Page 33 - Aircraft Stuctures for Engineering Student

18 Basic elasticity

which may be written when second-order terms are neglected

O’A’ = Sx( 1 + 2g)’

Applying the binomial expansion to this expression we have
O’A’ = Sx (1 +E) (1.17)

in which squares and higher powers of &/ax are ignored. Substituting for O‘A’ in
Eq. (1.16) we have
aY i1
E, = -

It follows that & =- (1.18)

The shear strain at a point in a body is defined as the change in the angle between
two mutually perpendicular lines at the point. Therefore, if the shear strain in the x-7
plane is T,~ then the angle between the displaced line elements O’A’ and O’C‘ in
Fig. 1.12 is 7r/2 - yxz radians.
Now cos A’O’C’ = cos(7r/2 - yxz) = sin yxz and as yxz is small then cos A 1“ 0 C =
T,=. From the trigonometrical relationships for a triangle
(O’A’)2 + (O‘C’)2 - (A’C’)2
COS A’O’C’ = (1.19)
2 (0’ A’ ) (O’C’)
We have previously shown, in Eq. (1.17), that

O’A‘= Sx(l+$)

Similarly
o’c‘ = Sz( 1 +E)

But for small displacements the derivatives of u, w and w are small compared with 1,
so that, as we are concerned here with actual length rather than change in length, we
may use the approximations
O’A’ M Sx, O’C’ M Sz
Again to a first approximation
(AC) = ( Sz--Sx )2 + ( Sx--Sz )?
“2

Substituting for O’A’, O’C‘ and A’C‘ in Eq. (1.19) we have
(ax2) + (q2 - [Sz - (aw/ax)SxI2 - [Sx - (au/az)Sz]2
COS A‘O‘C’ =
2SxSz