ELASTIC BUCKLING
So far we have discussed:
(1) the strength of the structure, i.e., its ability to support a specified load without
experiencing excessive stress;
(2) the ability of the structure to support a specified load without undergoing
unacceptable deformations.
we will study the stability of the structure,
structure i.e.,
i e its ability to support a given load
without experiencing a sudden change in its configuration.
Consider a long straight slender member subjected
to axial force. Such a member is called a
COLUMN. Assume both ends are pinned. Top end
is restricted to move side ways.
As the load P is increased, theoretically there should not be any transverse deflection perpendicular to the
member
b axis.
i
Due to a small disturbance in the transverse direction or due to some imperfection in the column the
member will bend as shown in figure.
On the release of load, if the column return to its straight vertical position, then it is in stable condition
under the applied load.
As the load is increased further, for a particular load, transverse deflection increases without any increase
off the
h axial
i l load.
l d (Unstable
(U
bl condition).
di i )
The load at which unstable condition is attained is called critical load.
We shall
W
h ll note that
h if its
i equilibrium
ilib i
is
i disturbed,
di
b d the
h system will
ill return to its
i original
i i l equilibrium
ilib i
position
ii
as long as P does not exceed a certain value Pcr, called the critical load. However, if P > Pcr, the system
will move away from its original position. In the first case, the system is said to be stable, and in the
second case, it is said to be unstable.
EULER'S FORMULA FOR PIN‐ENDED COLUMNS
The free body diagrams of the column and a section of the column at a distance x is shown above 2
d 2v M
P

 v  d v  P v0
2
dx
EI
EI
dx 2 EI
P
Setting
g 2 
we can write the above equation
q
as
EI
d 2v
  2v  0
dx 2
The above is the second order differential equation, the general solution for the above equation is given by
v  A sin  x  B cos  x
Boundary conditions are at x =0, v=0 and at x=L, v=0
B
d
diti
t 0 0 d t L 0
Therefore B= 0; we obtain
A sin  x  0
A = 0 is trivial solution; therefore sin L = 0
That leads to
n 2 2 EI
P
L2
The smallest of the values of P defined by above equation is that corresponding to n = 1. We thus have
Pcr 
 2 EI
L2
The expression obtained is known as Euler's formula, after the Swiss mathematician Leonhard Euler (1707 ‐1783). The Deflection equation is given by
v  A sin
i
x
L
which is the equation of the elastic curve after the column has buckled (see figure). which
is the equation of the elastic curve after the column has buckled (see figure)
We note that the value of the maximum deflection, vmax = A, is indeterminate. This is due to the fact that the differential equation is a linearized approximation of the actual governing differential equation for the elastic curve. If P < Pcr, the condition sin PL = 0 cannot be satisfied, and the solution does not exist. We must then have A = 0, and the only possible configuration for the column is a straight one Thus for P <Pcr the straight configuration of is stable.
straight one. Thus, for P <P
the straight configuration of is stable
• In the case of a column with a circular or square cross section, the moment of inertia I of the
cross section is the same about any centroidal axis, and the column is as likely to buckle in
one plane
l
as another,
h except for
f the
h restraints
i which
hi h may be
b imposed
i
d by
b the
h endd
connections.
• For other shapes
p of cross section,, the critical load should be computed
p
byy makingg I = Imin
• If buckling occurs, it will take place in a plane perpendicular to the corresponding principal
axis of inertia.
• The value of the stress corresponding to the critical load is called the critical stress and is
denoted by cr.
• Setting I = Ar2, where A is the cross-sectional area and r its radius of gyration, we have
Pcr  2 EAr 2
 2E
 cr 


2
A
AL2
L/ r
• The quantity L/r is called the slenderness ratio of the column.
• It is clear, that the minimum value of the radius of gyration r should be used in computing
the slenderness ratio and the critical stress in a column.
the critical stress is proportional to the modulus of elasticity of the
material, and inversely proportional to the square of the
slenderness ratio of the column.
The plot of cr versus L/r is shown in Figure for structural steel,
assuming E = 200 GPa and Sy = 250 MPa. We should keep in mind
that
h no factor
f
off safety
f has
h been
b
usedd in
i plotting
l i cr.
We also note that, if the value obtained for cr is larger than the
yield strength
y
g Sy, this value is of no interest to us,, since the column
will yield in compression and cease to be elastic before it has a
chance to buckle.
Our analysis of the behavior of a column has been based so far on
the assumption of a perfectly aligned centric load. In practice, this
is seldom the case.
Taking into account the effect of the eccentricity of the loading.
lead to a smoother transition from the buckling failure of long,
slender columns to the compression failure of short, stubby
columns It will also provide us with a more realistic view of the
columns.
relation between the slenderness ratio of a column and the load
which causes it to fail.
EXTENSION OF EULER'S FORMULA TO COLUMNS WITH OTHER END CONDITIONS
E l ' formula
Euler's
f
l was for
f a column
l
which
hi h was pin-connected
i
d at both
b h ends.
d
In the case of a column with one free end A supporting a load P and the other end fixed
pp half of a pin-connected
p
column.
we observe that the column will behave as the upper
The critical load for the column of is thus the same as for the
pin-ended column of Figure (b) and may be obtained from
Euler's formula by using a column length equal to twice the
actual length L of the given column.
We say that the effective length Le of the column of Figure (a)
is equal to 2L and substitute Le = 2L in Euler's formula:
Fixed‐Free column
Fixed‐Fixed Column
• The symmetry of the supports and of the loading about a horizontal axis through the
midpoint
id i t C requires
i that
th t the
th shear
h
att C andd the
th horizontal
h i t l components
t off the
th reactions
ti
att
A and B be zero.
• It follows that the restraints imposed upon the upper half AC of the column by the support
y the lower half CB are identical.
at A and by
• Portion AC must thus be symmetric about its midpoint D, and this point must be a point
of inflection, where the bending moment is zero.
• A similar reasoning shows that the bending moment at the mid-point E of the lower half
of the column must also be zero.
zero
• Since the bending moment at the ends of a pin-ended column is zero, it follows that the
portion DE of the column of must behave as a pin-ended column (b). We thus conclude
that the effective length of a column with two fixed ends is Le = L/2.
Fixed‐Pinned Column
• In the case of a column with one fixed end B and one pin-connected end A
supporting a load P we must write and solve differential equation of the elastic
curve to determine the effective length of the column.
• From the free-body diagram of the entire column we first note that a transverse
force V is exerted at end A, in addition to the axial load P, and that V is statically
indeterminate.
• Considering now the free-body diagram of a portion AQ of the column, we find
that the bending moment at Q is
M  Pv Vx
d 2v M
Pv Vx




dx 2 EI
EI EI
P
Set  2 
EI
d 2v
Vx
2
then




v
dx 2
EI
The solution for the above differential equation consist of general solution plus p
particular solution
Particular solution is given by
v
V
V
x x
2
 EI
P
 v  A sin  x  B cos  x 
V
x
P
The constants A, B and V can be solved
using Boundary conditions
dv
0
at x=0, v=0; at x= L, v=0 and
dx
Using these boundary conditions, we get
B 0
B=0;
V
A sin  L = L (a)
P
V
A cos  L=
(b)
P
From (a) and (b), we obtain
tan  L  PL
Solving by trial and error the above transcendal equation, we get
 L  4.4934
4 4934
P
20.19 EI
, we obtain Pcr 
EI
L2
Using effective length
length, we can write
Since  2 
 2 EI
2
e
L

20.19 EI
 Le  0.699 L  0.7 L
L2
ECCENTRIC LOADING ‐ THE SECANT FORMULA
The load P applied to a column is never perfectly centric.
‘e’ the eccentricity of the load, i.e., the distance between the line of action of P and the
axis of the column.
column
Replace the given eccentric load by a centric force P and a couple MA = Fe
Now, no matter how small the load P and the eccentricity e, the couple MA will cause
some bending of the column.
As the eccentric load is in-creased, both the couple MA and the axial force P
increase, and both cause the column to bend further.
Viewedd in
Vi
i this
thi way, the
th problem
bl
off buckling
b kli is
i nott a question
ti off determining
d t
i i how
h
long the column can remain straight and stable under an increasing load, but rather
how much the, column can be permitted to bend under the increasing load, if the
allowable stress is not to be exceeded and if the deflection vmax is not to become
excessive.
Secant Formula
Drawing the free‐body diagram of a portion AQ of the column choosing the coordinate axes as shown , we find that the bending moment at Q is
di t
h
fi d th t th b di
t tQi
M   Pv  M A   Pv  Pe
d 2v M
Pv Pe



2
dx
EI
EI EI
P
Set  2 
EI
2
d v
then
  2 v   2 e
dx 2
The solution for the above differential equation consist of general solution plus particular solution
Particular solution is given by v = ‐e
Therefore complete solution is given by
 v  A sin  x  B cos  x  e
The constants A, B can be obtained
using Boundary conditions
dv
0
at x=0, v=0; at x= L, v=0 and
dx
Using these boundary conditions, we get
B = e;
A sin  L = e(1  cos  L)
(a)
Using the trigonometric relations
sin  =2 sin( /2)cos( /2); and cos( )=1-2 2sin( /2)
A = e ttan
L
2
Substituting A and B we get
L


v  e  tan
sin  x  cos  x  1
2


The maximum deflection is at x=L/2
L


vmax  e  tan
sin( L / 2)  cos( L / 2)  1
2


 L  
=e  sec    1
  2  
  P L 
=e  sec 
  1

  EI 2  
From the above expression vmax becomes infinite when
P L 

EI 2 2
While the deflection does not actually become infinite, it nevertheless becomes
unacceptably large, and P should not be allowed to reach the critical value
Pcr 
 2 EI
2
Which is the same critical load under centric load
L
Solving for EI from the above equation and substituting in the equation for vmax we get
  
vmax  e  sec 
 2
 
P L  
  1
Pcr 2  
Maximum bending moment will be at mid‐span of the column
  
M max  P (e  vmax )  Pe  sec 

  2
P
Pcr
 
 
 
Maximum compressive stress is given by
Maximum compressive stress is given by
P Mc

A
I
 
P  ec
= 1  2 sec 
A  r
 2
 max 
P
Pcr
 P  ec

  1  2 sec 
2
 A  r
P L 

EI 2 
We note that, since max does not vary linearly with the load P, the principle of
superposition does not apply to the determination of the stress due to the simultaneous
application
li i
off severall loads;
l d the
h resultant
l
l d must first
load
fi
b computed,
be
d and
d
corresponding compressive stress must be computed.
For the same reason,, any
yg
given factor of safety
y should be applied
pp
to the load,, and not to
the stress.
Making I = A r2
P

A
and solving for the ratio P/A in front of the bracket, we write
 max
1
 1 P Le 
ec
sec


r2
 2 EA 2 
where the effective length is used to make the formula applicable to various end
conditions.
diti
This formula is referred to as the secant formula; it defines the force per unit area, P/A,
which causes a specified
p
maximum stress ((max in a column of g
given effective
slenderness ratio, Le/r, for a given value of the ratio ec/r2, where e is the eccentricity of
the applied load. We note that, since P/A appears in both members, it is necessary to
solve a transcendental equation by trial and error to obtain the value of P/A
corresponding to a given column and loading condition.
condition
The above equation was used to draw the curve shown in Figure below for a steel column,
assuming the values of E and Sy shown in the figure. These curves make it possible to
determine the load per unit area P/A, which causes the column to yield for given values of
the ratios Lelr and ec/r2
We note that, for small values of Le/r, the secant is almost equal to 1 in the equation for
P/A), and PIA may be assumed equal to
P  max

A 1  ec
r2
a value
al e which
hich could
co ld be obtained by
b neglecting the effect of the lateral deflection of the
column.
On the other hand, for large values of Le/r, the " curves corresponding to the various values
of the ratio ec/r2 get very close to Euler's curve, and thus that the effect of the eccentricity
of the loading on the value of P/A becomes negligible.
The secant formula is chiefly useful for intermediate values of Lelr.
lr
However, to use it effectively, we should know the value of the eccentricity e of the
loading, and this quantity, unfortunately, is seldom known with any , degree of accuracy.