Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

5.6 Flexibility Method 145

Table 5.7

F0,jF1,j F0,jF1,j F1,j(X 1 )

Member Lj F0,j F1,j(X 1 ) F1,j(R 2 ) (X 1 )Lj (R 2 )Lj F^2 1,j(X 1 )Lj F^2 1,j(R 2 )Lj F1,j(R 2 )Lj Fa,j

AB 1 10.0 −0.71 −2.0 −7.1 −20.0 0.5 4.0 1.41 0.67

BC 1.41 0 0 −1.41 0 0 0 2.81 0 −4.45

CD 1 0 0 1.0 0 0 0 1.0 0 3.15

DE 10 −0.71 1.0 0 0 0.5 1.0 −0.71 0.12

AD 1.41 0 1.0 0 0 0 1.41 0 0 4.28

BE 1.41 −14.14 1.0 1.41 −20.0 −28.11 1.41 2.81 2.0 −5.4

BD 10 −0.71 0 0 0 0.5 0 0 −3.03

=−27.1 =−48.11 =4.32 =11.62 =2.7

a 22 =

∑n

j= 1

F1,^2 j(R 2 )Lj

AE

=

11.62

AE

a 12 =a 21

∑n

j= 1

F1,j(X 1 )F1,j(R 2 )Lj

AE

=

2.7

AE

SubstitutinginEqs.(i)and(ii)andmultiplyingthroughbyAE,wehave

−27.1+4.32X 1 +2.7R 2 = 0 (iii)

−48.11+2.7X 1 +11.62R 2 =0(iv)

SolvingEqs.(iii)and(iv),weobtain

X 1 =4.28kN R 2 =3.15kN

Theactualforces,Fa,j,inthemembersofthecompletetrussarenowcalculatedbythemethodofjoints
andarelistedinthefinalcolumnofTable5.7.

5.6.1 Self-Straining Trusses

Staticallyindeterminatetrusses,unlikethestaticallydeterminatetype,maybesubjectedtoself-straining
inwhichinternalforcesarepresentbeforeexternalloadsareapplied.Suchasituationmaybecaused
byalocaltemperaturechangeorbyaninitiallackoffitofamember.Incasessuchasthese,theterm
on the right-hand side of the compatibility equations, Eq. (ii) in Example 5.8 and Eqs. (i) and (ii) in
Example5.9,wouldnotbezero.

Example 5.10
ThetrussshowninFig.5.22(a)isunstressedwhenthetemperatureofeachmemberisthesame,but
duetolocalconditions,thetemperatureinthememberBCisincreasedby30◦C.Ifthecross-sectional
areaofeachmemberis200mm^2 andthecoefficientoflinearexpansionofthemembersis7× 10 −^6 /◦C,
calculatetheresultingforcesinthemembers;Young’smodulusE=200000N/mm^2.