Chapter 7 Highlights:

7/28/2019 Chapter 7 Highlights:

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Chapter 7Highlights:

1. Understand the basic concepts of engineering stress and strain, yield strength, tensile strength,Young's(elastic) modulus, ductility, toughness, resilience, true stress and true strain, strain exponent,

and know the difference between elastic and plastic deformation.

2. Understand what a stress-strain curve is and what information it contains about materials properties.Be able to identify and/or calculate all the properties in #1 from a stress-strain curve, both in the

elastic and plastic (before necking) regions.

3. Understand how the mechanical behavior of ceramics differs from that of metals. Be able tonumerically manipulate the flexural strength and the effect of porosity on mechanical strength.

4. Understand how the mechanical behavior of polymers differs from that of metals. Understand andbe able to numerically manipulate the viscoelastic modulus.

Notes:

Show Figures 7.1 to 7.4.


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Define engineering stress and strain (tension and compression):

AF==stressgEngineerin0

l

l=

l

l-l==straingEngineerin

00

0i


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For shear stress,

A

F=stressShear

0

angle)strainis(=strainShear tan As shown in Figure 7.4, and described in Equations 7.4a and 7.4b, an applied axial force can be geometricallydecomposed into tensile and shear components.

Stress-strain test: slowly increase stress and measure strain until the material fractures (show Figures 7.2, 7.3,7.5, 7.10-12).

These tests are usually performed in tension. In the linear portion of the curve, Hooke's law is obeyed, =

E. E is called the modulus of elasticity, or Young's modulus, and is a property of the material. The units ofE are psi or MPa, remembering that Pa=N/m2. Typical values for E are given in Table 7.1 for metals,

ceramics and polymers.


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A similar relationship can be written for compressive, shear, and torsional loads. For shear stress,

G=G is the shear modulus and is a property of the material.

On an atomic scale, lengthening (compressing) a specimen during tensile (compressive) loading results in

lengthening (shortening) of atomic bonds. Young's modulus is a measure of the resistance to separation ofadjacent atoms.

dr

dFmodulusYoungs

r0

Show Figure 7.7 and Table 7.1.High E- W and Ni, low E- Mg, Al,Au, Ag.

Show Table 3.7. E can be differentin different directions, and thisreflects the different atomic densitiesin different planes and directions.

When there is not a significant linearportion of an engineering stress-strain diagram, the secant and tangent

modulus are sometimes employed.Show Figure 7.6. The tangent

modulus is the slope of the stress-strain curve at one particular point,

while the secant modulus is the slope of a secant drawn from the origin to a particular point on the stress-strain curve.

By common sense, tensile strain in the z direction should yield compressive strain in the x and y directions.Show Figure 7.9. x=y. Define Poisson's ratio

z

y

z

x= -= -

=0.25-0.35 for most metals, and is again a property of the material. It can be shown that

)+= 2G(1E G 0.4E for most metals.

The following analogies illustrate the differences between the linear and nonlinear portions of a stress-straincurve:


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Hooke's law Bonds stretched Elastic deformation No permanent damageNonlinear - Bonds broken Plastic deformation Permanent damage

Example Problem

A cylindrical steel bar 10 mm in diameter is to be elastically deformed. Using the data in Table 7.1,

determine the force needed to produce a reduction of 3x10 -3 mm in the diameter.

From Table 7.1, E = 207 GPa and = 0.30. We are given information about the desired radial strain:

43

10310

103 = xmm

mmxr

This can be related to the axial strain by the definition of Poissons ratio, eq. 7.8:

34

10130.0

103 = xxr

z

Now use Hookes law, eq. 7.5, to determine the stress that must be applied:

( ) MPaGPaxGPaE 207207.0101)207( 3 = From the definition of stress, eq. 7.3,

NxmPaxAF426

0 1063.1)01.0(4

10207 =

= Tensile Properties of Materials:

Show a stress-strain curve, Figure 7.12. We will now digress for a while but eventually return to this figureand try to understand it and extract useful information. The linear portion of this curve undergoes elasticdeformation (linear -), which disappears after the stress is removed. In other words, it is reversible. Whenthe stress exceeds the linear portion of the stress-strain curve, permanent (irreversible) plastic deformationoccurs.

Plastic deformation

Plastic deformation usually occurs near

0.005 for metals. When Hooke's law fails, the material undergoesplastic deformation. Show Figure 7.10. Either 7.10a or 7.10b can occur, usually 7.10a.

Since plastic deformation involves atomic-level breaking of bonds and reforming of new ones, it isirreversible.

We want to be able to define unambiguously when a material yields, when it is permanently damaged. Show

Figure 7.10a, focusing on point P. This is the proportional limit, where the stress-strain curve deviates fromlinearity. But this is ambiguous.


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Usually the stress-strain curve deviates from linearity quite gradually, so we need to define an arbitrary but

universal definition of the elastic region. Starting from point =0, =0.002, draw a parallel line to the linearportion of the stress-strain curve. The stress at which this line intersects with the stress-strain curve is called

the yield strength, y. This is one measure of a material's strength, its ability to resist plastic (permanent)deformation. In the case of Figure 7.10b the yield strength is the lower yield point.

Show Figure 7.11. The tensile strength is defined as the maximum stress prior to fracture. After M,"necking" occurs, the cross-sectional area shrinks abruptly. The ductility is defined as the % plasticdeformation at fracture.

x100l

l-l=Ductility

0

0f

A low ductility material is brittle. Show Figure 7.13.

Resilience is the capability of a material to absorb energy when it is

deformed elastically. The modulus of resilience Uris

d=U

y

0

r where y is the strain at fracture. Show Figure 7.15. For linearstress-strain curve,

2E=

2

1=U

y

yyr

2 Toughness is the ability of a material to absorb energy up to

fracture. It is defined by the same integral as Ur above, but theupper limit of integration is now fracture instead of yielding. ShowTable 7.3. High ductility corresponds somewhat with low strength.

Example Problem

For a given set of data, which I will not reproduce here, you shouldbe able to: a) Create a plot of engineering stress versus engineeringstrain, b) determine the elastic modulus, c) determine the yieldstrength, d) determine the tensile strength, e) determine the ductility,

and f) determine the modulus of resilience.

a) The plots below were obtained by taking:

( ) )(2.7771108.124

)(

4

)()(

232

NF

mx

NF

D

NFPa =


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8.50

8.50

0

0 == ii ll

ll

b) The elastic modulus is taken as the slope of the linear portion of the curve. Looking at the 2nd

plot,which focuses on the linear portion of the curve, take the average slope only from the 1 st four points.This yields:

00398.0

10362.2

00298.0

10795.1

00200.0

10173.1

00100.0

10696.5 8887 PaxPaxPaxPaxE =

Taking the average of these 4 values yields E =5.88x10

10

Pa.

c) On the 2ndplot, draw a line from = 0.002 with a slope of E = 5.88x1010 Pa. The intersection of this

line with the - curve gives y = 2.8x108 Pa.

d) From the 1st plot, the maximum in is the tensile strength = 3.7x108 Pa.

e) From the 1st

plot, the strain at failure is about 0.145, subtract out the elastic strain ( 0.005) to get theplastic strain of 0.14. Thus the ductility is 14%.

0.00 0.05 0.10 0.15

Strain

0.000e0

1.000e8

2.000e8

3.000e8

Stres

s

(Pa)


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f) Determine the modulus of resilience from equation 7.14:

( )Pax

Pax

Pax

E

Uy

r

5

10

282

1067.6

)1088.5(2

108.2

2

= Elastic recovery after plastic deformation:

Show figure 7.17 and discuss.

True Stress and Strain:

True stress(T) and true strain(T) are defined similarly to engineering stress and strain.

areaousInstantane

Force=

A

F=

i

T

lengthousinstantanetheisll

l= i

0

iT

ln

If volume is conserved, then Aili = A0l0 and

)+(1=T )+(1=T ln

The above two equations are only true until necking. After necking, true stress and true strain can only be


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determined by actual measurements. Show Figure 7.16, where the true stress is corrected to account for non-tensile components.

For some metals, from the onset of plastic deformation to the point at which necking begins, the true stress is

approximately:n

TT K= K and n, the strain hardening exponent, vary across different metals and alloys.

Example Problem

You are given that just prior to necking, (, ) = (235 MPa, 0.194) and (250 MPa, 0.296). What value of

will produce = 0.25. In order to use equation (7.19), we need to convert from engineering stress and strainto true stress and strain. From equations (7.18)

MPaT 59.280)1(1 =

MPaT 324)1(2 =

1773.0)1(ln1 =T

2593.0)1(ln2 =T 2231.0)1(ln3 =T

You need to use equation (7.19) to setup two equations with two unknowns. Taking the ln of both sides:

TT nK lnlnln + )2593.0(lnln)324(ln nKMPa +


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)1773.0(lnln)59.280(ln nKMPa + Subtracting the 2

ndfrom the 1

st

=

1773.0

2593.0

ln59.280

324

ln n

378.0n Substituting back into above:

)1773.0(ln)378.0(ln)59.280(ln +KMPa MPaKorK 540291.6ln =

Now find where T is 0.2231:n

TT K= MPaMPaT 3.306)2231.0()540(

378.0 = Convert back to engineering stress:

MPaMPaT 24525.01

3.306

1=++

Mechanical behavior: Ceramics

Ceramic materials are less mechanically useful than metals because they are generally quite brittle. Theirmechanical strength is not normally assessed by tensile stress-strain measurements. It is extremely difficultto prepare the proper sample geometry, the samples generally crack when gripped, and their strain at failure istoo small to accurately measure. Show Figure 7.19. For these reasons, the strength of ceramic materials is

normally assessed by a traverse bending test. Show Figure 7.18. The failure point is described by theflexural strength (fs), also known as the modulus of rupture(mr). For rectangular cross sections,

d2b

LF3=

2

f

fs

and for circular cross-sections:

R

LF= 3

f

fs


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Example Problem

A 3-point bending test is performed on Al2O3 with a circular cross-section of radius 3.5 mm. The specimenfractures at a load of 950 N when the support points are 50 mm apart. Consider a square sample of the same

material with a square cross-section of 12mm on each edge. If the support points are 40 mm apart, at whatload will the sample fracture?

First, determine the flexural strength of this material from equation (7.20b):

Paxmx

mxN

R

LFffs

8

33

3

31053.3

)105.3(

)1050)(950( =

Now determine the failure force from equation (7.20a):

N

mx

Paxmx

L

bdF

fs170,10

)1040(3

)1053.3()1012(2

3

2

3

8332 =

Ceramic solids are often fabricated by compaction or foaming of ceramic particles. In this case, the ceramicsolid may have considerable porosity, which degrades its strength. The modulus of elasticity (E) and the

flexural strength (fs) depend on the volume fraction porosity (P) according to:

( )20 9.09.11 PPEE +

)exp(0 nPfs where

0 and n are experimental constants.

Example Problem

Using the data in Table 7.2,a) Determine the flexural strength of nonporous MgO assuming n = 3.75

b) Calculate the volume fraction porosity when fs = 62 MPa.a) Taking the ln of both sides of equation 7.22:

nPfs 0lnln From Table 7.2, fs = 105MPa with 5% porosity, so

)05.0)(75.3(ln)105(ln 0 MPa 841.4ln 0 == MPa1270 ==


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b) Rearranging the above equation:

nP

fslnln 0 =

.%19190.075.3

)62(ln841.4volor

MPaP ==

Note that 5% porosity reducesfs by 17%, and 19% porosity reduces fs by 51%.


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Mechanical behavior: Polymers

Typical stress-strain diagrams for polymers are shown in Figure 7.22. The behavior ranges from strong andbrittle (curve A) to rubbery (curve C). The mechanical behavior of polymers is far more temperature-

dependent than that of metals and polymers. Show Figure 7.24. Polymethyl methylacrylate behaves like abrittle metal at low temperature, but like an elastomer (rubber band) at low temperature. At intermediate

temperatures, this polymer behaves like a viscoelastic material, which combines the behavior of a viscous andan elastic material. The most famous viscoelastic material is silly putty, which is intermediate between aliquid and a solid. Show Figure 7.26, which shows the strain behavior with time following instantaneous

application of a load.

Alternatively, one can study the mechanical behavior of a viscoelastic material at constant strain, in whichcase the stress that is initially applied will gradually decrease. Show Figure 7.27, which shows the stress as afunction of time and temperature. How can we compare different materials?? One way is to choose an

arbitrary time (10 sec is common) and define the relaxation modulus, E r(t), at that time:

0

)()(

ttEr =

Remember that the value of Er(10) depends on the temperature. Show Figure 7.27 again.


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Safety factor

Engineering with a safety factor accounts for material variability, nonideal conditions, human error, etc. and

is common engineering practice. The working stress (w) is reduced from the yield stress (y) by the safetyfactor (N) according to:

N

y

w

= The safety factor N is commonly taken as 2.

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Chapter 7 Highlights: