Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

23.3 Wing Ribs 627

Resolvingvertically

300 q 31 − 300 q 23 =15000N (ii)

Takingmomentsaboutflange3,

2 ( 50000 + 95000 )q 23 + 2 × 95000 q 12 =− 15000 ×300Nmm (iii)

SolutionofEqs.(i)through(iii)gives

q 12 =13.0N/mm q 23 =−7.0N/mm q 31 =43.0N/mm

ConsidernowthenoseportionoftheribshowninFig.23.10,andsupposethattheshearflowinthe
webimmediatelytotheleftofthestiffener24isq 1 .ThetotalverticalshearforceSy,1atthissectionis
givenby

Sy,1=7.0× 300 =2100N

Thehorizontalcomponentsoftheribflangeloadsresistthebendingmomentatthissection.Thus,

Px,4=Px,2=

2 × 50000 ×7.0

300

=2333.3N

Thecorrespondingverticalcomponentsarethen

Py,2=Py,4=2333.3tan15◦=625.2N

Fig.23.10

Equilibrium of nose portion of the rib.