Steel Designer''''s Manual Part 5 pptx

9.3.1 Elastic analysis of line elements under axial loading When a cross section is subjected to a compressive or tensile axial load, P, the ing stress is given by the load/area of the s

9.1.1 Equations of static equilibrium

From Newton’s law of motion, the conditions under which a body remains in staticequilibrium can be expressed as follows:

• The sum of the components of all forces acting on the body, resolved along anyarbitrary direction, is equal to zero This condition is completely satisfied if the

components of all forces resolved along the x, y, z directions individually add up

to zero (This can be represented by SP x = 0, SP y = 0, SP z = 0, where P x , P yand

P z represent forces resolved in the x, y, z directions.) These three equations

represent the condition of zero translation

• The sum of the moments of all forces resolved in any arbitrarily chosen planeabout any point in that plane is zero This condition is completely satisfied when

all the moments resolved into xy, yz and zx planes all individually add up to zero (SM xy = 0, SM yz = 0 and SM zx = 0.) These three equations provide for zero rotation about the three axes

If a structure is planar and is subjected to a system of coplanar forces, the

conditions of equilibrium can be simplified to three equations as detailed below:

• The components of all forces resolved along the x and y directions will ually add up to zero (SP x = 0 and SP y= 0)

individ-• The sum of the moments of all the forces about any arbitrarily chosen point in

the plane is zero (i.e SM = 0).

9.1.2 The principle of superposition

This principle is only applicable when the displacements are linear functions ofapplied loads For structures subjected to multiple loading, the total effect of severalloads can be computed as the sum of the individual effects calculated by applyingthe loads separately This principle is a very useful tool in computing the combinedeffects of many load effects (e.g moment, deflection, etc.) These can be calculatedseparately for each load and then summed

9.2 Element analysis

Any complex structure can be looked upon as being built up of simpler units orcomponents termed ‘members’ or ‘elements’ Broadly speaking, these can be

classified into three categories:

• Skeletal structures consisting of members whose one dimension (say, length) is much larger than the other two (viz breadth and height) Such a line element is

variously termed as a bar, beam, column or tie A variety of structures areobtained by connecting such members together using rigid or hinged joints.Should all the axes of the members be situated in one plane, the structures so

produced are termed plane structures Where all members are not in one plane, the structures are termed space structures.

• Structures consisting of members whose two dimensions (viz length and breadth)are of the same order but much greater than the thickness fall into the second

category Such structural elements are called plated structures Such structural

elements are further classified as plates and shells depending upon whether theyare plane or curved In practice these units are used in combination with beams

or bars Slabs supported on beams, cellular structures, cylindrical or sphericalshells are all examples of plated structures

• The third category consists of structures composed of members having all thethree dimensions (viz length, breadth and depth) of the same order The analysis of such structures is extremely complex, even when several simplifyingassumptions are made Dams, massive raft foundations, thick hollow spheres,caissons are all examples of three-dimensional structures

For the most part the structural engineer is concerned with skeletal structures.Increasing sophistication in available techniques of analysis has enabled the eco-nomic design of plated structures in recent years Three-dimensional analysis ofstructures is only rarely carried out Under incremental loading, the initial defor-mation or displacement response of a steel member is elastic Once the stressescaused by the application of load exceed the yield point, the cross section graduallyyields The gradual spread of plasticity results initially in an elasto-plastic responseand then in plastic response, before ultimate collapse occurs

9.3 Line elements

The deformation response of a line element is dependent on a number of

cross-sectional properties such as area, A, second moment of area (Ixx = Úy2dA;

Iyy = Úx2dA) and the product moment of area (I xy = Úxy dA) The two axes xx and yy

are orthogonal For doubly symmetric sections, the axes of symmetry are those for

which Úxy dA = 0 These are known as principal axes For a plane area, the principal

axes may be defined as a pair of rectangular axes in its plane and passing through

its centroid, such that the product moment of area Úxy dA = 0, the co-ordinates

referring to the principal axes If the plane area has an axis of symmetry, it is

obviously a principal axis (by symmetry Úxy dA = 0) The other axis is at right angles

to it, through the centroid of the area

Tables of properties of the section (including the centroid and shear centre of

the section) are available as published data (e.g SCI Steelwork Design Guide,

Vol 1).1

y

U

x

If the section has no axis of symmetry (e.g an angle section) the principal axes

will have to be determined Referring to Fig 9.1, if uOu and vOv are the principal axes, the angle a between the uu and xx axes is given by

(9.1)

(9.2)

The values of a, I uu and I vv are available in published steel design guides (e.g.Reference 1)

9.3.1 Elastic analysis of line elements under axial loading

When a cross section is subjected to a compressive or tensile axial load, P, the ing stress is given by the load/area of the section, i.e P/A Axial load is defined as

result-one acting at the centroid of the section When loads are introduced into a section

in a uniform manner (e.g through a heavy end-plate), this represents the state ofstress throughout the section On the other hand, when a tensile load is introducedvia a bolted connection, there will be regions of the member where stress concen-trations occur and plastic behaviour may be evident locally, even though the meanstress across the section is well below yield

If the force P is not applied at the centroid, the longitudinal direct stress bution will no longer be uniform If the force is offset by eccentricities of e x and e y

distri-measured from the centroidal axes in the y and x directions, the equivalent set of actions are (1) an axial force P, (2) a bending moment M x = Pe x in the yz plane and (3) a bending moment M y = Pe y in the zx plane (see Fig 9.2) The method of eval-

uating the stress distribution due to an applied moment is given in a later section

9.3.2 Elastic analysis of line elements in pure bending

For a section having at least one axis of symmetry and acted upon by a bendingmoment in the plane of symmetry, the Bernoulli equation of bending may be used

as the basis to determine both stresses and deflections within the elastic range Theassumptions which form the basis of the theory are:

• The beam is subjected to a pure moment (i.e shear is absent) (Generally thedeflections due to shear are small compared with those due to flexure; this is nottrue of deep beams.)

• Plane sections before bending remain plane after bending

• The material has a constant value of modulus of elasticity (E) and is linearly

radius of curvature of the beam at the neutral axis

M

I

f y

E R

Fig 9.2 Compressive force applied eccentrically with reference to the centroidal axis

-*cross section stress diagram

neutral axis

Line elements 291

From the above, the stress at any section can be obtained as

For a given section (having a known value of I) the stress varies linearly from zero

at the neutral axis to a maximum at extreme fibres on either side of the neutral axis:

(9.4)

The term Z is known as the elastic section modulus and is tabulated in section tables.1The elastic moment capacity of a given section may be found directly as the product

of the elastic section modulus, Z, and the maximum allowable stress.

If the section is doubly symmetric, then the neutral axis is mid-way between thetwo extreme fibres Hence, the maximum tensile and compressive stresses will be

equal For an unsymmetric section this will not be the case, as the value of y for the

two extreme fibres will be different

For a monosymmetric section, such as the T-section shown in Fig 9.4, subjected

to a moment acting in the plane of symmetry, the elastic neutral axis will be the

centroidal axis The above equations are still valid The values of ymax for the twoextreme fibres (one in compression and the other in tension) are different For anapplied sagging (positive) moment shown in Fig 9.4, the extreme fibre stress in theflange will be compressive and that in the stalk will be tensile The numerical values

of the maximum tensile and compressive stresses will differ In the case sketched in

Fig 9.4, the magnitude of the tensile stress will be greater, as ymaxin tension is greaterthan that in compression

Caution has to be exercised in extending the pure bending theory to asymmetricsections There are two special cases where no twisting occurs:

I

M Z

=where

ponents M uu and M vv about the principal axes uu and vv and combining the

result-ing longitudinal stresses with those resultresult-ing from axial loadresult-ing:

(9.5)

For a section having two axes of symmetry (see Fig 9.2) this simplifies to

Pure bending does not cause the section to twist When the shear force is applied

eccentrically in relation to the shear centre of the cross section, the section twistsand initially plane sections no longer remain plane The response is complex andconsists of a twist and a deflection with components in and perpendicular to theplane of the applied moment This is not discussed in this chapter A simplifiedmethod of calculating the elastic response of cross sections subjected to twistingmoments is given in an SCI publication.2

9.3.3 Elastic analysis of line elements subject to shear

Pure bending discussed in the preceding section implies that the shear force applied

on the section is zero Application of transverse loads on a line element will, ingeneral, cause a bending moment which varies along its length, and hence a shearforce which also varies along the length is generated

x y

xx xx

yy yy

A

M v I

M u I

u v

uu uu

vv vv

Fig 9.4 Monosymmetric section subjected to bending

shear stress

distribution rectangular

(b)

If the member remains elastic and is subjected to bending in a plane of symmetry (such as the vertical plane in a doubly symmetric or monosymmetricbeam), then the shear stresses caused vary with the distance from the neutral axis

For a narrow rectangular cross section of breadth b and depth d, subjected to a shear force V and bent in its strong direction (see Fig 9.5(a)), the shear stress varies parabolically from zero at the lower and upper surfaces to a maximum value, qmax,

at the neutral axis given by

i.e 50% higher than the average value

For an I-section (Fig 9.5(b)), the shear distribution can be evaluated from

(9.6)

where B is the breadth of the section at which shear stress is evaluated The

integration is performed over that part of the section remote from the neutral axis,

i.e from y = h to y = hmaxwith a general variable width of b.

Clearly, for the I- (or T-) section, at the web/flange interface the value of the integral will remain constant As the section just inside the web becomes the section

just inside the flange, the value of the vertical shear abruptly changes as B changes

from web thickness to flange width

yield point/0

strain hardening

<

strain hardening commences

strain

9.3.4 Elements stressed beyond the elastic limit

The most important characteristic of structural steels (possessed by no other material to the same degree), is their capacity to withstand considerable deforma-tion without fracture A large part of this deformation occurs during the process of

yielding, when the steel extends at a constant and uniform stress known as the yield stress.

Figure 9.6 shows, in its idealized form, the stress–strain curve for structural steelssubjected to direct tension The line 0A represents the elastic straining of the material in accordance with Hooke’s law From A to B, the material yields while the

stress remains constant and is equal to the yield stress, fy The strain occurring inthe material during yielding remains after the load has been removed and is called

plastic strain It is important to note that this plastic strain AB is at least ten times

as large as the elastic strain, ey, at yield point

When subjected to compression, various grades of structural steel behave in asimilar manner and display the same property of yield This characteristic is known

as ductility of steel.

9.3.5 Bending of beams beyond the elastic limit

For simplicity, the case of a beam symmetrical about both axes is considered first.The fibres of the beam subjected to bending are stressed in tension or compressionaccording to their position relative to the neutral axis and are strained as shown inFig 9.7

While the beam remains entirely elastic, the stress in every fibre is proportional

to its strain and to its distance from the neutral axis The stress, f, in the extreme fibres cannot exceed the yield stress, fy

When the beam is subjected to a moment slightly greater than that which firstproduces yield in the extreme fibres, it does not fail Instead, the outer fibres yield

Fig 9.6 Idealized stress–strain relationship for mild steel

occupy the whole area of the sections, which are then described as being fully plastic.

When the cross section of a member is fully plastic under a bending moment, anyattempt to increase this moment will cause the member to act as if hinged at that

point This point is then described as a plastic hinge.

Fig 9.7 Elastic distribution of stress and strain in a symmetric beam (a) Rectangular

section, (b) I-section, (c) stress distribution for (a) or (b), (d) strain distribution for (a) or (b)

Fig 9.8 Distribution of stress and strain beyond the elastic limit for a symmetric beam.

(a) Rectangular section, (b) I-section, (c) stress distribution for (a) or (b), (d) strain distribution for (a) or (b)

as well.

Shape factor

As described previously there will be two stress blocks, one in tension, the other incompression, each at yield stress For equilibrium of the cross section, the areas incompression and tension must be equal For a rectangular section the plasticmoment can be calculated as

which is 1.5 times the elastic moment capacity

It will be noted that, in developing this increased moment, there is large ing in the external fibres of the section together with large rotations and deflections.The behaviour may be plotted as a moment–rotation curve Curves for various sections are shown in Fig 9.10

strain-The ratio of the plastic modulus, S, to the elastic modulus, Z, is known as the shape factor,, and it will govern the point in the moment–rotation curve when non-linearity starts For the ideal section in bending, i.e two flange plates, this will have a value of unity The value increases for more material at the centre of the section For a universal beam, the value is about 1.15 increasing to 1.5 for a rectangle

Mp=2b d d fy= bd fy

2

Fig 9.9 Distribution of stress and strain in a fully plastic cross section (a) Rectangular

section, (b) I-section, (c) stress distribution for (a) or (b), (d) strain distribution for (a) or (b)

(u=1.50)

(v1 00) 1.00

0.87

0.67

M= jyMp

curvature

Plastic hinges and rigid plastic analysis

In deciding the manner in which a beam may fail it is desirable to understand theconcept of how plastic hinges form when the beam becomes fully plastic Thenumber of hinges necessary for failure does not vary for a particular structuresubject to a given loading condition, although a part of a structure may fail independently by the formation of a smaller number of hinges The member or structure behaves in the manner of a hinged mechanism, and, in doing so, adjacenthinges rotate in opposite directions

As the plastic deformations at collapse are considerably larger than elastic ones,

it is assumed that the line element remains rigid between supports and hinge positions i.e all plastic rotation occurs at the plastic hinges

Considering a simply-supported beam subjected to a point load at mid-span (Fig 9.11), the maximum strain will take place at the centre of the span where aplastic hinge will be formed at yield of full section The remainder of the beam willremain straight: thus the entire energy will be absorbed by the rotation of the plastichinge

Work done at the plastic hinge = Mp(2q)

Work done by the displacement of the load =

At collapse, these two must be equal:

The moment at collapse of an encastré beam with a uniformly distributed load

(w = W/L) is worked out in a manner similar to the above from Fig 9.12.

ˆ

¯

Fig 9.10 Moment–rotation curves

I L 1!

I IliiI I

plastic

w zone I yield zone stiff length

Equating the two,

The moments at collapse for other conditions of loading can be worked out by asimilar procedure

por

Fig 9.11 Centrally-loaded simply-supported beam

Fig 9.12 Encastré beam with a uniformly distributed load

9.3.6 Load factor and theorems of plastic collapse

The load factor at rigid plastic collapse, lp, is defined as the lowest multiple of thedesign loads which will cause the whole structure, or any part of it, to become amechanism

In the limit-state approach, the designer seeks to ensure that at the appropriate

factored loads the structure will not fail Thus the rigid plastic load factor, lp, mustnot be less than unity, under factored loads

The number of independent mechanisms, n, is related to the number of possible plastic hinge locations, h, and the degree of redundancy, r, of the skeletal structure,

by the equation

n = h - r

The three theorems of plastic collapse are given below for reference:

(1) Lower bound or static theorem

A load factor, ls, computed on the basis of an arbitrarily assumed bending moment diagram which is in equilibrium with the applied loads and where the fully plastic moment of resistance is nowhere exceeded, will always be less than,

or at best equal to, the load factor at rigid plastic collapse, lp

lpis the highest value of lswhich can be found

(2) Upper bound or kinematic theorem

A load factor, lk, computed on the basis of an arbitrarily assumed mechanism

will always be greater than, or at best equal to, the load factor at rigid plasticcollapse, lp

lpis the lowest value of lkwhich can be found

(3) Uniqueness theorem

If both the above criteria ((1) and (2)) are satisfied, then l = lp

9.3.7 Effect of axial load and shear

If a member is subjected to the combined action of bending moment and axial force,the plastic moment capacity will be reduced

The presence of an axial load implies that the sum of the tension and sion forces in the section is not zero (see Fig 9.13) This means that the neutral axismoves away from the equal area axis, providing an additional area in tension or compression depending on the type of axial load The presence of shear forces willalso reduce the moment capacity For the beam sketched in Fig 9.13,

compres-axial load resisted = 2a t fy

Defining axial force resisted

axial capacity of section

A

T B

f

total area = A total stresses = bending + axial compression

For a given cross section, the plastic moment capacity, Mp, can be evaluated as

explained previously The reduced moment capacity, M¢p, in the presence of the axialload can be calculated as follows:

(9.7)

where S is the plastic modulus of the section.

Section tables provide the moment capacity for available steel sections using theapproach given above.1Similar expressions will be obtained for minor axis bending

9.3.8 Plastic analysis of beams subjected to shear

Once the material in a beam has started to yield in a longitudinal direction, it is

unable to sustain applied shear When a shear, V, and an applied moment, M, are

applied simultaneously to an I-section, a simplifying assumption is employed toreduce the complexity of calculations; shear resistance is assumed to be provided

by the web, hence the shear stress in the web is obtained as a constant value of V

divided by the web area (see Fig 9.14) The longitudinal direct stress to cause yield,

f1, in the presence of this shear stress, q, is obtained by using the von Mises yield

jd f

shear stress longitudinal stress

where Mpw is the fully plastic moment of resistance of the web

The addition of an axial load to the above condition can be dealt with by shifting the neutral axis, as was done in Fig 9.13 The web area required to carry the

axial load is now given by P/f1and the depth of the web, da, corresponding to this

9.3.9 Plastic analysis for more than one condition of loading

When more than one condition of loading is to be applied to a line element, it maynot always be obvious which is critical It is necessary then to perform separate calculations, one for each loading condition, the section being determined by thesolution requiring the largest plastic moment

f1= (fy2-3q2)

t d f

y 1

y pw

w a 1

Unlike the elastic method of design in which moments produced by differentloading systems can be added together, the opposite is true for the plastic theory

Plastic moments obtained by different loading systems cannot be combined, i.e.

the plastic moment calculated for a given set of loads is only valid for that loading

condition This is because the principle of superposition becomes invalid when parts

of the structure have yielded

9.4 Plates

Most steel structures consist of members which can be idealized as line elements.However, structural components having significant dimensions in two directions(viz plates) are also encountered frequently In steel structures, plates occur as com-ponents of I-, H-, T- or channel sections as well as in structural hollow sections.Sheets used to enclose lift shafts or walls or cladding in framed structures are alsoexamples of plates

With plane sheets, the stiffness and strength in all directions is identical and the

plate is termed isotropic This is no longer true when stiffeners or corrugations are introduced in one direction The stiffnesses of the plate in the x and y directions are substantially different Such a plate is termed orthotropic.

The x and y axes for the analysis of the plate are usually taken in the plane of the plate, as shown in Fig 9.15, while the z axis is perpendicular to that plane An element of the plate will be subjected to six stress components: three direct stresses

(sx, syand sz ) and three shear stresses (t xy, tyzand tzx ) There are six corresponding

strains: three direct strains (ex, eyand ez) and three shear strains (gxy, gyz and gzx).These stresses and strains are related in the elastic region by the material proper-

ties Young’s modulus (E) and Poisson’s ratio (v).

When considering the response of the plate, the approach customarily employed

is termed plane stress idealization As the thickness, t, of the plate is small compared with its other two dimensions in the x and y directions, the stresses having compon- ents in the z direction are negligible (i.e s z, tyzand txzare all zero) This implies that

Fig 9.15 Stress components on an element

the out-of-plane displacement is not zero, and this condition is referred to as plane

stress idealization

For an isotropic plate, the general equation relating the displacement, w,

perpendicular to the plane of the plate element is given by

(9.10)

where q is the normal applied load per unit area in the z direction which will, in general, vary with x and y The term D is the flexural rigidity of the plate, given by

(9.11)

The main difficulty in using this approach lies in the choice of a suitable

displacement function, w, which satisfies the boundary conditions For loading

conditions other than the simplest, an exact solution of this differential equation isvirtually impossible Hence approximate methods (e.g multiple Fourier series) are

utilized Once a satisfactory displacement function, w, is obtained, the moments per

unit width of the plate may be derived from

(9.12)

For orthotropic plates, the stiffness in x and y directions is different and the

equations are suitably modified as given below:

(9.13)

where D x and D yare the flexural rigidities in the two directions

In view of the difficulty of using classical methods for the solution of plate problems, finite element methods have been developed in recent years to providesatisfactory answers

9.5 Analysis of skeletal structures

The evaluation of the stress resultants in members of skeletal frames involves thesolution of a number of simultaneous equations When a structure is in equilibrium,every element or constituent part of it is also in equilibrium This property is made

use of in developing the concept of the free body diagram for elements of a structure.

x

w y

y

w x

2

2

2 2

2

21

(0) (b)

The portal frame sketched in Fig 9.16 will now be considered for illustrating theconcept Assuming that there is an imaginary cut at E on the beam BC, the partABE continues to be in equilibrium if the two forces and moment which existed atsection E of the uncut frame are applied externally.The internal forces which existed

at E are given by (1) an axial force F, (2) a shear force V and (3) a bending moment

M These are known as stress resultants The external forces on ABE, together with the forces F, V and M, keep the part ABE in equilibrium; Fig 9.16(b) is called the

free body diagram On a rigid jointed plane frame there are three stress resultants

at each imaginary cut The part ECD must also remain in equilibrium This

consideration leads to a similar set of forces F, V and M shown in Fig 9.16(c) It

will be noted that the forces acting on the cut face E are equal and opposite If thetwo free body diagrams are moved towards each other, it is obvious the internal

forces F, V and M cancel out and the structure is restored to its original state of equilibrium As previously stated, equilibrium implies SP x = 0; SP y = 0; SM = 0 for

a planar structure These equations can be validly applied by considering the ture as a whole, or by considering the free body diagram of a part of a structure

struc-In a similar manner, it can be seen that a three-dimensional rigid-jointed frame has six stress resultants across each section These are the axial force, twoshears in two mutually perpendicular directions and three moments, as shown inFig 9.17

Fig 9.16 Free body diagram

Fig 9.17 Force and moments in x, y and z directions

The solution of forces in the frames is accomplished by relating the stress ants to the displacements The number of equations needed is governed by the

result-degrees of freedom, i.e the number of possible component displacements At one

end of the member of a pin-jointed plane frame, the member displacement has

translational components in the x and y directions only, and no rotational

displace-ment The number of degrees of freedom is two By similar reasoning it will beapparent that the number of degrees of freedom for a rigid-jointed plane framemember is three For a member of a three-dimensional pin-jointed frame it is alsothree, and for a similar rigid-jointed frame it is six

9.5.1 Stiffness and flexibility

Forces and displacements have a vital and interrelated role in the analysis of tures Forces cause displacements and the occurrence of displacements implies theexistence of forces The relationship between forces and displacements is defined in

struc-one of two ways, viz flexibility and stiffness.

Flexibility gives a measure of displacements associated with a given set of forces

acting on the structure This concept will be illustrated by considering the example

of a spring loaded at one end by a static load P (see Fig 9.18).

As the spring is linearly elastic, the extension, D, produced is directly

propor-tional to the applied load, P The deflection produced by a unit load (defined as the flexibility of the spring) is obviously D/P Figure 9.18(b) illustrates the deflection response of a beam to an applied load P Once again the flexibility of the beam is D/P.

In the simple cases considered above, flexibility simply gives the ment response at a point A more generalized definition applicable to the displace-

load–displace-Fig 9.18 Flexibility

ment response at a number of locations will now be obtained by considering thebeam sketched in Fig 9.19.

Considering a unit load acting at point 1 (Fig 9.19(b)), the corresponding

deflec-tions at points 1, 2 and 3 are denoted as f11, f21and f31(the first subscript denotesthe point at which the deflection is measured; the second subscript refers to the

point at which the unit load is applied) The terms f11, f21, f31are called flexibility ficients Figure 9.19(c) and (d) give the corresponding flexibility coefficients for load

coef-positions 2 and 3 respectively By the principle of superposition, the total deflections

at points 1, 2 and 3 due to P1, P2and P3can be written as

Î

ÍÍÍ

¸

˝Ô

Fig 9.19 Flexibility coefficients for a loaded beam

or {D} = [F ] {P}

where {D} = displacement matrix

[F ] = flexibility matrix relating displacements to forces {P} = force matrix

Hence {P} = [F ]-1{D}

Stiffness is the inverse of flexibility and gives a measure of the forces

cor-responding to a given set of displacements Considering the spring illustrated in Fig 9.18(a), it is noted that the deflection response is directly proportional to

the applied load, P The force corresponding to unit displacement is obviously P/D.

Likewise in Fig 9.18(b) the load to be applied on the beam to cause a unit

dis-placement at a point below the load is P/D In its simplest form, stiffness coefficient

refers to the load corresponding to a unit displacement at a given point and can beseen to be the reciprocal of flexibility The concept is explained further using Fig 9.20

First the locations 2 and 3 are restrained from movement and a unit displacement

is given at 1 This implies a downward force k11at 1, an upward force k21at 2 and a

downward force k31at 3 The forces at points 2 and 3 are necessary as otherwise

there will be displacements at the locations 2 and 3.

The forces k11, k21 and k31 are designated as stiffness coefficients In a similarmanner, the stiffness coefficients corresponding to unit displacements at points 2and 3 are obtained

Fig 9.20 Stiffness coefficients

The stiffness coefficients and the corresponding forces are linked by the ing equations

where [K] is the stiffness matrix relating forces and displacements.

9.5.2 Introduction to statically indeterminate skeletal structures

A structure for which the external reactions and internal forces and moments

can be computed by using only the three equations of statics (SP x = 0, SP y= 0 and

SM = 0) is known as statically determinate A structure for which the forces and moments cannot be computed from the principles of statics alone is statically indeterminate Examples of statically determinate skeletal structures are shown in

it ‘perfect’ (Fig 9.21(b)), the structure becomes statically indeterminate

The degree of indeterminacy (also termed the degree of redundancy) is obtained

by the number of member forces or reaction components (viz moments or forces) which should be ‘released’ to convert a statically indeterminate structure

to a determinate one If n forces or moments are required to be so released, the degree of indeterminacy is n We need n independent equations (in addition

to three equations of statics for a planar structure) to solve for forces and moments

at all locations in the structure The additional equations are usually written by considering the deformations or displacements of the structure This means that the section properties (viz area, second moment of area, etc.) have an importanteffect in evaluating the forces and moments of an indeterminate structure Also,the settlement of a support or a slight lack of fit in a pin-jointed structure con-tributes materially to the internal forces and moments of an indeterminate structure

˛

=È

Î

ÍÍÍ

¸

˝Ô

˛

DDD

Fig 9.21 Statically determinate and indeterminate skeletal structures (a), (b) and (c) are

determinate; (d), (e) and (f) are indeterminate

9.5.3 The area moment method

The simplest technique of analysing a beam which is indeterminate to a low degree

is by the area moment method The method is based on two theorems (see Fig 9.22):

• Area Moment Theorem 1: The change in slope (in radians) between two points

of the deflection curve in a loaded beam is numerically equal to the area under

the M/EI diagram between these two points.

• Area Moment Theorem 2: The vertical intercept on any chosen line between the

tangents drawn to the ends of any portion of a loaded beam, which was nally straight and horizontal, is numerically equal to the first moment of the area

origi-under the M/EI diagram between the two ends taken about that vertical line.

Mx

EI x

A B

qB qA area of diagram between A and B

d

=Ú

M EI M

EI x

A B

Fig 9.22 Area moment theorems

(Caution: The vertical intercept is not the deflection of the beam from its original

position.)

The area moment method can be used for solving problems like encastré beams,propped cantilevers, etc The procedure is as follows:

(1) The redundant supports are removed, thereby releasing the redundant forces

and moments The statically determinate M/EI diagram for externally applied

loads can then be drawn

(2) The externally applied loads are removed, the redundant forces and moments

are introduced one at a time, and the M/EI diagrams corresponding to each of

these forces and moments are drawn

(3) The slopes at supports and intercepts on a vertical axis passing through the ports are then calculated

sup-(4) A number of expressions are obtained These are then equated to known values

of slopes or displacements at supports The equations so obtained can then besolved for the unknown redundant reactions This enables the evaluation of theforces and moments in the structure

9.5.4 The slope–deflection method

The slope–deflection method can be used to analyse all types of statically minate beams and rigid frames In this method all joints are considered rigid andthe angles between members at the joints are considered not to change as the loadsare applied When beams or frames are deformed, the rigid joints are considered torotate as a whole

indeter-In the slope–deflection method, the rotations and translations of the joints arethe unknowns All end moments are expressed in terms of end rotations and trans-lations In order to satisfy the conditions of equilibrium, the sum of end momentsacting on a rigid joint must total zero Using this equation of equilibrium, theunknown rotation of each joint is evaluated, from which the end moments are computed

For the span AB shown in Fig 9.23, the object is to express the end moments MAB

arid MBAin terms of end rotations qAand qBand translation D.

With the applied loading on the member, fixed end moments MFABand MFBAarerequired to hold the tangents at the ends fixed in direction (Counter clockwise endmoments and rotations are taken as positive.) The slope–deflection equations forthe case sketched in Fig 9.23 are

CA IMFAB

9.5.5 The moment-distribution method

The moment-distribution method can be employed to analyse continuous beams orrigid frames Essentially it consists of solving the simultaneous equations in the slope–deflection method by successive approximations Since the solution is by succes-sive iteration, it is not even necessary to determine the degree of redundancy.Two facets of the method must be appreciated (see Fig 9.24):

• When a stiff joint in a structural system absorbs an applied moment with rotational movement only (i.e no translation), Fig 9.24(a), the moment resisted

by the various members meeting at the joint is in proportion to their respectivestiffnesses (Fig 9.24(b))

• When a member is fixed at one end and a moment, M, is applied at the other

freely supported end, the moment induced at the fixed end is half the applied

Fig 9.23 Slope–deflection method

Fig 9.24 Moment-distribution method

negative moment positive moment (counter clockwise) (clockwise)

convention: clockwise moments are positive

convention: downward settlement of B with

respect to A is positive

moment and acts in the same direction as M (This is frequently referred to as

the carry over.) (Fig 9.24(c).)

Figure 9.25 shows the sign conventions employed in the moment-distributionmethod and Fig 9.26 illustrates the moment-distribution procedure

The moment-distribution method consists of locking all joints first and thenreleasing them one at a time To begin with, all joints are locked, which implies thatthe fixed end moments due to applied loading will be applied at each joint Byreleasing one joint at a time, the unbalanced moment at each joint is distributed tothe various members meeting at the joint Half of these applied moments are thencarried over to the other end of each member This creates a further imbalance ateach joint and the unbalanced moments are once again distributed to all membersmeeting at each joint in proportion to their respective stiffnesses

This procedure is repeated until the totals of all moments at each joint are ciently close to zero At this stage the moment-distribution process is stopped andthe final moments are obtained by summing up all the numbers in the respectivecolumns

suffi-9.5.6 Unit load method

Energy methods provide powerful tools for the analysis of structures The unit loadmethod can be directly derived from the complementary energy theorem, which

Fig 9.25 Sign conventions used in moment-distribution method

as propped cantilever

step 2: calculate the fixed end moments caution: pay attention to signs

—1 6 step 3: release joint C

(simply supported)

—5 step 4: carry over to B

and distribute

(note: no carry over to

C is possible)

—13.72 +20.57 —20.57 0 step 7: sum up the momentsafter repeating

distribution and carry avers if necessary

states that for any elastic structure in equilibrium under loads P1, P2, , the

corresponding displacements x1, x2, are given by the partial derivatives of the

complementary energy, C, with respect to the loads P1, P2, etc In other words,

1

1 2 2

Fig 9.26 An example of moment-distribution procedure

The use of this will be illustrated by considering a simply-supported beam of

length l subject to an external loading (see Fig 9.27) Strain energy stored in the

beam is predominantly flexural and is given by

(9.21)

is the bending moment due to a unit load and is denoted by m.

Hence the procedure of the unit load method can be outlined (see Fig 9.27):

(1) diagram due to the external loading is obtained

(2) The external loads are now removed and the moment diagram (m) due to a

unit load applied at the point of required deflection is drawn

(3) These two diagrams should now be integrated; in other words, the ordinates ofthe two diagrams are multiplied to obtain the deflection, given by

The same principle can be employed to determine the displacement due to othercauses, viz axial load or shear or torsion.

9.6 Finite element method

The advent of high-speed electronic digital computers has given tremendousimpetus to numerical methods for solving engineering problems Finite elementmethods form one of the most versatile classes of such methods which rely strongly

on the matrix formulation of structural analysis The application of finite elementsdates back to the mid-1950s with the pioneering work by Argyris,4 Clough andothers

The finite element method was first applied to the solution of plane stress problems and subsequently extended to the analysis of axisymmetric solids,plate bending problems and shell problems A useful listing of elements developed

in the past is documented in text books on finite element analysis.5

Stiffness matrices of finite elements are generally obtained from an assumed placement pattern Alternative formulations are equilibrium elements and hybridelements A more recent development is the so-called strain based elements Theformulation is based on the selection of simple independent functions for the linearstrains or change of curvature; the strain–displacement equations are integrated toobtain expressions for the displacements

dis-The basic assumption in the finite element method of analysis is that the response

of a continuous body to a given set of applied forces is equivalent to that of a system

of discrete elements into which the body may be imagined to be subdivided Fromthe energy point of view, the equivalence between the body and its finite elementmodel is therefore exact if the strain energy of the deformed body is equal to that

of its discrete model

The energy due to straining of the element, U, written in two-dimensional form

where {d} is the nodal displacement vector, [f ] is a function defining the strain tribution and e refers to a typical finite element When the strain distribution within

dis-the model is exactly dis-the same as that prevailing within dis-the body, dis-then dis-the energyequation will be exactly satisfied

The exact determination of the strain distribution function [f ] in Equation (9.24)

presents considerable difficulties, since this can only be done by a rigorous solution

of the equations of linear elasticity It may not always be possible to obtain an exact shape function for the solution: however, a suitable function which is adequate to

model strains can usually be selected The derivation of simple membrane elementsfor plate problems is presented in the following pages

9.6.1 Finite element procedure

As mentioned above, the basic concept of the finite element method is the idealization of the continuum as an assemblage of discrete structural elements Thestiffness properties of each element are then evaluated and the stiffness properties

of the complete structure are obtained by superposition of the individual elementstiffnesses This gives a system of linear equations in terms of nodal point loads anddisplacements whose solution yields the unknown nodal point displacements.The idealization governs the type of element which must be used in the solution

In many cases only one type of element is used for a given problem, but sometimes

it is more convenient to adopt a ‘mixed’ subdivision in which more than one type

of element is used

The elements are assumed to be interconnected at a discrete number of nodalpoints or nodes The nodal degrees of freedom normally refer to the displacementfunctions and their first partial derivatives at a node but very often may includeother terms such as stresses, strains and second or even higher partial derivatives.For example, the triangular and rectangular membrane elements of Fig 9.28(a) and

(b) have 6 and 8 degrees of freedom respectively, representing the translations u and v at the corner nodes in the x and y directions.

A displacement function in terms of the co-ordinate variables x, y and the nodal displacement parameters (e.g u i , v i, or di) are chosen to represent the displacementvariations within each element By using the principle of virtual work or the prin-ciple of minimum total potential energy, a stiffness matrix relating the nodal forces

to the nodal displacements can be derived Hence the choice of suitable ment functions is the most important part of the whole procedure A good displacement function leads to an element of high accuracy with converging characteristics; conversely, a wrongly chosen displacement function yields poor ornon-converging results

displace-A displacement function may conveniently be established from simple nomials or interpolation functions The displacement field in each element must beexpressed as a function of nodal point displacements only, and this must be done insuch a way as to maintain inter-element compatibility, since this condition is neces-

poly-sary to establish a bound on the strain energy Therefore, the displacement patternand the nodal point degrees of freedom must be selected properly for each problemconsidered.

In general, it is not always necessary that the compatibility must be satisfied in

order to achieve convergence to the true solution If complete compatibility is notachieved, there exists an uncertainty as to the bound on the strain energy of thesystem Therefore, in order to justify the performance and the ability of these ele-ments to converge to the true solution, a critical test which indicates the perform-ance in the limit must be carried out: that is, a convergence test with decreasingmesh size If the performance is adequate, then convergence to the true solution isachieved within a reasonable computational effort and the requirement for thecomplete compatibility can be relaxed

Besides satisfying the compatibility requirement, the assumed displacement functions should include the following properties:

(1) Rigid body modes

(2) Constant strain and curvature states

(3) Invariance of the element stiffnesses

9.6.2 Idealization of the structure

The finite element idealization should represent the real structure as closely as sible with regard to geometrical shape, loading and boundary conditions The geo-metrical form of the structure is the major factor to be considered when decidingthe shape of elements to be used In two-dimensional analyses the most frequentlyused elements are triangular or rectangular shapes The triangular element has theadvantage of simplicity in use and the ability to fit into irregular boundaries Figure

pos-Fig 9.28 Triangular and rectangular membrane elements

— a

-9.29 shows an example using triangular elements and Fig 9.30 shows a combination

of triangular and rectangular elements These figures also demonstrate the need touse relatively small elements in areas where high stress gradients occur In manysuch cases the geometrical shape of the structure is such that a fine mesh of elements is required in order to match this shape

9.6.3 Procedure for evaluating membrane element stiffness

The formulation of the stiffness matrix [Kc] of the membrane element is briefly discussed below

The energy due to straining of the element is given by Equation (9.22) The stresscomponents are shown in Fig 9.31 and ex, ey, gxy are the corresponding strain

Fig 9.29 Circular hole in uniform stress field

Fig 9.30 Perforated tension strip (plane stress) – a quarter of a plate is analysed

components For the state of plane stress, the stress components are related to thestrain components as given below:

(9.25)

where E and n are Young’s modulus and Poisson’s ratio respectively.

The matrix [D] is referred to as the elasticity or property matrix The strain energy

of the element is given by Equation (9.22) as

The strains are obtained through making appropriate differentiations of the

displacement function with respect to the relevant co-ordinate variable x or y.

Thus,

(9.28)

e

eeg

{ }=

ÏÌÔÓ

¸

˝Ô

ÔÔ

Ó

ÔÔ

¸

˝

ÔÔ

˛

ÔÔ

x y xy

u x

u y

v x

x

y

xy

x y xy

¸

˝Ô

˛1

2n

nn

n2

Fig 9.31 Stress components on a plane element

where [B] is the transformation matrix.

Using Equations (9.26) and (9.29), the following expression for U in terms of the

nodal point displacement vector {de} is obtained:

(9.30)

Differentiation of U with respect to the nodal displacements yields the stiffness matrix [Ke]:

(9.31)

where t is the thickness of the element (assumed constant).

The calculation of the stiffness matrices is generally carried out in two stages Thefirst stage is to calculate the terms inside the square brackets of Equation (9.31) i.e.the integration part The second stage is to multiply the resulting integrations by the

inverse of the transformation matrix [C] and its transpose.

Equation (9.31) can now be written as

(9.32)

where

The simplest elements for plane stress analysis have nodal points at the corners

only and have two degrees of kinematic freedom at each nodal point, i.e u and v.

This type of element proves simple to derive and has been widely used The simplest elements of this type are rectangular and triangular in shape

A triangular element with nodal points at the corners is shown in Fig 9.28(a) Thedisplacement function of this element has two degrees of freedom at each nodalpoint and the displacements are assumed to vary linearly between nodal points Thisresults in constant values of the three strain components over the entire element;the displacement functions are

v = a4+ a5x + a6y

The rectangular element with sides a and b, shown in Fig 9.28(b), is used with the

following displacement functions:

x

9.6.4 Procedure for evaluating plate bending element stiffness

The energy due to straining (bending) of the element is

where w is the transverse displacement of the plate element.

The conventional relationship between curvatures and moment is

˛

=[ ]

ÏÌÔÓ

¸

˝Ô

˛

ccc

w x

w y

x y xy A

¸

˝Ô

˛

Ú Ú

1

12

c, ,

2 1

nn

nn

elements and analysing the structure.5–7These methods are used for solving a widerange of problems.

3 Timoshenko S (1976) Strength of Materials – Part 2, 3rd edn Van Nostrand &

Co., New York

4 Argyris J.H (1960) Energy Theorems and Structural Analysis Butterworths,

London

5 Zienkiewicz O.C & Cheung Y.K (2000) The Finite Element Method in Structural and Continuum Mechanics, 5th edn Butterworth-Heinemann, Oxford.

6 Coates R.C., Coutie M.G & Kong F.K (1988) Structural Analysis, 3rd edn.

Chapman & Hall, London

7 Nath B (1974) Fundamentals of Finite Elements for Engineers Athlone Press,

London

Further reading for Chapter 9

Brown D.G (1995) Modelling of Steel Structures for Computer Analysis The Steel

Construction Institute, Ascot, Berks