Aircraft structures for engineering students - part 3 docx

108 Energy methods of structural analysis

It 6z (1 + at)

R

(a) (b)

Fig. 4.29 (a) Linear temperature gradient applied to beam element; (b) bending of beam element due to

temperature gradient.

the element will increase in length to 6z( 1 + at), where a is the coefficient of linear

expansion of the material of the beam. Thus from Fig. 4.29(b)

R R+h

-=

Sz Sz(1 +at)

giving

R = h/at

Also

so that, from Eq. (4.32)

Szat 60 = - h

(4.32)

(4.33)

We may now apply the principle of the stationary value of the total complementary

energy in conjunction with the unit load method to determine the deflection A,, due

to the temperature of any point of the beam shown in Fig. 4.28. We have seen that the

above principle is equivalent to the application of the principle of virtual work where

virtual forces act through real displacements. Therefore, we may specify that the

displacements are those produced by the temperature gradient while the virtual

force system is the unit load. Thus, the deflection ATe,B of the tip of the beam is

found by writing down the increment in total complementary energy caused by the

application of a virtual unit load at B and equating the resulting expression to zero

(see Eqs (4.13) and (4.18)). Thus

References 109

or

wher the bending moment at any section due to th

de from Eq. (4.33) we have

at

ATe,B = IL dz

(4.34)

unit load. Substituting for

(4.35)

where t can vary arbitrarily along the span of the beam, but only linearly with depth.

For a beam supporting some form of external loading the total deflection is given by

the superposition of the temperature deflection from Eq. (4.35) and the bending

deflection from Eqs (4.27); thus

(4.36)

Example 4.17

Determine the deflection of the tip of the cantilever in Fig. 4.30 with the temperature

gradient shown.

Spanwise variation of t

Fig. 4.30 Beam of Example 4.1 1

Applying a unit load vertically downwards at B, MI = 1 x z. Also the temperature

t at a section z is to(I - z)/Z. Substituting in Eq. (4.35) gives

Integrating Eq. (i) gives

(i.e. downwards)

1 Charlton, T. M., Energy Principles in Applied Statics, Blackie, London, 1959.

2 Gregory, M. S., Introduction to Extremum Principles, Buttenvorths, London, 1969.

3 Megson, T. H. G., Structural and Stress Analysis, Arnold, London, 1996.

1 10 Energy methods of structural analysis

Argyris, J. H. and Kelsey, S., Energy Theorems and Structural Analysis, Butterworths, London,

Hoff, N. J., The Analysis of Structures, John Wiley and Sons, Inc., New York, 1956.

Timoshenko, S. P. and Gere, J. M., Theory of Elastic Stability, McGraw-Hill Book Company,

1960.

New York, 1961.

.-. . .

P.4.1 Find the magnitude and the direction of the movement of the joint C of the

plane pin-jointed frame loaded as shown in Fig. P.4.1. The value of LIAE for each

member is 1 /20 mm/N.

Ans. 5.24mm at 14.7" to left of vertical.

Fig. P.4.1

P.4.2 A rigid triangular plate is suspended from a horizontal plane by three

vertical wires attached to its corners. The wires are each 1 mm diameter, 1440mm

long, with a modulus of elasticity of 196 000 N/mm2. The ratio of the lengths of the

sides of the plate is 3:4:5. Calculate the deflection at the point of application due

to a lOON load placed at a point equidistant from the three sides of the plate.

Ans. 0.33mm.

P.4.3 The pin-jointed space frame shown in Fig. P.4.3 is attached to rigid

supports at points 0, 4, 5 and 9, and is loaded by a force P in the x direction and a

force 3P in the negative y direction at the point 7. Find the rotation of member 27

about the z axis due to this loading. Note that the plane frames 01234 and 56789

are identical. All members have the same cross-sectional area A and Young's

modulus E.

Ans. 382P19AE.

P.4.4 A horizontal beam is of uniform material throughout, but has a second

moment of area of 1 for the central half of the span L and 1/2 for each section in

Problems 111

Y

Fig. P.4.3

both outer quarters of the span. The beam carries a single central concentrated

load .P.

(a) Derive a formula for the central deflection of the beam, due to P, when simply

(b) If both ends of the span are encastrk determine the magnitude of the fixed end

Am. 3PL3/128EI, 5PL/48 (hogging).

P.4.5 The tubular steel post shown in Fig. P.4.5 supports a load of 250 N at the

free end C. The outside diameter of the tube is lOOmm and the wall thickness is

3mm. Neglecting the weight of the tube find the horizontal deflection at C. The

modulus of elasticity is 206 000 N/mm2.

supported at each end of the span.

moments.

Am. 53.3mm.

Fig. P.4.5

1 12 Energy methods of structural analysis

P.4.6 A simply supported beam AB of span L and uniform section carries a

distributed load of intensity varying from zero at A to wo/unit length at B according

to the law

w=””(l-&) L

per unit length. If the deflected shape of the beam is given approximately by the

expression

n-2 2x2

21 = u1 sin- + u2 sin- L L

evaluate the coefficients ul and a2 and find the deflection of the beam at mid-span.

Ans. a1 = 2w0L4(7? + 4)/EIr7, a2 = -w0L4/16EI2, O.O0918woL4/EI.

P.4.7 A uniform simply supported beam, span L, carries a distributed loading

which vanes according to a parabolic law across the span. The load intensity is

zero at both ends of the beam and wo at its mid-point. The loading is normal to a

principal axis of the beam cross-section and the relevant flexural rigidity is EI.

Assuming that the deflected shape of the beam can be represented by the series

im

Y = ai sin￾33

i=l ‘

find the coefficients ui and the deflection at the mid-span of the beam using the first

term only in the above series.

Am. ai = 32w0L4/EI7r7i7 (iodd), w0L4/94.4EI.

P.4.8 Figure P.4.8 shows a plane pin-jointed framework pinned to a rigid founda￾tion. All its members are made of the same material and have equal cross-sectional

area A, except member 12 which has area A&.

4a 5a

Fig. P.4.8

Problems 113

Under some system of loading, member 14 carries a tensile stress of 0.7N/mm2.

Calculate the change in temperature which, if applied to member 14 only, would

reduce the stress in that member to zero. Take the coefficient of linear expansion as

Q: = 24 x 10-6/"C and Young's modulus E = 70 000 N/mmz.

Ans. 5.6"C.

P.4.9 The plane, pin-jointed rectangular framework shown in Fig. P.4.9(a) has

one member (24) which is loosely attached at joint 2: so that relative movement

between the end of the member and the joint may occur when the framework is

loaded. This movement is a maximum of 0.25mm and takes place only in the

direction 24. Figure P.4.9(b) shows joint 2 in detail when the framework is

unloaded. Find the value of the load P at which member 24 just becomes an effective

part of the structure and also the loads in all the members when P is 10 000 N. All

bars are of the same material (E = 70000N/mm2) and have a cross-sectional area

of 300mm2.

Ans. P = 2947N: F12 = 2481.6N(T), F23 = 1861.2N(T), F34 = 2481.6N(T),

F41 = 5638.9N(C), F1, = 9398.1N(T), F24 = 3102.ON(C).

/0.25mrn

600 mm

I

I￾I

Fig. P,4.9

P.4.10 The plane frame ABCD of Fig. P.4.10 consists of three straight members

with rigid joints at B and C, freely hinged to rigid supports at A and D. The flexural

rigidity of AB and CD is twice that of BC. A distributed load is applied to AB, varying

linearly in intensity from zero at A to w per unit length at B.

Determine the distribution of bending moment in the frame, illustrating your

results with a sketch showing the principal values.

Ans. MB = 7w12/45, Mc = 8w12/45. Cubic distribution on AB, linear on BC

and CD.