Transcript Theory of Elasticity

Theory of Elasticity
Chapter 10
Three-Dimensional Problems
Content
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Introduction
Mathematical Preliminaries
Stress and Equilibrium
Displacements and Strains
Material Behavior- Linear Elastic Solids
Formulation and Solution Strategies
Two-Dimensional Problems
Three-Dimensional Problems
Introduction to Finite Element Method
Bending of Thin Plates
Chapter 10
Page 1
Three-Dimensional Problems
• 10.1 Review: stress Formulation (按应力求解空间问题)
• 10.2 Torsion of straight bars (Prismatical Bars)
（等截面直杆的扭转）
• 10.3 Elliptic Bars in torsion （椭圆面直杆的扭转）
• 10.4 Rectangular bars in torsion（矩形截面杆的扭转）
• 10.5 Membrane analogy of torsion（扭转的薄膜比拟）
Chapter 10
Page 2
10.1 Review： Stress Formulation（按应力求解）
Review: Stress Formulation
Equilibrium Equations
平衡方程
Geometrical Equations
几何方程
Physical Equations
物理方程
 ij 
1
(ui , j  u j ,i )
2
3
ε-u
1
 ij   kk ij  2 ij σ-ε
Eliminating the displacements and strains
Chapter 10
σ
Page 3
2
10.1 Review： Stress Formulation（按应力求解）
Eliminating the displacements
u
v
w

 x  u ,
 y  v ,
  w ,

 x  x ,
 y  y ,
 zz  z ,



x v
zv u 
w
uy w

 yz  w  v ,  zx  u  w ,  xy  v u 。

 yz  y  z ,  zx  z  x ,  xy  x  y。
y z
z x
x y 
2
  z   yz 


,
2
2
yz 
z
y
 2  z  2  x  2 zx 


,
2
2
zx 
x
z
2
2
 2  x   y   xy 


。

2
2
xy 
y
x
 2 y
Chapter 10
2
 2 x 
   yz  zx  xy 
 
  2


,
x  x
y
z 
yz 
 2 y 
   zx  xy  yz 
 
  2


,
y  y
z
x 
zx 

 2 z 
   xy  yz  zx 
 
  2


。
z  z
x
y 
xy 
Page 4
10.1 Review： Stress Formulation（按应力求解）
 2 y
z 2
 2 z
x 2
 2 x
y 2

,

 2  x  2 zx 


,
2
zx 
z
 2  y  2 xy 


。

2
xy 
x
2
 2 z   yz


2
yz
y
   yz  zx  xy   2 x 
 
  2


,
x  x
y
z  yz 
2

   zx  xy  yz    y 

2


,
y  y
z
x  zx 

   xy  yz  zx   2 z 
 
  2


。
z  z
x
y  xy 
Chapter 10
6 compatibility equations may
also be represented by the 3
independent fourth-order
equations
Page 5
10.1 Review： Stress Formulation（按应力求解）

 2
1     x  relations
0 
using Hooke’s law and eliminate the strains in the compatibility
2
x

2
incorporating the equilibrium equations into the system 2
1     y   2  0 
y
For the case with no body forces

2

 
2
1     z  2  0 

 x  yx  zx
z



 f x  0,
B.D

2
x
y
z
 

2



1






0
 y  yz  xy
yz


yz


 f y  0, 

y
z
x
2

 
2



1






0


 z  xz

zx
yz
Simple Connected

zx


 f x  0。

(单连通域)

z
x
y
2



2
 0
the necessary six relations to solve for the six unknown 1     xy 

xy
stress components for the general three-dimensional case.
Eliminating the strains
2
+
Chapter 10
Page 6
10.2 Torsion of straight bars （等截面直杆的扭转）
Examples:
Chapter 10
Page 7
10.2 Torsion of straight bars （等截面直杆的扭转）
Torsion of circular shaft
Assumptions( on the torsional deformation of cylinders of circular cross-section)
Each section rotates as a rigid body about the center axis.
For small deformation theory, the amount of rotation is a linear function of
the axial coordinate.
Because of symmetry, circular cross-sections remain plane after deformation.
 max 
T
R
Ip
The corss section of the bar remain plane and rotate without and distortion
Coulomb
Chapter 10
Page 8
10.2 Torsion of straight bars （等截面直杆的扭转）
For noncircular cross sections
Naivier, also applied above assumptions
Arrived at the erroneous conclusion.
The lateral surface of the bar is free form external forces
Naivier’s assumption in contradiction with above
x
τyz
τzx
y
Chapter 10
Page 9
10.2 Torsion of straight bars （等截面直杆的扭转）
assumptions the following for general cross-sections
The projection of each section on the x,y-plane rotates as a rigid body
about the central axis.
The amount of projected section rotation is a linear function of the axial
coordinate.
Plane cross-sections do not remain plane after deformation, thus leading
to a warping (翘曲) displacement
The Correct solution was ginven by Saint-Venant,1855
Saint-Venant’s Principle was proposed
and applied
Chapter 10
Page 10
10.2 Torsion of straight bars （等截面直杆的扭转）
Deformation field bases on above assumptions
The projection of each section on the x,y-plane
rotates as a rigid body about the central axis.
O ，center of twist, where u＝0， v＝0.
The amount of projected section rotation is a
linear function of the axial coordinate.
assume that the cylinder is fixed at z =0 and
 is the angle of twist per unit length.
Plane cross-sections do not remain plane after
deformation, thus leading to a warping (翘曲)
displacement
Chapter 10
Page 11
10.2 Torsion of straight bars （等截面直杆的扭转）
Stress Formulation
the strain-displacement relations
 x   y   z   xy  0
 xz
 yz
1 w
 (
 y )
2 x
1 w
 (
 x )
2 y
Chapter 10
Page 12
Hooke’s law
10.2 Torsion of straight bars （等截面直杆的扭转）
Stress Formulation (semi-inverse)
the equilibrium equations
（ with zero body forces）
 xz  yz

0
x
y
Poisson equation
introducing a stress function
(1) Prandtl stress function
compatibility equations
Hook’s law
governing equations for the stress formulation.
Chapter 10
Page 13
(2)
10.2 Torsion of straight bars （等截面直杆的扭转
Stress Formulation
Boundary Conditions (1)
the lateral surface is free of tractions
Chapter 10
Page 14
10.2 Torsion of straight bars （等截面直杆的扭转）
Stress Formulation
Boundary Conditions (2)
the ends of the cylinder
Chapter 10
Page 15
10.2 Torsion of straight bars （等截面直杆的扭转）
Summary
the equilibrium equations
（ with zero body forces）
compatibility equations
Simple Connected
s  0。
Boundary Conditions
Chapter 10
Page 16
10.3 Elliptic Bars in torsion （椭圆面直杆的扭转）
The boundary equation
2  Const  -2
x2 y2
 2 1  0
2
a
b
a stress function
s  0。
2 dxdy  T。
A
T x2 y2
 
( 2  2  1)
ab a b
Chapter 10
Page 17
10.3 Elliptic Bars in torsion （椭圆面直杆的扭转）
The boundary equation
2  Const  -2
x2 y2
 2 1  0
2
a
b
a stress function
s  0。
2 dxdy  T。
A
x2 y 2
  K ( 2  2  1)
a
b
a 2b 2 
K  2
a  b2
2a 2b2  1
T  2
( 2
2
a b a
T

 a 3b3 
a 2  b2
R
x 2 dxdy 
1
b2

R
y 2 dxdy   dxdy )
T (a 2  b2 )

 a3b3 
T x2 y2
 
( 2  2  1)
ab a b
Chapter 10
Page 17
R
10.3 Elliptic Bars in torsion （椭圆面直杆的扭转）
T x2 y2
 
( 2  2  1)
ab a b
Chapter 10
Page 18
10.3 Elliptic Bars in torsion （椭圆面直杆的扭转）
Contour lines of the stress function
Contour lines of the dipsplacement
a positive counterclockwise torque
Solid lines correspond to positive
values of w
dotted lines indicate negative values of
displacement
Along each of the coordinate axes the displacement is zero,
With a =b(circular section), the warping displacement vanishes everywhere.
If the ends restrained, normal stresses z are generated as a result of the torsion.
Chapter 10
Page 19
10.4 Rectangular bars in torsion（矩形截面杆的扭转）
Chapter 10
Page 20
10.4 Rectangular bars in torsion（矩形截面杆的扭转）
Chapter 10
Page 21
10.5 Membrane analogy of torsion（扭转的薄膜比拟）
Membrane analogy
Introduced by Prandtl,1903
A.A.Griffith and G.I.Taylor, Further development

z
Chapter 10
Page 22
10.5 Membrane analogy of torsion（扭转的薄膜比拟）

z
   Const  C
2 z  
2
z s  0。
s  0。
2  dxdy  M。
2V  2 zdxdy ,
A
Chapter 10
q
FT
A
Page 23
Homework
• 8-5
• 8-7
Chapter 10
Page 24