Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

15.2 Unsymmetrical Bending 439

Fig.15.13

Cross section of beam in Example 15.4.

Thepositionofthecentroidofthesectionmaybefoundbytakingmomentsofareasaboutsome
convenientpoint.Thus,

( 120 × 8 + 80 × 8 )y= 120 × 8 × 4 + 80 × 8 × 48

giving

y=21.6mm

and

( 120 × 8 + 80 × 8 )x= 80 × 8 × 4 + 120 × 8 × 24

giving

x=16mm

ThenextstepistocalculatethesectionpropertiesreferredtoaxesCxy(seeSection15.4)

Ixx=

120 ×( 8 )^3

12

+ 120 × 8 ×(17.6)^2 +

8 ×( 80 )^3

12

+ 80 × 8 ×(26.4)^2

=1.09× 106 mm^4

Iyy=

8 ×( 120 )^3

12

+ 120 × 8 ×( 8 )^2 +

80 ×( 8 )^3

12

+ 80 × 8 ×( 12 )^2

=1.31× 106 mm^4

Ixy= 120 × 8 × 8 ×17.6+ 80 × 8 ×(− 12 )×(−26.4)

=0.34× 106 mm^4