15.3 Deflections due to Bending 453fromwhich
C 1 =−
27 wL^3
2048Equations(iv)and(v)thenbecome
EIv′=3
64
wLz^2 −w
6[
z−L
2
] 3
+
w
6[
z−3 L
4
] 3
−
27 wL^3
2048(vi)and
EIv=wLz^3
64−
w
24[
z−L
2
] 4
+
w
24[
z−3 L
4
] 4
−
27 wL^3
2048z (vii)Inthisproblem,themaximumdeflectionclearlyoccursintheregionBCofthebeam.Thus,equating
theslopetozeroforBC,wehave
0 =
3
64
wLz^2 −w
6[
z−L
2
] 3
−
27 wL^3
2048whichsimplifiesto
z^3 −1.78Lz^2 +0.75zL^2 −0.046L^3 = 0 (viii)SolvingEq.(viii)bytrialanderror,weseethattheslopeiszeroatz 0.6L.HencefromEq.(vii),the
maximumdeflectionis
vmax=−4.53× 10 −^3 wL^4
EIExample 15.11
DeterminethedeflectedshapeofthebeamshowninFig.15.23.
Fig.15.23
Deflection of a simply supported beam carrying a point moment.