15.3 Deflections due to Bending 445
Thebendingmoment,M,atanysectionZisgivenby
M=
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
2
z+
wz^2
2
(i)