Page 37 - Aircraft Stuctures for Engineering Student

22 Basic elasticity

or substituting from Eqs (1.29)

(ED)2(1 + E,++)~ = (CD)2( 1 + E,)~ + (CE)2( 1 + E ~ ) ~
-2(CD)(CE)(l +~,)(1 +~,,)siny,,
Noting that (ED)2 = (CD)2 + (CE)2 and neglecting squares and higher powers of
small quantities this equation may be rewritten
~(ED)’E,+,/~ = 2(CD)2~, + 2(CE)2~y - 2(CE)(CD)yxy

Dividing through by 2(ED)’ gives
E, + r/2 = E, sin2 e + E,, cos2 e - COS e sin ey,, (1.30)

The strain E, in the direction normal to the plane ED is found by replacing the angle 0
in Eq. (1.30) by 8 - 7r/2. Hence

E, = E~ COS’ e + sin2 e + -sin20 (1.31)
TXY
2
Turning our attention now to the triangle C‘F’E’ we have
(C‘E’)? = (C’F’)’ + - 2(C’F’)(F’E’) cos(7r/2 - 7) (1.32)

in which
C’E’ = CE( 1 + E,,)
C’F‘ = CF( 1 + E,)

F’E’ = FE(l + E,+,/z)
Substituting for C’E’, C’F’ and F’E’ in Eq. (1.32) and writing cos(.rr/2 - y) = siny
we find

(CE)2(1 +E,,)’ = (CF)2(1 + En)2 + (FE)’(1 + En+,/2)2
-2(CF)(FE)(1 +&,)(I +~n+lr/dsiny (1.33)

All the strains are assumed to be small so that their squares and higher powers may be
ignored. Further, siny M y and Eq. (1.33) becomes

(CE)2(1 + 2~y) = (CF)’(1 + 2~n) + (FE)2(1 + 2~,+,/2) - 2(CF)(FE)y
From Fig. l.l3(a), (CE)2 = (CF)2 + (FE)2 and the above equation simplifies to
2(CE)2~,, ~(CF)’E, + ~(FE)’E,+,/~ - 2(CF)(FE)y
=
Dividing through by 2(CE)2 and transposing
2
E, sin’ e + E,+~/~ COS e - E~
7= sin 9 cos 8

~
~
~
+
Substitution of E, and E ~ from Eqs (1.31) and (1.30) yields
-_ - (E~-E~)sin2e-~,-os2e (1.34)
7
2 2 2