Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

15.3 Deflections due to Bending 447

Sincev=0whenz=0(orsincev=0whenz=L),itfollowsthatC 2 =0andthedeflectedshapeofthe
beamhastheequation

v=

w

24 EI

( 2 Lz^3 −z^4 −L^3 z) (iv)

Themaximumdeflectionoccursatmidspanwherez=L/2andis

vmidspan=−

5 wL^4

384 EI

(v)

Sofar,theconstantsofintegrationweredeterminedimmediatelyaftertheyarose.However,insome
cases,arelevantboundarycondition,say,avalueofgradient,isnotobtainable.Themethodisthento
carrytheunknownconstantthroughthesucceedingintegrationanduseknownvaluesofdeflectionat
twosectionsofthebeam.Thus,inthepreviousexample,Eq.(ii)isintegratedtwicetoobtain

EIv=

w

2

(

Lz^3

6

−

z^4

12

)

+C 1 z+C 2

Therelevantboundaryconditionsarev=0atz=0andz=L.ThefirstofthesegivesC 2 =0,whereas
fromthesecond,wehaveC 1 =−wL^3 /24.Thus,theequationofthedeflectedshapeofthebeamis

v=

w

24 EI

( 2 Lz^3 −z^4 −L^3 z)

asbefore.

Example 15.8
Figure 15.19(a) shows a simply supported beam carrying a concentrated loadWat midspan. Deter-
mine the deflection curve of the beam and the maximum deflection if the beam section is doubly
symmetrical.

ThesupportreactionsareeachW/2andthebendingmomentMatasectionZadistancez(≤L/2)
fromtheleft-handsupportis

M=−

W

2

z (i)

FromthesecondofEq.(15.32),wehave

EIv′′=

W

2

z (ii)

Integrating,weobtain

EIv′=

W

2

z^2

2

+C 1

Fromsymmetry,theslopeofthebeamiszeroatmidspanwherez=L/2.Thus,C 1 =−WL^2 /16and

EIv′=

W

16

( 4 z^2 −L^2 ) (iii)