Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

622 CHAPTER 23 Fuselage Frames and Wing Ribs

fromwhich

q 4 =23.3N/mm

Alternatively,fromtheequilibriumofstiffenerBF,

300 q 4 − 300 q 3 =5000N

fromwhich

q 4 =23.3N/mm

The flange and stiffener load distributions are calculated in the same way and are obtained from the
algebraicsummationoftheshearflowsalongtheirlengths.Forexample,theaxialloadPAatAinthe
flangeABCDisgivenby

PA= 250 q 1 + 250 q 3 + 250 q 4

or

PA= 250 ×11.3+ 250 ×6.7+ 250 ×23.3=10325N(tension)

Similarly,

PE=− 250 q 2 − 250 q 3 − 250 q 4

thatis,

PE= 250 ×2.6− 250 ×6.7− 250 ×23.3=−6850N(compression)

ThecompleteloaddistributionineachflangeisshowninFig.23.4.Thestiffenerloaddistributionsare
calculatedinthesamewayandareshowninFig.23.5.
ThedistributionofflangeloadinthebaysABFEandBCGFcouldhavebeenobtainedbyconsidering
thebendingandaxialloadsonthebeamatanysection.Forexample,atthesectionAE,wecanreplace
theactualloadingsystembyabendingmoment

MAE= 5000 × 250 + 2000 × 750 −3464.1× 50 =2576800Nmm

andanaxialloadactingmidwaybetweentheflanges(irrespectiveofwhetherornottheflangeareasare
symmetricalaboutthispoint)of

P=3464.1N

Thus,

PA=

2576800

300

+

3464.1

2

=10321N(tension)

and

PE=

− 2576800

300

+

3464.1

2

=−6857N(compression)