5.6 Flexibility Method 143Table 5.6
Member Lj(m) F0,j F1,j F0,jF1,jLj F1,^2 jLj Fa,j
AB L 0 −0.71 0 0.5L +0.40P
BC L 0 −0.71 0 0.5L +0.40P
CD L −P −0.71 0.71PL 0.5L −0.60P
BD 1.41L − 1.0 − 1.41L −0.56P
AC 1.41L 1.41P 1.0 2.0PL 1.41L +0.85P
AD L 0 −0.71 0 0.5L +0.40P
=2.71PL =4.82LTheactualforces,Fa,j,inthemembersofthecompletetrussofFig.5.20(a)arenowcalculatedusing
themethodofjointsandarelistedinthefinalcolumnofTable5.6.
Wenoteintheprecedingthat (^) BDispositive,whichmeansthat (^) BDisinthedirectionoftheunit
loads,BapproachesD,andthediagonalBDinthereleasedstructuredecreasesinlength.Therefore,
in the complete structure, the member BD, which prevents this shortening, must be in compression
as shown; alsoaBDwill always be positive, since it contains the termF1,^2 j. Finally, we note that the
cutmemberBDisincludedinthecalculationofthedisplacementsinthereleasedstructure,sinceits
deformation,underaunitload,contributestoaBD.
Example 5.9
Calculate the forces in the members of the truss shown in Fig. 5.21(a). All members have the same
cross-sectionalareaAandYoung’smodulusE.
Byinspection,weseethatthetrussisbothinternallyandexternallystaticallyindeterminate,since
it would remain stable and in equilibrium if one of the diagonals, AD or BD, and the support at C
wereremoved;thedegreeofindeterminacyistherefore2.UnlikethetrussinExample5.8,wecould
notremoveanymember,sinceifBCorCDwereremoved,theouterhalfofthetrusswouldbecomea
mechanism,whiletheportionABDEwouldremainstaticallyindeterminate.Therefore,weselectAD
andthesupportatCasthereleases,givingthestaticallydeterminatetrussshowninFig.5.21(b);we
shalldesignatetheforceinthememberADasX 1 andtheverticalreactionatCasR 2.
In this case, we shall have two compatibility conditions, one for the diagonal AD and one for
the support at C. We therefore need to investigate three loading cases: one in which the actual
loads are applied to the released statically determinate truss in Fig. 5.21(b), a second in which unit
loads are applied to the cut member AD (Fig. 5.21(c)), and a third in which a unit load is applied
at C in the direction ofR 2 (Fig. 5.21(d)). By comparing the previous example, the compatibility
conditionsare
(^) AD+a 11 X 1 +a 12 R 2 =0(i)
vC+a 21 X 1 +a 22 R 2 =0(ii)