Page 105 - Aircraft Stuctures for Engineering Student

4.7 Solution of statically indeterminate systems 89

Fig. 4.14 Framework of Example 4.5.
members is raised by 30°C. Calculate the resulting forces in all the members if the
coefficient of linear expansion Q! of the bars is 7 x 10-6/"C. E = 200000N/mm2.

Suppose that BC is the heated member, then the increase in length of
BC = 3000 x 30 x 7 x lop6 = 0.63 mm. Therefore, from Eq. (4.23)

1 dFi
-0.63 = FiLi -
200 x 200000. dR
2=1
Substitution from the summation of column @ in Table 4.4 into Eq. (i) gives
-0.63 x 200 x 200000
R= = -525N
48 000
Column @ of Table 4.4 is now completed for the force in each member.
So far, our analysis has been limited to singly redundant frameworks, although the
same procedure may be adopted to solve a multi-redundant framework of, say, nz
redundancies. Thus, instead of a single equation of the type (4.21) we would have
m simultaneous equations

from which the m unknowns R1, R2,. . . , R, would be obtained. The forces F in the
members follow, being expressed initially in terms of the applied loads and
R17R27. . . ,R,,-
Other types of statically indeterminate structure are solved by the application of
total complementary energy with equal facility. The propped cantilever of Fig. 4.15

Table 4.4 (Tension positive)
G3
Force (N)
AB 4000 4R/3 413 64 000R/9 -700
BC 3000 R 1 3 OOOR -525
CD 4000 4R/3 413 64 000R/9 -700
DA 3000 R 1 3 OOOR -525
AC 5000 -5R/3 -513 125 000R/9 875
DB 5000 -5R/3 -513 125 000R/9 875
C = 48 OOOR