Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

16.3 Shear of Closed Section Beams 495

ofsymmetry,thenIxy=0,andsinceSx=0,Eq.(16.15)simplifiesto

qs=−

Sy

Ixx

∫s

0

tyds+qs,0 (i)

inwhich

Ixx= 2

⎡

⎣

∫^10 a

0

t

(

8

10

s 1

) 2

ds 1 +

∫^17 a

0

t

(

8

17

s 2

) 2

ds 2

⎤

⎦

EvaluatingthisexpressiongivesIxx= 1152 a^3 t.
ThebasicshearflowdistributionqbisobtainedfromthefirstterminEq.(i).Then,forthewall41

qb,41=

−Sy

1152 a^3 t

∫s^1

0

t

(

8

10

s 1

)

ds 1 =

−Sy

1152 a^3

(

2

5

s^21

)

(i)

Inthewall12,

qb,12=

−Sy

1152 a^3

⎡

⎣

∫s^2

0

( 17 a−s 2 )

8

17

ds 2 + 40 a^2

⎤

⎦ (ii)

whichgives

qb,12=

−Sy

1152 a^3

(

−

4

17

s^22 + 8 as 2 + 40 a^2

)

(iii)

Theqbdistributionsinthewalls23and34followfromsymmetry.Hence,fromEq.(16.28),

qs,0=

2 Sy

54 a× 1152 a^3

⎡

⎣

∫^10 a

0

2

5

s^21 ds 1 +

∫^17 a

0

(

−

4

17

s^22 + 8 as 2 + 40 a^2

)

ds 2

⎤

⎦

giving

qs,0=

Sy

1152 a^3

(

58.7a^2

)

(iv)

Takingmomentsaboutthepoint2,wehave

Sy(ξS+ 9 a)= 2

∫^10 a

0

q 4117 asinθds 1

or

Sy(ξS+ 9 a)=

Sy 34 asinθ

1152 a^3

∫^10 a

0

(

−

2

5

s^21 +58.7a^2

)

ds 1 (v)