Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

22.3 Torsion 593

Fig.22.6

Wing section of Example 22.2.

Hence,

δ 12 ◦=

∫

12 ◦

ds

t∗

=

1650

1.07

= 1542

Similarly,

δ 12 i= 250 δ 13 =δ 24 = 725 δ 34 = 233 δ 35 =δ 46 = 736 δ 56 = 368

SubstitutingtheappropriatevaluesofδinEq.(22.6)foreachcellinturngivesthefollowing:

• ForcellI,

dθ

dz

=

1

2 × 258000 GREF

[qI( 1542 + 250 )− 250 qII](i)

• ForcellII,

dθ

dz

=

1

2 × 355000 GREF

[− 250 qI+qII( 250 + 725 + 233 + 725 )− 233 qIII](ii)

• ForcellIII,

dθ

dz

=

1

2 × 161000 GREF

[− 233 qII+qIII( 736 + 233 + 736 + 368 )] (iii)

Inaddition,fromEq.(22.4),

11.3× 106 = 2 ( 258000 qI+ 355000 qII+ 161000 qIII) (iv)

SolvingEqs.(i)through(iv)simultaneouslygives

qI=7.1N/mm qII=8.9N/mm qIII=4.2N/mm

Theshearstressinanywallisobtainedbydividingtheshearflowbytheactualwallthickness.Hence,
theshearstressdistributionisasshowninFig.22.7.