Thin Walled Structures

Thin-Walled Structures

v

Contents

CHAPTER 1

Airplane and ship structures

1

Structures and Engineering

1

Principal structural units

2

Design

4

Loads

6

Function of flight vehicle structural members

6

Ships structures

9

Key words and concepts from Chapter 1

16

References

16

CHAPTER 2

Bars Subjected to Axial Loads

17

Axially loaded bar

17

The tensile test

19

Effect of temperature on strain

23

Bar reference axis

24

Linear elastic response

26

EXAMPLE 2.1 Axial bar with a specified uniform distributed load and specified end displacements 27EXAMPLE 2.2 A bar with fixed ends and subjected to an axial point force. 28

Work and energy methods

30

Concept of virtual displacement

30

Principle virtual work

32

EXAMPLE 2.3 Approximating the response of a bar using PVW. 34

Contents

vi


Strain energy density

36

EXAMPLE 2.4 Strain energy of a bar with fixed ends and subjected to an axial point force. 38EXAMPLE 2.5 An elastic bar subjected to two forces and a thermal load 40

Castiglianos first theorem

41

EXAMPLE 2.6 Response of a stepped bar by Castiglianos first theorem 43

Complementary virtual work

45

Complementary strain energy

47

Relationship between the complementary strain energy and the strain energy densities

49

EXAMPLE 2.7 Application of complementary virtual work to an elastic bar 50

Generalized form of Castiglianos second theorem

52

EXAMPLE 2.8 Stepped bar response by Castiglianos second theorem 54EXAMPLE 2.9 A suspended bar subjected to self weight 55

Trusses

56

EXAMPLE 2.10 Three bar planar truss 60EXAMPLE 2.11 Three bar truss with lack of fit 61

References

62

Problems

63

CHAPTER 3

Axial Normal Stress in Pure Bending and Extension

67

Pure bending

67

Geometry of deformation

68

Bending normal stress flexure formula

75

EXAMPLE 3.1 Bending normal stress distribution in a cantilever beam with a thin-walled zee section. 78EXAMPLE 3.2 Lateral displacements of the zee section beam 80

Moments of areas

83

EXAMPLE 3.3 Thin-walled zee section properties by the composite body technique 86EXAMPLE 3.4 Semicircular section with two stringers 88

Extension, pure bending, and thermal effects for multi-material beams

89

EXAMPLE 3.5 A multi-material beam with a symmetric cross section 92

Problems

96

CHAPTER 4

Axial Force, Shear Force and Bending Moment Diagrams

99

Method of sections

99

Differential equation method

101

EXAMPLE 4.1 Cantilever wing with tip tank 105EXAMPLE 4.2 The air load acting on a wing given as discrete data. 109

Semi-graphical method

116

EXAMPLE 4.3 Uniform barge with symmetric load 116

Buoyancy Force Distribution on Ships

118


vii

Contents

References

121

Problems

121

CHAPTER 5

Bending of Beams under Transverse Loads

125

Approximations for slender beams

125

Beam displacements

126

EXAMPLE 5.1 A statically indeterminate beam with an unsymmetrical cross section. 131EXAMPLE 5.2 Contact between two cantilever beams. 135

Complementary virtual work and complementary energy

139

Complementary virtual work

140


142

Beam displacement by Castiglianos second theorem

145

EXAMPLE 5.3 End rotations of a simply supported beam subject to an end moment 145EXAMPLE 5.4 Tip displacement of a cantilever wing spar a under distributed load 147EXAMPLE 5.5 Strut-braced spar 149

Problems

153

CHAPTER 6

Shear Flow due to Shear Forces

155

Shear flows and shear stresses due to bending in a rectangular section beam

155

Shear flows due to transverse shear forces in open section beams

158

EXAMPLE 6.1 Shear flow distribution in a tee beam 160

Shear center of a thin-walled open section

163

EXAMPLE 6.2 Shear center location in an unsymmetrical section 167

Skin-stringer idealization

170

EXAMPLE 6.3 Shear flows in a stringer-stiffened C-section 173

Influence of transverse shear deformations on bending

177

Transverse shear strains, forces, and complementary energy density

178

Complementary energy density obtained from a two-dimensional element of the wall

180

Determination of the transverse shear compliances

183

EXAMPLE 6.4 Shear compliances of a stiffened blade section 184EXAMPLE 6.5 Deflection of a cantilevered beam due to bending and shear deformation. 186

Problems

188

CHAPTER 7

Bars Subjected to Torsional Loads

191

Uniform torsion of a circular tube

191

Uniform torsion of an open section

197

EXAMPLE 7.1 Torsional response of a thin-walled open section and an equivalent closed section 201

Non-uniform torsion; governing boundary value problem

202

Contents

viii


EXAMPLE 7.2 A uniform distributed torque acting on a bar with fixed ends. 203EXAMPLE 7.3 A point torque acting on a bar with fixed ends. 204

Virtual work and strain energy

205

Strain energy density

207

Complementary virtual work and energy

208


209

Unit twist of a single cell beam due to shear flow

210

Hookes law

211

Shear strain-displacement relation

212

Tangential displacement of a typical point on the contour

212

Relation between the shear flow and unit twist

214

Uniform torsion of a thin-walled closed section with a contour of arbitrary shape

215

EXAMPLE 7.4 Torsion of a circular, bi-material section 218

Shear center of a closed section

219

EXAMPLE 7.5 Shear center location of a single-cell, closed section having one axis of symmetry. 219

Uniform torsion of multi-cell closed sections

222

EXAMPLE 7.6 Uniform torsion of a two-cell section 225

Resultant of a constant shear flow in a curved branch

226

EXAMPLE 7.7 Torsion of a five cell closed section; circuit shear flow 229

Torsion of hybrid sections

231

References

232

Problems

233

CHAPTER 8

Criteria for Initial Yielding

237

Ductile and brittle behavior

237

Criteria for initial yielding of ductile materials

238

Stress transformation equations for generalized plane stress

241

Principal stresses and maximum shear stress

243

EXAMPLE 8.1 Maximum shear stress in tensile test 246EXAMPLE 8.2 Mohrs circle for hydrostatic stress state 246EXAMPLE 8.3 Principal stresses and maximum shear stress 247

Octahedral shear stress

249

Mises criterion for initiation of yielding

252

Maximum shear-stress criterion

254EXAMPLE 8.4 Factor of safety against initial yielding 256EXAMPLE 8.5 Stress responses of a stringer-stiffened, single cell beam. 257EXAMPLE 8.6 Minimum weight design of the beam in Example 8.5 subject to a constraint on initial yielding 261

References 264

Thin-Walled Structures ix

Contents

CHAPTER 9 Buckling 265

One-degree of freedom model 265Static equilibrium 266Stability analysis 267

Perfect Columns 269EXAMPLE 9.1 Critical load for clamped-free boundary conditions (B) 272

Imperfect columns 275Eccentric load 275Geometric imperfection 277

Column Design Curve 279Inelastic buckling 280

Bending of thin plates 283

Contents

x Thin-Walled Structures


1

CHAPTER 1


The objective of Thin-Walled Structures is to provide an understanding of the basic concepts of stress and defor-mation of stiffened-shell structures with applications to aerospace and ocean vehicles. In this chapter we discuss why a structure carries load, the types of loads acting on airplane and ship structures, the definitons and functions of principal structural units, and the relation of structural design to the overall design.

1.1 Structures and Engineering

A structure may be defined as any assemblage of materials which is intended to sustain loads. Structures function to protect people and things, and are so common and familiar to us that when we are informed of their use and form we conceive of ourselves as knowledgeable. After all, it does not take a structural engineering degree to build an ordinary shed. However, when asked to build an airplane or a ship we would probably be more hesitant, since if an airplane or a ship breaks many people are likely to be killed. The crux of the issue in the design of complex structures, like aircraft, is to not only know the use and form of the structure, but also to know

why

a structure is able to carry load.

Although Galileo (1564-1642) began an inquiry into the strength of materials, it is Robert Hooke (1635-1702) who is credited with providing the answer as to why structures carry load. A historical account of Hookes discoveries is discussed in chapter two of an informative and readable book on structures by Gordon (1978). Gor-dons text is the source for what is written here. By Newtons law of action and reaction we know that isolated forces do not exist in nature. A force acting on an inanimate solid is reacted by a force produced by the solid. But how does the solid produce such a reaction force? Hooke in 1676 recognized that every kind of solid changes its shape when subjected to a mechanical force and it is the change in shape which enables the solid to provide the reaction force. This shape change extends to the very fine scale of molecules where there is a large resistance to stretching and compressing of chemical bonds. Hookes measurements also showed that many solid materials recovered their original shape when they are unloaded; i.e, they are elastic. The science of elasticity is about the interactions between forces and deflections in materials and structures. The formulation of the mathematical the-ory of elasticity for a solid continuum is well established; e.g., see Sokolnikoff (1956). Exact elasticity solutions are known for a small, but important, number of structural problems. However, the mathematical solutions to elasticity problems are usually challenging. Approximations to elasticity theory which exploit the geometry and


2


material construction of practical structures is called structural mechanics. For example, the structural mechanics of beams, plates, and shells approximate elasticity by exploiting the fact that one or two dimensions of the solid body are very small with respect to the remaining dimensions. Structural mechanics reduces the mathematical complexity somewhat relative to three-dimensional elasticity, and it is the theory most often used by the practic-ing engineer.

Two important properties common to the analysis of any structure are

stiffness

and

strength

. Stiffness is a measure of the force required to produce a given deflection, and strength refers to the force, or force intensity, necessary to cause failure. A criterion for failure is required in order to determine the strength of a structure, and this depends upon the particular application. For example, failure can be defined when a stress (internal force intensity) exceeds the yield stress of the material, or failure can mean excessive displacements which occur dur-ing buckling. The stiffness and strength of a structure depend on its geometrical configuration, connections, and the stiffness and strength of the materials from which it is made.

It is important to recognize that structures are made from materials, and that the history of structures follows the development of materials and the development of tools to fabricate the materials. The evolution of the air-frame, for example, is tied closely to the introduction of materials and cost-effective means for their fabrication. For example, early aircraft were constructed of wire-braced wood frames with fabric covers. Currently, advanced composite materials are very attractive for weight-sensitive structures, like aircraft, because of their high stiff-ness-to-weight and strength-to-weight ratios.

The distinction between structures and materials is not always clear. It may be said that the forward-swept wing on Grumman's X-29 demonstrator airplane is a structure, and the material it is made from is an advanced composite. However, advanced fiber-reinforced composites are made from stiff, strong, continuous fibers embed-ded in a pliant matrix. The complex constitution of an advanced composite, therefore, may be considered either as materials or structures.

1.2 Principal structural units

The principal structural units of fixed-wing airplanes are the fuselage, wings, stabilizers, control surfaces, land-ing gear, nacelle and engine mounts. Light airplanes are shown in Fig. 1.1 and a heavier airplane (Douglas DC-3) is shown in Fig. 1.2. A cargo ship is depicted in Fig. 1.3, and its principal structural units will be discussed in more detail in Section 1.6. After some study of the structures shown in these figures, it is reasonable to suggest that some principal structural units such as the fuselage, wings, and ship hull have the features commonly attrib-utable to a beam. That is, two dimensions of the overall component, or the cross-sectional dimensions, are small with respect to the third, or longitudinal, dimension. Indeed, to simplify the analysis of such complex structures, we can approximate them as slender, built-up beams!

Hence, the basic conceptualizing we make for complex vehicle structures such as aircraft and ships is that a fuselage, a wing, or a hull is a thin-walled beam. That is, a vehicle structure as a whole is assumed to be a one-dimensional structural element in the mathematical sense that its response under load can be described by ordi-nary differential equations. Aero-hydrodynamic and other loads that act on the structure cause extensional, bend-ing, and torsional deformations of the structure. The cross section of the structure is built from many actual structural elements such as spars, frames, and panels. This beam assumption is particularly suited for the analysis required in preliminary design. Of course not all principal structural units can be modeled as a beam. In constrast to a high aspect ratio wing, a delta wing whose span and chord are of comparable value (low aspect ratio) is not modeled very well by using the beam assumption.


3

Principal structural units

Fig. 1.1 Principal structural units of light airplanes (from Aircraft Basic Science, 1948)

(b) Taylorcraft airplane

(a) Global Swift


4


1.3 Design

An aircraft or ship may be considered a system consisting of several subsystems; a structural subsystem, a con-trol subsystem, a propulsion subsystem, cargo handling subsystem, etc. Vehicle design must consider the total, or integrated, system to achieve optimal performance.

An important contribution to the overall vehicle performance is to minimize weight in the structural subsystem design.

A minimum weight vehicle structure can carry the same payload with less fuel consumption. In addition, a lighter structure can reduce operating costs through less main-tenance, and also may reduce initial cost by requiring less labor for fabrication. Modern engineering design has been revolutionized by the development of high-speed computers combined with optimization theory. As a math-ematical problem in optimization, structural design may be considered as the development of a computational algorithm for choosing member spacing and dimensions such that weight is minimized (objective) subject to constraints on strength and stiffness. The role of structural mechanics in this design process is to provide a description of the state of stress and deformation throughout the structure for a given structural configuration, such that the constraints can be evaluated. Structural mechanics provides the theory for the analysis of the struc-tural response (state of stress and deformation).

Fig. 1.2 Principal structural units of a Douglas airplane (from Aircraft Basic Science, 1948)


5

Design

The overall dimensions of a vehicle structure are usually determined by more general requirements rather than for structural considerations. Thus, the structural design begins with the overall dimensions given, and two levels of design may be distinguished. The first level is called

preliminary design

, and at this level the locations and dimensions of the principal structural members are determined. The second level is called

detail design

, and

Fig. 1.3 Levels of structural analysis for a large ship (from Hughes, 1983)


6


at this level the geometry and dimensions of the local structure like joints, cutouts, reinforcements, etc. are deter-mined.

1.4 Loads

The first step in preliminary design is to determine the external loads acting on the structure. Maneuvering flight vehicles are subjected to gravity, aerodynamic forces, and inertial loads. In addition, the landing loads, and wind gust loads during turbulent weather must be considered. Ships are subjected to gravity, buoyancy forces, and inertial loads. Also, wave loads and other hydrodynamic loads such as slamming, sloshing of liquid cargos, etc., must be considered in ship design. The calculation of aerodynamic and hydrodynamic forces are sufficiently complex such that their determination is done by specialists rather than by designers.

Loads on a vehicle structure may be classified as static or dynamic, and either deterministic or probabilistic. Gust loading conditions for aircraft and wave loading conditions for ships are not known with absolute certainty, so that these load magnitudes are estimated on a statistical basis using probability theory together with past expe-rience. The type of loading has a direct influence on the type of structural response analysis required. For exam-ple, dynamic loading requires a structural dynamics analysis.

In traditional structural design most of these loads are not affected by the structural configuration or dimen-sions of the members. They are a function of the wing shape, or hull shape, and other nonstructural factors. Hence, the determination of the loads, a very crucial aspect of the design process, is essentially a separate task typically performed by an aerodynamicist or hydrodynamicists. In modern flexible vehicles the loads greatly influence the shape and so aeroelastic or hydroelastic load analysis must be performed. That is, the structural and aero/hydrodynamic analysis must be combined to obtain the correct loads. This interaction between the structure and the shape is so important for flexible vehicles made from composite materials that it is expected that in the future the shape and structural design will be combined.

1.5 Function of flight vehicle structural members

The following description of the functions of flight vehicle structures is excerpted from Rivello (1969, Section 7-6).

The structure of a flight vehicle usually has a dual function: it transmits and resists the forces which are applied to the vehicle, and it acts a cover which provides the aerodynamics shape and protects the contents of the vehicle from the environment. This combination of roles is fortunate since, from the standpoint of structural weight, the most efficient location for the structural material is at the outer surface of the vehicle. As a result, the structures of most flight vehicles are essentially thin shells. If these shells are not supported by stiffening members, they are referred to as

monocoque

. When the cross-sectional dimensions are large, the wall of a monocoque structure must be relatively thick to resist bending, compressive, and torsional loads without buckling. In such cases a more efficient type of construction is one which con-tains stiffening members that permit a thinner covering shell. Stiffening members may also be required to diffuse concentrated loads into the cover. Constructions of this type are called

semimonocoque

. Typical examples of semimonocoque body structures are shown in Fig. 7-5. While at first glance these structures appear to differ considerably, functionally there are simi-larities. Both have thin-sheet coverings, longitudinal stiffening members, and transverse sup-


7

Function of flight vehicle structural members

porting elements which play similar structural roles.

In semimonocoque structures the cover, or skin, has the following functions:1. It transmits aerodynamic forces to the longitudinal and transverse supporting members by

plate and membrane action (Chap. 13).2. It develops shearing stresses which react the applied torsional moments (Chap. 8) and

shear forces (Chap. 9).3. It acts with the longitudinal members in resisting the applied bending and axial loads

(Chaps. 7, 15, and 16).4. It acts with the longitudinals in resisting the axial load and with the transverse members in

reacting the hoop, or circumferential, load when the structure is pressurized.

In addition to these structural functions, it provides an aerodynamic surface and cover for the content of the vehicle.

Spar webs

(Fig. 7-5b) play a role that is similar to function 2 of the skin.

The longitudinal members are known as

longitudinals

,

stringers

, or

stiffeners

. Longitudi-nals which have large cross-sectional areas are referred to as

longerons

. These members serve the following purposes:1. They resist bending and axial loads along with the skin (Chap. 7).2. They divide the skin into small panels and thereby increase its buckling and failing stresses

(Chaps. 15 and 16).3. They act with the skin in resisting axial loads caused by pressurization.

Longitudinalstringers

Transverseframes

Cover Skin

Cover skin

Spar web

Spar cap

Transverserib

Longitudinalstringers

(a)

(b)

Fig. 7-5 Typical semimonocoque construction. (a) Bodystructures: (b) aerodynamic surface structures.


8


The spar caps in aerodynamic surface perform functions 1 and 2.

The transverse members in body structures are called

frames

,

rings

, or if they cover all or most of the cross-sectional area,

bulkheads

. In aerodynamic surfaces they are referred to as

ribs

. These members are used to:1. Maintain the cross-sectional shape.2. Distribute concentrated loads into the structure and redistribute stresses around structural

discontinuities (Chap. 9).3. Establish the column length and provide end restraint for the longitudinals to increase their

column buckling stress (Chap. 14).4. Provide edge restraint for the skin panels and thereby increase the plate buckling stress of

these elements (Chap. 16).5. Act with the skin in resisting the circumferential loads due to pressurization.

The behavior of these structural elements is often idealized to simplify the analysis of the assembled component. The following assumptions are usually made:1. The longitudinals carry only axial stress.2. The webs (skin and spar webs) carry only shearing stresses.3. The axial stress is constant over the cross section of each of the longitudinals, and the

shearing stress is uniform through the thickness of the webs.4. The transverse frames and ribs are rigid within their own planes, so that the cross section is

maintained unchanged during loading. However, they are assumed to possess no rigid-ity normal to their plane, so that they offer no restraint to warping deformations out of their plane.

When the cross-sectional dimensions of the longitudinals are very small compare to the cross-sectional dimensions of the assembly, assumptions 1 and 3 result in little error. The webs in an actual structure carry significant axial stresses as well as shearing stresses, and it is therefore necessary to use an analytical model of the structure which includes this load-car-rying ability. This is done by combining the effective areas of the webs adjacent to a longitudi-nal with the area of the longitudinal into a

total effective area

of material which is capable of resisting bending moments and axial forces. A method for determining this effective area is given in Sec. 15-7. In the illustrative examples and problems on stiffened shells in this and suc-ceeding chapters it may be assumed that his idealization has already been made and that areas given for the longitudinals are the total effective areas. The fact that the cross-sectional dimensions of most longitudinals are small when compared with those of the stiffened-shell cross section makes it possible to assume without serious error that the area of the effective longitudinal is concentrated at a point on the midline of the skin where it joins the longitudinal. The locations of these idealized longitudinals will be indicated by small circles, as shown in Fig. 7-6b. In thin aerodynamic surfaces the depth of the longitudinals may not be small com-pared to the thickness of the cross section of the assembly, and more elaborate idealized model of the structure may be required.

The fewer the number of longitudinals, the simpler the analysis, and in some cases several longitudinal may be lumped into a single effective longitudinal to shorten computations. (Fig. 7-6). On the other hand, it is sometimes convenient to idealize a monocoque shell into an ideal-ized stiffened shell by lumping the shell wall into idealized longitudinals, as shown in Fig.7-7,and assuming that the skin between these longitudinals carries only shearing stresses.. The simplification of an actual structure into an analytical model represents a compromise, since elaborate models which nearly simulate the actual structure are usually difficult to analyze. A more complete discussion of the idealization of shell structures will be found in Ref. 4

Once the idealization is made, the stresses in the longitudinals due to bending moments,


9

Ships structures

axial load, and thermal gradients can be computed from the equations of this chapter if the structure is long compared to its cross-sectional dimensions and if there are no significant structural or loading discontinuities in the region where the stresses are computed. In many flight structures the cross section tapers; the effects of this taper upon the stresses are dis-cussed in Chap. 9. When discontinuities or other conditions arise which violate the analytical assumptions made in the Bernoulli-Euler theory, it is necessary to analyze the stiffened shell as an indeterminate structure (Chaps. 11 and 12).

1.6 Ships structures

The following description of the distortion and functions of ship structures is excerpted from Muckle (1967).

The Distortion of the ships structure

The study of the static forces on the ship has shown that the ship can bend in a longitudi-nal vertical plane like a beam. This is one of the most important types of distortion to which the ship is subjected, and is one in which the entire structure of the ship takes part. While consid-ering this longitudinal bending of the structure it should be mentioned that it is also possible for the ship to bend in horizontal plane. Consider a ship moving diagonally across a regular wave system as in Fig. 4. The crests are not perpendicular to the centre line of the ship and Fig. 5

Actual skin and webcarries axial andshear stresses

Effective longitudinals(axial stress only)

Idealized webs(shear stress only)

(a)

(b)Fig. 7-6 Idealization of semimonocoque structure. (a) Actualstructure; (b) idealized structure

Wall carries axialand shear stresses

(a)

Effective longitudinals(axial stress only)

Idealized web(shear stress only)

(b)

Fig. 7-7 Idealization of a monocoque shell. (a) Mono-coque shell; (b) idealization


10


shows that the slope of the waves at various points in the length of the ship varies, being sometimes positive and sometimes negative. This means that there are sideways forces acting on the ship which will not only cause swaying, but also bending in the horizontal plane. This bending has in the past been neglected and it is safe to say that the forces and moments gen-erated are likely to be of small amount.

Referring again to Fig. 5, it will be evident that, because of the variation in the wave slope at different sections in the length, not only will sideways forces be generated but there will also be moments applied at the various sections. As these may change sign along the length of the ship, twisting is possible with the consequent generation of torsional stresses. Once again it is perhaps doubtful whether this type of distortion is important from the point of view of the strength of the structure. The problem has been, partially investigated in the past, and at the present there appears to be some interest in it in view of the tendency to increase the size of hatch openings, thus reducing the torsional rigidity of the structure.

Consider now a transverse section of a ship as shown in Fig. 6. This section is subject first of all to static pressure due to the surrounding water. It will also be subjected to internal load-ing due to the weight of the structure itself and the weight of the cargo etc. which is carried. The effect of these static forces is to cause transverse distortion of the section, as shown by dotted lines in Fig. 6. It is worthy of note that this type of distortion would take place regardless of whether there was bending in the longitudinal direction. It is possible therefore to recognise an entirely independent study dealing with the transverse deformation of the ships structure.

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Fig. 4 Ship moving diagonally across waves

3/4 lengthF.P.A.P.

1/4 length

AmidshipsFig. 5 Wave surface at various

positions in length


11

Ships structures

Not only do the water pressure and the local internal loads cause transverse bending but it is possible to have local deformation of the structure due to these forces. A typical example of this is the bottom plating of a ship between floors or longitudinals. Fig. 7 shows a strip of such plating between two floors or longitudinals. The tendency is for the plating to bend as a beam

in between these members. Other parts of the structure which could be deformed under local loads are tank top plating, bulkheads, girders under heavy loads such as machinery etc. In this way it will be seen that there is another aspect of the strength of the structure which may be defined as local deformation.

Summarising this section, it is clear that the overall problem of the strength of the ships structure may be conveniently divided into three sections:

(1) Longitudinal strength (2) Transverse strength (3) Local strength

Since any given part of the structure of the ship may be subjected to one or more of the modes of distortion discussed, it will be seen that the resultant state of stress in that part could

Cargo load

Cargo load

Water pressure load

Fig. 6 Distortion of transverse section due to static loading

Floors

Outer bottom

Inner bottom

Water pressure

Fig. 7 Distortion of bottom plating due to water pressure


12


be very complex. It is for this reason that, in a first study at least of the strength of the ships structure, longitudinal bending, transverse bending and local bending are treated entirely inde-pendent, so that each of the three divisions of the subject of strength of ships quoted above can be investigated separately. This is the only realistic way of tackling the problem.

Function of the ships structure

It has been shown that the ship is capable of bending in a longitudinal vertical plane and it follows therefore that there must be material in the ships structure which will resist this bend-ing; or in other words there must be material distributed in the fore and after direction to fulfil this purpose. It follows that any material distributed over a considerable portion of the length of the ship will contribute to the longitudinal strength. Items which come into this category are the side and bottom shell plating, inner bottom plating and any decks which there may be. Fig. 8 is an outline section showing these items. As far as decks are concerned, it is usual to consider only the material abreast the line of openings, such as hatches and engine casings.

It will be clear that this longitudinal material forms a box girder of very large dimensions in relation to its thickness. Consequently, unless the plating was stiffened in some way it would be incapable of with standing compressive loads. For this reason therefore it becomes neces-sary to fit transverse rings of material spaced from 2 ft. to 3 ft. apart throughout the length of the ship. This is the procedure which is adopted in what is usually called a

transversely framed

ship. The transverse stiffening consists of three parts; in the bottom between the outer and inner bottoms there are several vertical plates called floors which have lightening and access holes cut in them as shown in Fig. 9; in the sides of the ship rolled sections called

side frames

, are welded to the plating (see Fig. 9); the decks are also supported by rolled sections welded to the plating, called

beams

. The floors, side frames and beams at the various decks are con-nected by means of brackets so that a continuous transverse ring of material is provided. As stated earlier, the spacing of these transverse rings, usually called the

frame spacing

, is between 2 ft. and 3 ft. and depends upon the length of the ship. It will be seen that the effect of

Upper deckplating

2nd deckplating

Sideshell

Inner bottomplating

Centregirder

Marginplate

Bottom shell

Fig. 8 Section through ship showing material resisting longitudinal bending


13

Ships structures

supporting the plating in this way is to reduce the unsupported span and hence to raise the buckling strength of the plating, to enable it to carry compressive loads.

Another function of these transverse rings is to prevent transverse distortion of the struc-ture, so that the floors, side frames and beams are the main items contributing to the trans-verse strength of the structure of the ship. The main force involved here is that due to water pressure and, as this will be greatest on the bottom of the ship, the bottom structure should be very heavy. This is in fact so, a very heavy girder being provide by the floor plate in conjunction with its associated inner and outer bottom plating. The side of the ship is also subjected to water pressure of rather lesser magnitude, and in this case adequate stiffening is provided by the girder consisting of the side frame welded to the side shell plating. As far as decks are con-cerned, here again the beam with its associated deck plating forms an effective built-up girder. The main factor determining the sizes of the beams is the load which they have to carry. This load may be a cargo load, a load due to passengers or, in the case of a weather deck some weather load.

Other items of the structure which contribute to transverse strength are watertight bulk-heads. Their primary object is, of course, to divide the ship into a series of watertight compart-ments, but since they consist of transverse sheets of plating they have very considerable transverse rigidity and hence contribute greatly to the prevention of transverse deformation of the structure.

The structure shown in Fig. 9 is typical of a transversely framed ship. It is common practice nowadays to adopt a different form of construction in which the sides of the ship are stiffened transversely whilst the decks and bottom are stiffened by means of longitudinals. This type of construction is shown in Fig. 10. As will be shown later, the effect of stiffening the deck and bottom by longitudinal members instead of transverse members is to increase very greatly the buckling strength of the plating, and it is largely for this reason that this method of construction has been adopted.Since these longitudinals are effectively attached to the plating they contrib-ute also to the general longitudinal strength of the structures. The longitudinals have to carry cargo and water pressure loads and so, in order to reduce their scantlings, they must be sup-ported at positions other than at bulkheads. This is achieved by introducing deep transverse beams in the decks spaced some 6 to 12 ft. apart and by having transverse plate floors in the

Upper deckbeam Tween deckpillar

Tween deckframe

Second deckbeam Hold

pillar

Floorplate

Tank sidebracket

Side girder

Centregirder

Fig. 9 Section through ship showing transverse structure


14


bottom at the same spacing. These widely spaced transverse members, in conjunction with closely spaced side framing, then provide the transverse strength of the structure.

The longitudinal system of framing has often also been extended to the sides of the ship as well as the decks and bottom. In fact when initially developed for use in oil tankers this was the method which was adopted. This was called the Isherwood System. At a later stage in the

development of the tanker the combined system of longitudinals in the bottom and deck with transverse side framing was employed. In many of the larger oil tankers of the present day, however, the complete longitudinal framing system has been used. Figure 11 shows the mid-ship section of such a tanker.

Where transverse beams are employed in the decks of ships it would be impracticable to

Tween deckframe

Upper decklongitudinals

2nd decklongitudinals

Side frame

Deep transversesspaced 6 to 12 ft.apart

Inner bottomlongitudinals

Outer bottomlongitudinals

Fig. 10 Section through ship with longitudinally stiffened decks and bottom

Flat bar decklongitudinals

Sidegirder

Centre girder

Wing bulkheadFlat bar longitudinals

Side girderCenter girder

Flat bar bottomlongitudinals

Transverses spacedabout 10 ft. apart

Flat bar side longitudinals

Fig. 11 Section through large modern oil tanker


15

Ships structures

run these from side to side of the ship without intermediate support. It is therefore necessary to introduce pillars to support the beams. In the early development of the iron and steel ship these pillars were closely spaced, generally being on alternate beams with a longitudinal angle runner fitted under the beams to spread the load to those beams not supported by pillars. This meant that access to the sides of cargo holds could only be made between two pillars, so that the available space was only about 5 ft. The later development was to support the deck beams by one or more heavy longitudinal girders and to support these girders by means of wide-spaced pillars. With this arrangement there would be probably two girders in the breadth of the ship each supported by two pillars in the length of the hold. Such an arrangement is shown in Fig. 12. By supporting the deck beams with lines of pillars or heavy longitudinals, the scant-

lings of the beams are greatly reduced and, further, by the use of wide-spaced pillars access to the holds is made easy. When longitudinal stiffening of decks is used, the system of con-struction just described can be imagined to have been turned around, The longitudinals replace the beams and the deep transverse beams replace the longitudinal deck girders in the transversely framed ship.

In addition to their functions in resisting longitudinal and transverse bending, many of the parts of the structure referred to in this section have also to support local loads. Thus beams and girders will often be subjected to loads due to machinery and loads produced by lifting

Hatch coaming

Hatch coamingPillars

Girder

Girder

Bulk

head

Bulk

headPillars

Inner bottom

Outer bottomElevation through hold

Girder

Girder

Deckbeams

Pillars

Pillars

Plan at deck

Fig. 12 Wide-spaced pillar and girder arrangement in transversely framed ship


16


equipment such as derricks and the like. The outside plating of the ship has also to withstand water pressure, and this could produce local bending of the plating between the stiffening members such as floors and frames. In general it could be said that nearly every structural member in the ship is a local strength member.

The foregoing discussion has shown briefly the functions which the various parts of the ships structure have to perform. It can be seen that particular part of the structure may have to perform several functions at the same time. In succeeding chapters methods for determining the stresses in the various parts will be dealt with in detail.

1.7 Key words and concepts from Chapter 1

structureelasticitystructural mechanicsstiffnessstrengthpreliminary designtypes of loadsmonocoque & semimonocoquespar, spar caps, spar webbulkheads, ribs, ringsstructural functions of the skin, longitudinals, and framesidealization of semimonocoque structurestresses due to bending and torsionlongitudinal, transverse, and local strength of ship structurestransversely framed, longitudinally framed, and Isherwood system of framing of shipsgirder, pillar, beam, floor platehatch, hatch coamingbuckling strengthscantlings

1.8 References

Anon.,

Aircraft Basic Science

, 1948, First Edition, Northrop Aeronautical Institute, Charles E. Chapel, Chief Editor, McGraw-Hill Book Company, Inc, p. 59 & 60.

Gordon, J.E., 1978,

Structures: or, Why things dont fall down

, (A Da Capo paperback) Reprint. Originally published by Harmondsworth: Penguin Books, pp. 33-44.

Hughes, O.F., 1983,

Ship Structural Design

, John Wiley and Sons, New York, N.Y., p. 8.

Muckle, W., 1967,

Strength of Ships' Structures

, E. Arnold Inc., pp. 5-12.

Rivello, R. M., 1969,

Theory and Analysis of Flight Structures

, McGraw-Hill, pp. 143-147.

Sokolnikoff, I.S., 1956,

Mathematical Theory of Elasticity

, Second Edition, McGraw-Hill Book Company, New York.


17

CHAPTER 2


A bar is a structural member that is relatively long along one axis and relatively compact in cross section in planes perpendicular to the axis. Bars can be straight or curved. Bars are among the most widely use structural elements. In this chapter only straight bars are considered that are subjected to loads directed along the reference axis of the bar. The reference axis is parallel to the long axis of the bar and will be defined in Section 2.4. Axial loads applied along the reference axis of a straight bar cause extensional and/or compressive deformations. A slender bar in compression is likely to buckle and in that case the bar is called a column. Buckling results in a combination of bending and compressive deformations of the column. Loads applied perpendicular to the refer-ence axis cause the bar to bend, and in that case the bar is called a beam. Beams are the subject of the next chap-ter.

The three basic steps to analyzed the static response of any structure are discussed for a bar in Section 2.1 to Section 2.5. These three fundamental steps of static structural mechanics are

equilibrium conditions,

strain-displacement conditions, or conditions of geometric fit, and

a material law, or constitutive behavior.

Work and energy methods are presented in Section 2.6 to Section 2.11, which includes the topics of virtual work, strain energy, complementary virtual work, complementary strain energy, and Castiglianos theorems. Applica-tions of the energy method to trusses is presented in Section 2.12.

2.1 Axially loaded bar

Consider a straight bar of length

L

, whose cross section is uniform along its length with its cross-sectional area denoted by

A

. The bar is referred to a Cartesian coordinate system

x

,

y

, and

z

with the

z

-axis parallel to the length and the

x

and

y

axes in a plane parallel to the cross section. The origin of the

z

-axis is taken at the left end of bar, so . The bar is subjected to the following loads: a distributed force per unit length of intensity ,

either an axial force or axial displacement at the left end, and to either an axial force or axial dis-

0 z L pz z( )Q1 q1 Q2


18


placement at the right end. The distributed force intensity , forces

Q

1

and

Q

2

, and the corresponding

displacements

q

1

and

q

2

, respectively, are defined positive if they act in the positive in the positive

z

-direction. See Fig. 2.1. Under the imposed loads, the bar is in tension and/or compression.

Equilibrium

The internal normal force, or axial force, acting in the

z

-direction is denoted by function , and

N

is positive if tensile and negative if compressive. See Fig. 2.2. If we consider an interior element of the bar

as shown in the center sketch of Fig. 2.2 and a positive normal force is defined to act in the positive

z

-direction on a positive

z

-face, then the action-reaction law requires a positive normal force acting on the negative z-face to act in the negative z-direction. A positive

z

-face of this interior element is the face whose normal pointing away from the material inside the element is in the positive

z

-direction. Conversely, a negative

z

-face of this interior element is the face whose normal pointing away from the material inside the element is in the negative

z

-direc-tion. Force equilibrium in the

z

-direction of the differential interior element of the bar shown in the figure, in the limit as , gives the following differential equation of equilibrium.

(2.1)

(In Fig. 2.2 note that coordinate

z

* where the distributed load intensity is evaluated on the differential element

approaches

z

in the limit as the element length goes to zero; i.e., in the limit.)

Let the axial displacement function be denoted by

w(z)

. The function

w(z)

is the displacement in the

z

-direc-tion of a particle located at

z

in the undeformed bar due to the imposed loads as is shown in Fig. 2.3. The axial

q2 pz z( )

x

y

z w,

y

L

Q2 q2,Q1 q1,

Fig. 2.1 Axially loaded bar

pz z( )

Cross section

N z( )

z

L

Q2Q1

dz

N dN+

pz z*( )dz

N ( )

N

z

N L ( )

z z* z dz+<

1 21 y= 0 2 y< < y

Thin-Walled Structures 255


Fig. 8.14.) If , then the role of and interchange on the Mohrs circles, so the maximum shear

stress is now . Therefore, maximum shear-stress criterion plots as the horizontal line for

.

In the second quadrant of Fig. 8.14 and , so that Mohrs circles

appear as shown in the figure to the right. The maximum shear stress is . Therefore, the maximum shear-stress criterion, eq. (8.55), gives

in the second quadrant. This is a straight line with a one-to-one slope

intersecting the at . Plotting the maximum shear-stress criterion in quad-

rants three and four in Fig. 8.14 proceed in a similar manner.

The maximum shear-stress envelope in Fig. 8.14 is contained within the Mises envelope. Hence, the maxi-mum shear stress criterion is conservative from a design perspective, with the largest differences between the predictions being about 15%. However, in analysis the Mises criterion is easier to implement than the maximum shear stress criterion. Mises criterion is a single equation, see eq. (8.50), but the maximum shear stress criterion requires that we compute the principal stresses and find their numerical order. Also note that the maximum shear stress acts on four planes at the material point, refer to Fig. 8.10, while the octahedral shear stress acts on eight planes at the material point. Laboratory tests on thin-walled tubes subject to an axial force, torque, and internal pressure are often used to study yielding under combined stress states. The experimental data for ductile metal tubes fall between the maximum shear-stress criterion and the Mises criterion on a plot such as Fig. 8.14, with the data closer to the Mises prediction (Dowling, 1993, pp. 251 and 252).

In design, the limit state for no yielding by maximum shear-stress criterion is simply

-1 -0.5 0.5 1

-1

-0.5

0.5

1

1y-----

2 y

3 0=

Mises criterion

Maximumshear-stress criterion

Fig. 8.14 Criteria for yield initiation in the principal stress plane.1 2

2 1 0> > 1 22 2 2 y=

0 1 y<

2 1( ) 22 y 1+=

2-axis y


256 Thin-Walled Structures

(8.56)

and we can define a factor of safety against yielding as

(8.57)

EXAMPLE 8.4 Factor of safety against initial yielding

Compute the factor of safety against the initiation of yield by Mises criterion and the maximum shear-stress criterion for 2024-T6 aluminum alloy that has a yield stress of 325 MPa.

Solution From Example 8.3, the principal stresses are , , and , and the

maximum shear stress is . If we calculate the Mises stress from eq. (8.47), we get

If we calculate the Mises stress via eq. (8.51), we get

which is the same value as obtained from eq. (8.47). Hence the factor of safety against yield, eq. (8.49), is

The factor of safety against yield using the maximum shear stress criterion, eq. (8.57), is

In linear structural analysis where the stresses are proportional to the load, the factor of safety means that the load can be increased by 5.61, in the case of Mises criterion, before the material initiates yielding. In the case of the maximum shear-stress criterion, the load can be increased by 5.16 before the material begins to yield. The lower factor of safety predicted by the maximum shear stress criterion illustrates it is slightly conservative with respect Mises prediction (in this case by about 8%).

max y 2= for no yielding

1 63MPa= 2 12MPa= 3 0=31.5MPa

M12--- 63 12( )2 12 0( )2 0 63( )2+ +( ) 57.94MPa= =

M 602 152 60 15 0 0 3 12( )2+ + 57.94MPa= =

FS 32557.9396------------------- 5.61= =

FS 325 231.5

---------------- 5.16= =



EXAMPLE 8.5 Stress responses of a stringer-stiffened, single cell beam.

The thin-walled, prismatic beam of length L shown in Fig. 8.15 is clamped at z = 0. It is subjected to a linearly distributed load acting through the locus of shear centers, and a torque T applied at z = L.

The cross-sectional contour is an isosceles triangle with branches one and three having the same length b and the same thickness . The vertical branch, or second branch, has length h and thickness . The beam is stiff-

ened by two longitudinal stringers of cross-sectional area . The overall dimensions are specified as L = 1800

mm, h = 100 mm, and b = 130 mm. The material is 2024-T4 aluminum alloy whose properties are listed in the table below.

Take , , , and

. For the cross section at z = 0,

d. plot the axial normal stress in MPa versus the contour coordinate s in

mm, and

e. the shear stress in MPa versus the contour coordinate s in mm.

The contour coordinate is related to the branch contour coordinates by

Aluminum alloy 2024-T4

Property Value Units

E, modulus of elasticity 73 GPa

, Poissons ratio 0.33 noneG, Shear modulus 27 GPa

, density 2800 kg/m3

yield, yield strength in tension 325 MPa

py z( ) p0 1 z L( )=

t1

s3

s1s2

t1t2

t1

h2---

h2---

b

x

y

py

CS.C.

T

Cross-section

As

AsL

py z( )

zT

Fig. 8.15 A stringer-stiffened, cantilevered beam. The contour in the cross section is an isosceles triangle.

t2

As

zsz

s

s 0=

p0 3.0 N/mm= T 750 N-m= t1 t2 1.0 mm= =

As 45 mm2=

z

zs



Equations for the stress analysis

Geometry

( a)

( b)

Axial normal stress due to bending

( c)

Shear stress tangent to the contour

( d)

Shear flow due to torque

( e)

Shear flow due to transverse the shear force

( f)

( g)

s

s1

s2 b+

s3 b h+ +

=

0 s1 b 0 s2 h 0 s3 b

x

y

s1

s2

s3

Ch

c

contour origin s

q1

q2 q3

As

As

y1 s1( )h2---

s1b----= 0 s1 b

y2 s2( )h2--- s2= 0 s2 h

y3 s3( )h2---

h2---

s3b----+= 0 s3 b

I xx y12 s1( )t1 s1d

0

b

y22 s2( )t2 s2d0

h

y32 s3( )t1 s3d0

b

h2--- 2

Ash2---

2 As+ + + +=

z z s,( )Mx z( )

I xx---------------y s( )=

zs z s,( )q z s,( )

t s( )---------------= q z s,( ) qb z s,( ) qt z( )+=

qtT

2--------= enclosed area of the cell

12---hc= =

qb1 z s1,( ) q0V y z( )

I xx------------- y1 s1( )t1 s1d

0

s1

=

qb2 z s2,( ) qb1 b( )V y z( )

I xx------------- h

2--- As

V y z( )

I xx------------- y2 s2( )t2 s2d

0

s2

=

upper stringer



( h)

To determine the shear flow at the contour origin, , due to the transverse shear force acting through the

shear center, we invoke the condition of no twist. That is,

( i)

or

( j)

Equilibrium

( k)

( l)

To locate the centroid (the point labeled C)

( m)

( n)

( o)

To locate the shear center (the point labeled S.C.)

From the solution of equations (f) to (j), we can find the shear flow in branch two due to the shear force

only. Now we use torque equivalence to locate the S.C. Sum torques about the apex to get

( p)

qb3 z s3,( ) qb2 h( )V y z( )

I xx------------- h

2---

AsV y z( )

I xx------------- y3 s3( )t1 s3d

0

s3

=

lower stringer

q0 V y

zd

dz q s( ) sdGt s( )---------------- 0= =

q s( )t s( )----------ds 0=

qb1 s1( )

t1----------------- s1d

0

b

qb2 s2( )

t2----------------- s2d

0

h

qb3 s3( )

t1----------------- s3d

0

b

+ + 0=

V y

Mx

py

dV ydz

--------- py z( )= and V y L( ) 0= V y z( ) py z( ) zd

z

L

=

dMxdz

----------- V y z( )= and Mx L( ) 0= Mx z( ) V y z( ) zd

z

L

=

x

y

s1

s2

s3

Ch

c

As

As

xc

A area 2bt1 ht2 2As+ += =

Qy x1 s1( )t1 s1d

0

b

x3 s3( )t1 s3d0

b

+= xc Qy A=

x1 s1( ) c 1 s1 b( )= x3 s3( ) c s3 b( )=

V y

c qb2 s2( ) s2d

0

h

c xsc( )V y=



Equation (p) yields a relation to find , or the location of the shear center. The location of the shear center is

independent of the magnitude of the shear force.

Results The shear force and bending moment attain maximum magnitudes at the root. A the root

and . The torque at the root section is

. Note that 1 N/mm2 equals 1 MPa

a) The axial normal stress is plotted as a function of the contour coordinate in graph below.

b) The shear stress is plotted as a function of the contour coordinate in the following graph.

xsc

x

y

S.C.

c

xsc

V y

c xsc

x

y

s2h

c

qb2 s2( )

V y 0( ) 2700 N= Mx 0( ) 1.62610( ) N-mm=

T 0( ) 750 310 N-mm=

50 100 150 200 250 300 350s,mm

-150

-100

-50

50

100

150

sz,Nmm2



EXAMPLE 8.6 Minimum weight design of the beam in Example 8.5 subject to a constraint on initial yielding

Consider the design for minimum weight of the aluminum alloy beam in Example 8.5, which is shown in Fig. 8.15. The design is constrained by material yielding with the factor of safety specified as 1.5. Use the material data for 2024-T4 aluminium as listed in the Example 8.5 problem statement, Mises yield criterion, and take the

value of the local acceleration due to gravity as 9.81 m/s2. The specified dimensions of the beam are , , and . Take the value of the applied loads as and

. (Note the value of the torque is changed with respect to its value in Example 8.5.)

The objective is to minimize the weight subject to no yielding given the loads and . That is, what are

the thicknesses and and the stringers cross-sectional area for minimum weight? Parameters , and

are called design variables. This is a problem in constrained optimization, which is stated mathematically as

where is the objective function, or weight in this problem, and are constraint func-

tions. For design against yielding the constraint functions are defined as

50 100 150 200 250 300 350s,mm

10

20

30

40

50

60

70

80tzs,Nmm2

L 1800 mm= h 100 mm= b 130 mm= p0 3 N/mm=

T 250 310 N-mm=

p0 T

t1 t2 As t1 t2

As

minimize W t1 t2 As, ,( )

such that gi t1 t2 As, ,( ) 0>

W t1 t2 As, ,( ) gi t1 t2 As, ,( )

giall M( )i

all----------------------------=



where the allowable stress, , is defined as

and the Mises stress is . These par-

ticular constraint functions are called static margins, and positive values indicate the degree of safety against exceed-ing the allowable stress. Due to symmetry about the x-axis you only need to calculate the margin of safety for yielding at four points in the cross section at the root as indicated in Fig. 8.16. That is, compute the margins of safety at the points labeled 1, 2, 3, and 4 in Fig. 8.16. In addition, the shear force and bending moment attain their largest magni-

tude simultaneously at z = 0, so the four constraints are evaluated at z = 0.

The intent of this exercise is to study the influence of the stringers on the design for minimum weight. For each stringer area given in the table below, determine the values of thicknesses t1 and t2 for minimum weight. List these values along with the weight, and the four margins of safety in the table.

To calibrate the computations, the beam weight is 57.87 N and the margins of safety at points 1 to 4 are 50.3716, 1.92285, 1.92539, and 8.56734, respectively, for the design variable values of t1 = 2.84 mm, t2 = 3.42

mm, and As = 45 mm2.

Influence of the stringer area on the minimum weight designs

Stringer area As in

mm2

Beam weight in

N

Thicknesses in mm Margins of safety, dimensionless

t1 t2 point 1 point 2 point 2 point 4

50

60

70

80

90

100

xy

C

point 1, s1 0=

point 2, s1 b=

point 3, s2 0=

point 4, s2 h 2=

Fig. 8.16 Critical points for yield evaluation in the cross section at the root

allall yield F.S.= M



Results s

0.2 0.3 0.4 0.5

0.3

0.4

0.5

0.6

0.7

0.8

As Wt. t1 t2 M1 M2 M3 M4

50 14.3955 0.437253 0.774719 3.2 9.4 10-10 0.0029 1.1

60 13.5688 0.342651 0.653471 2.1 -1.7 10-12 0.0008 0.82

70 13.3053 0.279341 0.564789 1.4 0.001 -5.810-8 0.57

80 13.4942 0.239829 0.505728 0.97 0.0024 -2.410-11 0.41

90 13.9704 0.214902 0.466853 0.72 0.0034 3.1 10-12 0.3

100 14.6189 0.198474 0.440723 0.57 0.0039 -1. 10-10 0.23

14 N16 N

18 N

As 100 mm2=

t1, mm

t2, mm

Minimum static margin 0=

least weight design

constant weight lines

Design plane for

feasible designs

infeasible designs



8.8 References

Ashby, M.F., 1992, Materials Selection in Mechanical Design, Butterworth-Heinemann, Ltd., Oxford.

Dowling, N.E., 1993, Mechanical Behavior of Materials, Prentice-Hall, Inc., Englewood Cliffs, New Jersey.


251

CHAPTER 9

Buckling

Buckling of a structure means

failure due to excessive displacements (loss of structural stiffness), and/or

loss of stability of an equilibrium configuration of the structure

Stability of equilibrium

means that the response of the structure due to a small disturbance from its equilib-rium configuration remains small; the smaller the disturbance the smaller the resulting magnitude of the displace-ment in the response. If a small disturbance causes large displacement, perhaps even theoretically infinite, then the equilibrium state is unstable. Practical structures are stable at no load. Now consider increasing the load slowly. We are interested in the value of the load, called the

critical load

, at which buckling occurs. That is, we are interested in when a a sequence of equilibrium stable states as a function of the load, one state for each value of the load, ceases to be stable.

If buckling occurs before the elastic limit of the material, which is roughly the yield stress of the material, then it is called

elastic buckling

. If buckling occurs beyond the elastic limit, it is called

inelastic buckling

, or plas-tic buckling if the material exhibits plasticity during buckling (mainly metals). Most thin-walled structural com-ponents buckle in compression below the elastic limit. Therefore, buckling determines the limit state in compression rather than material yielding. In fact, about 50% of an airplane structure is designed based on buck-ling constraints.

9.1 One-degree of freedom model

To illustrate the physical nature of buckling as a stability problem and failure by excessive displacements, it is instructive to analyze the response of a simple structural model to a compressive force. This model is shown in Fig. 9.1 and has one coordinate

,

, to describe the configuration of the model under the deadweight load

P

. The model consists of a rigid rod of length l, connected by smooth hinge to a rigid base. The rod can rotate about the hinge but it is restrained by a linear elastic torsional spring of stiffness K (dimensional units of F-L/ radian). The spring is unstretched at

= 0. Neglect the weight of the rod with respect to the applied load P.

<

Buckling

252


From the free body diagram of the rod shown in Fig. 9.1, the equation of motion for rotation about the fixed hinge is

(9.1)

where

I

0

is the moment of inertia of the rod about the fixed point and

t

is time.

9.1.1 Static equilibrium

Consider equilibrium states under the static, downward load

P

which are characterized by the angle

being independent of time

t

. Hence, the inertia term in eq. (9.1) vanishes and we have

(9.2)

The solutions to eq. (9.2) are

(9.3)

and

(9.4)

Recall from the calculus using lHpitals rule that the limit of the indeterminate form as is one. The two equilib-rium paths are plotted in the load-deflection diagram shown in Fig. 9.2. Equilibrium path

P

1

coincides with the load axis in the plot and is called the primary equilibrium path, or the trivial equilibrium path. Equilibrium path

P

2

is called the secondary path and we note it is symmetric about

= 0. The two equilibrium paths intersect at (

,

P

) = (0,

K/l

). This intersection of the two paths is called a bifurca-tion point. At no load the rod is vertical and this corresponds to the origin in the load-deflection diagram. As the load

P

is slowly increased from zero the rod remains vertical (

= 0), and at

P = K/l

adjacent equilibrium states exists on the secondary path. The exist-ence of adjacent equilibrium states in the vicinity of the primary

l

K

l

KOx

Oy

P P

initial deflected

FBD

Fig. 9.1 One degree of freedom structural model

l

Pl sin K I0 t2

2

dd

= t( )= t 0>

Pl sin K 0= K/l

if there are infinitesimal disturbances present (there always are), but will rotate either to the left or right depending on type of infinitesimal disturbance. We note that the magnitude of the angle

becomes large as the load is increased from

K/l

on the secondary path. The load at the bifurcation point is called the critical load and is denoted as . Thus,

(9.5)

Small

analysis

(9.6)

Consider the small angles of rotation such that for

measured in radians. Equilibrium eq. (9.2) becomes

(9.7)

And the solutions of this equation are

, and

(9.8)

(9.9)

These solutions are shown in the load-deflection plane in Fig. 9.3. The equilibrium path coincides with path , but path is not a good approximation to path

unless

is very small. However, the bifurcation point is the same as obtained in

the large

-analysis. Hence, the critical load from the small

-analysis is the same as obtained in eq. (9.5) from the large

-analysis.

9.1.2 Stability analysis

Let the rotation angle

(9.10)

where is independent of time and satisfies the equilibrium eq. (9.2); i.e.,

(9.11)

Consider the additional rotation angle to be small in magnitude but a function of time. Thus, we are considering small oscillations about an equilibrium state

as shown in Fig. 9.4. Substitute eq. (9.10) for q in the equation of motion,

eq. (9.1), to get

(9.12)

where the dots denote derivatives with respect to time; e.g., . Using the

trigonometric identity for the sine of the sum of two angles and performing some minor rearrangements the last equation becomes

Pcr

Pcr K l=

sin

Pl K 0=

P1: 0 for any P=

P2: P K l= for any small

PK l----------

0

1

P1

P2

Fig. 9.3 Small analysis

P1 P1 P2 P2

t( ) 0 t( )+=

0

Pl 0sin K0 0=

l

P

0

t( ) t( )

Fig. 9.4 Rotations in the stability analysis

t( )

P 0,( )

I0 K 0 +( ) Pl 0 +( )sin+ 0=

t2

2

dd =

Buckling


(9.13)

Now expand the trigonometric functions of angle in a Taylor Series about to get

(9.14)

in which means terms of order and higher. Arrange eq. (9.14) in powers of to get

(9.15)

Note that "coefficient" of the term vanishes because of the equilibrium condition given by eq. (9.11).

For very small additional rotation angles about the equilibrium configuration, eq. (9.15) is approxi-mated by

(9.16)

where

(9.17)

The solution of the second order differential equation, eq. (9.16), for is

(9.18)

in which constants and are determined by initial conditions for and . The solution given by

eq. (9.18) is a harmonic oscillation about the equilibrium configuration and is interpreted as the natural fre-

quency in radians per second. Initial conditions and are considered to be very small but arbitrary to simulate and arbitrary small initial disturbance. The smaller the initial disturbance, the smaller the maximum

amplitude of the oscillation in . Thus, is a condition for a stable equilibrium configuration with respect to infinitesimal disturbances.

The solution of the second order differential equation, eq. (9.16), for is

(9.19)

For arbitrary initial conditions, the term with the positive exponent in the dominates the solution. This corre-

sponds to large values of the no matter how small the initial disturbance. Hence, is a condition of unstable equilibrium configuration with respect to infinitesimal disturbances. The dynamic criterion for structural stability is

Dynamic criterion for stability of an equilibrium state

The equilibrium state is stable if

The equilibrium state is critical if

The equilibrium state is unstable if

I0 K0 K Pl 0sin cos 0cos sin+[ ]+ + 0=

0=

I0 K0 K Pl 0sin 112---2 O 4( )+ Pl 0cos

16---3 O 5( )++ + 0=

O n( ) n

I0 K0 Pl 0sin( ) K Pl 0cos( )Pl2----- 0sin 2 Pl

6----- 0cos 3 O 4( )+ + + + + 0=

0=

0

t( )

I0 K Pl 0cos( )+ 0= or 2+ 0=

2 K Pl 0cos( ) I0=

2 0>

t( ) A1 t( )sin A2 t( )cos+= 2 0>

A1 A2 0( ) 0( )

0( ) 0( )

2 0>

2 0

P Pcr= Pcr

z,w

y,v

L

P

Fig. 9.5 A straight column subjected to a centric, compressive axial force.

dN dz 0= N EAz= zz dw dz=

z dz

z

y

w z dz+( )

w z( )

dz w z dx+( ) w z( )+ 1zd

dw+ dz

N N dN+

Fig. 9.6 An element of the column in the pre-buckling equilibrium state

w 0( ) 0= N L( ) P=

N P= w z( ) PzEA-------= v z( ) 0= 0 z L

Buckling


The end shortening under the compressive load is , and this is plotted on the load-end shortening plot shown in Fig. 9.7. The equilibrium configuration of pure compression of the perfect column is called the trivial

equilibrium state. Note that in the trivial equilibrium state the lateral displacement of the column, is zero for all values of the compressive load P. Researchers in structural stability recognized from experience that buck-ling of the column is associated with the appearance of second, non-trivial, equilibrium configuration at the buck-ling load. This observation is the basis of the adjacent equilibrium method of stability analysis. The question characterizing the method of adjacent equilibrium is

What is the value of the load for which the perfect system admits non-trivial equilibrium configura-tions?

To answer this question we consider equilibrium of a slightly deflected element of the column at the same value of the external load P. The free body diagram of this element is shown in Fig. 9.8. The displacement due to

buckling is denoted by v1(z), and all quantities due to buckling are labeled with the subscript 1. Vertical force equilibrium gives

(9.22)

where is the y-direction shear force due to buckling. Moment equilibrium about the x-axis gives

where is the bending moment due to buckling. In general, the axial strain in the equilibrium configuration

is very small in magnitude compared to unity and is then neglected with respect to unity in this equation.

w L( )

w L( )0

EAL

-------

1

P

Fig. 9.7 Load-end shortening plot in pre-buckling

v z( )

1 dwdz-------+

dz

dv1dz

--------dzx1

v1v1

dv1dz

-------- dz( )+

z

y Mx1Mx1 dMx1+

V y1

V y1 dV y1+P

P

Fig. 9.8 Free body diagram of an element of the column in the buckled state.

dV y1dz

------------ 0=

V y1

dMx1dz

------------- 1 dwdz-------+

V y1dv1dz--------

P+ 0=

Mx1

dw dz


Perfect Columns

Then, the moment equation becomes

(9.23)

Neglecting with respect to unity, and assuming the rotation due to buckling is small, implies that the

rotation is given by . A very important term in eq. (9.23) is the contribution of the axial

compressive load P through the buckling displacement v1 to moment equilibrium. This term that couples the axial compressive load in equilibrium state to the buckling displacement arises only because we took equilibrium on the slightly deflected column element. Hookes law for the bending moment is

(9.24)

where Ixx is the second area moment of the cross section about the x-axis. (We assume the cross section is sym-metric about either the x-axis or y-axis, or that these axes are principal axes of the cross section if no symmetry is present. In design we use the minimum second are moment of the cross section). If we take the derivative of eq. (9.23), use eq. (9.22), and then substitute eq. (9.24) for the bending moment due to buckling we get

(9.25)

This is the governing fourth order, ordinary differential equation for the buckling displacement v1(z). For conve-nience in writing, we will drop the subscripts on the second area moment in the following developments. Also note that in the buckling theory the vertical shear force is determined in terms of the buckling displacement v1 by substituting eq. (9.24) into eq. (9.23) to get

(9.26)

To determine the buckling displacement v1 we need boundary conditions at z = 0 and z =L in addition to the o.d.e. given by eq. (9.25). There are four standard boundary conditions. These are

dMx1dz

------------- V y1dv1dz--------

P+ 0=

dw dz x1x1 z( ) dv1 dz( )=

Mx1 EI xx z2

2

d

d v1

=

z2

2

dd

EI xx z2

2

d

d v1

z2

2

d

d v1

P+ 0=

V y1 zdd

EIz2

2

d

d v1

dv1dz--------P=

zL

P

A. Pinned-pinned

v1 0( ) 0= v1 L( ) 0=

Mx1 0( ) 0= Mx1 L( ) 0=

zL

PB. Clamped-free

v1 0( ) 0= Mx1 L( ) 0=

x1 0( ) 0= V y1 L( ) 0=

Buckling


One solution to the o.d.e., eq. (9.25), subject to boundary conditions A-D is for all values of the

load P. This is the trivial solution. Are there any other solutions? Can we get them? The answer is yes to both questions if EI = constant. For EI = constant, eq. (9.25) becomes

or

(9.27)

where

(9.28)

The general solution of eq. (9.27) for is

(9.29)

where A1, A2, A3, and A4 are arbitrary constants to be determined by boundary conditions.

EXAMPLE 9.1 Critical load for clamped-free boundary conditions (B)

Consider the clamped-free boundary conditions; i.e. b.c.s (B) above. Determine the buckling load for

which the perfect column ceases to be stable.

Solution The boundary conditions in this case become

where the primes denote derivatives with respect to z. Taking derivatives of eq. (9.29) we have

zL

P

C. Clamped-clamped

v1 0( ) 0=

x1 0( ) 0=v1 L( ) 0=

x1 L( ) 0=

zL

P

D. Clamped-pinned

v1 0( ) 0=

x1 0( ) 0=v1 L( ) 0=

Mx1 L( ) 0=

v1 z( ) 0=

EIz4

4

d

d v1 Pz2

2

d

d v1+ 0=

z4

4

d

d v1 k2z2

2

d

d v1+ 0= 0 z L<

v1 z( ) A1 kz( )sin A2 kz( )cos A3z A4+ + +=

Pcr

v1 0( ) 0= v1 0( ) 0= EIv1 L( ) 0= v1 k2v1+[ ]z L= 0=


Perfect Columns

Substitute these solutions into the four boundary conditions to get

(9.30)

A non-trivial solution for A1 to A4 requires the determinate of coefficients to vanish

(9.31)

After expanding this determinate we get

(9.32)

which gives n-values, n = 1, 2, 3,..., of kL; or

(9.33)

For , the fourth row of matrix eq. (9.30) gives ; using this result in the sec-

ond row of matrix eq. (9.30) gives ; the first row of matrix eq. (9.30) gives . Note that the

third row of matrix eq. (9.30) is identically satisfied. So eq. (9.33) implies

, (9.34)

where Pn are the buckling loads. Equation (9.32) is called the characteristic equation, and the roots of this equa-tion determine the buckling loads. For each value of n we have an associated buckling mode (A1 = A3 = 0, A2 = A4)

(9.35)

where coefficient A4 is arbitrary. The first three buckling modes are shown in Fig. 9.9. Note that the amplitude

of the buckling mode is not known. However, we can plot its shape. The critical load, denoted by , is the low-

est buckling load. That is

v1 A1 kz( )sin A2 kz( )cos A3z A4+ + +=

v1 A1k kz( )cos A2k kz( )sin A3+=v1 A1k2 kz( )sin A2k2 kz( )cos=

v1 A1k3 kz( )cos A2k3 kz( )sin+=

0 1 0 1

k 0 1 0

k2 kL( )sin k2 kL( )cos 0 0

0 0 k2 0

A1A2A3A4

0=

det

0 1 0 1

k 0 1 0

k2 kL( )sin k2 kL( )cos 0 0

0 0 k2 0

0=

k5 kL( )cos 0=

kn2n 1( )

L--------------------

2---

PnEI------= = n 1 2 3 , , ,=

knL 2n 1( ) 2( )= A3 0=A1 0= A2 A4=

Pn 2n 1( )2---

2EIL2------= n 1 2 3 , , ,=

v1n z( ) A4 1 knz( )cos[ ]=

Pcr

Pcr P124-----EI

L2------= =

Buckling


Remember that in design we use the minimum EI for the cross section.

0.2 0.4 0.6 0.8 1 z/L0.51

1.52

v1/A40.2 0.4 0.6 0.8 1 z/L

0.51

1.52

v1/A40.2 0.4 0.6 0.8 1 z/L

0.20.40.60.81

v1/A4P1

24-----EI

L2------=

P292

4---------EI

L2------=

P3252

4------------EI

L2------=

n = 1

n = 2

n = 3

Fig. 9.9 First three buckling modes for the clamped-free column.


Imperfect columns

The critical loads for boundary conditions A through D and for EI = constant are given in Fig. 9.10 .

9.3 Imperfect columns

9.3.1 Eccentric load

Consider a uniform (EI = constant), pinned-pinned column subjected to an eccentric axial load P. Let e denote the perpendicular distance between the line of action of load P and the z-axis. This situation is statically equiva-lent to a centric axial load P and a moment of magnitude eP applied to the ends of the column. See Fig. 9.11. Hence, the eccentric axial load will simultaneously subject the column to compression and bending in the equi-librium state. The analysis for the equilibrium response of the column in compression (axial force N and axial displacement w(z)) is identical to the case of the perfect column, since the z-axis passes through the centroid of each cross section (decoupling the axial compression from bending in the material law). The equilibrium response of the column in bending, which includes the influence of axial compression on bending, is determined by the same analysis that led to eqs. (9.27) to (9.29), except that we drop the subscript 1 on the lateral displace-ment, since in the eccentric load case the lateral displacement refers to an equilibrium state and not to a buckling mode. The differential equation and boundary conditions for equilibrium displacement v(z) are

L

D. Clamped-pinned Pcr 2.0462EIL2------ 4.49( )2

EIL2------= =

L

C. Clamped-clamped Pcr 42EIL2------=

L

B. Clamped-free

Pcr24-----EI

L2------=

L

A. Pinned-pinned Pcr 2EIL2------=

Fig. 9.10 Buckling loads for the standard boundary conditions A to D.

Buckling


(9.36)

(9.37)

where k2 is as given in (9.28). Note that the boundary conditions, eqs. (9.37), are inhomogeneous. Thus, eqs. (9.36) and (9.37) do not have the trivial solution v(z) = 0 for all values of the load P. Using the general solution

form given by (9.29) for the solution of differential equation (9.36), subject to the boundary conditions (9.37), we find

(9.38)

By a trigonometric identity, . Let denote the midspan displacement; i.e., = v(L/2).

From eq. (9.38) we get

The factor kL/2 can be written in terms of the eccentric axial load P and the critical load Pcr for the pinned-pinned uniform column subjected to centric load as

.

Thus, the center deflection of the eccentrically loaded column becomes

z4

4

dd v

k2z2

2

dd v

+ 0 0 z L<


Imperfect columns

(9.39)

The load-displacement response is shown in Fig. 9.12. Note that as for . That is, no

matter the magnitude of the eccentricity as the value of the center deflection gets very large.

9.3.2 Geometric imperfection

Consider a uniform, pinned-pinned column that is slightly crooked under no load. The initial shape under no load is described by the function . The column is subjected to a centric, axial compressive load P. The lateral

displacement of the column is denoted by , so that when P = 0. Also, the bending moment

in the column is zero under no load. Thus, we write the material law for bending as

(9.40)

Vertical force equilibrium of the deflected column leads to a differential equation similar to eq. (9.22) except that we drop the 1 subscript since it is the configuration of the column is one of equilibrium and not a buckling mode. Similarly, moment equilibrium of the imperfect column leads to a differential equation similar to eq. (9.23) with the subscript 1 dropped. Combining these differential equations of vertical force equilibrium and moment equilibrium via elimination of the vertical shear force gives

(9.41)

The pinned-pinned boundary conditions are

(9.42)

e 12--- P

Pcr-------

cos------------------------------ 1=

P/Pcr

0

1

increasing e

e = 0

Pcr 2EIL2------=

Fig. 9.12 Load-deflection curves for an eccentrically loaded column

P Pcr e 0

P Pcr

v0 z( )

v z( ) v z( ) v0 z( )=

Mx EI z2

2

dd v

z2

2

d

d v0

=

z2

2

d

d Mxz2

2

dd v

P+ 0=

v 0( ) 0= Mx 0( ) 0= v L( ) 0= Mx L( ) 0=

Buckling


Consider the imperfection shape , where denotes the amplitude at midspan of the

slightly crooked column. Substitute eq. (9.40) into eq. (9.41) to eliminate the moment to get

(9.43)

where is given by eq. (9.28). The boundary conditions, eqs. (9.42), for this imperfection shape lead to

(9.44)

The solution of the differential equation (9.43) subject to boundary conditions (9.44) is

(9.45)

It is convenient to measure the deflection of the imperfect column under load with respect to its original unloaded state. That is, let define the additional displacement at midspan by . Hence,

(9.46)

The load-displacement response is sketched in Fig. 9.13. Note that as for . That is, for a non-

zero value of the imperfection amplitude, the displacement gets very large as the axial force approaches the buckling load of the perfect column. Also, the imperfect column deflects in the direction of imperfection; e.g., if , then

.

Collectively the eccentric load and the geometric shape imperfection are called imperfections. All real columns are imperfect. Even for a well manufactured column whose geo-metric imperfections are small and with the load eccentricity small, the displacements become excessive as the axial com-pressive force P approaches the critical load of the per-

fect column. Hence, the critical load determined from the analysis of the perfect column is meaningful in practice.

v0 z( ) a1zL----- sin= a1

z4

4

dd v

k2z2

2

dd v

+z2

2

d

d v0 a1L--- 2 z

L----- sin= =

k2

vz2

2

dd v 0= = at z = 0 and z = L

v z( )a1

1kL------

2----------------------- z

L----- sin= 0 z L

v L 2( ) v0 L 2( )=

a1

PPcr-------

1P

Pcr-------

----------------------=

0

P Pcr

1.0

Pcr 2EIL2------=

increasing a1

Fig. 9.13 Load-deflection response plots for geometrically imperfect columns

P Pcr a1 0

a1 0>

0>

Pcr


Column Design Curve

9.4 Column Design Curve

Consider the pinned-pinned uniform column whose critical load is given by . Let A denote the

cross-sectional area of the column. At the onset of buckling the critical stress is defined as

(9.47)

We write the second area moment as , where r denotes the minimum radius of gyration of the cross section. For the rectangular section shown in the adjacent sketch,

and , so that , where . Thus, the critical

stress becomes

(9.48)

and is called the slenderness ratio. The slenderness ratio is the column length divided by a cross-sectional dimension significant to bending.

For any set of boundary conditions we define the effective length by the formula

(9.49)

The effective lengths for the four standard boundary conditions are as follows:

The definition of effective length uses case A boundary conditions as a reference. the concept of effective length accounts for boundary conditions other than simple support, or pinned-pinned end conditions.

The column curve is a plot of the critical stress versus the effective slenderness ratio; i.e., .

For elastic column buckling under all boundary conditions

(9.50)

which is a hyperbola that depends only on the modulus of elasticity E of the material. This equation governing elastic buckling is called the Euler curve, and columns that buckle in the elastic range are called long columns. See Fig. 9.14

A pinned-pinned

B clamped-free

C clamped-clamped

D clamped-pinned

Pcr 2 EI L2( )=

cr Pcr A 2EI(

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