Presentation on theme: "Chapter 2 Fundamentals of the Mechanical Behavior of Materials"— Presentation transcript:
1 Chapter 2 Fundamentals of the Mechanical Behavior of Materials Fetweb.ju.edu.jo/staff/ie/mbarghashChapter 2 Fundamentals of the Mechanical Behavior of Materials
2 Types of StrainFIGURE Types of strain. (a) Tensile, (b) compressive, and (c) shear. All deformation processes in manufacturing involve strains of these types. Tensile strains are involved in stretching sheet metal to make car bodies, compressive strains in forging metals to make turbine disks, and shear strains in making holes by punching.
29 Mechanical Properties of Materials Table 2.1 Typical mechanical properties of various materials at room temperature.
30 Loading and UnloadingFIGURE Schematic illustration of loading and unloading of a tensile-test specimen. Note that during unloading, the curve follows a path parallel to the original elastic slope.
31 True Stress-True-Strain Curves in Tension FIGURE (a) True stress-true-strain curve in tension. Note that, unlike in an engineering stress-strain curve, the slope is always positive, and the slope decreases with increasing strain. Although stress and strain are proportional in the elastic range, the total curve can be approximated by the power expression shown. On this curve, Y is the yield stress and Yf is the flow stress. (b) True-stress true-strain curve plotted on a log-log scale. (c) True stress-true-strain curve in tension for 1100-O aluminum plotted on a log-log scale. Note the large difference in the slopes in the elastic and plastic ranges. Source: After R. M. Caddell and R. Sowerby.
32 Power Law Material Behavior K: strength coefficientn: strain hardening coefficientTable 2.3 Typical values of K and n in Eq. (2.11) at room temperature.
33 True Stress - True Strain Curves for Various Metals FIGURE True-stress-true-strain curves in tension at room temperature for various metals. The point of intersection of each curve at the ordinate is the yield stress Y; thus, the elastic portions of the curves are not indicated. When the K and n values are determined from these curves, they may not agree with those given in Table 2.3, because of the different sources from which they were collected. Source: S. Kalpakjian.
34 Strain Rate Effects C: strength coefficient Table 2.5 Approximate range of values for C and m in Eq. (2.16) for various annealed materials at true strains ranging from 0.2 to 1.0.C: strength coefficientM: strain rate sensitivity exponent
35 Effect of temperature Power Law Creep One of the most common forms of plastic flow isPower-Law Creep, given by the formula: Strain Rate = C (Stress)n exp(-Q/RT)Let's take each part of the formula in turn:C is a scaling constant.n means that the strain rate increases much faster than stress. Typically n is about 3 but can range from a bit less than 2 to 8. Recall that with viscous deformation stress is proportional to strain rate (n=1). With power-law creep it's faster: the effective viscosity drops with stress.Q is the activation energy required to get crystal dislocations moving. It's typically kilojoules per mole, sometimes up to 500.R is the Universal Gas Constant that turns up everywhere in physical chemistry. In SI units it equals joules/mole-degree Kelvin.T is the temperature in degrees Kelvin. As T increases, Q/RT decreases and thus exp(-Q/RT) increases, though much more slowly than exp(T). At very large T, Q/RT approaches zero and the exponential term approaches 1. This does not happen, though, at geologically realistic temperatures.
36 Barreling In Compression Plane Strain CompressionFIGURE Barreling in compression of a round solid cylindrical specimen (7075-O aluminum) between flat dies. Barreling is caused by interfaces, which retards the free flow of the material. See also Figs. 6.1 and 6.2. Source: K. M. Kulkarni and S. Kalpakjian.FIGURE Schematic illustration of the plane-strain compression test. The dimensional relationships shown should by satisfied for this test to be useful and reproducible. This test give the yield stress of the material in plane strain, Y’. Source: After A. Nadai and H. Ford.
66 Plane StrainA state of plane strain exits when the strains are confined to a single plane, such as the x-y plane.This generally means that the stresses in the other direction; eg., the z direction, are non-zero.Plane Strain occurs in thick sections that “constrain” out of plane deformations
67 Transformations in plane strain ,e,gThe state of plane strain at a point p is given by.Determine thexyxyprincipal strains and the maximum in-plane shear strain and show theorientations of the elements subjected to these strains. Also determine theabsolute maximum shear strain.e =
68 g xy tan 2 q p e e x y Solution g 1 xy 1 -0.006 . . q atan atan p ( )2ee2xy
69 Principal Strains -0.006 0.004 0.002 -79.6º -79.6º =0.005162 eeeegxyxyxy....e'cos2qsin2qxpp222eeeegxyxyxy....e'cos2qsin2qypp222-0.0060.0040.002.-79.6ºe'cos-79.6ºsinx222=e’y=
70 Maximum Shear Strain = gmax = 0.006325 'e'xy=ge'testxe'ygmax =Use diameters of circles!, Mohr’s circle plots g/2, so g is the diameter(=2x radius).
72 Mohr’s Circle Plot normal strain on the x-axis Plot ½ the shear strain on the y-axisSolve as you would for plane stress problemg/2e
73 Strain energyFor the elastic regionFor 3D caseFor principle stress case
74 Plane StrainA state of plane strain exits when the strains are confined to a single plane, such as the x-y plane.This generally means that the stresses in the other direction; eg., the z direction, are non-zero.Plane Strain occurs in thick sections that “constrain” out of plane deformations
75 Transformations in plane strain ,e,gThe state of plane strain at a point p is given by.Determine thexyxyprincipal strains and the maximum in-plane shear strain and show theorientations of the elements subjected to these strains. Also determine theabsolute maximum shear strain.e =
76 g xy tan 2 q p e e x y Solution g 1 xy 1 -0.006 . . q atan atan p ( )2ee2xy
77 Principal Strains -0.006 0.004 0.002 -79.6º -79.6º =0.005162 eeeegxyxyxy....e'cos2qsin2qxpp222eeeegxyxyxy....e'cos2qsin2qypp222-0.0060.0040.002.-79.6ºe'cos-79.6ºsinx222=e’y=
78 Maximum Shear Strain = gmax = 0.006325 'e'xy=ge'testxe'ygmax =Use diameters of circles!, Mohr’s circle plots g/2, so g is the diameter(=2x radius).
84 We want to determine the normal and shear stresses on the plane We want to determine the normal and shear stresses on the plane. The normal stresses are easiest. Unfortunately, we can't add stresses, only forces, so we have to determine the forces the stresses exert, add them up, then convert back to stress.Consider stress S1. It acts along the X1 axis, .but the stress "sees" only the area of the plane visible along the X1 axis, which is c1. So F1 = S1c1. Similarly, F2 = S2c2 and F3 = S3c3. The force normal to the plane exerted by F1 is F1c1, and the total force normal to the plane is F1c1 + F2c2 + F3c3. Since F1 = S1c1, we find:Fn = S1c12 + S2c22 + S3c32Furthermore, stress = force/area, but the area of the plane is one, so we haveSn = S1c12 + S2c22 + S3c32Determining shear stress can be a lot messier, if we do things the brute force way. Or we can do it the easy way.
85 Here we are looking in the plane of the normal and shear forces Here we are looking in the plane of the normal and shear forces. It's obvious from the vector diagram that F2 = Fn2 + Fs2.Since the plane has an area of one and stress = force per unit area, we have F2 = Sn2 + Ss2. Note that it's only the magnitudes of the stresses that we are adding. Stresses do not add vectorially!The total force F can be found from the three vectors F1, F2 and F3 above. Since these three components are mutually perpendicular, we haveF2 = F12 + F22 + F32 orSn2 + Ss2 = S12c12 + S22c22 + S32c32 (this will be very useful a bit later)
89 Triaxial Stress State (+ve sense shown) Remember: txy=tyx, txz=tzx, tyz=tzyInclined planes will have a combination of normal and shear stresses that do not line up with any of the axes. As with the plane stress element this can be solved using static equilibrium. The formulae are rather long and complicated and are beyond the scope of this course.(+ve sense shown)
90 3D Principal – Triaxial Stress It is possible to determine the unique orientation of an element having only principal stresses acting on its faces. As shown. These principal stresses are assumed to have magnitudes of maximum, intermediate, and minimum intensity, i.e.:We will assume that the orientation and magnitudes of the principal stresses are known (3D transformations are beyond the scope of the course). This is a condition known as triaxial stress.
91 3D Stress – Principal Stresses The three principal stresses are obtained as the three real roots of the following equation:whereOf particular importance are the three principal stresses, which are obtained as the three real roots of the following equation:I1, I2, and I3 are known as the stress invariants, because they do not change in value when the axes are rotated to new positions.The process of finding the principal stresses is carried out as follows:determine the normal and shear stresses on the element in the x, y, and z frame of reference;calculate the stress invariants;solve the cubic equation for its three roots. (these three roots are the three principal stresses (s1, s2, s3)I1, I2, and I3 are known as stress invariants as they do not change in value when the axes are rotated to new positions.
92 Stress Invariants for Principal Stress Zero shear stress onprincipal planes
93 Derive the two dimensional principle stress case using the Equations for the tri-axial stress case
94 Mohr’s Circle?There is no Mohr’s circle solution for problems of triaxial stress stateSolution for maximum principal stresses and maximum shear stress is analyticalEither closed form solution or numerical solution (or computer program) are used to solve the eigenvalue problem.
95 Maximum Shear Stresses Absolute max shear stress is the numerically larger of:ty’z’, tabs maxtx’y’ty’z’s1s2s3After finding the principal stresses, it is relatively easy to obtain the maximum shear stresses. Because no shear stresses act on the principal planes, it follows that an element oriented to the principal directions is in a state of triaxial stress. Therefore the three maximum shear stresses are:NormalStress, s
96 3D Mohr’s Circle – Plane Stress A Case Study – The two principal stresses are of the same signs1s2s3-st
97 3D Mohr’s Circle – Plane Stress A Case Study – The two principal stresses are of opposite signs1s2s3st
98 y 80 MPa 50 MPa 120 MPa Tensor shows that: sz = 0 and t xz = t yz = 0 Example:For the following state of stress, find the principal and critical values.y80 MPa120 MPa50 MPaTensor shows that:sz = 0 and t xz = t yz = 0x
100 3-D Mohr’s Circlest max = 77 MPaShear Stress, MPa
101 Example: triaxial stress state, not plane stress Determine the maximum principal stresses and the maximum shear stress for the following triaxial stress state. (+ve values as defined in slide 1)s=MPa
105 2. Max Principal Stress Criterion: smax= 63.5 MPa Safety Factor?Not RequiredIf the stress state was determined on a steel crankshaft, made of forged SAE1045 steel with a yield strength of 300 MPa, what is the factor of safety against yield?Tresca Criterion: tmax= 58.5 MPa2. Max Principal Stress Criterion: smax= 63.5 MPa
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