Page 26 - Aircraft Stuctures for Engineering Student

1.7 Principal stresses 1 1

Again dividing through by ED and simplifying
(sx - Cy) .
7= sin 28 - rXy cos 28 (1.9)
2
..- ... - .
1.7 Prir icipal stresses

For given values of a,, ay and T,,, in other words given loading conditions, an varies
with the angle 8 and will attain a maximum or minimum value when dan/d8 = 0.
From Eq. (1.8)

5 = -2a, cos 8 sin 8 + 20, sin 8 cos 8 + 2ryy cos 28 = 0
do
Hence
-(u.~ - u,,) sin 28 + 2TsJ cos 28 = 0
or

(1.10)

Two solutions, 8 and 8 + n/2, are obtained from Eq. (1.10) so that there are two
mutually perpendicular planes on which the direct stress is either a maximum or a
minimum. Further, by comparison of Eqs (1.10) and (1.9) it will be observed that
these planes correspond to those on which there is no shear stress. The direct stresses
on these planes are called principal stresses and the planes themselves, principal planes.
From Eq. (1.10)
2Tcy a, - ay
sin28 =
Jm.’ JW
28=
‘Os
and
sin 2(8 + n/2) = dv, - 2?xy cos 2(8 + n/2) = 4U.Y - Uy)
(0, - cy) + 4?xy JW
Rewriting Eq. (1.8) as

U.
an =%(l+cos20)+-~(1 -cos28)+rx,sin28
2 2
ahd substituting for (sin28, cos 28} and {sin2(8 + 7r/2), cos 2(8 + n/2)} in turn gives
0.Y + 0.v
01 = ~ (1.11)
and

where aI is the maximum or major principal stress and olI is the minimum or minor
principal stress. Note that a1 is algebraically the greatest direct stress at the point