buckling
TRANSCRIPT
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Buckling AnalysisBuckling Analysis
THE UNIVERSITY OF HONG KONGTHE UNIVERSITY OF HONG KONGDr. Ray SuDr. Ray Su
Department of Civil EngineeringDepartment of Civil Engineering
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Buckling of TrussBuckling of Truss
Before buckling After buckling
Maximum Load = 58N(Buckling load)
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A crane collapsed on a Hennessy Centre demolition site in Causeway Bay, Hong Kong in 2007. The accident killed two construction workers.
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Buckling InstabilityBuckling Instability• Buckling is an instability phenomenon where a
structural system or a structural member under predominant compressive forces is unduly sensitive to changes.
• Buckling load is the maximum load that the structure can be withstood before buckling.
• The aim of this course is to use matrix method to determine the buckling load and the buckling displacements (buckling modes) of frame structures
Neutral Equilibrium Unstable
A small disturbance will lead to excessive movement
A curved surface
Equilibrium positionat local min. Stable
Concept of Stability
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M
θ
M
Bending Stiffness Kθ
=M/θ
The additional axial force increases the end rotations ,
lateral deflection x
and internal moment (Madd
=Px) and reduces the bending stiffness.
MM
θPP
Under combined axial and bending loads
x
The Effect of Axial Load on the The Effect of Axial Load on the Bending StiffnessBending Stiffness
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Force approach
Consider a simple structural system with two struts only
Roller
Hinge
Hinge
Spring
L
L
L = the length of each portion of the columnk = spring stiffness
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S Joint equilibrium
S
N
N/S = L/xS = (x/L) N
By the geometric argument
compression
compression
Force approach
L
L
F
N = axial load, compression taken to be positiveF = the applied lateral loadX = lateral deflection
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S
= (x/L) N
If F0 kx
k
2S
Horizontal force equilibrium
F= kx
– (2N/L)
x (1)
Spring force
Lateral force generated by N
Rearranging
(k-2N/L) x
= F (2)
Additional stiffness(geometric stiffness)
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Notes:Notes:• The effect of the axial force N
is equivalent to an
additional stiffness of value (-2N/L) acting laterally at the pin location. We term it as geometric stiffness.
• When the axial force is tensile (-ve), the geometric stiffness is positive. The effects on the system tend to be beneficial, the lateral deflection and the restraining force in the spring are both reduced.
• When the axial force is compressive (+ve), the geometric stiffness is negative. The effects on the system tend to be detrimental, the lateral deflection and the restraining force in the spring are both increased.
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when N=Ncr , the negative geometric stiffness will cancel out the spring stiffness, causing the lateral stiffness of the system to become zero.
When the compressive force is kept increasing, N > 0 and assuming that F=0 (no horizontal applied force)
When buckling happen, the lateral (or bending) stiffness of the member becomes ZERO
For an axially loaded beam, the lateral stiffness comes from the bending stiffness. Buckling implies that the bending stiffness is cancelled out by the geometric stiffness
Critical load (Ncr)
k-2Ncr /L = 0
Ncr = kL/2 (3)
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If the analysis is to be accurate to within 10%, it may need to allow for the effects of buckling if N/Ncr >0.1.
Stiffness K
with geometric effects
Decreasing in the lateral stiffness K=
k
(1-N/Ncr
)
increasing in lateral displacement
Buckling equation Substituting (3) into (2)
k(1-N/Ncr ) x
= F (4)
F
Nk (1-N/Ncr
)k
==k/K
Effects of displacement amplification
10.01.110.9
5.001.250.8
3.331.430.7
2.501.670.6
2.002.000.5
1.672.500.4
1.503.330.3
1.255.000.2
1.1110.000.1
1.000
Displacement Amplification
crN/Ncr
10.01.110.9
5.001.250.8
3.331.430.7
2.501.670.6
2.002.000.5
1.672.500.4
1.503.330.3
1.255.000.2
1.1110.000.1
1.000
Displacement Amplification
crN/Ncr
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Energy approachWe will use the energy approach to derive the stiffness and geometric matrices.
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y
=2Δ= 2 [ L -
d ]
Ignore the higher order terms
(of the spring)
L
Δ
2
Δ=y
Energy gain +ve
Energy loss -ved
x
L
kx
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W
By the theory of minimum potential energy (see page B7)
U(with geometric effect)
A curve surface
Equilibrium positionat local min. i.e. δП=0
Stable
Concept of Stability
П
x
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U+W= =U
=W
δП≠0
δП≠0δП=0
N
this implies
x
can be any
value. i.e. buckling! N=kL/2
U=
=W
=U+W
U = 0П
= W
F = 0 -F x =0П
= W = 0
=WδП=0for all x
a line
No local minimum
x=F/(k-2N/L)
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Notes on TomorrowNotes on Tomorrow’’s Tests Test• The test venue is at Knowles Building 419• The design part is on slab design.• The theory part is on sketch of BMD and assembly
of system matrix and vectors• Please bring along the HK code and a calculator.• Table for reinforcement areas will be provided.
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Buckling Problems with Multiple Degrees of Freedom
Energy Approach
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Assumed disp.
shape
Have been derived in Matrix Method for Static Load Analysis
(ai
=Nodal displacements)
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Consider a beam segment of length dx
Longitudinal deformations1.
Hooke’s law (axial shortening)2.
Geometric
u’
= du/dxv’
= dv/dx
Binomial expansionSee page B5
(due to geometric deformation)
Hooke’s law Geometric
dxdu dx
dx=ds
du
dv
v
ignored
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Since v = Σ
ai
Ni
4
i=1
(see page B6)
-ΔP=
044
33
22
11
aa
aa
aa
aa
Hence:
Not a function of xa function of x
Π
is a function of ai
Strain energy axial load nodal forcesθ=dv/dx=v’dθ/dx=v’’dθ=v”dxM=EI v”
dU=½Mdθ
Note:
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4x4
scalar
Geometric stiffness affecting the total stiffness of the beam element by the applied axial load P
PP
=kij
=gij
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These two matrices will be given in the examination.
Stiffness matrix
Geometric stiffness matrix
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22
22
22
22
3
433336336
343336336
304626
6126122646612612
LLLLLL
LLLLLL
LP
LLLLLL
LLLLLL
LEIPgk
The stiffness matrix and geometric stiffness matrix
It will be provided in the Examination paper.
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Consider the last two eqns.
(Buckling Analysis)
Only considering the deformation due to geometric change
There is no end shear force and moment.
2
22
vd
=0
u2
PWhen buckling just occurred, u2 =0
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Detailed solution of the Detailed solution of the eigenvalueseigenvalues
043
3363046
612det 223
ll
ll
Pll
llEI
043336
3046612
det2
EI
Pl
EIPl
30
2
043336
46612
det
04436363612
det
013515612 2
222 1807.32or4860.230lEI
lEI
lEIP
0)36()44)(3612( 2
2
2
2
2
lvv
043336
3046612
det 3
l
PlEI
Set θ=lθl l l l
=0.0829 or 1.0727
By formula02 cbxax
aacbbx
242
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by the first equation9.01685/l×1-
5.7514θ2
=0 or θ2
=9.01685/5.7514l=1.5677/lor by the second equation-5.7514×1
+ 3.66854lθ2
=0 or θ2
=5.7514/3.66854l=1.5677/l
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2DOF only
Subdivided into more line segments /elements
Knowing the nodal displacements, the column deformation can be calculated by interpolation.a3 a4
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Sketch the mode shapes
Mode 1, v2 =1 and θ2 =1.5677/l
v
x/lθ2
θ1 =0
v
x/lθ2
θ1 =0
Mode 2, v2 =1 and θ2 =-9.567/l
v2 1 v2 1
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=0
XY
=0
u2
PWhen buckling just occurred, u2 =0
C
=0 Axial force=0
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Stiffness matrix of a beam Stiffness matrix of a beam elementelement
Stiffness equations for element 1
212
121
42
24
ML
EI
ML
EI
v1
=0 v2
=0u1
=0 u2
=0
2
2
2
1
1
1
22
22
22
22
3
0
0
4626612612
2646612612
uv
uv
LLLLLL
IAL
IAL
LLLLLL
IAL
IAL
LEI
0
0
removed
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1 2
For Element 1
NO AXIAL FORCE (P=0)
0
Node 1 is fixed and vertical displacement at node 2 =0
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1
2
For Element 2
Node 1 is fixed
2
v2
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The stiffness matrix of element 2 is (the same as that in Example 1)
For element 2
XY
Element from example 1
Rotate the element and the local coordinate axes by 90o clockwise
The coordinate system is the same as that of element 1 in example 1.
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Eqn(3)
Involving one DOF only.
Stiffness from element 1
To get accurate results, we have to use more elements
C
System
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≠0Sliding only, no rotation
System
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=0By static condensation reduced 3DOFs to 2DOFs
System &
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Matrix condensation 2DOF 1DOF
=0By static condensation reduced 3DOFs to 2DOFs
221
121
2
1
2
1
42
24
4224
MLEI
MLEI
MM
LEI
The effects of pinned joint i.e. M1
’=0 has been considered.
&System
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No transformθ1
u1
With stiffness matrix only
Stiffness+ geometric
matrices
Ignore the axial deformations v1
=v2
=0, u1
=u2
2
1
YX
Fixed end
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θ2
u1
Take away l
in the matrices of the above equation
System
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With only numerical values in the matrices
(details see next slide)
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Expansion of DeterminantExpansion of DeterminantFor a 2×2 determinant
For a 3×3 determinant
2bcacbba
edcb
dfdeb
bfeec
afedecbdba
= acf
+ 2bde -
ae2 -
b2f –
cd
2
It will be provided in the Examination paper.
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P
>
Considering the symmetric buckling mode of the 1Considering the symmetric buckling mode of the 1stst
mode, the mode, the problem can be simplified as a cantilever with a length problem can be simplified as a cantilever with a length LL/2 subjected /2 subjected to a point load to a point load
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Past examination question, year 2000
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EndEnd
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(Shear Building)
The flexural stiffness of the horizontal beam elements is infinitely high and the beams would not be bent and rotated. This type of buildings
is called shear building. Each story has one horizontal DOF only.
Undeformed
shape Deformed shape
2DOF=0
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+
SkippedIllustration of Energy Method
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≠0
k Deq
dD
(internal W.D.)+ -P dD
(external W.D.)0
Skipped
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Formulation for the stiffness Formulation for the stiffness matrix of element 1matrix of element 1
No use Stiffness for element 1
removed
212211
121111
420
240
ML
EIFv’L
EAv’L
EA
ML
EIFv’L
EAv’L
EA
v’1
= -v1 v’2
= -v1
2
2
1
1
2
12
1
1
22
22
22
22
3
42
24
MFMF
v’v’
v’
LLI
ALI
ALLL
IAL
IAL
LEI
v1 v1
u1
=0 u2
=0
2
2
2
1
1
1
22
22
22
22
3
0
0
4626612612
2646612612
uv
uv
LLLLLL
IAL
IAL
LLLLLL
IAL
IAL
LEI