ch8 air craft structures by th megson

8/15/2019 ch8 air craft structures by th megson

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Solutions to Chapter 8 Problems 117

S.7.12

From Eq. (7.36) the deection of the plate from its initial curved position is

w1 = B11 sinπ xa sin

π yb

in which

B11 = A11 N x

π 2 Da 2

1 +a 2

b2

2

− N xThe total deection, w, of the plate is given by

w = w1 +w0i.e.

w = A11 N x

π 2 Da 2

1 +a 2

b2

2

− N x+ A11 sin

π xa

sinπ yb

i.e.w =

A11

1 − N xa 2

π 2 D1 +

a 2

b2

2 sinπ xa

sinπ yb

Solutions to Chapter 8 Problems

S.8.1

The forces on the bar AB are shown in Fig. S.8.1 where

M B =K dvd z B

(i)

and P is the buckling load.

From Eq. (8.1)

EI d2vd z2 = −P v (ii)

The solution of Eq. (ii) is

v = A cos µ z + B sin µ z (iii)


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118 Solutions Manual

P A

B P

y

z

l

M BV B

Fig. S.8.1

where µ 2 =P/EI .When z =0, v =0 so that, from Eq. (iii), A =0. Hencev = B sin µ z (iv)

Thendvd z = µ B cos µ z

and when z = l, dv /d z= M B / K from Eq. (i). Thus

B

=

M B

µ K cos µ land Eq. (iv) becomes

v = M B

µ K cos µ lsin µ z (v)

Also, when z = l, Pv B = M B from equilibrium. Hence, substituting in Eq. (v) for M BvB =

P vBµ K cos µ l

sin µ l

from which

P =µ K

tan µ l(vi)

(a) When K →∞, tan µ l →∞and µ l →π /2, i.e.

P EI l →π2from which

P →π 2 EI 4l2

which is the Euler buckling load of a pin-ended column of length 2 l.(b) When EI →∞, tan µ l →µ l and Eq. (vi) becomes P =K / l and the bars remainstraight.


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S.8.2

Suppose that the buckling load of the column is P . Then from Eq. (8.1) and referringto Fig. S.8.2, in AB

EI d2vd z2 = −P v (i)

and in BC

4 EI d2vd z2 = −P v (ii)

B

DA

P P

EI EI

v z

y 4 EI C

l /2l /4 l /4

Fig. S.8.2

The solutions of Eqs (i) and (ii) are, respectively

vAB = A cos µ z + B sin µ z (iii)vBC =C cos

µ

2 z + D sin

µ

2 z (iv)

in which

µ 2 =P

EI

When z =0, vAB =0 so that, from Eq. (iii), A =0. ThusvAB = B sin µ z (v)

Also, when z = l /2, (dv /d z)BC =0. Hence, from Eq. (iv)

0 = −µ

2C sin

µ l4 +

µ

2 D cos

µ l4

whence

D = C tanµ l4

Then

vBC =C cosµ

2 z + tan

µ l4

sinµ

2 z (vi)


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When z = l /4, vAB =vBC so that, from Eqs (v) and (vi)

B sinµ l4 = C cos

µ l8 + tan

µ l4

sinµ l8

which simplies to

B sinµ l4 = C sec

µ l4

cosµ l8

(vii)

Further, when z = l /4, (dv /d z)AB =(dv /d z)BC . Again from Eqs (v) and (vi)

µ B cosµ l4 = C −

µ

2sin

µ l8 +

µ

2tan

µ l4

cosµ l8

from which

B cosµ l4 =

C 2

secµ l4

sinµ l8

(viii)

Dividing Eq. (vii) by Eq. (viii) gives

tanµ l4 = 2 tan

µ l8

or

tanµ l4

tanµ l8 = 2

Hence

2tan 2 µ l/ 81 − tan 2 µ l/ 8 =

2

from which

tanµ l8 =

1√ 2

and

µ l8 = 35 .26◦ =0.615 rad

i.e.

P EI l8 = 0.615so that

P =24 .2 EI

l2


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S.8.3

With the spring in position the forces acting on the column in its buckled state areshown in Fig. S.8.3. Thus, from Eq. (8.1)

EI d2vd z2 =4P (δ −v) −k δ(l − z) (i)

The solution of Eq. (i) is

v = A cos µ z + B sin µ z +δ

4P[4P +k ( z − l)] (ii)

4 P y

l

v z

δ k δ

Fig. S.8.3

where

µ 2 =4P EI

When z

=0, v

=0, hence, from Eq. (ii)

0 = A +δ

4P(4P −kl)

from which

A =δ(kl −4P )

4P

Also when z =0, dv /d z=0 so that, from Eq. (ii)

0 = µ B +δk 4P

and

B = −δk

4P µ


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Eq. (ii) then becomes

v =δ

4P(kl −4P )cos µ z −

k µ

sin µ z +4P +k ( z − l) (iii)

When z = l, v =δ. Substituting in Eq. (iii) gives

δ =δ

4P(kl −4P )cos µ l −

k µ

sin µ l +4P

from which

k =4P µ

µ l − tan µ l

S.8.4

The compressive load P will cause the column to be displaced from its initial curvedposition to that shown in Fig. S.8.4. Then, from Eq. (8.1) and noting that the bendingmoment at any point in the column is proportional to the change in curvature produced(see Eq. (8.22))

EI d2v

d z2 − EI

d2v0

d z2 = −P v (i)

Now

v0 =a4 zl2

(l − z)so that

d2v0d z2 = −

8al2

P P

y

z

l

v

v 0

Fig. S.8.4

and Eq. (i) becomes

d2vd z2 +

P EI

v = −8al2

(ii)


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v = A cos λ z + B sin λ z −8a / (λ l)2 (iii)

where λ 2

=P / EI .

When z =0, v =0 so that A =8a /(λ l)2 . When z = l /2, dv /d z=0. Thus, from Eq. (iii)

0 = −λ A sinλ l2 + λ B cos

λ l2

whence

B =8a

(λ l)2 tan

λ l2

Eq. (iii) then becomes

v =8a

(λ l)2cos λ z + tan

λ l2

sin λ z −1 (iv)

The maximum bending moment occurs when v is a maximum at z= l /2. Then, fromEq. (iv)

M (max) = −P vmax = −8aP(λ l)2

cosλ l2 + tan

λ l2

sinλ l2 −1

from which

M (max) = −8aP(λ l)2

secλ l2 −1

S.8.5

Under the action of the compressive load P the column will be displaced to the positionshown in Fig. S.8.5. As in P.8.4 the bending moment at any point is proportional to thechange in curvature. Then, from Eq. (8.1)

EI d2vd z2 − EI

d2v0d z2 = −P v (i)

y

z

v

v 0

δ

l /2 l /2

P P

Fig. S.8.5


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In this case, since each half of the column is straight before the application of P ,d2v0 /d z2 =0 and Eq. (i) reduces to

EI d2v

d z2 = −P v (ii)


v = A cos µ z + B sin µ z (iii)in which µ 2 =P / EI . When z =0, v =0 so that A =0 and Eq. (iii) becomes

v = B sin µ z (iv)The slope of the column at its mid-point in its unloaded position is 2 δ/ l. This must bethe slope of the column at its mid-point in its loaded state since a change of slope overzero distance would require an innite bending moment. Thus, from Eq. (iv)

dvd z =

2δl = µ B cos

µ l2

so that

B = 2δµ l cos( µ l/ 2)and

v =2δ

µ l cos ( µ l/ 2)sin µ z (v)

The maximum bending moment will occur when v is a maximum, i.e. at the mid-pointof the column. Then

M (max) = −P vmax = −2P δ

µ l cos ( µ l/ 2)sin

µ l2

from which

M (max) = −P2δl EI P tan P EI l2

S.8.6

Referring to Fig. S.8.6 the bending moment at any section z is given by

M =P (e +v) −wl2

z +w z2

2


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z

y

P P

l

e e

w

wl 2

wl 2

v

Fig. S.8.6

or

M =P (e +v) +w2

( z2 − lz) (i)Substituting for M in Eq. (8.1)

EI d2vd z2 +P v = −Pe −

w2

( z2 − lz)or

d2vd z2 +µ

2v = −µ2e −

wµ 2

2P( z2 − lz) (ii)


v = A cos µ z + B sin µ z − e +w

2P(lz − z

2) +w

µ 2P(iii)

When z =0, v =0, hence A =e −w / µ2P . When z = l /2, dv /d z=0 which gives

B = A tan µ l2 = e − wµ 2P tan µ l2Eq. (iii) then becomes

v = e −w

µ 2Pcos µ ( z − l/ 2)

cos µ l/ 2 −1 +w

2P(lz − z

2) (iv)

The maximum bending moment will occur at mid-span where z= l/ 2 and v =vmax .From Eq. (iv)vmax = e −

EIwP 2

secµ l2 − 1 +

wl2

8P

and from Eq. (i)

M (max) = Pe +P vmax −wl2

8


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whence

M (max) = Pe −w

µ 2sec

µ l2 +

wµ 2

(v)

For the maximum bending moment to be as small as possible the bending moment atthe ends of the column must be numerically equal to the bending moment at mid-span.Thus

Pe + Pe −wµ 2

secµ l2 +

wµ 2 =0

or

Pe 1 + secµ l2 =

wµ 2

secµ l2 −1

Then

e =w

P µ 21 − cos µ l/ 21 + cos µ l/ 2

i.e.

e = wP µ 2

tan 2 µ l

4(vi)

From Eq. (vi) the end moment is

Pe =w

µ 2 tan2

µ l4 =

wl2

16tan µ l/ 4

µ l/ 4tan µ l/ 4

µ l/ 4

When P →0, tan µ l /4 →µ l /4 and the end moment becomes wl2 /16.

S.8.7

From Eq. (8.21) the buckling stress, σ b , is given by

σ b =π 2 E t(l/ r )2

(i)

The stress–strain relationship is

10 .5 ×106ε = σ +21 000

σ

49 000

16(ii)

Hence10 .5 ×10

6 dεdσ = 1 +

16 ×21 000(49 000) 16

σ 15

from which

E t =dσ dε =

10 .5 × 106× (49 000)

16

(49 000) 16 +16 × 21 000( σ )15


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y

z P P

a

v

b

A

B

B

z 1

v 1

y 1M B

M BC

Fig. S.8.8

When z1 =b, dv1 /d z1 =0. Thus A = M Bb and EI

dv1d z1 = − M B ( z1 −b) (i)

At B, when z1 =0, Eq. (i) givesdv1d z1 =

M Bb EI

(ii)

In BC Eq. (8.1) gives

EI d2vd z2 = −P v + M B

or

EI d2vd z2 +P v = M B (iii)

The solution of Eq. (iii) is

v = B cos λ z +C sin λ z + M B / P (iv)When z =0, v =0 so that B =− M B / P .When z

=a /2, dv /d z

=0 so that

C = B tanλ a2 = −

M BP

tanλ a2

Eq. (iv) then becomes

v = − M B

Pcos λ z + tan

λ a2

sin λ z −1


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so that

dvd z = −

M BP −λ sin λ z + λ tan

λ a2

cos λ z

At B, when z =0,dvd z = −

M BP

λ tanλ a2

(v)

Since d v1 /d z1 =dv /d z at B then, from Eqs (ii) and (v)b

EI = −λP

tanλ a2

whence

λ a2 = −

12

ab

tanλ a2

S.8.9

In an identical manner to S.8.4

EI d2vd z2 − EI

d2vd z2 = −P v

where v is the total displacement from the horizontal. Thus

d2vd z2 +

P EI

v =d2vd z2

or, since

d2vd z2 = −

π 2

l2 δ sin

πl z and µ 2 =

P EI

d2vd z2 +µ

2v = −π 2

l2 δ sin

π zl

(i)

The solution of Eq. (i) is

v = A cos µ z + B sin µ z +π 2δ

π 2 −µ 2 l2 sin

π zl (ii)

When z =0 and l, v =0, hence A = B=0 and Eq. (ii) becomes

v =π 2δ

π 2 −µ 2 l2 sin

π zl


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The maximum bending moment occurs at the mid-point of the tube so that

M (max) = P v =Pπ 2δ

π 2 −µ 2 l2 =P δ

1 −Pl 2 /π 2 EI i.e.

M (max) =P δ

1 −P / P e =P δ

1 −αThe total maximum direct stress due to bending and axial load is then

σ (max) =P

π dt + Pδ1 −α

d / 2π d 3 t / 8

Henceσ (max) =

Pπ dt

1 +1

1 −α4δd

S.8.10

The forces acting on the members AB and BC are shown in Fig. S.8.10

A

B

P P

v

y

a b

V

P

z v B

V B

C

Fig. S.8.10

Considering rst the moment equilibrium of BC about C

P vB =Vbfrom which

vB =VbP

(i)

For the member AB and from Eq. (8.1)

EI d2vd z2 = −P v −Vz

or

d2vd z2 +

P EI

v = −Vz EI

(ii)


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v = A cos λ z + B sin λ z −VzP

(iii)

When z =0, v =0 so that A =0. Also when z =a , dv /d z=0, hence

0 = λ B cos λ a −V P

from which

B =V

λ P cos λ a

and Eq. (iii) becomes

v =V P

sin λ zλ cos λ a − z

When z =a , v =vB =Vb/P from Eq. (i). ThusVb

P =V

P

sin λ a

λ cos λ a −a

from which

λ (a +b) = tan λ a

S.8.11

The bending moment, M , at any section of the column is given by

M =P CR v = P CR k (lz − z2) (i)

Also

dvd z = k (l −2 z) (ii)

Substituting from Eqs (i) and (ii) in Eq. (8.47)

U +V =P 2CR k

2

2 E 1

I 1 a

0(lz − z

2)2d z +1

I 2 l−a

a(lz − z

2)2d z +1

I 1 l

l−a(lz − z

2)2d z

−P CR k 2

2 l

0(l −2 z)

2d z


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i.e.

U +V =P 2CR k

2

2 E 1

I 1

l2 z3

3 −lz4

2 + z5

5

a

0

+1

I 2

l2 z3

3 −lz4

2 + z5

5

l−a

a

+1

I 1

l2 z3

3 −lz4

2 +l5

5

l

l−a−

P CR k 2

2l2 z −2lz

2+

4 z3

3

l

0

i.e.

U +V =P 2CR k

2

2 EI 2

I 2 I 1 −1

l2a 3

3 −la 4

2 +a 5

5 −l2(l − a )

3

3 +l(l − a )

4

2 −(l − a )

5

5

+ I 2 I 1

l5

30 −P CR k 2 l3

6

From the principle of the stationary value of the total potential energy

∂(U +V )∂k =

P 2CR k

EI 2

I 2 I 1 −1

l2a 3

3 −la 4

2 +a 5

5 −l2(l − a )

3

3

+l(l − a )

4

2 −(l − a )

5

5 + I 2 I 1

l5

30 −P CR kl3

3 = 0

Hence

P CR = EI 2 l3

3 I 2 I 1 −1

l2a 3

3 −la 4

2 +a 5

5 −l2(l − a )

3

3

+l(l − a )4

2 −(l − a )5

5 + I 2 I 1

l5

30

(iii)

When I 2 =1.6 I 1 and a =0.2l, Eq. (iii) becomes

P CR =14 .96 EI 1

l2 (iv)

Without the reinforcement

P CR =π 2 EI 1

l2 (v)

Therefore, from Eqs (iv) and (v) the increase in strength is

EI 1l2

(14 .96 −π2)


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Thus the percentage increase in strength is

EI l2

(14 .96 −π2)

l2

π 2 EI ×100 = 52%

Since the radius of gyration of the cross-section of the column remains unchanged

I 1 = A1r 2 and I 2 = A2r

2

Hence A2 A1 =

I 2 I 1 = 1.6 (vi)

The original weight of the column is lA1ρ where ρ is the density of the material of thecolumn. Then, the increase in weight =0.4 lA1ρ +0.6 lA2ρ − lA1ρ =0.6lρ ( A2 − A1).Substituting for A2 from Eq. (vi)

Increase in weight = 0.6lρ (1 .6 A1 − A1) = 0.36 lA1ρi.e. an increase of 36%.

S.8.12

The equation for the deected centre line of the column is

v =4δl2

z2 (i)

in which δ is the deection at the ends of the column relative to its centre and the originfor z is at the centre of the column. Also, the second moment of area of its cross-sectionvaries, from the centre to its ends, in accordance with the relationship

I = I 1 1 −1.6 zl (ii)

At any section of the column the bending moment, M , is given by

M =P CR (δ −v) = P CR δ 1 −4 z2

l2 (iii)

Also, from Eq. (i)

dvd z =

8δl2

z (iv)

Substituting in Eq. (8.47) for M , I and dv /d z

U +V = 2 l/ 2

0

P 2CR δ2(1 −4 z

2 / l2)2

2 EI 1(1 −1.6 z/ l)d z −

P CR2

2 l/ 2

0

64δ2

l4 z2d z


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or

U +V =P 2CR δ

2

EI 1 l3 l/ 2

0

(l2 −4 z2)2

(l −1.6 z)d z −

64P CR δ2

l4 l/ 2

0 z2d z (v)

Dividing the numerator by the denominator in the rst integral in Eq. (v) gives

U +V =P 2CR δ

2

EI 1 l3 l/ 2

0(−10 z

3−6.25 lz

2+1.09 l

2 z +0.683 l3)d z

+0.317 l3

l/ 2

0

d z(1 −1.6 z/ l) −

64P CR δ2

l4 z3

3

l/ 2

0

Hence

U +V =P 2CR δ

2

EIl 3 −10 z4

4 −6.25 l z3

3 +1.09 l2 z

2

2 +0.683 l3 z

−0.317

1.6l4 log e 1 −

1.6 zl

l/ 2

0 −8P CR δ2

3l

i.e.

U

+V

=0.3803 P 2CR δ

2 l

EI 1 −8P CR δ2

3lFrom the principle of the stationary value of the total potential energy

∂(U +V )∂δ =

0.7606 P 2CR δl EI 1 −

16P CR δ3l = 0

Hence

P CR

=

7.01 EI 1l2

For a column of constant thickness and second moment of area I 2 ,

P CR =π 2 EI 2

l2 (see Eq. (8.5))

For the columns to have the same buckling load

π 2 EI 2l2 =

7.01 EI 1l2

so that

I 2 =0.7 I 1Thus, since the radii of gyration are the same

A2 =0.7 A1


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The total strain energy of the column will be the sum of the strain energy due tobending and the strain energy due to the resistance of the elastic foundation. For thelatter, consider an element, δ z, of the column. The force on the element when subjectedto a small displacement, v, is k δ zv. Thus, the strain energy of the element is 12 kv

2δ z and

the strain energy of the column due to the resistance of the elastic foundation is

l

0

12

k v2d z

Substituting for v from Eq. (v)

U (elastic foundation) =12

k a 22l4

l

0 z4 −

10 z5

3l +37 z6

9l2 −20 z7

9l3 +4 z8

9l4d z

i.e. U (elastic foundation) =0.0017 ka22 l.

Now substituting for d 2v /d z2 and d v /d z in Eq. (8.48) and adding U (elastic founda-tion) gives

U +V = EI 2

l

0

4a 22l4

1 −10 z

l +33 z2

l2 −40 z3

l3 +16 z4

l4d z +0.0017 ka

22 l

−P CR

2 l

0

a 22l4 4 z

2

−20 z3

l +107 z4

3l2 −80 z5

3l3 +64 z6

9l4 d z (viii)

Eq. (viii) simplies to

U +V =0.4 EI

l3 a22 +0.0017 ka

22 l −

0.019 a 22P CRl

From the principle of the stationary value of the total potential energy

∂(U +V )∂a 2 =

0.8 EI l3

a2 +0.0034 ka2 l −0.038 a 2P CR

l

whence

P CR =21 .05 EI

l2 +0.09kl2

S.8.14

The purely exural instability load is given by Eq. (8.7) in which, from Table 8.1le =0.5l where l is the actual columnlength. Also it is clear that the least secondmomentof area of the column cross-section occurs about an axis coincident with the web. Thus

I = 2 ×2tb3

12 =tb3

3


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22/25


Eqs (i) and (iii) may be written, respectively, as

P CR =1.33C 1

l2

and

P CR( θ ) =C 2 +1.175 C 1

l2

where C 1 and C 2 are constants. Thus, if l were less than the value found, the increasein the last term in the expression for P CR( θ ) would be less than the increase in the valueof P CR , i.e. P CR( θ ) < P CR for a decrease in l and the column would fail in torsion.

S.8.15

In this case Eqs (8.77) do not apply since the ends of the column are not free to warp.From Eq. (8.70) and since, for the cross-section of the column, xs = ys =0,

E d4θ d z4 + I 0

P A −GJ

d2θ d z2 =0 (i)

For buckling, P =P CR , the critical load and PCR / A=σ CR , the critical stress. Eq. (i)may then be writtend4θ d z4 +λ

2 d2θ d z2 =0 (ii)

in which

λ 2 =( I 0σ CR −GJ )

E (iii)


θ

= A cos λ z

+ B sin λ z

+ Dz

+F (iv)

The boundary conditions are:

θ =0 at z = 0 and z = 2ldθ d z = 0 at z = 0 and z = 2l (see Eq. (18.19))

Then B = D =0, F =− A and Eq. (iv) becomesθ

= A( cos λ z

−1) (v)

Since θ =0 when z =2lcos λ 2l = 1

or

λ 2l = 2nπ


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Hence, for n =1

λ 2 =π 2

l2

i.e. from Eq. (iii) I 0σ CR −GJ

E =π 2

l2

so that

σ CR =1

I 0GJ +

π 2 E l2

(vi)

For the cross-section of Fig. P.8.15

J =st 3

3(see Eq. (18.11))

i.e.

J =8bt 3

3 =8 ×25 .0 × 2.5

3

3 = 1041 .7 mm4

and

I xx =4bt (b cos 30 ◦)2+2

(2b)3 t sin2 60◦12

(see Section 16.4.5)

i.e.

I xx =4b3 t =4 × 25 .0

3×2.5 = 156 250 .0 mm

4

Similarly

I yy =4bt 3

12 +btb2

+2(2b)3 t cos2 60◦

12 =14b3 t

3

so that

I yy =14 ×25 .03×2.5/ 3 = 182 291 .7 mm

4

Then

I 0 = I xx + I yy =338 541 .7 mm4

The torsion-bending constant, , is found by the method described in Section 27.2 andis given by

= b5 t =25 .0

5× 2.5 = 24 .4 ×10

6 mm 4

Substituting these values in Eq. (vi) gives

σ CR =282 .0 N / mm2


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S.8.16

The three possible buckling modes of the column are given by Eqs (8.77) i.e.

P CR( xx) = π2 EI xx L2

(i)

P CR( yy) =π 2 EI yy

L2 (ii)

P CR( θ ) = A I 0

GJ +π 2 E

L2 (iii)

From Fig. P.8.16 and taking the x axis parallel to the anges

A = (2 × 20 +40) ×1.5 = 120 mm2

I xx =2 × 20 × 1.5 × 202+1.5 × 40

3 / 12 = 3.2 ×104 mm 4

I yy =1.5 ×403 / 12 = 0.8 × 10

4 mm 4

I 0 = I xx + I yy =4.0 ×104 mm 4

J = (20 +40 +20) ×1.53 / 3 = 90 .0 mm

4 (see Eq. (18.11))

= 1.5 × 203

×402

122 × 40 +2040 +2 ×20

= 2.0 ×106 mm 6 (see Eq. (ii) of Example 27.1)

Substituting the appropriate values in Eqs (i), (ii) and (iii) gives

P CR( xx) =22 107 .9 NP CR( yy) =5527 .0 NP CR( θ ) =10 895 .2 N

Thus the column will buckle in bending about the y axis at a load of 5527.0 N.

S.8.17

The separate modes of buckling are obtained from Eqs (8.77), i.e.

P CR( xx) =P CR( yy) = π2 EI

L2 ( I xx = I yy = I , say) (i)

and

P CR( θ ) = A I 0

GJ +π 2 E

L2 (ii)


25/25


In this case

I xx = I yy =π r 3 t =π ×40

3×2.0 = 4.02 × 10

5 mm 4

A

=2π rt

=2π

×40

×2.0

=502 .7 mm 2

J =2π rt 3 / 3 = 2π × 40 ×2.0

3 / 3 = 670 .2 mm4

From Eq. (8.68)

I 0 = I xx + I yy + Ax2s (note that ys =0)

in which xs is the distance of the shear centre of the section from its vertical diameter;it may be shown that xs =80 mm (see S.17.3). Then

I 0

=2

×4.02

×10 5

+502 .7

×80 2

=4.02

×10 6 mm 4

The torsion-bending constant is found in a similar manner to that for the sectionshown in Fig. P.27.3 and is given by

= π r 5 t

23

π 2 −4

i.e.

=π

×405

×2.0

2

3π 2

−4

=1.66

×109 mm 6

(a) PCR( xx) =P CR( yy) =π 2 ×70 000 ×4.02 × 10

5

(3 .0 ×10 3)2 =3.09 ×10

4 N

(b) PCR( θ ) =502 .7

4.02 × 10622 000 ×670 .2 +

π 2 ×70 000 ×1.66 ×109

(3 .0 ×103)2

=1.78

×10 4 N

The exural–torsional buckling load is obtained by expanding Eq. (8.79). Thus

(P −P CR( xx))(P −P CR( θ )) I 0 / A −P2 x2s =0

from which

P 2(1 − Ax2s / I 0) −P (P CR( xx) +P CR( θ )) +P CR( xx)P CR( θ ) =0 (iii)

Substituting the appropriate values in Eq. (iii) gives

P 2 −24 .39 × 104P +27 .54 ×10 8 = 0 (iv)The solutions of Eq. (iv) are

P = 1.19 × 104 N or 23 .21 × 10

4 N

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ch8 air craft structures by th megson