Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

496 CHAPTER 16 Shear of Beams

We may replace sinθ by sin(θ 1 −θ 2 )=sinθ 1 cosθ 2 −cosθ 1 sinθ 2 , where sinθ 1 = 15 /17,cosθ 2 =
8 /10,cosθ 1 = 8 /17,andsinθ 2 = 6 /10.SubstitutingthesevaluesandintegratingEq.(v)gives

ξS=−3.35a

whichmeansthattheshearcenterisinsidethebeamsection.

Reference

[1] Megson,T.H.G.,StructuralandStressAnalysis,2ndedition,Elsevier,2005.

Problems..............................................................................................

P.16.1 Abeamhasthesinglysymmetrical,thin-walledcrosssectionshowninFig.P.16.1.Thethicknesstofthe
wallsisconstantthroughout.Showthatthedistanceoftheshearcenterfromthewebisgivenby

ξS=−d

ρ^2 sinαcosα

1 + 6 ρ+ 2 ρ^3 sin^2 α

where

ρ=d/h

Fig. P.16.1

P.16.2 Abeamhasthesinglysymmetrical,thin-walledcrosssectionshowninFig.P.16.2.Eachwallofthesection
isflatandhasthesamelengthaandthicknesst.Calculatethedistanceoftheshearcenterfromthepoint3.

Ans. 5 acosα/8.