megson solution manual aircraft structures ch01

7/22/2019 Megson Solution Manual Aircraft Structures CH01

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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

x = 80 N/mm2,y = 0 (or vice versa) andxy = 45 N/mm2. Thus, from Eq. (1.11)

I=80

2+ 1

2

802 + 4 452

i.e.

I=

100.2 N/mm2

From Eq. (1.12)

II=80

2 1

2

802 + 4 452

i.e.

II= 20.2 N/mm2

The directions of the principal stresses are defined by the angle in Fig. 1.8(b) in

whichis given by Eq. (1.10). Hence

tan 2= 2 4580 0 = 1.125

which gives

= 2411 and = 11411

It is clear from the derivation of Eqs (1.11) and (1.12) that the first value of

corresponds toI while the second value corresponds toII.

Finally, the maximum shear stress is obtained from either of Eqs (1.14) or (1.15).Hence from Eq. (1.15)

max=100.2 (20.2)

2= 60.2 N/mm2

and will act on planes at 45 to the principal planes.


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S.1.2

The principal stresses are given directly by Eqs (1.11) and (1.12) in which x =50 N/mm2,y

=35 N/mm2 andxy

=40 N/mm2. Thus, from Eq. (1.11)

I=50 35

2+ 1

2

(50 + 35)2 + 4 402

i.e.

I= 65.9 N/mm2

and from Eq. (1.12)

II=50

35

2

1

2

(50 + 35)2 + 4 402

i.e.

II= 50.9 N/mm2

From Fig. 1.8(b) and Eq. (1.10)

tan 2= 2 4050+

35= 0.941

which gives

= 2138(I) and = 11138(II)The planes on which there is no direct stress may be found by considering the

triangular element of unit thickness shown in Fig. S.1.2 where the plane AC represents

the plane on which there is no direct stress. For equilibrium of the element in a direction

perpendicular to AC

0 = 50AB cos 35BC sin + 40AB sin+ 40BC cos (i)

A

BC

35N/mm2

40N/mm2

50N/mm2

Fig. S.1.2


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Solutions to Chapter 1 Problems 5

Dividing through Eq. (i) by AB

0 = 50 cos 35 tan sin + 40 sin + 40 tan cos

which, dividing through by cos , simplifies to

0 = 50 35 tan2 + 80 tan

from which

tan = 2.797 or 0.511Hence

=7021 or

275

S.1.3

The construction of Mohrs circle for each stress combination follows the procedure

described in Section 1.8 and is shown in Figs S.1.3(a)(d).

Fig. S.1.3(a)

Fig. S.1.3(b)


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Fig. S.1.3(c)

Fig. S.1.3(d)

S.1.4

The principal stresses at the point are determined, as indicated in the question, by

transforming each state of stress into a x, y, xy stress system. Clearly, in the

first case x = 0, y = 10 N/mm2, xy = 0 (Fig. S.1.4(a)). The two remaining casesare transformed by considering the equilibrium of the triangular element ABC in

Figs S.1.4(b), (c), (e) and (f). Thus, using the method described in Section 1.6

and the principle of superposition (see Section 5.9), the second stress system of

Figs S.1.4(b) and (c) becomes the x, y, xy system shown in Fig. S.1.4(d) while


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10 N/mm2

Fig. S.1.4(a) Fig. S.1.4(b)

Fig. S.1.4(c)

Fig. S.1.4(d)

the third stress system of Figs S.1.4(e) and (f) transforms into thex,y,xy system of

Fig. S.1.4(g).

Finally, the states of stress shown in Figs S.1.4(a), (d) and (g) are superimposed

to give the state of stress shown in Fig. S.1.4(h) from which it can be seen that

I = II = 15 N/mm2 and that thex and y planes are principal planes.


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Fig. S.1.4(e) Fig. S.1.4(f)

Fig. S.1.4(g)

Fig. S.1.4(h)

S.1.5

The geometry of Mohrs circle of stress is shown in Fig. S.1.5 in which the circle is

constructed using the method described in Section 1.8.

From Fig. S.1.5

x= OP1= OB BC + CP1 (i)


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OC

max

max

P1

P2

Q1(x, xy)

Q2(sy,xy)

B (I)

Fig. S.1.5

In Eq. (i) OB= I, BC is the radius of the circle which is equal to max andCP1 =

CQ21 Q1P21 =

2max 2xy. Hence

x= I max +2max 2xySimilarly

y= OP2= OB BC CP2 in which CP2= CP1Thus

y= I max 2max 2xy

S.1.6

From bending theory the direct stress due to bending on the upper surface of the shaft

at a point in the vertical plane of symmetry is given by

x=My

I= 25 10

6 75

1504/64

= 75 N/mm2

From the theory of the torsion of circular section shafts the shear stress at the same

point is

xy=Tr

J= 50 10

6 75 1504/32 = 75 N/mm

2


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Substituting these values in Eqs (1.11) and (1.12) in turn and noting thaty = 0

I=75

2+ 1

2752 + 4 752

i.e.

I= 121.4 N/mm2

II=75

2 1

2

752 + 4 752

i.e.

II= 46.4 N/mm2

The corresponding directions as defined byin Fig. 1.8(b) are given by Eq. (1.10)i.e.

tan 2= 2 7575 0 = 2

Hence

= 3143(I)and

= 12143(II)

S.1.7

The direct strains are expressed in terms of the stresses using Eqs (1.42), i.e.

x=1

E[x (y + z)] (i)

y= 1E

[y (x + z)] (ii)

z=1

E[z (x + y)] (iii)

Then

e = x + y + z=1

E[x + y + z 2(x + y + z)]

i.e.

e = (1 2)E

(x + y + z)

whence

y + z=Ee

(1 2) x


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Substituting in Eq. (i)

x=1

E

x

Ee

1 2 x

so that

Ex= x(1 + ) Ee

1 2Thus

x=Ee

(1 2)(1 + ) +E

(1 + )x

or, sinceG=

E/2(1+

) (see Section 1.15)

x= e + 2Gx

Similarly

y= e + 2Gyand

z

=e

+2Gz

S.1.8

The implication in this problem is that the condition of plane strain also describes the

condition of plane stress. Hence, from Eqs (1.52)

x=1

E(x y) (i)

y=1

E(y x) (ii)

xy=xy

G= 2(1 + )

Exy (see Section 1.15) (iii)

The compatibility condition for plane strain is

2xy

x y =2y

x2 +2x

y2 (see Section 1.11) (iv)

Substituting in Eq. (iv) for x,y andxy from Eqs (i)(iii), respectively, gives

2(1 + )2xy

x y=

2

x2(y x) +

2

y2(x y) (v)


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Also, from Eqs (1.6) and assuming that the body forces XandYare zero

x

x+ zy

y= 0 (vi)

y

y+ xy

x= 0 (vii)

Differentiating Eq. (vi) with respect toxand Eq. (vii) with respect toyand adding gives

2x

x2+

2xy

y x+

2y

y2+

2xy

x y= 0

or

22xy

x y=

2x

x2+

2y

y2

Substituting in Eq. (v)

(1 + )2x

x2+

2y

y2

=

2

x2(y x) +

2

y2(x y)

so that

(1 + )2x

x2+

2y

y2

=

2y

x2+

2x

y2

2x

x2+

2y

y2

which simplifies to

2y

x2+

2x

y2+

2x

x2+

2y

y2 = 0

or 2

x2+

2

y2

(x + y) = 0

S.1.9

Suppose that the load in the steel bar isPst and that in the aluminium bar isPal. Then,

from equilibriumPst + Pal= P (i)

From Eq. (1.40)

st=Pst

AstEstal=

Pal

AalEal


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Since the bars contract by the same amount

Pst

AstEst= Pal

AalEal(ii)

Solving Eqs (i) and (ii)

Pst=AstEst

AstEst +AalEalP Pal=

AalEal

AstEst +AalEalP

from which the stresses are

st=Est

AstEst

+AalEal

P al=Eal

AstEst

+AalEal

P (iii)

The areas of cross-section are

Ast= 752

4= 4417.9 mm2 Aal=

(1002 752)4

= 3436.1 mm2

Substituting in Eq. (iii) we have

st

=

106 200 000

(4417.9 200 000 + 3436.1 80 000) =172.6 N/mm2 (compression)

al=106 80 000

(4417.9 200 000 + 3436.1 80 000) = 69.1 N/mm2 (compression)

Due to the decrease in temperature in which no change in length is allowed the strain

in the steel isstTand that in the aluminium isalT. Therefore due to the decrease in

temperature

st

=EststT

=200 000

0.000012

150

=360.0 N/mm2 (tension)

al= EalalT= 80 000 0.000005 150 = 60.0 N/mm2 (tension)

The final stresses in the steel and aluminium are then

st(total) = 360.0 172.6 = 187.4 N/mm2 (tension)al(total) = 60.0 69.1 = 9.1 N/mm2 (compression).

S.1.10

The principal strains are given directly by Eqs (1.69) and (1.70). Thus

I=1

2(0.002 + 0.002) + 1

2

(0.002 + 0.002)2 + (+0.002 + 0.002)2


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i.e.

I= +0.00283Similarly

II= 0.00283The principal directions are given by Eq. (1.71), i.e.

tan 2= 2(0.002) + 0.002 0.0020.002 + 0.002 = 1

Hence

2

= 45 or

+135

and

= 22.5 or +67.5

S.1.11

The principal strains at the point Pare determined using Eqs (1.69) and (1.70). Thus

I=

12

(222 + 45) + 12

(222 + 213)2 + (213 45)2

106

i.e.

I= 94.0 106

Similarly

II= 217.0 106

The principal stresses follow from Eqs (1.67) and (1.68). Hence

I=31 000

1 (0.2)2 (94.0 0.2 271.0) 106

i.e.

I= 1.29 N/mm2

Similarly

II= 8.14 N/mm2

Since Plies on the neutral axis of the beam the direct stress due to bending is zero.

Therefore, at P,x = 7 N/mm2 andy = 0. Now subtracting Eq. (1.12) from (1.11)

I II=2x+ 42xy


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i.e.

1.29 + 8.14 =

72 + 42xyfrom whichxy

=3.17 N/mm2.

The shear force at Pis equal toQ so that the shear stress at Pis given by

xy= 3.17 =3Q

2 150 300from which

Q = 95 100 N = 95.1 kN.

Solutions to Chapter 2 Problems

S.2.1

The stress system applied to the plate is shown in Fig. S.2.1. The origin, O, of the axes

may be chosen at any point in the plate; let P be the point whose coordinates are (2, 3).

2p

2p

3p3p

4p

4p

4p

4p

y

xO

P (2,3)

Fig. S.2.1

From Eqs (1.42) in whichz = 0

x= 3p

E 2p

E= 3.5p

E(i)

y=2p

E+ 3p

E= 2.75p

E(ii)

Hence, from Eqs (1.27)

u

x= 3.5p

Eso that u = 3.5p

Ex +f1(y) (iii)

megson solution manual aircraft structures ch01

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