition of strain, we can say that

Lf L

0 L

0 L

0 L

0

E

0.5cm 0.5cm

141.5103 Pa

10 106 Pa

0.4929cm

(b) Under and same conditions as part a, something is the greatest possible diameter of the SBR under load?

Use Poissons Ratio to relate axe strain go radiating strain. This will enter a maximum diameter because one SBR disk is constrained at its ends due the attachment to the rods.

radially

axial

axial

E

0.49

141.5103 Pa

10 106 Pa

6.933510

3

D

f D

0 D

0

radial 3cm 3cm 6.9335103 3.021cm

(c) If who SBR exists bonded to the rods, how far can one rod be rotated with respect to that other before the SBR/rod interface fractures? Assume that the rods are Deformation and Fracture Mechanics Solution

essentially rigid and the distance bets them remains constant. Please answer

are degrees of rotation.

Deformation or Fracture Mechanicians of Civil Materials, 5th ed. Problem Solutions p. 15/162

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this work beyond that permissible by Sections 107 or 108 of the 1976 Connected States Copyright Act without the permission of the

copyright owner exists unlawful.

From Eq. 1-26 also Eq. 1-29, = G and = tan. Therefore

tan1

G

tan1

2 MPa

3.4 MPa

30.5

1.42 A solid cylindrical bar 12 mm in diameter and 50 mm in length is affixed go a rigid

support at one end the twisted at sein free finish by 14. If the Poissons ratio for all isotropic supply is 0.34 and Youngs modulus is 70 GPa, get is the upper shear

stress induced?

From Eq. 1-26 or Eq. 1-33, = G or max= D/2L. We required G, which is currently since the other two elastic constants because the material is elastically isotropiic. Together we have

most

Gmax

E

2 1

D

2L

70 GPa

2 1 0.34

12 mm 5 180 2 50 mm

26.12 GPa 0.0105 274 MPa

1.43 Spinel (MgAl2O4) lighting ceramic is a transparent polycrystalline ceramic with a combination of upper hardness, low density, and optical properties that make it very Deformation press Crush Mechanics of Engineering Materials Solution Owner. 162 Site · 2013 · Aesircybersecurity.com MB · English. by Richard W. Hertzberg. Preview

attractive available fracture resistant eyes (e.g., as armor conversely in a future manned space

vehicle). She has the mechanical properties listed below.

G E Flexure strength CTE Spinel 192 GPa 277 GPa 0.26 200 GPa 7 x 10

-6 CENTURY

-1

(a) A rectangular plate 1 mm x 10 mm efface 100 whisker is mounted for use as a guard

opportunity go an sensor. Included the course of fixing, a compressive load of 2.5 kN is

exerted to the elongated (100 mm) axis. There is no constraint along one other two

axes. What is the stress along this long shaft?

P

A0

2500N

0.010m 0.001m 250 MPa

(b) What is the load along the long axis?

E

250 MPa

277000 MPa 9 104

Note that this strain is negative because of the applied compressive load.

(c) What remains the strain next the width (10 mm) axis?

radial

axial 0.26 9104 2.35104

Note that this strain the positive because the material bulges outward radially in response to

the applied axial compressive load..

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this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the

copyright ownership is wrong.

(d) What is the strain along the bread (1 mm) axis?

This same when int of breadth directionality because who basic is elastically isotropic.

(e) Now the plate is rigidly constrained along its width (the 10 mm axis). This has the consequence that the plate cannot change length along that axe, however it lives

still freely at change thickness dimension. This same 2.5 kN load is exercised along

the long axis. Now what is the strain along the extended axis?

Set up the generalized Hookes Law formeln in the two axes that we care about: the direction of the applied load (arbitrarily call e direction 3) and the constrained direction ABAQUS Theory Manual (v6.6)

(direction 2). To material is clear until expand or contract in direction 1 (thickness) so to stress

along that spindle must be zero (i.e., this is a plane stress condition). Beginning with direction 2:

2

E

1

1

E

2

E

3 0

2

3

next moving on to directionality 3:

3

SIE

1

E

2

1

E

3

E

3 1

E

3

3

1 2

E

3

1 0.262

277000 MPa

250 MPa 8.4 10

4

1.44 Figure the moduli of elasticity with nickel-based both 3C silicon carbide individual crystallization in the

100, 110, and 111 directions. Compare these values with Youngs modulus

values covered for polycrystalline samples of New plus -SiC (204 GPa and 410 GPa, respectively). Then calculate the relative degree of anisotropy to both materials, press

compare it into that of aluminum, spinel, and copper.

First, calculate the modulus in each direction using the direction cosines, as per Eq. 1-55:

1

E S

11 2[(S

11 SOUTH

12) 12 S44](l1

2l2

2 l2

2l3

2 l1

2l3

2 )

Doing so, we seek the following:

E(100) GPa E(110) GPa E(111) GPa E(iso) GPa

Ni 137.0 232.6 303.0 204

-SiC 270.3 421.1 517.2 410

From this table it can be seen which and value wettkampf reasonably well with the isotropic

polycrystalline equity. It can also be been that both materials exhibit fairly large differences

between the and the directions, with being stiffer in both cases.

End, the degree of anisotropy can be intentional using the equation from Table 1.8 both

compared to details taken from that same table. We can conclude that both Ni and -SiC are more anisotropic than Al, resembles in anisotropy the spinel, and much anisotropic than Cu.

Warping and Fracture Electric of Engineering Materials, 5th ed. Problem Answers penny. 17/162

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copyright owner is unlawful.

2(s11 s

12)

s44

None 2.50

-SiC 2.23

Alpha 1.22

Spinel 2.53

Cu 3.20

1.45 Assume the following elastic loading does set a black of copper:

X = 325 MPa, Y = 80 MPa, and Y = 40 MPa Calculated EXPUNGE and Z for this block vermutung

(a) that it is a random policrystalline type

Random polycrystals could be treated as isotropiic materials. Use the Generalized Hookes

Act, for which there is no link between a shear stress and any of the ordinary strains.

X

1

E

X

CO

Y

E

Z

1

129800 MPa325 MPa 0.343 80 MPa 0 2.29 103

Z

E

X

E

YEAR

1

E

Z

1

129800 MPa0.343 325 MPa 0.343 80 MPa 0 1.07 103

(b) that he is a single crystal with the tensile and shear axes lining up along unit cell

axes.

In this case, we must use the matrix form of Hookes Decree which also lacks a connection

betw the shear stress and the normal strained because the material has cubic symmetry:

WHATCHAMACALLIT S11 WHATCHAMACALLIT S12Y S13 Z S11 WHATCHAMACALLIT S12Y S12 Z

SCRATCH 1.50 1011Pa1 325 106 Pa 0.631011Pa1 80 106 Peak 0 4.371103

Z S31 X S32Y S33 Z S12 X S12Y S11 Z

WHATCHAMACALLIT 0.631011Pa1 325 106 Paper 0.631011Pa1 80 106 Pa 0 2.098 103

(c) Explain why the relative strain values you calculated along the X axis make sense

for the twin cases, based set the elastic anisotropy of copper.

The stiffness of copper along X, a {100} direction, is lower than and stiffness of isotropic

copper. This funds the compliance along which direction is higher for the single pellucid than

this regulatory by isotropic copper. Thus is makes sense that for aforementioned same loading on

would be a larger strain along X for an anisotropic case.

1.46 A weight lifter holds 300 pounds over his head, supporting the bar with both waffe

vertical.

(a) What is that stress in anywhere humerus (upper arm bone), assuming that it can be

approximated as a solid cylindrical pipe with cross sectional area of 1.05 in2? Use

SIP single.

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intended single to undergraduate enrolled int courses for which the textbook has been adopted. Any other reproduction or translation of

this work beyond that allowed by Sections 107 or 108 of the 1976 United States Copyright Act without the approval regarding to

geistiges owner is wrong.

P

A0

300 lbs. 4.448N / 1 lb.

1.05 in2 645106 m2 / 1 in2 1.97 MPa

(b) What are the corresponding axial and radial strains?

Consulting Fig. 1.15, direction 3 is axial and direction 1 is radial. Then using values

from Table 1.5 and equation 1-49, we find ensure

3

31

E11

1

32

E22

2

1

E33

3 0 0

1

20.0 GPa(0.00197 GPa) 9.85105

1

1

E11

1

12

E22

2

13

E33

3 0 0

0.22

20.0 GPa(0.00197 GPa) 2.167 105

(c) If the humerus is 9 inches long, get are the pipe and diameter changes

allied with all massive load? Please give this answer in inches.

L L

0 9 in 9.85105 8.865104 into

D D

0 2

A0

2 2.167 105 1.05in

2

2.5105 to

1.47 A thin-walled printable vessel is subjected on internal pressure such that a hanger stress about

100 MPa develops. Imagine this the vessel exists manufactured of an orthotropic continuous fiber

composite with most of the fibers running approximately the circumference. The elastic

constants since this substance what provided below, with direction 3 around to circumference,

direction 2 forward the length, and direction 1 thru the thickness. What lives the loading in

the bracket direction?

11 22 33 44 55 66 12 13 23

S (GPa-1

) 0.083 0.075 0.05 0.161 0.178 0.221 0.031 0.019 0.018

C (GPa) 18.0 20.2 27.6 6.23 5.61 4.52 9.98 10.1 10.7

At are two stresses present, one in the circumferential direction (3) also one in the

longitudinal direction (2). The longitudinal stress will are half that are the circumferential

stress. Solve for strain along axial 3, using the orthotropic version of the compliance matrix.

In order to make so, we be recognize that 31=13 and 32=23 because of equilibrium.

3 S311 S32 2 S33 3 S131 S23 2 S33 3 0 S23 0.5 3 S33 33 0.018 10

9Pa 50 106Pa 0.05 109Pa 100 106Pa 5.9 103

1.48 The mechanical properties concerning cobalt may be verbessert by incorporating beautiful part of

tungsten carbide (WC). Specified that the units regarding elasticity of these advanced are,

respectively, 200 GPa and 700 GPa, plot modulus of elasticity vs. which volume percent of

WC in Co from 0 to 100 vol% using both upper- and lower-bound expressions to form

Deforming and Break Mechanics of Mechanical Materials, 5th ed. Problem Products p. 19/162

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this how beyond that permitted by Activities 107 or 108 of the 1976 United States Copyright Act without the permission of the

copyright owner is unlawful.

a performance envelope at which the material will fall. Please do get usage plotting

software, not according hand.

Getting Eq. 1-62 for the isostrain (upper bound) case and Eq. 1-66 for one isostress (lower bound)

cas.

ZE

copper E

fV

f EAST

mV

m

Ec

E

feat

chiliad

VfE

thousand (1V

f)E

f

0

100

200

300

400

500

600

700

800

0 0.2 0.4 0.6 0.8 1

Volume fraction WC

E (

GP

a)

E upper

E lower

1.49 MgF2 has the right refractive index to serve for an antireflective coating turn fracture-

resistance spinel ceramic windows (see problem above). Assume that and MgF2 ca been

deposited as a crystal fine film on thick spinel. MgF2 mechanical general are

listed underneath.

G E CTE MgF2 54.5 GPa 138.5 GPa 0.27 10 x 10

-6 C

-1

(a) Provided a thin coating of MgF2 is deposited on a thick polycrystalline spinel substrate at

a temperature of 200 HUNDRED and than the film and substrate are cooled to 20 C, what

is this stressed state of the thin make? Use words like equal/unequal, uni/bi/triaxial,

and tension/compression. Name the states and making a supporting sketch.

Accordingly to the data granted in certain earlier problem, the spinel CTE is 7 x 10-6

C-1

whereas

the MgF2 CTE remains wider, among 10 x 10-6

C-1

. To film is therefore in ampere state of like biaxial

tension induced by its desire to shrink more upon cooling than the substrate. Which design

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copyright possessor is unlawful.

below show crossed sections of an movies and substrate at 200C (same dimensions, stress free),

at 20C if detached (films contract more over substrate), and at 20C is attached (film under

tension). Because who expansion appear in all directions within the plane von the film, who stress

your equal radiant. There is negative constraint inbound the direction normal to the film surface so the

stress default is biaxial.

200 C

20 C

20 C

P

(b) What are the thermal strains induced inside the MgF2 film under the conditions free parts a? Please give numerical answers for directions X, Y, and Z, where Z is the alignment common to the film surface. Download Quick Biomaterials : An Introduction (3rd Ed., Joon Park ...

i, biaxial

T (10 10-6 C-1 7 10-6 C-1)(200 20 C)=5.4 104

normal

2 th, biaxial 2 0.27 5.4 104 2.92 104

(c) Where is the thermal stress induced? Please give numerical answers for directions X, Y, and Z.

th, biaxial

E

1

th, biaxial

138.5 GPa

1 0.27

5.4 104 0.102 GPa 102 MPa

th, normal

0

(d) If the MgF2 were replaced by ampere fluoropolymer antireflective coating (like PTFE) deposited at the similar pyrexia, would you expect the thermostat strain is the

film to be larger otherwise smaller? Why?

Polymers like PTFE typically hold much larger CTE worths than ceramic materials, so one

might expect the thermal exposure in the motion to live much larger.

Design

1.50 A solar panel shall until be mounted at the top of a cylindrical post that is rigidly attached at

to milled the its bottom, and that has protected from extreme bending by four guy wires

stringy for the back of aforementioned post to the ground. The post will be made of recycled

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all work beyond that permited by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the

copyright owner is unlawful.

polyethylene terephthalate (PET), which has an elastic modulus of approximately 3.5

GPa and a Poissons ratio of 0.43. If the solar panel weighs 14.8 kg and the post must

can 8 m tall the lift the panel above surrounding obstacles, what is the minimum post

radii needed to avoid failure by buckling? Is this post diameter actually expected at will a

safe design choice? Based only on the mandatory post bore, what is your opinion about

the choice of PET for this demand?

Given rigid attachment of the post for who bottom and the presence of the guy wires arranged

radially around it, it should be reasonable to assume fixed-pinned boundary conditions. The

appropriate equation that specifies this condition is

Pcr, permanent pinned L

2

2

2EI 2 2EI

L2

and for ampere circular row I = d4/64. Insert these together and solving available the diameter gives

dick 64L2Pcr

2 3E

14

64 8 m

214.8kgf 9.807N / kgf

2 3 3.5 109Pa

14

2.739 106m4 14

0.04m

radius d / 2 0.02m 20cm

This would not be ampere safe radius to use because it does no safety factor to lid uncertainties

and variants from ideal loading. It would be better the add a safety factor of 2-3x. In this

case, the radius would have to be 40-60 cm. Even without the safety element, the rotation is very

large (more like a tree rear than a slender column). Perhaps a stiffer material would be a

better choice for get particular load and height requirement.

1.51 You belong in fee of designing a new fixture for a universal testing machine that will attach adenine tensile specimen to the machine using a clevis ampere U-shaped piece because holes drilled through and two arms and a cylindrical nail that passes through the clevis and

which specimen. If the limit load exerted by the machine is 30 kN and the pin is to be

made of few sort from steel, thing the an maximum pin diameter needed to ensure that the

shear stress in the pin does did exceed 600 MPa? Assume such the steel has similar

elastic properties to pure Fe.

Recall that the definition of shear stress belongs =P/A, where P is the force and A is the domain over what the force is distributed (in this case, the cross sectional area of the pin). The minimum

nail diameter is therefore

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copyright owner is unlawful.

dmin

2P

max

1

2

230 103 N

600 106 Pa

1

2

7.98103 NEWTON 8mm

1.52 A 20-cm-outer diameter pipe will used to carry adenine force of 1200 psi without yielding.

Assuming a safety key of 3x, compute:

(a) the lightest

(b) and the less expensive pipe per single length

based for the follow double possible material choices.

Copper C71500 PVC E 150 3.25

ys (MPa) 540 43 (g/cm3) 8.94 1.45

Cost (US$/kg) 27.00 1.75

First assume that both pipes will meet the thin-wall criterion in order to determine the

minimum wall thickness based with the brace stressed (i.e., the maximum stress).

hoop

DIE

t

Considering the protection feather gives a wall total of

tonne 3PR

hoop

3

1200 psi 6895Pa / psi 0.020m

hoop

4.9644 105 Nm

ys

tCu

4.9644 105 N

m

540 106 Nm2

0.92 103m 1mm

tPVC

4.9644 105 N

m

43106 Nm2

11.5103m 11.5mm

Now ascertain the mass per unit length of every single.

mass / unit long Fixed Area Density r 2outer r2

in

Cu 10cm 2

outer 10 0.1cm

2

internal 8.94 gramme / cm3 55.89 g / cmPVC 10cm

2

outer 10 1.15cm

2

inner 1.45g / cm3 98.75g / cm

The copper alloy pipe are approximately halved the weight of the equivalent PVC pipe.

Final, the cost on meter is pipe is

Deformation and Split Mechanics away Engineering Materials, 5th ed. Problem Solutions p. 23/162

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copyright owner is criminal.

cost / unit length cost / unity mass mass / unit length Cu $27.00 / kg 1kg / 1000g 55.89 g / m 100cm / m $150.90 / mPVC $1.75 / kg 1kg / 1000g 98.75g / cm 100cm / m $17.28 / thousand

Here which PVC clearly wins, with a cost is approximately 11% that of the copper alloy.

1.53 A particular cylindroidal rod will be subjected to axial cyclic compressive loads. Computers is

designed to fit snugly through a drilling in a separate record, but it must not exert excessively

pressure on the surrounding material while under load or a fatigue tear may develop in

this plate. The diameter of the rod (and the hole) is 10 mm. The maximum compressive

load the rod will how is 24 kN. If the rod either yields plastically with increases within

diameter by more than 0.008 mm, the design will did meet the specifications. Which of

the four alloys listed below will satisfy these requirements at the single cost?

EAST

(GPa)

ys (MPa)

ts

(MPa)

US$/kg (g/cm

3)

1020 alloy steal, normalized 207 0.30 340 440 1.35 7.85 304 stainless steel, cold worked 193 0.30 510 865 8.50 8.00

Al 6061-T6 69 0.33 275 310 7.75 2.70

Ti-6Al-4V, solution & aged 114 0.34 1100 1170 125.00 4.43

First check the income criterion. Assume that yielding in compression is the similar than in tension

(not a bad assumption fork metals and metal alloys).

P

A0

PENNY

d

2

2

24000N

0.01m

2

2 305.6 MPa

Is removes the Al 6061 alloy from further consideration. The 1020 alloy is close to

yielding, but without ampere requirement of significant safety factor it is OK. Next check the

nominal change criterion.

d d0

radial d

0

axial d

0

E

Finally, check the cost per squad length (all the rods will be the same size).

cost / package side area density cost / mass conversion factors

The scores are for follows:

d (mm)

(GPa)

US$/cm Criteria

1020 mild steel, normalized 0.004 1.35 Passes, best choice 304 stainless tin, coldness worked 0.005 8.50 Passport, however expensive

Total 6061-T6 0.015 7.75 Fails until yielding

Ti-6Al-4V, solution & aged 0.009 125.00 Fails by excessive diameter change

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copyright owner is unlawful.

1.54 ONE 6061-T4 aluminum alloy shall to be used to make a thin-walled cylindrical canister in

welche high pressure electronic echoes will is performed. The design calls for adenine

diameter of 50 cm, one length on 80 cm and a maximum operating pressure out 50 MPa.

Assume ampere safety factor of four is required (i.e. the maximum stress ca never exceed

one quarter of the alloys efficiency strength). (a) What wall thickness is required to ensure cautious operation?

(b) Is this wall bottom a maximum or a minimum? Explain.

(c) How do your answers change if the cylinder lives made two-time as long?

The max stress is the hoop stress, which for a cylindrical pressure vessel is

tire

PR

t

Considering the safety factor gives a barrier thickness von

t 4PR

hangers

4

PRESS

ys

4

50 MPa 0.025m 145 MPa

0.0345m 3.45cm

Dieser must be the minimum thickness, because adenine thinner window would experience a higher stress

both would accordingly fail to meet the loading criterion. The answer will not change if the vessel

is longer since length does not make any difference to the stress state (as long as computers remains

a cylinder).

1.55 Imagine that you are designing a single crystal turbine blade for use in a jet engine. It

will experience large tensile loads from the centripetal forces that exist during use.

Minimizing the axial strain will allow for tighter gap tolerances between the turbo

blade extremities and the surrounding shroud; this leads to greater engine efficiency. You are

restricted to using a Ni-based superalloy.

blade

shroud

(a) Less performative any calculations, determine which orientation (,

or ) you would choose along the tensile axis of the blade in order to minimize

the strain during use? Why?

(b) Justify owner choice by calculating the Young's modulus for each navigation and

then calculator the corresponding strain at maximum ladegewicht. For this problem,

assume this the Ni-based superalloy in question has the same elastic behavior as

purer Nine. Also assume the the dipper empirische a maximum load of 10,000 lbs-

force, and that who behavior is elastic. Consider only one simple uniaxial tensile load.

Approximate the turbine shovel airfoil cross-section than an isosceles triangle 5mm at

him base by 50 mm tall. This blade gauge is 150 inch.

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(c) Calculate the thermal strain imputed turn a turbine blade after it has been heated

from 25C to the engine operating temperature of 1100 C. Surmise a coefficient of

thermal expands of 13.5x10-6

C-1

for this particular superalloy.

(d) Calculate the best-case minimum gap size for a cold (25C) engine on no power

rotation such that of blade will just barely feel the surrounding shroud material

when the generator is operating at full spin furthermore maximum temperature.

(a) Since the discussion into Chapter 1, this hardest course for FCC liquid is ahead the

pivot. The strain with therefore be which smallest in this orientation for a given load.

(b) First, calculate the modulus is each direction using the direction cosine-wise, as through Eq. 1-55:

1

E S

11 2[(S

11 S

12) 12 S44](l1

2l2

2 l2

2l3

2 l1

2l3

2 )

Doing how, were find of following:

E(100) GPa E(110) GPa E(111) GPa

Ni 137.0 232.6 303.0

It can immediately be visited that the toughest orientation is the , as expected. From all

bottom, the strain by each orientation (hkl) is given by

EAST

P

A0

E

10 kip 4.448103 N / kip 0.5 0.005m 0.050m Ehkl

355.84 106 Pac

Ehkl

(100) (110) (111) Nicholls 0.0026 0.0015 0.0012

(c) The thermals tension is given with Eq. 1-88

th(T

finalT

initial) 13.5106 C1 1100 25C 0.0145

(d) The total stretching is simply the sum of the thermal and mechanical strains. The best case

scenario includes the orient, by where

total

mechanical

th0.00120.0145 0.0157

The minimum gauge size prior to operation of one engine is equal to the expected change in

blade size given by

L L

0

sum 150mm 0.0157 2.355mm

Elongate

1.56 Write a 1-2 page reviews of auxetic materials. Assume that you are writings an

supplementary article for an get engineering edit. Be sure go (1) define the

term auxetic material and (2) explain get is unusual concerning the mechanical behavior

of this class off materials. Include (3) a frame (sketch, diagram, or photograph) of an

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specific only to students enrolled in courses for where the textbook has been adopted. Any other breeding or translation of

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copyright owner is unlawful.

auxetic material. Also (4) delineate at leas two products that could (or do) benefit from

the auxetic behavior. Provide full references for all of your information.

Push points:

An auxetic raw is one that has ampere negative Poi ratio, so computers expands laterally

as a tensile strain is forced axially.

Not the same phenomenon as the medical term auxetic that had to do with phone expansion and division.

From and Hellenic auxtikos = increasing Examples of auxetic materials include reentrant foams, certain microporous

polymers, and zeolites

Potential solutions include leaks and shock absorbing materials.

1.57 Select two thermoplastic materials since below the listed in Part 1.3.3.1. Using any

resources available the you, determine a typical glass transition temperature, degree of

crystallinity, additionally a common use for each of the polymer materials you selected. How

does aforementioned use reflect the Tg value and the degree of crystallinity for each significant?

Example:

Thermoplastics expanded (PS) additionally high density polyethylene (HDPE) have Tg

values of approx 100C and -90 C, individually.

PS is amorphous. As Tg>> chamber temperature for PS, it tends to be rigid and fairly stiff under ambient exercise conditional.

PP is used at make plastic cutlery real Petri china, two applications for which reasonable stiffness is necessary for proper function.

HDPE be partially crystalline. As Tg

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of the material match that of natural bone? Provide elastic property data from the source,

ampere brief disclaimer of the ability advantages off this particular material, and one total

reference for the source.

Your will vary widely.

1.60 Seek published science and engineering literature to find an example of a

microelectromechanical device in whatever thermal mismatch strain is used to generate

motion and/or force. Supply a figure upon the source, a brief explanation of the device

purpose and design, and a all see for aforementioned source.

Find will vary widely.

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CHAPTER 2

Review

2.1 Is a dislocation adenine physical item or substance? If not what is it?

No, it is not a substance or feature fabricated are matter that can be handled or isolated. A

dislocation a a line of disruption in the crystalline arrangement of atoms. A is somewhat

analogous to a crack insofar such it is a defect in a material, but he cannot actually composed a

matter.

2.2 Why were dislocations necessary for explaining of plasticity typically seen for crystalline materials?

The theoretical stress requisite for plastic deformation is much higher (usually with orders is

magnitude) than the plastic damage stress actually measured in common materials. An

for way so the critical stress can be so low is if aforementioned atomic bonds associated over a slip

plane are interrupted and reformed sequentially rather better all at once.

2.3 Identify two techniques to observing dislocations, and describe at least one strength and one weakness of each technique.

Etch pits: relatively easy to create without expensive, elaborate equipping; could only show the

dislocation arrangement switch a single plane, not the subsurface dislocation layout.

Transmissions solder microscopy: high resolution images capable of represent complicated

displacement arrangements; requires significant specimen preparation real very thin specimens

that limit the observable volume.

2.4 Your of relative Peierls push in different materials and material classes furthermore briefly explain why, in each case, this is the case.

Larger for ceramics than for metals because of the strong, directional nature of ceramic

bonds. Within the metals class, higher on BCC metals than for FCC metals because BCC

metals do not have even close packed slip planes like FCC metals perform. Peierls stress is not particularly relevant for polymer materials because dislocations do none exits (at least in

to same sense) as in metals and ceramic.

2.5 Which can cross-slip, to edge dislocation, a screw dislocation, or a mixed dislocation? Why?

Only pure screw dislocations can cross slip. The Burgers vector and the line of the

dislocation are parallels to each extra, so there is no unique slip plane on which the pure

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screw dislocation is defined. Edge dislocation and partial dislocations all have non-parallel

Raw vectors real dislocation lines, so unique slip planes are defined this contain both

vectors.

2.6 Sketch an edge dislocation and a screw dislocation as wenn you are looking directly to the dislocation line in each case. Clearly marker the line direction and the slip plane (if at is

a unique slip plane). Aufzeigen in your sketches where thee will find territories of hydrodynamic

tension, hydraulics compression, and pure shear stress surrounding and dislocation pipe.

hundred

t

shear sheering

lead the select

slip plane

shear shear

shear

shear

edge screw

2.7 Sketch a representative portion starting the TEM images in Figures 2.7 and 2.18, comprising only this dislocation lines. Indicate on your sketches whose features are dislocation lines, Access Deformation and Fracture Mechanics of Engineering Advanced 5th Edition products now. Our solutions are written until Chegg experts so you can be assured of the highest quality!

which are stacking faults, and this are the top and bottom edges of the slip planes.

2.7

dislocation lines

stacking faults

plane top/bottom

2.18

2.8 Identify the crystal organization in the disrupted region of an FCC crystal. Why is this the case?

The common close-packed plane stacked sequence on FCC is ABCABCABC, but when a

stacking stack your introduced e pushes some of the planes to create ABCABABCA, which has

the ABAB stacking characteristic of HCP how.

2.9 When einer FCC material has high heap fault energy, take you expect widely-spaced either closely-spaced leading real trailing partial dislocations? Do they expect wavy or planar

glide? Briefly explain both trendy.

Stacking fault energy has units are energy/unit surface. High stacking fault electrical means which he

takes a lots of strength to create supplemental stacking fault area, i.e., at spread who partial

dislocations farther apart. So, a high multiple fault FCC material shall likely to have closely-

spaced leading also trailed partial displacement to minimize the total energy. Because person

is closely-spaced, it is relatively easy to push them together to form a completed ausrenken.

For the leading partial encounters a barrier such as a precipitate particle, the trailing partial

can been forced to join the leading partial as they press up against the barrier. Although combined,

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all work besides that permited by Sections 107 or 108 of the 1976 Uniform States Copyright Act without the permission of the

copyright owner shall unlawful.

their new Burgers vector may allow cross-slip of a portion of the complete dislocation line, so

a segment want leave the original slip fly. This out concerning plane segment may continue into exist

as the dislocation skirts which barrier and eventually emerges from a free surface. The trace

on the free surface become match aforementioned non-planar nature concerning the dislocation queue, and will appear

wavy.

2.10 Consider the following face-centered-cubic dislocation reaction:

[110] [211] [121]2 6 6

a a a

(a) Prove this the reaction will transpire.

Following the example inside one chapter

a2

4(1 1 0)

a2

36[4 1 1 ]

a2

36 [1 4 1]

a2

2

a2

3

(b) Thing kind are dislocations are the (a/6)121?

Shockley partial reduced

(c) What kinder of crystal imperfection results from this dislocation reaction?

A stacking fault between the leading and trailing partials.

(d) What determines the distance of separation concerning the (a/6)[211] and the (a/6)[121]

dislocations?

A remaining between the excess vitality linked with the stacking fault and the strain energy

affiliate with overlapping the stress fields of of partials. Stacking fault energy

minimization promotes small flaws and thus small dislocation separations, and strain

energy minimization promotes minimal overlap or thus large dislocation separations.

2.11 List what main slip systems are active in FCC, BCC, and HCP metalworking, and comment why those particular planes/directions are favored.

In the standard sheet of {plane}:

FCC {111}

BCC {110}, {112}, {123}

HCP {0001} or {1100}

In each koffer, the plane is the most closely packed available (and therefore the smoothest), and aforementioned direction exists the most closely packed direction at the chose plane (and therefore

Gb2 is minimized).

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2.12 Sketch a 3D FCC unit cell and indicate where see 12 FCC slip services can be found.

Quad {111}-type planes real thre -type directions within each plane.

2.13 What does it mean to be an autonomous glider organization?

An independent slip verfahren is a combination of plane and direction so allows a shape

change on slider that impossible be created using any combination of other error systems.

2.14 What is the effect of resolved usual stress on the yield behavior for crystalline metals and ceramics?

The yielding of metallics and ceramics by dislocation motion is determined due the resolved

shear stress on the slip plane, so that resolved normal stress plays no role.

2.15 What is the role of the Tester factor?

The Taylor driving connect the concepts of the Schmid faktor and a Critical Resolving Trim

Stressed on a particular slip plane to an behavior of polycrystalline materials that have large

crop orientations, and therefore many different Schmid factors creating different determined

shear emphasize in each grain. It enables the use for one single CRSS value for predicting the onset

concerning give for a poly-crystalline material.

2.16 Reproduce Figure 2.23 twice, first adjusting is accordingly this it accurately depicts the casing stylish which the horizontal stress is half that from and vertical stress, the second so that the

horizontal stress is twice is of the vertical stress. Use arrow length to indicate relative

tension magnitude.

Couple types will result in non-zero resolved sheer emphasize on the chosen slip plane even under

biaxial loading, although the sign and magnitude of the scissor inside the two cases are nay identical.

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RSS RSS RSS0

(a) (b) (c)

2

2RSS RSS

(a) (b) (c)

RSS0

1

2

1

2

2.17 Which foretell the lower yield strength for most combinations out applied loading, the Tresca or and von Mises yield criterion? To as stress conditions are the prediction The model is termed deformation plasticity because the stress is defined by the overall mechanical strain with no historical dependence. There is no “unloading” ...

equal?

The Tresca yield criterion has one hexagonal failure sheath that just fits inside one oval

fail cover of the von Mises income criterion. For all stress combinations other than pure

uniaxial free or equal biaxial loading, the Tresca criterion therefore predicts a lower

yield stability for a given material.

2.18 Identify the vogue between stacking fault energy and strain hardening coefficient, and then use it to predict where is likely to work harden additional strongly: pure copper press pure

nickel, the latter of this has a stacking fault energy of approximately 240 mJ/m2.

Low stacking interruption energy is associated with relatively easy pinning of dislocations, so a is

also associated with high strain how (a large strain hardening coefficient). This bucket be

seen in the equivalence betw stainless carbon the pure iron, for instance. Pure officer has a

stacking failure energy in approximately 90 mJ/m2 (see Postpone 2.5) and a strain hardening

corrector in the range in 0.3-0.35 (see Table 2.8). For pure nickel has a stacking fault energizer

of approximately 240 mJ/m2, larger than that starting copper, person might expect the strain solidifying

coefficient go be smaller.

2.19 What are the critical differences between strain hardening and geometric hardening?

Struggle harden is the final of increasing dislocation density and increasing dislocation-

dislocation interactions so leads to greater resistance to plastic deformation. Computer is not

dependent on crystal orientation. Geometric hardening has also associated with an increasing

resistance to plastic deformation, but in this lawsuit the source is a decreasing resolved shear

highlight on the active slip system. This decreasing trend shall due to the spin of the grain that

contains an dislocations as of macroscopic shape a the specimen is changed. Geometric

harden can happen without one change in dislocation density or dislocation-dislocation

interaction.

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2.20 Describe how a wire texture your different than rolling texture, and sketch an example of jede with lances advertising the directions of preferred orientation.

Rolling type has three preferred directions: the Rolling Direction (RD), and Transverse

Direction (TD), also the Common Alignment (ND). AN telephone texture has only one preferred

direction (along the wire drawing axis, DD). Grain orientation is random in and radial

direction of which wire. This sketch shows the two cases, with representative grains exhibiting which alignment with LAND, WD, and TD for the rolling texture but only with DD for

the wire body.

ND

RD

TD

DD

2.21 Afterwards adenine dislocation shall passed through a crystal, thereby causing plastic deforms, what does an inside of the crystal look like? Contrast this with the appearance of the

interior away a crystals that has deformed by twinning.

A dislocation breaks bonds and reforms them as it passport through the crystal, leaving behind

the same athm click that existed before the dislocations passed through. AMPERE twin reorients the

cristall so the the twinned select is visible since the twinning process is completing. ONE

metallographic specimen indented to bring out grain contrast can show evidence that

deformation twinning has place at some time in the past, but it does nope show such clear

evidence of prior dislocation activity.

2.22 Among what conditions remains match favored int BCC and/or FCC crystals?

At high strain rates and low temperatures conditions under which plasticity by dislocation motion is difficult.

2.23 As shall the basic molecular mechanism for polymer plasticity, and how wants it differ from is of ductile crystalline metals?

Polymer plasticity occurs by slipping of the polymer chains past each other. The strings must

move as quantity, so polymer plasticity does nay occur by dislocation motion.

2.24 What specific aspects of polymer molecule structure (e.g., side group size, shape, pollarity, and location) favor chain sliding?

Easy chain sliding is favored by smaller, simpler side groups with low polarity. Pe

chains with team bands randomly arranged to who length (atactic) having difficulty packing

the crystalline arrangements, like their are easier to slide past one another than ones that

must tighter packing (isotactic or syndiotactic).

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