1 Introduction to Engineering Materials ENGR2 Chapter 6: Mechanical Properties of Metals Dr. Coates
2 6.2 Concepts of Stress and Strain tension compression shear torsion
3 Tension Tests The specimen is deformed to fracture with a gradually increasing tensile load
4 Fracture & Failure Fracture occurs when a structural component separates into two or more pieces. fracture represents failure of the component Failure is the inability of a component to perform its desired function. failure may occur prior to fracture e.g. a car axle that bends when you drive over a pothole
5 Engineering Stress & Engineering Strain Engineering stress : σ = F A Engineering strain : ε = l l = l l l F-instantaneous load applied to specimen cross section A -original cross section area before any load is applied l i -instantaneous length l -original length
6 If you double the cross sectional area, would your load needed to produce a given elongation be the same? Why not just use force F and deformation l to study mechanical behavior?
7 Compression Tests Similar to the tensile tests Force is compressive Specimen contracts along direction of the stress
8 Engineering Stress & Engineering Strain Engineering stress : F σ = A Engineering strain : ε = l l = l l l <
9 Shear stress : τ = F A Shear γ = tanθ θ Shear and Torsional Tests strain (by definition) : ( for small θ ) F-instantaneous load applied // to specimen cross section
10 Shear and Torsional Tests Shear stress due to applied torque : τ = f ( T ) Shear strain due to applied torque : γ = f ( ) φ
11 Stress Transformation A force balance in the y direction yields: A force balance in the tangential direction yields:
12 Deformation All materials undergo changes in dimensions in response to mechanical forces. This phenomenon is called deformation. If the material reverts back to its original size and shape upon removal of the load, the deformation is said to be elastic deformation. If application and removal of the load results in a permanent change in size or shape, the deformation is said to be plastic deformation.
13 Elastic Deformation 6.3 Stress-Strain Behavior Hooke's law : σ = Eε Modulus of E = σ ε Shear modulus : G = τ γ elasticity: ELASTIC DEFORMATION!
14 Modulus of Elasticity Young s modulus Elastic modulus Stiffness Material s resistance to elastic deformation greater the elastic modulus, the stiffer the material or smaller the elastic strain that results from the application of a given stress.
15 Tangent or Secant Modulus Materials with non-linear elastic behavior Gray cast iron, concrete, many polymers
16 Elastic modulus of metals, ceramics & polymers Ceramics high elastic modulus diamond, graphite (covalent bonds) Metals high elastic modulus Polymers low elastic modulus weak secondary bonds between chains Increasing E
17 Elastic Modulus on an Atomic Scale Elastic strain small changes in inter-atomic spacing stretching of inter-atomic bonds Elastic modulus resistance to separation of adjacent atoms inter-atomic bonding forces
18 Review 2.5 Bonding Forces and Energies The net force : FN = FA + FR At equilibrium : F - the centers of the two atoms are r = F F N A + R = is known as the bond length. r apart.
20 Elastic Modulus on an Atomic Scale How might E be related to df dr r? Why?
21 Effect of Temperature on the Elastic Modulus
23 Example 6.1 A piece of Cu originally 35 mm long is pulled in tension with a stress of 276 Mpa. If the deformation is entirely elastic, what will be the resultant elongation?
24 Example 6.1 Given : l = 35mm σ = 276 MPa Using Table 6.1: E Cu l =? = 11Gpa Using Hooke's law : σ = Eε = E l l l l = σ =.77 mm E
25 6.4 Anelasticity - time dependent elastic behavior Time dependent elastic strain component. Elastic deformation continues after the stress application. Upon load release, finite time is required for complete recovery.
28 Poisson s Ratio Poisson's ratio (a material constant) : ε x υ = ε υ > υ υ z theoretical max =.25 υ ε y = ε = = For most metals: z ( no net volume change).35.25
29 Elastic Properties of Materials Elastic constants (material properties) : E, G, υ For isotropic materials (same properties in all directions) : E G = 2 1+υ ( )
30 Example 6.2 A tensile stress is to be applied along the long axis of a cylindrical brass rod that has a diameter of 1 mm. Determine the magnitude of the load required to produce a 2.5 x 1-3 mm change in diameter if the deformation is entirely elastic.
31 Given : d d = 1mm = mm Using Table 6.1: E υ brass brass F =? = 97Gpa =.34
32 Compute the strain : ε x = d d = Using the Poisson's ratio : ε x ε z = = υ Using Hooke's law : σ = ε ze = 71.3MPa Compute the force : 2 d F = σ π 4 = 4 56 N
33 Plastic Deformation Elastic deformation Up to strains of.5 Plastic deformation Hooke s law is no longer valid Non-recoverable or permanent deformation Phenomenon of yielding
34 σ ε σ ys yp uts yield strength yield point strain ultimate tensile strength 6.6 Tensile Properties ( stress corresponding to the elastic limit) ε u uniform strain σ f fracture strength (stress at fracture) strain at fracture ε f ( maximum engineering stress)
35 Yielding and Yield Strength Gradual elastic-plastic transition Proportional limit (P) Initial departure from linearity Point of yielding Yield strength Determined using a.2 strain offset method
36 Yielding and Yield Strength Steels & other materials Yield point phenomenon Yield strength Average stress associated with the lower yield point
38 Example 6.3 From the tensile stress-strain behavior for the brass specimen shown, determine the following The modulus of elasticity The yield strength at a strain offset of.2 The maximum load that can be sustained by a cylindrical specimen having an original diameter of 12.8 mm The change in length of a specimen originally 25 mm long that is subjected to a tensile stress of 345 Mpa.
41 Given : d F max = 12.8mm =? Using the tensile strength : σ F uts = 45 MPa πd 4 2 max = σ uts = 57,9 N
42 Given : l = 25 mm σ = 345 MPa l =? Determine the strain (point A) : ε.6 Compute the elongation : l = εl 15mm =
43 Ductility Measure of the degree of plastic deformation that has been sustained at fracture. Brittle materials Little or no plastic deformation prior to fracture Ductile materials Significant plastic deformation prior to fracture
44 Percent elongation : l f l % EL = 1 = ε f 1 l % RA = where and where l is the original length and l is the length at fracture. f Percent reduction in area : A f A A A A f 1 Ductility is the original cross - sectional area is the final cross - sectional area of the necked region.
45 Ductility Why do we need to know the ductility of materials? The degree to which a structure will deform plastically prior to fracture (design & safety measures) The degree of allowable deformation during fabrication
47 Mechanical properties of metals Depend on prior deformation, presence of impurities heat treatments
48 Engineering stress-strain behavior for Fe at different temperatures brittle ductile
49 Resilience Capacity of a material to absorb energy when it is deformed elastically and then, upon unloading, to have this energy recovered. Modulus of resilience Strain energy per unit volume required to stress a material from an unloaded state up to the point of yielding.
50 Modulus of resilience: U r U r ε y = σdε Resilience Assuming a linear elastic region : 1 = σ yε y σ y σ y = σ y = 2 E 2E
51 Resilience Materials used in spring applications High resilience High yield strength Low elastic modulus
52 Toughness Measure of the ability of a material to absorb energy up to fracture. Area under the stress-strain curve up to the point of fracture.
55 Example Which material has the highest elastic modulus? Which material has the highest ductility? Which material has the highest toughness? Which material does not exhibit any significant plastic deformation prior to fracture?
56 6.7 True Stress and Strain Decline in stress necessary to continue deformation past the maximum point M. Is the material getting weaker?
57 6.7 True Stress and Strain Short-coming in the engineering stress-strain curve Decreasing cross-sectional area not considered
58 6.7 True Stress and Strain True stress : F = Ai where A σ t True strain : ε t = l l i dl is l l the instantaneous area. l = ln
59 6.7 True Stress and Strain Assuming no volume change : Al i i = A l Prior to necking : ε t = σ = σ t ( + ε ) ( 1+ ε ) ln 1
60 6.7 True Stress and Strain Necking introduces a complex stress state
61 Strain-hardening exponent From the onset of plastic deformation to the point of necking : σ = t K ε n t where n is the strain - hardening exponent and K is the strength coefficient.
63 Example 6.4 A cylindrical specimen of steel having an original diameter of 12.8 mm is tensile tested to fracture and found to have an engineering fracture stress of 46 Mpa. If its cross-sectional diameter at fracture is 1.7 mm, determine The ductility in terms of percent reduction in area The true stress at fracture
64 Given : = = = f f mm d MPa mm d σ 3% 1 1 % Ductility : 2 2 = = = d d A A A RA f f?? % 1.7 = = = tf f RA mm d σ 3% 1 2 = = d d d f
65 Engineering stress at fracture: σ F f = 46 MPa = The applied load : tf = σ A f F A f F A = 59,2 N True stress at fracture: σ = = 66 MPa
66 Example 6.5 Compute the strain-hardening exponent n for an alloy in which a true stress of 415 Mpa produces a true strain of.1. Assume a value of 135 Mpa for K.
67 Example 6.5 Given : σ = 415 MPa t ε t =.1 K = 135MPa n =? Strain - hardening exponent : σ = K n t ε K t n =.4
68 6.1 Hardness Measure of a materials resistance to localized plastic deformation (small dent or scratch)
69 Rockwell Hardness Tests A hardness number is determined Difference in depth of penetration from an initial minor load followed by a major load Combinations of indenters & loads are used
70 Hardness Testing Techniques
71 Minor load=1 kg - used for thick specimens
72 Minor load = 3 kg - thin specimens
73 Correlations between hardness & tensile strength Both are measures of resistance to plastic deformation
74 Property Variability & Design/Safety Factors Reading assignment Section 6.11 on variability of material properties Example 6.6
75 6.12 Design/Safety Factors Design stress : σ = d N' σ c σ is the calculated stress (on the basis of c N' > 1 is the design factor. Material selection criteria : σ ys σ d the estimated max load)
76 6.12 Design/Safety Factors Safe stress or working stress : σ w σ ys σ ys = N is the yield strength of the material and N > 1 is the factor of safety. Factor of safety : 1.2 N 4.
77 Design Example 6.1 A tensile-testing apparatus is to be constructed that must withstand a maximum load of 22 kn. The design calls for two cylindrical support posts, each of which is to support half of the max load. Furthermore, plain carbon steel ground and polished shafting rounds are to be used; the minimum yield and tensile strengths of this alloy are 31 Mpa and 565 Mpa respectively. Specify a suitable diameter for these support posts.
78 Given : F max σ ys σ uts d = =? MPa = Assume a N = 5 ( 22kN ) = 565 MPa factor of = 11kN safety : Working stress : ys σ w = σ = 62 MPa N Minimum diameter required : σ w F = π d max ( 2 d / 4) = 47.5mm
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