310 Ordinary Differential Equations
A+D=O (From the condition y(O) = 0)
B+>.C = 0 (From the condition y'(O) = 0)
A+ BL+ Csin>.L + Dcos>.L = 0 (From the condition y(L) = 0)
- >.^2 C sin >.L - >.^2 D cos >.L = 0 (From the condition y"(L)=O)
Written in matrix-vector notation, this system becomes
( l
0
1
L
00
>.
sin>.L
->.^2 sin>.L1
0
cos>.L
->.^2 cos>.L H~) m(If you are unfamiliar with matrix-vector notation, see section 8.1.) Or sym-
bolically, Mu= 0 where NJ denotes the 4 x 4 matrix; u denotes the column
vector with entries A, B, C, D; and 0 is the vector of zeroes. Notice that
u = 0 is always a solution of Mu = 0. The solution u = 0 is called the
trivial solution and means A = B = C = D = 0. So corresponding to the
trivial so lut ion, u = 0 , of Mu= 0 is the solution y(x) = 0 of equation (48).
The equation Mu= 0 has a nontrivial solution (a nonzero solution) if and
only if the determinant of the matrix M ( det M) is zero. For this example,
the condition det M = 0 is the condition
->.^2 (->.Lcos>.L+sin>.L) =0.
Since >.^2 = P/ EI -j. 0, for det M to be zero >.must satisfy
->.Leos >.L +sin >.L = 0.Letting z = >.L and dividing by cos >.L = cos z, we see z must satisfy
J( z ) = - z + tanz = 0.There are an infinite number of positive solutions of the equation f(z) = 0.
To verify this fact, graph w = z and w = tan z and notice that the graphs
intersect exactly once in each of the intervals ( (2n-1 )7r / 2, (2n+ 1 )7r /2) for n =
1, 2, 3, .... So the small est positive root of f(z) lies in the interval (7r/2, 37r/2).
Using Newton's method, we find the smallest positive root, accurate to six
decimal places, to be z = >.L = 4.493409. Since >.^2 = P /EI, the critical load
(Euler load) for a column which is clamped at the base and pinned at the top
is
Per= EJ>.^2 = EI(4.493409/ L)^2 = 20 .190724EI/ L^2.
Observe that a shorter column can support a larger load without buckling.