15.3 Deflections due to Bending 453
fromwhich
C 1 =−
27 wL^3
2048
Equations(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
2048
z (vii)
Inthisproblem,themaximumdeflectionclearlyoccursintheregionBCofthebeam.Thus,equating
theslopetozeroforBC,wehave
0 =
3
64
wLz^2 −
w
6
[
z−
L
2
] 3
−
27 wL^3
2048
whichsimplifiesto
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
EI
Example 15.11
DeterminethedeflectedshapeofthebeamshowninFig.15.23.
Fig.15.23
Deflection of a simply supported beam carrying a point moment.