Exam (elaborations) TEST BANK FOR Deformation and Fracture Mechanics of Engineering Materials 5th Edition By Richard W. Hertzberg (Solution Manual)

Exam (elaborations) TEST BANK FOR Deformation and Fracture Mechanics of Engineering Material 5Edition By Richard W. Hertzberg (Solution Manual) Deformation and Fracture Mechanics of Engineering Materials, 5th ed. Problem Solutions p. 1/162 Draft document, Copyright R. Hertzberg, R. Vinci, J. Hertzberg 2009 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. CHAPTER 1 Review 1.1 In your own words, what are two differences between product testing and material testing? Possible answers include: (a) The goal of the two procedures is different. Whereas product testing is design to determine the lifetime of a component under conditions that mimic realworld use, material testing is intended to extract fundamental material properties that are independent of the material’s use. (b) The specimen shape is different. Product testing must use the material in the shape in which it will be used in the real product. Material testing uses idealized specimen shapes designed to unambiguously determine one or more properties of the material with the simplest analysis possible. 1.2 What are the distinguishing differences between elasticity, plasticity, and fracture? Elasticity involves only deformation that is fully reversible when the applied load is removed (even if it takes time to occur). Plasticity is permanent shape change without cracking, even when no load exists. Fracture inherently involves breaking of bonds and the creation of new surfaces. Often two or more of these processes take place simultaneously, but the contribution of each can be separated from the others. 1.3 Write the definitions for engineering stress, true stress, engineering strain, and true strain for loading along a single axis.  eng  engineering stress  load initial cross-sectional area  P A0 (1-1a)  true  true stress  load instantaneous cross-sectional area  P Ai (1-2a)  eng  engineering strain  change in length initial length  l f  l0 l0 (1-1b)  true  true strain  ln final length initial length  ln l f l0 (1-2b) 1.4 Under what conditions is Eq. 1-4 valid? What makes it no longer useful if those conditions are not met? Deformation and Fracture Mechanics of Engineering Materials, 5th ed. Problem Solutions p. 2/162 Draft document, Copyright R. Hertzberg, R. Vinci, J. Hertzberg 2009 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.  true  P A0 (li / l0 )   eng (li / l0 )   eng (1 eng ) (1-4) This expression is true when volume is conserved. However, it is only useful if the crosssectional area is the same everyone on the test specimen. If this isn’t the case then the stress and strain will vary from one part of the specimen to another. 1.5 Sketch Figure 1.3, curve ‘b’ (a ductile metal). Label it with the following terms, indicating from which location on the curve each quantity can be identified or extracted: elastic region, elastic-plastic region, proportional limit, tensile strength, onset of necking, fracture stress. strain stress fracture stress elastic region elastic-plastic region proportional limit tensile strength onset of necking 1.6 On a single set of axes, sketch approximate atomic force vs. atom-separation curves like the one shown in Fig. 1.4b for tungsten at temperatures of 200, 600, and 1000 K. Pay close attention to the point x0 and the slope dF/dx for each of the curves you draw. The key features of the plot are the increasing x0 spacing with increasing temperature (i.e., with thermal expansion) and the decreasing slope associated with decreased elastic modulus. The plot is exaggerated but the trends are reasonable. F x x0 (1000 K) x0 (600 K) x0 (200 K) dF dx 200 K 600 K 1000 K Deformation and Fracture Mechanics of Engineering Materials, 5th ed. Problem Solutions p. 3/162 Draft document, Copyright R. Hertzberg, R. Vinci, J. Hertzberg 2009 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 1.7 State the critical difference in the processing behavior of thermoplastics vs. thermosets. Thermoplastics can be melted and resolidified multiple times, so processing often involves several heating, forming, and cooling steps. Thermosets harden by a one-time chemical reaction so there cannot be any additional forming operations after the cross-linking operation takes place. 1.8 What happens to the stiffness of a polymer as the temperature Tg is exceeded? For what group of polymers is this change the greatest? The smallest? The stiffness of a polymer decreases above the glass transition temperature, sometimes dramatically. The effect is the largest for amorphous, uncross-linked polymers. It is the smallest for highly cross-linked polymers (such as certain epoxies). 1.9 Write typical values of E for diamond, steel, aluminum, silicate glass, polystyrene, and silicone rubber subjected to small strains (note that the latter value is not included in this chapter, but is widely available). Clearly indicate the units for each value. The following values are not intended to represent any particular processing method or alloy composition; they are rounded average values for certain material families. Diamond ~ 1000 GPa Steel ~ 200 GPa Aluminum ~ 70 GPa Silicate glass ~ 70 GPa Polystyrene ~ 3 GPa Silicone rubber ~ 10 MPa (0.010 GPa) 1.10 What is the purpose of a plasticizer, and what specific effect on room temperature behavior is likely when a plasticizer is added? A plasticizer is added to a polymer to break up the molecular interactions, allowing more chain mobility than would otherwise be possible for that particular polymer at the temperature of interest. At room temperature, therefore, the polymer is more likely to have a low elastic modulus (i.e., a ordinarily-hard polymer may become flexible). 1.11 Identify a minimum of two structural characteristics and two mechanical characteristics that set elastomers apart from other classes of materials (including other polymers). Elastomers are amorphous and moderately cross-linked. They tend to display significant changes in stiffness as their use temperate exceeds Tg, but they do not melt at even higher temperature. 1.12 Define what is meant by uniaxial, biaxial and triaxial loading. Uniaxial loading occurs along a single direction, biaxial along two directions, and triaxial along three. Note that there may be multiaxial strains even when the loading is restricted to one or two directions. 1.13 State one advantage and disadvantage of compression testing. Deformation and Fracture Mechanics of Engineering Materials, 5th ed. Problem Solutions p. 4/162 Draft document, Copyright R. Hertzberg, R. Vinci, J. Hertzberg 2009 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. An advantage may be to avoid failure due to tensile cracking at low loads (as in the case for ceramics and glasses), and therefore to allow exploration of degrees of plasticity impossible to achieve under tensile loading. One disadvantage would be the difficulty in achieving ideal friction-free conditions between the specimen and the loading platen. 1.14 Is buckling failure initiated by an elastic, plastic, or cracking process? Explain. Buckling failure is initially an elastic process in which the member deflects in a direction perpendicular to the loading axis. This failure may then be followed by plasticity or fracture, but these processes are not inherent in buckling. 1.15 What is the difference between the resilience and the strain energy density of a material under load? Illustrate your answer by reproducing Figure 1.3, curve ‘b’ (a ductile metal), and annotating it appropriately. Resilience is a measure of the maximum elastic strain energy stored in the material before the onset of plasticity. The strain energy density is a more general term that is a measure of the stored elastic energy at any point during a mechanical test. It may be greater or less than the resilience, depending on the hardening or softening behavior that takes place after plastic deformation begins. strain stress strain energy density at the point of necking resilience 1.16 Sketch Figure 1.3, curve ‘b’ (a ductile metal) and show on the figure the difference between the proportional limit and the offset yield strength. Deformation and Fracture Mechanics of Engineering Materials, 5th ed. Problem Solutions p. 5/162 Draft document, Copyright R. Hertzberg, R. Vinci, J. Hertzberg 2009 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. strain stress proportional limit offset yield strength 0.2% 1.17 Describe when and why bend testing (flexural testing) is most advantageous. Bend testing may be used for any class of materials. It can be used to assess elastic or plastic properties. It is particularly useful when the material is only available in the shape of a rectangular prism, or when the material would be likely to fail prematurely due to extreme flaw sensitivity (as is usually the case for brittle ceramic and glass materials). 1.18 Where can the maximum stress be found for a rectangular bar undergoing 3-point bending? 4-point bending? The maximum stress in 3-point bending is found in two locations: at the top and bottom surfaces directly aligned with the central load point. In 4-point bending, the maximum stress is also found on the top and bottom surfaces, but it exists at a constant level between the inner (closer) load points. 1.19 Write the basic isotropic form of Hooke’s law relating stress and strain for uniaxial tension/compression loading and shear loading. Define all quantities. Linear elastic uniaxial tension/compression is described by   E , where  is stress, E is Young’s modulus, and  is strain. An analogous form exists for shear loading, for which  = G. In this expression,  is the shear stress, G is the shear modulus, and  is the shear strain. 1.20 Why do we define engineering and true stresses for tension/compression loading but not for shear loading? In tension and compression the cross-sectional area bearing the load changes during deformation, so it is often necessary to account for this change. In shear loading there is distortion of the material but the area over which the force is distributed does not change, so there is no need for a true stress definition. 1.21 Sketch a pair of pliers squeezing an object and use it to show why the hinge pin is under shear loading. When the clamping force is applied to an object, a reaction force must exist at the pin. The two jaw faces experience equal and opposite clamping forces, so the pin must experience equal and opposite reaction forces at its two ends. These create shear stress within the pin. Deformation and Fracture Mechanics of Engineering Materials, 5th ed. Problem Solutions p. 6/162 Draft document, Copyright R. Hertzberg, R. Vinci, J. Hertzberg 2009 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. applied force clamping force reaction force top jaw bottom jaw 1.22 Write out the most general expression for tension or compression strain along a single axis resulting from all possible applied stresses, assuming that the material is elastically isotropic.  xx  1 E  xx   E  yy         E  zz         xx  v( yy  zz ) E 1.23 Write out the most general expression for shear strain along a single axis resulting from all possible applied stresses, assuming that the material is elastically isotropic.  xy   xy G Sketch and name the stress state present in the skin of a cylindrical thin-walled pressure vessel. Repeat for the strain state. biaxial stress triaxials train 1.24 What is the name of the matrix, Sij? The Compliance Matrix. 1.25 Why can the compliance and stiffness tensors for cubic and orthotropic materials be greatly simplified from the general case? In cubic and orthotropic materials there are several directions that are structurally identical, and therefore have identical elastic properties. Furthermore, the high degree of symmetry reduces the degree of coupling between applied stresses and induced strains (e.g., the absence of XY, YZ, or XZ shear strains generated by XX, YY, and ZZ stresses). 1.26 Describe the geometric criteria that differentiate orthotropic and cubic symmetry. Orthotropic materials have three distinct a, b, and c axes that are each separated by interior angles of ===90°. Cubic materials also have three axes separated by ===90°, but the three axes are identical. 1.27 Define hydrostatic stress state. Deformation and Fracture Mechanics of Engineering Materials, 5th ed. Problem Solutions p. 7/162 Draft document, Copyright R. Hertzberg, R. Vinci, J. Hertzberg 2009 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. A hydrostatic stress state is one in which the three normal stress components, XX, YY, and ZZ, are all equal, and there are no shear stresses. 1.28 What is the primary purpose of the fibers in a composite material? Of the matrix? The fibers usually act as the reinforcement phase, supporting the majority of the load. They provide most of the stiffness and stress of the composite material. The matrix hold the fibers together, and serves to transfer the load to the fibers. 1.29 What does it mean for a fiber-reinforced composite to be quasi-isotropic, and how is this typically achieved? A quasi-isotropic layered composite has elastic properties that are essentially identical in all directions within the plane of the material. This is typically achieved by orienting an equal volume fraction of fibers in each of the 0°, 90°, and ±45° directions. 1.30 Which is the stiffer orientation for a unidirectional fiber-reinforced composite, the isostress orientation or the isostrain orientation? Explain, and provide a sketch to support your answer. The stiffer direction is the isostrain orientation, in which the fibers and the matrix must strain by equal amounts. In this orientation, the stiffer fibers typically bear the majority of the load, and so the composite stiffness is maximized for a given volume fraction of fibers. In the isostress orientation, the compliant matrix is free to strain to a larger degree than the stiff fibers, so the overall stiffness is lower. Ec, isostrain Ec, isostress load load 1.31 Why are pairs of materials more likely to experience thermal stress problems when they represent two different material classes? Pairs of materials are more likely to experience thermal stress problems when they represent two different material classes because the differences in atomic bonding character that exist between material classes can lead to large differences in their coefficients of thermal Deformation and Fracture Mechanics of Engineering Materials, 5th ed. Problem Solutions p. 8/162 Draft document, Copyright R. Hertzberg, R. Vinci, J. Hertzberg 2009 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. expansion. If the materials are joined together, this thermal expansion incompatibility can cause larger thermal stresses to develop. Practice 1.32 Sketch a tensile member with (a) a rectangular cross-section, (b) a solid circular crosssection, and (c) a circular tube cross-section, and label the dimensions symbolically (e.g., label the radius for the solid circular case). For each member, write out the definition of engineering stress in terms of the actual dimensions of the component. If the rectangular member has dimensions of width and thickness equal to 1 cm x 0.3 cm, what would be the radius of a solid circular member such that the stress is equal for an equal tensile load? If a tube has an outer radius equal to that of this same solid cylinder, what is the maximum inner radius such that the stress does not exceed 50% of the stress in the solid cylinder? w t r r1 r2  eng  F A  F tw  eng  F A  F  r 2  eng  F A  F  r1 2   r2 2  F  r1 2  r2 2   For the rectangular bar, Arect  (1cm)(0.3 cm)  0.3 cm2 . To create equal stress under an identical load, the solid cylinder must have the same cross sectional area, so Arect  Acyl   r 2 which requires a radius of approximately 0.309 cm. The tube can have a load-bearing cross section that is equal to half that of the solid cylinder since the maximum stress is half. Setting the outer radius r1 equal to 0.309 cm and the tube area equal to 0.5(0.3 cm2) gives the inner radius r2=0.218 cm. 1.33 A commercially-pure copper wire originally 10.00 m long is pulled until its final length is 10.10 m. It is annealed, then pulled again to a final length of 10.20 m. What is the engineering strain associated with each of the two steps in the process? What is the true strain for each step? What are the total engineering and true strains for the combined steps? Finally, what agreement (if any) is there between the total strains calculated as the sum of two steps of 0.10 m vs. a single step of 0.20 m? The engineering strain for each step is: Deformation and Fracture Mechanics of Engineering Materials, 5th ed. Problem Solutions p. 9/162 Draft document, Copyright R. Hertzberg, R. Vinci, J. Hertzberg 2009 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 1:  eng1  L L0  0.1m 10m  0.010 and  true1  ln Lfinal L0        ln 10  0.1m 10m        0.00995 2 :  eng2  L L1  0.1m 10.1m  0.0099 and  true2  ln Lfinal L1        ln 10.1 0.1m 10.1m        0.00985 So the sum of the two steps in each case is  eng   eng1   eng2  0.010  0.0099  0.0199 and  true   true1   true2  0.00995  0.00985  0.0198 Calculating the total strain as if the elongation were performed in a single step:  eng  L L0  0.2m 10m  0.020 and  true  ln Lfinal L0        ln 10  0.2m 10m        0.01980 So, it can be seen that the sum of the individual engineering strains does not agree with the strain calculated for a single elongation step of 0.20 m, whereas the total true strains is the same regardless of whether the 0.20 m elongation is applied in one increment or two. 1.34 A 3-mm-long gold alloy wire intended to electrically bond a computer chip to its package has an initial diameter of 30 μm. During testing, it is pulled axially with a load of 15 grams-force. If the wire diameter decreases uniformly to 29 μm, compute the following: (a) The final length of the wire. (b) The true stress and true strain at this load. (c) The engineering stress and strain at this load. The final length of the wire is lf  l0 A0 Af        3mm  30106m 2  2  29106m 2  2          3.21mm The true stress and true strain are  true  P Afinal  (15 g  f ) 9.807103 N 1g f    29106m 2  2  223MPa true  ln Lf L0        ln 3.21mm 3mm       0.0677 The engineering stress and engineering strain are Deformation and Fracture Mechanics of Engineering Materials, 5th ed. Problem Solutions p. 10/162 Draft document, Copyright R. Hertzberg, R. Vinci, J. Hertzberg 2009 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.  eng  P A0  (15 g  f ) 9.807103 N 1g f    30106m 2  2  208MPa  eng  Lf  L0 L0  3.21 3mm 3mm  0.070 1.35 A cylindrical rod of Ni 200 alloy has the following properties: E = 204 GPa,  = 0.31. It is loaded elastically in compression at 12.5 kN. If the original rod length and diameter are 20 mm and 15 mm, respectively, determine the rod length and diameter under load. We need to determine the axial strain, then the axial and radial dimensions can be calculated. Because the loading is elastic, we can apply the Young’s modulus and Hooke’s law as follows  eng   E  P A0E  12.5 103N  15103m 2  2 204 109 N m2    3.47 104 L  L0eng  20mm 3.47 104   6.94 103 mm Lf  L0  L  19.9931mm Then relate the axial strain to the radial strain using the Poisson’s ratio  Poisson   applied  0.31 3.47 104   1.0757 104 d  d0  15mm 1.0757 104   1.614 103 mm df  d0  d  15.0016mm 1.36 A 0.5 m long rod of annealed 410 stainless steel was loaded to failure in tension. The rod originally had a square cross section measuring 1.25 cm on a side. What was the load necessary to break the sample? If 85% of the total elongation occurred prior to the onset of localized deformation, compute the true stress at the point of incipient necking. We need to know the tensile strength (the maximum strength before necking commences) and the total elongation at failure. From Table 1.2a, these values are 515 MPa and 35%, respectively. Thus P  A0  1.25 102 m2 515 106 Pa 80.5 kN To determine the true stress at necking, we need the actual cross sectional area, which can be determined by assuming constant volume deformation throughout the period of uniform elongation. Uniformelongation  0.85(35%)  29.75% Deformation and Fracture Mechanics of Engineering Materials, 5th ed. Problem Solutions p. 11/162 Draft document, Copyright R. Hertzberg, R. Vinci, J. Hertzberg 2009 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. so the instantaneous length at necking was 129.75% of the original, or 0.64875 m. Then Ai  A0L0 Li  1.5625 104m20.5m 0.64875m  1.204 104m2  true  P Ai  80.5 103N 1.204 104m2  668.6MPa 1.37 Natural rubber is tested in tension to a maximum extension ratio of =3. The Mooney- Rivlin constants for this material are found to be C1=0.069 MPa and C2= 0.125 MPa. Plot the corresponding uniaxial stress vs. extension ratio behavior over the tested range. Derive an expression for the slope of the function, then determine the secant and tangent moduli at 100% strain. Plot equation 1-14 first, then determine the slope of the line between =1 and =2 (i.e., at 100% strain). The slope is the secant modulus, which is 0.460 MPa for this particular rubber material. To derive an expression for the slope of the function, take the first derivative of stress with respect to extension ratio. The value of the slope at =2 is the tangent modulus, which in this case is 0.219 MPa.   2 C1  C2          1  2        2C1  2C2  2C1  2  2C2  3 d d        2C1  4C1  3  6C2  4 0 100 200 300 400 500 600 700 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3  Stress (kPa) 1.38 Compare the resilience of annealed alloy Ti-6Al-4V with that of annealed stainless steel alloy 304. Then compare the elastic strain energy density just prior to the onset of necking for both alloys. Assume that the Young’s moduli are 114 GPa for the