Page 585 - Aircraft Stuctures for Engineering Student

566 Elementary aeroelasticity

where

A4 = -EI- d2V (see Eq. (13.45))
azz
Substituting for from Eq. (1 3.48) gives
d2 V
M = -EZ-sin(wt + E)
#
so that from Eq. (13.51)

1
U=-sin2(wt+E) (13.52)
2
For a non-uniform beam, having a distributed mass pA(z) per unit length and
carrying concentrated masses, ml, m2, m3,. . . , m, at distances zl, z2, z3,. . . , z, from
the origin, the kinetic energy KE may be written as

Substituting for v(z) from Eq. (13.48) we have

1
KE = -u2 cos2(wt + E) [ JLpA(z) V2 dz + 2m.r V(Z~)}~]
2 ( 13.53)
r=l
Since KE + U = constant, say C, then
1
-sin2(wt+E)jLEI(-) d2V dz+2dcos2(wt+E) 1
7 dz2
I
= C
x [jLpA(i)v2dz+gmr{V(~r)}2] (13.54)
r=
1
Inspection of Eq. (13.54) shows that when (wt + E) = O,T, 27r,.
1
-w2 [ JL pA(z) V2 dz + 2 mr{ V(Z~)}~] = C (13.55)
2
r= 1
and when
.
.
(ut + E) = 7r/2,3~/2,5~/2;.
then
ijLEI( =) dz = C
d2V
( 13.56)
In other words the kinetic energy in the mean position is equal to the strain energy in
the position of maximum displacement. From Eqs (13.55) and (13.56)

(13.57)