3. STRAIN
3.1. Deformation and Strain tensor
In prent chapter we examine the deformation geometry of the deformable solid without regard for the actual forces required to produce it. The most obvious and direct method of describing the motion of a continuum solid is to consider the motion of each and every particle making up the solid. If the relative position of every particle is not changed, there is only rigid moving and rotation, then we may consider it as a rigid displacement. If the relative position of every particle is changed, in the same time the initial shape of the body is distorted, then we called there is a deformation. In the following, we will discuss the deformation of elastic-plastic body.
Suppo the distance between two points P o(x o, y o) and P(x,y) is P o P in plane Oxy before deformation. After deformation the two ends of gment P o P moved to P o′(x o′y o′) and P′(x′, y′). Let P o P =s, P o′P′= s ′then the components of vectors s′and s along the x , y axes are:
s x′=s x+ s x
s y′=s y′+s y
The displacement component at point P o is
u o =x o′‑x o
v o =y o′‑y o (3.1) Similarly, at point P the displacement component is(Fig.3.1):
u =x′– x
v =y′– y (3.2) Suppo the displacement u and v are the single-value continuously functions of x and y, then we can expand the displacement at point P in an infinite Taylor ries about point P o, that is:
u = u o + s x + s y + 0 (s x2, s y2 )
v =v o + s x + s y+ 0(s x2, s y2) (3.3) Becau point P is in the neighbourhood of the point P o, therefore the quantity s is sufficiently small, so that we obtain the formula积玉堆金
s x =s x′–s x = (x′-x ) – (x o′-x o ) = s x+s y
s y =s y′–s y = (y′-y) – (y o′-y o )= s x+
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Using the indicial notation and summation convention, the equations
may be written more compactly as
s i = u i ,j s j (3.4) where (i,j =x, y ) in two dimensional ca, ( i,j =x, y, z ) in three dimensional ca, this gives
u i , j = (3.5) which is so called relative displacement tensor.
Fig. 3.1 displacement of a gment
In order to get a clear idea of strain, we shall return by considering the figure 3.1,from which we know there is a rigid displacement only, when the gment s moved to s′, without any deformation. The length of s is not changed, then we obtain
s2 =s′ 2 = ( s x + s x)2 +( s y + s y)2 (3.6)
Expanding above formula (3.6) and neglecting the high-order small quantity s i , we obtain
s2= s2 + 2 (s x s x + s y s y )
From which we have
s x s x + s y s y= 0墨守陈规
or
s i s i = 0 (3.7) From equation (3.4) we obtained
s i u i , j s j = 0 (3.8) or
s x2 +s x s y + s y2 = 0
12341In view of the arbitrary of s x , s y , we have
(3.9) Similar to above discussion for planes Oyz and Oxz we can obtain another three conditions:
= 0
+ = = 0
早上起来恶心干呕Therefore we have
u i , j = --u j ,i (3.10) It means that the relative displacement tensor of rigid-body motion must be a skew-symmetric tensor
Every cond-order tensor can be decompod into two parts uniquely, one is a symmetric tensor and another is a skew-symmetric. Hence we can write u i ,j as the following form:
u i , j = ( u i , j + u j , i ) + ( u i , j – u j , i ) ( 3.11) or
u i , j =i j + i j (3.11a) where
= (3.12)
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ij
= (3.13)
ij
and ij are the strain tensor and rotation tensor respectively for two-ij
dimensional ca.
From equation (3.11) we can easy write the expression of strain tensor and rotation tensor for three-dimensional cas.
In this ca, equation (3.4) may be written as
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s i = ij s j (3.14) If gment s is parallel x-axis, then we have
s x = s, s y = 0
and
= x = = (3.15)
xx
It is clear that x is the unit elongation ( compression) of a unit line element vector paralleled x-axis, and the same to y , z , so called line strain or normal strain.
If vector s1and vector s2are parallel axes O x, O y respectively before deformation,i, j are the unit vector along the direction of O x, O y axes, respectively. ( Fig. 3.2 ) Therefore we have
s1 = i s1 s2 = i s2它们是怎么来的
After deformation s1, s2became to s1’, s2’ , respectively. Then we have
s1′ = i (s1x + s1) + j s1y
