Mechanics
of
Thin Structure
Lecture 15 Wrapping Up the Course
Shunji Kanie
Mechanics of Thin Structure
What you learned are;
Introduction for Linear Elasticity
Stress and Strain with 3D General Expressions
Plane Stress and Plane Strain
Principle of Energy
Principle of Virtual Work
Calculus of Variations
Theory of Beams
Theory of Plates
Example
Pure bending problem
Solve the problem
Equilibrium
Compatibility
Energy
Governing Equation
Example
Pure bending problem
Solve the problem
Equilibrium
Compatibility
Vertical
dQ

 q( x)  0
dx
Moment
Q 
dx M

 Q 
dx dx  q( x)dx

dx  0
x 
2
x

dM
d 2M
Q
 q ( x)
2
dx
dx
Example
Pure bending problem
Solve the problem
Equilibrium
Compatibility
X direction
 x  yx  zx


0
x
y
z
h
h
 yx
 x
2
2  zx
 h 2 x zdz   h 2 y zdz   h 2 z zdz  0
h

2
h
2
h
2
h
h
 zx
zdz   zx z h2   h2  zx dz


z
2
2
dM x
Qx 
dx
Example
Pure bending problem
Solve the problem

Equilibrium
dv u
dx
y
Compatibility
dv
u( y)   y   y
dx
du
d 2v
 x ( y) 
 y
dx
dx 2

d 2v 
d 2v
M x   E x ydA   E   y 2  ydA   EI 2
dx 
dx

A
A
Example
Pure bending problem
Solve the problem
Equilibrium
Compatibility
2 
2

d
v
d
v
d 2M


  q ( x) M x   E x ydA   E  y 2 ydA   EI 2
2
dx 
dx
dx
A
A 
Governing Equation
d 4v
EI 4  q ( x)  0
dx
Example
Pure bending problem
Compatibility
Energy
dv
u( y)   y   y
dx
du
d 2v
 x ( y) 
 y
dx
dx 2

dv u
dx
y
Energy should be Minimum,
so that Energy is calcualted
such as;
Example
Pure bending problem
Energy for section
2
1
du
d
v


bdy
 x ( y) 
 y
2 x x
dx
dx 2
l
l 1
1


2
U      x  x bdy dx     E x bdy dx
0 2
0 2


2
2


2
2


l 1

EI l  d v 
2 d v 
    Ey
bdy dx 
dx

2
0
0
 dx 2 
2
dx
 2







Example
Pure bending problem
Potential Energy
l
W   q( x)vdx
0

l  EI
  U W  
0 2


d v

  q( x)v  dx
 dx 2 




2
You have the Functional !!
  J v   F ( x, v, v' ' )dx
l
0
2
Example
Pure bending problem
You have the Functional !!
  J v   F ( x, v, v' ' )dx
l
0
 EI

2
  U  W    v' '  q( x)v  dx
0
 2

l
Apply the Euler’s Equation
F d  F  d 2
 
 2
v dx  v'  dx
 F 

0
 v' ' 
Example
Pure bending problem
F d  F  d
 
 2
v dx  v'  dx
2
 F 

0
 v' ' 
F
F
F
 q
 EIv ' '   l EI v' '2  q( x)v  dx
0
0  2

v
v' '
v'

q
d
2
dx

EIv ' '  EI
2
4
d v
dx
4
q 0
Governing Equation for pure bending beam
from the Euler’s Equation
Example
Pure bending problem
Boundary Condition
See eq. (6-31)
F d  F 
d
v
 
   EIv ' '  Qx Shearing force
v' dx  v' ' 
dx
F
dv
 EIv ' '  M x
 
Bending Moment v'  
dx
v' '
Example
Pure bending problem
Mechanical
Boundary
Condition
Geometrical
Boundary
Condition
Qx
Mx
or
dv
v'  
 
dx
or
v
J  Q x v x  M x 
x2
x2
1
x1

x2
x1
 d2

 2 EIv ' '  q vdx
 dx

Example
Solution 1 : Direct Integration
d 4v
EI 4  q  0
dx
EI
EI
d 3v
dx
3
x
d 2v
dx
  q ( x)dx  C1
2
   q ( x)dxdx  C1 x  C 2
xx
5 qL4
 0.01302
384 EI
Example
Solution 2 : Virtual Work
Equilibrium
d 4v
EI 4  q  0
dx
No Energy produced if the Equilibrium is satisfied
Virtual Displacement
 d 4v

x EI dx 4  q vdx  0 Virtual Work
What is the requirement for the virtual
displacement?
Example
Solution 2 : Virtual Work
 d 4v

x EI dx 4  q vdx  0

~
v  a sin 
L
 
EI  
L
4
 Assumption
x


  
 a sin L x vdx   qvdx  0
x
x
 
v  sin  x 
L 
Virtual Displacement
Example
Solution 2 : Virtual Work

~
v  a sin 
L
 Assumption
x

 
v  sin  x 
L 
4
Virtual Displacement
2
   1          

EI   a
sin
vdx
xdx
vdx
xx2dx
qq
sin
x0dxx0dx  0
qsin
asin
1 cos




 L  x x 2   LL  L x x  x  L L 
Example
Solution 2 : Virtual Work

~
v  a sin 
L
 Assumption
x

   Virtual Displacement
v  sin  x 
L 
4
4 
4




1    1
 L 
q L 
EI  aEI
4  x 5dx  0
 a   1Lcos
2q 2  x 0dx  aq
sin
  
EI
L
2L   xL2 
  L 

 
x 
4 qL4
qL4
a
 0.01307
306 EI
EI
5 qL4
qL4
 0.01302
384 EI
EI
Example
Solution 2 modified
 d 4v

x EI dx 4  q vdx  0
 m
~
v   a m sin 
 L
m
4

x

Assumption
4
 Lm    m 
EI EI   a sin
x vdx xqvdx
vdx 0qvdx  0
am sin


L  L x
m    L
x  x 
x
 n 
sin  x  x 
vnsin
 L L  
Virtual Displacement
Example
Solution 2 modified
 m 
 EI  L 
m
4
 n
vn  sin
 L
 m 
 EI  L 
m
4

 m 
 am sin L x vdx   qvdx  0
x
x

x

 n
 q sin L
x

x dx
Fourier Transform

 m   n 
 n 
 am sin L x  sin L x dx   q sin L x dx  0
x
x
It’s same as the solution introduced for Plate Theory
Example
Solution 3 :Point Collocation
 d 4v

x EI dx 4  q vdx  0
Assume you would like to have the value at the
center
 
v   L 2
Dirac Delta Function
4
4
L





v x 
d v
 1d 
2   EI

q

 L 2EI 4  q dx
4
L
 dx
 x  L
0dx x 
 

x
2

2
 
Example
Solution 3 :Point Collocation
4
4




d v
 d v

 EI dx 4  qdx   EI dx 4  q  0


 xL
x
2
   Assumption
~
v  a sin  x 
L 
4

 d v

 
  
 aEI   sin x   q 
0
 EI 4 4 q 
4 
4 
4
L
L

 qL

 dx
L  q x  L
5 qL
L qL

x

2
a 
 0.01027
 0.013022
EI 384 EI
EI
   EI
4
Example
Solution 4 :Least Square Method
EI
d 4v
dx 4
q  0
Equilibrium
Error should be minimum
   Assumption
~
v  a sin  x 
L 
4
4~
d v
 
   Estimated
EI 4  EIa   sin x 
dx
L
L 
Example
Solution 4 :Least Square Method
4~
Error
4
 
 
EI 4  q  EIa   sin x   q
dx
L
L 
d v
2
   
   
    EIa   sin x   q  dx
 L  
x
 L
4
Squared Error
In order to expect the Error minimized,
an appropriate value for a must be
 
4
      


EIa   sin


a a  x 
L
L




 
x   q  dx

 


2
Example
Solution 4 :Least Square Method
2
 

4

      
  


EIa
sin
x

q
dx









a a x 
L
L






 

4



  
 2  EIa   sin
a
L
x
 L
4

   

x   q  EI   sin
   L 
L
4


      
  
   EIa   sin x   q  sin x dx  0
a x   L 
 L    L 

x dx  0

Example
Solution 2 : Virtual Work

~
v  a sin 
L
 Assumption
x

 
v  sin  x 
L 
 
EI  
L
4
Virtual Displacement

  
 a sin L x vdx   qvdx  0
x
x
Example
Solution 4 :Least Square Method
4


      
  
   EIa   sin x   q  sin x dx  0
a x   L 
 L    L 
4
2
   
   
 
EI   a  sin x  dx   q sin x dx  0
 L  x   L 
 L 
x 
Example
Solution 5 :Galerkin Method 1
 d 4v

x EI dx 4  q vdx  0
Starting Equation is Same
   Assumption
~
v  a sin  x 
L 

v  sin 
L

x

Weighting Function
Example
Solution 5 :Galerkin Method 2
 d 4v

x EI dx 4  q vdx  0
Starting Equation is Same
L


d v
d 3v v
 d v

 EI dx4  qvdx  EI dx3 v   EI dx3 x dx   qvdx

x
x
x
0
4
 EI
3
d v
dx
3
3
L
v  EI
0
d v  v
2
dx
2
2
x
L
  EI
2
0
x
d 2 v  2v
dx
2
x
2
dx   qvdx
x
Example
Solution 5 :Galerkin Method 2
 EI
3
d v
dx
  EI
x
3
L
v  EI
0
d 2 v  2v
dx
2
x
2
d v  v
2
dx
2
2
x
  EI
2
0
x
d 2 v  2v
dx
2
x
2
dx   qvdx
dx   qvdx  0
x
v~  ax3  bx 2  cx  d
3
2
~
v  aL  bL  cL  0

L

v~  ax3  bx 2  cx
c  aL  bL
v~  ax x 2  L2  bxx  L
2
x
Example
Solution 5 :Galerkin Method 2
d 2 v  2v
 EI dx 2
x
x

2
dx   qvdx  0

x
2
2
~
v  ax x  L  bxx  L
dv~
 3ax 2  2bx  c
dx
d 2v
v1  x x  L
v2  xx  L 

d v1
2
dx 2
2
 6x
2

dx
2
 6ax  2b
d 2v2
dx
2
2
Example
Solution 5 :Galerkin Method 2
d 2 v  2v
 EI dx 2
x
2
d v
dx
2
x
2
d 2v1
dx   qvdx  0
 6ax  2b
x

v1  x x  L
2
2
v2  xx  L 

dx 2
d 2v2
dx 2
2
2


dx  0




EI
6
ax

2
b
6
x
dx

qx
x

L


x
x
 EI 6ax  2b2dx   qxx  Ldx  0
x
x
 6x
2
Example
Solution 5 :Galerkin Method 2
24
2
4 4










EI
6
ax

2
b
6
x
dx

qx
x

L
dx

0
L


L
L
3
2
a

0




3
2


EI
12 aL  6bL  q q
   0
x
x
2 
 4 4
3 3
3  

2








EI
6
ax

2
b
2
dx

q
x
x
L
dx

0
L

2
L
L




2
qL


EI 6aL  4bL  q q    0b 
x
x

6
2 
  3
24 EI


4
4
22
2
~
qL
qL
qL
vv~ 
 ax x 
 vx L  0.0104166
xLx  Lbx
96 EI
EI
24 EI
5 qL4
qL4
 0.01302
384 EI
EI
Example
Solution 5 :Galerkin Method 3
v~  ax3  bx 2  cx  d
v~  d  v~
dv~
2
 3ax  2bx  c
dx
1
1   2
v~  aL3  bL2  cL  0  v~2
a
c  aL  bL  1
 21   2 
b
L
2
3aL  2bL  c  2aL  bL   2
2
2
3
2 2    23
2



3
~
1
2
1
2
vv~ 
  x 2 2xx  x  xx x1 x 
 L2
L1  2
 2
L
L



L

L

2
L
Example
Solution 5 :Galerkin Method 3
3
2
3
2



x
x
x
x




~
v   2 2
 x1   2   2
L
L



L

L

d 2 v~  x
1
1
 x
 6 2  4 1  6 2  2  2
2
L
L
dx
 L
 L
d v  v
2
 EI dx 2
x
2
x 2
dx   qvdx  0
x
Example
Solution 5 :Galerkin Method 3
2

2  L
4
EI  1   2   
L  
L
 12


q  0


2
4  
2
 L
EI  1   2   
L  
L
 12
2

6
3
L

EI 1  
L

 12


q  0


3
3

qL
   qL   
q 1
2
24
EI
24
EI


Example
Solution 5 :Galerkin Method 3
3
2
3
2



x
x
x
x




~
v   2 2
 x1   2   2
L
L



L

L

3
3
qL
qL
2  
1 
24 EI
24 EI
qL
L L L
L L
~
v     1     2 
96 EI
8 2 2
8 4
4
qL4
qL4
qL4 5 qL4
 0.01302
v
 0.0104166
EI
96 EI
EI 384 EI
Mechanics of Thin Structure
What you learned are;
Introduction for Linear Elasticity
Stress and Strain with 3D General Expressions
Plane Stress and Plane Strain
Principle of Energy
Principle of Virtual Work
Calculus of Variations
Theory of Beams
Theory of Plates