Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

5.7Total Potential Energy 147

Then,fromEq.(i),

X 1 =−525N

The forces,Fa,j, in the members of the complete truss are given in the final column of Table 5.8.
ComparetheprecedingwiththesolutionofExample5.4.

5.7 TotalPotentialEnergy...............................................................................

In the spring–mass system shown in its unstrained position in Fig. 5.23(a), we normally define the
potentialenergyofthemassastheproductofitsweight,Mg,anditsheight,h,abovesomearbitrarilyfixed
datum.Inotherwords,itpossessesenergybyvirtueofitsposition.Afterdeflectiontoanequilibrium
state (Fig. 5.23(b)), the mass has lost an amount of potential energy equal toMgy.Thus,wemay
associatedeflectionwithalossofpotentialenergy.Alternatively,wemayarguethatthegravitational
forceactingonthemassdoesworkduringitsdisplacement,resultinginalossofenergy.Applyingthis
reasoningtotheelasticsystemofFig.5.1(a)andassumingthatthepotentialenergyofthesystemis
zerointheunloadedstate,thenthelossofpotentialenergyoftheloadPasitproducesadeflectionyis
Py.Thus,thepotentialenergyVofPinthedeflectedequilibriumstateisgivenby

V=−Py

WenowdefinetheTPEofasysteminitsdeflectedequilibriumstateasthesumofitsinternalorstrain
energy and the potential energy of the applied external forces. Hence, for the single member–force
configurationofFig.5.1(a),

TPE=U+V=

∫y

0

Pdy−Py

For a general system consisting of loads P 1 ,P 2 ,...,Pnproducing corresponding displacements

(i.e.,displacementsinthedirectionsoftheloads;seeSection5.10) 1 , 2 ,..., (^) n,thepotentialenergy
Fig.5.23
(a) Potential energy of a spring–mass system and (b) loss in potential energy due to change in position.