15.3 Deflections due to Bending 441or,whensecond-ordertermsareneglected
Sy=∂Mx
∂zWemaycombinetheseresultsintoasingleexpression
−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.