Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

22.4 Shear 597

IfthemomentcenterischosentocoincidewiththepointofintersectionofthelinesofactionofSxand
Sy,Eq.(22.11)becomes

0 =

∑N

R= 1

∮

R

qbp 0 ds+

∑N

R= 1

2 ARqs,0,R (22.12)

Example 22.3
ThewingsectionofExample22.1(Fig.22.3)carriesaverticallyupwardshearloadof86.8kNinthe
planeoftheweb572.Thesectionhasbeenidealizedsuchthattheboomsresistallthedirectstresses,
whilethewallsareeffectiveonlyinshear.Iftheshearmodulusofallwallsis27600N/mm^2 exceptfor
thewall78forwhichitisthreetimesthisvalue,calculatetheshearflowdistributioninthesectionand
therateoftwist.Additionaldataaregiveninthetable.

Wall Length (mm) Thickness (mm) Cell area (mm^2 )

12, 56 1023 1.22 AI= 265000

23 1274 1.63 AII= 213000

34 2200 2.03 AIII= 413000

483 400 2.64

572 460 2.64

61 330 1.63

78 1270 1.22

ChoosingGREFas27600N/mm^2 then,fromEq.(22.9),

t∗ 78 =

3 × 27600

27600

×1.22=3.66mm

Hence,

δ 78 =

1270

3.66

= 347

Also,

δ 12 =δ 56 = 840 δ 23 = 783 δ 34 = 1083 δ 38 = 57 δ 84 = 95 δ 87 = 347

δ 27 = 68 δ 75 = 106 δ 16 = 202

We now “cut” the top skin panels in each cell and calculate the “open section” shear flows using
Eq.(19.6),which,sincethewingsectionisidealized,singlysymmetrical(asfarasthedirectstress-
carryingareaisconcerned)andissubjectedtoaverticalshearloadonly,reducesto

qb=

−Sy

Ixx

∑n

r= 1

Bryr (i)