15.3 Deflections due to Bending 441
or,whensecond-ordertermsareneglected
Sy=
∂Mx
∂z
Wemaycombinetheseresultsintoasingleexpression
−wy=
∂Sy
∂z
=
∂^2 Mx
∂z^2
(15.23)
Similarlyforloadsinthexzplane,
−wx=
∂Sx
∂z
=
∂^2 My
∂z^2
(15.24)
15.3 DeflectionsduetoBending.........................................................................
Wehavenotedthatabeambendsaboutitsneutralaxiswhoseinclinationrelativetoarbitrarycentroidal
axes is determined from Eq. (15.22). Suppose that at some section of an unsymmetrical beam the
deflectionnormaltotheneutralaxis(andthereforeanabsolutedeflection)isζ,asshowninFig.15.15.
Inotherwords,thecentroidCisdisplacedfromitsinitialpositionCIthroughanamountζtoitsfinal
positionCF.SupposealsothatthecenterofcurvatureRofthebeamatthisparticularsectionisonthe
oppositesideoftheneutralaxistothedirectionofthedisplacementζandthattheradiusofcurvature
isρ. For this position of the center of curvature and from the usual approximate expression for cur-
vature,wehave
1
ρ
=
d^2 ζ
dz^2
(15.25)
Fig.15.15
Determination of beam deflection due to bending.