Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

630 CHAPTER 23 Fuselage Frames and Wing Ribs

Hence,

P 1 =P 3 =

√

13533.3^2 +3626.2^2 =14010.7N

Thetotalshearforceatthissectionis15000+ 300 ×7.0=17100N.Therefore,theshearforceresisted
bythewebis17100− 2 ×3626.2=9847.6Nsothattheshearflowq 3 inthewebatthissectionis

q 3 =

9847.6

300

=32.8N/mm

Problems..............................................................................................

P.23.1 ThebeamshowninFig.P.23.1issimplysupportedateachendandcarriesaloadof6000N.Ifalldirect
stresses are resisted by the flanges and stiffeners and the web panels are effective only in shear, calculate the
distributionofaxialloadintheflangeABCandthestiffenerBEandtheshearflowsinthepanels.

Ans. q(ABEF)=4N/mm,q(BCDE)=2N/mm

PBEincreaseslinearlyfromzeroatBto6000N(tension)atE

PABandPCBincreaselinearlyfromzeroatAandCto4000N(compression)atB.

Fig. P.23.1

P.23.2 Calculatetheshearflowsinthewebpanelsanddirectloadintheflangesandstiffenersofthebeamshown
inFig.P.23.2ifthewebpanelsresistshearstressesonly.

Ans. q 1 =21.6N/mm q 2 =−1.6N/mm q 3 =10N/mm

PC= 0 PB=6480N(tension) PA=9480N(tension)

PF= 0 PG=480N(tension) PH=2520N(compression)

PEinBEG=2320N(compression) PDinED=6928N(tension)

PDinCD=4320N(tension) PDinDF=320N(tension).