MIT OpenCourseWare http://ocw.mit.edu For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms3.22 Mechanical Properties of MaterialsSpring 2008.Quiz #1 Example Solutions Spring 2008 counterweighthook (~ no mass)40 m10 mgroundq = beam weight/length = 1 kN/m30 mABCD(a)(b)weight due to hoisted I-beamFigure 1: Schematic of crane used to complete construction along the Charles River, indicated with the horizontal element of the crane (a trussed beam) parallel to the ground. (a) Entire crane; (b) Hook-end of the horizontal element. 1. (a) Draw a free body diagram of the horizontal element of the crane (the horizontal span in Fig. 1a), treating it as a beam and indicating all known and unknown applied / reaction forces. You will need to make assumptions about how the crane is connected to the real world, and these should be briefly justified. Solution: The purpose of this question is to confirm the basics of structural mechanics required to implement understanding of material mechanical behavior in context of de-vices. The moment about the vertical support point v, Mv equals zero if we treat it as a pin. However it does not equal zero if we treat it as a clamp. We definitely know that it is not a roller because it is not free to translate. (b) Consider the hook-end of the crane’s horizontal element (Fig. 1b), now loaded by the weight of the steel I-beam. Can member AC be made from a thin steel cable (essentially 3.22 Mechanical Behavior of Materials 1 Prof. K.J. Van VlietMvPvPcwqan elastic rope)? Here, you’re considering the elements of the truss explicitly, so their weight as a whole beam structure (i.e., q) can be neglected. Explain your answer fully. Solution: Because of the tensile applied load at the hook, segment AB is in tension. The sum of the forces in the y-direction must be balanced at node A (see figure below). Therefore, the force along segment AC is in compression to balance the tensile force on AB. Since buckling may occur when the thin steel cable is loaded in compression, it would not be advisable to make the AC member out of cable. Node AFADFACFABθ3.22 Mechanical Behavior of Materials 2 Prof. K.J. Van Vliet������� �������������� ������� 2. (a) We know σij and �ij can be expressed as quadrics, three-dimensional surfaces defined by a cubic polynomial equation of σp or �p. Explain concisely, but fully and accurately, what this fact immediately implies about tensorial stress and strain. Solution: The equation defining the quadratic has three principal roots (principal stresses or strains) along the axes defining normals to planes on which no shear components ex-ist. This absence of shear components in the principal stress state is one of the most important concepts in stress/strain tensor transformations, and is defined explicitly by the quadric or principal root equation that includes the invariants I1, I2, and I3. (It is also true that the quadric form demonstrates symmetry of shear elements in the 2nd rank stress and strain tensors.) (b) You are asked to confirm that your thesis advisor’s following calculations are correct. (He is an expert in electromagnetic material behavior, and often leans on you for help with mechanical behavior.) He has fabricated a metal interconnect line from vapor de-posited Cu, and x-ray diffraction has indicated that all the grains in the polycrystalline line are oriented with all grain boundaries inclined 45o with respect to the Si substrate. When the device is heated, your advisor expects the material at the center of the line to expand uniformly in all directions (neglecting the edge effects at the substrate and free surface). He therefore predicts that this means the line will probably fail because the grains will shear apart at their boundaries. Respectfully explain to him why this predic-tion is incorrect, even if we allow for neglect of edge effects. Solution: In thermal expansion, a hydrostatic strain/stress (depending on the material constraints) forms if edge effects are neglected. By definition, a hydrostatic strain/stress has no shear components. Therefore, the grains will not be under a driving force for shear failure. Some students noted that elastic isotropy is unlikely in this case, which is true because the lines are likely textured (have the same unit cell orientation within each grain). That is probably true, though not explicitly stated. Even so, the shear component for an lin-ear elastic / anisotropic material would be minimal under an isotropic thermal field. (c) He then asks you to check a stress transformation onto the grain boundary planes, cal-culated by your labmate, as shown below. Is this transformation correct? (You do not need to repeat the transformation to answer this.) σij = 1 3 10 4 5 7 8 1 12 → σi�j� = 3 7 4 1 11 8 6 5 2 Solution: No. The trace or first invariant I1 of the original “tensor” and the transformed “tensor” (which is also asymmetric) are not equal. Also, the original stress tensor is not actually a tensor in the first place because it is asymmetric (e.g. σij =� σji). 3.22 Mechanical Behavior of Materials 3 Prof. K.J. Van Vliet� � 3. A polycrystalline, pure silver (Ag) slug is placed in a rigid die as shown, to create custom jewelry. Uniaxial compressive stress σzz = σo is applied normal to the slug face. We will assume the die is well lubricated at the material/die interface. zzAgrigid steel dieFigure 2: Ag slug compressed within rigid die. (a) Determine the orthogonal stresses σxx and σyy in terms of the applied stress and appro-priate elastic constants of the material. Here, note that there are constrained and free surfaces created by the geometry of the die, such that the applied stress is uniaxial but the stress state in the material is more complex. Solution: The applied stress σ is compressive. The slug is constrained in what I as-◦signed to be the y-direction. Therefore, the strain in the y-direction �y and the stress in the x-direction σx are zero. We can then determine the stress in the y-direction by � � �y = σy E − ν σx E + σz E 0 = σy 0 + σ ◦ E − νE σy = −νσ ◦ Finally because the slug is unconstrained in the x-direction, σx = 0. (b) Determine the axial strain �zz in terms of the applied stress and appropriate elastic con-stants of