Aircraft Structures for Engineering Students , Seventh Edition [7th Ed] (Instructor's Edu Resource 1 of 2, Instructor's Solution Manual) (Solutions)

390 Pages·2021·15.405 MB·English

by  T.H.G. Megson

Aircraft Structures
for Engineering Students
Seventh Edition
Solutions Manual
T. H. G. Megson

Solutions Manual
Solutions to Chapter 1 Problems
S.1.1
The principal stresses are given directly by Eqs (1.11) and (1.12) in which σ ¼80N/mm2, σ ¼0
x y
(or vice versa), and τ ¼45N/mm2.Thus,from Eq. (1.11),
xy
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
80 1
σ ¼ + 802+4(cid:2)452
I 2 2
i.e.,
σ ¼100:2N=mm2
I
From Eq.(1.12),
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
80 1
σ ¼ (cid:3) 802+4(cid:2)452
II 2 2
i.e.,
σ ¼(cid:3)20:2N=mm2
II
ThedirectionsoftheprincipalstressesaredefinedbytheangleθinFig.1.8(b)inwhichθisgivenby
Eq.(1.10).Hence,
2(cid:2)45
tan2θ¼ ¼1:125
80(cid:3)0
which gives
θ¼24°110 and θ¼114°110
ItisclearfromthederivationofEqs(1.11)and(1.12)thatthefirstvalueofθcorrespondstoσ whilethe
I
second value correspondsto σ .
II
Finally, the maximum shear stress is obtained from either of Eqs (1.14) or (1.15). Hence from
Eq.(1.15),
100:2(cid:3)ð(cid:3)20:2Þ
τ ¼ ¼60:2N=mm2
max 2
and will act onplanes at 45°tothe principal planes.
S.1.2
The principal stresses are given directly by Eqs (1.11) and (1.12) in which σ ¼50N/mm2,
x
σ ¼–35N/mm2,andτ ¼40N/mm2.Thus,fromEq.(1.11),
y xy
3
4 Solutions Manual
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
50(cid:3)35 1
σ ¼ + ð50+35Þ2+4(cid:2)402
I 2 2
i.e.,
σ ¼65:9N=mm2
I
and fromEq.(1.12),
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
50(cid:3)35 1
σ ¼ (cid:3) ð50+35Þ2+4(cid:2)402
II 2 2
i.e.,
σ ¼(cid:3)50:9N=mm2
II
From Fig. 1.8(b) and Eq. (1.10),
2(cid:2)40
tan2θ¼ ¼0:941
50+35
which gives
θ¼21°380ðσÞ and θ¼111°380ðσ Þ
I II
Theplanesonwhichthereisnodirectstressmaybefoundbyconsideringthetriangularelementofunit
thicknessshowninFig.S.1.2wheretheplaneACrepresentstheplaneonwhichthereisnodirectstress.
For equilibrium ofthe element ina directionperpendicular toAC,
0¼50ABcosα(cid:3)35BCsinα+40ABsinα+40BCcosα (i)
Dividingthrough Eq. (i) byAB,
0¼50cosα(cid:3)35tanαsinα+40sinα+40tanαcosα
which, dividing throughby cos α, simplifiesto
0¼50(cid:3)35tan2α+80tanα
FIG.S.1.2
Solutions to Chapter 1 Problems 5
from which
tanα¼2:797 or (cid:3)0:511
Hence,
α¼70°210 or (cid:3)27°50
S.1.3
The construction of Mohr’s circle for each stress combination follows the procedure described in
Section1.8 andis shown inFigs S.1.3(a)–(d).
FIG.S.1.3(a)
FIG.S.1.3(b)
FIG.S.1.3(c)
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FIG.S.1.3(d)
S.1.4
Theprincipalstressesatthepointaredetermined,asindicatedinthequestion,bytransformingeach
stateofstressintoaσ ,σ ,τ stresssystem.Clearly,inthefirstcase,σ ¼0,σ ¼10N/mm2,τ ¼0
x y xy x y xy
(Fig.S.1.4(a)).Thetworemainingcasesaretransformedbyconsideringtheequilibriumofthetrian-
gularelementABCinFigsS.1.4(b),(c),(e),and(f).Thus,usingthemethoddescribedinSection1.6
andtheprincipleofsuperposition(seeSection5.9),thesecondstresssystemofFigsS.1.4(b)and(c)
becomestheσ ,σ ,τ systemshowninFig.S.1.4(d)whilethethirdstresssystemofFigsS.1.4(e)and
x y xy
(f) transformsinto the σ ,σ , τ system of Fig.S.1.4(g).
x y xy
Finally,thestatesofstressshowninFigsS.1.4(a),(d),and(g)aresuperimposedtogivethestateof
stress shown in Fig. S.1.4(h) from which it can be seen that σ ¼σ ¼15N/mm2 and that the x and
I II
y planes are principal planes.
FIG.S.1.4(a)
Solutions to Chapter 1 Problems 7
FIG.S.1.4(b)
FIG.S.1.4(c)
FIG.S.1.4(d)
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FIG.S.1.4(e)
FIG.S.1.4(f)
FIG.S.1.4(g)
Solutions to Chapter 1 Problems 9
FIG.S.1.4(h)
FIG.S.1.5
S.1.5
ThegeometryofMohr’scircleofstressisshowninFig.S.1.5inwhichthecircleisconstructedusing
the method described inSection 1.8.
From Fig. S.1.5,
σ ¼OP ¼OB(cid:3)BC+CP (i)
x 1 1
IqnffiffiffiEffiffiffiqffiffiffi.ffiffiffiffiffi(ffiiffiffi)ffiffiffiffiffiOffiffi B¼qσffiffiIffiffi,ffiffiffiffiffiBffiffiffiCffiffiffiffiffiffiffiis the radius of the circle that is equal to τmax and CP1¼
CQ2(cid:3)Q P2¼ τ2 (cid:3)τ2 . Hence,
1 1 1 max xy
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
σ ¼σ (cid:3)τ + τ2 (cid:3)τ2
x 1 max max xy
Similarly,
σ ¼OP ¼OB(cid:3)BC(cid:3)CP inwhichCP ¼CP
y 2 2 2 1
Thus,
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
σ ¼σ (cid:3)τ (cid:3) τ2 (cid:3)τ2
y I max max xy
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S.1.6
Frombendingtheorythedirectstressduetobendingontheuppersurfaceoftheshaftatapointinthe
vertical plane ofsymmetry is given by
My 25(cid:2)106(cid:2)75
σ ¼ ¼ ¼75N=mm2
x I π(cid:2)1504=64
From the theoryof the torsion of circular sectionshafts, the shear stress at the same point is
Tr 50(cid:2)106(cid:2)75
τ ¼ ¼ ¼75N=mm2
xy J π(cid:2)1504=32
Substituting these valuesin Eqs(1.11) and(1.12) in turn andnotingthat σ ¼0,
y
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
75 1
σ ¼ + 752+4(cid:2)752
I 2 2
i.e.,
σ ¼121:4N=mm2
I
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
75 1
σ ¼ (cid:3) 752+4(cid:2)752
II 2 2
i.e.,
σ ¼(cid:3)46:4N=mm2
II
The corresponding directions asdefinedby θ in Fig.1.8(b) are given by Eq. (1.10)
i.e.,
2(cid:2)75
tan2θ¼ ¼2
75(cid:3)0
Hence,
θ¼31°430ðσÞ
I
and
θ¼121°430ðσ Þ
II
S.1.7
The direct strainsare expressed interms ofthe stressesusingEqs(1.42),i.e.,
(cid:5) (cid:3) (cid:4)(cid:6)
1
ε ¼ σ (cid:3)v σ +σ (i)
x E x y z
(cid:5) (cid:6)
1
ε ¼ σ (cid:3)vðσ +σ Þ (ii)
y E y x z
(cid:5) (cid:3) (cid:4)(cid:6)
1
ε ¼ σ (cid:3)v σ +σ (iii)
z E z x y