弹塑性力学第三章
岩土弹塑性力学教学课件(共13章)第3章_应变状态
§3.1 应变状态11
• 三个刚性转动分量及6个应变分量合在一起,才全 面反映了物体变形
xyz x y z xy yz zx
B
B’’ 刚性转动
B’’’
B’
变形
A 刚性平动 A`
§3.1 应变状态12
• 工程应变: ln l0
l0
变形后长度 原始长度
不适用于大变形
• 自然应变/对数应变:
在塑性变形较大时,用-曲线不能真正代表加载和变形的状态。
x y z
• ——弹性体一点的体积改变量
• 引入体积应变有助于简化公式。
• 大于零表示体积膨胀,小于零体积压缩。
• 注意:土力学中塑性体应变符号约定相反。
§3.2 主应变与应变主方向8
应变Lode参数: 为表征偏量应变张量的形式,引入应变Lode参数:
22 3 1 3
1
(1.66)
如果两种应变状态με 相等,表明它们所对应的应变莫尔圆 相似,也即偏应变张量的形式相同。
Vz y
;
zx
Vz x
Vx z
;
§3.3 应变率张量 2
小变形情况下,应变速率分量与应变分量间存在如下关系:
x
Vx x
du x dt
d dt
u x
x
u x
y
Vy y
dv y dt
d v
dt
y
y
v y
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Vz z
z
dw dt
d w dt z
z
w z
线应变速率
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(1.56)
§3.3 主应变与应变主方向 4
由于时间度量的绝对值对塑性规律没有影响,因
弹塑性力学课件第三章
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第三章 本构关系
一、线性弹性体的本构方程——具有一个弹性对称面的线
性弹性体
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第三章 本构关系
一、线性弹性体的本构方程——各向同性弹性体
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第三章 本构关系 一、线性弹性体的本构方程——各向同性弹性体
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第三章 本构关系 一、线性弹性体的本构方程——正交各向异性弹性体
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弹塑性力学课件第三章
第三章 本构关系
本章学习要点:
掌握各项同性材料的广义Hooke定律 掌握弹性应变能密度函数的概念及计算 理解初始屈服、后继屈服以及加卸载的概 念 掌握几个常用的屈服条件 理解弹塑性材料的增量和全量本构关系的 基本概念
第三章 屈服准则
• 下一章来解决材料屈服后的应力应变的本构关系.
弹塑性力学基础---主讲:韩志仁
1. 屈服
物体受到荷载作用后,
随着荷载增大,由弹性状
态到塑性状态的这种过渡,
叫做屈服。
加载路径
2. 屈服条件
屈服点
物体内某一点开始产 生塑性应变时,应力或应 变所必需满足的条件,叫 做屈服条件。
only twist
Twist and extension
著名的Taylor和Quinney铜管拉扭 屈服试验(1931)
弹塑性力学基础---主讲:韩志仁
3. 屈服函数
一般情况下,屈服条件 与应力、应变、时间、温度 等有关,而且是它们的函数, 这个函数F称为屈服函数。
在不考虑时间效应(如应 变率)和温度的条件下:
在不考虑应力主轴旋转 情况下,可以用三个主应力 分量或应力不变量表示:
F( ij ,ij ,t,T ) 0
弹塑性力学基础---主讲:韩志仁
第三章 屈服准则
(yield criteria)
弹塑性力学基础---主讲:韩志仁
塑性模型三要素
屈服条件 流动法则
硬化规律
判断何时 达到屈服
屈服后塑性应变 增量的方向,也 即各分量的比值
决定给定的应力 增量引起的塑性 应变增量大小
弹塑性计算分 析的首要条件
弹塑性力学基础---主讲:韩志仁
这条曲线如图所示的红色曲线. 如果一个应力状态在这条曲线
第三章-弹塑性断裂力学PPT课件
(20)
对弹塑性情况, δ可由弹性的δe和塑性的δp两部分
组成,即:
.
27
e P
(21)
式中, δe为对应于载荷P的裂纹尖端弹性张开位移,
(1)D-B模型假设:裂纹尖端的塑性区沿裂纹线两边 延伸呈尖劈带状;塑性区的材料为理想塑性状态,整 个裂纹和塑性区周围仍为广大的弹性区所包围;塑性
区与弹性区交界面上作用有均匀分布的屈服应力σs 。
.
9
于是,可以认为模型在远场均匀拉应力σ作用下
裂纹长度从2a延长到2c,塑性区尺寸R=c-a,当以带 状塑性区尖端点c为“裂尖”点时,原裂纹(2a)的 端点的张开量就是裂纹尖端张开位移。
按等效原则,令非贯穿裂纹的等于无限大板中心穿透裂纹
的,则等效穿透裂纹长度为:. a*= α2 a
(17)
22
(c)材料加工硬化修正
考虑材料的加工硬化修正,可用流变应力σf代替 屈服点,对于σs =200~400MPa的低碳钢,一般取:
σf =0.5( σs + σb)
(18)
式中σb为材料的抗拉强度。
δ与应变e、裂纹几何和材料性能之间的关系,即引入 应变这一物理量。
由含中心穿透裂纹的宽板拉伸试验,可绘出无量 钢COD即/2esa 与标称应变 e / e s 之间的关系曲线 。
.
16
其中es是相应于材料屈服点σs的屈服应变,a是裂 纹尺寸,标称应变e是指一标长下的平均应变,通常 两个标点取在通过裂纹中心而与裂纹垂直的线上。
R
a
sec
2
s
1
若将 s e c 按级数展开,则 2 s
12 54 sec2s 122s242s
2
当
弹塑性力学第三章
3. STRAIN3.1. Deformation and Strain tensorIn present 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.Suppose 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 segment 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 xs y′=s y′+s yThe displacement component at point P o isu o =x o′‑x ov o =y o′‑y o (3.1) Similarly, at point P the displacement component is(Fig.3.1):u =x′– xv =y′– y (3.2) Suppose 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 series 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) Because point P is in the neighbourhood of the point P o, therefore the quantity s is sufficiently small, so that we obtain the formulas x =s x′–s x = (x′-x ) – (x o′-x o ) = s x+s ys y =s y′–s y = (y′-y) – (y o′-y o )= s x+Using the indicial notation and summation convention, these equationsmay be written more compactly ass i = u i ,j s j (3.4) where (i,j =x, y ) in two dimensional case, ( i,j =x, y, z ) in three dimensional case, this givesu i , j = (3.5) which is so called relative displacement tensor.Fig. 3.1 displacement of a segmentIn 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 segment s moved to s′, without any deformation. The length of s is not changed, then we obtains2 =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 obtains2= s2 + 2 (s x s x + s y s y )From which we haves x s x + s y s y= 0ors i s i = 0 (3.7) From equation (3.4) we obtaineds i u i , j s j = 0 (3.8) ors x2 +s x s y + s y2 = 0In 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+ = = 0Therefore we haveu i , j = --u j ,i (3.10) It means that the relative displacement tensor of rigid-body motion must be a skew-symmetric tensorEvery second-order tensor can be decomposed 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) oru i , j =i j + i j (3.11a) where= (3.12)ij= (3.13)ijand ij are the strain tensor and rotation tensor respectively for two-ijdimensional case.From equation (3.11) we can easy write the expression of strain tensor and rotation tensor for three-dimensional cases.In this case, equation (3.4) may be written ass i = ij s j (3.14) If segment s is parallel x-axis, then we haves x = s, s y = 0and= x = = (3.15)xxIt 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 haves1 = i s1 s2 = i s2After deformation s1, s2became to s1’, s2’ , respectively. Then we haves1′ = i (s1x + s1) + j s1ys2′ = i s2x+j (s2y +s2 )Fig.3.2 The change of angle between two vectorsLet the angle between vectors s1’and s2’, then from the definition of inner product of two vectors, we obtains1′s2′= s1′ s2′ cos (3.16) and s1′s2′= (s1x′ i+s1y′ j) (s2x′ i+s2y′ j )=s1x′ s2x′ +s1y′ s2y′where s1y′ =s1y s2x′ = s2xNeglecting the two-order small quantity of s, we haves1′s2′ = s1s2x + s2s1y (3.17)Substituting this relation into the expression (3.16) and neglecting high-order small quantity again, after some reduction that we can obtaincos(s1s2x+s2s1y) / s1s2 =+Let the change in the angle between line elements s1, s2 , after deformation is, considering that is a small quantity too, then we havecos= cos (/2—) =sin=From xy = yx , we find=xy + yx= 2xySo that the strain component xy is seen to represent simply one- half the change in the angle between line elements lying initially in the x and y axes directions. A similar interpretation also applies to the strain components xz and yz, that isLet the angle between the two lines which is perpendicularly each other and coincident with positive direction of axes O x, O y before deformation have been changed after deformation, if the changed quantity is xy namely shear strain, then= = 2xy = + (3.19)xyor (3.19’ )In the following, we customarily take positive sign (or negative sign ) of shear strain, when the perpendicular angle between two coordinate axes is decreased ( or negative sign ). The strain components are clearly in two-dimensional case are= , y =, xy= + (3.20)xand in three-dimensional case are (in compact form )=u i , i ( i , j= x,y,z ) (3.21)ii= u i, j+u j , i (3.21′)ijwhere u x u, u y v, u z w, and as before by using the von. Karman’s notation:, yyy , zz=z , andxxx= ji (3.22)ijEquation (3.21) is the strain displacement relation:=1/2 ( u i ,j + u j , i ) (i ,j =x,y,z ) (3.23)ijIt is clearly when i =j ,we get a normal strain , ij , we get a shear strain. In addition we have= ji (3.24)ij3.2. Principal Strain and Octahedral StrainIn last chapter, it has been demonstrated that there exist at least three mutually orthogonal planes, which have no shear stress acting on them, that is, the principal planes and their associated principal directions, known as principal axes. In the analysis of strain at a point, such principal axes also exist. As we known, that the principal axis or direction is the direction for which the shear strains would be zero. The corresponding strains are called principal strains. Obviously, the direction of principal strain is coincides with principal direction. This strain is the normal strain. Other word, if the outward normal of a plane is coincides with the principal direction, then this plane is just the principal plane.Suppose vectors s n, coincide with the outward normal line of plane ABC ( Fig. 3.3 ).In the process of deformation the direction of s n is unchanged and only changed its length. If the quantity of changing is s n , then we haves n / s n =s x / s x = s y / s y = s z / s zWhere s x , s y , s z and s x , s y , z are the projection of s n , and s n on axes Ox, Oy, Oz respectively. Considering thatFig. 3.3 Principal strainn(3.25)Then we haves n = n s n (3.26) As we have discussed before, the relative displacement vector corresponding to pure deformation is called the strain vector, then we obtain from equation (3.14)s i= ij s j (3.27) From equations (2.5) and (2.6) we have(ij –ijn ) s j = 0 (3.28) Similarly, we have a cubic algebra equation of n:n 3 – I1′n2 +I2′n – I3′ = 0 (3.29)whereI1′ =I2′ = xy +yz +zx – (xy2 +yz2 +zx2 ) (3.30)I3′==xyz+2xyyzzx –(xyz2+yzx2+zyx2)Where I1′ I2′ and I3′are called the first, second, and third invariant of strain tensor. In terms of the principal strain 1, 2, and3, we haveI1′ =1+2+ 3I2′ =12+23+31 (3.31)I3′ = 123The shear strain for some directions at a point that assume stationary values are called the principal shear strains. The principal shear strain and the corresponding directions can therefore be obtained in exactly the same manner as for stresses. Thus, designating the maximum shear strain by 1, 2, and 3 , we can write1= (2 - 3)2= (1 - 3) (3.32)3= (1 - 2)The octahedral shear strain is= [(x –y)2 +(y –z)2 +(z –x)2 +6(xy2 +yz2 +zx2 )]1/2 (3.33)8or= (I1′ 2 – 3I2′ )1/2 (3.34)83.3. Strain Deviator TensorIn the same manner as for stresses we can obtain the strain deviator tensor and their invariant . As in the case of the stress tensor, the strain tensor can be decomposed into two parts too, a spherical part associated with a change in volume, and a deviatoric part associated with a change in shape distortion. that is ,= e ij +kk ij (3.35)ijwhere e ij is defined here as the strain deviator tensor and 1/3(kk ) is the mean strain or the hydrostatic strain. Thus, the strain deviator tensor e ij becomes(3.36)The invariants of the strain deviator tensor e ij are analogous to those obtained for stress deviator tensor s ij, these deviatoric strain invariants appear in the cubic equation of the determinant equation [e ij- e ij]=0.e3- J1′e2 – J2′e – J3′ = 0 (3.37)whereJ1′ =e ii =0J2′ =e1e2+e2e3+e3e1 (3.38)J3′= e1e2e3orJ1′ = 0J2′ = 1/3 (I1′ – 3I2′ )J3′= 1/27 (2I1′ 3 – 9I1′I2′ +27I3′ ) (3.39) It can be seen that the octahedral shear strain 8is related to the second invariant of the deviator strain tensor J’2, as was the case for stresses:(3.40)Now we introduce two special quantities, which is useful in theory of plasticity, the effective strain(3.41) and shear strain strength (or equivalent shear strain )(3.42) 3.4. Equations of Strain CompatibilityFor static elasticity in the last chapter, it has been pointed out that we must establish the equilibrium equations to ensure that the body is always in a equilibrium state. In the analysis of strain , however, there must be some conditions to be imposed on the strain components so that the deformed body remains continuous. In mathematical review, that the displacement function u, v, and w must be the continuum single-value function in their definite region. After deformation, there are no cracks, no rips, and no folds in whole body. The mathematical restriction on strain components which ensure that this is the case are known as the compatibility equations.To establish these equations, let us first suppose that displacement components do indeed exist. In this case, we have that the strain components2ij = u i,j + u j,I (3.44) It can be shown that the compatibility equations for a simply connected region may be written in the form+ kl , ji – ik , jl – jl ,ik = 0 (3.45)ij , klIn other hand, if we take the second derivative of x with respect to y, and y with respect to x, then after added together we can obtained(3.46) Equation (3.46) is the compatibility equation for two-dimensional problem.Expanding equation (3.45), we get(3.47)These 6 compatibility equations are the necessary and sufficient conditions required to ensure that the strain components give single-valued continuous displacements for a simply connected region. It is clear that the compatibility equations are introduced from the equations of strain-displacement relation. Therefore, if we can find the displacement functions u, v, w and strain components correctly according to strain-displacement relations, then the strain compatibility equations will be satisfied spontaneously.Problems1. Expand , that is write it in conventional notation.2. Given the strain tensorFind : (a) principal strains; (b) principal directions; (c) octahedral shear strain; (d) invariance of strain.Anwser: (a) -2.764 -7.236; (b) inclined to x-axis; (c) (d)3. Prove I1′is a invariant while coordinate system is rot。
考博弹塑性力学,第三章
h/2 h/2
M
x
y
l ( l >>h)
第三章
平面问题的直角坐标解答
本题属于平面应力问题,且为单连体, Φ 若按 Φ 求解, 应满足相容方程及 S = Sσ 上的应力边界条件。 求解步骤: ⑴ 由逆解法得出,可取 Φ = ay ,且满足 4 ∇ Φ=0 ⑵ 求出应力分量 σ x = 6ay, σ y = τ xy = 0 (a)
3Ah 3Ah
o
3Ah
h/2 h/2 l 3Ah (a)
x
y
第三章
平面问题的直角坐标解答
(2)对于坐标轴不同,可以解决不同的问 题。对于图(b)所示的坐标系,可解决矩形截 面梁的偏心受拉问题;对于图(c)所示的坐标 系,则可解决偏心受压问题。
o h 6Ah y l (b) 6Ah y x 6Ah o h l (c) x 6Ah
3
第三章
平面问题的直角坐标解答
⑶ 检验应力边界条件,原则是: a.先校核主要边界(大边界),必须 精确满足应力边界条件。 b.后校核次要边界(小边界),若不 能精确满足应力边界条件,则应用圣维南 原理,用积分的应力边界条件代替。
第三章
平面问题的直角坐标解答
对于主要边界 y = ± h / 2
(σ y ) y=± h/2 = 0, (τ xy ) y =± h / 2 = 0
(a)
( lσ
x
+ mτ xy ) = f x ,
S
( lτ
xy
+ mσ y ) = f y
S
(b )
⑶ 多连体中的位移单值条件。
(c)
第三章
平面问题的直角坐标解答
对于单连体,(c)是自动满足的。只 须满足条件(a)和(b)。 由 Φ 求应力分量的公式:
第三章 弹塑性本构关系
d ij d 0 dσ n 0
p ij
加载准则
意义:只有当应力增量指向加载面的外部时才能产生塑性变形。
3德鲁克塑性公设的评述
德鲁克公设的适用条件:
(1)应力循环中外载所作 的真实功与ij0起点无关;
p ij
ij d ij 0
(2)附加应力功不符合功的 定义,并非真实功
1 屈服曲面的外凸性
0 ( ij ij )dijp | A0 A || d p | cos 0
ij
此式限制了屈服面的形状: 对于任意应力状态,应力增量方向 与塑性应变向量之间所成的夹角不应 该大于90° 稳定材料的屈服面必须是凸的.
(a)满足稳定材 料的屈服面
0 ij
由得屈服条件流动法则硬化规律判断何时达到屈服屈服后塑性应变增量的方向也即各分量的比值决定给定的应力增量引起的塑性应变增量大小本节内容屈服后塑性应变增量的方向也即各分量的比值1加载曲面后继屈服面由单向拉伸试验知道对理想塑性材料一旦屈服以后其应力保持常值屈服应力卸载后再重新加载时其屈服应力的大小也不改变没有强化现象
3.1.4 塑性位势理论与流动法则
与弹性位势理论相类似,Mises于1928年提出塑性 位势理论。他假设经过应力空间的任何一点M,必有 一塑性位势等势面存在,其数学表达式称为塑性位势 函数,记为:
g I1, J 2 , J3 , H 0
g ij , H 0
或
式中, H 为硬化参数。 塑性应变增量可以用塑性位势函数对应力微分的表达 式来表示,即: g p
残余应力增量与塑性 应变增量存在关系:
p p d ij D d ij
式中,D为弹性矩阵。 根据依留申公设,在 完成上述应变循环中, 外部功不为负,即
弹塑性力学第三章
b
y
b
x
图 3-1b
§ 3-1
多项式解答
♦ 同理,应力函数
cy 2
c 0
O
能解决矩形板在 x 方向受 均布拉力(设 c> 0 )或均 布压力 (设 c < 0 ) 的问 题,图3-1c 。
2
2 2Φ 12kxy Φ x 2 3 y 2 0 y h x 2Φ 6ky 2 3k 3 xy xy h 2h
O l y
h x
(2)边界条件:
上下边界
y y h 2
0
2
xy y h 2
h 6k 3k 2 0 3 h 2h
y
图 3-1 a
§ 3-1
多项式解答
可见,应力函数 ax 能
2
2a
O
解决矩形板在y方向受均布 拉力(设a > 0)或均布压 力(设a < 0)的问题。
2a
y 图 3-1a
x
§ 3-1
多项式解答
(2) bxy
b 0
b b
O
x 0, y 0, xy yx b
12 M x 3 y, y 0, xy yx 0 代入式(a),得: h
M x y, y 0, xy yx 0 I 结果与材料力学中完全相同。 对于长度l 远大于深度h 的梁,上面答案 是有实用价值的;对于长度l与深度h 同等大 小的所谓深梁,这个解答是不准确的。
清华大学研究生弹塑性力学讲义 4弹塑性_弹性材料的广义胡克定律
具有单值关系的弹性范围内,σ ∼ ε ′ 也同样具有单值关系,而且当σ ∼ ε 具有线性关系
的线弹性范围内,σ ∼ ε ′ 也同样具有线性关系。也就是说,上述比例极限、弹性极限都
是针对整个均匀变形状态的,而不是针对变形状态的某个应力、应变分量的。试验还
表明,在线弹性范围内横向收缩应变与轴向伸长应变之比是一个常数,即
3. 由于线弹性材料的应力张量与应变张量之间满足线性关系,因此应变能密度函数不 仅可以用应变分量来表示,还可以用应力分量来表示,试导出各向同性弹性材料用 应力分量表示应变能密度函数的公式。
4. 对于线弹性材料,试证明如下卡氏公式:
∂W ∂σ ij
= εij
5. 将应力张量和应变张量分别分解为球形张量和偏斜张量之和,即
⎪⎪σ
22
⎪ ⎪
⎢ ⎢
E E 2222
2233
0
0
0
⎥⎪ ⎥⎪
ε 22
⎪ ⎪
⎪⎪σ ⎨⎪σ
33 23
⎪⎪ ⎬ ⎪
=
⎢ ⎢ ⎢
E3333
0
0
E2323
0
0 0
⎥ ⎥ ⎥
⎪⎪⎨⎪2εε3233
⎪⎪ ⎬ ⎪
(14)
⎪σ ⎪⎪⎩σ
31 12
⎪ ⎪ ⎪⎭
⎢ ⎢ ⎣⎢
sym.
E3131
0 E1212
⎥ ⎥ ⎦⎥
E
2(1 +ν
)
⎛ν ⎜⎝ 1 − 2ν
ε iiε
jj
+
ε ijε ij
⎞⎤ ⎟⎠⎥⎦
(18)
独立常数
E= ν= λ= μ =G = K=
表 1 各向同性弹性体弹性常数间的关系
弹塑性力学第3章
设一点应力:
四面体在所有力的作用下保持力的平衡
px = x l x yx l y + zx lz
py = xy l x y l y + zy lz pz = xz l x yz l y + z lz pi ij l j
x0 y0 z 0
px A= x l x A yx l y A+ zx lz A
sx x m
s1 1 m
sy y m
s2 2 m
sz z m
s3 3 m
偏应力的主轴方向与应力张量的主轴方向一致
J1 sx s y sz 0
2 2 J 2 s x s y s y sz sz s x s xy s2 s yz zx
对应的三个主应力的方向称之为主轴. 求解一点的主应力及主应力方向的基本公式
已知一点的应力为:
x xy xz ij yx y yz zx zy z
3.2.1 一点的应力状态
x xy xz ij yx y yz zy z zx
l x x l x xy l y xz l z l y yx l x y l y yz l z l z zx l x zy l y z l z
分别将 1 , 2 , 3 代入:
1 l x x l x xy l y 13 l z 1 l y yx l x y l y yz l z 1 l z zx l x zy l y z l z
弹塑性力学第三章
(1)11=k(x12+x22) x3 , 22=kx22x3 , 33=0
12=2k x1 x2 x3, 23= 13=0
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32
作业:
(2) 11=k(x12+x22) , 22=kx22 , 33=0, 12=2kx1x2, 23= 13=0
2x32222x22332x222x33
22
u2 x2
33
u3 x3
23
1(u3 2 x2
u2 x3
)
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§3-5 变形协调条件(相容条件)
2x12332x32112x233x11
33
u3 x3
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§3-3 应变张量和转动张量的坐 标变换式
在 xk 坐标系中,已知变形体内任一点应ij和 ’ij 均可以通
过二阶张量的坐标转换式求出它们。
即:
' ij
Qi'kQ
j'l
kl
i'j Qi'kQ j'l kl
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4
§3-1 位移和(工程)应变
工程应变共有六个分量:
三个正应变,正应变以伸长为正,
三个剪应变,剪应变以使直角变小为正。
x3
dx1
dx2
x3
dx3 P
x1
2020/1/23
x2
22dx2
P x1
x2
23
5
§3-2 应变张量和转动张量
应变张量和转动张量是描述一点变形 和刚体转动的两个非常重要的物理量,本 节将讨论一下它们与位移之间关系,在讨 论之前,先介绍一下相对位移矢量和张量.
弹塑性力学第三章 应力与应变讲解
式中:n和s分别为微分面的法线和切线方向的单位 矢量。全应力和应力分量之间有
n pn n
n pn s
pn2
2 n
(3.3)
研究具体问题时,总是在一个可以选定坐标系里进 行。对给定的直角坐标系,全应力还可以沿坐标系 方向进行分解。
p 的单位法向量,它与三个坐标轴之间的夹角余弦为 l1、l2、l3
则该主平面上的应力矢量 n 可表示为
pn n (3.14)
或
px py
l1 l2
(3.15)
pz
l3
式中: 表示主应力
将应力分量表达式(3.7)代入上式,经移项并整理后得
(
x
)l1
设给定的坐标系Oxyz下,某点M的应力张量为
ij yxx
xy y
xz yz
zx zy z
现让该坐标系原点不动,坐标轴任意旋转一个角度而得 到新坐标系Ox’y’z’,新旧坐标关系如下表:
x
y
z
X’ l11 cos(x ', x) l12 cos(x ', y) l13 cos(x ', z)
要使主方向存在,也即要使方程组(3.17)或(3 .18)有 非零解,则其系数行列式必须为零。
x yx zx
xy y
zy
xz yz 0 z
(3.19a)
方程组(3.19)也可以写成
det ij ij 0
(3.19b)
式(3.19)展开后,得
对面)上有9个应力分量。这9个应力分量的整
弹塑性力学第三章
(9)
将(9)式代回(2)式第二项得八面体剪应变 2 γ xz = γ 8 = ± (ε1 − ε 2 ) 2 + (ε 2 − ε3 ) 2 + (ε3 − ε1 ) 2 3
应变张量
确定物体上一点有应变状态的九个应变分量构成应变张量。 即
εx 1 εij = γ yx 2 1 γzx 2
γ xz 2 =0 2 2 )m + n = 0 (εij − εδij )n j = 0, 且ni ni = 1 2 2 γ zy γ zx l + m + n(ε z − ε) = 0 2 2 ⇒ εij − εδij = 0
γ xy
rlε x
(1)
若取x轴与三个主轴1,2,3成等角,这样的方向共有八个, 以此八个方向为法线的平面在该点组成八面体。 1 八面体上有 l1 = m1 = n1 =
3
1 ε x = ε8 = (ε1 + ε 2 + ε3 ) 3 2 这时γ xy = (ε1l2 + ε 2 m2 + ε 3 n2 ) 3 (2) 2 同理γ xz = (ε1l3 + ε 2 m3 + ε 3n3 ) 3 其中 ε8 为八面体正应变,即在矢量x方向上单 位长度的长度变化。 而 γ 8为八面体剪应变,即矢量x和八面体平面 之间原来所夹直角的变化。当以x轴为轴旋转xyz 坐标系时,yz轴即在八面体上变动, γ xy , γ xz也将随 之变动,当其中一个为零时,另一个就是要求的 八面体剪应变。
几何方程
刚体位移
六个应变为零时的位移为刚体位移,即:
ε x = ε y = ε z = γ xy = γ yz = γ zx = 0
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§3-5 变形协调条件(相容条件)
在本章第二节中我们讨论了一点的应变 张量,它包含了一点的变形信息,应变张量
与位移微分关系称为几何方程(共六个)。 u 如果已知变形体的位移 状态, 则由这六
弹塑性力学第三章
第三章 应变分析
§3-1 位移和(工程)应变 §3-2 应变张量和转动张量 §3-3 应变张量和转动张量的坐标变换式 §3-4 主应变、主应变方向、应变张量
的三个不变量
§3-5 变形协调条件(相容条件)
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2
§3-1 位移和(工程)应变
在第二章我们研究了应力张量本身和 体力、面力之间的关系式,即平衡规律。 本章将讨论变形体研究的另一个基本关系: 变形与位移之间的关系。当然要以小变形 假设为基础,位移和形变相对于变形体几 何尺寸是微小的。
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§3-1 位移和(工程)应变
1.1位移
x3
P
P
u
P’
o r x2
x1
变形体任意点P的位移矢量 uuiei
u有三个分量。
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§3-1 位移和(工程)应变
1.2 (工程)应变
工程应变是通常工程中描述物体局部几何 变化,分为正应变和剪应变。
夹角的l l,改(变角量变。形)=两微元线段 (工程)正应变:11、22、33 , (工程)剪应变:12=xy、23=yz、31=zx
或 Uijui,j ijij
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11
§3-2 应变张量和转动张量
其中
ij
1 2
(ui,
j
uj,i )
ij 12(ui, j uj,i)
ij = ji(对称张量), ij = -ji (反对称张量)
而 ij 表示变形体的形变,ij 表示了刚体转动。
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§3-2 应变张量和转动张量
24.11.2020
5
§3-1 位移和(工程)应变
工程应变共有六个分量:
三个正应变,正应变以伸长为正,
三个剪应变,剪应变以使直角变小为正。
x3
dx1
dx2
x3
dx3 P
x1
24.11.2020
x2
22dx2
P x1
x2
23
6
§3-2 应变张量和转动张量
应变张量和转动张量是描述一点变形 和刚体转动的两个非常重要的物理量,本 节将讨论一下它们与位移之间关系,在讨 论之前,先介绍一下相对位移矢量和张量.
以在平面x1 —x2的两个垂直线段PQ、PR 的相对位移来说明并直观看一下ij,ij二阶张
量表示了形变和刚体转动。
x2
R
dx2=1
P
Q
dx1=1
x1
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§3-2 应变张量和转动张量
x2 R
dx2=1
x2 u2 ,1 u2 ,2 R’’ R’
u 1,1 u 1,2 u 2 ,1 u 2 ,2
过二阶张量的坐标转换式求出它们。
即:
' ij
Qi'kQ
j'l
kl
i'j Qi'kQ j'l kl
Q i'k e i'e k Q k' i
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19
§3-4主应变、应变方向应变张量的三 个不变量
确定一点的主应变和应变主方向方法与 求主应力和应力主方向的方法完全一致,求 主应变的方程
3 Ⅰ 2 Ⅱ Ⅲ 0
解出1、2、3 (实根)
、Ⅱ、Ⅲ
分别为应变张量的三个不变量。
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§3-4 主应变、应变方向应变张量的三个不
变量
Ⅰ = 1 1 2 2 3 3 1 2 3 e
——体积应变
Ⅱ = 1 22331
Ⅲ 123
当 1 2 3 时(三个主应变不相等), 三个主方向相互垂直。
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§3-2 应变张量和转动张量
2.1 相对位移矢量和相对位移张量
PQ 平 移P'Q'' 伸 长 + 转 P'Q 动 '
Q ''Q ' d u d r ' d r x3
dr
Q
u+du
——相对位移矢量
P
PuΒιβλιοθήκη rox2x1
Q’’ Q’
P’
P’
dr
24.11.2020
令
U u i,je iej U ie jiej
为一个二阶张量——相对位移张量
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10
§3-2 应变张量和转动张量
2.2 应变张量和转动张量
相对位移张量 ui,j 包含了变形和刚体转动, 为了将两者分开,对 ui,j 进行整理,张量分成 对称和反对称张量之和。
U ij u i,j1 2 (u i,j u j,i) 1 2 (u i,j u j,i)
dx2=1
相对位移
Q’
P’
Q’’
u1 ,2 x1
dx1=1 u1 ,1
Q P dx1=1
x1
u1、u2
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§3-2 应变张量和转动张量
x2
22=u2 ,2 21= (u2 ,1 +u1 ,2 )/ 2
(+)/2
+
x1
12=(u1 ,2 +u2 ,1 ) /2 11=u1 ,1
17
§3-2 应变张量和转动张量
比较工程应变定义和应变张量,可得:
11 12 13 11 212 213 21 22 23221 22 223 31 32 33 231 232 33
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§3-3 应变张量和转动张量的坐 标变换式
在 xk 坐标系中,已知变形体内任一点应 变张量 kl 和转动张量 kl ,则在新笛卡尔坐 标系x’i中此点应变张量’ij和 ’ij 均可以通
为 ,其大小 3:
31 2(1 2 2)1 1 2(e 1213 2e 212 3 )1
类似可得,其它两个坐标平面转动矢量,
2e2
1e1
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§3-2 应变张量和转动张量
综合三个坐标面的转动矢量 :
kek 12eij kijek
为转动张量的对偶矢量。
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x2
21=(u2 ,1 -u1 ,2 ) /2
12= (u1 ,2 -u2 ,1 ) /2
x1
11,12= 21,22 纯变形 12= -21 纯转动
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§2.33-转2 动应张变量的张对量偶和矢转量动张量
于一由个纯沿刚x体3 轴转方动向可的见转,动矢12=量-321e,3,正方好向相当e3
8
§3-2 应变张量和转动张量
2. 1 相u 对位移u矢ie 量i和相对位移张量
du ei
ui x j
dxj
——( a)
而
r xjej drdjxej
djxejdr——(b)
将(b)式代入(a)式,得
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9
§3-2 应变张量和转动张量
duui,jeiejdr
根据商法则 duUdr