Aircraft Structures 3E Episode 6 pps

It is assumed that the flanges resist the internal bending moment at any section of the beam while the web, of thickness t , resists the vertical shear force.. FT and FB are produced by

Fig 6.22 Column seclion of Example 6.1

= o (6.91)

6.1 2 Flexural-torsional buckling of thin-walled columns 187

If the column has, say, Cx as an axis of symmetry, then the shear centre lies on this

axis and y s = 0 Equation (6.91) thereby reduces to

(6.92)

The roots of the quadratic equation formed by expanding Eqs (6.92) are the values of

axial load which will produce flexural-torsional buckling about the longitudinal and

x axes If PCR(,,,,) is less than the smallest of these roots the column will buckle in pure

bending about the y axis

Example 6.2

A column of length l m has the cross-section shown in Fig 6.23 If the ends of the

column are pinned and free to warp, calculate its buckling load; E = 70 OOON/mm2,

G = 30 000 N/mm2

Fig 6.23 Column section of Example 6.2

In this case the shear centre S is positioned on the C x axis so that y s = 0 and

Eq (6.92) applies The distance X of the centroid of area C from the web of the section

is found by taking first moments of area about the web Thus

2( 100 + 100 + 1OO)X = 2 x 2 x 100 x 50 which gives

i = 33.3mm The position of the shear centre S is found using the method of Example 9.5; this gives

x s = -76.2mm The remaining section properties are found by the methods specified

in Example 6.1 and are listed below

From Eqs (6.90)

P ~ ~= 4.63 ( ~x io5 ~ N, P ~ ~ ( ~ ~ ~ ) ) = 8.08 x io5 N, P ~ ~ ( ~ ) = 1.97 x io5 N

Expanding Eq (6.92)

( P - P C R ( ~ ~ ) ) ( P - P C R ( 8 ) ) z O / A - p2xg = 0 (i) Rearranging Eq (i)

P2(1 - A x t / z O ) - P ( p C R ( ~ ~ ) + P C R ( B ) ) + PCR(s.~)pCR(8) = (ii) Substituting the values of the constant terms in Eq (ii) we obtain

P 2 - 29.13 x 105P + 46.14 x 10" = 0 (iii) The roots of Eq (iii) give two values of critical load, the lowest of which is

P = 1.68 x 10'N

It can be seen that this value of flexural-torsional buckling load is lower than any of the uncoupled buckling loads PCR(xx), PCR(yy) or PcR(e) The reduction is due to the interaction of the bending and torsional buckling modes and illustrates the cautionary remarks made in the introduction to Section 6.10

The spans of aircraft wings usually comprise an upper and a lower flange connected

by thin stiffened webs These webs are often of such a thickness that they buckle under shear stresses at a fraction of their ultimate load The form of the buckle is shown in Fig 6.24(a), where the web of the beam buckles under the action of internal diagonal compressive stresses produced by shear, leaving a wrinkled web capable of supporting diagonal tension only in a direction perpendicular to that of the buckle; the beam is

then said to be a complete tensionJield beam

6.1 3 Tension field beams 189

Y l l l

6.1 3.1 Complete diagonal tension

is - _. * _-

The theory presented here is due to H Wagner'"4

The beam shown in Fig 6.24(a) has concentrated flange areas having a depth d

between their centroids and vertical stiffeners which are spaced uniformly along the

length of the beam It is assumed that the flanges resist the internal bending

moment at any section of the beam while the web, of thickness t , resists the vertical

shear force The effect of this assumption is to produce a uniform shear stress

distribution through the depth of the web (see Section 9.7) at any section Therefore,

at a section of the beam where the shear force is S , the shear stress r is given by

S

td

Consider now an element ABCD of the web in a panel of the beam, as shown in

Fig 6.24(a) The element is subjected to tensile stresses, at, produced by the diagonal

tension on the planes AB and CD; the angle of the diagonal tension is a On a vertical

plane FD in the element the shear stress is r and the direct stress a, Now considering

the equilibrium of the element FCD (Fig 6.24(b)) and resolving forces vertically, we

have (see Section 1.6)

a,CDt sin a = TFDt which gives

27 sin 2a

Further, resolving forces horizontally for the element

azFDt = atCDt cos a

through the depth of the beam

The direct loads in the flanges are found by considering a length z of the beam as

shown in Fig 6.25 On the plane m m there are direct and shear stresses az and r acting

Fig 6.25 Determination of flange forces

in the web, together with direct loads FT and FB in the top and bottom flanges respectively FT and FB are produced by a combination of the bending moment Wz

at the section plus the compressive action (a,) of the diagonal tension Taking moments about the bottom flange

The diagonal tension stress a, induces a direct stress a,, on horizontal planes at any

point in the web Thus, on a horizontal plane HC in the element ABCD of Fig 6.24 there is a direct stress a,, and a complementary shear stress 7, as shown in Fig 6.26

B

Fig 6.26 Stress system on a horizontal plane in the beam web

6.13 Tension field beams 191

From a consideration of the vertical equilibrium of the element HDC we have

ayHCt = a,CDt sin a

The tensile stresses a,, on horizontal planes in the web of the beam cause compression

in the vertical stiffeners Each stiffener may be assumed to support half of each

adjacent panel in the beam so that the compressive load P in a stiffener is given by

P = a,tb which becomes, from Eq (6.101)

Wb

P = ana

If the load P is sufficiently high the stiffeners will buckle Tests indicate that they

buckle as columns of equivalent length

I, = d

forb < 1.5d

for b > 1.5d (6.103)

In addition to causing compression in the stiffeners the direct stress a,, produces

bending of the beam flanges between the stiffeners as shown in Fig 6.27 Each

flange acts as a continuous beam carrying a uniformly distributed load of intensity

aut The maximum bending moment in a continuous beam with ends fixed against

rotation occurs at a support and is wL2/12 in which w is the load intensity and L

the beam span In this case, therefore, the maximum bending moment M,,, occurs

Fig 6.27 Bending of flanges due to web stress

at a stiffener and is given by

Midway between the stiffeners this bending moment reduces to Wb2 tan a/24d

The angle a adjusts itself such that the total strain energy of the beam is a minimum

If it is assumed that the flanges and stiffeners are rigid then the strain energy comprises

the shear strain energy of the web only and a = 45" In practice, both flanges and stiffeners deform so that a is somewhat less than 45", usually of the order of 40"

and, in the type of beam common to aircraft structures, rarely below 38" For beams having all components made of the same material the condition of minimum strain energy leads to various equivalent expressions for Q, one of which is

which may be solved for a An alternative expression for a, again derived from a consideration of the total strain energy of the beam, is

an axis in the plane of the web is 2000 mm4; E = 70 000 N/mm2

6.1 3 Tension field beams 193

The maximum flange stress will occur in the top flange at the built-in end where the

bending moment on the beam is greatest and the stresses due to bending and diagonal

tension are additive Thus, from Eq (6.98)

moment and the diagonal tension is 17.7 x 103/350 = 50.7N/mm2 In addition to

this uniform compressive stress, local bending of the type shown in Fig 6.27

occurs The local bending moment in the top flange at the built-in end is found

using Eq (6.104), i.e

5 x lo3 x 3002 tan42.6"

12 x 400 = 8.6 x 104Nmm

M n a x =

The maximum compressive stress corresponding to this bending moment occurs at

the lower extremity of the flange and is 8.6 x 104/750 = 114.9N/mm2 Thus the

maximum stress in a flange occurs on the inside of the top flange at the built-in end

of the beam, is compressive and equal to 114.9 + 50.7 = 165.6N/mm2

The compressive load in a stiffener is obtained using Eq (6.102), i.e

5 x 300 tan 42.6"

400 = 3.4 kN

P =

Since, in this case, b < 1.5d, the equivalent length of a stiffener as a column is given by

the first of Eqs (6.103) Thus

1, = 400/d4 - 2 x 300/400 = 253 mm

From Eqs (6.7) the buckling load of a stiffener is then

= 22.0 kN

7? x 70000 x 2000

2532

P C R =

Clearly the stiffener will not buckle

In Eqs (6.107) and (6.108) it is implicitly assumed that a stiffener is f d y effective in resisting axial load This will be the case if the centroid of area of the stiffener lies in

the plane of the beam web Such a situation arises when the stiffener consists of two members symmetrically arranged on opposite sides of the web In the case where the web is stiffened by a single member attached to one side, the compressive load P is offset from the stiffener axis thereby producing bending in addition to axial load

For a stiffener having its centroid a distance e from the centre of the web the combined bending and axial compressive stress, a,, at a distance e from the stiffener centroid is

-

6.13.2 Incomplete diagonal tension

In modern aircraft structures, beams having extremely thin webs are rare They retain, after buckling, some of their ability to support loads so that even near failure they are in a state of stress somewhere between that of pure diagonal tension and the pre-buckling stress Such a beam is described as an incomplete diagonal tensionfield

beam and may be analysed by semi-empirical theory as follows

It is assumed that the nominal web shear T ( = S / t d ) may be divided into a 'true shear' component T~ and a diagonal tension component TDT by writing

TDT = k7, T~ = (1 - k)7 (6.110)

where k, the diagonal tension factor, is a measure of the degree to which the diagonal

tension is developed A completely unbuckled web has k = 0 whereas k = 1 for a web

in complete diagonal tension The value of k corresponding to a web having a critical

6.13 Tension field beams 195

stress TCR may be calculated from the formula

(6.1 12) where k,, is the coefficient for a plate with simply supported edges and & and Rb are

empirical restraint coefficients for the vertical and horizontal edges of the web panel

respectively Graphs giving k,,, Rd and Rb are reproduced in Kuhn14

The stress equations (6.106) and (6.107) are modified in the light of these assump-

tions and may be rewritten in terms of the applied shear stress r as

direction of a given by

2kr sin 2a

(TI =- + r(l - k) sin2a (6.115) and CQ perpendicular to this direction given by

a, = -r(1 - k) sin2a (6.116)

The secondary bending moment of Eq (6.104) is multiplied by the factor k, while the

effective lengths for the calculation of stiffener buckling loads become (see Eqs

(6.103))

or

where d, is the actual stiffener depth, as opposed to the effective depth d of the web,

taken between the web/flange connections as shown in Fig 6.29 We observe that

Eqs (6.1 13)-(6.116) are applicable to either incomplete or complete diagonal tension

Fig 6.30 Effect of taper on diagonal tension field beam calculations

field beams since, for the latter case, k = 1 giving the results of Eqs (6.106), (6.107)

Timoshenko, S P and Gere, J M., Theory of Elastic Stability, 2nd edition, McGraw-Hill

Book Company, New York, 1961

Gerard, G., Introduction to Structural Stability Theory, McGraw-Hill Book Company,

New YQrk, 1962

Murray, N W., Introduction to the Theory of Thin-walled Structures, Oxford Engineering

Science Series, Oxford, 1984

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

Aeronautical Society, 1952

Bleich, F., Buckling Strength of Metal Structures, McGraw-Hill Book Company, New

York, 1952

Gerard, G and Becker, H., Handbook of Structural Stability, Pt I, Buckling of Flat Plates,

NACA Tech Note 3781, 1957

Rivello, R M , Theory and Analysis of Flight Structures, McGraw-Hill Book Company,

New York, 1969

Stowell, E Z., Compressive Strength of Flanges, NACA Tech Note 1323, 1947

Mayers, J and Budiansky, B., Analysis of Behaviour of Simply Supported Flat Plates Com-

pressed Beyond the Buckling Load in the Plastic Range, NACA Tech Note 3368, 1955

Gerard, G and Becker, H., Handbook of Structural Stability, Pt I V , Failure of Plates and

Composite Elements, NACA Tech Note 3784, 1957

Gerard, G., Handbook of Structural Stability, Pt V , Compressive Strength of Flat Stiffened

Panels, NACA Tech Note 3785, 1957

Gerard, G and Becker, H., Handbook of Structural Stability, Pt V U , Strength of Thin

Wing Construction, NACA Tech Note D-162, 1959

Gerard, G., The crippling strength of compression elements, J Aeron Sci 25(1), 37-52

Jan 1958

Kuhn, P., Stresses in Aircraft and Shell Structures, McGraw-Hill Book Company, New

York, 1956

P.6.1 The system shown in Fig P.6.1 consists of two bars A B and BC, each of

bending stiffness EZ elastically hinged together at B by a spring of stiffness K (i.e

bending moment applied by spring = K x change in slope across B)

Regarding A and C as simple pin-joints, obtain an equation for the first buckling

load of the system What are the lowest buckling loads when (a) K + 00, (b)

EZ + 00 Note that B is free to move vertically

P.6.2 A pin-ended column of length 1 and constant flexural stiffness EZ is

Considering symmetric modes of buckling only, obtain the equation whose roots

Ans

reinforced to give a flexural stiffness 4EZ over its central half (see Fig P.6.2)

yield the flexural buckling loads and solve for the lowest buckling load

tanp1/8 = l / d , P = 24.2EZ/12

Fig P.6.2

P.6.3 A uniform column of length 1 and bending stiffness EZ is built-in at one end

and free at the other and has been designed so that its lowest flexural buckling load

is P (see Fig P.6.3)

Subsequently it has to carry an increased load, and for this it is provided with a

lateral spring at the free end Determine the necessary spring stiffness k so that the

buckling load becomes 4P

A m k = 4 P p / ( p l - tan p l )

Fig P.6.3

P.6.4 A uniform, pin-ended column of length I and bending stiffness EZ has an

initial curvature such that the lateral displacement at any point between the

column and the straight line joining its ends is given by

Problems 199

tv I vo

Fig P.6.4

P.6.5 The uniform pin-ended column shown in Fig P.6.5 is bent at the centre so

that its eccentricity there is 6 If the two halves of the column are otherwise straight

and have a flexural stiffness EI, find the value of the maximum bending moment

when the column carries a compression load P

Fig P.6.5

P.6.6 A straight uniform column of length I and bending stiffness EI is subjected

to uniform lateral loading w/unit length The end attachments do not restrict rotation

of the column ends The longitudinal compressive force P has eccentricity e from the

centroids of the end sections and is placed so as to oppose the bending effect of the

lateral loading, as shown in Fig P.6.6 The eccentricity e can be varied and is to be

adjusted to the value which, for given values of P and w , will result in the least

maximum bending moment on the column Show that

P.6.7 The relation between stress u and strain E in compression for a certain material is

10.5 x 1 0 6 ~ = a+ 21 000 -

(49:OO) l6

Assuming the tangent modulus equation to be valid for a uniform strut of this material, plot the graph of ab against l / r where fsb is the flexural buckling stress, 1

the equivalent pin-ended length and r the least radius of gyration of the cross-section

Estimate the flexural buckling load for a tubular strut of this material, of 1.5 units

outside diameter and 0.08 units wall thickness with effective length 20 units

Ans 14 454 force units

P.6.8 A rectangular portal frame ABCD is rigidly fixed to a foundation at A and

D and is subjected to a compression load P applied at each end of the horizontal

member BC (see Fig P.6.8) If the members all have the same bending stiffness EI

show that the buckling loads for modes which are symmetrical about the vertical centre line are given by the transcendental equation

P.6.9 A compression member (Fig P.6.9) is made of circular section tube,

diameter d, thickness t The member is not perfectly straight when unloaded, having a slightly bowed shape which may be represented by the expression

Show that when the load P is applied, the maximum stress in the member can be expressed as

Problems 201

Fig P.6.9

where

(Y = P / P ~ , P, = ~ E I / P

Assume r is small compared with d so that the following relationships are applicable:

Cross-sectional area of tube = rdt

Second moment of area of tube = r d 3 t / 8

P.6.10 Figure P.6.10 illustrates an idealized representation of part of an aircraft

control circuit A uniform, straight bar of length a and flexural stiffness EI is built-

in at the end A and hinged at B to a link BC, of length b, whose other end C is

pinned so that it is free to slide along the line ABC between smooth, rigid guides

A, B and C are initially in a straight line and the system carries a compression

force P, as shown

Fig P.6.10

Assuming that the link BC has a sufficiently high flexural stiffness to prevent its

buckling as a pin-ended strut, show, by setting up and solving the differential

equation for flexure of AB, that buckling of the system, of the type illustrated in

Fig P.6.10, occurs when P has such a value that

tan Xa = X(a + b)

where

X2 = P / E I

P.6.11 A pin-ended column of length I has its central portion reinforced, the

second moment of its area being I2 while that of the end portions, each of length a,

is I I Use the energy method to determine the critical load of the column, assuming

that its centre-line deflects into the parabola w = kz(1- z ) and taking the more

accurate of the two expressions for the bending moment

In the case where I2 = 1.611 and a = 0.21 find the percentage increase in strength

due to the reinforcement, and compare it with the percentage increase in weight on the basis that the radius of gyration of the section is not altered

Ans P C R = 14.96EII/l2, 52%, 36%

P.6.12 A tubular column of length I is tapered in wall-thickness so that the area and the second moment of area of its cross-section decrease uniformly from AI and Il

at its centre to 0.2A1 and 0.211 at its ends

Assuming a deflected centre-line of parabolic form, and taking the more correct form for the bending moment, use the energy method to estimate its critical load when tested between pin-centres, in terms of the above data and Young’s modulus

E Hence show that the saving in weight by using such a column instead of one having the same radius of gyration and constant thickness is about 15%

that all the kinematic (geometric) and static boundary conditions are satisfied, allowing for one arbitrary constant only

Using the result thus obtained, find an approximation to the lowest flexural buckling load P C R by the Rayleigh-Ritz method

Problems 203

Fig P.6.14

If the plate is subjected to a uniform compressive stress a in the x-direction (see

Fig P.6.14), find an expression for the elastic deflection w normal to the plate

Show also that the deflection at the mid-point of the plate can be presented in the

form of a Southwell plot and illustrate your answer with a suitable sketch

Ans w = [ ( ~ t s / ( 4 2 ~ / 2 - at)] sin-sin- T X Ty

P.6.15 A uniform flat plate of thickness t has a width b in the y direction and

length I in the x direction (see Fig P.6.15) The edges parallel to the x axis are

clamped and those parallel to the y axis are simply supported A uniform compressive

stress (T is applied in the x direction along the edges parallel to the y axis Using an

energy method, find an approximate expression for the magnitude of the stress a

which causes the plate to buckle, assuming that the deflected shape of the plate is

For the particular case I = 2b, find the number of half waves m corresponding to the

lowest critical stress, expressing the result to the nearest integer Determine also the lowest critical stress

A ~ s

P.6.16 A panel, comprising flat sheet and uniformly spaced Z-section stringers, a

part of whose cross-section is shown in Fig P.6.16, is to be investigated for strength under uniform compressive loads in a structure in which it is to be stabilized by frames

a distance 1 apart, 1 being appreciably greater than the spacing b

(a) State the modes of failure which you would consider and how you would determine appropriate limiting stresses

(b) Describe a suitable test to verify your calculations, giving particulars of the specimen, the manner of support, and the measurements you would take The latter should enable you to verify the assumptions made, as well as to obtain the load supported

m = 3, CTCR = [6E/( 1 - G)] ( t / b ) 2

Fig P.6.16

P.6.17 Part of a compression panel of internal construction is shown in Fig P.6.17 The equivalent pin-centre length of the panel is 500mm The material has a Young’s modulus of 70 000 N/mm2 and its elasticity may be taken as falling catastrophically when a compressive stress of 300 N/mm2 is reached Taking coefficients of 3.62 for buckling of a plate with simply supported sides and of 0.385 with one side simply supported and one free, determine (a) the load per mm width

of panel when initial buckling may be expected and (b) the load per mm for ultimate failure Treat the material as thin for calculating section constants and assume that after initial buckling the stress in the plate increases parabolically from its critical value in the centre of sections

Problems 205

‘li

T

Fig P.6.18

P.6.18 Figure P.6.18 shows the doubly symmetrical cross-section of a thin-walled

column with rigidly fixed ends Find an expression, in terms of the section dimensions

and Poisson’s ratio, for the column length for which the purely flexural and the purely

torsional modes of instability would occur at the same axial load

In which mode would failure occur if the length were less than the value found? The

possibility of local instability is to be ignored

A m I = ( 2 7 r b 2 / t ) d m Torsion

P.6.19 A column of length 21 with the doubly symmetric cross-section shown in

Fig P.6.19 is compressed between the parallel platens of a testing machine which

fully prevents twisting and warping of the ends

Using the data given below, determine the average compressive stress at which the

column first buckles in torsion

I = 500 mm; b = 25.0 mm, t = 2.5 mm, E = 70 000 N/mm2, EIG = 2.6

AYE OCR = 282 N / m 2

Fig P.6.19