Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

1.7 Principal Stresses 15

Twosolutions,θandθ+π/2,areobtainedfromEq.(1.10)sothattherearetwomutuallyperpen-
dicular planes on which the direct stress is either a maximum or a minimum. Further, by comparing
Eqs. (1.9) and (1.10), it will be observed that these planes correspond to those on which there is no
shearstress.Thedirectstressesontheseplanesarecalledprincipalstresses,andtheplanesthemselves
arecalledprincipalplanes.
FromEq.(1.10),

sin2θ=

2 τxy

√

(σx−σy)^2 + 4 τxy^2

cos2θ=

σx−σy

√

(σx−σy)^2 + 4 τxy^2

and

sin2(θ+π/ 2 )=

− 2 τxy

√

(σx−σy)^2 + 4 τxy^2

cos2(θ+π/ 2 )=

−(σx−σy)

√

(σx−σy)^2 + 4 τxy^2

RewritingEq.(1.8)as

σn=

σx

2

( 1 +cos2θ)+

σy

2

( 1 −cos2θ)+τxysin2θ

andsubstitutingfor{sin2θ,cos2θ}and{sin2(θ+π/ 2 ),cos2(θ+π/ 2 )}inturngives

σI=

σx+σy

2

+

1

2

√

(σx−σy)^2 + 4 τxy^2 (1.11)

and

σII=

σx+σy

2

−

1

2

√

(σx−σy)^2 + 4 τxy^2 (1.12)

whereσIisthemaximumormajorprincipalstressandσIIistheminimumorminorprincipalstress.
NotethatσIisalgebraicallythegreatestdirectstressatthepoint,whileσIIisalgebraicallytheleast.
Therefore,whenσIIisnegative—thatis,compressive—itispossibleforσIItobenumericallygreater
thanσI.
Themaximumshearstressatthispointinthebodymaybedeterminedinanidenticalmanner.From
Eq.(1.9),

dτ

dθ

=(σx−σy)cos2θ+ 2 τxysin2θ= 0

giving

tan2θ=−

(σx−σy)

2 τxy

(1.13)