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Materials Science and Engineering: An Introduction

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Unformatted text preview: REVISED PAGES EQA 184 - Chapter 7 I Dislocations and Strengthening Mechanisms (b) If slip occurs on a (110) plane and in a [111] direction, and the critical resolved shear stress is 30 MPa (4350 psi), calculate the magnitude of the applied tensile stress necessary to initiate yielding. Solution (a) A BCC unit cell along with the slip direction and plane as well as the direction of the applied stress are shown in the accompanying diagram. In order to solve this problem we must use Equation 7.2. However, it is first nec— essary to determine values for 95 and A, where, from the diagram below, (35 is the angle between the normal to the (110) slip plane (i.e., the [119] direc— tion) and the [010] direction, and It represents the angle between [111] and [010] directions. In general, for cubic unit cells, an angle 9 between directions 1 and 2, represented by [uivlwl] and [112921.02], respectively, is equal to (7.6) 6 = cos_1[ uluz + 0102 + wlwz ] V (in2 + of + wf){u§ + v5 + wé) For the determination of the value of 45, let [ululwl] = [110] and [ugvzwzl = [010] such that ¢ _ ms_1{ {1X0} + (1X1) + (0X0) } War + (1)2 + (error + (1)2 + (0):] = cos"(%) = 45° Normal to Slip direction [111] Direction of applied stress [010] However, for )1, we take [u1vlwl] : [111] and [H2U2W2] : [010], and (-1.J(0) + (1X1) + (1X0) V [(—1)2 + (1)2 + (1)2][(0)2 + (1)2 + (0)2] = cos“(%) = 54.7° Thus, according to Equation 7.2, 7R = 0 cos 4: cos A = (52 MPa)(cos 45“)(cos 54.7") : (52 MPa)(%)(%) = 21.3 MPa (3060 PS” )1 — cos_1[ ...
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