Transcript 平面问题的直角坐标求解
Theory of Elasticity
弹性力学
Chapter 8
Two-Dimensional Solution
平面问题的直角坐标求解
Content(内容)
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Introduction(概述)
Mathematical Preliminaries (数学基础)
Stress and Equilibrium(应力与平衡)
Displacements and Strains (位移与应变)
Material Behavior- Linear Elastic Solids(弹性应力应变关系)
Formulation and Solution Strategies(弹性力学问题求解)
Two-Dimensional Formulation (平面问题基本理论)
Two-Dimensional Solution (平面问题的直角坐标求解)
Two-Dimensional Solution (平面问题的极坐标求解)
Three-Dimensional Problems(三维空间问题)
Bending of Thin Plates (薄板弯曲)
Plastic deformation – Introduction(塑性力学基础)
Introduction to Finite Element Mechod(有限元方法介绍)
Chapter 1
Page
1
Two-Dimensional Solution in
Cartesian Coordinate
• 8.1 Cartesian Coordinate Solutions Using
Polynomials(直角坐标下的多项式解答)
• 8.2 Uniaxial Tension of a Beam(梁的单轴拉
伸问题)
• 8.3 Bending of a Beam by Uniform
Transverse Loading(受均匀横向载荷的梁弯
曲问题)
Chapter
8
Page
2
8.1 Cartesian Coordinate Solutions Using
Polynomials (直角坐标下的多项式解答)
ui , ij , ij ; , , Fi 0
3D
15 unknowns including 3 displacements, 6 strains, and 6 stresses.
x xy
Fx 0 2
2
x
y
( 2 2 )( x y ) 0
x y
xy y
Fy 0
x
y
2 2 2 4 0
4
x
x y
y
4
4
4
1 unknowns
Chapter
8
Page
3
2D
8.1 Cartesian Coordinate Solutions Using
Polynomials (直角坐标下的多项式解答)
General Solution Strategies(求解方法)
1、Direct Method(直接法)
Direct integration of the field equations(直接积分场方程)
ui , ij , ij ; , , Fi 0
Or stress and/or displacement formulations (得到应
力/位移方程.)
Chapter
8
Page
4
8.1 Cartesian Coordinate Solutions Using
Polynomials (直角坐标下的多项式解答)
Example:
Stretching of Prismatic Bar Under Its Own Weight (受自重
的等截面杆)
Fx Fy 0, Fz g
x yx zx
x
y f xxy 0, yz zx 0
x
y
z
The
equilibrium
equations
reduce to
y yz xy
f y 0,
(平衡方程简化为)
y
z
x
z xz yz
z
Fz 0。
Fz g
z
x
y
z
Chapter
8
Page
4
8.1 Cartesian Coordinate Solutions Using
Polynomials (直角坐标下的多项式解答)
Example:
z
Fz g
z
boundary condition
σz |z=0=0
z ( z) gz
using Hooke’s law
gz
v gz
z
, x y
E
E
xy xz yz 0
Chapter
8
Page
5
8.1 Cartesian Coordinate Solutions Using
Polynomials (直角坐标下的多项式解答)
Example:
gz
v gz
z
, x y
E
E
xy xz yz 0
integrating the strain-displacement relations:
with boundary conditions of zero displacement
and rotation at point A(积分应变-位移方程,加上A点
位移和转动都为0)
v gxz
v gzy
u
,v
E
E
g 2
z v( x 2 y 2 ) l 2
w
2E
Chapter
8
Page
5
8.1 Cartesian Coordinate Solutions Using
Polynomials (直角坐标下的多项式解答)
2、Inverse Method(逆解法)
particular displacements or stresses are selected that satisfy the
basic field equations. A search is then conducted to identify a
specific problem that would be solved by this solution field.(选择满
足相容方程的应力函数,再根据应力边条和几何边条找出能用所选取的应
力函数解决的问题.)
φ
2
x 2 FX x
y
2
y 2 FY y
x
2
xy
xy
Chapter
8
4
4
4
2 2 2 4 0
4
x
x y
y
Boundary Conditions
geometry
Page
6
8.1 Cartesian Coordinate Solutions Using
Polynomials (直角坐标下的多项式解答)
3、Semi-Inverse Method(半逆解法)
part of the displacement and/or stress field is specified, and the other
remaining portion is determined by the fundamental field equations
(normally using direct integration) and the boundary conditions.(假定部
分或全部应力分量为某种形式的双调和函数,从而导出应力函数,再考察由这个
应力函数得到的应力分量是否满足全部边界条件)
φ
σ(part)
2
x 2 FX x
y
FY y
2
x
2
xy
xy
y
Chapter
2
8
σ
4
4
4
2 2 2 4 0
4
x
x y
y
Page
7
B. C.?
8.1 Cartesian Coordinate Solutions Using
Polynomials (直角坐标下的多项式解答)
Inverse Method in terms of Polynomials:(多
项式解法)
1:one-order Polynomials (一次多项式)
f
2
x
y
y
2
2 f
FX x
Assume Fx=Fy=0
x y xy 0
FY y
x
2 f
xy
xy
2
f C0 C1x C2 y
Chapter
One-order Polynomials is fit for zero body force,
zero stress state.(适用于零体力,零面力情况)
8
Page
8
8.1 Cartesian Coordinate Solutions Using
Polynomials (直角坐标下的多项式解答)
2:two-order Polynomials (二次多项式)
f C1x2 C2 xy C3 y2
x
y
2 f
y
2
2 f
FX x
FY y
x
2 f
xy
xy
2
x 2C3
y 2C1
Assume Fx=Fy=0
Chapter
xy C2
two-order Polynomials is fit for a uniformity
distribution of stress(适用于均匀应力分布)
8
Page
8
8.1 Cartesian Coordinate Solutions Using
Polynomials (直角坐标下的多项式解答)
3:general states(推广到无穷阶)
f ( x, y ) Amn x y
m
n
y
m 0 n 0
m + n 1
x=0, y=0,,xy=0
second-order
a constant stress field
third-order
a linear distribution of stress
higher-order
……
Chapter
8
x
Page
9
2 f
y 2
2 f
x 2
2 f
xy
xy
8.1 Cartesian Coordinate Solutions Using
Polynomials (直角坐标下的多项式解答)
f A40 x4 A22 x2 y2 A40 y4
满足双调各方程:
4
4
4
2 2 2 4 0
x 4
x y
y
A40 x4 A22 x2 y2 A40 y4 0
8.1
specifies one constant in terms of the other two leaving two constants
to be determined by the boundary conditions.(由上式8.1再加上边界条件
就可求出所有系数)
Chapter
8
Page
9
8.1 Cartesian Coordinate Solutions Using
Polynomials (直角坐标下的多项式解答)
f ( x, y ) Amn x m y n
m 0 n 0
满足双调各方程:
4
4
4
2 2 2 4 0
x 4
x y
y
( m 2)( m 1)m(m 1) Am 2,n 2 2m(m 1)n( n 1) Amm
( n 2)( n 1)n( n 1) Am2,n 2 0
the general relation that must be satisfied to ensure that
the polynomial grouping is biharmonic(对于任意阶多项式要
满足双调和方程)
Chapter
8
Page
10
8.2 Uniaxial Tension of a Beam
(单轴拉伸梁)
Problem:
plane stress case
Saint Venant approximation to the more general case with
nonuniformly distributed tensile forces at the ends x =±l. (由圣维南
原理可知对于在x =±l处拉力分布不均但静力等效的情况也适用)
Solution( inverse method) (逆解法):
The boundary conditions
(边界条件):
Chapter
8
( y ) y c 0, ( xy ) y c 0
( x ) x l T , ( xy ) x l 0
Page
11
8.2 Uniaxial Tension of a Beam
(单轴拉伸梁)
求应力函数Φ
constant stresses on each of the beam’s boundaries:
A02 y2
2
2
2
x 2 , y 2 , xy
y
x
xy
polynomial is biharmonic
Boundary condition
x 2 A02 , y xy 0
( x ) x l T , ( xy ) x l 0
Therefore, this problem is given by:
A02 T 2
x T , y xy 0
Chapter
8
Page
12
8.2 Uniaxial Tension of a Beam
(单轴拉伸梁)
求u ,ε
u
1
T
x ( x v y )
x
E
E
v
1
T
y ( y v x ) v
y
E
E
Integral(积分)
xy
u v
2 xy
0 f ( y) g ( x) 0
y x
f g 0
T
u x
E
T
v y
E
Chapter
T
u x f ( y)
E
T
v v y g ( x )
E
g ( x) f ( y) cons tan t
f ( y) o y uo
g ( x) o x vo
x T / E
y T / E
Page
x T , y xy 0
8.2 Uniaxial Tension of a Beam
(单轴拉伸梁)
inverse method(逆解法)
A02 y
x 2 A02 , y xy 0
2
Physical Equations
e
Geometrical
Equations
4
4
4
2 2 2 4 0
4
x
x y
y
Chapter
8
( x ) xl T , ( xy ) xl 0
Page
14
u
8.3 Bending of a Beam by Uniform Transverse
Loading(受均匀横向载荷的梁弯曲问题)
q
1
ql
ql
h/2
h/2
z
y
x
l
l
y
Airy stress function(艾里应力函数)
stress field(应力场)
Boundary Conditions(边界条件)
compare with elementary strength of materials(和材力结果相比)
Chapter
8
Page
15
8.3 Bending of a Beam by Uniform Transverse
Loading(受均匀横向载荷的梁弯曲问题)
q
1
ql
ql
h/2
h/2
z
x
y
l
y
∵ q =const.
y f ( y)
2) Format of Stress function
l
2
y 2 f ( y)
x
plane stress conditions
(semi-inverse method 半逆解法)
1, Airy stress function
Integrate above:
xf ( y ) f1 ( y )
Stress Component
Force
x2
x
M (主要由弯矩引起;)
x
f ( y ) xf1 ( y ) f 2 ( y )
Q (主要由剪力引起;)
xy
2
f ( y), f1 ( y), f 2 ( y) to be determined
q (由 q 引起)
y
1) Stress components
Chapter
8
Page
16
8.3 Bending of a Beam by Uniform Transverse
Loading(受均匀横向载荷的梁弯曲问题)
xf ( y ) f1 ( y )
x 2
x
f ( y) xf1 ( y) f 2 ( y)
2
3)satisfy the biharmonic equation
4 0
4
4
4
4
4
2
x 4
x 2y 2 y 4
4
0
4
x
Chapter
8
4 x 2 ( 4)
( 4)
( 4)
f
(
y
)
xf
(
y
)
f
1
2 ( y)
4
y
2
4
2 2 2 2 f ( 2) ( y)
x y
Page
x 2 ( 4)
f ( y ) xf1( 4) ( y ) f 2( 4) ( y )
2
2 f ( 2) ( y ) 0
17
8.3 Bending of a Beam by Uniform Transverse
Loading(受均匀横向载荷的梁弯曲问题)
x ( 4)
f ( y ) xf1( 4) ( y ) f 2( 4) ( y ) 2 f ( 2) ( y ) 0h/
h/
2
2
2
q
1
2
y
关于 x 的二次方程,且要求 -l≤ x ≤ l 内方程均成立。
f
( 4)
( y) 0
( 4)
1
f
( y) 0
f ( y) Ay3 By 2 Cy D
f1 ( y) Ey3 Fy 2 Gy
2
此处略去了f1(y)中的常数项
x
f ( y) xf1 ( y) f 2 ( y)
2
Chapter
8
Page
f 2 ( y) 2 f
( 4)
f 2 ( y)
q
lx
q
zl
l y
( 2)
l
( y) 0
A 5 B 4
y y Hy 3 Ky 2
10
6
x2
( Ay 3 By 2 Cy D) x( Ey 3 Fy 2 Gy)
2
A
B
( y 5 y 4 Hy 3 Ky 2 )
10
6
18
8.3 Bending of a Beam by Uniform Transverse
Loading(受均匀横向载荷的梁弯曲问题)
x2
( Ay 3 By 2 Cy D) x( Ey 3 Fy 2 Gy)
2
A 5 B 4
(
y y Hy 3 Ky 2 )
10
6
9 unknown
coefficients
2, stress field
2 x 2
x 2
(6 Ay 2 B) x(6 Ey 2 F ) 2 Ay 3 2 By 2 6 Hy 2 K
y
2
2
y 2
x
Ay3 By 2 Cy D
2
xy
xy
x(3 Ay 2 2By C ) (3Ey 2 2Fy G)
Chapter
8
Page
19
8.3 Bending of a Beam by Uniform Transverse
Loading(受均匀横向载荷的梁弯曲问题)
3, Boundary Conditions
2
x 2
y
1) Symmetry Condition
2
y 2
x
Ay3 By 2 Cy D
2
xy
xy
x(3 Ay 2 2By C ) (3Ey 2 2Fy G)
q
q
lx
q
l
x , y
xy
x2
(6 Ay 2B) x(6 Ey 2 F ) 2 Ay 3 2 By 2 6 Hy 2 K
2
6 Ey 2 F 0
3Ey 2 2Fy G 0
l
l y
E F G 0
—— x 的偶函数
x2
x (6 Ay 2 B) 2 Ay 3 2 By 2 6 Hy 2 K
2
—— x 的奇函数
y Ay3 By 2 Cy D
xy x(3 Ay 2 2By C )
Chapter
8
Page
20
8.3 Bending of a Beam by Uniform Transverse
Loading(受均匀横向载荷的梁弯曲问题)
q
2) Boundary Conditions
ql
ql
a) Top and bottom(上下面,主要)
h
y , y q;
2
h
y , y 0;
2h
y , xy 0;
2
x
h2
3A
Bh C 0
42
h
3A
Bh C 0
4
h3
h2
h
A B C D q
8
4
2
h3
h2
h
A B C D 0
8
4
2
x2
x (6 Ay 2 B) 2 Ay 3 2 By 2 6 Hy 2 K
2
y Ay3 By 2 Cy D
8
l
2q
A 3 ,
h
B 0,
3q
C
2h
q
D
2
xy x(3 Ay 2 2By C )
Chapter
l y
Page
21
8.3 Bending of a Beam by Uniform Transverse
Loading(受均匀横向载荷的梁弯曲问题)
6q 2
4q 3
x 3 x y 3 y 6 Hy 2 K
h
h
2q 3 3q
q
ql
y 3 y
y
h
2h
2
xy
6q 2 3q
3 xy
x
h
2h
q
ql
x
l y
b) End conditions(左右边界,次要)
x l,
x
x l
h
h
y
2
2
0
xy
x l
h
h
y
2
2
未知
Using Saint-Venant’s Principle
轴力 N = 0;
statically equivalent force
弯矩 M = 0;
剪力 Q = -ql;
Chapter
8
Page
22
l
8.3 Bending of a Beam by Uniform Transverse
Loading(受均匀横向载荷的梁弯曲问题)
轴力 N = 0;
弯矩 M = 0;
剪力 Q = -ql;
h
2
h
2
h
2
h
2
h
2
h
2
N x x l dy 0
M x x l ydy 0
Q xy dy ql
6q
4q
x 3 x 2 y 3 y 3 6 Hy 2 K
h
h
y
2q 3 3q
q
y
y
h3
2h
2
6q 2 3q
xy 3 xy x
h
2h
Chapter
8
Page
x l
K 0
ql 2
q
H 3
h
10h
自动满足
2
6q 2
y
y
3
2
x 3 (l x ) y q (4 2 )
h
h h 5
2
q y 2 y
y 1 1
2 h
h
6q h 2
2
xy 3 x y
h 4
23
8.3 Bending of a Beam by Uniform Transverse
Loading(受均匀横向载荷的梁弯曲问题)
4, compare with elementary strength of materials
2
6q 2
y
y
3
x 3 (l x 2 ) y q (4 2 )
h
h 5
2h
M
y y2 3
x
y q 4 2
I
h h 5
2
q y 2 y
y 1 1
2 h
h
q y 2 y
y 1 1
2 h
h
6q h 2
2
xy 3 x y Q qx
h 4
2
q
ql
ql
x
l y
Chapter
l
8
h
y2
S
8
2
1 3
I h
12
xy
q 2
M (l x 2 )
2
Page
24
QS
bI
8.3 Bending of a Beam by Uniform Transverse
Loading(受均匀横向载荷的梁弯曲问题)
4, compare with elementary strength of materials
M
y y 3
x
y q 4 2
I
h h 5
2
q y 2 y
y 1 1
2 h
h
2
xy
QS
bI
Chapter
8
第一项与材力结果相同,为主要项。
第二项为修正项。当 h / l<<1,该项误差
很小,可略;当 h / l较大时,须修正。
为梁各层纤维间的挤压应力,材力中不
考虑。
与材力中相同。
Page
25
8.3 Bending of a Beam by Uniform Transverse
Loading(受均匀横向载荷的梁弯曲问题)
M
y y2 3
x
y q 4 2
I
h h 5
2
q y 2 y
y 1 1
2 h
h
q
ql
ql
x
l y
l
y τ xy
x ()
xy
( )
Chapter
8
Page
26
QS
bI
Vocabulary(词汇)
多项式
单轴的
梁
均匀的
逆解法
半逆解法
双调和的
艾里应力函数
Polynomials
Uniaxial
Beam
Uniform
Inverse Method
Semi-Inverse Method
Biharmonic
Airy stress function
Chapter
8
Page
27
Homework
o
x
b
1:设有矩形截面的长竖柱,密度为ρ,在一边侧面上受
均布剪力q,如图1,试求应力分量.
q
提示:可假设σx=0,或假设τxy=f(x),或假设σy如材料力
学中偏心受压公式所示.上端边界条件如不能精确满
足,可应用圣维南原理.
ρg
y
图1
o
x
α
ρg
2:设图2中的三角形悬臂梁只
受重力作用,而梁的密度为ρ,
试用纯三次式的应力函数求
解.
y
图2
Chapter
8
Page
27
Homework
y
o
3:挡水墙的密度为ρ1,厚度为b,图3,水的密
度为ρ2,试求应力分量.
b/2 b/2
ρ2 g
ρ1g
(提示:可假设σy=xf(y)上端的边界条件如不
能精确满足,可应用圣维南原理,求出近似的
解答)
x
图3
Chapter
8
Page
27