Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

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20.2 Open and Closed Section Beams 569

inwhich,forthebeamsectionshowninFig.20.6(b),


Ixx= 4 × 900 × 3002 + 2 × 1200 × 3002 =5.4× 108 mm^4

Then,


σz,r=

− 200 × 106

5.4× 108

yr

or


σz,r=−0.37yr (ii)

Hence,


Pz,r=−0.37yrBr (iii)

The value ofPz,ris calculated from Eq. (iii) in column②in Table 20.1;Px,randPy,rfollow from
Eqs.(20.10)and(20.9),respectively,incolumns⑤and⑥.TheaxialloadPr,column⑦,isgivenby
[②^2 +⑤^2 +⑥^2 ]^1 /^2 andhasthesamesignasPz,r(seeEq.(20.12)).ThemomentsofPx,randPy,rare
calculatedforamomentcenteratthecenterofsymmetrywithanticlockwisemomentstakenaspositive.
NotethatinTable20.1,Px,randPy,rarepositivewhentheyactinthepositivedirectionsofthesectionx
andyaxes,respectively;thedistancesηrandξrofthelinesofactionofPx,randPy,rfromthemoment
centerarenotgivensigns,sinceitissimplertodeterminethesignofeachmoment,Px,rηrandPy,rξr,
byreferringtothedirectionsofPx,randPy,rindividually.


Fromcolumn⑥


∑^6

r= 1

Py,r=33.4kN

Fromcolumn⑩


∑^6

r= 1

Px,rηr= 0

Fromcolumn


∑^6

r= 1

Py,rξr= 0

FromEq.(20.15),


Sx,w= 0 Sy,w= 100 −33.4=66.6kN

Table 20.1
① ② ③ ④ ⑤ ⑥ ⑦ ⑧ ⑨ ⑩
Pz,r δxr/δz δyr/δz Px,r Py,r Pr ξr ηr Px,rηr Py,rξr
Boom (kN) (kN) (kN) (kN) (m) (m) (kNm) (kNm)

1 − 100 0.1 −0.05 −10 5 −101.3 0.6 0.3 3 − 3
2 − 133 0 −0.05 0 6.7 −177.3 0 0.3 0 0
3 − 100 −0.1 −0.05 10 5 −101.3 0.6 0.3 − 33
4 100 −0.1 0.05 −10 5 101.3 0.6 0.3 − 33
5 133 0 0.05 0 6.7 177.3 0 0.3 0 0
6 100 0.1 0.05 10 5 101.3 0.6 0.3 3 − 3
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