Orthogonality of Mode Shapes

A most significant property of the mode shapes derived for the string is that they form a set of orthogonal mathematical functions. If the mass distribution is nonuniform along x, then the mode shapes are no longer sin(i nx/t); instead, they must be found by solving the first of Eqs. (3.6). The resulting mode shapes, however, may not be expressible in closed form. Nonetheless, they are orthogonal but with respect to the mass distribution as a weighting function. In such a case, this condition of functional orthogonality can be described analytically as

ft

/ т(х)фі (х)фj (x) dx = 0 (i = j) h (3.28)

= 0 (i = j)

To prove that the mode shapes obtained for the string problem are orthogonal, an individual modal contribution given by

Vi (x, t) = фі (x)%i (t)

where фі(x) is a normalized solution of the first of Eqs. (3.6). Substituting vi(x, t) into the governing differential equation (i. e., wave equation), we obtain

d 2 Vi d 2 Vi

T dx2 = m d t2

For example, with SI units, one has the units of T as N, m as kg/m, and t as m. With English units, one has the units of T as lb, m as slugs/ft, and t as ft.

or

Тф’/(х)& (t) = т(х)фі (x)h (t)

Because the general (i. e., homogeneous) solution for the generalized coordinate is a simple harmonic function, then we may write

ft = —2 ft

Thus, the wave equation becomes

ТФШі (t) = ~т(х)фі (х)о2& (t)

so that

Тф-(х) = —т(х)фі (х)о2

If this procedure is repeated by substituting the jth modal contribution into the wave equation, a similar result

Тф’-(х) = —т(х)ф j (х)а>2

is obtained. After multiplying Eq. (3.34) by ф j and Eq. (3.35) by фі, subtracting, and integrating the result over the length of the string, we obtain

fe fe

(o2 — a>2) J т(х)фі (х)фj(х)йх = TJ [фі (х)ф’-(х) — ф"(х)фj(х)] йх (3.36)

The integral on the right-hand side can be integrated by parts using

by letting

for the second. The result becomes

(o2 — a)2) т(х)фі (х)ф j (х)йх

Jo

fг

= T {фіф] — ф[фj) |0 — ^ (ф[ф] — ф[ф])йх = 0

Every term on the right-hand side is zero: the first and second because the mode shape is zero at both ends by virtue of the boundary conditions given by Eqs. (2.27),
and the integral because of cancellation. It may now be concluded that when i = j, because юі = юj, it follows that

[ т(х)фі(х)фj(x)dx = 0 (3.41)

J0

This relationship thus demonstrates that the mode shapes for a string that is fixed at both ends form an orthogonal set of functions. However, when i = j

fe

/ m(x)ф’2^(x)dx = Mi (3.42)

J0

The value of this integral, Mi, is referred to as the “generalized mass” of the ith mode. The numerical values of the generalized masses depend on the normalization scheme used for the mode shapes фі (x).

This development is for a string of nonuniform mass per unit length and con­stant tension force. It is important to note that it readily can be generalized to more involved developments for beam torsional and bending deformation. In such cases, the structural stiffnesses—which are analogous to the tension force in the string problem—also may be nonuniform along the span. Although the struc­tural stiffnesses may not be taken outside the integrals in such cases, the rest of the development remains similar. See Problems 8(a) and 10(a) at the end of this chapter.

For uniform strings and the mode shapes normalized as in Eq. (3.24), it is shown easily for all і and j that the orthogonality condition and generalized mass, Eqs. (3.41) and (3.42), respectively, reduce to