Optimal placement of supports against elastic instability
We consider the system illustrated in Figure 1, in which a beam of flexural
rigidity EI is built in at z = 0, simply supported at z = a and loaded by an
axial force P at z = L. The problem is to determine the optimal location
for the support (and hence value of a) such that the first critical load of the
supported system is as large as possible.
L
P
z
a
Figure 1
We shall approach the problem in two ways. Firstly, we shall assume that
the following theorem is true:“Suppose the stability problem for an elastic structure under a load P
is formulated as a small-displacement linear eigenvalue problem with critical
loads P1 , P2 , ... such that P1 < P2 < P3 ... etc with corresponding eigenmodes
u1 , u2 , .... If a single support is now placed at the (presumably unique) node
of the eigenmode u2 , so as to constrain the displacement there to zero, the
critical load of the new system will be larger than that obtained by placing a
single support at any other point in the structure.”
We shall then give a solution of the problem without making this assumption and verify that it gives the same result.
Solution using the proposed theorem
The theorem states that the optimal location for the support will be at the
node of the second mode of the unsupported system illustrated in Figure 2.
z
P
L
Figure 2
1
The problem of Figure 2 is elementary, since only the part 0 < z < L
of the beam is loaded and that merely by the axial force P . If the lateral
displacement at z = L is denoted by d, we have a bending moment
M(z) = P (d − u(z)) = EI
and hence
d2 u
dz 2
d2 u P u
Pd
+
=
,
2
dz
EI
EI
with general solution
u = d + A cos(λz) + B sin(λz)
where
λ=
s
P
.
EI
The boundary conditions are
u(0) = u′ (0) = 0 ;
u(L) = d ,
giving
d+A=0;
Bλ = 0 ;
A cos(λL) + B sin(λL)
Thus, B = 0 and the third condition gives
A cos(λL) = 0
with non-trivial solution
(2n − 1)π
,
2
where n is an integer. The second mode corresponds to n = 2 and
λL =
λ=
3π
; ; d = −A
2L
and
u(z) = −A[1 + cos(λz)] ; 0 < z < L .
In z > L, the beam is unloaded and hence straight. We can therefore find the
point at which this straight line cuts the undeformed line u = 0 by finding
the slope at z = L. We have
u′(z) = Aλ sin(λz) ;
u′(L) = Aλ sin(λL) =
2
3π
3πA
3πA
sin( ) = −
2L
2
2L
and
u(L) = d = −A .
Thus, the straight line will cut the axis a distance
−A
3πA
−
2L
=
2L
3π
to the right of the support. The theorem therefore predicts that the optimal
value of a is
2L
2
= 1.2122L .
a=L+
=L 1+
3π
3π
Solution from first principles
We now consider the same problem but without invoking the theorem. Once
again, we denote the displacement at the load
u(L) = d ,
but this time there will also be a reaction at the support which we denote as
R. The bending moment in 0 < z < L is
M = P (d − u) + R(a − z)
and in L < z < a,
M = R(a − z) .
The governing equation in in 0 < z < L is
EI
d2 u
= P (d − u) + R(a − z)
dz 2
or
d2 u P u
P d R(a − z)
+
=
+
dz 2
EI
EI
EI
with solution
u=d+
Ra Rz
−
+ A cos(λz) + B sin(λz) ,
P
P
u′ = −
R
− Aλ sin(λz) + Bλ cos(λz) .
P
In L < z < a we have
d2 u
EI 2 = R(a − z)
dz
or
3
d2 u
R(a − z)
=
dz 2
EI
with solution
Rz 3
Raz 2
−
+ Cz + D ,
2EI
6EI
Raz
Rz 2
u′ =
−
+C .
EI
2EI
The boundary conditions are
u=
u(0)
u′ (0)
u(L+ )
u(L− )
u′(L+ )
u(a)
=
=
=
=
=
=
0
0
d
d
u′(L− )
0.
This provides six equations for the six unknowns A, B, C, D, R, d. The resulting simultaneous equations are
Ra
+A = 0
P
R
− + Bλ = 0
P
d+
Ra RL
−
+ A cos(λL) + B sin(λL)
P
P
RaL2 RL3
−
+ CL + D − d
2EI
6EI
RaL RL2
R
+
−C
− − Aλ sin(λL) + Bλ cos(λL) −
P
EI
2EI
Ra3
+ Ca + D
3EI
= 0
= 0
= 0
= 0.
Routine algebraic eliminations then yield the eigenvalue equation
f (x, β) ≡ 2βx + sin(x) + x2 β 2 sin(x) − (2β + 1)x cos(x) +
where
x = λL ;
β=
4
a
−1.
L
x3 β 3 cos(x)
=0,
3
Numerical solution of this equation for various values of β confirms that the
maximum value of x occurs when
β = 0.2122 ,
as in the solution using the theorem. Alternatively, for the optimal location
of the support, we require
dx
=0
dβ
and since f (x, β) = 0 for all β, we must have
∂f (x(β), β)
∂f dx ∂f
=
+
=0
∂β
∂x dβ ∂β
and hence
∂f
=0
∂β
as well as
f (x, β) = 0 .
It is easily verified that the solution
β=
2
3π
; x=
3π
2
satisfies both of these conditions exactly.
5