Chapter 6:Mechanical Properties

Why mechanical properties? Need to design materials that can withstand applied load… e.g. materials used in building bridges that can hold up automobiles, materials for pedestrians… skyscrapers in the Windy City… materials for space exploration…

NASA

materials for and designing MEMs and NEMs… Space elevators?Chapter 6 - 1 ISSUES TO ADDRESS...

• Stress and strain : What are they and why are they used instead of load and deformation? • Elastic behavior: When loads are small, how much deformation occurs? What materials deform least? • Plastic behavior: At what point does permanent deformation occur? What materials are most resistant to permanent deformation? • Toughness and ductility : What are they and how do we measure them?

Chapter 6 - 2 Stress and Strain

Stress: Pressure due to applied load.

tension, compression, shear, torsion, and their combination. force stress = σ = area

Strain : response of the material to stress (i.e. physical deformation such as elongation due to tension).

Chapter 6 - 3 Compression Tension

Shear Torsion Chapter 6 - 4 COMMON STATES OF STRESS

• Simple tension: cable F F Ao = cross sectional Area (when unloaded) F σ = σ σ Ao

Ski lift (photo courtesy P.M. Anderson)

From Callister 6e resource CD.

Chapter 6 - 5 COMMON STATES OF STRESS

• Simple compression:

Ao

Canyon Bridge, Los Alamos, NM (photo courtesy P.M. Anderson)

F Note: compressive Balanced Rock, Arches σ = structure member National Park A (photo courtesy P.M. Anderson) o (σ < 0 here).

From Callister 6e resource CD. Chapter 6 - 6 COMMON STATES OF STRESS

• Hydrostatic compression:

Fish under water (photo courtesy P.M. Anderson)

σ < 0 From Callister 6e resource CD. h Chapter 6 - 7 Tension and Compression

Tension F Engineering stress = σ = Ao l − l ∆l Engineering strain = ε = i o = lo lo

Ao = original cross sectional area li = instantaneous length lo = original length Note: strain is unitless.

Compression Same as tension but in the opposite direction (stress and strain defined in the same manner). By convention, stress and strain are negative for compression. Chapter 6 - 8 Shear

F Pure shear stress = τ = Ao

Pure shear strain = γ = tan θ

Strain is always dimensionless.

Chapter 6 - 9 Elastic Deformation 1. Initial 2. Small load 3. Unload

bonds stretch

return to initial δ F F Linear- elastic Elastic means reversible ! Non-Linear- -a non-permanent deformation where the elastic material completely recovers to its original δ state upon release of the applied stress. Chapter 6 - 10 Plastic Deformation (Metals) 1. Initial 2. Small load 3. Unload bonds stretch planes & planes still shear sheared

δplastic δelastic + plastic F F

Plastic means permanent ! linear linear elastic elastic δ δplastic Chapter 6 - 11 Stress-Strain Testing • Typical tensile test • Typical tensile machine specimen

Adapted from extensometer specimen Fig. 6.2, Callister 7e.

gauge length

Adapted from Fig. 6.3, Callister 7e. (Fig. 6.3 is taken from H.W. Hayden, W.G. Moffatt, and J. Wulff, The Structure and Properties of Materials , Vol. III, Mechanical Behavior , p. 2, John Wiley and Sons, New York, 1965.) Chapter 6 - 12 Linear Elastic Properties F • Modulus of Elasticity, E: (also known as Young's modulus) • Hooke's Law : σ = E ε σ F stress strain simple tension Modulus of elasticity E test (Young’s modulus) ε Measure of material’s resistance to Linear- elastic deformation (stiffness). elastic

For metals, typically E ~ 45 – 400 GPa Chapter 6 - 13 Note: some materials do not have linear elastic region (e.g. cast iron, concrete, many polymers…)

Define secant modulus and tangent modulus.

Tangent modulus = slope of the tangent line σ

σ2 σ1 ∆σ σ1 Secant modulus = = ∆ε ε1

ε1 ε

Chapter 6 - 14 Silicon (single crystal) 120 - 190 (depends on crystallographic direction) Glass (pyrex) 70 SiC (fused or sintered) 207 - 483 Graphite (molded) ~12 High modulus C-fiber 400 Carbon Nanotubes ~1000

If we normalize to density: ~20 times that of steel wire

Density normalized strength is ~56X thatChapter of steel6 - 15 wire Poisson Ratio

So far, we’ve considered stress only along one dimension…

Along z: tension ∆l ε z = lo

Along x: compression z ∆d ε x = do lo x do lo+∆l Isotropic x and y: ε y = ε x do+∆d ε x ε y Poisson ratio =υ = − = − ε z ε z

Elongation Compression

Relation between elastic and shear moduli: E = 2G(1+ ν) Chapter 6 - 16 Poisson Ratio

• Poisson Ratio has a range –1 ≤ ν ≤ 1/2

Look at extremes l l • No change in aspect ratio: ∆ w /w = ∆ / l w ∆w /w ν = − = −1 ∆l /l • Volume (V = AL) remains constant: ∆V =0 or l∆A = - A ∆l σ

Hence, ∆V = ( l ∆A+A ∆l) = 0.

In terms of width, A = w2, and ∆A = w2 -(w+∆w)2 = 2 w ∆w + ∆w2

then ∆A/A = 2 ∆w/w + ∆w2/w2

in the limit of small changes ∆w / w (− 1 ∆l / l) ∆A/A = 2 ∆w/w ν = − = − 2 = 2/1 ∆l / l ∆l / l then 2 ∆w/w = -∆l/l Chapter 6 - 17 Poisson's ratio, ννν ε • Poisson's ratio, ννν: L

ε ν = − L ε ε

metals : ν ~ 0.33 -ν ceramics : ν ~ 0.25 polymers : ν ~ 0.40

Units: –ν > 0.50 density increases E: [GPa] or [psi] –ν < 0.50 density decreases ν: dimensionless (voids form)

Chapter 6 - 18 Poisson Ratio: materials specific

Metals: Ir W Ni Cu Al Ag Au 0.26 0.29 0.31 0.34 0.34 0.38 0.42 generic value ~ 1/3 Solid Argon: 0.25

Covalent Solids: Si Ge Al 2O3 TiC 0.27 0.28 0.23 0.19 generic value ~ 1/4

Ionic Solids: MgO 0.19

Silica Glass: 0.20

Polymers: Network (Bakelite) 0.49 Chain (PE) 0.40

Elastomer: Hard Rubber (Ebonite) 0.39 (Natural) 0.49

Chapter 6 - 19 Mechanical Properties • Slope of stress strain plot (which is proportional to the elastic modulus) depends on bond strength of metal

Adapted from Fig. 6.7, Callister 7e.

Chapter 6 - 20 Other Elastic Properties M • Elastic Shear τ modulus, G: G simple γ torsion τττ = G γγγ test M • Elastic Bulk P P modulus, K: ∆V ∆V PP P = -K V K Vo pressure o test: Init. vol = Vo. • Special relations for isotropic materials: Vol chg. = ∆V E E G = K = 2(1 +++ ν) 3(1 −−− 2ν) Chapter 6 - 21 Young’s Moduli: Comparison

Graphite Metals Composites Ceramics Polymers Alloys /fibers Semicond 1200 1000 800 Diamond 600 Si carbide 400 Tungsten Al oxide Carbon fibers only Molybdenum Si nitride Steel, Ni CFRE(|| fibers)* E(GPa) 200 Tantalum <111> Platinum Si crystal Cu alloys <100> Aramid fibers only 100 Zinc, Ti Silver, Gold 80 Glass -soda AFRE(|| fibers)* Based on data in Table B2, 60 Aluminum Glass fibers only Magnesium, Callister 7e . 40 Tin GFRE(|| fibers)* Concrete Composite data based on 10 9 Pa 20 GFRE* reinforced epoxy with 60 vol% CFRE* of aligned GFRE( fibers)* 10 Graphite carbon (CFRE), 8 CFRE( fibers) * aramid (AFRE), or 6 AFRE( fibers) * Polyester glass (GFRE) 4 PET PS fibers. 2 PC Epoxy only PP 1 HDPE 0.8 0.6 Wood( grain) PTFE 0.4

0.2 LDPE Chapter 6 - 22 Useful Linear Elastic Relationships • Simple tension: • Simple torsion: 2ML δ = FLo δ = −νFw o α = o L 4 EAo EA o πro G F M = moment δ/2 α = angle of twist Ao

Lo Lo wo

2ro δL/2 • Material, geometric, and loading parameters all contribute to deflection. • Larger elastic moduli minimize elastic deflection.

Chapter 6 - 23 Plastic (Permanent) Deformation

Adapted from Fig. 6.10 (a), Callister 7e. • Simple tension test: Elastic+Plastic engineering stress, σ at larger stress

Elastic initially permanent (plastic) after load is removed

εp engineering strain, ε plastic strain

•A permanent deformation (usually considered for T

A. Yield strength (σσσy): the strength required to produce a very slight yet specified amount of plastic deformation. What is the specified amount of strain? Strain offset method 1. Start at 0.002 strain (for most metals). 2. Draw a line parallel to the linear region. σ 3. σy = where the dotted line crosses the σσσy stress-strain curve. P P = proportional limit (beginning of deviation from linear behavior.

Mixed elastic-plastic behavior

For materials with nonlinear elastic region: σy is 0.002 ε defined as stress required to produce specific amount of strain (e.g. ~0.005 for most metals). Elastic region Chapter 6 - 25 Tensile properties

Yield point phenomenon occurs when elastic-plastic transition is well- defined and abrupt.

No offset methods required here.

Fig. 6.10 Callister Chapter 6 - 26 Yield Strength : Comparison Graphite/ Metals/ Composites/ Ceramics/ Polymers Alloys fibers Semicond 2000 σ Steel (4140) qt y(ceramics) >>σ 1000 y(metals) Ti (5Al-2.5Sn) a 700 W (pure) >> σy(polymers) 600 Cu (71500) cw 500 Mo (pure) Steel (4140) a (MPa) 400 Steel (1020) cd y

300 , σ Al (6061) ag Room T values hr 200 Steel (1020) ¨ Ti (pure) a Ta (pure) Cu (71500) hr Based on data in Table B4, Callister 7e . 100 a = annealed dry

70 measure, to Hard hr = hot rolled Hard to measure to Hard PC 60 Al (6061) a Nylon 6,6 ag = aged 50 PET cd = cold drawn PVC humid 40 cw = cold worked PP 30 HDPE qt = quenched & tempered in tension, fracture usually occurs before yield. before occurs fractureusually in tension, Yield strength,

20 matrix composites, since epoxy and matrix in ceramic since in tension, fracture usually occurs before yield. before occurs fracture usually tension, in since

LDPE Tin (pure) Chapter 6 - 27 10 Tensile Strength, TS • Maximum stress on engineering stress-strain curve. Adapted from Fig. 6.11, TS Callister 7e. F = fracture or σσσ y ultimate strength stress Typical response of a metal Neck – acts as stress engineering concentrator strain engineering strain • Metals : occurs when noticeable necking starts. • Polymers : occurs when polymer backbone chains are aligned and about to break. Chapter 6 - 28 True stress and strain

TS Notice that past maximum stress point, σ decreases. Does this mean that the material is becoming weaker?

stress Typical response of a metal engineering Necking leads to smaller cross sectional strain area! F Recall: Engineering Stress = σ = Ao Original cross sectional area!

F Ai = instantaneous area True Stress = σ T = li = instantaneous length Ai If no net volume change (i.e. A l = A l ) l i i o o ε = ln i True Strain = T σ = σ 1( + ε ) lo T Only true at the onset of necking εT = ln( 1+ ε ) Chapter 6 - 29 Example problem

Calculate/determine the following for a brass specimen that exhibits stress-strain behavior shown on the left. 1) Modulus of elasticity. 2) Yield strength. 3) Maximum load for a cylindrical specimen with d = 12.8mm. 4) Change in length at 345MPa if the initial length is 250mm.

Chapter 6 - 30 Tensile Strength : Comparison Graphite/ Metals/ Composites/ Ceramics/ Polymers Alloys fibers Semicond 5000 C fibers Aramid fib 3000 E-glass fib 2000 Steel (4140) qt AFRE(|| fiber) 1000 W (pure) Diamond GFRE(|| fiber) Ti (5Al-2.5Sn) a Steel (4140) a CFRE(|| fiber) (MPa) Cu (71500) cw Si nitride Cu (71500) hr Steel (1020) Al oxide 300 Al (6061) ag Ti (pure) a 200 TS Ta (pure) Room Temp. values Al (6061) a 100 Si crystal wood(|| fiber) Based on data in Table B4, <100> Nylon 6,6 Glass-soda PC PET Callister 7e . 40 Concrete PVC GFRE( fiber) a = annealed 30 PP CFRE( fiber) AFRE( fiber) hr = hot rolled HDPE ag = aged Graphite 20 LDPE cd = cold drawn strength, cw = cold worked 10 qt = quenched & tempered AFRE, GFRE, & CFRE = aramid, glass, & carbon fiber-reinforced epoxy wood ( fiber) composites, with 60 vol%

Tensile fibers. 1 Chapter 6 - 31 Tensile properties

C. Ductility ::: measure of degree of plastic deformation that has been sustained at fracture. • Ductile materials can undergo significant plastic deformation before fracture. • Brittle materials can tolerate only very small plastic deformation.

Chapter 6 - 32 Ductility L − L • Plastic tensile strain at failure: %EL = f o x 100 L o smaller % EL Engineering tensile stress, σ larger % EL A L o o Af Lf Adapted from Fig. 6.13, Callister 7e.

Engineering tensile strain, ε

A - A • Another ductility measure: %RA = o f x 100 A o

Chapter 6 - 33 Toughness

• Energy to break a unit volume of material • Approximate by the area under the stress-strain curve. Engineering small toughness (ceramics) tensile large toughness (metals) stress, σ

Adapted from Fig. 6.13, very small toughness Callister 7e. (unreinforced polymers)

Engineering tensile strain, ε

Brittle fracture: elastic energy Ductile fracture: elastic + plastic energy

Chapter 6 - 34 Resilience, Ur

• Ability of a material to store energy – Energy stored best in elastic region

ε U = y σdε r ∫0 If we assume a linear stress-strain curve this simplifies to 1 U ≅ σ ε r 2 y y

Adapted from Fig. 6.15, Callister 7e. Chapter 6 - 35 Elastic recovery after plastic deformation

This behavior is exploited to increase yield strengths of metals: strain hardening (also called cold working ).

Chapter 6 - 36 Hardness • Resistance to permanently indenting the surface. • Large hardness means: --resistance to plastic deformation or cracking in compression. --better wear properties. apply known force measure size e.g., of indent after 10 mm sphere removing load

Smaller indents D d mean larger hardness.

most brasses easy to machine cutting nitrided plastics Al alloys steels file hard tools steels diamond

increasing hardness

Chapter 6 - 37 Hardness: Measurement

• Rockwell – No major sample damage – Each scale runs to 130 but only useful in range 20-100. – Minor load 10 kg – Major load 60 (A), 100 (B) & 150 (C) kg • A = diamond, B = 1/16 in. ball, C = diamond

• HB = Brinell Hardness – TS (psia) = 500 x HB – TS (MPa) = 3.45 x HB

Chapter 6 - 38 Hardness scales

Indentation with diamond pyramid tip Indentation with spherical hardened steel and conical diamond (for hardest materials)

Indentation with spherical hardened steel or tungsten carbide tip. Chapter 6 - 39 Qualitative scale Hardness: Measurement

Table 6.5

Chapter 6 - 40 True Stress & Strain Note: S.A. changes when sample stretched

• True stress σT = F Ai σT = σ(1+ ε) l l • True Strain εT = ln ( i o ) εT = ln ()1+ ε

Adapted from Fig. 6.16, Callister 7e.

Chapter 6 - 41 Hardening • An increase in σ due to plastic deformation. σ y large hardening σy 1 σ small hardening y0

ε • Curve fit to the stress-strain response: hardening exponent: n n = 0.15 (some steels) σT = K()εT to n = 0.5 (some coppers)

“true” stress ( F/A) “true” strain: ln( L/Lo) Chapter 6 - 42 Variability in Material Properties

• Elastic modulus is material property • Critical properties depend largely on sample flaws (defects, etc.). Large sample to sample variability. • Statistics n Σ x – Mean x = n n

1 n  2  2 Σ()xi − x  – Standard Deviation s =  n −1   

where n is the number of data points

Chapter 6 - 43 Design or Safety Factors • Design uncertainties mean we do not push the limit. • Factor of safety, N Often N is σy between σ = 1.2 and 4 working N • Example: Calculate a diameter, d, to ensure that yield does not occur in the 1045 carbon steel rod below. Use a factor of safety of 5. σ d σ = y 1045 plain working carbon steel: N L σy = 310 MPa o 220 ,000 N 5 TS = 565 MPa π()d 2 / 4 F = 220,000N d = 0.067 m = 6.7 cm Chapter 6 - 44 Summary

• Stress and strain : These are size-independent measures of load and displacement, respectively. • Elastic behavior: This reversible behavior often shows a linear relation between stress and strain. To minimize deformation, select a material with a large elastic modulus ( E or G). • Plastic behavior: This permanent deformation behavior occurs when the tensile (or compressive)

uniaxial stress reaches σy. • Toughness : The energy needed to break a unit volume of material. • Ductility : The plastic strain at failure.

Chapter 6 - 45