KB6005 Advanced Structural Systems: 2019/20 Semester 1
KB6005 Advanced Structural Systems: 2019/20 Semester 1
KB6005 Advanced Structural Systems: 2019/20 Semester 1
2019/20 Semester 1
0191-349-5121
sherry.chen@northumbria.ac.uk
1
Lecture 3: Buckling
• Basic concepts
• Critical buckling loads of columns
• Finite element analysis
2
Basic Concepts
3
General mechanical failure modes I
Ultimate
stress
4
General mechanical failure modes II
Yield
stress
5
General mechanical failure modes III
6
General mechanical failure modes IV
lgS
Endurance
limit stress
lgSe
Infinite life
lgN
7
General mechanical failure modes V
Buckling
• Instability in columns, shells, frames and
beams
• Often leads to a collapse, as a buckled
member loses the entire or part of its load
bearing capacity
8
Failure due to buckling
9
Experiment
Video: https://www.youtube.com/watch?v=jNwvub87l8o
10
Critical buckling loads
of columns
11
Critical load
Safe Failed
12
Euler’s Formula for critical load
F
For beams with both ends
pinned
𝝅𝟐 𝑬𝑰
𝑭𝒄𝒓 = 𝟐
𝑳
• E : Young’s modulus
• L : beam length
• I : second moment of area
13
Second moments of area of some
particular shapes
𝝅𝑫𝟒
𝑰𝒙 = 𝑰𝒚 =
𝟔𝟒
𝒃𝒉𝟑 𝒃𝟑 𝒉
𝑰𝒙 = 𝑰𝒚 =
𝟏𝟐 𝟏𝟐
𝝅
𝑰𝒙 = 𝑰𝒚 = 𝑫𝟒 − 𝒅𝟒
𝟔𝟒
14
Generalised Euler’s Formula for
critical load and stress
𝝅𝟐 𝑬𝑰
𝑭𝒄𝒓 =
(𝑳𝒆 )𝟐
• Le : effective length
Le depends on the boundary conditions of
the beam.
15
Effective length for different boundary
conditions I
16
Effective length for different boundary
conditions II
17
Example I
F = 2 kN
A force F of 2 KN is applied on the free
end (A) of a steel beam AB. The other end
(B) is fixed. The beam length L is 1.5 m.
The Young’s modulus of steel is E = 200
GPa.
Find the Factor of Safety (FoS) for the
beam.
50 mm
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Example I – continued I
𝝅𝟐 𝑬𝑰
𝑭𝒄𝒓 =
(𝑳𝒆 )𝟐
• E = 200 GPa
= 𝟐𝟎𝟎 × 𝟏𝟎𝟗 𝐏𝒂
19
Example I – continued II
𝝅𝟐 𝑬𝑰
𝑭𝒄𝒓 =
(𝑳𝒆 )𝟐
• E = 𝟐𝟎𝟎 × 𝟏𝟎𝟗 𝐏𝒂
𝝅𝑫𝟒
• 𝑰 =
𝟔𝟒
𝝅 −𝟑 𝟒
= 𝟓𝟎 × 𝟏𝟎 𝐦
𝟔𝟒
−𝟕 𝟒 𝝅𝑫𝟒
= 𝟑. 𝟎𝟕 × 𝟏𝟎 𝐦 𝑰𝒙 = 𝑰 𝒚 =
𝟔𝟒
20
Example I – continued III
𝝅𝟐 𝑬𝑰
𝑭𝒄𝒓 =
(𝑳𝒆 )𝟐
• E = 𝟐𝟎𝟎 × 𝟏𝟎𝟗 𝐏𝒂
• 𝑰 = 𝟑. 𝟎𝟕 × 𝟏𝟎−𝟕 𝐦𝟒
• Le = 2L
= 𝟐 × 𝟏. 𝟓 𝐦
=𝟑𝐦
21
Example I – continued IV
• E = 𝟐𝟎𝟎 × 𝟏𝟎𝟗 𝐏𝒂
• 𝑰 = 𝟑. 𝟎𝟕 × 𝟏𝟎−𝟕 𝐦𝟒
• Le = 𝟑 𝐦
⇒ 𝑭𝒄𝒓 = 𝟔𝟕 × 𝟏𝟎𝟑 𝐍
22
Example I – continued V
𝟔𝟕×𝟏𝟎𝟑 𝐍
=
𝟐×𝟏𝟎𝟑 𝐍
≈ 𝟑𝟒
23
Example II
a
a
24
Example II – continued I
= 𝟏𝟎𝟎 𝐤𝐍 × 𝟐. 𝟓
= 𝟐𝟓𝟎 𝐤𝐍
= 𝟐𝟓𝟎 × 𝟏𝟎𝟑 N
25
Example II – continued II
a
𝟏
𝑰𝒙 = 𝒂𝒃𝟑
𝟏𝟐
𝟏
𝑰𝒚 = 𝒃𝒂𝟑
𝟏𝟐
26
Example II – continued III
𝝅𝟐 𝑬𝑰
𝑭𝒄𝒓 =
(𝑳𝒆 )𝟐
𝟏 𝟒
• 𝑰= 𝒂
𝟏𝟐
• 𝑳𝒆 = 𝑳
27
Example II – continued IV
𝝅𝟐 𝑬𝑰
𝑭𝒄𝒓 =
(𝑳𝒆 )𝟐
𝟏 𝟒
• 𝑰= 𝒂
𝟏𝟐
• 𝑳𝒆 = 𝑳
𝟏
𝝅𝟐 𝑬×𝟏𝟐𝒂𝟒
⟹ 𝑭𝒄𝒓 =
(𝑳)𝟐
𝝅𝟐 𝑬
= × 𝒂𝟒
𝟏𝟐𝑳𝟐
28
Example I – continued V
= 𝟎. 𝟎𝟗𝟖𝟑 𝐦
29
Finite Element Analysis
30
Comparison
31
Buckling analysis in ABAQUS
32
Buckling modes of polymer sheet
33
Compression-induced buckling
modes in nature
34
Buckling modes in spheroidal
film/substrate systems I: modelling
35
Buckling modes in spheroidal film/substrate
systems II: comparison with natural
biological system
FEA
results
Butterfly’s egg Human
neutrophil cell
FEA
results 36
Strain transducers in MEMS (Micro-
Electro-Mechanical Systems)
Bi-layer elastomer on pre-stretched
Fluorophore spreading onto
substrate
plasma treated surface
Fluorophore coating
Measured thickness
of coating at 0 strain
~ 600 nm
Compression
Compression d
f
Guided morphology of
folding-in
Lateral (y-direction) line patterns of surface creasing under uniaxial (x
direction) compression
Compression
38
Finite element model
39
Post-buckling analysis (video)
40
Effect of geometry
41
Summary I
Safe Failed
42
Summary II
Generalised Euler’s Formula for critical
buckling load of columns:
• E : Young’s modulus
𝝅𝟐 𝑬𝑰 • Le : effective length
𝑭𝒄𝒓 =
(𝑳𝒆 )𝟐 • A : cross-sectional area
• I : second moment of area
One fixed end, Both ends One fixed end, Both ends
one free end pinned one pinned end fixed 43
Summary III
44
Further studies:
Column buckling
45
Beer, F. P., Johnston, E. R. Jnr.,
DeWolf, J. T. and Mazurek, D. F.
(2015). Mechanics of Materials,
7th Edition in SI units. McGraw-
Hill.
46