Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

116 CHAPTER 5 Energy Methods

We note that the support reactions do not appear in Eq. (5.9). This convenient absence derives

fromthefactthatthedisplacements 1 , 2 ,..., (^) naretherealdisplacementsoftheframeandfulfill
the conditions of geometrical compatibility and boundary restraint. The complementary energy of
the reaction at A and the vertical reaction at B is therefore zero, since both of their corresponding
displacementsarezero.IfweexamineEq.(5.11),wenotethatλiistheextensionoftheithmemberof
theframeworkduetotheappliedloadsP 1 ,P 2 ,...,Pn.Therefore,theloadsFiinthesubstitutionforλiin
Eq.(5.11)arethosecorrespondingtotheloadsP 1 ,P 2 ,...,Pn.Theterm∂Fi/∂P 2 inEq.(5.11)represents
therateofchangeofFiwithP 2 andiscalculatedbyapplyingtheloadP 2 totheunloadedframeand
determining the corresponding member loads in terms ofP 2. This procedure indicates a method for
obtainingthedisplacementofeitherapointontheframeinadirectionnotcoincidentwiththelineof
actionofaloador,infact,apointsuchasCwhichcarriesnoloadatall.Weplaceatthepointandin
therequireddirectionafictitiousordummyload,sayPf,theoriginalloadsbeingremoved.Theloads
in the members due toPfare then calculated and∂F/∂Pfobtained for each member. Substitution in
Eq.(5.11)producestherequireddeflection.
It must be pointed out that it is not absolutely necessary to remove the actual loads during the
applicationofPf.Theforceineachmemberwouldthenbecalculatedintermsoftheactualloading
andPf.FifollowsbysubstitutingPf=0,and∂Fi/∂PfisfoundbydifferentiationwithrespecttoPf.
Obviously the two approaches yield the same expressions forFiand∂Fi/∂Pf, although the latter is
arithmeticallyclumsier.
Example 5.1
Calculate the vertical deflection of the point B and the horizontal movement of D in the pin-jointed
framework shown in Fig. 5.4(a). All members of the framework are linearly elastic and have cross-
sectionalareasof1800mm^2 .Eforthematerialofthemembersis200000N/mm^2.
ThemembersoftheframeworkarelinearlyelasticsothatEq.(5.11)maybewrittenas

=

∑k

i= 1

FiLi

AiEi

∂Fi

∂P

(i)

orsinceeachmemberhasthesamecross-sectionalareaandmodulusofelasticity,

=

1

AE

∑k

i= 1

FiLi

∂Fi

∂P

(ii)

The solution is completed in Table 5.1, in whichFare the member forces due to the actual loading
of Fig. 5.4(a),FB,fare the member forces due to the fictitious loadPB,fin Fig. 5.4(b), andFD,fare
theforcesinthemembersproducedbythefictitiousloadPD,finFig.5.4(c).Wetaketensileforcesas
positiveandcompressiveforcesasnegative.
TheverticaldeflectionofBis

(^) B,v=

1268 × 106

1800 × 200000

=3.52mm