15.3 Deflections due to Bending 445Thebendingmoment,M,atanysectionZisgivenbyM=
w
2(L−z)^2 (i)SubstitutingforMinthesecondofEq.(15.32)andrearranging,wehave
EIv′′=−w
2(L−z)^2 =−w
2(L^2 − 2 Lz+z^2 ) (ii)IntegrationofEq.(ii)yields
EIv′=−w
2(
L^2 z−Lz^2 +z^3
3)
+C 1
Whenz=0atthebuilt-inend,v′=0,sothatC 1 =0and
EIv′=−w
2(
L^2 z−Lz^2 +z^3
3)
(iii)IntegratingEq.(iii),wehave
EIv=−w
2(
L^2
z^2
2−
Lz^3
3+
z^4
12)
+C 2
andsincev=0whenx=0,C 2 =0.Thedeflectioncurveofthebeam,therefore,hastheequation
v=−w
24 EI( 6 L^2 z^2 − 4 Lz^3 +z^4 ) (iv)andthedeflectionatthefreeendwherex=Lis
vtip=−wL^4
8 EI(v)whichisagainnegativeanddownward.
Example 15.7
Determine the deflection curve and the midspan deflection of the simply supported beam shown in
Fig.15.18(a);thebeamhasadoublysymmetricalcrosssection.
ThesupportreactionsareeachwL/2andthebendingmoment,M,atanysectionZ,adistancezfrom
theleft-handsupportis
M=−
wL
2z+wz^2
2(i)