Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

628 CHAPTER 23 Fuselage Frames and Wing Ribs

Fig.23.11

Equilibrium of rib forward of intermediate stiffener 56.

Thus,theshearforcecarriedbythewebis2100− 2 ×625.2=849.6N.Hence,

q 1 =

849.6

300

=2.8N/mm

Theaxialloadsintheribflangesatthissectionaregivenby

P 2 =P 4 =(2333.3^2 +625.2^2 )^1 /^2 =2415.6N

The rib flange loads and web panel shear flows, at a vertical section immediately to the left of the
intermediatewebstiffener56,arefoundbyconsideringthefreebodydiagramshowninFig.23.11.At
thissection,theribflangeshavezeroslopesothattheflangeloadsP 5 andP 6 areobtaineddirectlyfrom
thevalueofbendingmomentatthissection.Thus,

P 5 =P 6 =2[( 50000 + 46000 )×7.0− 49000 ×13.0]/ 320 =218.8N

Theshearforceatthissectionisresistedsolelybytheweb.Hence,

320 q 2 =7.0× 300 +7.0× 10 −13.0× 10 =2040N

sothat

q 2 =6.4N/mm

Theshearflowintheribimmediatelytotherightofstiffener56isfoundmostsimplybyconsidering
theverticalequilibriumofstiffener56asshowninFig.23.12.Thus,

320 q 3 =6.4× 320 + 15000

whichgives

q 3 =53.3N/mm