Page 103 - Aircraft Stuctures for Engineering Student

4.7 Solution of statically indeterminate systems 87

Hence from Eq. (4.22)
4.83lU + 2.707PL = 0
or
R = -0.56P
Substitution for R in column @ of Table 4.2 gives the force in each member. Having
determined the forces in the members then the deflection of any point on the frame-
work may be found by the method described in Section 4.6.
Unlike the statically determinate type, statically indeterminate frameworks may be
subjected to self-straining. Thus, internal forces are present before external loads are
applied. Such a situation may be caused by a local temperature change or by an initial
lack of fit of a member. Suppose that the member BD of the framework of Fig. 4.12 is
short by a known amount AR when the framework is assembled but is forced to fit.
The load R in BD will then have suffered a displacement AR in addition to that caused
by the change in length of BD produced by the load P. The total complementary
energy is then

and
dC dFi
=
- Xi- - AR = 0
dR r=l dR
.
or
1 8Fi
dR
AE.
AR = -E ~~1,~- (4.23)
r=l
Obviously the summation term in Eq. (4.23) has the same value as in the previous case
so that
AE
R = -0.56P + -
4.831, AR
Hence the forces in the members are due to both applied loads and an initial lack of
fit.
Some care should be given to the sign of the lack of fit AR. We note here that the
member BD is short by an amount AR so that the assumption of a positive sign for AR
is compatible with the tensile force R. If BD were initially too long then the total
complementary energy of the system would be written
C=xj:AidFi-PA-R(-AR)
k

i= 1
giving

1 aFi
-AR = -E FiLi-
AE.
1=1