rees - basic engineering plasticity - introduction with applications (elsevier, 2006)

[...]... Contracted forms of eq(1.23c) appear, in the two alternative notations, as det(or9 - £r4) = 0 or det(T - aJ) = 0 Expanding eq(1.23c), leads to a cubic (or characteristic) equation (ax- e%[(af - a)(at - d )- r^r w ] - T^IX^O, - o) - ryju] + ^[r^r^ ra a3 - {ax + ay + 0,)^ + (axay + 0^ + ofix - v^ - t£ - tj- )a Ir^r^ - axt^ - ayvj - atr^) = 0 (1.24a) The three roots (the eigen values) to eq( 1.24a) give the... eq( 1.24b) becomes T3-J1T2 + / j T - I / s = 0 (1.25d) Substituting from eqs(1.25a-c), the theorem states that T must satisfy: T 3 - T 2 tr T + V6T [ (tr T) 2 - tr T 2 ] - I detT = 0 (1.25e) Taking the trace of eq( 1,25d) gives an alternative expression for J3: tr T ' — J tr T 2 + / tr T — 3 / = 0 and substituting from eqs( 1.25a and b) gives J3 = - ( t r T ) 3 - - t r T t r T 2 + - t r T 3 3 6 2 3 1.4.2... (1.12c) supplies the shear stress on this plane as r = y/ (r2 - a2) = v' (1402 - 98.962) = 99.03 MPa STRESS ANALYSIS II and eqs(1.13a,b,c) gives its direction cosines as I, = (rx - la)/T= (102.39 - 98.96 cos 67°13')/ 99.03 = 0.647 (a, = 49°41') ms = {ry - ma)/T= (36.24 - 98.96 cos 30°)/ 99.03 = - 0.500 (fls = 120°) ns = (rz - no~)/r= (88.33 - 98.96 cos 71 o 34 r y99.03 = 0.576 (f, = 54°50') from which... = 0.579 Thus a unit vector (eq(L26a)) may be identified with the 1-direction as: i% = 0.516U, + 0.63 lii, + 0.579ut Similarly setting c^ = - 2.37 MPa in eq(L23a) leads to a unit vector for the 2-direction as % = Q.815U, - 0.15311, - 0.560u, Finally, setting c = - 5.83 MPa in eq(1.23a), leads to a unit vector for the 3-direction as % % = 0.265% - 0.761U,, + Q.592us Taking the dot products of ux, 1 and... parallel to each co-ordinate directions This equilibrium condition will appear later with the alternative engineering stress notation (see BASIC ENGMffiRING PLASTICITY eqs(l.lla,b,c)) The Cauehy stress tensor, T, with components &y (where i,j = 1,2, 3), is defined in from eq(i.6b) when the co-ordinates xt are referred to the deformed configuration 1.2.2 General Stress State WitMn a general three-dimensional... substitution into eq( 1.24a) These give ff 3- 1 8 ^ - 2 0 1 0 - 3 6 2 = 0 from which the invariants in eq(1.24b) are identified as: Jy = 18 MPa, J2 = - 201 (MPa)1 and / 3 = 362 (MPa)3 The roots to this cubic are identified with the principal stresses according toflf,> £ > £ as follows: at = 26.2, oj = - 2.37 and a% = - 5.83 MPa Thethreedirection % % cosines for the 1-direction 0j, nt,, MJ, are found from... stress aligned with the 1-direction and the normal to its plane is also in the xr direction Normal stresses will always appear with two similar subscripts in this notation STRESS ANALYSE 7 Different subscripts denote shear stresses, e.g al2 is aligned with the ^-direction but acts on the plane whose normal is aligned with the x,-direction Great care must be token not to confuse generalised co-ordinates:... from the substitution of rx, ry and rz into eq(l 12b) as r 2 = (azcos &+ c^sin a f + (e^sin a + r^cos df - {e^ = (of - axay + o*) sin2*cos2ar+ r w sin 2«[ax(l 2 2 2 sin 1 2a) - r v (oi - 6 0 sin 2arcos = [^(ax00 s i n 2 « - i ^ c o s 2 s ] 2 r = % (oi - o p sin 2ar - r^cos 2m ^sin2 a+ r^sin 2a)1 0 ( 1 - 2sin 2 ^] The respective matrix forms for this plane reduction is found from the transformation eq(1.22a):... invariants in eqs(1.25a-e) become: 1 x y, 2 x y t ^ and J3 = 0 and the principal stress cubic eq( 1.24b) reduces to a quadratic a2 - (ax + a-y)ff+ (trx ay- r v 2 ) = 0 The solution gives the two principal stresses as roots a%2 = ¥i(ax+ cry) ± ¥i A (ax - ayf + 4 r ^ ] where the positive discriminant applies to ov 1.5 Principal Stresses as Co-ordinates Let the co-ordinate axes become aligned with the orthogonal... classical plasticity with experimental data Differences between them have been attributed to a strain history dependence lying within non-radial loading paths Chapter 5 compiles solutions to a number of elastic-perfect plastic structures Ultimate loads, collapse mechanisms and residual stress are among the issues considered from a loading beyond the yield point In Chapter 6 it is shown how large scale plasticity . alt="" BASIC ENGINEERING PLASTICITY BASIC ENGINEERING PLASTICITY This page intentionally left blank BASIC ENGINEERING PLASTICITY An Introduction with Engineering and Manufacturing Applications D Cataloging-in-Publication Data A catalog record for this book is available from the Library of Congress ISBN-13: 97 8-0 -7 50 6-8 02 5-7 ISBN-10: 0-7 50 6-8 02 5-3 For information on all Butterworth-Heinemann. Cataloging-in-Publication Data A catalog record for this book is available from the Library of Congress ISBN-13: 97 8-0 -7 50 6-8 02 5-7 ISBN-10: 0-7 50 6-8 02 5-3 For information on all Butterworth-Heinemann