Lecture 9
Stability of Elastic Structures
Lecture 10
Advanced Topic in Column Buckling
Problem 9-1:
A clamped-free column is loaded at its tip by a load P. The issue here is to find the critical
buckling load.
a) Suggest a simple form of the buckled of the column, satisfying kinematic boundary
conditions.
b) Use the Rayleigh-Ritz quotient to find the approximate value of the buckling load.
c) Come up with another buckling shape which would give you a lower value for the
buckling load.
d) Find the exact solution of the problem and show the convergence of the approximate
solution to the exact solution.
Follow the example of a pin-pin column, which is presented in the notes of Lecture 9.
1
Problem 9-1 Solution:
a)
Kinematic boundary condition, in term of shape function I ( x ) , for a clamped-free
column is
(0)
'(0)
0
Choose a buckling shape
b)
( x)
x2 '( x)
2x
''( x)
2
Use Rayleigh-Ritz Quotient, the critical buckling load is
l
Nc
EI
I ''I '' dx
0
l
0
I 'I ' dx
l
EI
0
l
0
Nc
2 u 2dx
2x u 2xdx
3
EI
L2
2
c)
Choose a buckling shape similar to a cantilever beam
I ( x)
x 3 3Lx 2
I '( x) 3x 2 6Lx
I ''( x) 6x 6L
l
EI
Nc
I ''I '' dx
0
l
0
I 'I 'dx
l
³ 6x 6L dx
EI
³
3x 6Lx dx
2
0
l
2
2
0
EI
12L3
24 L5 5
Nc
Compare to the result in b), N c
d)
3
2.5
EI
L2
EI
, this buckling shape givers a lower value
L2
Choose buckling shape
I ( x) 1 cos
S
2L
x 3
S
I '(x)
2L
sin
Nc
EI
cos
2L
x
I 'I 'dx
ª§ S
³0 ««© 2L
¬
2
l
2
º
cos
x » dx
2L ¼»
2
S
S
§
sin
x ¸ dx
¨
0 2L
2L
©
l
EI
x
I ''I '' dx
0
l
0
EI
2L
2
§S
I ''( x) ¨
© 2L
l
S
S2
4L2
Nc
2.47
EI
L2
Check for local equilibrium of the solution
EIw
IV
N c w ''
ª
§S
A « EI ¨
© 2L
¬«
4
§S
cos
x EI 2 ¨
2L
4L © 2L
2
2
cos
S
x
2L »¼
0
This is the exact solution to the clamped-free buckling
4
Problem 9-2:
Consider a clamped-free column loaded by a compressive force at the free end.
a) Determine the critical slenderness ratio E crit distinguishing between the elastic and
plastic buckling response. What is the buckling stress and strain?
b) Calculate the critical plastic buckling load for E
0.5E crit and the corresponding
stress and strain.
c) Calculate the critical elastic buckling load for E
2E crit and the corresponding
stress and strain.
d) Compare all three results.
Problem 9-2 Solution: a)
First, find the bending load˖
For Clamed-Free column
Pcr
S 2 EI
2L
2
S 2 EI
4L2
Second, find the buckling stress and strain
V cr buckling
Recall that
Pcr
A
S 2 EI
4AL2
5
I
Ÿ r2
A
r
I
A
Then
V cr
2
Er 2
4L2
buckling
2
E
4L r2
2
Recall that
L2 r 2
2
Er 2
4L2
V cr
H cr
Third, find when V buckling
E
E crit when V cr
y
V cr
E
2
S2
4E 2
, which is
S 2E
4V y
E
2 Vy
E
4
E
V yield
S 2E
4E crit 2
b)
2
Vy
E crit 2
E crit
0.5E crit , the column yields + hits plastic buckling
6
S 2 Et
crpl
2
§1
4 E crit ¸
¹
©2
S 2 Et
(From Lecture Note, equation 9.73)
E crit
2
n
pl
c)
§1
4 E crit ¸
©2
S2
n
E crit 2
2
2E crit , the column will buckle elastically
V cr
2
2
E
4
V cr
E
2
E
4 2E crit
2
H cr
d)
S2
E
16E crit 2
2
S2
16E crit 2
Compare the three results
Yield
Elastic Buckling
y
0.67
E
2
y
0.07
0.67
y
2
crit
crit
Plastic Buckling
19.74 E
2
4
crit
To simply our comparison, assume n
0.5
0.2 , Et
7.9
y
4
1.98
2
crit
0.5E (*) and recall
2
crit
E
Vy
E crit
(*)In order to compare plastic buckling to elastic+yield, we need to make future
assumption about the material properties.
7
Problem 9-3:
a)
Consider the pin-pin column.
( x) to improve the approximate
Suggest a polynomial buckling shape function
solution derived in lecture note.
Note that the one used in class was the parabolic
shape.
b)
Determine the accuracy relative to the exact solution.
P
EI
L
x
Problem 9-3 Solution:
a)
The exact solution is w
§Sx
sin ¨
¸ , use the none-dimensioned value F
L
x
, the Taylor
L
series expansion is
SF sin SF SF
3
6
So we know the shape function must be I ( F ) C1 F C2 F 3
For 0
x
L 2 , the boundary conditions are ­I 0 0
®
°I ' F
¯
The first boundary condition gives
1
¸ 0
2 ¹
I 0 C1 0 C2 0 this doesn’t help.
The second boundary condition gives 8
2
C1 3C2
'
1
2
'
0
3
C2
4
C1
0
4
C1
3
C2
So we have
4
C1
3
( ) C1
4
3
( ) C1
3
3
We can us the Rayleigh-Ritz Quotient
2
EI
N cr
'' dx
2
' dx
2
'( x) C1 1 4
'( )
2
C12 1 8
''( x)
2
''( )
where d F dx
d x l
dx
2
d
16
8C1
d
64C12
2
dx
4
d
2
2
dx
d
dx
4
dx
1
l
Since we have considered the shape function for 0 d x d L 2 , we must adjust the limits on
the integral
N cr
2
EI
'' dx
2
' dx
2
EI
2
EI
l
2
0
l
2
0
C12 1 8
l
2
0
l
2
0
64C12
2
2
16
64 x l
1 8 xl
2
2
4
d
dx dx
4
2
d
dx dx
4
1 l dx
16 x l
4
2
1 l dx
after lengthly algebra
10
EI
L2
9
N cr
10
EI
L2
b)
The result are compared with the polynomial used in class and the exact solution
Coefficient
Error
Exact Solution
Parabolic
C sin
C1
2
N/A
9.87
C2
Cubic
2
C1
12
10
21.5%
1.3%
C2
3
Notice how we significantly reduce the error by including a higher order term.
10
Problem 9-4:
Present a step-by-step derivation of the buckling solution of the pin-clamped column from the
local equilibrium equation.
P
EI
L
x
Problem 9-4 Solution:
Boundary condition for this problem
w0
w L 0
w' 0 0
EIw'' L 0
Start with 4th order ODE
EIwIV Pw'' 0
P
wIV
w'' 0
EI
We have an eigenvalue problem
P
EI
4
2
1
Define
P
EI
KŸ
3
4
2
0
P
EI
2
0,
2
3
0
4
i
P
EI
iK
w C1 C2 x C3 sin Kx C4 cos Kx
11
Use the boundary conditions to solve for constants C1 , C2 , C3 and C4
w 0 0
w 0
0 C1 C4
C1
C4
w ' 0 0
w' x
C2
KC3 cos Kx KC4 sin Kx
w' 0
C2
KC3
C2
0
KC3
w L 0
w L C1 C2 L C3 sin KL C4 cos KL 0
Substitute C1 , C2 into the above expression
C3
KL sin KL
C4
1 cos KL
0
w '' L 0
w '' x
K 2C3 sin Kx K 2C4 cos Kx
w '' L
K 2C3 sin KL K 2C4 cos KL
KL sin KL
2
2
K sin KL
det >
C3
1 cos KL
C4
K cos KL
0
0
0
K 2 cos KL
KL sin KL
K 2 sin KL
KL cos KL sin KL
KL
sin KL
cos KL
1 cos KL
0
0
tan KL
So the equation to solve in order to find Pcr is
tan KL KL 0
The smallest roots are KL
we choose KL
0 and KL
4.49 ,
4.49
12
Pcr
P
L 4.49
EI
2
20.16
EI
EI
2
2
L
0.7 L
2
Pcr
EI
0.7 L
2
13
Problem 9-5:
a) Derive the solution for an imperfect clamped-free column (like that considered in
problem 9-1, following a similar derivation given in the notes for a pin-pin column in
the notes.
b) Find the ratio of current deflection amplitude to the amplitude of the initial
imperfection such that the resulting load is 80% of the theoretical buckling load of a
perfect column.
Problem 9-5 Solution:
a)
w x : shape of initial imperfection
w x : actual buckled shape
wo x : amplitude of initial imperfection
wo : end amplitude of actual imperfection
Moment equilibrium of imperfect column
EI w w '' P w w o 0
Perfect column
w x 0
Assume that the initial imperfection is in the same shape as the buckling shaper
w x
wo 1 cos O x w x
wo 1 cos O x
From boundary condition
w L 0
14
wo O 2 cos O L
0
2n 1
2
L
From moment equilibrium of imperfect column
EI
2
w wo cos x Pwo 1
Pwo
Perfect column
2
EI
1 cos x
0
w wo
wo
P
0
S 2 EI
EI O 2
4L2
Imperfect column
Pcr wo wo Pwo
2
P
b)
When
P
Pcr
wo
EI
1
2
wo
4L
wo
wo
P
Pcr
1
P
Pcr
0.2
0.8
wo
wo
1
wo
wo
5
15
Problem 9-6:
The pin-pin elastic column of length L (shown below) is an “I” section can buckle in either plane.
a) Determine the buckling load in terms of L, b1,b2, t and E. Assume that t<<b.
b) What should the ratio of b1/b2 be in order for the probability of buckling in either of
the buckling planes to be the same?
Bonus: What could happen for very large width to thickness ratio?
P
b2
b1
EI
L
x
Problem 9-6 Solution:
a) The moment s of inertia for an “I” shape cross-section is
I yy
I zz
1 3
§b ·
tb2 2b1t ¨ 2 ¸
12
2¹
1 2
tb2 b2 6b1 12
2
1 3
1
t b2 2 tb13
12
12
1 3
tb1
6
16
If I yy
I zz , the column will buckle in x-z plane
Pcr
If I yy
S 2 EI yy
l
2
2
12l
tb2 2 b2 6b1 I zz , the column will buckle in x-y plane Pcr
b)
S 2E
S 2 EI zz
l
2
S 2E
6l
2
tb13
For the probability of buckling in either of the planes to be the same , we want
I yy
I zz
1 2
tb2 b2 6b1 12
b1
b2
3
3
b1
b2
1 3
tb1
6
1
2
0
The only physical solution is
b1
b2
c)
1.81
If b1 !! t, b2 !! t , then local plate buckling my develop.
17
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