Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

15.1 Symmetrical Bending 427

sothat

εz=−

y

R

(15.1)

ThenegativesigninEq.(15.1)indicatesthatfibersintheregionwhereyispositivewillshortenwhen
thebendingmomentisnegative.Then,fromEq.(1.40),thedirectstressσzinthefiberSTisgivenby

σz=−E

y

R

(15.2)

ThedirectornormalforceonthecrosssectionofthefiberSTisσzδA.However,sincethedirectstress
inthebeamsectionisduetoapurebendingmoment,inotherwords,thereisnoaxialload,theresultant
normalforceonthecompletecrosssectionofthebeammustbezero.Then

∫

A

σzdA=0, (15.3)

whereAistheareaofthebeamcrosssection.
SubstitutingforσzinEq.(15.3)from(15.2)gives

−

E

R

∫

A

ydA= 0 (15.4)

inwhichbothEandRareconstantsforabeamofagivenmaterialsubjectedtoagivenbendingmoment.
Therefore,

∫

A

ydA= 0 (15.5)

Equation(15.5)statesthatthefirstmomentoftheareaofthecrosssectionofthebeamwithrespect
totheneutralaxis—thexaxis—isequaltozero.Thus,weseethattheneutralaxispassesthroughthe
centroidofareaofthecrosssection.Sincetheyaxisinthiscaseisalsoanaxisofsymmetry,itmust
alsopassthroughthecentroidofthecrosssection.Hencetheorigin,O,ofthecoordinateaxescoincides
withthecentroidofareaofthecrosssection.
Equation (15.2) shows that for a sagging (i.e., negative) bending moment, the direct stress in the
beam section is negative (i.e., compressive) whenyis positive and positive (i.e., tensile) whenyis
negative.
ConsidernowtheelementalstripδAinFig.15.4(b);thisis,infact,thecrosssectionofthefiberST.
Thestripisabovetheneutralaxissothattherewillbeacompressiveforceactingonitscrosssectionof
σzδAwhichisnumericallyequalto(Ey/R)δAfromEq.(15.2).Notethatthisforcewillactatallsections
alongthelengthofST.AtS,thisforcewillexertaclockwisemoment(Ey/R)yδAabouttheneutralaxis,
whileatT,theforcewillexertanidenticalanticlockwisemomentabouttheneutralaxis.Considering