Chapter 6 - Mechanical Properties.pdf

CHAPTER 6: MECHANICAL PROPERTIES 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 do dislocations cause permanent deformation? What materials are most resistant to permanent deformation? • Toughness and ductility: What are they and how do we measure them? Chapter 6 - 1

ELASTIC DEFORMATION 1. Initial

2. Small load

3. Unload

bonds stretch return to initial

 F Elastic means reversible!

Chapter 6 - 2

PLASTIC DEFORMATION (METALS) 1. Initial

2. Small load

3. Unload

F Plastic means permanent!

linear elastic

linear elastic

plastic



Chapter 6 - 3

ENGINEERING STRESS () • The instantaneous load applied to a specimen divided by the cross-sectional are before any deformation.

Where F = instantaneous load applied perpendicular to the specimen cross section (N or lbf) A0 = the original cross sectional before any load is applied (m2 or in2)

Chapter 6 - 4

ENGINEERING STRESS • Tensile stress, :

• Shear stress, t:

Ft  Ao original area before loading

Stress has units: N/m2 or lb/in2 Chapter 6 - 5

COMMON STATES OF STRESS • Simple tension: cable

F  Ao • Simple shear: drive shaft

Ski lift

(photo courtesy P.M. Anderson)

Fs t  Ao Note: t = M/AcR here. Chapter 6 - 6

OTHER COMMON STRESS STATES (1) • Simple compression:

Ao

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

Balanced Rock, Arches National Park (photo courtesy P.M. Anderson)

Note: compressive structure member ( < 0 here).

Chapter 6 - 7

OTHER COMMON STRESS STATES (2) • Bi-axial tension:

Pressurized tank (photo courtesy P.M. Anderson)

• Hydrostatic compression:

Fish under water

 > 0 z > 0

(photo courtesy P.M. Anderson)

h< 0 Chapter 6 - 8

ENGINEERING STRAIN () • The change in gauge length of a specimen (in the direction of an applied stress) divided by its original gauge length

Where: l0 = original length li – l0 = denoted as l, is the deformation elongation or change in length at some instant, as referenced to the original length. Chapter 6 - 9

ENGINEERING STRAIN • Tensile strain:

• Lateral strain:

/2

wo • Shear strain:

L /2

Lo /2 L /2

/2

 = tan  /2 -  /2

/2

Strain is always dimensionless.

Chapter 6 - 10

STRESS-STRAIN TESTING • Typical tensile specimen

• Typical tensile test machine

Adapted from Fig. 6.2, Callister 6e.

• Other types of tests: --compression: brittle materials (e.g., concrete) --torsion: cylindrical tubes, shafts.

Adapted from Fig. 6.3, Callister 6e. (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 - 11

ELASTIC DEFORMATION • Deformation in which stress and strain are proportional

Chapter 6 - 12

STRESS-STRAIN BEHAVIOR • The degree to which a structure deforms or strains depends on the magnitude of the imposed stress. • For metals, that are stressed in tension and at relatively low levels, stress and strain are proportional to each other as stated in Hooke’s Law. Chapter 6 - 13

HOOK’S LAW • Stress is proportional to strain =Ee Where:   stress (N/m2) E = modulus of elasticity (MPa,) e  strain Chapter 6 - 14

SAMPLE PROBLEM HL1 Copper wire which has a rectangular cross section of 15.2 mm x 19.1 mm is pulled in tension with 44,500 N force producing only elastic deformation. Calculate the resulting strain.

Chapter 6 - 15

SOLUTION TO HL1 GIVEN: A0 = L x W = (0.0152 m)(0.0191 m) = 2.9 x 10-4 m2 F = 44,500 N From Table of E for copper: 110 x 103 Mpa REQUIRED: e

SOLUTION:   F/A0   (44,500 N)/(2.9 x 10-4 m2)   1.53 x 10-4 N/m2 or 153 MPa   Ee e  /E  (153 MPa)/(110 x 103 MPa) e = 0.00139

Chapter 6 - 16

SAMPLE PROBLEM HL2 • A cylindrical specimen of a nickel alloy having a modulus of elasticity of 207 GPa and an original diameter of 10.7 mm will experience only elastic deformation when a tensile load of 8000 N is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.25 mm Chapter 6 - 17

SOLUTION TO HL2 GIVEN: D0 = 10.2 mm or 0.0102 m F = 8000 N E = 207 GPa = 207 x 109 N/m2 L = 0.00025 m REQUIRED: L0

Chapter 6 - 18

SAMPLE EXERCISE HL3 Consider a cylindrical titanium wire 3 mm in diameter and 2.5 x 104 mm long. Calculate its elongation when a load of 500 N is applied. Assume that the deformation is entirely elastic. ANSWER : 16.527 m Chapter 6 - 19

POISSON’S RATIO (n)

Where: n = Poisson’s ratio e = tensile strain eL  lateral strain Poisson's ratio, n: metals: n ~ 0.33 ceramics: ~0.25 polymers: ~0.40 Chapter 6 - 20

OTHER ELASTIC PROPERTIES • Elastic Shear modulus, G:

t=G

M

t

G 1

simple torsion test

 M

• Elastic Bulk modulus, K:

• Special relations for isotropic materials: E E G K 2(1  n) 3(1  2n)

P P

P pressure test: Init. vol =Vo. Vol chg. = V

Chapter 6 - 21

YOUNG’S MODULI: COMPARISON Metals Alloys 1200 1000 800 600 400

E(GPa)

200 100 80 60 40

10 9 Pa

Graphite Ceramics Polymers Semicond

Composites /fibers

Diamond Tungsten Molybdenum Steel, Ni Tantalum Platinum Cu alloys Zinc, Ti Silver, Gold Aluminum Magnesium, Tin

Si carbide Al oxide Si nitride

Carbon fibers only

CFRE(|| fibers)*

<111>

Si crystal

Aramid fibers only

<100>

AFRE(|| fibers)*

Glass-soda

Glass fibers only

GFRE(|| fibers)*

Concrete GFRE*

20 10 8 6 4 2

1 0.8 0.6 0.4 0.2

CFRE* GFRE( fibers)*

Graphite

Polyester PET PS PC

CFRE( fibers) * AFRE( fibers) *

Epoxy only

Based on data in Table B2, Callister 6e. Composite data based on reinforced epoxy with 60 vol% of aligned carbon (CFRE), aramid (AFRE), or glass (GFRE) fibers.

PP HDPE PTFE LDPE

Wood( grain)

Chapter 6 - 22

USEFUL LINEAR ELASTIC RELATIONS • Simple tension:

• Simple torsion:

M=moment  =angle of twist

Lo

2r o • Material, geometric, and loading parameters all contribute to deflection. • Larger elastic moduli minimize elastic deflection.

Chapter 6 - 23

ANELASTICITY • Time dependent elastic behavior • Elastic deformation will continue after the stress application , and upon load release some finite time is required to complete recovery.

Chapter 6 - 24

SAMPLE PROBLEM PR1 A tensile stress is to be applied along the long axis of a cylindrical brass rod that has a diameter of 10 mm. Determine the magnitude of the load required to produce a 2.5 x 10-3 mm change in diameter if the deformation is entirely elastic. n = 0.34, E = 97 GPa. Chapter 6 - 25

SAMPLE PROBLEM PR2 A cylindrical specimen of steel having a diameter of 15.2 mm and a length of 250 mm is deformed elastically in tension with a force of 48,400 N. Determine a.) the amount by which this specimen will elongate. b.) the change in diameter of the specimen. n = 0.30, E = 207 GPa Chapter 6 - 26

PLASTIC DEFORMATION • The stress is no longer proportional to strain and permanent, non recoverable.

Chapter 6 - 27

PLASTIC (PERMANENT) DEFORMATION (at lower temperatures, T < Tmelt/3)

• Simple tension test:

Chapter 6 - 28

YIELD STRENGTH, y • Stress at which noticeable plastic deformation has occurred. when ep = 0.002 tensile stress,



y

engineering strain,

e

ep = 0.002 Chapter 6 - 29

YIELD STRENGTH: COMPARISON y(ceramics) >>y(metals) >> y(polymers) Room T values Based on data in Table B4, Callister 6e. a = annealed hr = hot rolled ag = aged cd = cold drawn cw = cold worked qt = quenched & tempered

Chapter 6 - 30

TENSILE STRENGTH, TS • Maximum possible engineering stress in tension. Adapted from Fig. 6.11, Callister 6e.

• Metals: occurs when noticeable necking starts. • Ceramics: occurs when crack propagation starts. • Polymers: occurs when polymer backbones are aligned and about to break.

Chapter 6 - 31

TENSILE STRENGTH: COMPARISON TS (ceram) ~TS (met) ~ TS (comp) >> TS (poly) Room T values Based on data in Table B4, Callister 6e. a = annealed hr = hot rolled ag = aged cd = cold drawn cw = cold worked qt = quenched & tempered AFRE, GFRE, & CFRE = aramid, glass, & carbon fiber-reinforced epoxy composites, with 60 vol% fibers. Chapter 6 - 32

DUCTILITY • It is a measure of the degree of plastic deformation that has been sustained at fracture. • BRITTLE – a material that experiences very little or no plastic deformation upon fracture.

Chapter 6 - 33

DUCTILITY, %EL

Adapted from Fig. 6.13, Callister 6e.

L f  Lo %EL  x100 Lo Chapter 6 - 34

DUCTILITY, %AR • Another ductility measure:

Ao  A f %AR  x100 Ao • Note: %AR and %EL are often comparable. --Reason: crystal slip does not change material volume. --%AR > %EL possible if internal voids form in neck.

Chapter 6 - 35

A knowledge of the ductility of material is important for at least two reasons. 1. It indicates to a designer the degree to which the structure will deform plastically before fracture. 2. It specifies the degree of allowable deformation during fabrication operation.

Chapter 6 - 36

RESILIENCE • The capacity of a material to absorb energy when it is deformed elastically and then, upon unloading, to have this energy recovered. • MODULUS OF RESILIENCE, Ur – the strain energy per unit volume required to stress a material from an unloaded state up to a point of yielding. Chapter 6 - 37

TOUGHNESS • Energy to break a unit volume of material • Approximate by the area under the stress-strain curve. Engineering tensile stress, 

smaller toughness (ceramics) larg er toughness (metals, PMCs) smaller toughnessunreinforced polymers

Engineering tensile strain,

e

Chapter 6 - 38

TRUE STRESS AND STRAIN • TRUE STRESS (T) - is defined as the load F divided by the instantaneous cross-sectional area Ai over which deformation is occuring (i.e., the neck, past the tensile point) T = F/Ai

• TRUE STRAIN (eT) – it is defined by eT = ln (li/l0) Chapter 6 - 39

TRUE STRESS AND STRAIN • If no volume change occurs during deformation, that is, if Aili = A0l0 • True and engineering stress and strain are related according to T = (1 + e) eT = ln(1 + e) • For some metals and alloys the region of the true stressstrain curve from the onset of plastic deformation to the point at which necking begin maybe approximated by T = KeTn) Chapter 6 - 40

SAMPLE PROBLEM D1 A cylindrical specimen of steel having an original diameter of 12.8 mm is tensile tested to fracture and found to have engineering fracture strength of 460 MPa. If it cross-sectional diameter at fracture is 10.7 mm, determine: (a) the ductility in terms of percent reduction in area (b) the true stress at fracture.

Chapter 6 - 41

HARDNESS • Resistance to permanently indenting the surface. • Large hardness means: --resistance to plastic deformation or cracking in compression. --better wear properties.

Adapted from Fig. 6.18, Callister 6e. (Fig. 6.18 is adapted from G.F. Kinney, Engineering Properties Chapter 6 - 42 and Applications of Plastics, p. 202, John Wiley and Sons, 1957.)

HARDENING • An increase in y due to plastic deformation.

• Curve fit to the stress-strain response:

Chapter 6 - 43

DESIGN OR SAFETY FACTORS • Design uncertainties mean we do not push the limit. • Factor of safety, N Often N is between y  working  1.2 and 4 N • Ex: 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.

 working 

220,000N    d2 / 4   

y N

5

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 y. • Toughness: The energy needed to break a unit volume of material. • Ductility: The plastic strain at failure. Note: For materials selection cases related to mechanical behavior, see slides 22-4 to 22-10.

Chapter 6 - 45