120 CHAPTER 5 Energy Methods
where the symbols take their previous meanings and is the vertical deflection of any point on the
beam.Then,
∂C
∂Pf=
∫L
0dθ∂M
∂Pf− (^) T= 0 (5.14)
Asbefore
dθ=
M
EI
dzbut
M=Pfz+wz^2
2(Pf= 0 )Hence,
∂M
∂Pf=zSubstitutinginEq.(5.14)fordθ,Mand∂M/∂Pf,andrememberingthatPf=0,wehave
(^) T=
∫L
0wz^3
2 EIdzgiving
(^) T=
wL^4
8 EI
It will be noted that here, unlike the method for the solution of the pin-jointed framework, the
fictitiousloadisappliedtotheloadedbeam.Thereis,however,noarithmeticaladvantagetobegained
bytheformerapproachalthoughtheresultwouldobviouslybethesame,sinceMwouldequalwz^2 / 2
and∂M/∂Pfwouldhavethevaluez.
Example 5.2
Calculate the vertical displacements of the quarter and the midspan points B and C of the simply
supportedbeamoflengthLandtheflexuralrigidityEIloaded,asshowninFig.5.7.
ThetotalcomplementaryenergyCofthesystemincludingthefictitiousloadsPB,fandPC,fis
C=
∫
L∫M
0dθdM−PB,f (^) B−PC,f (^) C−
∫L
0wdz (i)Hence,
∂C
∂PB,f=
∫
Ldθ∂M
∂PB,f− (^) B=0(ii)