Plastic analysis and design of steel structures

The plastic method has been used extensively by engineers for the design of steel structures, including simple beams, continuous beams, and simple portal frames. Traditionally, the analysis is based on the rigidplastic theory whereby the plastic collapse load is evaluated through virtual work formulation in which elastic deflection is ignored. For more complex frames, specialist computer packages for elastoplastic analysis are usually employed. Current publications on plastic design method provide means of analysis based on either virtual work formulation or sophisticated plastic theory contained in specialist computer packages. This book aims to bridge this gap.

Plastic Analysis and

Design of Steel Structures

Plastic Analysis and

Design of Steel Structures

M Bill Wong

Department of Civil Engineering

Monash University, Australia

AMSTERDAM • BOSTON • HEIDELBERG • LONDON

NEW YORK • OXFORD • PARIS • SAN DIEGOSAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO

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Library of Congress Cataloging-in-Publication Data

Wong, Bill.

Plastic analysis and design of steel structures/by Bill Wong 1st ed.

p cm.

Includes bibliographical references and index.

ISBN 978-0-7506-8298-5 (alk paper)

1 Building, Iron and steel 2 Structural design 3 Plastic analysis

(Engineering) I Title.

TA684.W66 2009

624.1’821 dc22

2008027081 British Library Cataloguing-in-Publication Data

A catalogue record for this book is available from the British Library.

ISBN: 978-0-7506-8298-5

For information on all Butterworth–Heinemann publications

visit our Web site at www.elsevierdirect.com

Printed in the United States of America

08 09 10 11 12 10 9 8 7 6 5 4 3 2 1

1.3 Statically Indeterminate Structures—Direct Stiffness

2.6 Comparison of Linear Elastic and Plastic Designs 73

2.8 Overview of Design Codes for Plastic Design 752.9 Limitations of Plastic Design Method 76

3 Plastic Flow Rule and Elastoplastic Analysis 813.1 General Elastoplastic Analysis of Structures 813.2 Reduced Plastic Moment Capacity Due to Force

3.5 Derivation of General Elastoplastic Stiffness Matrices 923.6 Elastoplastic Stiffness Matrices for Sections 953.7 Stiffness Matrix and Elastoplastic Analysis 99

4.7 Distributed Loads in Elastoplastic Analysis 126

5.7 Calculation of Member Forces at Collapse 1555.8 Effect of Axial Force on Plastic Collapse Load 157

6.4 General Description of the Discrete Plane Frame

8.2 Serviceability Limit State Requirements 219

Contents vii

The plastic method has been used extensively by engineers for thedesign of steel structures, including simple beams, continuous beams,and simple portal frames Traditionally, the analysis is based on therigid-plastic theory whereby the plastic collapse load is evaluatedthrough virtual work formulation in which elastic deflection isignored For more complex frames, specialist computer packages forelastoplastic analysis are usually employed Current publications onplastic design method provide means of analysis based on either vir-tual work formulation or sophisticated plastic theory contained inspecialist computer packages This book aims to bridge this gap.The advent of computers has enabled practicing engineers toperform linear and nonlinear elastic analysis on a daily basis usingcomputer programs widely available commercially The results fromcomputer analysis are transferred routinely to tools with automatedcalculation formats such as spreadsheets for design The use of thisroutine procedure is commonplace for design based on elastic, geo-metrically nonlinear analysis However, commercially available com-puter programs for plastic analysis are still a rarity among theengineering community

This book emphasizes a plastic analysis method based on thehinge by hinge concept Frames of any degree of complexity can beanalyzed plastically using this method This method is based on theelastoplastic analysis procedure where a linear elastic analysis, per-formed either manually or by computers, is used between the forma-tion of consecutive plastic hinges The results of the linear elasticanalysis are used in a proforma created in a spreadsheet environmentwhere the next plastic hinge formation can be predicted automaticallyand the corresponding culmulative forces and deflections calculated

In addition, a successive approximation method is described to takeaccount of the effect of force interaction on the evaluation of the col-lapse load of a structure This method can be performed using resultsfrom analysis obtained from most commercially available computerprograms

The successive approximation method is an indirect way toobtain the collapse load of structures using iterative procedures For

direct calculation of the collapse load without using iterative dures, special formulations, possibly with ad-hoc computer program-ming, according to the plastic theory must be used Nowadays, thestiffness method is the most popular and recognized method for struc-tural analysis This book provides a theoretical treatment for deriva-tion of the stiffness matrices for different states of plasticity in anelement for the stiffness method of analysis The theory is based onthe plastic flow rule and the concept of yield surface is introduced.

proce-An introduction to the use of the linear programming techniquefor plastic analysis is provided in a single chapter in this book Thispowerful and advanced method for plastic analysis is described indetail using optimization procedures Its use is important in an auto-mated computational environment and is particularly important forresearchers working in the area of nonlinear structural plastic analysis.This chapter was written by Professor Francis Tin-Loi, a prominentresearcher in the use of mathematical programming methods forplastic analysis of structures

In this book, new insights into various issues related to plasticanalysis and design are given, such as the effect of high temperature

on plastic collapse load and the use of plastic rotation capacity as alimit state for plastic design Based on the elastoplastic approach, aninterpolation procedure is introduced to calculate the design forcesand deflections at the design load level rather than at the collapse loadlevel

In the final chapter of this book, a comparison among designcodes from Australia, Europe, and the United States for plastic designmethod is given This comparison enables practicing engineers tounderstand the issues involved in the plastic design procedures andthe limitations imposed by this design method

Bill Wong

if any, are being used for daily routine design in engineering offices.This may be because of the perception by many engineers that theplastic design method is used only for certain types of usually simplestructures, such as beams and portal frames This perception dis-courages commercial software developers from developing computerprograms for plastic analysis because of their limited applications.Contrary to the traditional thinking that plastic analysis is per-formed either by simple manual methods for simple structures or bysophisticated computer programs written for more general applica-tions, this book intends to introduce general plastic analysis methods,which take advantage of the availability of modern computationaltools, such as linear elastic analysis programs and spreadsheet applica-tions These computational tools are in routine use in most engi-neering design offices nowadays The powerful number-crunchingcapability of these tools enables plastic analysis and design to be per-formed for structures of virtually any size.

The amount of computation required for structural analysis islargely dependent on the degree of statical indeterminacy of the

structure For determinate structures, use of equilibrium conditionsalone will enable the reactions and internal forces to be determined.For indeterminate structures, internal forces are calculated by consid-ering both equilibrium and compatibility conditions, through whichsome methods of structural analysis suitable for computer applica-tions have been developed The use of these methods for analyzingindeterminate structures is usually not simple, and computers areoften used for carrying out these analyses Most structures in practiceare statically indeterminate.

Structural analysis, whether linear or nonlinear, is mostly based

on matrix formulations to handle the enormous amount of numericaldata and computations Matrix formulations are suitable for computerimplementation and can be applied to two major methods of struc-tural analysis: the flexibility (or force) method and the stiffness (or dis-placement) method

The flexibility method is used to solve equilibrium and ibility equations in which the reactions and member forces areformulated as unknown variables In this method, the degree of stat-ical indeterminacy needs to be determined first and a number ofunknown forces are chosen and released so that the remaining struc-ture, called the primary structure, becomes determinate The pri-mary structure under the externally applied loads is analyzed andits displacement is calculated A unit value for each of the chosenreleased forces, called redundant forces, is then applied to the pri-mary structure (without the externally applied loads) so that, fromthe force-displacement relationship, displacements of the structureare calculated The structure with each of the redundant forces iscalled the redundant structure The compatibility conditions based

compat-on the deformaticompat-on between the primary structure and the redundantstructures are used to set up a matrix equation from which theredundant forces can be solved

The solution procedure for the force method requires selection ofthe redundant forces in the original indeterminate structure and thesubsequent establishment of the matrix equation from the compati-bility conditions This procedure is not particularly suitable for com-puter programming and the force method is therefore usually usedonly for simple structures

In contrast, formulation of the matrix equations for the stiffnessmethod is done routinely and the solution procedure is systematic.Therefore, the stiffness method is adopted in most structural analysiscomputer programs The stiffness method is particularly useful forstructures with a high degree of statical indeterminacy, although

it can be used for both determinate and indeterminate structures.The stiffness method is used in the elastoplastic analysis described

in this book and the basis of this method is given in this chapter

In particular, the direct stiffness method, a variant of the general ness method, is described For a brief history of the stiffness method,refer to the review by Samuelsson and Zienkiewicz.5

stiff-1.2 Degrees of Freedom and Indeterminacy

Plastic analysis is used to obtain the behavior of a structure at collapse

As the structure approaches its collapse state when the loads are ing, the structure becomes increasingly flexible in its stiffness Itsflexibility at any stage of loading is related to the degree of statical inde-terminacy, which keeps decreasing as plastic hinges occur with theincreasing loads This section aims to describe a method to distinguishbetween determinate and indeterminate structures by examining thedegrees of freedom of structural frames The number of degrees of free-dom of a structure denotes the independent movements of the structuralmembers at the joints, including the supports Hence, it is an indication

increas-of the size increas-of the structural problem The degrees increas-of freedom increas-of a ture are counted in relation to a reference coordinate system

struc-External loads are applied to a structure causing movements atvarious locations For frames, these locations are usually defined

at the joints for calculation purposes Thus, the maximum number

of independent displacements, including both rotational and tional movements at the joints, is equal to the number of degrees offreedom of the structure To identify the number of degrees of freedom

transla-of a structure, each independent displacement is assigned a number,called the freedom code, in ascending order in the global coordinatesystem of the structure

Figure 1.1 shows a frame with 7 degrees of freedom Note that thepinned joint at C allows the two members BC and CD to rotate indepen-dently, thus giving rise to two freedoms in rotation at the joint

In structural analysis, the degree of statical indeterminacy isimportant, as its value may determine whether the structure

1

7 3

2

4 6 5

D A

FIGURE 1.1 Degrees of freedom of a frame

Structural Analysis—Stiffness Method 3

is globally unstable or stable If the structure is stable, the degree ofstatical indeterminacy is, in general, proportional to the level of com-plexity for solving the structural problem.

The method described here for determining the degree of staticalindeterminacy of a structure is based on that by Rangasami andMallick.6

Only plane frames will be dealt with here, although the methodcan be extended to three-dimensional frames

1.2.1 Degree of Statical Indeterminacy of Frames

For a free member in a plane frame, the number of possible ments is three: horizontal, vertical, and rotational If there are n mem-bers in the structure, the total number of possible displacements,denoted by m, before any displacement restraints are considered, is

For two members connected at a joint, some or all of the cements at the joint are common to the two members and these com-mon displacements are considered restraints In this method fordetermining the degree of statical indeterminacy, every joint is con-sidered as imposing r number of restraints if the number of commondisplacements between the members is r The ground or foundation

displa-is considered as a noncounting member and has no freedom Figure 1.2indicates the value of r for each type of joints or supports in a planeframe

For pinned joints with multiple members, the number of pinnedjoints, p, is counted according to Figure 1.3 For example, for a four-member pinned connection shown in Figure 1.3, a first joint iscounted by considering the connection of two members, a secondjoint by the third member, and so on The total number of pinnedjoints for a four-member connection is therefore equal to three In gen-eral, the number of pinned joints connecting n members is p ¼ n – 1.The same method applies to fixed joints

(e) Rigid ( fixed)

FIGURE 1.2 Restraints of joints

For a connection at a roller support, as in the example shown inFigure 1.4, it can be calculated that p ¼ 2.5 pinned joints and that thetotal number of restraints is r ¼ 5.

The degree of statical indeterminacy, fr, of a structure is mined by

a If fr¼ 0, the frame is stable and statically determinate

b If fr< 0, the frame is stable and statically indeterminate to thedegree fr

c If fr> 0, the frame is unstable

Note that this method does not examine external instability orpartial collapse of the structure

Example 1.1 Determine the degree of statical indeterminacy for thepin-jointed truss shown in Figure 1.5

No of pins, p = 1 No of pins, p = 2 No of pins, p = 3

FIGURE 1.3 Method for joint counting

No of pins, p = 2.5

FIGURE 1.4 Joint counting of a pin with roller support

FIGURE 1.5 Determination of degree of statical indeterminacy in Example 1.1

Structural Analysis—Stiffness Method 5

Solution For the truss in Figure 1.5a, number of membersn ¼ 3; ber of pinned joints p ¼ 4.5.

num-Hence, fr¼ 3  3  2  4:5 ¼ 0 and the truss is a determinatestructure For the truss in Figure 1.5b, number of members n ¼ 2;number of pinned joints p ¼ 3

Hence, fr¼ 3  2  2  3 ¼ 0 and the truss is a determinatestructure

Example 1.2 Determine the degree of statical indeterminacy for theframe with mixed pin and rigid joints shown in Figure 1.6

Solution For this frame, a member is counted as one between twoadjacent joints Number of members ¼ 6; number of rigid (or fixed)joints ¼ 5 Note that the joint between DE and EF is a rigid one,whereas the joint between BE and DEF is a pinned one Number ofpinned joints ¼ 3

Hence, fr¼ 3  6  3  5  2  3 ¼ 3 and the frame is an terminate structure to the degree 3

inde-1.3 Statically Indeterminate Structures—Direct

Stiffness Method

The spring system shown in Figure 1.7 demonstrates the use of thestiffness method in its simplest form The single degree of freedomstructure consists of an object supported by a linear spring obeyingHooke’s law For structural analysis, the weight, F, of the object andthe spring constant (or stiffness), K, are usually known The purpose

AB

C

ED

F

FIGURE 1.6 Determination of degree of statical indeterminacy in Example 1.2

of the structural analysis is to find the vertical displacement, D, andthe internal force in the spring, P.

From Hooke’s law,

F

f gn1 ¼ K½ nnf gD n1 (1.7)where Ff gn1 is the load vector of size n  1ð Þ containing the externalloads, K½ nn is the structure stiffness matrix of size ðn  nÞcorresponding to the spring constant K in a single degree systemshown in Figure 1.7, and Df gn1 is the displacement vector of size

FIGURE 1.7 Load supported by linear spring

Structural Analysis—Stiffness Method 7

The unknown displacement vector can be found by solvingEquation (1.7) as

D

Details of the formation of Ff g, K½ , and Df g are given in the followingsections

1.3.1 Local and Global Coordinate Systems

A framed structure consists of discrete members connected at joints,which may be pinned or rigid In a local coordinate system for a mem-ber connecting two joints i and j, the member forces and thecorresponding displacements are shown in Figure 1.8, where the axialforces are acting along the longitudinal axis of the member and theshear forces are acting perpendicular to its longitudinal axis

In Figure 1.8, Mi,j, yi,j ¼ bending moments and correspondingrotations at ends i, j, respectively; Ni,j, ui,j are axial forces andcorresponding axial deformations at ends i, j, respectively; and Qi,j,

vi,j are shear forces and corresponding transverse displacements atends i, j, respectively The directions of the actions and movementsshown in Figure 1.8 are positive when using the stiffness method

As mentioned in Section 1.2, the freedom codes of a structure areassigned in its global coordinate system An example of a memberforming part of the structure with a set of freedom codes (1, 2, 3, 4,

5, 6) at its ends is shown in Figure 1.9 At either end of the member,the direction in which the member is restrained from movement isassigned a freedom code “zero,” otherwise a nonzero freedom code isassigned The relationship for forces and displacements between localand global coordinate systems will be established in later sections

1.4 Member Stiffness Matrix

The structure stiffness matrix K½  is assembled on the basis of theequilibrium and compatibility conditions between the members For

a general frame, the equilibrium matrix equation of a member is

P

where Pf g is the member force vector, K½  is the member stiffnessematrix, and df g is the member displacement vector, all in the mem-ber’s local coordinate system The elements of the matrices in Equa-tion (1.9) are given as

37775

where the elements of Pf g and df g are shown in Figure 1.8

1.4.1 Derivation of Elements of Member Stiffness Matrix

A member under axial forces Ni and Nj acting at its ends producesaxial displacements ui and uj as shown in Figure 1.10 From thestress-strain relation, it can be shown that

4

5 6

FIGURE 1.9 Freedom codes of a member in a global coordinate system

Structural Analysis—Stiffness Method 9

where E is Young’s modulus, A is cross-sectional area, and L is length

of the member Hence, K11¼ K14¼ K41 ¼ K44¼EAL

For a member with shear forces Qi, Qj and bending moments

Mi, Mj acting at its ends as shown in Figure 1.11, the end ments and rotations are related to the bending moments by theslope-deflection equations as

(1.11b)

Hence, K62¼ K65¼6EIL2, K63¼2EIL , and K66¼4EIL

By taking the moment about end j of the member in Figure 1.11,

Also, by taking the moment about end i of the member, we obtain

1.5.1 Load Transformation

The forces in the global coordinate system shown with superscript “g”

in Figure 1.12 are related to those in the local coordinate system by

Hgi ¼ Nicos a  Qisin a (1.14a)

1.5.2 Displacement Transformation

The displacements in the global coordinate system can be related tothose in the local coordinate system by following the procedure simi-lar to the force transformation The displacements in both coordinatesystems are shown in Figure 1.13

Dg e

1.6 Member Stiffness Matrix in Global Coordinate SystemFrom Equation (1.15),

Fg e

FIGURE 1.13 Displacements in the local and global coordinate systems

Structural Analysis—Stiffness Method 13

An explicit expression for K½  isg

(1.19)where C = cos a; S = sin a

1.7 Assembly of Structure Stiffness Matrix

Consider part of a structure with four externally applied forces, F1, F2,F4, and F5, and two applied moments, M3 and M6, acting at thetwo joints p and q connecting three members A, B, and C as shown inFigure 1.14 The freedom codes at joint p are {1, 2, 3} and at joint q are{4, 5, 6} The structure stiffness matrix [K] is assembled on the basis oftwo conditions: compatibility and equilibrium conditions at the joints

1.7.1 Compatibility Condition

At joint p, the global displacements are D1 (horizontal), D2 (vertical),and D3 (rotational) Similarly, at joint q, the global displacements areD4 (horizontal), D5 (vertical), and D6 (rotational) The compatibilitycondition is that the displacements (D1, D2, and D3) at end p of mem-ber A are the same as those at end p of member B Thus,

ðugjÞA¼ u gi B¼ D1, ðvjgÞA¼ v ig B¼ D2, and ðygjÞA¼ ðygiÞB¼ D3 Thesame condition applies to displacements (D4, D5, and D6) at end q

of both members B and C

The member stiffness matrix in the global coordinate systemgiven in Equation (1.19) can be written as

37775

FIGURE 1.14 Assembly of structure stiffness matrix [K]

Structural Analysis—Stiffness Method 15

F1 ¼ Hgj

Aþ H gi B (1.22a)F2 ¼ Vjg

Aþ V ig B (1.22b)M3 ¼ M gj

Aþ M giB (1.22c)Also, at joint q,

F4 ¼ Hgj

Bþ H gi C (1.22d)F5 ¼ Vjg

Bþ V ig C (1.22e)M6 ¼ M gj

Bþ M giC (1.22f)

By writing Equations (1.22a) to (1.22f) in matrix form using Equations(1.21a) to (1.21l) and applying this operation to the whole structure,the following equilibrium equation of the whole structure is obtained:

D1D2D3D4D5D4

where the “l” stands for matrix coefficients contributed from theother parts of the structure In simple form, Equation (1.23) can bewritten as

F

f g ¼ K½  Df gwhich is identical to Equation (1.7) Equation (1.23) shows how thestructure equilibrium equation is set up in terms of the load vectorF

f g, structure stiffness matrix K½ , and the displacement vector Df g.Close examination of Equation (1.23) reveals that the stiffnesscoefficients of the three members A, B, and C are assembled into K½ 

in a way according to the freedom codes assigned to the members.Take member A as an example By writing the freedom codes in theorder of ends i and j around the member stiffness matrix in the globalcoordinate system shown in Figure 1.15, the coefficient, for example,

k54, is assembled into the position [2, 1] of K½  Similarly, the cient k45 is assembled into the position [1, 2] of K½  The coefficients

coeffi-in all member stiffness matrices coeffi-in the global coordcoeffi-inate system can

be assembled into K½  in this way Since the resulting matrix is metric, only half of the coefficients need to be assembled

sym-A schematic diagram showing the assembly procedure for thestiffness coefficients of the three members A, B, and C into K½  isshown in Figure 1.16 Note that since K½  is symmetric, Kg ½  is alsosymmetric Any coefficients in a row or column corresponding to zerofreedom code will be ignored

1.8 Load Vector

The load vector Ff g of a structure is formed by assembling the ual forces into the load vector in positions corresponding to the direc-tions of the freedom codes For the example in Figure 1.14, the loadfactor is given as that shown in Figure 1.17

1 2 3

53

FIGURE 1.15 Assembly of stiffness coefficients into the structure stiffnessmatrix

1.9 Methods of Solution

The displacements of the structure can be found by solving Equation(1.23) Because of the huge size of the matrix equation usually encoun-tered in practice, Equation (1.23) is solved routinely by numericalmethods such as the Gaussian elimination method and the iterativeGauss–Seidel method It should be noted that in using these

F

6 5 4 3 2 1

F F F F F F

Freedom codes 1

2 3 4 5 6

FIGURE 1.17 Assembly of load vector

1 2 3 4 5 6

1 2 3 4 5 6

[K g] of member A

[K g] of member C

[K g] of member B

[K ] of structure

FIGURE 1.16 Assembly of structure stiffness matrix

Structural Analysis—Stiffness Method 19

numerical methods, the procedure is analogous to inverting the ture stiffness matrix, which is subsequently multiplied by the loadvector as in Equation (1.8):

struc-D

The numerical procedure fails only if an inverted K½  cannot befound This situation occurs when the determinant of K½  is zero,implying an unstable structure Unstable structures with a degree ofstatically indeterminacy, fr, greater than zero (see Section 1.2) willhave a zero determinant of K½  In numerical manipulation by compu-ters, an exact zero is sometimes difficult to obtain In such cases, agood indication of an unstable structure is to examine the displace-ment vector Df g, which would include some exceptionally largevalues

1.10 Calculation of Member Forces

Member forces are calculated according to Equation (1.9) Hence,

P

f g ¼ K½  def g

¼ K½  Te ½ tf gDg (1.24)where Df g is extracted from Dg f g for each member according to itsfreedom codes and

Ke

½  T½ t¼

S12EIL3 C12EIL3 6EIL2 S12EIL3 C12EIL3 6EIL2

S6EIL2 C6EIL2 4EIL S6EIL2 C6EIL2 2EIL

For the example in Figure 1.14,

2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4

3 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5

displace-2 Assign an arrow to each member so that ends i and j aredefined Also, the angle of orientation a for the member isdefined in Figure 1.18 as:

3 Assemble the structure stiffness matrix K½  from each of themember stiffness matrices

4 Form the load vector Ff g of the structure

i j

FIGURE 1.18 Definition of angle of orientation for member

Structural Analysis—Stiffness Method 21

5 Calculate the displacement vector f g by solving forDD

f g ¼ K½ 1f g.F

6 Extract the local displacement vector Df g from Dg f g and culate the member force vector Pf g using Pf g ¼ K½  Te½ tf g.Dge1.10.1 Sign Convention for Member Force Diagrams

cal-Positive member forces and displacements obtained from the stiffnessmethod of analysis are shown in Figure 1.19 To plot the forces in con-ventional axial force, shear force, and bending moment diagrams, it isnecessary to translate them into a system commonly adopted forplotting

The sign convention for such a system is given as follows.Axial Force

For a member under compression, the axial force at end i is positive(from analysis) and at end j is negative (from analysis), as shown inFigure 1.20

Shear Force

A shear force plotted positive in diagram is acting upward (positivefrom analysis) at end i and downward (negative from analysis) atend j as shown in Figure 1.21 Positive shear force is usually plotted

in the space above the member

FIGURE 1.19 Direction of positive forces and displacements using stiffnessmethod

Compressive

FIGURE 1.20 Member under compression

Bending Moment

A member under sagging moment is positive in diagram (clockwiseand negative from analysis) at end i and positive (anticlockwise and pos-itive from analysis) at end j as shown in Figure 1.22 Positive bendingmoment is usually plotted in the space beneath the member In doing

so, a bending moment is plotted on the tension face of the member

Example 1.3 Determine the member forces and plot the shear forceand bending moment diagrams for the structure shown inFigure 1.23a The structure with a pin at D is subject to a vertical force

of 100 kN being applied at C For all members, E ¼ 2  108 kN/m2,

6 5

8 1

3 2

9 10

A

C

E 0

0 0

0 0 0

indicate end i (tail of arrow) and end j (head of arrow) Thus, the tations of the members are

0 0 0 1 2 3

123456

4 5 6 7 8 9

7 8 10 0 0 0

By assembling from K½  of all members, the structure stiffnessegmatrix is obtained:

The load vector is given by

The member forces can be calculated using Pf g ¼ K½  Te½ tf g.Dge

For member 1, where C ¼ cos 90¼ 0 and S ¼ sin 90¼ 1,

S12EIL3 C12EIL3 6EI

L 2 S12EIL3 C12EIL3 6EI

L 2

S6EIL2 C6EIL2 4EIL S6EIL2 C6EIL2 2EIL

CEAL SEAL 0 CEAL SEAL 0

S12EIL3 C12EIL3 6EIL2 S12EIL3 C12EIL3 6EIL2

S6EIL2 C6EIL2 2EIL S6EIL2 C6EIL2 4EIL

0 0 0 1:354  10 3

73:9 6:5

; and Pf g4¼

26:16:50

26:1

6:532:5

The shear force and bending moment diagrams are shown inFigure 1.24.

1.11 Treatment of Internal Loads

So far, the discussion has concerned externally applied loads actingonly at joints of the structure However, in many instances, externallyapplied loads are also applied at locations other than the joints, such

as on part or whole of a member Loads being applied in this mannerare termed internal loads Internal loads may include distributedloads, point loads, and loads due to temperature effects In such cases,the loads are calculated by treating the member as fixed-end, andfixed-end forces, including axial forces, shear forces, and bendingmoments, are calculated at its ends The fictitiously fixed ends ofthe member are then removed and the effects of the fixed-end forces,now being treated as applied loads at the joints, are assessed usingthe stiffness method of analysis

In Figure 1.25, fixed-end forces due to the point load and the formly distributed load are collected in a fixed-end force vector Pf gF

uni-for the member as

6.5 6.5 73.

Shear force diagram

Bending moment diagram

FIGURE 1.24 Shear force and bending moment diagrams

Structural Analysis—Stiffness Method 27

MEi, QEj, MEj, being applied at the joints pertaining to both ends iand j of the member The equivalent force vector is expressed as

P

f g ¼ K½  de f g þ Pf gF (1.28)Fixed-end forces for two common loading cases are shown inTable 1.1

Example 1.4 Determine the forces in the members and plot the ing moment and shear force diagrams for the frame shown inFigure 1.26a The structure is fixed at A and pinned on a roller support

Bending moment at end i, MFi MFi¼12Lw2hðL  aÞ3ðL þ 3aÞ  c3ð4L  3cÞiBending moment at end j, MFj MFj¼ 12Lw2hðL  cÞ3ðL þ 3cÞ  a3ð4L  3aÞiP

Shear force at end j, QFj QFj¼ P  QFi

Fj¼PabL22Bending moment at end j, MFj MFj¼PaL22b

... under sagging moment is positive in diagram (clockwiseand negative from analysis) at end i and positive (anticlockwise and pos-itive from analysis) at end j as shown in Figure 1.22 Positive bendingmoment... fixed-end, andfixed-end forces, including axial forces, shear forces, and bendingmoments, are calculated at its ends The fictitiously fixed ends ofthe member are then removed and the effects of the...

4 Form the load vector Ff g of the structure

i j

FIGURE 1.18 Definition of angle of orientation for member

Structural Analysis? ??Stiffness Method 21