Yx0 0 yx0 EI yxL 0 yxL et

to give

EIk2 sin kL

Ma EIk2

Ma EIk

Substituting Equations 2.299a to d into the deflection function Equation 2.297 and rearranging gives y =

EIk cos kL x

sin kL L

EIk2

sin kL L

0 0

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