Aircraft Structures 3E Episode 8 doc

Structural members such as these are known as open section beams, while the cellular components are termed closed section beams; clearly, both types of beam are subjected to axial, bendi

266 Airworthiness and airframe loads

in which llo is a function of height h and

Suppose that the aircraft is climbing at a speed V with a rate of climb ROC The time taken for the aircraft to climb from a height h to a height h + Sh is Sh/ROC during which time it travels a distance VShIROC Hence, from Eq (8.55) the fatigue damage experienced by the aircraft in climbing through a height Sh is

The total damage produced during a climb from sea level to an altitude H at a constant speed V and rate of climb ROC is

(8.56) Plotting l/llo against h from ESDU data sheets for aircraft having cloud warning radar and integrating gives

From the above

per cent of the total damage in the climb occurs in the first 3000 m

example, the change in wing stress produced by a gust may be represented by

dh/llo = 320.4, from which it can be seen that approximately 95

An additional factor influencing the amount of gust damage is forward speed For

= klueV, (see Eq (8.24)) (8.57)

in which the forward speed of the aircraft is in EAS From Eq (8.57) we see that the

gust velocity uf required to produce the fatigue limit stress S , is

The gust damage per km at different forward speeds V, is then found using Eq (8.54)

with the appropriate value of uf as the lower limit of integration The integral may be

evaluated by using the known approximate forms of N(S,,J and E(u,) from Eqs

(8.48) and (8.50) From Eq (8.48)

m

s a = su,e = K, (1 + c / J G ) from which

where Su?, = kl Veu, and S;,, = kl VeuF Also Eq (8.50) is

8.7 Fatigue 267

or, substituting for r(ue) from Eq (8.51)

3.23 105~;5.’6

E(ue) = 1000/~0 Equation (8.54) then becomes

It can be seen from Eq (8.59) that gust damage increases in proportion to V:/e5.26 so

that increasing forward speed has a dramatic effect on gust damage

The total fatigue damage suffered by an aircraft per flight is the sum of the damage

caused by the ground-air-ground cycle, the damage produced by gusts and the

damage due to other causes such as pilot induced manoeuvres, ground turning and

braking, and landing and take-off load fluctuations The damage produced by

these other causes can be determined from load exceedance data Thus, if this extra

damage per fight is De,,, the total fractional fatigue damage per flight is

Dtotal = DGAG + D g R a v + Dextra

We have seen that the concept of fail-safe structures in aircraft construction relies on

a damaged structure being able to retain sufficient of its load-carrying capacity to

268 Airworthiness and airframe loads

Shear, normal

to crack front

in plane of crack (edge sliding mode)

Shear, parallel to crack front (tearing mode)

Tension, normal

to faces of crack (opening mode)

1

Fig 8.19 Basic modes of crack growth

prevent catastrophic failure, at least until the damage is detected It is therefore essential that the designer be able to predict how and at what rate a fatigue crack will grow The ESDU data sheets provide a useful introduction to the study of crack propagation; some of the results are presented here

The analysis of stresses close to a crack tip using elastic stress concentration factors breaks down since the assumption that the crack tip radius approaches zero results in the stress concentration factor tending to infinity Instead, linear elastic fracture mechanics analyses the stress field around the crack tip and identifies features of the field common to all cracked elastic bodies

There are three basic modes of crack growth, as shown in Fig 8.19 Generally, the

stress field in the region of the crack tip is described by a two-dimensional model which may be used as an approximation for many practical three-dimensional loading cases Thus, the stress system at a distance I ( I < u ) from the tip of a crack of length

2 4 shown in Fig 8.20, can be expressed in the form

(8.62)

in whichf(8) is a different function for each of the three stresses and K is the stress

intensity factor; K is a function of the nature and magnitude of the applied stress

levels and also of the crack size The terms (2xr)f andf(8) map the stress field in

the vicinity of the crack and are the same for all cracks under external loads that cause crack openings of the same type

Equation (8.62) applies to all modes of crack opening, with K having different

values depending on the geometry of the structure, the nature of the applied loads and the type of crack However, if K has the same value for different types of crack and applied stress levels the stress fields around each crack will be identical

Since the mode of cracking shown in Fig 8.19(a) is the most common the remaining

analysis applies to this type of crack

Experimental data show that crack growth and residual strength data are better correlated using K than any other parameter K may be expressed as a function of

the nominal applied stress S and the crack length in the form

8.7 Fatigue 269

s

S

Fig 8.20 Stress field in the vicinity of a crack

in which a is a non-dimensional coefficient usually expressed as the ratio of crack

length to any convenient local dimension in the plane of the component; for a

crack in an infinite plate under an applied uniform stress level S remote from the

crack, a = 1 O Alternatively, in cases where opposing loads P are applied at points

close to the plane of the crack

(8.64)

in which P i s the load/unit thickness Equations (8.63) and (8.64) may be rewritten as

where &, is a reference value of the stress intensity factor which depends upon the

loading For the simple case of a remotely loaded plate in tension

and Eqs (8.65) and (8.63) are identical so that for a given ratio of crack length to plate

width a is the same in both formulations In more complex cases, for example the in-

plane bending of a plate of width 2b and having a central crack of length 2u

KO = 3 ( T U ) ? (8.67)

in which M is the bending moment per unit thickness Comparing Eqs (8.67) and

(8.63), we see that S = 3Ma/4b3 which is the value of direct stress given by 5asic

bending theory at a point a distance f a / 2 from the central axis However, if S was

specified as the bending stress in the outer fibres of the plate, i.e at &b, then

S = 3M/2b2; clearly the different specifications of S require different values of a

On the other hand the final value of K must be independent of the form of presentation

used Use of Eqs (8.63), (8.64) and (8.65) depends on the form of the solution for KO and

270 Airworthiness and airframe loads

care must be taken to ensure that the formula used and the way in which the nominal stress is defined are compatible with those used in the derivation of a

There are a number of methods available for determining the value of K and a In one method the solution for a component subjected to more than one type of loading

is obtained from available standard solutions using superposition or, if the geometry

is not covered, two or more standard solutions may be compounded7 Alternatively, a finite element analysis may be used

In certain circumstances it may be necessary to account for the effect of plastic flow

in the vicinity of the crack tip This may be allowed for by estimating the size of the plastic zone and adding this to the actual crack length to form an effective crack length 2al Thus, if rp is the radius of the plastic zone, al = a + rp and Eq (8.63)

becomes

(8.68)

in which K p is the stress intensity factor corrected for plasticity and a1 corresponds to

al Thus for rp/t > 0.5, i.e a condition of plane stress

Under constant amplitude loading the rate of crack propagation may be repre- sented graphically by curves described in general terms by the law

References 271 The curves represented by Eq (8.71) may be divided into three regions The first

corresponds to a very slow crack growth rate (< m/cycle) where the curves

approach a threshold value of stress intensity factor AK,, corresponding to

4 x lO-”m/cycle, i.e no crack growth In the second region (10-8-10-6m/cycle)

much of the crack life takes place and, for small ranges of A K , Eq (8.71) may be

represented by

du

- = C ( A K ) ”

in which C and n depend on the material properties; over small ranges of da/dN and

A K , C and n remain approximately constant The third region corresponds to crack

growth rates >

An attempt has been made to describe the complete set of curves by the relationship

m/cycle, where instability and final failure occur

(Ref 13)

da - C ( A K ) ”

in which K, is the fracture toughness of the material obtained from toughness tests

Integration of Eqs (8.74) or (8.75) analytically or graphically gives an estimate of

the crack growth life of the structure, that is, the number of cycles required for a

crack to grow from an initial size to an unacceptable length, or the crack growth

rate or failure, whichever is the design criterion Thus, for example, integration of

Eq (8.74) gives, for an infinite width plate for which Q = 1.0

for n > 2 An analytical integration may only be carried out if n is an integer and Q is

in the form of a polynomial, otherwise graphical or numerical techniques must be

Zbrozek, J K., Atmospheric gusts ~ present state of the art and further research, J R ~ J

Aero Soc., Jan 1965

Cox, R A,, A comparative study of aircraft gust analysis procedures, J Roy Aero Soc

Oct 1970

Bisplinghoff, R L., Ashley, H and Halfman, R L., Aeroelasticity, Addison-Wesley

Publishing Co Inc., Cambridge, Mass., 1955

Babister, A W., Aircraft Stability and Control, Pergamon Press, London 1961

Zbrozek, J K., Gust Alleviation Factor, R and M No 2970, May 1953

Handbook of Aeronautics No 1 Structural Principles and Datu, 4th edition, The Royal

Aeronautical Society, 1952

ESDU Data Sheets, Fatigue, No 80036

Knott, J F., Fundamentals of Fracture Mechanics, Butterworths, London, 1973

McClintock, F A and Irwin, G R., Plasticity aspects of fracture mechanics In: Fructure

Toughness Testing and its Applications, American Society for Testing Materials, Phila-

delphia, USA, ASTM STP 381, April, 1965

Paris, P C and Erdogan, F., A critical analysis of crack propagation laws, Trans An? Soc

Mech Engrs, 85, Series D, No 4 Dec 1963

272 Airworthiness and airframe loads

11 Rice, J R., Mechanics of crack tip deformation and extension by fatigue In: Fatigue Crack Propagation, American Society for Testing Materials, Philadelphia, USA, ASTM STP 415, June, 1967

12 Paris, P C., The fracture mechanics approach to fatigue In: Fatigue - An Znterdisciplinaty Approach, Syracuse University Press, New York, USA, 1964

13 Forman, R G., Numerical analysis of crack propagation in cyclic-loaded structures,

Trans Am Soc Mech Engrs, 89, Series D, No 3, Sept 1967

Freudenthal, A M., Fatigue in Aircraft Structures, Academic Press, New York, 1956

P.8.1 The aircraft shown in Fig P.8.l(a) weighs 135 kN and has landed such that

at the instant of impact the ground reaction on each main undercarriage wheel is

200 k N and its vertical velocity is 3.5 mjs

Fig P.8.1

If each undercarriage wheel weighs 2.25 kN and is attached to an oleo strut, as shown in Fig P.8.l(b), calculate the axial load and bending moment in the strut; the strut may be assumed to be vertical Determine also the shortening of the strut when the vertical velocity of the aircraft is zero

Finally, calculate the shear force and bending moment in the wing at the section

AA if the wing, outboard of this section, weighs 6.6 kN and has its centre of gravity 3.05 m from AA

193.3 kN, 29.0 k N m (clockwise); 0.32m; 19.5 kN, 59.6 kN m (anticlockwise) Determine, for the aircraft of Example 8.2, the vertical velocity of the nose

Ans

P.8.2

wheel when it hits the ground

Ans 3.1 mjs

P.8.3 Figure P.8.3 shows the flight envelope at sea-level for an aircraft of wing

span 27.5 m, average wing chord 3.05 m and total weight 196 000 N The aerodynamic centre is 0.91 5 m forward of the centre of gravity and the centre of lift for the tail unit

Problems 273

is 16.7m aft'of the CG The pitching moment coefficient is

CM,o = -0.0638 (nose-up positive) both CM.o and the position of the aerodynamic centre are specified for the complete

aircraft less tail unit

"t

V m/s

Fig P.8.3

For steady cruising fight at sea-level the fuselage bending moment at the CG is

600 000 Nm Calculate the maximum value of this bending moment for the given

flight envelope For this purpose it may be assumed that the aerodynamic loadings

on the fuselage itself can be neglected, i.e the only loads on the fuselage structure

aft of the CG are those due to the tail lift and the inertia of the fuselage

Ans i 549500Nm at n = 3.5, V = 152.5m/s

P.8.4 An aircraft weighing 238000N has wings 88.5m2 in area for which

C , = 0.0075 + 0.045C; The extra-to-wing drag coefficient based on wing area is

0.0128 and the pitching moment coefficient for all parts excluding the tailplane

about an axis through the CG is given by C M - c = (0.427C~ - 0.061) m The

radius from the CG to the line of action of the tail lift may be taken as constant at

12.2 m The moment of inertia of the aircraft for pitching is 204 000 kg m2

During a pull-out from a dive with zero thrust at 215 m/s EAS when the flight path

is at 40" to the horizontal with a radius of curvature of 1525 m, the angular velocity of

pitch is checked by applying a retardation of 0.25 rad/sec2 Calculate the manoeuvre

load factor both at the C G and at the tailplane CP, the forward inertia coefficient and

the tail lift

Ans IZ = 3.78(CG), IZ = 5.19 at TP, f = -0.370, P = 18 925N

P.8.5 An aircraft flies at sea level in a correctly banked turn of radius 610 m at a

speed of 168 m/s Figure P.8.5 shows the relative positions of the centre of gravity,

aerodynamic centre of the complete aircraft less tailplane and the tailplane centre

of pressure for the aircraft at zero lift incidence

274 Airworthiness and airframe loads

Calculate the tail load necessary for equilibrium in the turn The necessary data are given in the usual notation as follows:

Weight W = 133 500N dCL/da! = 4.5/rad

Wing area S = 46.5m2 CD = 0.01 + 0.05Ci Wing mean chord C = 3.0m CM,o = -0.03

Ans 73 160N

P.8.6 The aircraft for which the stalling speed V, in level flight is 46.5m/s has a

maximum allowable manoeuvre load factor nl of 4.0 In assessing gyroscopic effects

on the engine mounting the following two cases are to be considered:

(a) pull-out at maximum permissible rate from a dive in symmetric flight, the angle (b) steady, correctly banked turn at the maximum permissible rate in horizontal

Find the corresponding maximum angular velocities in yaw and pitch

Ans (a) Pitch, 0.37 rad/sec; (b) Pitch, 0.41 rad/sec, Yaw, 0.103 rad/sec

P.8.7 A tail-first supersonic airliner, whose essential geometry is shown in Fig P.8.7, flies at 610m/s true airspeed at an altitude of 18300m Assuming that thrust and drag forces act in the same straight line, calculate the tail lift in steady straight and level flight

of the flight path to the horizontal being limited to 60" for this aircraft; flight

Problems 275

If, at the same altitude, the aircraft encounters a sharp-edged vertical up-gust of

18m/s true airspeed, calculate the changes in the lift and tail load and also the

resultant load factor n

The relevant data in the usual notation are as follows:

Wing: S = 280m’, Tail: S T = 28m2,

aCL/aa = 1.5

a C ~ , T / a a = 2.0 Weight W = 1600000N

CM,O = - 0.01 Mean chord E = 22.8 m

At 18 300m

p = 0.116kg/m3

A m P = 267 852N, AP = 36257N, AL = 271 931 N, n = 1.19

P.8.8 An aircraft of all up weight 145 000 N has wings of area 50 m2 and mean

chord 2.5m For the whole aircraft CD = 0.021 + O.O41C;, for the wings

dCL/da = 4.8, for the tailplane of area 9.0m2, dCL,T/da = 2.2 allowing for the

effects of downwash: and the pitching moment coefficient about the aerodynamic

centre (of complete aircraft less tailplane) based on wing area is C,, = -0.032

Geometric data are given in Fig P.8.8

During a steady glide with zero thrust at 250m/s EAS in which CL = 0.08: the

aircraft meets a downgust of equivalent ‘sharp-edged’ speed 6 m/s Calculate the

tail load: the gust load factor and the forward inertia force, po = 1.223 kg/m3

P = -28 902 N (down), n = -0.64, forward inertia force = 40 703 N

9

Bending, shear and torsion

of open and closed,

thin-walled beams

In Chapter 7 we discussed the various types of structural component found in air- craft construction and the various loads they support We saw that an aircraft is basically an assembly of stiffened shell structures ranging from the single cell closed section fuselage to multicellular wings and tail-surfaces each subjected to bending, shear, torsional and axial loads Other, smaller portions of the structure consist of thin-walled channel, T-, Z-, 'top hat' or I-sections, which are used to stiffen the thin skins of the cellular components and provide support for internal loads from floors, engine mountings etc Structural members such as these are

known as open section beams, while the cellular components are termed closed section beams; clearly, both types of beam are subjected to axial, bending, shear and torsional loads

In this chapter we shall investigate the stresses and displacements in thin-walled open and single cell closed section beams produced by bending, shear and torsional loads, and, in addition, we shall examine the effect on the analysis of idealizing such sections when they have been stiffened by stringers

We shall show that the value of direct stress at a point in the cross-section of a beam subjected to bending depends on the position of the point, the applied loading and the geometric properties of the cross-section It follows that it is of no consequence whether or not the cross-section is open or closed We therefore derive the theory for a beam of arbitrary cross-section and then discuss its application to thin-walled open and closed section beams subjected to bending moments

The basic assumption of the theory is that plane sections of beams remain plane after displacement produced by the loading We shall, in addition, make the simplify- ing assumptions that the material of the beam is homogeneous and linearly elastic However, before we derive an expression for the direct stress distribution in a beam subjected to bending we shall establish sign conventions for moments, forces

9.1 Bending of open and closed section beams 277

and displacements, investigate the effect of choice of section on the positive directions

of these parameters and discuss the determination of the components of a bending

moment applied in any longitudinal plane

I - I Y * I- " - P I C I

-9.1 I Sign conventions and notation

Forces, moments and displacements are referred to an arbitrary system of axes O x y ~ ,

of which Oz is parallel to the longitudinal axis of the beam and O x y are axes in the

plane of the cross-section We assign the symbols M , S, P, T and w to bending

moment, shear force, axial or direct load, torque and distributed load intensity respec-

tively, with suffixes where appropriate to indicate sense or direction Thus, M , is a

bending moment about the x axis, S, is a shear force in the x direction and so on

Figure 9.1 shows positive directions and senses for the above loads and moments

applied externally to a beam and also the positive directions of the components of

displacement u, w and w of any point in the beam cross-section parallel to the x, y

and z axes respectively A further condition defining the signs of the bending moments

M , and M y is that they are positive when they induce tension in the positive x y quad-

rant of the beam cross-section

Fig 9.1 Notation and sign convention for forces, moments and displacements

If we refer internal forces and moments to that face of a section which is seen

when viewed in the direction z 0 then, as shown in Fig 9.2, positive internal forces

and moments are in the same direction and sense as the externally applied loads

whereas on the opposite face they form an opposing system The former system,

which we shall use, has the advantage that direct and shear loads are always positive

in the positive directions of the appropriate axes whether they are internal loads or

not It must be realized, though, that internal stress resultants then become

equivalent to externally applied forces and moments and are not in equilibrium

with them

278 Open and closed, thin-walled beams

A bending moment M applied in any longitudinal plane parallel to the z axis may be

resolved into components M , and M y by the normal rules of vectors However, a visual appreciation of the situation is often helpful Referring to Fig 9.3 we see that a bending moment M in a plane at an angle 8 to Ox may have components of differing sign depending on the size of 8 In both cases, for the sense of M shown

M , = Msin8

which give, for 8 < ~ 1 2 , M , and M y positive (Fig 9.3(a)) and for 8 > ~ 1 2 , M,

positive and My negative (Fig 9.3(b))

Fig 9.3 Resolution of bending moments

9.1 Bending of open and closed section beams 279

9.1.3 Direct stress distribution due to bending

-a

-p- I - - Y - I _ I - - Y I y Y I I _ y I - - ~ ~ , m * IICILI "~.CIII

Consider a beam having the arbitrary cross-section shown in Fig 9.4(a) The beam

supports bending moments M , and My and bends about some axis in its cross-section

which is therefore an axis of zero stress or a neutral axis (NA) Let us suppose that the

origin of axes coincides with the centroid C of the cross-section and that the neutral

axis is a distancep from C The direct stress a: on an element of area SA at a point

( x , )I) and a distance E from the neutral axis is, from the third of Eqs (1.42)

a= = E&? (9.1)

If the beam is bent to a radius of curvature p about the neutral axis at this particular

section then, since plane sections are assumed to remain plane after bending, and by a

comparison with symmetrical bending theory

The beam supports pure bending moments so that the resultant normal load on any

section must be zero Hence

280 Open and closed, thin-walled beams

i.e the first moment of area of the cross-section of the beam about the neutral axis is zero It follows that the neutral axis passes through the centroid of the cross-section

The moment resultants of the internal direct stress distribution have the same sense as

the applied moments Mx and M y Thus

Substituting for a= from Eq (9.4) in Eqs (9.5) and defining the second moments of

area of the section about the axes Cx, Cy as

Zxx = 1 y2dA, gives

E sin a E cos a E sina E cos a

so that, from Eq (9.4)

Alternatively, Eq (9.6) may be rearranged in the form

Mx(ZyyY - Ixyx) My (ZXXX - ZxyY) IxxZyy - y z: Zx.xZyy - c y

From Eq (9.7) it can be seen that if, say, M y = 0 the moment Mx produces a stress which varies with both x and y; similarly for M y if Mx = 0

9.1 Bending of open and closed section beams 281

In the case where the beam cross-section has either (or both) Cx or Cy as an axis of

symmetry the product second moment of area Ixy is zero and Cxy are principal axes

Equation (9.7) then reduces to

Further, if either M,, or M , is zero then

(9.9) Equations (9.8) and (9.9) are those derived for the bending of beams having at least a

singly symmetrical cross-section It may also be noted that in Eqs (9.9) a= = 0 when,

for the first equation, y = 0 and for the second equation when x = 0 Therefore, in

symmetrical bending theory the x axis becomes the neutral axis when M,, = 0 and

the y axis becomes the neutral axis when M x = 0 Thus we see that the position of

the neutral axis depends on the form of the applied loading as well as the geometrical

properties of the cross-section

There exists, in any unsymmetrical cross-section, a centroidal set of axes for which

the product second moment of area is zero These axes are then principal axes and the

direct stress distribution referred to these axes takes the simplified form of Eqs (9.8) or

(9.9) It would therefore appear that the amount of computation can be reduced if

these axes are used This is not the case, however, unless the principal axes are obvious

from inspection since the calculation of the position of the principal axes, the princi-

pal sectional properties and the coordinates of points at which the stresses are to be

determined consumes a greater amount of time than direct use of Eqs (9.6) or (9.7) for

an arbitrary, but convenient set of centroidal axes

9.1.4 Position of the neutral axis

The neutral axis always passes through the centroid of area of a beam’s cross-section but its inclination a (see Fig 9.4(b)) to the x axis depends on the form of the applied

loading and the geometrical properties of the beam’s cross-section

At all points on the neutral axis the direct stress is zero Therefore, from Eq (9.6)

where xh:A and J J ~ A are the coordinates of any point on the neutral axis Hence

JJNA - - MJxx - M.Jx.v

XNA M.rIy.v - M/.xy

or, referring to Fig 9.4(b) and noting that when a is

opposite sign

My?rx - MxIxy MJyy - MJxy

t a n a =

positive x ~ A and y N A are of

(9.10)

282 Open and closed, thin-walled beams

Example 9.1

A beam having the cross-section shown in Fig 9.5 is subjected to a bending moment

of 1500 N m in a vertical plane Calculate the maximum direct stress due to bending

stating the point at which it acts

Fig 9.5 Cross-section of beam in Example 9.1

The position of the centroid of the section may be found by taking moments of

areas about some convenient point Thus

( 1 2 0 ~ 8 + 8 0 ~ 8 ) J = 1 2 0 ~ 8 ~ 4 + 8 0 ~ 8 ~ 4 8 giving

J = 21.6mm and

(120 x 8 +80 x 8 ) X = 80 x 8 x 4 + 120 x 8 x 24 giving

X = 1 6 ~ The next step is to calculate the section properties referred to axes Cxy Hence

9.1 Bending of open and closed section beams 283

Since M , = 1500 N m and M y = 0 we have, from Eq (9.7)

In some cases the maximum value cannot be obtained by inspection so that values of

CJ- at several points must be calculated

9.1.5 Load intensity, shear force and bending moment

relationships, general case

Consider an element of length Sz of a beam of unsymmetrical cross-section subjected

to shear forces, bending moments and a distributed load of varying intensity, all in

the yz plane as shown in Fig 9.6 The forces and moments are positive in accordance

with the sign convention previously adopted Over the length of the element we

may assume that the intensity of the distributed load is constant Therefore, for

equilibrium of the element in the y direction

284 Open and closed, thin-walled beams

or, when second-order terms are neglected

We may combine these results into a single expression

We have noted that a beam bends about its neutral axis whose inclination relative to

arbitrary centroidal axes is determined from Eq (9.10) Suppose that at some section

of an unsymmetrical beam the deflection normal to the neutral axis (and therefore an absolute deflection) is C, as shown in Fig 9.7 In other words the centroid C is displaced from its initial position CI through an amount C to its final position CF

Suppose also that the centre of curvature R of the beam at this particular section is

on the opposite side of the neutral axis to the direction of the displacement C and that the radius of curvature is p For this position of the centre of curvature and from the usual approximate expression for curvature we have

9.1 Bending of open and closed section beams 285

Differentiating Eqs (9.14) twice with respect to z and then substituting for C from

Eq (9.13) we obtain

In the derivation of Eq (9.6) we see that

(9.15)

(9.16)

Substituting in Eqs (9.16) for sinalp and cosalp from Eqs (9.15) and writing

ut' = d2u/d3, v" = d2v/d3 we have

(9.17)

It is instructive to rearrange Eqs (9.17) as follows

{ zi} = - E [ 2 t] { :I} (see derivation of Eq (9.6)) (9.18)

i.e

(9.19)

The first of Eqs (9.19) shows that M , produces curvatures, that is deflections, in

both the xz and yz planes even though M - 0; similarly for M y when M x = 0

Thus, for example, an unsymmetrical beam will deflect both vertically and horizon-

tally even though the loading is entirely in a vertical plane Similarly, vertical and

horizontal components of deflection in an unsymmetrical beam are produced by

are L, I,T and LY

Determine the horizontal and vertical components of the tip deflection of the canti-

lever shown in Fig 9.8 The second moments of area of its unsymmetrical section

From Eqs (9.17)