0.3 Boundary Value Problems 33
It is known that the internal bending moment (positive when counterclock-
wise) in a column is given by the product
EId(^2) u
dx^2 ,
whereEis Young’s modulus andIis the moment of inertia of the cross-
sectional area. (The momentI=b^4 /12 for a column whose cross section is
asquareofsideb.) Thus equating the external moment to the internal mo-
ment gives the differential equation
EId
(^2) u
dx^2
=−Pu, 0 <x<a, (19)
which, together with the boundary conditions
u( 0 )= 0 , u(a)= 0 , (20)
determines the functionu(x).
In order to study this problem more conveniently, we set
P
EI=λ
2
so that the differential equation becomes
d^2 u
dx^2
+λ^2 u= 0 , 0 <x<a. (21)
Now, the general solution of this differential equation is
u(x)=c 1 cos(λx)+c 2 sin(λx).
Asu( 0 )=0, we must choosec 1 =0, leavingu(x)=c 2 sin(λx). The second
boundary condition requires that
u(a)=0: c 2 sin(λa)= 0.
If sin(λa)is not 0, the only possibility is thatc 2 =0. In this case we find that
the solution is
u(x)≡ 0 , 0 <x<a.
Physically, this means that the column stands straight and transmits the load
to its support, as it was probably intended to do.
Something quite different happens if sin(λa)=0, for then any choice of
c 2 gives a solution. The physical manifestation of this case is that the column
assumes a sinusoidal shape and may then collapse, or buckle, under the axial
load. Mathematically, the condition sin(λa)=0meansthatλais an integer