Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

450 CHAPTER 15 Bending of Open and Closed, Thin-Walled Beams

and

EIv=

1

8

Wz^3 −

W

6

[z−a]^3 −

W

6

[z− 2 a]^3 +

W

3

[z− 3 a]^3 +C 1 z+C 2 (iv)

inwhichC 1 andC 2 arearbitraryconstants.Whenz=0(atA),v=0,andhenceC 2 =0.Notethatthe
second, third, and fourth terms on the right-hand side of Eq. (iv) disappear forz<a.Also,v=0at
z= 4 a(F)sothat,fromEq.(iv),wehave

0 =

W

8

64 a^3 −

W

6

27 a^3 −

W

6

8 a^3 +

W

3

a^3 + 4 aC 1

whichgives

C 1 =−

5

8

Wa^2

Equations(iii)and(iv)nowbecome

EIv′=

3

8

Wz^2 −

W

2

[z−a]^2 −

W

2

[z− 2 a]^2 +W[z− 3 a]^2 −

5

8

Wa^2 (v)

and

EIv=

1

8

Wz^3 −

W

6

[z−a]^3 −

W

6

[z− 2 a]^3 +

W

3

[z− 3 a]^3 −

5

8

Wa^2 z, (vi)

respectively.
To determine the maximum upward and downward deflections, we need to know in which bays
v′=0 and thereby which terms in Eq. (v) disappear when the exact positions are being located. One
methodistoselectabayanddeterminethesignoftheslopeofthebeamattheextremitiesofthebay.
Achangeofsignwillindicatethattheslopeiszerowithinthebay.
ByinspectionofFig.15.20,itseemslikelythatthemaximumdownwarddeflectionwilloccurin
BC.AtB,usingEq.(v)

EIv′=

3

8

Wa^2 −

5

8

Wa^2

whichisclearlynegative.AtC,

EIv′=

3

8

W 4 a^2 −

W

2

a^2 −

5

8

Wa^2

whichispositive.Therefore,themaximumdownwarddeflectiondoesoccurinBCanditsexactposition
islocatedbyequatingv′tozeroforanysectioninBC.Thus,fromEq.(v)

0 =

3

8

Wz^2 −

W

2

[z−a]^2 −

5

8

Wa^2

or,simplifying,

0 =z^2 − 8 az+ 9 a^2 (vii)