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Aero Structures-Column Buckling
1
By
Dr. Mahdi Damghani
2016-2017

Suggested Readings
Reference 1
2
Reference 2
Chapter 9 of Ref [1]
Chapters 8 and 9 of Ref [2]

Topics
3
• Familiarisation with buckling of columns with various
boundary conditions
• Spar function
• Spar loading
• Buckling of web of spars

Introduction
• Buckling is a stability issue (not
strength issue)
• Compressive stress is high enough
to trigger sudden sideways
deflection of slender structures
• Buckling stress is less than yield
stress of material (so material does
not fail but becomes unstable)
• A large proportion of an aircraft’s
structure comprises thin webs
stiffened by slender stringers
(stiffeners, longerons wing skin,
spar webs, etc)
4

Buckling of columns
10
• The first significant contribution to the theory of the
buckling of columns was made as early as 1744 by
Euler
• His classical approach is still valid, and likely to
remain so, for slender columns possessing a variety
of end restraints.

Reminder
z
z2
z1
The bending moment
M causes the length of
beam to bend about a
centre of curvature C
Element is small in length and a pure moment is
applied. The curved shape can be assumed to be
circular with a radius of curvature R measured to
the neutral plane
StressesSatrnedssetrsaiannsdare
zero on n
se
tru
atir
n
a
s
l ar
xe
iszs
e
o
ro
its
length d o
e
nsn
n
e
o
u
ttr
c
a
h
la
a
n
x
g
is
e
Planes at an
angle of 
z
12

Reminder
z
z2
z1
z
Fibre ST has shortened in
length whilst NQ
increased in length so
they have gone through
strains
Remember that the length
of neutral axis does not
change and remains as z
Original Length
z
 
Change in Length

z
R R
 
R  y  R
 
y
Positive y gives negative
strain, i.e. compression
z z
 
R  y z
  
R  y  R 
z z
z z
R
z
  E    E
y

13

Reminder
• Look at the beam cross section now;
dA
Neutral axis
y
R
z
  E
y


A
A A A
  z  R
E
R
y 2
M  Fy   ydA   E ydA   y dA 
M  
EI

R
dA
y

 1 
 dz2

 d 2
y 
 R 
M  EI   EI
14

Reminder
• Solving second order homogenous differential equations;
15

Reminder
m2

PCR
 0 
EI




EI
EI
EI
P P

PCR
EI
m   
PCR
 i
CR
 i CR
2
m1  
PCR
2

PCR

EI EI
  v  e0z
Acosz  Bsin z
v  Acosz  Bsin z

16

Buckling of columns (hinged-hinged)
A,B  ?
@ z  0, z  l  v  0
A  0
B sin l  0  sin l  0  l  n;n  (1,2,3,...)
l2
 2
EI
PCR
17

• These higher values of buckling load cause more
complex modes of buckling
• If no restraints are provided, then these forms of
buckling are unstable and have little practical
meaning
18

• We like to work with stresses
so;
 2
EI
PCR 
l2
CR

PCR
A

 
 
 A
 I 
l
CR
2

2
E

 
 
 r 
 l 
2
 r2

 l2

CR

 2
E

 2
E

• r is radius of gyration
• l/r is slenderness ratio
I
A
r 
• Taking
19

Column buckling (clamped-clamped)
• In practice, columns usually have
their ends restrained against
rotation so that they are, in effect,
fixed
• An axial compressive load that
has reached the critical value,
PCR, so that the column is in a
state of neutral equilibrium
• The ends of the column are
subjected to clamping moments,
MF, in addition to axial load
20

@ x  0  v  0
@ x  l  v  0
21

v is indeterminate as we
do not know the value of
MF
But we know slope
(dv/dx) at x=l is zero
22

Boundary condition effects
• So far we have seen that for a hinged-hinged column
the buckling load is;
• For a clamped-clamped column is;
• What do you make of this?
 2
EI
PCR 
l2
4 2
EI
PCR 
l2
23

Column buckling (general)
• In a more general form buckling stress can be written
as;
• le is called effective length which is multiple of the
length of structure depending on the boundary
conditions
le  kl
24

Boundary conditions
k 1
k  0.5
le  kl
k  0.7
k  2
k 1
k  2
25

Example 1
• A 7m long steel tube having
the cross section shown is
to be used as a pin-ended
column. Determine the
maximum allowable axial
load the column can
support so that it does not
buckle or yield. Take the
yield stress of 250MPa.
26

Example 2
28
• A 2m long pin ended column with a square cross
section is to be made of wood. Assuming E=13GPa
and allowable stress of 12MPa (σall=12MPa) and
using a factor of safety of 2.5 to calculate Euler’s
critical load for buckling. Determine the size of the
cross section if the column is to safely support a 100
kN load.

Solution
For a square of side a;
Now let’s check stress in the column;
P PCR  FS 100  2.5100  250kN
 2
EI P l2
PCR   I  CR

l2
 2
E
250103
N2m2

I 
   7.79410 6
m4
 2
13109
Pa
3 6 4
12 12
1 a4
I  a a   7.794 10 m  a  98.3mm 100mm
A
 10MPa 
100kN
0.100m2
 
P
   all 10 12
29

Example 3
• A uniform column of length L and flexural stiffness EI
is simply supported at its ends and has an additional
elastic support at mid-span. This support is such that
if a lateral displacement vc occurs at this point, a
restoring force kvc is generated at the point. Derive an
equation giving the buckling load of the column.
30

Solution
M
v  0
@ z  0
@ z  L / 2
v  vc dv/ dz  0
@ z  L / 2
A  0
31

Example 4
• A uniform column of length l and bending stiffness EI
is built-in at one end and free at the other and has
been designed so that its lowest flexural buckling load
is P. Subsequently, it has to carry an increased load,
and for this, it is provided with a lateral spring at the
free end. Determine the necessary spring stiffness k
so that the buckling load becomes 4P.
32

Solution
• Buckled state of the
column and acting force
of spring
M
33

Tutorial 1
• The structural member shown is to be used as a pin-connected
column. Determine the largest axial load it can support before it
either begins to buckle or the steel yields. Section properties are
as;
35

Tutorial 2
• A 3m column with the following
cross section is constructed of
material with E=13GPa and is
simply supported at its two
ends.
• Determine Euler buckling load
• Determine stress associated with
the buckling load
36

Tutorial 3
• A uniform, pin-ended
column of length l and
bending stiffness EI has an
initial curvature such that
the lateral displacement at
any point between the
column and the straight line
joining its ends is given by
• Show that the maximum
bending moment due to a
compressive end load P is
given by
37

Spars
38
Span-wise members that carry
shear loads.
 Fuel Tank Boundary.
Provide mounting for WLG
Fittings and Leading and Trailing
edge fittings.

Spars loading
F
F
d
M
The distance between the
centroid of spar caps
39

Spar loading
• Spar caps (boom) is
assumed to carry axial
load only
• Spar web is assumed to
carry constant shear
stresses (average
shear) only
d
40

Spar buckling
• Spar web is so thin that it is susceptible to
buckling under shear loading
• The web of the beam buckles under the
action of internal diagonal compressive
stresses produced by shear
• This leads to a wrinkled web capable of
supporting diagonal tension only in a
direction perpendicular to that of the buckle
• The beam is then said to be a complete
tension field beam
41

(t ,0)

Reminder
• How does compression come about?

(0,)
(c ,0)
42

Complete diagonal tension
• At any section where shear force is S, average shear
stress for spar web is;
43

• Let’s isolate element ABCD at a chosen
section
• Tensile stresses σt are present on faces
AB and CD
• The angle of diagonal tension is α
• On vertical plane FD we have both shear and direct stresses
• Let’s write equation of equilibrium for element FDC;
Fy  0 CD  FDcos
S W
44

• Let’s write equation of equilibrium for
element FDC;
Fx  0 CD  FDcos
S W
45

46
• Both tensile stress and shear stress are constant
through the depth of the beam
• σz is also constant

• Direct forces in the flanges (booms) can be calculated
by drawing a free body diagram at length z of the beam;
Fx  0
47

• The diagonal tensile stress (σt) induces a direct stress
(σy) on horizontal planes at any point in the web
• On a horizontal plane
HC in the element
ABCD, there is a direct
stress (σy) and a
complementary shear
stress 
48

(t ,0)

2
49
Reminder
• Why do we have horizontal stress on plane HC?

(y ,)
(0,0)

Fy  0
• σy causes compression in the vertical stiffeners so we
can find axial compressive force in the stiffener P as
of the next slide;
50

b

b
 b
2 2
y
d
51
Wb
td
y
y
P  tb  P  tan
 
W
tan

Buckling of stiffeners
• If P is high then stiffeners can buckle similar to
columns
• Effective length of stiffener is then taken as;
• Load P also brings about another affect. Can any
body indicate as to what could happen next?
52

Bending of flanges
y
yt
Mmid  0.5M max
53

Angle α
Adjusts itself for
potential energy to
become minimum
If stiffeners and
flanges are assumed
to be rigid then α=45o
In real life nothing is
rigid and therefore
54
38o<α< 45o

Angle α
Uniform direct
compressive stress
in the stiffener
• The angle can be calculated as follow;
Uniform direct
compressive stress
in the flange
We know
F S
55
A and A are cross
sectional area of flange and
stiffener, respectively

Example 5
56
• The beam is assumed to have a complete tension field
web. If the cross-sectional areas of the flanges and
stiffeners are, respectively, 350mm2 and 300mm2 and
the elastic section modulus of each flange is 750mm3
• Determine the maximum stress in a flange
• Determine whether or not the stiffeners will buckle
• The thickness of the web is 2mm, and the second
moment of area of a stiffener about an axis in the plane
of the web is 2,000mm4; E = 70,000N/mm2.

Solution
2
1
17700
350
 50.7N / mm
 
2
4
2
750
114.9N / mm
8.610
 
2
Total  1 2 165.6N / mm
58

Solution
b

b
 b
2 2
y
d
P 
Wb
tan
P  PCR  No buckling
60

Tutorial 1
61
• The beam shown in the next slide is clamped at one end
and has a point load of 7.0kN at the other end. The beam is
assumed to have a complete tension field web. The cross-
sectional areas of the flanges and stiffeners are,
respectively, 385mm2 and 300mm2 and the elastic section
modulus of each flange is 750mm3. The thickness of the
web is 2mm, and the second moment of area of a stiffener
about an axis in the plane of the web is 1,500mm4. Assume
modulus of elasticity of material E = 71,700N/mm2.
• Determine the maximum stress in a flange
• Determine whether or not the stiffeners will buckle

Tutorial 2
63
• A simply supported beam has a span of 2.4m and carries
a central concentrated load of 10 kN. The flanges of the
beam each have a cross-sectional area of 300mm2, while
that of the vertical web stiffeners is 280mm2. If the depth of
the beam, measured between the centroids of area of the
flanges, is 350mm and the stiffeners are symmetrically
arranged about the web and spaced at 300mm intervals,
determine the maximum axial load in a flange and the
compressive load in a stiffener. It may be assumed that
the beam web, of thickness 1.5mm, is capable of resisting
diagonal tension only.

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Buckling and tension.pptx

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Buckling and tension.pptx