Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

15.1 Symmetrical Bending 431

Now

Mx=100cos30◦=86.6kNm

My=100sin30◦=50.0kNm

}

(iii)

Mxis,inthiscase,apositivebendingmomentproducingtensionintheupperhalfofthebeamwhere
yispositive.Also,Myproducestensionintheleft-handhalfofthebeamwherexisnegative;weshall
thereforecallMyanegativebendingmoment.SubstitutingthevaluesofMxandMyfromEq.(iii)but
withtheappropriatesigninEq.(ii)togetherwiththevaluesofIxxandIyyfromExamples15.1and15.2,
weobtain

σz=

86.6× 106

193.7× 106

y−

50.0× 106

27.0× 106

x (iv)

or

σz=0.45y−1.85x (v)

Equation (v) gives the value of direct stress at any point in the cross section of the beam and may
alsobeusedtodeterminethedistributionoveranydesiredportion.Thus,ontheupperedgeofthetop
flangey=+150mm,100mm≥x≥−100mm,sothatthedirectstressvarieslinearlywithx.Atthetop
left-handcornerofthetopflange,

σz=0.45×(+ 150 )−1.85×(− 100 )=+252.5N/mm^2 (tension)

Atthetopright-handcorner,

σz=0.45×(+ 150 )−1.85×(+ 100 )=−117.5N/mm^2 (compression)

The distributions of direct stress over the outer edge of each flange and along the vertical axis of
symmetryareshowninFig.15.7.Notethattheneutralaxisofthebeamsectiondoesnotinthiscase
coincide with either thexoryaxis, although it still passes through the centroid of the section. Its
inclination,α,tothexaxis,say,canbefoundbysettingσz=0inEq.(v).Then,

0 =0.45y−1.85x

or

y

x

=

1.85

0.45

=4.11=tanα

whichgives

α=76.3◦

NotethatαmaybefoundingeneraltermsfromEq.(ii)byagainsettingσz=0.Hence,

y

x

=−

MyIxx

MxIyy

=tanα (15.12)