theory of elasticity 弹性力学
DESCRIPTION
Theory of Elasticity 弹性力学. Chapter 8 Two-Dimensional Solution 平面问题的直角坐标求解. Theory of Elasticity. Chapter. Page. Content (内容). 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) - PowerPoint PPT PresentationTRANSCRIPT
Theory of Elasticity弹性力学Chapter 8
Two-Dimensional Solution
平面问题的直角坐标求解
Chapter Page
Content (内容)
1 1
1. Introduction (概述)2. Mathematical Preliminaries (数学基础)3. Stress and Equilibrium (应力与平衡)4. Displacements and Strains (位移与应变)5. Material Behavior- Linear Elastic Solids (弹性应力应变关
系 )6. Formulation and Solution Strategies (弹性力学问题求
解)7. Two-Dimensional Formulation (平面问题基本理论)8. Two-Dimensional Solution (平面问题的直角坐标求解)9. Two-Dimensional Solution (平面问题的极坐标求解)10. Three-Dimensional Problems (三维空间问题)11. Bending of Thin Plates (薄板弯曲)12. Plastic deformation – Introduction (塑性力学基础)13. Introduction to Finite Element Mechod (有限元方法介
绍)
Chapter Page
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 )
8 2
Chapter Page
8.1 Cartesian Coordinate Solutions Using Polynomials (直角坐标下的多项式解答)
8 3
15 unknowns including 3 displacements, 6 strains, and 6 stresses.
3 D
2 D02
4
4
22
4
4
4
yyxx
1 unknowns
, , ; , , 0i ij ij iu F
0
0
xyxx
xy yy
Fx y
Fx y
2 2
2 2( )( ) 0x yx y
Chapter Page
8.1 Cartesian Coordinate Solutions Using Polynomials (直角坐标下的多项式解答)
8 4
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 )
Or stress and/or displacement formulations ( 得到应力 / 位移方程 .)
, , ; , , 0i ij ij iu F
Chapter Page
8.1 Cartesian Coordinate Solutions Using Polynomials (直角坐标下的多项式解答)
8 4
Example:Stretching of Prismatic Bar Under Its Own Weight ( 受自重的等截面杆 )
The equilibrium equations reduce to( 平衡方程简化为 )
0
0
x y z
x y xy yz zx
F F F g
,
zzF g
z
Chapter Page
8.1 Cartesian Coordinate Solutions Using Polynomials (直角坐标下的多项式解答)
8 5
boundary conditionσz |z=0=0
using Hooke’s law
Example: zzF g
z
( )z z gz
0
z x y
xy xz yz
gz v gz
E E
,
Chapter Page
8.1 Cartesian Coordinate Solutions Using Polynomials (直角坐标下的多项式解答)
8 5
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)
Example:
0
z x y
xy xz yz
gz v gz
E E
,
2 2 2 2
2
v gxz v gzyu v
E Eg
w z v x y lE
,
( )
Chapter Page 8 6
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 .)
024
4
22
4
4
4
yyxx
φ2
2
2
2
2
x X
y Y
xy
F xy
F yx
x y
Boundary Conditionsgeometry
8.1 Cartesian Coordinate Solutions Using Polynomials (直角坐标下的多项式解答)
Chapter Page
8.1 Cartesian Coordinate Solutions Using Polynomials (直角坐标下的多项式解答)
8 7
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 )
3 、 Semi-Inverse Method( 半逆解法 )
σ ( part) φ σ
024
4
22
4
4
4
yyxx
B. C.?2
2
2
2
2
x X
y Y
xy
F xy
F yx
x y
8.1 Cartesian Coordinate Solutions Using Polynomials (直角坐标下的多项式解答)
0 1 2f C C x C y 2
2
2
2
2
fx X
fy Y
fxy
F xy
F yx
x y
Chapter Page 8 8
Inverse Method in terms of Polynomials:( 多项式解法 )
1:one-order Polynomials ( 一次多项式 )
Assume Fx=Fy=0
0x y xy
One-order Polynomials is fit for zero body force, zero stress state.( 适用于零体力 , 零面力情况 )
8.1 Cartesian Coordinate Solutions Using Polynomials (直角坐标下的多项式解答)
2 21 2 3f C x C xy C y
2
2
2
2
2
fx X
fy Y
fxy
F xy
F yx
x y
Chapter Page 8 8
2:two-order Polynomials ( 二次多项式 )
Assume Fx=Fy=0 3
1
2
2
2x
y
xy
C
C
C
two-order Polynomials is fit for a uniformity distribution of stress( 适用于均匀应力分布 )
8.1 Cartesian Coordinate Solutions Using Polynomials (直角坐标下的多项式解答)
Chapter Page 8 9
m + n 1 x=0, y=0,,xy=0
second-order a constant stress field
third-order a linear distribution of stress
higher-order ……
0 0
m nf mn
m n
x y A x y
( , )
2
2
2
2
2
fx
fy
fxy
y
x
x y
3:general states( 推广到无穷阶 )
8.1 Cartesian Coordinate Solutions Using Polynomials (直角坐标下的多项式解答)
Chapter Page 8 9
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 )
4 2 2 440 22 40f A x A x y A y
4 2 2 440 22 40 0A x A x y A y
024
4
22
4
4
4
yyxx
满足双调各方程 :
8.1
8.1 Cartesian Coordinate Solutions Using Polynomials (直角坐标下的多项式解答)
0 0
m nf mn
m n
x y A x y
( , )
Chapter Page 8 10
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 )
024
4
22
4
4
4
yyxx
满足双调各方程 :
2 2
2 2
2 1 1 2 1 1
2 1 1 0m n mm
m n
m m m m A m m n n A
n n n n A
,
,
( )( ) ( ) ( ) ( )
( )( ) ( )
8.2 Uniaxial Tension of a Beam( 单轴拉伸梁 )
Chapter Page
plane stress case
Saint Venant approximation to the more general case withnonuniformly 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 ):
0)(,)(
0)(,0)(
lxxylxx
cyxycyy
T
8 11
Problem:
Chapter Page
constant stresses on each of the beam’s boundaries:
求应力函数 Φ
0)(,)( lxxylxx T
Therefore, this problem is given by:
8 12
202A y
2 2 2
2 2x y xyy x x y
, ,
022 , 0x y xyA
02 2A T0x y xyT ,
Boundary condition
polynomial is biharmonic
8.2 Uniaxial Tension of a Beam( 单轴拉伸梁 )
Chapter Page
求 u ,ε
/
/x
y
T E
T E
y
E
Tv
xE
Tu
1
1
x x y
y y x
u Tv
x E Ev T
v vy E E
( )
( )
Tu x f y
ET
v v y g xE
( )
( )
2 0 ( ) ( ) 0xyxy
u vf y g x
y x
( ) ( ) cons tan tg x f y
( )
( )o o
o o
f y y u
g x x v
0f g
0x y xyT ,
Integral( 积分 )
8.2 Uniaxial Tension of a Beam( 单轴拉伸梁 )
022 , 0x y xyA
Chapter Page
inverse method( 逆解法 )
0)(,)( lxxylxx T
Physical Equations
e
u
Geometrical Equations
024
4
22
4
4
4
yyxx
8 14
202A y
8.2 Uniaxial Tension of a Beam( 单轴拉伸梁 )
Chapter Page
8.3 Bending of a Beam by Uniform Transverse Loading( 受均匀横向载荷的梁弯曲问题 )
Airy stress function( 艾里应力函数 )
x
yl l
ql ql1
y
zh/2h/2
q
stress field( 应力场 )
Boundary Conditions( 边界条件 )
compare with elementary strength of materials( 和材力结果相比 )
8 15
Chapter Page 8 16
plane stress conditions(semi-inverse method 半逆解法 )
x
yl l
ql ql1
y
zh/2h/2
q
1, Airy stress function
y
xxy
Stress Component Force
M ( 主要由弯矩引起; )
Q ( 主要由剪力引起; )
q ( 由 q 引起 )
∵ q =const.
)(yfy
1) Stress components
2) Format of Stress function
)(2
2
yfxy
Integrate above:
)()( 1 yfyxfx
)()()(2 21
2
yfyxfyfx
)(),(),( 21 yfyfyf to be determined
8.3 Bending of a Beam by Uniform Transverse Loading( 受均匀横向载荷的梁弯曲问题 )
Chapter Page 8 17
)()( 1 yfyxfx
)()()(2 21
2
yfyxfyfx
3 ) satisfy the biharmonic equation
04
4 4
4
22
4
4
4
2yyxx
04
4
x
)()()(2
)4(2
)4(1
)4(2
4
4
yfyxfyfx
y
)(22 )2(22
4
yfyx
4
0)(2
)()()(2
)2(
)4(2
)4(1
)4(2
yf
yfyxfyfx
8.3 Bending of a Beam by Uniform Transverse Loading( 受均匀横向载荷的梁弯曲问题 )
Chapter Page 8 18
0)(2)()()(2
)2()4(2
)4(1
)4(2
yfyfyxfyfx
x
yl l
ql
ql
1
y
zh/2h/2
q
关于 x 的二次方程,且要求 - l≤ x ≤ l 内方程均成立。
0)()4( yf 0)(2)( )2()4(2 yfyf0)()4(1 yf
GyFyEyyf 231 )(
DCyByAyyf 23)(
此处略去了 f1(y) 中的常数项
23452 610)( KyHyy
By
Ayf
)()()(2 21
2
yfyxfyfx
)610
(
)()(2
2345
23232
KyHyyB
yA
GyFyEyxDCyByAyx
8.3 Bending of a Beam by Uniform Transverse Loading( 受均匀横向载荷的梁弯曲问题 )
Chapter Page 8 19
)610
(
)()(2
2345
23232
KyHyyB
yA
GyFyEyxDCyByAyx
9 unknown coefficients
2, stress field
2
2
yx
KHyByAyFEyxBAy
x2622)26()26(
223
2
2
2
xy
DCyByAy 23
yxxy
2
)23()23( 22 GFyEyCByAyx
8.3 Bending of a Beam by Uniform Transverse Loading( 受均匀横向载荷的梁弯曲问题 )
Chapter Page 8 20
3, Boundary Conditions
1) Symmetry Condition
yx , —— x 的偶函数
xy —— x 的奇函数
x
yl l
ql
ql
q
2
2
yx
KHyByAyFEyxBAy
x2622)26()26(
223
2
2
2
xy
DCyByAy 23
yxxy
2
)23()23( 22 GFyEyCByAyx
026 FEy
023 2 GFyEy
0 GFE
KHyByAyBAyx
x 2622)26(2
232
DCyByAyy 23
)23( 2 CByAyxxy
8.3 Bending of a Beam by Uniform Transverse Loading( 受均匀横向载荷的梁弯曲问题 )
Chapter Page
2) Boundary Conditions
a) Top and bottom( 上下面,主要 ) x
yl l
ql qlq
;0,2
xy
hy
;,2
qh
y y
;0,2
y
hy
KHyByAyBAyx
x 2622)26(2
232
DCyByAyy 23
)23( 2 CByAyxxy
,23h
qA
,0B
2
qD
h
qC
2
3
8 21
04
32
CBhh
A
04
32
CBhh
A
0248
23
DCh
Bh
Ah
qDCh
Bh
Ah
248
23
8.3 Bending of a Beam by Uniform Transverse Loading( 受均匀横向载荷的梁弯曲问题 )
Chapter Page
KHyyh
qyx
h
qx 26
46 33
23
22
32 33
qy
h
qy
h
qy
xh
qxy
h
qxy 2
36 23
x
yl l
ql qlq
b) End conditions( 左右边界,次要 )
,lx 未知
22
hy
hlx
xy022
h
yhlx
x
Using Saint-Venant’s Principle
statically equivalent force
轴力 N = 0 ;弯矩 M = 0 ;剪力 Q = - ql ;
8 22
8.3 Bending of a Beam by Uniform Transverse Loading( 受均匀横向载荷的梁弯曲问题 )
Chapter Page
轴力 N = 0 ;弯矩 M = 0 ;剪力 Q = - ql ;
qldyQh
h lxxy 2
2
02
2
dyN
h
h lxx
02
2
dyyM
h
h lxx
0K
h
q
h
qlH
103
2
自动满足
KHyyh
qyx
h
qx 26
46 33
23
22
32 33
qy
h
qy
h
qy
xh
qxy
h
qxy 2
36 23
)5
34()(
62
222
3
h
y
h
yqyxl
h
qx
22
112
h
y
h
yqy
2
2
3 4
6y
hx
h
qxy
8 23
8.3 Bending of a Beam by Uniform Transverse Loading( 受均匀横向载荷的梁弯曲问题 )
Chapter Page 8 24
4, compare with elementary strength of materials
)5
34()(
62
222
3
h
y
h
yqyxl
h
qx
22
112
h
y
h
yqy
2
2
3 4
6y
hx
h
qxy
x
yl l
ql qlq
3
12
1hI28
22 yhS
)(2
22 xlq
M
qxQ
5
34
2
2
h
y
h
yqy
I
Mx
22
112
h
y
h
yqy
bI
QSxy
8.3 Bending of a Beam by Uniform Transverse Loading( 受均匀横向载荷的梁弯曲问题 )
Chapter Page
5
34
2
2
h
y
h
yqy
I
Mx
22
112
h
y
h
yqy
bI
QSxy
4, compare with elementary strength of materials
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。
8 25
8.3 Bending of a Beam by Uniform Transverse Loading( 受均匀横向载荷的梁弯曲问题 )
xyτx y
)(
)(
Chapter Page 8 26
x
yl l
ql qlq
5
34
2
2
h
y
h
yqy
I
Mx
22
112
h
y
h
yqy
bI
QSxy
8.3 Bending of a Beam by Uniform Transverse Loading( 受均匀横向载荷的梁弯曲问题 )
Chapter Page 8 27
Vocabulary( 词汇 )
Polynomials 多项式Uniaxial 单轴的Beam 梁Uniform 均匀的Inverse Method 逆解法Semi-Inverse Method 半逆解法Biharmonic 双调和的Airy stress function 艾里应力函数
Homework
Chapter Page 8 27
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 .
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 .
x
y
o
q
b
ρg
图 1
y
xoα
ρg
图 2
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 .
Homework
Chapter Page 8 27
x
图 3
y o
ρ1g
b/2 b/2
ρ2g
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 .
( 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 )