MCEN 2024, Spring 2008 The week of Feb 25 HW 6–Final Quiz #6 will be based upon this HW, and will be open from Saturday morning (Feb 29) until 1:30 PM, Monday, Mar 03. Please have this homework by you side when you open the quiz. References to A&J: The reading list for this week: Chapter 8: The Yield Strength, Tensile Strength, and Ductility Overview: We now move over to learn about plastic deformation in solids. Such deformation is characterized by permanent deformation, that is, a change of shape. The topic is studied from the point of view of bridging the length scales, from the atomistic to the macroscopic. For example the design of a diving board so that it does not “yield” when a person stands on the edge is related to the yield stress of the material. The yield strength is then related to the atomic structure. The objective is to model the yield stress in terms of atomistics (atoms spacings and bonding), and then to relate the yield stress to component design. 1. Draw a distinction between the two phenomena expressed in each of the sub–questions given below: (a) Bending a copper tube, and hot–forging of a horse–shoe. (b) Hot–forging of a horse–shoe and forming of glass into a chandelier. (c) Deformation of a ductile metal versus the deformation of a "brittle material". (d) Deformation of a metal vs. the time dependent deformation of a polymer. (a) Because the yield stress decreases with increasing temperature. The yield stress of iron–alloys is too high at room temperature to forge into shape with a hammer. (b) Deformation of metals involves a distinct yield stress. Glass, is different it flows like a liquid albeit with a very high viscosity. (c) A brittle material breaks when stressed to its elastic limit, while a ductile deforms plastically at the elastic limit. (d) The yield stress of metals is essentially time independent, while deformation of polymers is time dependent, for example they continue to deform slowly if loaded for an extended period of time. 2. Make a plot of the yield stress (vertical axis) and the elastic modulus of the following materials, using the data given in the tables above: aluminum alloys, mild steel, stainless steels, cast iron, nickel alloys, beryllium alloys, copper alloys, and titanium alloys. Show the range in the data with error bars. (a) Why are the error bars greater for the yield stress than for the elastic modulus? (b) Draw two straight lines through the data which border the upper bound and the lower bound for the data. The slope of these lines gives the yield stress as a fraction of the elastic modulus. What are these fractions? (c) Show that the fractions given above are equal to elastic strain at the onset of plastic yielding. The plot is given on the next page: (a) The error bars are greater for the yield stress because alloying and heat–treatments are used to "engineer" the yield behavior, while the elastic constants depend only on the composition and can be roughly calculated using the rule–of–mixtures of the values for E for the constituents. (b) The slopes give 1.7*10-3 < σ y / E < 6.7*10-3. (c) The elastic strain at the onset of yielding is the same as σ y / E , that is 0.17% to 0.67%. 3. Plastic strains are often much larger than elastic strains. In some situations, like in forging and rolling the strains can be very large indeed. Therefore the stresses, and strains in large deformations have to be defined more carefully than the stresses and strains for elastic deformation. For large deformations, these quantities are described as "true stress" and "true strain", written as σ and ε . The nominal stresses and strains, σ n and ε n , are defined with respect to the initial cross–section, Ao, and initial length, Lo of the specimen. If the force at any point of deformation is written as P, and the length as L, then: σn = P L − Lo , εn = . Ao Lo (1) For plastic deformation, the strain is defined by the incremental deformation, so that: dε = dL L Show that, for a simple uniaxial deformation: (2) ε = n(1 + ε n ) (3) Then using the condition of constant volume for plastic deformation show that: σ = σ n eε (4) The volume of the specimen at any state of plastic deformation is given by the product of the cross sectional area and the gage length of the specimen. Since the volume remains constant we have that: AL = Ao Lo (A) Now, substituting from Eq. (1) into Eq. (3), and then again from Eq. (A) we have: ε = n(1 + L − Lo L A ) = n( ) = n( o ) Lo Lo A (B) Raising the left most and the right most terms in (B) to the power of "e", we have: Ao = eε A (C) Now, σ= P P Ao = . A Ao A (D) Substituting from (C) into (D) gives the desired result. 4. Show that the area under the stress–strain curve, where the y–axis is the true stress, and the x–axis is the true strain, is a measure of the total work done per unit volume of the material that is deformed in a uniaxial test. (In class we considered nominal stress and nominal strain. Here you need to extend that integral to true stress and strain. Use the constant volume condition that Ao Lo = AL ). The total work done on the specimen is the integral of the force times the displacement, which can be written as: WTOT = ∫ Pd(ΔL) (E) Dividing both sides by the volume of the specimen which is equal to AL, we get: WTOT P d(ΔL) (F) =∫ = ∫ σdε AL A L which is equal to the area under the true-stress true–strain curve. W perVOL = 5. A plate of aluminum which is 5 mm thick and 25 cm wide is rolled in one pass to a sheet of the same width but a thickness of only 0.1 mm. The yield stress of the aluminum (in uniaxial compression) may be assumed to have a constant value of 300 MPa. The work capacity of the roller–system is 100 HP (horse–power). Calculate the maximum speed at which the plate can be put through the roller machine. (Hints: 1 HP = 746 watts, 1 watt = 1 J s-1. First calculate the true strain in the sheet, and then the work done per unit volume. This will tell you how much volume of the plate can be put through the machine in one second). Answer: 0.5 m s-1. Work done per unit volume of the material is simply equal to the product of true-stress and true–strain (since the stress–strain curve is assumed to be square shaped, that is, the yield stress is assumed to be constant). True strain in the plate is given by Eq. (B): ε = n( L 0.1 ) = n( ) = −3.9 Lo 5 (G) Therefore the work done per unit volume of throughput will be given by: 300*106*3.9 J m-3 (the minus sign means that the strain is negative, not that the work done is negative). If the feed velocity for the plate is V m/s, and the input cross section is 25cm x 5mm, then the work done per second on the metal, which must be equal to the HP capacity of the machine, leads to the following equation (keep track of units - use J and m): (300 *10 6 * 3.9) * (25 *10 −2 * 5 *10 −3 *V ) = 100 * 746 which gives: 0.05 m/s of 10) (the answer in the book is too large by a factor 6. and 7. H (MPa) TrueStress NomStrain NomStrain+H TrueStrain NomStress 423 141 0.01 0.09 0.09 129.36 606 202 0.1 0.18 0.17 171.19 756 252 0.2 0.28 0.25 196.88 870 290 0.3 0.38 0.32 210.14 957 319 0.4 0.48 0.39 215.54 1029 343 0.5 0.58 0.46 217.09 1080 360 0.6 0.68 0.52 214.29 1116 372 0.7 0.78 0.58 208.99 1170 390 1 1.08 0.73 187.50 Notes: • Hardness measures the true value of the yield stress, and is three times the tensile yield stress • The fourth column is simply the nominal strain +0.08, as said in the problem • The true strain is calculated from Eq. (3) above. • The nominal stress can now be calculated from Eq. (4). The nominal stress-strain plot is obtained from the 3rd and the 6th columns, and is given below. The solutions to problems (a), (b), and (d) are explained by the above figure. For the reduction in area at the point of tensile strength, problem (c), we proceed as follows: RA = Ao − A A L = 1− = 1− o Ao Ao L (H) But we know that: εn = L − Lo L = −1 Lo Lo Therefore, RA = 1− 8. 9. and 10. (I) 1 = 0.38 1 + εn 8. Copper can be deformed in rolling because strain localization in the neck which leads to failure is prevented in compressive deformation. 9. The tensile strength is achieved where the applied load is a maximum, which occurs at 101kN. The answer to (a) therefor is 0.25*101E+3/(160E-6) Pa = 158 MPa. The extension at 0.1% strain will be equal to 0.001*50=0.05 mm. The load at this extension is 12kN which gives a nominal stress of 75 MPa. With a safety factor of 0.6 the stress is 45 MPa (the answer in the book appears to be wrong). 10. The area of the footprint for a sphere, of radius r indented into a surface to a depth of h is equal to 2πrh , hence the answer.