? 16.21 Techniques of Structural Analysis and Design Spring 2003 Unit #3 - Kinematics of deformation Figure 1: Kinematics of deformable bodies Deformation described by deformation mapping : x = ?(x) (1) 1 ? ? ? ? ? We seek to characterize the local state of deformation of the material in a neighborhood of a point P . Consider two points P and Q in the undeformed: P : x = x 1 e 1 + x 2 e 2 + x 3 e 3 = x i e i (2) Q : x + dx = (x i + dx i )e i (3) and deformed P ? : x = ? 1 (x)e 1 + ? 2 (x)e 2 + ? 3 (x)e 3 = ? i (x)e i (4) Q ? : x ? + dx ? = ? i (x) + d? i e i (5) con?gurations. In this expression, dx ? = d? i e i (6) Expressing the di?erentials d? i in terms of the partial derivatives of the functions ? i (x j e j ): ?? 1 ?? 1 ?? 1 d? 1 = dx 1 + dx 2 + dx 3 , (7) ?x 1 ?x 2 ?x 3 and similarly for d? 2 ,d? 3 , in index notation: ?? i d? i = dx j (8) ?x j Replacing in equation (5): Q ? : x ? + dx ? = ? i + ?? i dx j e i (9) ?x j dx ? i = ?? i dx j e i (10) ?x j ?→ We now try to compute the change in length of the segment PQ which ??→ deformed into segment P ? Q ? . Undeformed length (to the square): ds 2 = ?dx? 2 = dx · dx = dx i dx i (11) 2 ? ?? ? Deformed length (to the square): (ds ? ) 2 = ?dx ? ? 2 = dx ? · dx ? = ?? i dx j ?? i dx k (12) ?x j ?x k ?→ The change in length of segment PQ is then given by the di?erence between equations (12) and (11): (ds ? ) 2 ? ds 2 = ?? i dx j ?? i dx k ? dx i dx i (13) ?x j ?x k We want to extract as common factor the di?erentials. To this end we observe that: dx i dx i = dx j dx k δ jk (14) Then: (ds ? ) 2 ? ds 2 = ?? i dx j ?? i dx k ? dx j dx k δ jk ?x j ?x k ? ?? i ?? i ? = ? δ jk dx j dx k (15) ?x j ?x k 2? jk : Green-Lagrange strain tensor Assume that the deformation mapping ?(x) has the form: ?(x) = x + u (16) where u is the displacement ?eld. Then, ?? i ?x i ?u i ?u i = + = δ ij + (17) ?x j ?x j ?x j ?x j and the Green-Lagrange strain tensor becomes: ? ?u m ?? ?u m ? 2? ij = δ mi + ?x i δ mj + ?x j ? δ ij (18) ?u i ?u j =? δ ij + ?x j + ?x i + ?u m ?u m ?? δ ij ?x i ?x j 3 ? ? Green-Lagrange strain tensor : (19) When the absolute values of the derivatives of the displacement ?eld are much smaller than 1, their products (nonlinear part of the strain) are even smaller and we’ll neglect them. We will make this assumption throughout this course (See accompanying Mathematica notebook evaluating the limits of this assumption). Mathematically: ? ?u i ? ?u m ?u m ? ? ? 1 ? ～ 0 (20) ?x j ?x i ?x j We will de?ne the linear part of the Green-Lagrange strain tensor as the small strain tensor : ? ij = 1 2 ? ?u i ?x j + ?u j ?x i ? (21) Transformation of strain components Given: ? ij , e i and a new basis ?e k , determine the components of strain in the new basis ?? kl ? ? ?? ij = 1 ? ?u i + ?u j ? (22) ? ?2 ?x j ?x i We want to express the expressions with tilde on the right-hand side with their non-tilde counterparts. Start by applying the chain rule of di?erentia- tion: ? ??u i ?u i ?x k = (23) ? ??x j ?x k ?x j Transform the displacement components: u m e m = u l e l (24)u = ? ? u m (? e i ) = u l (e l · ?? e m · ? e i ) (25) u m δ mi = u l (e l · ?? e i ) (26) u i = u l (e l · ?? e i ) (27) ? ij = 1 2 ? ?u i ?x j + ?u j ?x i + ?u m ?x i ?u m ?x j ? 4 take the derivative of u i with respect to x k , as required by equation (23):? ? ?u i ?u l = (e l · ?e i ) (28) ?x k ?x k and take the derivative of the reverse transformation of the components of the position vector x: x = x j e j = x k ?? e k (29) x j (e j · e i ) = x k (?? e k · e i ) (30) x j δ ji = x k (?? e k · e i ) (31) x i = x k (?? e k · e i ) (32) ?x i ? ?x k = (? e k · e i ) = (?e k · e i ) = δ kj (? e j · e i ) (33) ? ? ?x j ?x j Replacing equations (28) and (33) in (23): ? ?u i ?u i ?x k ?u l e i )(?= = (e l · ? e j · e k ) (34) ? ??x j ?x k ?x j ?x k Replacing in equation (22): 1 ? ?u l ?u l ? e i )(? e j )(??? ij = 2 ?x k (e l · ? e j · e k ) + ?x k (e l · ? e i · e k ) (35) Exchange indices l and k in second term: 1 ? ?u l ?u k ? e i )(? e j )(??? ij = 2 ?x k (e l · ? e j · e k ) + ?x l (e k · ? e i · e l ) 1 ? ?u l ?u k ? (36) = 2 ?x k + ?x l (e l · ? e j · e k )e i )(? Or, ?nally: e i )(??? ij = ? lk (e l · ? e j · e k ) (37) Compatibility of strains Given displacement ?eld u, expression (21) allows to compute the strains components ? ij . How does one answer the reverse question? Note analogy with potential-gradient ?eld. Restrict the analysis to two dimensions: ?u 1 ?u 2 ?u 1 ?u 2 ? 11 = , ? 22 = , 2? 12 = + (38) ?x 1 ?x 2 ?x 2 ?x 1 5 Di?erentiate the strain components as follows: ? 2 ? ? ? 3 u 1 ? 3 u 2 ?x 1 ?x 2 2? 12 = ?x 1 ?x 2 + ?x 2 (39) 2 1 ?x 2 ? 2 ? 11 ? 3 u 1 = ?x 2 ?x 1 ?x 2 (40) 2 2 ? 2 ? 22 ? 3 u 2 = ?x 2 ?x 2 ?x 2 (41) 1 1 and conclude that: ? 2 ? 12 ? 2 ? 11 ? 2 ? 22 2 = + ?x 1 ?x 2 ?x 2 ?x 2 (42) 2 1 6