![ME3101-Materials Sc. and Engr.
Dr. Atiqullah. Compiled from: Callister and
others 4
3
1. Initial 2. Small load 3. Unload
Plastic means permanent!
F
δ
linear
elastic
linear
elastic
δplastic
planes
still
sheared
F
δelastic + plastic
bonds
stretch
& planes
shear
δplastic
ATOMIC MODEL OF PLASTIC DEFORMATION
Consider what
happens at the
atomic/grain level
7
• Bi-axial tension:2D • Hydrostatic compression:3D
Fish under waterPressurized tank
σz > 0
σθ > 0
σ < 0h
(photo courtesy
P.M. Anderson)
(photo courtesy
P.M. Anderson)
OTHER COMMON STRESS STATES (2)
• Submarine ? Offshore Pipeline?
• Modulus of Elasticity, E:
(also known as Young's modulus)
10
• Hooke's Law:
σ = E ε
• Poisson's ratio, ν:
metals: ν ~ 0.30-0.34
ceramics: ~0.25
polymers: ~0.40 compaction?
εν = − L
ε
εL
ε
1
-ν
F
F
simple
tension
test
σ
Linear-
elastic
1
E
ε
Units:
E: [GPa] or [psi]
ν: ?
LINEAR ELASTIC PROPERTIES
Materials properties data:
See Table 6.1 AGAIN (page 137) 6th ed. P-118).
Verify the range of ν values.](https://image.slidesharecdn.com/ch06mechprop-fall2016-sent-161102134149/85/Ch06-mech-prop-fall2016-sent-4-320.jpg)
![ME3101-Materials Sc. and Engr.
Dr. Atiqullah. Compiled from: Callister and
others 14
Table 6.4 : n and K values for some alloys (page-171)
n
TT K εσ ⋅=
True vs. Engineering Stress/Strain Curves
M tensile strength
For Some metals and alloys,
Approximate true stress/strain curve between yield point and max
strength (M), is given by: n
TT Kεσ =
Table of n and K: see previous page
Class Exercise
I: True Stress 500 MPa (72,50 psi) true strain 0.16 (unit?)
K=825 MPa (120,000 psi)
II: If true stress is 600 MPa ( 87,000 psi) true strain = ?
n
TT K εσ ⋅=
Solution strategy:
1. For part I – calculate n. [take log, solve
for n.]
2. In part II use this value of n to calculate
true strain.](https://image.slidesharecdn.com/ch06mechprop-fall2016-sent-161102134149/85/Ch06-mech-prop-fall2016-sent-14-320.jpg)
Ch06 mech prop fall2016-sent
- 1. ME3101-Materials Sc. and Engr. Dr. Atiqullah. Compiled from: Callister and others 1 ME3101: Materials Sc. & Engr. Ch 6: Mech. Properties How many of these terms do we know? • Anelasticity - viscoelasticity, polymers • Elastic recovery • Engineering Stress/strain – materials testing • True stress/strain – mfg. • Poisson’s Ratio – mfg. • Proportional limit – up to which .. • Toughness – bumper design for absorbing ? energy • Resilience (p-168) – spring design- energy absorbing but with no permanent ..?... Dr. Mir Atiqullah Includes materials adopted from Callister text Tensile and Compressive Deformations 0A F =σ Engineering Stress (considers only the original dimensions) Engineering Strain (considers only the original dimensions) 00 0 l l l lli ∆ = − =ε ve A F −= − = 0 σ In compression Poisson’s Ratio: ratio of the lateral and axial strains: 00 00 /)( /)( lll ddd z y z x − − −=−=−= ε ε ε ε ν Shear and Torsional Deformations 0 StressShear A F ==τ J Tc ==τ(Torque)TorsiontodueStressShear GJ Tl == ϕ(Torque)Torsiontodue(rad.)istAngular tw
- 2. ME3101-Materials Sc. and Engr. Dr. Atiqullah. Compiled from: Callister and others 2 Tensile Tests Most common among mechanical tests Standard test methods and samples: ASTM Standard E8 and E8M (metric) Circular and flat specimens with standard gauge lengths (50 mm or 2”) and areas. Testing machine usually elongate the specimen at constant rate. Simultaneously measures load (stress-using load cells) and elongation (strain- using extensometer). Engineering stress and engineering strain, from tensile tests are determined by: 0A F =σ and 00 0 l l l lli ∆ = − =ε More LATER…. Gage length ? Compression Tests Preferred test method if service loads are compressive-e.g. ? Standard test methods and samples: ASTM Standard E9 and E9M ? (metric) Compressive force and stress are NEGATIVE. Tensile tests are easier to perform and compressive tests rarely reveals any additional information. Engineering stress and engineering strain, from compressive tests are determined by same equations as for tensile tests. Compressive tests are specially necessary for determining materials behavior for large strains e.g. in manufacturing (forging, rolling, bending, stamping,..) Shear and Torsional Tests 0A F =τ Shear caused by parallel forces tending to slide a layer over another. Shear stress (transverse): Shear strain γ =tangent of the strain angle θ Torsion (twisting) produces pure shear E.g. : drive shaft, vehicle axle, twist drill, screw driver,.. J Tc =τ
- 3. ME3101-Materials Sc. and Engr. Dr. Atiqullah. Compiled from: Callister and others 3 Geometric Considerations of the Stress State State of Stress is also a function of the orientation of the plane along which the stresses are considered. For a plane at an angle θ: Resolved stresses are- + ⋅=⋅= 2 2cos1 cos2' θ σθσσ ⋅=⋅= 2 2sin cossin' θ σθθστ o45ismax' θτ when= F δ bonds stretch return to initial 2 1. Initial 2. Small load 3. Unload Elastic means reversible! F δ Linear- elastic Non-Linear- elastic ATOMIC MODEL OF ELASTIC DEFORMATION εσ ⋅= E εσ ⋅= E Consider happens at the atomic/grain level Materials properties data: See Table 6.1 NOW (page 137). Similarly shear stress and strain are related by: γτ ⋅= G γτ ⋅= G Table 6.1 Page 174 9th ed. G =~35-40%~ 8.5E6 psi For a metal/alloy, E=22E6 psi, ν=0.34 Estimate G= ? Use )1(2 ν+= GE
- 4. ME3101-Materials Sc. and Engr. Dr. Atiqullah. Compiled from: Callister and others 4 3 1. Initial 2. Small load 3. Unload Plastic means permanent! F δ linear elastic linear elastic δplastic planes still sheared F δelastic + plastic bonds stretch & planes shear δplastic ATOMIC MODEL OF PLASTIC DEFORMATION Consider what happens at the atomic/grain level 7 • Bi-axial tension:2D • Hydrostatic compression:3D Fish under waterPressurized tank σz > 0 σθ > 0 σ < 0h (photo courtesy P.M. Anderson) (photo courtesy P.M. Anderson) OTHER COMMON STRESS STATES (2) • Submarine ? Offshore Pipeline? • Modulus of Elasticity, E: (also known as Young's modulus) 10 • Hooke's Law: σ = E ε • Poisson's ratio, ν: metals: ν ~ 0.30-0.34 ceramics: ~0.25 polymers: ~0.40 compaction? εν = − L ε εL ε 1 -ν F F simple tension test σ Linear- elastic 1 E ε Units: E: [GPa] or [psi] ν: ? LINEAR ELASTIC PROPERTIES Materials properties data: See Table 6.1 AGAIN (page 137) 6th ed. P-118). Verify the range of ν values.
- 5. ME3101-Materials Sc. and Engr. Dr. Atiqullah. Compiled from: Callister and others 5 • Elastic Shear modulus, G: 11 τ 1 G γ τ = G γ • Elastic Bulk modulus, K: P= -K ∆V Vo P ∆V 1 -K Vo • Special relations for isotropic materials: P P P M M G = E 2(1+ ν) K = E 3(1− 2ν) simple torsion test pressure test: Init. Vol. =Vo. Vol. chg.= ∆V OTHER ELASTIC PROPERTIES 120.2 8 0.6 1 Magnesium, Aluminum Platinum Silver, Gold Tantalum Zinc, Ti Steel, Ni Molybdenum Graphite Si crystal Glass-soda Concrete Si nitride Al oxide PC Wood( grain) AFRE( fibers)* CFRE* GFRE* Glass fibers only Carbon fibers only Aramid fibers only Epoxy only 0.4 0.8 2 4 6 10 20 40 60 80 100 200 600 800 1000 1200 400 Tin Cu alloys Tungsten <100> <111> Si carbide Diamond PTFE HDPE LDPE PP Polyester PS PET CFRE( fibers)* GFRE( fibers)* GFRE(|| fibers)* AFRE(|| fibers)* CFRE(|| fibers)* Metals Alloys Graphite Ceramics Semicond Polymers Composites /fibers E(GPa) Eceramics > Emetals >> Epolymers 109 Pa 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. YOUNG’S MODULI: COMPARISON • Simple tension: 13 δ = FLo EAo δ L = −νFwo EAo δ/2 δ/2 δL/2δL/2 Lo wo F Ao • Simple torsion: M=moment α =angle of twist 2ro Lo • Material, geometric, and loading parameters all contribute to deflection. • Larger elastic moduli minimize elastic deflection. USEFUL LINEAR ELASTIC RELATIONS JG LM ⋅ ⋅ = 0 α
- 6. ME3101-Materials Sc. and Engr. Dr. Atiqullah. Compiled from: Callister and others 6 Example. Rod diameter 10 mm undergoes a tensile loading Calc the load necessary to produce a 2.5E-3 mm change in diameter within elastic region. units?45.2 0 −−= ∆ = E d d xε ν ε ε x z −= G = E 2(1+ ν) Ezεσ = 0AF ⋅= σ PLASTIC (PERMANENT) DEFORMATION Elastic limit= ? Max strain at Elastic limit= ? e.g. 0.005 or 0.5% Plastic deformation ~ breaking the bond with old neighbors and building new bonds next… In crystalline solids deformation ~ slip between adjacent crystal planes. Noncrystalline solids (also liquids)- deformation by viscous flow of material. Material is not an organized array of atoms, to begin with. 14 • Simple tension test: (at lower temperatures, T < Tmelt/3) PLASTIC (PERMANENT) DEFORMATION tensile stress, σ engineering strain, ε Elastic initially Elastic+Plastic at larger stress permanent (plastic) after load is removed εp plastic strain Elastic strain/ recovery
- 7. ME3101-Materials Sc. and Engr. Dr. Atiqullah. Compiled from: Callister and others 7 15 • Stress at which noticeable plastic deformation has occurred. when εp = 0.002 (standard) Other standards exist: e.g. 0.4% etc. Other values may also be used.tensile stress, σ engineering strain, ε σy εp = 0.002 YIELD STRENGTH, σy Partial plastic deformation No distinct point where yield starts 16 Graphite/ Ceramics/ Semicond Metals/ Alloys Composites/ fibers Polymers Yieldstrength,σy(MPa) PVC Hardtomeasure, sinceintension,fractureusuallyoccursbeforeyield. Nylon 6,6 LDPE 70 20 40 60 50 100 10 30 200 300 400 500 600 700 1000 2000 Tin (pure) Al (6061)a Al (6061)ag Cu (71500)hr Ta (pure) Ti (pure)a Steel (1020)hr Steel (1020)cd Steel (4140)a Steel (4140)qt Ti (5Al-2.5Sn)a W (pure) Mo (pure) Cu (71500)cw Hardtomeasure, inceramicmatrixandepoxymatrixcomposites,since intension,fractureusuallyoccursbeforeyield. HDPE PP humid dry PC PET ¨ Room T values σy(ceramics) >>σy(metals) >> σy(polymers) Based on data in Table B4, Callister 6e. a = annealed hr = hot rolled ag = aged cd = cold drawn cw = cold worked qt = quenched & tempered YIELD STRENGTH: COMPARISON 17 • Maximum possible engineering stress in tension. • Metals: occurs when noticeable necking starts. • Ceramics: occurs when crack propagation starts. • Polymers: occurs when polymer backbones are aligned and about to break. TENSILE STRENGTH, TS strain engineering stress TS Typical response of a metal Practice Example 6.3
- 8. ME3101-Materials Sc. and Engr. Dr. Atiqullah. Compiled from: Callister and others 8 18 Room T values Si crystal <100> Graphite/ Ceramics/ Semicond Metals/ Alloys Composites/ fibers Polymers Tensilestrength,TS(MPa) PVC Nylon 6,6 10 100 200 300 1000 Al (6061)a Al (6061)ag Cu (71500)hr Ta (pure) Ti (pure)a Steel (1020) Steel (4140)a Steel (4140)qt Ti (5Al-2.5Sn)a W (pure) Cu (71500)cw LDPE PP PC PET 20 30 40 2000 3000 5000 Graphite Al oxide Concrete Diamond Glass-soda Si nitride HDPE wood( fiber) wood(|| fiber) 1 GFRE(|| fiber) GFRE( fiber) CFRE(|| fiber) CFRE( fiber) AFRE(|| fiber) AFRE( fiber) E-glass fib C fibers Aramid fib TS(ceram) ~TS(met) ~ TS(comp) >> TS(poly) 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. TENSILE STRENGTH: COMPARISON • Plastic tensile strain at failure: 19 Engineering tensile strain, ε Engineering tensile stress, σ smaller %EL (brittle if %EL<5%) larger %EL (ductile if %EL>5%) • Another ductility measure:% reduction in area: 100% x A AA RA o fo − = • Note: %RA and %EL are often comparable. --Reason: crystal slip does not change material volume. --%RA > %EL possible if internal voids form in neck. Lo Lf Ao Af %EL = Lf − Lo Lo x100 Adapted from Fig. 6.13, Callister 6e. DUCTILITY, %EL Resilience Defn: Capacity of a material to absorb energy when it is deformed elastically, which releases this energy upon unloading. Modulus of Resilience EE U yy yyyr 22 1 2 1 2 σσ σεσ = == Resilient material high in ? Resilient material low in ? Yield strength Elasticity Application ? springssprings
- 9. ME3101-Materials Sc. and Engr. Dr. Atiqullah. Compiled from: Callister and others 9 • Energy to break a unit volume of material • Approximate by the area under the stress-strain curve. 20 smaller toughness- unreinforced polymers Engineering tensile strain, ε Engineering tensile stress, σ smaller toughness (ceramics) larger toughness (metals, PMCs) TOUGHNESS Test Procedures: Notch/Impact Toughness (Charpy or Izod methods) Fracture Toughness: Resistance to the growth of an existing crack. Open p-260 eq. 8.4 and P- 262 Table 8.1 KIC Values. • Resistance to permanently indenting the surface. • Large/high hardness means: --resistance to plastic deformation or cracking in compression. --better wear properties. --higher E or stiffness. 21 e.g., 10mm sphere apply known force (1 to 1000g) measure size of indent after removing load dD Smaller indents mean larger hardness. increasing hardness most plastics brasses Al alloys easy to machine steels file hard cutting tools nitrided steels diamond HARDNESS Principle of hardness tests CBN Why Hardness Tests so Common? • Simple and inexpensive- no special preparation of workpiece/specimen. Testers are relatively inexpensive. • (Usually) test is non destructive. Only a small indentation is left. Microhardness tests leave an indentation, which is so small, it requires a microscope to locate and measure hardness. • Other mechanical properties may be estimated from hardness values. e.g. tensile strength: see figure 6.19 Hardness test is performed more frequently than other methods, because:
- 10. ME3101-Materials Sc. and Engr. Dr. Atiqullah. Compiled from: Callister and others 10 Hardness vs tensile strength Source : Metals Handbook Some linear relationship: TS(Mpa) = 3.45X HB TS(Psi) = 500 X HB HB-Brinnel hardness number. Can we set up such a Linear relationship between TS and HRC? Why? Why Not? HRC scale is not linear.. Standard Hardness Measuring Methods Brinell Test: Oldest system, uses hardened steel ball upto 3000 kg of load. Rockwell Test: Most common, Uses diamond cone and various loads (60, 100, 150 kg). Minor load 10 kg. Indenters: Diamond ‘brale’, steel balls( 1/16, 1/8 inch diam.). Superficial Rockwell Test: Similar to Rockwell but using lower loads (15, 30, and 45 kg), Uses same indenters. Minor load 3 kg. Knoop and Vickers Microhardness tests: uses very small diamond tip, produces almost invisible indentation, which is later measured by a cross-hair under microscope. Loads are much smaller between 1 and 1000 grams. These methods are suitable for selective hardness measurements in a small region of a specimen. Knoop test (uses elongated diamond indenter) is suitable for ceramic materials. Microhardness test is suitable for ALL type of materials. Brinell Tester-applications The Brinell test is generally used for bulk metal hardness measurements - the impression is larger than that of the Vickers or Rockwell test and this is useful as it averages out any local heterogeneity and is affected less by surface roughness. 80 HRB =?
- 11. ME3101-Materials Sc. and Engr. Dr. Atiqullah. Compiled from: Callister and others 11 Brinell Hardness Test-Limitations However, because of the large ball diameter the test cannot be used to determine the hardness variations in a welded joint for which the Vickers test is preferred. Very hard metals, over 450BHN may also cause the ball to deform resulting in an inaccurate reading. To overcome this limitation a tungsten carbide ball is used instead of the hardened steel ball but there is also a hardness limit of 600BHN with this indentor. Rockwell Tester A modern Rockwell Hardness tester ( Dream) Specimen Indenter 80 HRC =? 60 HR45T =? A Rockwell Hardness tester Micro hardness Testers There are two types of indenters, a square base pyramid shaped diamond for testing in a Vickers tester and a narrow rhombus shaped indenter for a Knoop tester. Vickers indentation 250 HK =? 200 VHN =?
- 12. ME3101-Materials Sc. and Engr. Dr. Atiqullah. Compiled from: Callister and others 12 Summary Hardness Testing Methods. Errors in Hardness Testing Flatness and surface finish -flatness is most important - a maximum angle of approximately ± 1° would be regarded as acceptable. To achieve the required flatness tolerance and surface finish surface grinding or machining may be necessary. The correct load must be applied and to achieve this there must be no friction in the loading system otherwise the impression will be smaller than expected - regular maintenance and calibration of the machine is therefore essential. The condition of the indentor is crucial - whilst the Vickers diamond is unlikely to deteriorate with use unless it is damaged or loosened in its mounting by clumsy handling, the Brinell ball will deform over a period of time and inaccurate readings will result. This deterioration will be accelerated if a large proportion of the work is on hard materials. The length of time that the load is applied is important and must be controlled to some uniform value. Errors in Hardness Testing(contd.) The specimen dimensions are important - if the test piece is too thin the hardness of the specimen table will affect the result. As a rule of thumb the specimen thickness should be ten times the depth of the impression for the Brinell test and twice that of the Vickers diagonal. If the impression is too close to the specimen edge then low hardness values will be recorded - again as a rule the impression should be some 4 to 5 times the impression diameter from any free edge. Performing hardness testing on cylindrical surfaces eg pipes and tubes, the radius of curvature will affect the indentation shape and can lead to errors. It may be necessary to apply a correction factor - this is covered in, ISO 6507 Part 1.
- 13. ME3101-Materials Sc. and Engr. Dr. Atiqullah. Compiled from: Callister and others 13 Errors in Hardness Testing(contd.) The specimen table should be rigidly supported and must be in good condition - burrs or raised edges beneath the sample will give low readings. Impact loading must be avoided. It is very easy to force the indentor into the specimen surface when raising the table into position. This can strain the equipment and damage the indentor. Operator (or student) training is crucial and Regular validation or calibration is essential if hardness rest results are to be accurate and reproducible. (material compiled from various sources) True Stress and True Strain Defn: True Stress= applied load divided by instantaneous area. Defn: True Stress= applied load divided by instantaneous area. 0 ln l l T =ε i T A F =σ )1( εσσ +=T )1ln( εε +=T For some metals/alloys, true stress-strain relationship during plastic deformation meaning after yielding and until necking begins, is approximated by: n TT K εσ ⋅= n=strain hardening exponent. (0.12-0.5) See Table 6.4 page 171 NOW What is the value of n for 2024- T3 Al ? • What is the equation relating engr. Stress and engr strain? • An increase in σy due to plastic deformation. • Curve fit to the stress-strain response –yield to necking : σ ε large hardening small hardening unload reload σy0 σy 1 σT = C εT( )n “true” stress (F/A) “true” strain: ln(L/Lo) hardening exponent: n=0.15 (some steels) to n=0.5 (some copper) HARDENING Elastic Recovery: Return to original shape/size (slope) New yield point Who said it works?
- 14. ME3101-Materials Sc. and Engr. Dr. Atiqullah. Compiled from: Callister and others 14 Table 6.4 : n and K values for some alloys (page-171) n TT K εσ ⋅= True vs. Engineering Stress/Strain Curves M tensile strength For Some metals and alloys, Approximate true stress/strain curve between yield point and max strength (M), is given by: n TT Kεσ = Table of n and K: see previous page Class Exercise I: True Stress 500 MPa (72,50 psi) true strain 0.16 (unit?) K=825 MPa (120,000 psi) II: If true stress is 600 MPa ( 87,000 psi) true strain = ? n TT K εσ ⋅= Solution strategy: 1. For part I – calculate n. [take log, solve for n.] 2. In part II use this value of n to calculate true strain.
- 15. ME3101-Materials Sc. and Engr. Dr. Atiqullah. Compiled from: Callister and others 15 Another Example #6.47 page 212 • True Stress 60 ksi at true strain 0.15 • True Stress 70 ksi at true strain 0.25 • True stress = 65 ksi At true strain 0.20 – is it true? Why? • Strategy: – Set up 2 equations – Sove for K and n. – Use these K, n to calculate True stress at true strain 0.21. • Design uncertainties mean we do not push the limit. • Factor of safety, N 23 σworking = σy N Often N is between 1.2 and 4, but depends on application • 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. 1045 plain carbon steel: σy=310MPa TS=565MPa F = 220,000N d Loσworking = σy N 220,000N π d2 /4 /5=310 MPa DESIGN OR SAFETY FACTORS d= ? Key Terms this chapter Anelasticity Elastic limit Elastic recovery Engineering Stress VS True Stress Engineering Strain VS True Strain True stress True strain .. Hardness Test Types – Brinell, Rockwell, Micro(Vickers, knoop), Superficial (rockwell), Durometer..(for ____) Rockwell Scales: A, B, C, ....many. Typical hardness for CR steels HRC 35-40 HR lower. Resilience – useful property for design of ___?