452 CHAPTER 15 Bending of Open and Closed, Thin-Walled Beams
FollowingthemethodofExample15.9,wedeterminethesupportreactionsandfindthebending
moment,M,atanysectionZinthebayfurthestfromtheoriginoftheaxes.Then
M=−RAz+wL
4
[
z−5 L
8
]
(i)ExaminingEq.(i),weseethatthesingularityfunction[z− 5 L/8]doesnotbecomezerountilz≤ 5 L/8
althoughEq.(i)isonlyvalidforz≥ 3 L/4.Toobviatethisdifficulty,weextendthedistributedloadto
thesupportDwhilesimultaneouslyrestoringthestatusquobyapplyinganupwarddistributedloadof
thesameintensityandlengthastheadditionalload(Fig.15.22).
AtthesectionZ,adistancezfromA,thebendingmomentisnowgivenby
M=−RAz+w
2[
z−L
2
] 2
−
w
2[
z−3 L
4
] 2
(ii)Equation(ii)isnowvalidforallsectionsofthebeamifthesingularityfunctionsarediscardedasthey
becomezero.SubstitutingEq.(ii)intothesecondofEqs.(15.32),weobtain
EIv′′=3
32
wLz−w
2[
z−L
2
] 2
+
w
2[
z−3 L
4
] 2
(iii)Integrating,Eq.(iii)gives
EIv′=3
64
wLz^2 −w
6[
z−L
2
] 3
+
w
6[
z−3 L
4
] 3
+C 1 (iv)EIv=wLz^3
64−
w
24[
z−L
2
] 4
+
w
24[
z−3 L
4
] 4
+C 1 z+C 2 ,(v)whereC 1 andC 2 arearbitraryconstants.Therequiredboundaryconditionsarev=0whenz=0and
z=L.FromthefirstoftheseweobtainC 2 =0,whilethesecondgives
0 =
wL^4
64−
w
24(
L
2
) 4
+
w
24(
L
4
) 4
+C 1 L
Fig.15.22
Method of solution for a part span uniformly distributed load.