Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

1.15 Stress–Strain Relationships 33

FromEqs.(1.52),

εx=

1

200000

( 60 +0.3× 40 )= 360 × 10 −^6

εy=

1

200000

(− 40 −0.3× 60 )=− 290 × 10 −^6

FromEq.(1.50),theshearmodulus,G,isgivenby

G=

E

2 ( 1 +ν)

=

200000

2 ( 1 +0.3)

=76923N/mm^2

Hence,fromEqs.(1.52),

γxy=

τxy

G

=

50

76923

= 650 × 10 −^6

NowsubstitutinginEq.(1.35)forεx,εy,andγxy,

εI= 10 −^6

[

360 − 290

2

+

1

2

√

( 360 + 290 )^2 + 6502

]

whichgives

εI= 495 × 10 −^6

Similarly,fromEq.(1.36),

εII=− 425 × 10 −^6

FromEq.(1.37),

tan2θ=

650 × 10 −^6

360 × 10 −^6 + 290 × 10 −^6

= 1

Therefore,

2 θ= 45 ◦or225◦

sothat

θ=22.5◦or112.5◦

ThevaluesofεI,εII,andθareverifiedusingMohr’scircleofstrain(Fig.1.17).AxesOεandOγ
are set up, and the points Q 1 ( 360 × 10 −^6 ,^12 × 650 × 10 −^6 )and Q 2 (− 290 × 10 −^6 ,−^12 × 650 × 10 −^6 )