Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

40 CHAPTER 1 Basic Elasticity

Substitutingthevaluesofεa,εb,andεcinEq.(1.69),

εI=

10 −^6

2

( 1000 − 300 )+

10 −^6

√

2

√

( 1000 + 200 )^2 +(− 200 + 300 )^2

whichgives

εI= 1202 × 10 −^6

Similarly,fromEq.(1.70),

εII=− 502 × 10 −^6

NowsubstitutingforεIandεIIinEq.(1.67),

σI=

70000 × 10 −^6

1 −(0.3)^2

(− 502 +0.3× 1202 )=−80.9N/mm^2

Similarly,fromEq.(1.68),

σII=−10.9N/mm^2

Sinceσy=0,Eqs.(1.11)and(1.12)reduceto

σI=

σx

2

+

1

2

√

σx^2 + 4 τxy^2 (i)

and

σII=

σx

2

−

1

2

√

σx^2 + 4 τxy^2 (ii)

respectively.AddingEqs.(i)and(ii),weobtain

σI+σII=σx

Thus,

σx=80.9−10.9=70N/mm^2

ForanaxialloadP,

σx=70N/mm^2 =

P

A

=

P

π× 502 / 4

fromwhich

P=137.4kN

SubstitutingforσxineitherofEq.(i)orofEq.(ii)gives

τxy=29.7N/mm^2