This text book covers the principles and methods of load effect calculations that are necessary for engineers and design
1,198 314 23MB
English Pages XVI, 354 [368] Year 2020
Table of contents :
Front Matter ....Pages i-xvi
Introduction to the Theory of Forces (Einar N. Strømmen)....Pages 1-26
Statically Determinate Systems (Einar N. Strømmen)....Pages 27-44
The Theory of Stress and Strain (Einar N. Strømmen)....Pages 45-86
Design Criteria (Einar N. Strømmen)....Pages 87-96
Deformations of Beams, Trusses and Frames (Einar N. Strømmen)....Pages 97-110
Statically Indeterminate Systems (Einar N. Strømmen)....Pages 111-148
Stresses in Composite Beams (Einar N. Strømmen)....Pages 149-156
Non-Symmetric Beam Cross Sections (Einar N. Strømmen)....Pages 157-168
Some Special Structural Systems (Einar N. Strømmen)....Pages 169-192
The Theory of Torsion (Einar N. Strømmen)....Pages 193-228
Bending of Plates (Einar N. Strømmen)....Pages 229-246
Elastic Buckling (Einar N. Strømmen)....Pages 247-304
Back Matter ....Pages 305-354
Einar N. Strømmen
Structural Mechanics The Theory of Structural Mechanics for Civil, Structural and Mechanical Engineers
Structural Mechanics
Einar N. Strømmen
Structural Mechanics The Theory of Structural Mechanics for Civil, Structural and Mechanical Engineers
123
Einar N. Strømmen Trondheim, Norway
ISBN 978-3-030-44317-7 ISBN 978-3-030-44318-4 https://doi.org/10.1007/978-3-030-44318-4
(eBook)
© Springer Nature Switzerland AG 2020 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland
In loving memory of Alf & Signe
Preface The theory of structural mechanics enables engineers and designers to evaluate the safety of structural systems. In doing so, the process starts with establishing the relevant external loads that the system may be subject to in the environment it is intended to function during its lifetime. Then, it is necessary to quantify the action effects these loads may inflict upon the system, first by the identification of possible overall or local stability limits, such that the safety against collapse may be evaluated, after which, it is necessary to investigate material straining at all critical points in the system. It is the designer’s duty to find the most onerous point of material straining and compare this to material strength. Thus, the theory of structural mechanics is about seeing the system from an external environmental point of view and moving in the direction of a closer and closer look at how the system in detail responds to whatever burden it may be subject to. It is also the designer’s duty to limit structural displacements; such that the system may fulfil its intended functions. This text book is intended for studies of the principles and methods of load effect calculations and corresponding strength considerations, with focus on civil engineering structures that may be described by line-like beam or beam-columns, or by systems of rectangular or circular plates. It covers the mathematical development from basic assumptions to the final equations ready for practical use. It has been my intention to start off at a basic level and step by step bring the reader up to a level where the necessary design safety considerations to static load effects can be performed, i.e. to a level where displacements and cross sectional forces with corresponding stresses can be calculated and compared to the strength of the system. Relevant load descriptions are only included in so far as it has been necessary for the performance of illustrative examples. For more comprehensive descriptions of environmental loads the reader should consult the relevant national or international standards, e.g. ISO or Eurocode. It is taken for granted that the reader possess good knowledge of calculus, differential equations and basic matrix operations. The finite element method for line-like systems has been covered, but not the finite element method for shells and plates. This may readily be found elsewhere. The writing of this book has been supported by The Structural Engineering Department at NTNU. Anne Gaarden has prepared the illustrations; her contribution is greatly appreciated. Trondheim, February 2019
Einar N. Strømmen [email protected]
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CONTENTS Preface Notation
vii xii
1 INTRODUCTION TO THE THEORY OF FORCES 1.1 1.2 1.3 1.4 1.5 1.6 1.7
Basic definitions The equilibrium conditions in a two-dimensional format The free-body-diagram Friction force The modelling of structural systems The principle of super-position The equilibrium conditions in a three-dimensional format
1 5 8 12 13 17 20
2 STATICALLY DETERMINATE SYSTEMS 2.1 2.2 2.3
The classification of structural systems External and internal equilibrium of line-like systems External and internal equilibrium of trusses
27 30 41
3 THE THEORY OF STRESS AND STRAIN 3.1 3.2 3.3 3.4 3.5 3.6
Introduction to the theory of elasticity The effects of thermal variations Stress components from axial force and bending Stress components from shear forces Decomposition of stress components, principal stresses Decomposition of strain components, principal strains
45 52 54 63 73 80
4 DESIGN CRITERIA 4.1 4.2 4.3
Basic material properties Tresca and von Mises yield criteria Safety considerations
87 89 93
5 DEFORMATIONS OF BEAMS, TRUSSES AND FRAMES 5.1 5.2 5.3
Deformations due to axial force The differential equation of beam bending The use of virtual unit force
97 98 101
ix
x
CONTENTS
6 STATICALLY INDETERMINATE SYSTEMS 6.1 6.2 6.3 6.4 6.5
Introduction The use of deformation compatibility for simple trusses The use of the superposition principle for simple beams The matrix method for simple frames Background to the finite element method
111 111 115 118 132
7 STRESSES IN COMPOSITE BEAMS 7.1 7.2 7.3
Introduction Normal stresses in composite beams Shear stresses in composite beams
149 150 152
8 NON-SYMMETRIC BEAM CROSS SECTIONS 8.1 8.2
Main axes and normal stress calculations The calculation of shear stresses
157 164
9 SOME SPECIAL STRUCTURAL SYSTEMS 9.1 9.2 9.3 9.4
The theory of simple arches The shallow cable theory Beams on elastic foundation Cylindrical shells
169 175 182 186
10 THE THEORY OF TORSION 10.1 10.2 10.3 10.4 10.5 10.6 10.7
Introduction Torsion of thin-walled tube or compact circular section St. Venant torsion of compact cross section Single cell thin walled hollow cross section Multiple cells thin walled hollow cross section Warping torsion The differential equations of beam-columns
193 195 197 203 206 210 224
11 BENDING OF PLATES 11.1 Introduction to the theory of plate bending 11.1 The differential equation of rectangular plates 11.3 The differential equation of circular plates
229 235 242
12 ELASTIC BUCKLING 12.1 Columns 12.2 Beams
249 260
CONTENTS
12.3 12.4 12.5 12.6
Beam-columns Frames Rectangular plates The numeric eigenvalue problem
xi 269 277 285 298
Appendix A: BACKGROUND TO THE ENERGY METHODS A.1 A.2 A.3 A.4 A.5 A.6 A.7
The concept of energy conservation Strain energy in beam-columns and plates The principle of virtual work The principle of virtual displacements The principle of virtual forces Rayleigh-Ritz method Galerkin’s principle of weighted residuals
305 308 315 319 323 326 328
Appendix B: SOLUTION TO SYSTEMS OF LINEAR EQUATIONS B.1 Matrix inversion by Gauss-Jordan elimination B.2 Solution by Cholesky decomposition
331 333
Appendix C: COLLECTION OF IMPORTANT FORMULAS C.1 C.2 C.3
Definitions and formulas in structural mechanics Some useful mathematical formulas Basic matrix operations
References Index
335 343 346 349 351
NOTATION Matrices and vectors: Matrices are in general bold upper case Latin or Greek letters, e.g. K . Vectors are in general bold lower case Latin or Greek letters, e.g. q . diag[] is a diagonal matrix whose content is written within the bracket.
det is the determinant of the matrix within the bracket. Superscripts and bars above symbols: Super-script T is the transposed of a vector or a matrix. Primes on a variable is its derivative, e.g. y dy dx , y d 2 y dx 2 …. Hat ( ) on a symbol (e.g. iˆ ) is a non-dimensional quantity. Bar on a letter (e.g. r ) is its average value. A curved bar on a letter (e.g. r ) mean a chosen approximate vector/function. ) is a virtual quantity. or M Tilde on a letter (e.g. F , w Abbreviations:
CC is short for centre of cross-sectional neutral axis bending. SC is short for shear centre. tot is short for total. max, min are short for maximum and minimum. Integrals:
or
L
means integration over the length L or the area A .
A
means integration along joint walls between cells i and j .
means integration along closed circumferential curve R .
ij
R
Latin letters:
A, A0 Am , As
Area, cross sectional area Cross sectional cell area, outer slice of shear area xii
NOTATION
A, An a, a0 , ai a B , B b, b0 , bi b c, ci , Ci c D d di , d E e, en , e p e y , ez ej F , Fi , F f f f y , fu
Connectivity matrix associated with element n Distance, coefficient, vector component, amplitude Coefficient vector Cross sectional width, warping bi-moment Distance, width, coefficient, vector component Coefficient vector Distance, coefficient, i 1,2, , distance or length Coefficient vector Cross sectional depth, plate stiffness, distance Length, coefficient Element degrees of freedom, degrees of freedom vector Modulus of elasticity Distance, length, eccentricity, n 1,, N , polar radius Shear centre coordinates Unit vector in direction of j x, y orz Force, force in direction of j x, y orz , force vector Function of variable within brackets Arch height or cable sag Yield stress, ultimate stress Shear modulus of elasticity G g Gravity constant H Friction force, horizontal cable force component, height h H eight, distance I i , Iij Second area moment, i, j y orz It , I w St. Venant and warping torsion constants I Identity matrix i, j Index variables im , ik Cross sectional asymmetry parameter, buckling radial parameter i y , iz , i Cross sectional radial parameters K, K j Stiffness, stiffness coefficient K, K G Stiffness matrix, geometric stiffness Index variable, node or sample number k k x , k y , k xy Buckling coefficients k, k G , k 0 Element, geometric and cross sectional stiffness matrices
xiii
xiv
NOTATION
L, L0 , Ln Length (of structural system), n 1,2, Lx , Ly Length and width of rectangular plate Lk Buckling length Cable length M , Mi Bending or cross sectional bending moment, i, j x, y orz M r , M , M r Bending and torsion moments in polar coordinates M E , M E y Lateral torsional elastic buckling moment m Index variable Normal force, normal force in i x, y orz directions N , Nj N E , N Ei Elastic stability limit load i y, z or Nk Ultimate stability limit Element number, unit length vector, element axial load matrix n, n Arbitrary point, external load energy P p Node number, inner pressure p Probability of occurrence q, qz Distributed force, distributed force in z direction q Shear flow qE Elastic (Euler) distributed stability limit R Radius, electric resistance Rn Reaction force n A, Rp ,R Nodal load p 1,2, , load vector r Radial distance ri , rp , r Displacement degree of freedom i, p 1,2, , displacement vector r Chosen displacement vector S y , S z , S The static moment of As about y or z axis, warping static moment Sn Axial force in truss n 1, 2, s Sector coordinate T , T Temperature (in Centigrade), stress period T Transformation matrix t Thickness, time Strain energy stored in the material fibres of a system U u , u0 Cross sectional displacement in x direction Shear force, voltage V Vi Cross sectional shear forces, i y orz
NOTATION
Vr ,V r v , v0 w , w0 x, y , z
Shear forces in polar coordinates Cross sectional displacement in y direction Cross sectional displacement in z direction Cartesian structural element main axis (with origo in neutral bending centroid, x in span-wise direction and z vertical)
Greek letters:
, z , xs
xv
Coefficient, angle Angle, coefficient, safety coefficient Shear strain, Displacement Virtual displacement , x 0 Shape imperfection Derivative operator , ε , j Strain, strain vector, strain component ( j x, y or z ) y , u Yield strain, ultimate strain Cross sectional rotation Coefficient Coefficient, wave length, eigenvalue , k k Non-dimensional buckling length parameter , s , k Gauge factor, coefficient, friction coefficients , Poisson’s ratio, coefficient Total energy , i Density, density of component number j , i Normal stress, normal stress component j x, y orz E , k Elastic buckling stress, ultimate buckling stress , ij Shear stress, shear stress component i, j x, y orz y , u Yield stress, ultimate stress , Angle or angle of force direction, Prandtl stress function Angle, shape function , x Shape function matrix ψ Warping sector parameter ,
xvi
NOTATION
Combinations of Greek and Latin letters:
B , H L, R , T u, w, r , d, r
Dimensional changes Elongation, cable elongation Change of electric resistance, Change of temperature (in Centigrade) Incremental cross sectional displacement
Incremental displacement, incremental displacement vectors Change of energy
Chapter 1 Equation Chapter 1 Section 1
INTRODUCTION TO THE THEORY OF FORCES 1.1 Basic definitions In addition to the gravity load of itself and loads from human activity, the built environment is subject to a variety of force effects, such as
wind pressure (proportional to the wind velocity squared), the gravity and pressure effects of snow (dependent of depth and density), water pressure, currents and waves,
all dependent on location, position and shape of the system. These load effects are usually regulated by law, i.e. they may be taken from public standard documents, and they are most often presented in the form of equivalent static forces, although there is no environmental load that is static, they all to some extent vary with time. Unless otherwise stated, we shall in the following limit ourselves to nominally static forces acting on systems that may be regarded as two-dimensional, e.g. as illustrated in Fig. 1.1. I.e., it is taken for granted that a two-dimensional body has a unit extension in the out-of-plane direction. The effects of dynamic forces, such as wind turbulence, ground motion due to earthquake, land and snow slides are not included (see e.g. Strømmen [18]). The basic concept of force comes from Isaac Newton’s idea [2] that the gravity field may be regarded as a force between two bodies. Based on this simple hypothesis and the concept of body equilibrium he formulated three basic physical laws, which have now survived more than three hundred years: 1. If no net force is acting on a body it will remain at rest or, if already in motion, it will continue in a straight line at constant speed. 2. The acceleration ( a ) of a body is proportional to the net force ( F ) acting on the body, and inversely proportional to its mass ( m ).
© Springer Nature Switzerland AG 2020 E. N. Strømmen, Structural Mechanics, https://doi.org/10.1007/978-3-030-44318-4_1
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1 INTRODUCTION TO THE THEORY OF FORCES
2
3. The force acting from one body on another is equal to and opposite of the force acting from the second body to the first. The first law is called the law of inertia, the second is the law of motion and the third is the law of reciprocal force effects. It is the second law F ma
(1.1)
that has provided a convenient definition of force ( F ), carrying Newton’s name. Thus, adopting the SI units of mass (kg), length (meter) and time (second), the force unit of 1 N is defined as the force that is required to give a mass of 1 kg the acceleration of 1 m/s2. Thus, basic units (as defined by ISO [22]) are as follows:
displacement: meter (m ) time: second ( s ) mass: kilogram ( kg )
force:
Newton ( N kg m s 2 ).
Fig. 1.1 Free-body motion It should be noted that if the net force is acting through the mass center of the body (as shown in Fig. 1.1), then the force and the acceleration will take place in a straight line in the same direction. If the force acts outside of the mass center the motion will also contain a body rotation. The mass center C of a body is defined by (see Fig. 1.1)
1.1 BASIC DEFINITIONS
rdA 0
3
(1.2)
A
where dA is an infinitesimal element at distance r from the mass center of the body and is its density ( kg m2 ). If the body is homogeneous then is constant and the mass center will coincide with area center as defined by
rdA 0
(1.3)
A
In two-dimensional cases the system may be described in Cartesian coordinates x, z , in which case Eq. 1.2 may be replaced by
x
0
z dA 0
(1.4)
A
Thus, the mass center may readily be determined by defining an arbitrary point of integration C * , whose distance from C is x0 , z0 and whose distance from
the element dA is x * , z * , see Fig. 1.1. Then x x* x0 and z z* z0 , and thus the position of C is defined by
x
x* x0 x* x 0 dA dA 0 dA * 0 z0 A z z0 A z *
z dA A
rendering
A
x* x0 1 z dA 0 dA A z *
(1.5)
(1.6)
A
which, if is constant across the body, reduces into x0 1 x* z A dA 0 A z *
(1.7)
For a straight line motion (without any body rotation) it is not a requirement that forces (e.g. F1 and F2 in Fig. 1.1) are acting at the mass center. The only requirement is that they point through the mass center, because, as illustrated in Fig. 1.2, a force will render the same rectilinear motion if it is acting on the front face of the body or on its back face, as long as they both work through the mass
1 INTRODUCTION TO THE THEORY OF FORCES
4
center. We call such forces collinear. If the force is acting outside of the mass center (as illustrated by the force acting on the body at the right hand side of Fig. 1.2) it will not only cause acceleration in the direction of the force but it will also cause an accelerating rotation of the body. Thus, a force is a vector defined by its magnitude, its position and its direction.
Fig. 1.2 Collinear and non-rectilinear forces acting on a rigid body
Fig. 1.3 The summation of collinear forces
a) Axis and force vector definitions Fig. 1.4
b) The resultant of two vectors
Basic definitions of force vectors and vector addition
1.2 THE EQUILIBRIUM CONDITIONS IN ATWO-DIMENSIONAL FORMAT
5
1.2 The equilibrium conditions in a two-dimensional format In the following the equilibrium conditions of force and moment vectors in a two-dimensional format is presented. The corresponding condition in a threedimensional format is presented in Chapter 1.7. The effect of two collinear forces is additive, and thus, the two forces F1 and F2 in Fig. 1.3 may be replaced by F F1 F2 . The combined effects of the two non-collinear forces F1 and F2 in Fig. 1.4 are also additive, but only in a vector description. Assume
that the motion of a body is the result of only a single force, and that the force is acting through the mass center of the body. Then the motion of the body will take place in the direction of this force. If the body is subject to several noncollinear forces, all acting at or through the mass center of the body (e.g. as illustrated in Fig. 1.4), then the body will still move in one singular direction, which inevitably means that these forces have a common resultant in that particular direction. I.e., several forces acting through the mass center of a body will have a single force resultant that is the vector sum of all the other forces. Let us consider the two forces F1 and F2 in Fig. 1.4 in Cartesian axis x and z :
F1x F1 F1z
and
F2 x F2 F2 z
(1.8)
Assuming that these two forces both act at the same point, they may be replaced by a single force resultant F which is the vector sum of the two forces:
F1x F2 x F1x F2 x F F1 F2 F1z F2 z F1z F2 z
(1.9)
Its effect is identical to the effect of two forces acting together on the body. The vector summation may be performed as shown in Eq. 1.9, or geometrically as illustrated in Fig. 1.4.b, the resultant is the diagonal in a parallelogram with adjacent sides F1 and F2 . Vector summation of forces is convenient in many cases, but in some cases the opposite operation may also be convenient. The decomposition of forces into other directions may render the solution to a problem far more obvious, as illustrated by the rectangular body sliding along a tilting surface in Fig. 1.5, where the gravity force mg is replaced by a force mg cos perpendicular to the surface and mg sin parallel to the surface.
Clearly, mg sin is the force driving the motion along the surface, while
1 INTRODUCTION TO THE THEORY OF FORCES
6
friction and mg cos are the stabilizing forces, and thus, a friction force larger than mg sin will be required if the body is to be at rest.
Fig. 1.5
Fig. 1.6
Rectangular body sliding on a tilting surface
Definition of moments, M (with unit Nm )
In addition to vector forces, a body may also be subject to rotational or twisting moments. As illustrated in Fig. 1.6, the moment M (with unit Nm ) is equivalent to two equal forces F acting in opposite directions at a distance c , such that M Fc .The effect of moments may be illustrated by the case of a rigid rectangular body (see Fig. 1.7), resting on a surface with infinite friction, and subject to two forces F1 and F2 at distances c1 and c2 above the surface. The assumption of infinite friction implies that the body is at rest and restricted from a sliding motion in the direction of F1 and F2 , a condition that is fulfilled if the friction force F between the body and the horizontal surface is equal to F1 F2 . We say that the body is in a state of equilibrium if the sum of all forces
acting in the horizontal x -direction is equal to zero, i.e. if:
1.2 THE EQUILIBRIUM CONDITIONS IN ATWO-DIMENSIONAL FORMAT
Fx 0
F1 F2 F 0
F F1 F2
7 (1.10)
But, what can be seen (Fig. 1.7) is that there is also a the possibility the body may tilt due to moment effects F1c1 and F2 c2 . The force effect that can prohibit tilting is the weight mg of the body, such that if no tilting occurs then the total moment effect with respect to the tilting point must be equal to zero, i.e. that
M 0
F1c1 F2 c2 mg b 2 0
(1.11)
rendering the non-tilting condition that
mg b
Fig. 1.7
2 F1c1 F2 c2
(1.12)
Tilting body
Thus, a body is at rest if it is in a condition of force and moment equilibrium, i.e. if the sum of all forces acting on the system in any direction is equal to zero, and if the sum of all moments acting about an arbitrary point P is also equal to zero. Or, as mathematically expressed in a Cartesian equilibrium (and at rest) if
x,z
system: a body is in
8
1 INTRODUCTION TO THE THEORY OF FORCES
Fx 0 Fz 0
and
MP 0
(1.13)
where P is an arbitrary chosen point of moment summation. (To convince yourself that P is arbitrary, try another point, e.g. somewhere else on the body, first acknowledging that Fx 0 and Fz 0 .)
Fig. 1.8
Merging force and parallel forces
While the combined effects of the two non-collinear forces F1 and F2 in Fig. 1.4 are vector additive, it is worth noting that the two parallel forces F1 and F2 in Fig. 1.7 are also additive in the sense that their joint horizontal and tilting effects may be replaced by a single force F F1 F2 acting at a height c F1c1 F2 c2 F1 F2 , simply because
Fc F1c1 F2 c2 .
Thus, we
distinguish between forces that have a common point of attack and forces that are parallel, as illustrated in Fig. 1.8.
1.3 The free-body-diagram As shown in Eq. 1.13, we choose to express the two-dimensional equilibrium conditions in a Cartesian coordinate system x , z . In doing so it is convenient to commence the equilibrium considerations by establishing a free-body-diagram of the system. Let us consider a simple mass m suspended from two points A and B by wires which are interlinked at point C, as illustrated in Fig. 1.9.a.
1.3 THE FREE-BODY-DIAGRAM
Fig. 1.9
9
Equilibrium of a suspended mass m
Fig. 1.10 Force equilibrium
With respect to bending the wires are flexible such that they can only sustain stretching forces. Thus, the gravity force mg exerted to the system is supported by wire forces F1 and F2 . The application of Eq. 1.13 is then most conveniently applied to the free-body-diagram as illustrated in Fig. 1.9.b, where supports at A and B have been replaced by the relevant force components F1 and F2 . Then it is easily seen there are two equilibrium requirements; one that the horizontal
1 INTRODUCTION TO THE THEORY OF FORCES
10
component of F1 is equal to the horizontal component of F2 , such that there is no net force acting in the horizontal direction on the body (i.e. that
Fx 0 ).
Second, that the joint resultant of F1 and F2 in the vertical direction is equal to the gravity force of the body mg (i.e. that
Fz 0 ), see Fig. 1.10. Applying
Eq. 1.13, this may be expressed by:
Fx 0 Fz 0 MC 0
F1 sin 1 F2 sin 2 0
(1.14)
F1 cos 1 F2 co 2 mg 0
(1.15) (1.16)
In this particular case it is convenient to choose moment equilibrium about C because all forces merge at this point, and thus M C 0 is automatically fulfilled. Since sin 1 c1 L1 , sin 2 c2 L2 , cos 1 d L1 , cos 2 d L2 , L1 c12 d 2 and L2 c22 d 2 , Eqs. 1.14 and 1.15 may conveniently be
expressed in a matrix format: c1 L1 d L 1
c2 L2 F1 0 d L2 F2 mg
(1.17)
from which the following is obtained:
F1 c2 L1 mg F d c c c L 1 2 1 2 2
(1.18)
A body at rest (or at a constant velocity) we call a static system. A static system is in a force condition of static equilibrium. I.e., in a free body condition, where all reaction forces are included, there are no net forces acting on the system. The reaction forces are in equilibrium with external forces. This equilibrium of forces may always be expressed similar to that of the example above. Let us consider a see-saw (Fig. 1.11) subject to gravity forces m1 g and m2 g at eccentricities e1 and e2 . Obviously, equilibrium in the horizontal ( x ) direction is obsolete, as there are no forces acting in this direction. Equilibrium in the vertical ( z ) direction is obtained by (see free-body-diagram in Fig. 1.11.b)
Fz 0
F m1 g m2 g 0
F m1 m2 g
(1.19)
1.3 THE FREE-BODY-DIAGRAM
11
However, this equilibrium condition will only secure no motion in the vertical z direction. To prevent the bar to be set into rotation about the center point C, it will be necessary to require moment equilibrium
MC 0
m1 g e1 m2 g e2 0 e1 e2 m2 m1
(1.20)
The sign convention in the moment equilibrium condition above may perhaps be best understood by regarding the two moment effects M 1 m1 g e1 and M 2 m2 g e2 as vectors in an x, y , z system as illustrated in Fig. 1.12.
Fig. 1.11
The equilibrium conditions of a see-saw
Fig. 1.12 Moment vectors (identified by double arrow)
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1 INTRODUCTION TO THE THEORY OF FORCES
1.4 Friction force
Fig. 1.13
Two bodies joined by friction
Two bodies held together by friction is illustrated in Fig. 1.13. Each of the two bodies is subject to a pulling force N . What is characteristic to the friction force H between the two bodies (see Fig. 1.14) is that it is a direct reaction force to N and it is proportional to the force F holding the two bodies together. Thus, it is an illustrative example on how the equilibrium condition may be understood. Its magnitude is determined by a proportionality factor called the friction coefficient between H and F , i.e. H F . If the two bodies are at rest, then
s , which is what we call the static friction coefficient. If the two bodies are in a state of motion along their common friction surface, then k , which is what we call a kinematic (or dynamic) friction coefficient, see Fig. 1.15. Thus, if the two bodies are at rest, then N s F
(1.21)
If N H s F then the two bodies will start moving along the friction surface, and as long as N k F the bodies will move at an increasing velocity (dependent of the mass of each of the two bodies). For instance, the body in Fig. 1.5 sliding along a tilting surface is at rest if mg sin s mg cos
if: tan s
(1.22)
1.5 THE MODELLING OF STRUCTURAL SYSTEMS
Fig. 1.14
Fig. 1.15
13
The friction force H F
Static and kinematic friction coefficients
1.5 The modelling of structural systems The most basic structural system is a bar or a beam, which is a structure or a structural member (element) whose length in relation to its cross sectional dimensions is such that the system may for all practical purposes be regarded as line-like, i.e. it may be regarded as a body with the length as its single dimension. We shall begin with considering its internal load effects due to axial and transverse load (e.g. its distributed gravity load) as described in coordinates x, y, z , see Fig. 1.16, and later we shall also include its elastic deformations. The load effects of such a system may best be illustrated by a single beam with
1 INTRODUCTION TO THE THEORY OF FORCES
14
length L resting on two wheels, as illustrated in Fig. 1.17. If the system is symmetric and subject to its gravity load alone, q gA (with unit N m ), where is its density and A is the area of its cross section, then the system is in equilibrium if vertical support reactions are equal and given by qL 2 .
Fig. 1.16
Fig. 1.17
The beam, a line-like structural element
Single beam with length L resting symmetrically on two wheels
Fig. 1.18
Typical cross section of line-like beams or columns
1.5 THE MODELLING OF STRUCTURAL SYSTEMS
Fig. 1.19
15
Structural systems built up by line-like elements
The idea with a single beam is that it either spans a length L or it is embedded at one end and cantilevered a certain length L . Some typical cross sections often chosen for beam or column type of elements are shown in Fig. 1.18. We distinguish between open and closed cross sections and also between symmetric and non-symmetric cross sections. Some structural systems built up by line-like elements are illustrated in Fig. 1.19. While most structural systems may be
1 INTRODUCTION TO THE THEORY OF FORCES
16
subdivided into beams and columns between joints, the arch will need some special treatment simply because it is curved, and so will be the case for any system of suspended cables. The cable element (the catenary) has the special property that it can only sustain stretching. Depending on its position in a structural system and on its structural purpose, a beam or a column may at either end have a point of bearing that is
fixed from in-plane motion, but free to rotate, free to in-plane motion as well free to rotate, or fixed from in-plane motion as well as fixed from any rotation,
as illustrated in Fig. 1.20. The support conditions for the simply supported beam and truss or lattice type of girders as illustrated in Fig. 1.19 have been chosen because they will render the system free from any constraining forces due to
sagging from gravity or external forces, temperature variations, differential settlement of foundations, or unwanted internal forces due to unavoidable overall deformations.
Fig. 1.20
Typical support conditions for structural systems
1.6 THE PRINCIPLE OF SUPER-POSITION
17
Fig. 1.21 Modelling of internal connections between beams and columns
1.6 The principle of super-position We shall take it for granted that equilibrium refers to un-deformed geometry and linear material behaviour; then any effect of two or more forces is additive, e.g. the reaction forces due to external forces F1 and F2 together is equal to the reaction forces due to F1 alone plus the reaction forces due to F2 alone, the structural displacements due to external forces F1 and F2 together is equal to the deformations due to F1 alone plus the deformations due to F2 alone. This is what we call the principle of super-position. At this stage it is best illustrated by the simple beam on two wheels, and let us now assume that it is subject to its own distributed gravity load q gA plus the gravity load F mg from a separate body at position a and b from either of the wheels, as illustrated in Fig. 1.22. At the upper right hand side a free-body-diagram is shown with external loads and relevant reaction forces RA and RB . Certainly, R A and RB may readily be determined from the free-body-diagram.
1 INTRODUCTION TO THE THEORY OF FORCES
18
Fig. 1.22
Fig. 1.23
Beam with length L resting on two wheels at distance a b
The principle of super-position applied to a simple beam on wheels
However, it may in many cases be a desirable simplification and a quicker way at arriving to the solution if the problem is split into the sum of
q gA alone, plus b) F mg alone, a)
as illustrated in Fig. 1.23. The total effect of q gA and F mg may then be obtained by first considering a)
q gA alone, where it is seen that horizontal equilibrium is fulfilled because there is no horizontal load, vertical equilibrium requires R A RB qL 2 , while moment equilibrium about any point requires R A RB (the system is symmetric), rendering R A RB qL 2 ,
and then considering
1.6 THE PRINCIPLE OF SUPER-POSITION
19
b) F mg alone, where again it is seen that horizontal equilibrium is fulfilled, while vertical equilibrium requires R A RB F and moment equilibrium e.g. about point A requires RB ( a b ) Fa 0 , rendering R A Fb a b and RB Fa a b . Thus, from applying the principle of super-position the following combined effect is obtained: qL Fb qL Fa and RB (1.23) RA 2 ab 2 ab In Fig. 1.24 the principle of super-position is illustrated for a simply supported beam. The total effect of the two forces F1 and F2 may be handled by adding the effects of each; for the calculation of reaction forces (illustrated in Fig. 1.25) as well as for the calculation of deformations (illustrated in Fig. 1.26).
Fig. 1.24 The principle of super-position applied to a simple beam
Fig. 1.25
The principle of super-position as applied to a free-body-diagram
1 INTRODUCTION TO THE THEORY OF FORCES
20
Fig. 1.26
The principle of super-position as applied to structural displacements
1.7 The equilibrium conditions in a three-dimensional format In a general three-dimensional Cartesian format, forces and moments may be described as x , y and z component vectors (see Fig. 1.27): Fx M x F Fy and M M y F M z z
with absolute values
Fig. 1.27
M M x2 M y2 M z2 F Fx2 Fy2 Fz2
Forces and moments in a three-dimensional format
(1.24)
(1.25)
1.7 THE EQUILIBRIUM CONDITIONS IN A THREE-DIMENSIONAL FORMAT
21
Consequently, the corresponding general equilibrium conditions are given by: Fx Fy 0 F z
and
M x M y 0 M z
(1.26)
in which case the equilibrium conditions in Eq. 1.13 are simply replaced by
Fi 0 Mi 0
Fxi 0, Fyi 0 and Fzi 0 M xi 0, M yi 0, M zi 0
(1.27)
where the index i on the F and M vectors indicate relevant force and moment components in x, y or z coordinates. Thus, force and moment equilibrium is just a matter of demanding the sum of all components in x , y and z directions equal to zero.
Example 1.1: Support reaction forces of crane
Fig. 1.28 Crane subject to vertical load F
1 INTRODUCTION TO THE THEORY OF FORCES
22
Let us then consider the more complex system of a crane type of system in Fig. 1.28. As can be seen, the reaction forces on (see right hand side free-body diagram in Fig. 1.28) may be described by the internal reaction force vectors:
Dx Dy D z
and
Nx A N AC N yA L N z A
a b h
and
N N 2 N 2 N 2 xA yA zA AC where N BC N x2B N y2B N z2B
Nx B N BC N yB L N zB
and L
a b h
a 2 b2 h 2
Thus, force and moment equilibrium is given by (please note that there are five unknown support reaction forces and six equilibrium equations available; hence, one is obsolete):
Fx 0 : Dx N AC a L N BC a L 0 Fy 0 : D y N AC b L N BC b L 0 Fz 0 : Dz N AC h L N BC h L F 0 M x 0 : N AC h L b N BC h L b Fc 0 M y 0 : N AC h L a N BC h L a 0 M x 0 : N AC a L b N AC b L a N BC a L b N BC b L a 0 M y 0
Mx 0 Fx 0
Fy 0
Fz 0
N AC N BC (Thus,
N AC N BC
M z 0 is also fulfilled),
F cL 2 hb
Dx 0
b c F L h h c Dz F 2 N AC F 1 L b D y 2 N AC
The same result is developed in a strictly mathematical procedure in Elaboration 1.1 below. It should be noted that in this particular case it will simplify moment equilibrium conditions greatly if origo in the axis system is rather chosen at point C , because both N AC and N BC merge at this point (and their moment contribution would be zero).
1.7 THE EQUILIBRIUM CONDITIONS IN A THREE-DIMENSIONAL FORMAT
23
Elaboration 1.1: A strictly mathematical procedure
Fig. 1.29
The moment product of a force vector F and its radial vector r
Fig. 1.30
The cross product of two vectors
The problem of three dimensional moment equilibrium may in complex cases more conveniently be handled by the moment product of a force vector F and its radial vector r as illustrated on the left hand side of Fig. 1.29, a problem consistently defined by the mathematical operations of cross products, where the cross product of two vectors
a a x is defined by
ay
a z
T
and b bx
c a b a b sin n
by
bz
T
1 INTRODUCTION TO THE THEORY OF FORCES
24
where is the angle between the two vectors and n is a unit length vector perpendicular to the joint plane of a and b . The cross product of two vectors will then be a third vector perpendicular to both of them, as illustrated in Fig. 1.30. To enable a consistent vector cross product directly from their components, it is convenient to define the coordinate system unit vectors
e x 1 0 0 , e y 0 1 0 T
T
and e x 0 0 1
T
a a x e x a y e y a z e z b bx e x by e y bz e z
and describe vectors a and b by: Thus
a b a x e x a y e y a z e z bx e x by e y bz e z
a x bx e x e x a y bx e y e x a z bx e z e x a x by e x e y a y by e y e y a z by e z e y a x bz e x e z a y bz e y e z a z bz e z e z Since
ex e y ez e y ez ex , ez ex e y then
e y e x e z e z e y e x e x e z e y
and
ex ex 0 e y e y 0 ez ez 0
a b a y bz a z by e x a x bz a z bx e y a x by a y bx e z e x det a x b x
ey ay by
ez az bz
That the moment product of a force vector and its radial vector is identical to the cross product may be illustrated by the two dimensional case shown at the right hand side in Fig. 1.29, where
F F cos 0 F sin
T
and r a 0 b
T
Thus
ex r F det a F cos
ey 0 0
a b e y det F sin F sin ez
b F cos
1.7 THE EQUILIBRIUM CONDITIONS IN A THREE-DIMENSIONAL FORMAT
0 0 r F 1 aF sin bF cos aF sin bF cos 0 0
As can be seen, r F is the moment vector effect M M x
x, y , z
25
My
M z
T
of F in
coordinates. This procedure may also be illustrated by the crane example in
Fig. 1.28. As can be seen, all the force effects acting on a free-body diagram of this system (see right hand side illustration in Fig. 1.28) may be described by the load vector
F 0 0 F and the internal reaction force vectors D , N A and NB : T
Nx Nx Dx A B D D y , N A N y A N AC e AC and NB N yB N BC e BC D z N z A N zB
N N N 2 N 2 N 2 A xA yA zA AC where N BC NB N x2B N 2yB N z2B and where L
and
1 T e AC L a b h e BC 1 a b h T L
a 2 b 2 h 2 e AC e BC .
Thus, force equilibrium is given by:
Dx N x N x 0 A B Fi 0 D N A NB F 0 D y N y A N yB 0 0 Dz N z A N zB F while moment equilibrium in is defined by
r 0 0 h T C M 0 r N r N r F 0 where 0 C A C B F T rF 0 c d While the first is expressing a straight forward force equilibrium condition, the moment equilibrium condition may be further developed into
1 INTRODUCTION TO THE THEORY OF FORCES
26
rC N A rC NB rF F 0 N 0 AC L h N det AC L
a 0 b 0 N BC L h h
e x 0 a
ez N h BC 0 L b h
ey
a 0 0 b c F 0 h d 1 e x ex e y ez h F 0 0 0 a b h 0
ey c 0
ez d 1
N AC N e x bh e y ah BC e x bh e y ah F e x c L L
N BC N AC L bh L bh Fc 0 N AC N BC ah ah 0 L L 0 0 This is the moment vector equilibrium condition of the system, i.e.
N AC N bh BC bh Fc 0 L L F cL N AC N BC M y L ah L ah 0 N AC N BC 2 bh Mz 0
Mx
Chapter 2 Equation Chapter 2 Section 1
STATICALLY DETERMINATE SYSTEMS 2.1
The classification of structural systems
a) Simply supported beam
b) Gantry
c) Simply supported truss Fig. 2.1
Examples of statically determinate systems
© Springer Nature Switzerland AG 2020 E. N. Strømmen, Structural Mechanics, https://doi.org/10.1007/978-3-030-44318-4_2
27
2 STATICALLY DETERMINATE SYSTEMS
28
a) Beam fixed against rotation at one end and simply supported at the other
b) Gantry with three supporting cables
c) Single bay three storey frame Fig. 2.2
Examples of statically indeterminate systems
In this chapter we shall focus on what may be called simple statically determinate line-like systems. These are structural systems that can be modelled by lines in a Cartesian coordinate system where their three-dimensional cross sectional properties have been concentrated to a line through a spanwise axis at the centroid of pure bending (see Chapter 3). Similar indeterminate systems are handled in Chapter 6. Some special systems (e.g. arches, cables and cylindrical
2.1 THE CLASSIFICATION OF STRUCTURAL SYSTEMS
29
shells) are presented in Chapter 9. Plates are handled in Chapter 11. To enable any safety considerations of structural systems it is necessary first to obtain the overall reaction forces, and then all the inner axial, bending, shear and torsion components. What is available for the determination of reaction forces in twodimensional systems (e.g. as shown in Figs. 2.1 and 2.2) is the three equilibrium conditions given in Eq. 1.13: Fx 0 , Fz 0 and M P 0 . In all the examples shown in Fig. 2.1 there are three reaction forces, and thus, these may uniquely be determined by these three equilibrium conditions. We say that any structural system with three reaction forces is statically determinate. The systems in Fig. 2.2.a and b both have four unknown reaction forces. Thus, none of them can uniquely be determined by the three equilibrium equations that are available. We say that these systems are statically indeterminate of order one. Thus, we will need one more condition to determine the reaction forces of the system. This additional condition is provided by the distribution of stiffness in the system, as stiffer parts carry more load than more flexible parts. E.g., for the gantry in Fig. 2.2.b, N BD and N BC depend on the stiffness ratio between cables BD and BC. The condition used to determine N BD and N BC is that the stretching of cables BD and BC is coherent with a joint vertical displacement of point B. This is what we call the compatibility condition of the system. The single bay three storey frame in Fig. 2.2.c has six unknown reaction forces. Thus, is it three times statically indeterminate, and hence, to determine all reaction forces we will need three compatibility conditions in addition to the three equilibrium conditions. We shall later show (Chapter 6) that a more general approach is usually adopted for complex systems such as the one shown in Fig. 2.2.c. It is based on the observation that if a system is linear in all its behaviour, then the reaction forces and inner force distribution are uniquely identified if the elastic deformation of all members is known. It should be noted that some systems are statically determinate with respect to reaction forces, but statically indeterminate with respect to their inner force distribution, see for instance the truss in Fig. 2.3, where external reaction forces Ax , Az and Bz may readily be determined, while the axial forces in the individual truss members cannot be determined by simple equilibrium conditions alone (see Chapter 2.3 below). Such systems are usually not recommended for practical design. The reason is that the weakest members may be subject to unwanted deformation induced constraint forces. In this chapter we focus on statically determinate systems.
30
Fig. 2.3
2 STATICALLY DETERMINATE SYSTEMS
Truss, externally statically determinate, internally indeterminate
2.2 External and internal equilibrium of line-like systems Since it is the designer’s duty to find the most onerous point of material straining and compare this to material strength, all structural analysis commence with the determination of reaction forces, followed by the determination of internal forces. In the following some simple line-like systems are presented. Let us first consider the gantry shown in Fig. 2.4.a. It consists of a bending stiff horizontal beam (AB), and a cable type of element (BC). It is subject to a force F at its tip B. A free-body-diagram is shown in Fig. 2.4.b, assuming that the cable BC can only sustain stretching. We set out to calculate the reaction forces Ax , Az and
N by using the equilibrium conditions given in Eq. 1.13.
Fig. 2.4 Simple gantry
2.2 EXTERNAL AND INTERNAL EQUILIBRIUM OF LINE-LIKE SYSTEMS
31
Without further ado we choose the moment equilibrium in relation to point A. Thus:
Fx 0 Fz 0 M A 0
Az N sin 30 F 0 FL N L sin 30 0 Ax N cos30 0
(2.1)
from which the following is obtained
Ax F cos30 sin 30 F 3 Az 0 N F sin 30 2 F
(2.2)
There are three important lessons to be learnt from this simple example. First, as illustrated at the upper right hand side of Fig. 2.4, the forces Ax , N and F are in equilibrium at point B, such that
Ax2 F 2 N
(2.3)
Second, it would in this case have been more convenient to choose the moment equilibrium in relation to point B, in which case
MB 0
Az L 0
(2.4)
and we immediately would have seen that Az 0 . Third, the conclusion that
Az 0 is one we could have seen only by acknowledging that a simply supported bar with no spanwise external load cannot sustain a force in its transverse direction (like the bar AB). Thus, already to start off with Az could simply have been omitted from the free-body-diagram in Fig. 2.4, in which case any moment equilibrium could also have been omitted, as all forces merge in point B. Let us then consider the more demanding problem of a simply supported beam subject to a distributed load q and two concentrated forces, F at quarterpoint and 2F at mid-point, as illustrated in Fig. 2.5.a. The free-body-diagram of the system is shown in Fig. 2.5.b. The relevant unknown support forces Ax , Az and Bz may be obtained from the equilibrium conditions
2 STATICALLY DETERMINATE SYSTEMS
32
Fx 0 Fz 0 MA 0
Az Bz F 2 F qL 0 Bz L F L 4 2 F L 2 qL L 2 0
Ax 0
(2.5)
Ax 0
Az 7 F 4 qL 2 Bz 5F 4 qL 2
Fig. 2.5
(2.6)
Simply supported beam
The model of a catenary mast is shown in Fig. 2.6.a. It is embedded (i.e. rigidly connected) at ground level point A, and subject to horizontal force F at its top and vertical forces F and 2F at the tip of its outliers at points C and D. The corresponding free-body-diagram is shown in Fig. 2.6.b. As can be seen, the only support is point A, where the unknown reaction forces Ax and Az occur together with an unknown moment M A . The following equilibrium conditions apply:
Fx 0 Fz 0 M A 0
Az F 2 F 0 M A FH Fe1 2 Fe2 0 Ax F 0
(2.7)
2.2 EXTERNAL AND INTERNAL EQUILIBRIUM OF LINE-LIKE SYSTEMS
33
Ax F Az 3F M F H e 2e 1 2 A
(2.8)
from which the following is obtained:
Fig. 2.6
Fig. 2.7
Cantilevered catenary mast
Single bay portal frame with one support tilting 45
2 STATICALLY DETERMINATE SYSTEMS
34
A single bay portal frame with one support tilting 45 is illustrated in Fig. 2.7.a with corresponding free-body-diagram in Fig. 2.7.b. The frame is subject to vertical force F at mid-span of the beam. To facilitate equilibrium conditions it is convenient to decompose the reaction force C45 into x and z direction components C x C45 cos 45 C45
2 and C z C45 sin 45 C45
2 . Thus,
it is readily seen that the equilibrium conditions are defined by
2 L 0
(2.9)
FL 2 FH FL FL , Az and C45 2 H L 2 H L 2 H L
(2.10)
Fx 0 Fz 0 M A 0
Ax
Ax C45
2 0
Az C45
2F 0
F L 2 C45
2 H C45
So far we have only been concerned with the calculation of support reaction forces. But, to enable a complete safety evaluation, it is necessary that the designer is in due control also of the internal forces in the system, as maximum load effect may occur on the cross section anywhere along the span of the system, e.g. as illustrated on Fig. 2.8.
Fig. 2.8
Beam subject to concentrated force at mid span
2.2 EXTERNAL AND INTERNAL EQUILIBRIUM OF LINE-LIKE SYSTEMS
Fig. 2.9
35
The definition of a cross section with positive surface normal vector
Therefore, it is necessary to determine the internal forces all over the system such that the most critical point may be identified and checked against exceedance of material strength. The definition of cross sectional force components are given on a cross section with positive surface normal vector, as defined in Fig. 2.9. Then, as illustrated in Fig.2.10, there are
N three force V y and moment V z
M x M y components M z
(2.11)
of which N is the normal force, Vy and Vz are the shear forces, M x is the torsion (twisting) moment, while M y and M z are the cross sectional bending moments. (In Fig. 2.10 it is convenient already here to include the relevant displacements components u v
w , as well as cross sectional rotation .) T
It is in the following taken for granted that the problem is two-dimensional, i.e. that the transverse loading is acting exclusively in x or z direction and that the cross section is symmetric about the vertical z axis. We shall also assume that there are no torsion (twisting) effects, which means that the relevant cross
36
2 STATICALLY DETERMINATE SYSTEMS
section is symmetric about z axis, and that the load is acting through the origo of the axis system. In that case it is only N , Vz and M y that are the relevant cross sectional forces in the system.
Fig. 2.10 The definition of positive cross sectional forces and displacements
Let us first consider the simple case of the cantilevered beam in Fig. 2.11.a. At its tip the beam is subject to concentrated vertical and horizontal forces F1 and
F2 . A free-body diagram of the system is shown in Fig. 2.11.b, from which the external support reaction forces Ax F2 , Az F1 and M A F1L is easily obtained. Let us then move to an arbitrary position x along the span, make a transverse cut across the cross section, and consider the equilibrium conditions of the beam segment the to the left of this point, i.e. the segment with a positive surface normal as illustrated in Fig. 2.11.c. At the left hand side of this beam segment there are the support reaction forces Ax F2 , Az F1 and
M A F1L , which must be balanced out by cross sectional forces N x , Vz x and M y x , all functions of the position x , and positively defined as shown on Fig. 2.11.c (in accordance with basic definitions in Fig. 2.10).
2.2 EXTERNAL AND INTERNAL EQUILIBRIUM OF LINE-LIKE SYSTEMS
37
Fig. 2.11 Cantilevered beam with concentrated forces F1 and F2
From the equilibrium conditions of the free-body beam segment in Fig. 2.11.c, the following is obtained: F2 N x 0 Fx 0 F1 Vz x 0 Fz 0 M A 0 F1L Vz x x M y 0
N x F2
Vz x F1 M y x F1 L x
(2.12)
Thus, as illustrated in Fig. 2.11.d, the normal force N x is constant and equal to F2 , the shear force Vz x is constant and equal to F1 , while the bending moment M y x varies linearly from F1L at the clamped end to zero at the tip of the beam. We call these cross sectional force variations the axial-, shear forceand bending moment diagrams of the system. Another simple example is shown in Fig. 2.12. A simply supported beam is subject to distributed load q (with unit
38
2 STATICALLY DETERMINATE SYSTEMS
N m ). From the free-body diagram Ax 0 and Az Bz qL 2 is readily
obtained. At arbitrary position x (see lower left hand side illustration in Fig. 2.12) the following equilibrium conditions apply:
Ax N x 0 Fz 0 Az Vz x qx 0 x M A 0 qx 2 M y x Vz x 0 N x 0 Vz x q x L 2 qx M y x L x 2
Fx 0
(2.13)
(2.14)
Shear- and moment diagrams are shown at the lower right hand side in Fig. 2.12 (normal force diagram has been omitted as it is equal to zero). Bending moment and shear force diagrams for some basic examples are shown in Fig. 2.13.
Fig. 2.12 Simply supported beam with distributed load q
2.2 EXTERNAL AND INTERNAL EQUILIBRIUM OF LINE-LIKE SYSTEMS
Fig. 2.13
39
Bending moment and shear force diagrams for some basic cases
40
2 STATICALLY DETERMINATE SYSTEMS
Example 2.1: Two span cantilevered beam
Fig. 2.14 Continuous statically determinate beam
A simply supported beam (Fig. 2.14), cantilevered across one of its supports, and subject to vertical distributed force q in span L1 and concentrated force F at tip of span L2 . From the free-body diagram the following support reaction forces are obtained:
Ax 0 Fx 0 : Az Bz qL1 F 0 Fz 0 : L1 M B L qL F L L 0 : 0 A 1 1 2 z 1 2
Ax 0 Az qL1 2 F L2 L1 B qL 2 F 1 L L 1 2 1 z
Then, internal cross section shear force and bending moments are determined by:
Az Vz x qx 0 Vz x qL1 2 FL2 L1 2 At 0 x L1 qx q L2 2 M y x Vz x 0 M y x x 2 x L1 F L 1
Az Bx Vz x qL1 0 Vz x F 2 At L1 x L2 qL 1 M y x Bz L1 Vz x 0 M y x F L1 L2 x 2 Shear force Vz x and bending moment M y x diagrams are shown in Fig. 2.15.
2.3 EXTERNAL AND INTERNAL EQUILIBRIUM OF TRUSSES
41
Fig. 2.15 Bending moment and shear force diagrams
2.3
External and internal equilibrium of trusses
Fig. 2.16 Typical trusses subject to concentrated forces in their joints
For long span beams or beams subject to unusually large forces it is structurally sound as well as economically beneficial to abandon the idea of a compact design, and rather resort to a lattice type of system, e.g. as illustrated in Fig. 2.16. We call this type of design trusses or lattice girders. It is based on the observation that in a system of bars joined together in series of triangles and merging at each corners of the triangle, they will all carry any load as pure tension or compression. If all external loads may be ascribed to the joints (as illustrated in Fig. 2.16), then, apart from gravity, unfavourable bending may largely been avoided. Let k be the number of joints in the truss (including
42
2 STATICALLY DETERMINATE SYSTEMS
support joints), p the number of bar elements and j the number of support reaction forces. Then, acknowledging that for each joint there are two equilibrium conditions Fx 0 and Fz 0 (and no moment equilibrium requirement necessary, as all forces merge at the joint), the system will be statically determinate if
2k p j
(2.15)
Let us consider the simple truss shown in Fig. 2.17. The number of joints k 5 , the number of bars p 7 and the number of support reaction forces j 3 .
Fig. 2.17 Simple truss subject to concentrated forces at its joints
2.3 EXTERNAL AND INTERNAL EQUILIBRIUM OF TRUSSES
43
Thus, 2k 10 and p j 10 , implying that the system is statically determinate. Starting with overall equilibrium of the entire system (Fig. 2.17.b), it is seen that
Fx 0 Fz 0 MA 0
Ax 0 Ax 0 Az E z 4 F 0 Az 2 F (2.16) F L 4 2 F L 2 F 3L 4 E z L 0 E z 2 F
Then, taking it for granted that tensile force is positive, the normal force in each member is simply obtained by demanding equilibrium in one joint after the other (Fig. 2.17.c):
Fx 0 N1 Joint A: Fz 0 N1
Fx 0 N1 Joint B: Fz 0 N1
2 N2 0 2 Az 0
N 2 2 F N1 2 F 2
2 N3
2 N4 0
2 N3
2F 0
(2.17)
N 3 F 2 (2.18) N 4 3F
Fx 0 N 2 N 3 2 N 5 2 N 6 0 N 6 2 F Joint C: (2.19) N 5 F 2 Fz 0 N 3 2 N 5 2 2 F 0 Fx 0 N 4 N 5 2 N 7 2 0 N 2 F 2 7 Joint D: (2.20) Fz 0 N 5 2 N 7 2 F 0
Fx 0 N 6 N 7 2 0 Joint E: Fz 0 N 7 2 E z 0
(2.21)
As can be seen, there are three equations unused; these are the equations that have been replaced by the overall equilibrium conditions in Eq. 2.16, i.e. one is free to choose whichever is most convenient. In general, the following applies:
N pk cos pk Fxk Rxk 0 pk Joint k : N pk sin pk Fzk Rzk 0 pk
(2.22)
44
2 STATICALLY DETERMINATE SYSTEMS
where N pk is the axial force in bar p associated with joint k , pk is the angle between bar p and the horizontal x axis, Fxk and Fzk are external forces associated with joint k , see Fig. 2.18, while Rxk and Rzk are possible reaction forces associated with joint k . If K is the total number of joints, this will render 2 K equilibrium conditions for the determination of N pk member forces as well as all Rxk and Rzk reaction forces.
Fig. 2.18 Arbitrary bar element p associated with joint k
Chapter 3 Equation Chapter 3 Section 1
THE THEORY OF STRESS AND STRAIN 3.1 Introduction to the theory of elasticity Unless otherwise stated we shall in the following assume two-dimensional systems and loads, material homogeneity (materials have same properties all over) and isotropy (materials have same properties in all directions). We shall limit ourselves to axial, bending and shear problems of beam elements with single or double symmetric cross sections. Composite beams are handled in Chapter 7, non-symmetric cross sections in Chapter 8, plates in Chapter 11.
Fig. 3.1 Rectangular bar subject to axial stretching
The background to the theory of strength of materials comes from model experiments, such as the simple rectangular bar shown in Fig. 3.1. The results of these experiments are made generally applicable (i.e. in any geometric scale) by defining the concepts of stress and non-dimensional displacement strain
x N A0 x L L0
© Springer Nature Switzerland AG 2020 E. N. Strømmen, Structural Mechanics, https://doi.org/10.1007/978-3-030-44318-4_3
(3.1)
45
46
3 THE THEORY OF STRESS AND STRAIN
where N is the axial force, A0 is the area of the unloaded cross section, L0 is the unloaded length of the bar, and L L L0 is the stretching of the bar. For an elastic-plastic material the characteristic stress-strain relationship from such a test is illustrated in Fig. 3.2. As can be seen, initially the curve is (approximately) linear elastic with slope E , i.e. in this region
x E x
(3.2)
where E is the modulus of elasticity. This is called Hooke’s law [3].
Fig. 3.2 Characteristic stress-strain relationship of elastic-plastic material
Fig. 3.3
Element subject to pure shear
3.1 INTRODUCTION TO THE THEORY OF ELASTICITY
47
A similar linear relationship may be established for pure shear force tests as illustrated in Fig. 3.3, i.e. z G (3.3) where z Vz A0 is the shear stress intensity, G is the shear modulus and
1 2 is the non-dimensional shear angle. As illustrated in Fig. 3.1, during stretching the bar will also have its cross section narrowed, i.e. at a certain value of N the cross sectional dimensions have been reduced from h0 , b0 to smaller values h , b . This transverse strain
y b b0 b0 h h0 h0
(3.4)
is negative (contraction) if the bar is subject to stretching, it is positive (extension) if the bar is subject to compression. Under elastic conditions y is linearly connected to x by Poisson’s ratio such that
y x
(3.5)
From Fig. 3.2 it is seen that at a certain value of strain the stress-strain relationship is no longer linear, the material starts yielding. The stress f y where this transition occurs is the characteristic material yield stress. The stressstrain curve reaches its apex at ultimate stress fu and final fracture occurs at ultimate strain u . All these quantities are key characteristic material properties in the safety considerations of any structural system. Exceedance of yield stress f y will be accompanied by large plastic deformations, and at ultimate strain the material will rupture and cause total structural collapse. The properties of some materials typically encountered in structural engineering are given in Table 4.1. Since stress is intensity and strain is non-dimensional deformation, independent of geometric scale, it is convenient to operate with these quantities at an infinitesimal level. Thus, a two-dimensional version of H ooke’s law may readily be developed by considering an infinitesimal element dx, dy subject to orthogonal stresses x and y as illustrated in Fig. 3.4. Assuming that x and
y are both in the elastic region (less than the yield stress f y ), the system is linear, and thus, the superposition principle applies.
3 THE THEORY OF STRESS AND STRAIN
48
Fig. 3.4
Material element subject to two-dimensional stress
Thus, let x be applied to the element dx, dy alone, in which case
x x E y x x E
(3.6)
Similarly, let y be applied to the element dx, dy alone, in which case
x y y E
(3.7)
y y E
According to the superposition principle these two effects are additive, and thus, if x and y are applied simultaneously, then
1 x y E 1 y x y E
x
Defining the stress and strain vectors σ x y it is seen that ε Aσ where
A
11 E
1
(3.8)
T
and ε x
T
y , then (3.9) (3.10)
3.1 INTRODUCTION TO THE THEORY OF ELASTICITY
49
From this it is seen that
x E 1 x σ A 1ε 1 2 1 y y
(3.11)
which is the two-dimensional version of Hooke’s law. (See also Chapter 11.1.)
Elaboration 3.1: Three-dimensional version of Hooke’s law
Fig. 3.5
Material element subject to three-dimensional stress
A three-dimensional infinitesimal element
dx , dy , dz
is illustrated in Fig. 3.5. For
such a general case it is important to keep stringent rules for the notation of each component: j where j x , y or z is its direction . Normal stress components: i x, y or z is the surface normal direction Shear stress component: ij where j x, y or z is its direction Following the same procedure as for the two-dimensional case above,
3 THE THEORY OF STRESS AND STRAIN
50
first applying
x alone
then applying y alone
and finally, applying
z alone
x x E y x x E z x x E
x y y E y y E z y y E
x z z E y z z E z z E
Then, according to the superposition principle, all stress components applied simultaneously will render the following normal stress strain relationship:
x 1 1 y E z
1
x y 1 z
Solving with respect to the stress components, renders the following three-dimensional version of Hooke’s law
x x 1 E y 1 y 1 1 2 1 z z Altogether there are six shear components: xy , xz , yx , yz , zx , zy . However, as later shown in Chapter 3.4 (see Eq. 3.33) they occur in opposite pairs such that
xy xy ,
yz zy and zx xz , and since xy , yz and zx are independent, then x 1 0 0 xy 1 yz G 0 1 0 yz 0 0 1 zx zx rendering the following three-dimensional shear stress-strain version of Hook’s law
xy 1 0 0 x yz G 0 1 0 yz 0 0 1 zx zx
3.1 INTRODUCTION TO THE THEORY OF ELASTICITY
Fig. 3.6 External actions rendering normal and shear stresses in beam
Fig. 3.7 Typical normal and shear stress distributions in beams
51
52
3 THE THEORY OF STRESS AND STRAIN
We shall in the following develop the necessary tools to determine normal stresses x and shear stresses xz and xy in beams. As illustrated in Fig. 3.6, normal stresses x may arise from axial force N and bi-axial or mono-axial bending M y and M z , while shear stresses xz and xy may arise from shear forces V y or Vz . Typical normal and shear stress distributions are illustrated in Fig. 3.7. In this chapter we limit ourselves to considering homogeneous and symmetric cross sections (i.e. single material systems and cross sections that are symmetric about the z axis). H ybrid (composite) and non-symmetric cross sections are dealt with in Chapters 7 and 8, shear stresses due to torsion are dealt with in Chapter 10. Normal stresses may also arise from thermal changes, as first dealt with in Chapter 3.2 below.
3.2 The effects of thermal variations
Fig. 3.8
Fig. 3.9
Thermal expansion
Bar subject to thermal expansion
3.2 TE EFFECTS OF THERMAL VARIATIONS
53
As illustrated in Fig. 3.8, the effect of a temperature increase T (in degrees Celsius) on a two-dimensional x, y system with width B and height H is that it will expand B and H in each (all) directions such that
x B B y H H T
(3.12)
where is the linear thermal coefficient of the material (with unit C 1 ). The properties of some typical materials are included in Table 4.1. For a line-like bar or a beam-column with length L (see Fig. 3.9), the most important effect of a temperature increase T is that it will be lengthened L x L T L
(3.13)
Example 3.1: Beam fixed between two rigid walls
Fig. 3.10 Constrained bar subject to thermal expansion A linear elastic bar with length L and cross section A is fixed between two walls as illustrated on the left hand side of Fig. 3.10. The stress effect of a temperature increase of
T (in C ) may be obtained by using the superposition principle as shown at middle and right hand side illustrations in Fig. 3.10. Let the bar first be free from its constraint at one end and subject to the temperature increase, rendering an elongation
LT T L . Then exert an axial force N at the free end, such that the
compression due to N brings the elongated bar back to fit between the two walls, such that (see Eq. 3.2): L LT LN L where LN x L x E L NL and thus
T L NL EA 0
AE
N EA T , from which it is
seen that the effect of T is an axial stress x N A E T .
3 THE THEORY OF STRESS AND STRAIN
54
3.3 Stress components from axial force and bending
Fig. 3.11
Normal stress x from axial force and bending
As illustrated in Fig. 3.11 there are three cross sectional force components that contribute to normal (axial) stress x : axial force N and bending moments
M y and M z about y and z axes. Since M y and M z are intended to render pure bending components, they will not render stresses that contribute to the axial force. Thus, origin in the axis system x, y , z is defined such that M y alone is rendering pure bending about the y -axis, and M z alone is rendering
3.3 STRESS COMPONENTS FROM AXIAL FORCE AND BENDING
55
pure bending about the z axis. Vice versa, axial force N alone must cause no bending, and thus, it is acting at the origin of the axis system and in the direction of its x -axis. Thus, N alone will render a constant axial stress distribution on the cross section: x N N A (3.14) where A is the area of the cross section. The effect of
N is a cross sectional
axial deformation u . The connection between N and u may be expressed in a calculus format (as first developed by Leibnitz [4]).
Fig. 3.12 Infinitesimal element dx subject to axial force N
Let the axial deformation at x be u , and at an incremental additional position x dx be u du , see Fig. 3.12, then the axial strain is given by
x
u du u du u dx
(3.15)
dx
and thus,
x x E uE
N xdA uE dA EAu A
(3.16)
A
rendering the connection between axial force and axial displacements. Let us then turn to the stress contribution from pure bending about the y axis alone, and again, we resort to a calculus approach, as illustrated in Fig. 3.13. In such an approach it is necessary to comply strictly with the positive definition of forces as well as deformations and their derivatives. The procedure is as follows:
3 THE THEORY OF STRESS AND STRAIN
56
Fig. 3.13
Fig. 3.14
Axial stress from pure bending M y
Beam subject to distributed transverse load q
Consider the beam in Fig. 3.14, subject to transverse distributed load q , rendering transverse displacements w (in z direction) at an arbitrary position x along its span, accompanied by cross sectional pure bending M y and shear force Vz . At an incremental added position x dx the corresponding bending, shear force and displacements are M y dM y , Vz dVz and w dw . Let us first consider the equilibrium requirements for the incremental element dx (see right hand side illustration in Fig. 3.14). Horizontal ( x direction) equilibrium is obsolete, as there are no forces in this direction. Thus, what remains is to demand vertical and moment equilibrium of the element. Vertical ( z direction) equilibrium requires
3.3 STRESS COMPONENTS FROM AXIAL FORCE AND BENDING
Vz Vz dVz qdx 0
dVz Vz q dx
57 (3.17)
Moment equilibrium about the mid-point of the element requires
Vz dx 2 M y Vz dVz dx 2 M y dM y 0
Vz dx dVz
dx dM 0 2
(3.18) (3.19)
where the higher order term dVz dx 2 may be omitted, and thus dM y dx
M y Vz
(3.20)
Taking the derivative once and introducing Eq. 3.17, then it is seen that M y Vz q
(3.21)
In Fig. 3.15 the same incremental element is shown in its displaced position, w at one end and w dw at the other. As can be seen, in agreement with the principle of calculus, it is given a formal displacement where dw , dw and dw are all positive quantities. It is taken for granted that displacements are small, such that sinus and tangent to any angle is approximately equal to the angle itself, and that cosine is approximately equal to one. Furthermore, it is assumed that Navier’s hypothesis [5] applies, i.e. that
a plane cross section perpendicular to the beam axis will remain plane and perpendicular to the beam axis after the deformation has taken place.
Thus, during displacement from zero to w , an infinitesimal element dA with length dx at arbitrary cross sectional position z (see Fig. 3.15) has gained the axial strain wz w dw z dw x z wz (3.22) dx dx The corresponding stress x E x Ewz , and thus (see Fig. 3.16)
M y z x dA z Ewz dA E z 2dA w A A A
(3.23)
3 THE THEORY OF STRESS AND STRAIN
58
Fig. 3.15 Infinitesimal element dx in its displaced position
Fig. 3.16
Bending moment M y and corresponding normal stress x
Defining the second area moment with respect to bending about the y axis
3.3 STRESS COMPONENTS FROM AXIAL FORCE AND BENDING
59
I y z 2 dA
(3.24)
it is seen that
M y EI y w
(3.25)
and thus (see Eq. 3.21)
EI y w =q
(3.26)
A
which is what we call the differential equation of beam bending. The normal stress distribution is then given by
x E x Ewz
My Iy
(3.27)
z
Before we proceed it should be noted that the development above was based on the assumption that the beam cross section was subject to pure M y bending, i.e.
N x dA Ew zdA 0 A
(3.28)
A
zdA 0
which can only be fulfilled if
(3.29)
A
from which the vertical position of the y axis is defined. Similarly, since the cross section is subject to pure bending about the y axis, any bending stress resultant about the z axis must be zero, i.e. we must demand that M z y x dA y EwzdA Ew yzdA 0 A
This can only be fulfilled if
A
(3.30)
A
yzdA 0
(3.31)
A
which defines the orientation of what we call the main axis system. Cross sections symmetric about the z axis will automatically satisfy Eq. 3.31.
Example 3.2 A simple rectangular cross section is shown in Fig. 3.17. It is symmetric about both y and
z
axes. Eqs. 3.29 and 3.31 are satisfied with origo at its area centre.
3 THE THEORY OF STRESS AND STRAIN
60
Fig. 3.17 Rectangular cross section
I y z 2dA
Then
A
h2
z 2 bdz
h 2
bh3 12
Elaboration 3.2: Bi-axial bending Similarly, in the case of pure bending M z about
z axis, then
x v z v dv z dx dv dx z v z where v is the displacement in y direction (see Fig. 2.10) and thus
x E x Evy , from which (see Fig. 3.11): M z y x dA y Evy dA E y 2dA v A A A Defining
I z y 2 dA
M z EI z v
it is seen that
and thus
A
x E x Evy
Mz y Iz
and
N x dA 0 A
ydA 0 A
In case of bi-axial bending M y and M z , the system is linear and the superposition principle applies. Thus
x
My Iy
z
Mz y Iz
3.3 STRESS COMPONENTS FROM AXIAL FORCE AND BENDING
61
Elaboration 3.3: Steiner’s theorem For practical purposes it is often convenient to develop Eqs 3.24 and 3.29 in a displaced coordinate system
y , z where *
*
y* is parallel to y and z* z c . According to
Eq. 3.29, the origo of the pure bending coordinate system
y, z
may then be
determined by:
zdA z A
A
*
c dA z*dA cA 0 c A
1 * z dA A A
coordinates the second area moment is given by (see Eq. 3.24): 2 2 I * z* dA z c dA z 2dA 2c zdA c2 A y *
Similarly, in y , z
*
A
Since
A
zdA 0 it is seen that: A
A
I
y*
A
2
Iy c A
or
which is what we call Steiner’s theorem. See Example 3.3 below.
Example 3.3: Bending of T-section
Fig. 3.18
Thin walled T cross section
Iy I
y*
c2 A
3 THE THEORY OF STRESS AND STRAIN
62
Let us consider the T cross section in Fig. 3.18. It is taken for granted that it is thin-
z
walled, i.e. that t b 1 . Due to symmetry the
axis is positioned through the web as
illustrated in Fig. 3.18. Thus, Eq. 3.31 is automatically fulfilled if also Eq. 3.29 is:
zdA z A
A
*
c dA z*dA cA 0 c A
bt b 2 btb 3b 1 * z dA 2bt 4 AA
The second area moment with respect to bending about y axis (using Example 3.2 and Steiner’s theorem, as developed in Elaboration 3.3): 2
tb3 b bt 3 5tb3 2 I y z dA c bt b c bt 12 2 12 24 A 2
flange
web
Thus, the normal stress distribution due to pure bending about the y axis is defined by:
x
My Iy
5tb2 5tb2
x x z b c 6M y 1 z: x2 x z c 18M y
as illustrated in Fig. 3.18.
Elaboration 3.4: Relevant cross sectional centre-points
Fig. 3.19 Cross sectional centre points Origo of the
y, z
axis system of a cross section is chosen to be that of pure bending,
it is chosen to fulfil Eqs. 3.29 and 3.31. It should not come as a surprise that a cross section may have other centre points that may be of interest in other and equally relevant cases. Generally, one is free to choose whatever axis system is most convenient from a computational point of view. We have previously seen (Chapter 1.1) that
3.4 STRESS COMPONENTS FROM SHEAR FORCES
63
an area centre (Fig. 3.19.a) and a mass centre (Fig. 3.19.b) may be of interest, as well as the axis system of pure bending (Fig. 3.19.c) However, in some cases (e.g. torsion) it may be favourable to define what we call the shear centre (Fig. 3.19.d, see also Chapter 10) defined as the axis system through which a force will cause no twisting, e.g. as illustrated in Fig. 3.20 below.
Fig. 3.20 Cantilevered beam subject to force through shear centre
3.4 Stress components from shear forces
Fig. 3.21 The equilibrium conditions of infinitesimal element dx, dz
Since we are now limited to two-dimensional problems of line-like beams, let us simplify the shear stress notation into x and z , where index indicates its direction. Let us first consider the equilibrium conditions for an infinitesimal element dx, dz , see Fig. 3.21:
3 THE THEORY OF STRESS AND STRAIN
64
Fx 0 Fz 0
d x dx d x dz 0 z dz z d z dz 0
MA 0
d x d x dz dx d z 0
(3.32)
z dz dx 2 z d z dz dx 2 x dx dz 2 x d x dx dz 2 0 z x 0
where higher order terms have been omitted. Thus, shear stresses on orthogonal planes occur in pairs, i.e. x z (3.33) From the first equilibrium condition in Eq. 3.32 it is also seen that it is the axial stress changes d x dx that cause shear stress changes d x dz , i.e. it is the normal stresses that feed in changes to the shear stresses. This is what we in the following shall use to quantify the shear stress variation. Consider an infinitesimal beam element dx subject to bending and shear, see Fig. 3.22.a. On its left hand side, the cross section is subject to a linear variation of bending induced normal stresses x z . On its right hand side, these normal stresses have increased into x z d x z , an infinitesimal increase that has to be balanced out by shear stresses x z z z . Thus, let us set out to determine the
z that provides horizontal equilibrium across the element. The part of the cross section outside of z is As , as
shear stress z at arbitrary position
illustrated in Fig. 3.22.b. Considering the slice of the element with length dx and cross sectional area As , it is seen that it is subject to normal stresses x and
x d x as well as shear stresses z z on either side. As shown above, shear stresses occur in identical pairs, and therefore, there is also a horizontal shear component x z z z on the underside of the slice. On its outer (upper)
surface there cannot be any shear stresses, they must vanish at the outer edges of the cross section. Assuming cross sectional symmetry about the z axis, there is no reason why the shear stress should vary in the y direction of the cross section, and thus, demanding x direction equilibrium of element dx, As : x dA As
x d x dA z z bdx 0
As
where b is the width of the cross section at
z . Thus, it follows that
(3.34)
3.4 STRESS COMPONENTS FROM SHEAR FORCES
z z
1 d x dA b A dx
65
(3.35)
s
a) Infinitesimal beam element dx subject to bending and shear
b) Equilibrium condition of outer segment As
Fig. 3.22 Beam element dx subject to bending and shear
3 THE THEORY OF STRESS AND STRAIN
66
It shall in the following be assumed that the cross section does not change along the span of the beam, and then
d x d M y dx dx I y
M y V z z z z I Iy y
(3.36)
Thus, the shear stress at an arbitrary cross sectional position z is given by
z z where S y
Vz bI y
zdA
As
Vz S y
(3.37)
bI y
zdA is called the static moment of
As with respect to the y axis.
As
Example 3.4: Shear stresses in rectangular cross section
Fig. 3.23 Upper slice As of beam element dx Let us first consider the simple case of a rectangular cross section (Fig. 3.23). Then
Sy
As
h2
zdA
z
zs bdzs
2 b h 2 z 2 2
z
Vz V Sy z 2I y bI y
h 2 2 z 2
3.4 STRESS COMPONENTS FROM SHEAR FORCES
67
Since S y is the static moment of As it is worth noting that it could more quickly have been calculated by the product between As b h 2 z and its centre-distance from the y axis h 2 z 2 , i.e. 2 h 1 h b h S y b z z z 2 2 2 2 2 2
The largest value of the shear stress occurs at z 0 : zmax Vz 2 I y
h 2 2 .
Example 3.5: Shear stresses in T-section
Fig. 3.24 Thin-walled T cross section Let us consider the more complex problem of a T cross section (Fig. 3.24). It is taken for granted that it is thin-walled, i.e. that t b 1 . Due to symmetry the
z
axis is
positioned through the web as illustrated on Fig. 3.24. The position of the y axis as well as the second area with respect to bending about y axis have been calculated in Example 3.3 above ( c 3b 4 and I y 5tb3 24 ). To determine the shear stress distribution on this type of cross sections it is necessary to consider several positions, in this case one on each side of its upper flange and one on its web. It should also be noted that the b in Eq. 3.38 is the width or plate thickness across which the shear stress is constant.
3 THE THEORY OF STRESS AND STRAIN
68
Fig. 3.25 Relevant positions for the determination of y and z on a T-section
In this case the shear flow is acting across the thickness t , as illustrated in Fig. 3.25. First we consider an arbitrary position pos. 1 on Fig. 3.25:
bt b b S y y t b c y 4 2 2
y
Vz bt b y tI y 4 2
Thus, it is seen that in the horizontal flange a vertical shear force Vz is responsible for a horizontal shear stress component y , i.e. a shear stress that is perpendicular to Vz .
bt b V bt b b y t b c y y z y tI 2 4 2 4 2 y
At pos. 2: S y
Thus, it is seen that the shear stresses on the flanges are identically distributed but of opposite signs. Finally, we consider pos. 3 on Fig. 3.25:
3.4 STRESS COMPONENTS FROM SHEAR FORCES
2 1 b2t t b z 2 S y bt b c b c z t b c z 2 4 2 4
2 Vz Vz b2 1 b S y z 2 z tI y I y 4 2 4
The shear stress distribution is shown on Fig. 3.26. As can be seen, z z c 0 .
Fig. 3.26 Shear stress distribution
Example 3.6: Shear stresses in single symmetric I-section
Fig. 3.27 Thin walled single-symmetric I section
69
3 THE THEORY OF STRESS AND STRAIN
70
Let us further expand in complexity by considering the single-symmetric I type of cross section shown in Fig. 3.27. Then
A bt(4 2 41 2 2) 16bt
c b 2t 4 1 2 4 2 4 A 5b 2
and
t 4b 5 5 148 3 5 I y b 2 b 2t b 2b 4bt 4 b 4b2t bt 12 2 3 2 2 3
2
The normal stress distribution
2
2
x M y I y z is shown in Fig. 3.28. For the
determination of S y all over the cross section it is necessary to consider the three positions shown in Fig. 3.29 (as the cross section is symmetric about arbitrary upper flange position 3b 2 , y : 2b
Sy
3
2 b 2tdys 3bt 2b y
y
y
3 2
At web position 0,z : S y b 4b2t
z
3b 2
z
Vz 3bt 2b y 2t 148 3 b3t
z
axis). At
9 Vz y 2 b 296 bt
2 t 3 z tdzs 12b2t b z 2 . 2 2
2 2 2 Vz t 3 3 Vz 3 z 12b2t b z 2 24 148 3 2 2 2 b 296 bt b t 2t 3
3 b, y : S y 2
At lower flange position
y
2b
5
2 b 2tdys 5bt b y . y
Vz 15 Vz y 5bt b y 1 296 bt b 2t 148 3 b3t
The shear stress distribution is shown in Fig. 3.30.
Fig. 3.28 Normal stress distribution
3.4 STRESS COMPONENTS FROM SHEAR FORCES
71
Fig. 3.29 Relevant positions for the determination of z
Fig. 3.30 Shear stress distribution for a thin walled single-symmetric I section
Example 3.7: Circular tube So far the development has been based on the condition that the shear stress at an outer edge is zero (see Eq. 3.34). Thus, the theory is strictly spoken only applicable to open cross sections, and can in general not be used for closed or partly closed cross sections, unless symmetry may be exploited, as illustrated below for a circular cross section.
3 THE THEORY OF STRESS AND STRAIN
72
Fig. 3.31 Circular cross section subject to bending and shear It is taken for granted that the cross section is thin-walled, i.e. that t r 1 . As this cross section is symmetric about any pair of Cartesian coordinates
y, z
with origo at
its centre, there is no doubt about the position of the main axis of pure bending. Then the second area moment with respect to bending about the y axis may most conveniently be calculated in polar coordinates (right hand side illustration in Fig. 3.31):
I y z 2 dA
2
2
0
0
A
2 2 3 3 r sin trd r t sin d r t
Since the cross section is symmetric about
z
axis, the
z distribution must also be
symmetric, and thus, as illustrated on right hand side in Fig. 3.32, the equilibrium condition in Eq. 3.34 may be replaced by
x dA As
x d x dA 2 z tdx 0
As
z
and thus
z
Vz Sy 2tI y
where (same as before)
1 2t
Sy
As
d x dA dx
zdA
As
As illustrated on left hand side of Fig. 3.32, S y may be obtained by:
Sy
zdA
As
2
2
r cos trd r 2t
2
cos d 2 r 2t cos
2
Vz V 2V S y z r 2 cos z cos 2tI y Iy A
where A 2 rt
3.5 DECOMPOSITION OF STRESS COMPONENTS, PRINCIPAL STRESSES
73
The distribution of is illustrated in Fig. 3.33. As can be seen max 2Vz A .
Fig. 3.32 Equilibrium condition (right hand side) and calculation of S y (left)
Fig. 3.33 Shear stress distribution for a circular thin-walled cross section
3.5 Decomposition of stress components, principal stresses In the preceding chapters we have developed the relevant tools to determine normal and shear stress components in structural main bending axes x , z . These components are shown on an infinitesimal element dx , dz in Fig. 3.34.
3 THE THEORY OF STRESS AND STRAIN
74
Fig. 3.34
Stress components in structural axes x , z
Fig. 3.35 Stress components at angle with structural x axis
This has been necessary because it is the designer’s duty to check the stress condition everywhere in his system. H owever, it is also his responsibility to check stresses in other directions, because it is not necessarily the stresses in the direction of structural main axis that are most critical for the system. Therefore, given stress components x , z and xz (where double indexing of shear stress is retained for more generality), let us set out to determine the corresponding stresses at angle and 2 with the structural x axis, as illustrated in Fig. 3.35. To do this it is convenient to consider an infinitesimal
3.5 DECOMPOSITION OF STRESS COMPONENTS, PRINCIPAL STRESSES
75
right angled triangular element with hypotenuse and corresponding catheti sin and cos . It is taken for granted that all stress components are constant across the element thickness in y direction. The equilibrium conditions in x and z directions for this element is then defined by:
Fx 0 Fz 0
: x dy cos xz dy sin dy cos dy sin 0 (3.38) : z dy sin xz dy cos dy sin dy cos 0
Multiplying the first of these equations by cos , the second by sin , and adding them together, the following is obtained:
cos2 sin 2 x cos2 z sin 2 2 xz sin cos
(3.39)
Multiplying the first of the equations in Eq. 3.38 by sin , the second by cos , and subtracting them from each other, the following is obtained:
sin 2 cos2 x sin cos z sin cos xz sin 2 cos2
Introducing
2sin cos sin 2 sin 2 cos2 1
and
cos2 1 cos2 2 sin2 1 cos2 2
(3.40)
(3.41)
then the unknown normal and shear stress components are obtained:
x z
x z
cos 2 xz sin 2 x z sin 2 xz cos 2 2 2
2
(3.42)
From these two expressions the stresses at an arbitrary attitude may be calculated if the stresses x , z and xz in x , z coordinates are known. Obviously, it
would be of interest to find the extreme (maximum and minimum) values of and . This may be obtained by setting their first derivatives equal to zero: d d x z sin 2 2 xz cos 2 0 d d x z cos 2 2 xz sin 2 0
from which it is seen that has an extreme value at the condition that
(3.43)
3 THE THEORY OF STRESS AND STRAIN
76
tan 2 2 xz x z
(3.44)
while has an extreme value at the condition that cotan 2 2 xz x z
(3.45)
Since cotan tan 2 , it is seen that the extreme values of and are orthogonal. The extreme value of may be obtained by taking the square of the first lines in Eqs. 3.42 and 3.43 2 2 x z x z cos 2 sin 2 xz 2 2 2 x z 0 sin 2 xz cos 2 2
(3.46)
and adding them together, in which case is eliminated from the expressions, and the following is obtained
x z 2
2
max z 2 x xz 2 min
(3.47)
Similarly, taking the square of the second lines in Eqs. 3.42 and 3.43 2 x z sin 2 xz cos 2 2 2 x z 0 cos 2 xz sin 2 2
2
(3.48)
and adding them together, in which case the following is obtained 2
max x z 2 xz 2 min
(3.49)
It is readily seen from Eqs. 3.47 and 3.49 that the extreme values of may also be obtained from the extreme values of :
3.5 DECOMPOSITION OF STRESS COMPONENTS, PRINCIPAL STRESSES
max 1 max min min 2
77 (3.50)
The extreme values max , min , max , min are called principal stresses; because it is from these stresses safety against material failure is quantified. The corresponding axes are called principal stress axes.
Example 3.8:
Fig. 3.36 Infinitesimal element subject to stress components x , z and xz
Given an element subject to stress components with unit N mm 2 ). Extreme values of
x 100 , z 20 and xz 80 (all
will then occur at:
53 2 53 180 233 100 20 100 20 cos53 80sin 53 140 2 2 At 53 2 26.5 100 20 sin 53 80cos53 0 2 100 20 100 20 cos 233 80sin 233 60 2 2 At 233 2 116.5 100 20 sin 233 80cos 233 0 2 2 80 4 tan2 100 20 3
78
3 THE THEORY OF STRESS AND STRAIN
143 4 2 80 2 Extreme values of : cotan 2 100 20 3 143 180 323
which is equal to the values above plus 90 . The extreme values of by:
are then given
100 20 100 20 cos143 80sin143 40 2 2 At 143 2 71.5 100 20 sin143 80cos143 100 2 100 20 100 20 cos323 80sin 323 40 2 2 At 323 2 161.5 100 20 sin 323 80cos323 100 2 See Fig. 3.37.
Fig. 3.37
Extreme values of and
3.5 DECOMPOSITION OF STRESS COMPONENTS, PRINCIPAL STRESSES
79
Elaboration 3.3: Mohr’s stress circle Mohr’s stress circle [6] is a geometric interpretation of the expressions of the stress components in Eq. 3.42. Taking the square of both expressions: 2
x z x z cos 2 xz sin 2 2 2 2
x z 2
sin 2 xz cos 2
and
2
and adding them, then the following is
2
obtained:
2
2
x z x z 2 2 xz 2 2 x z ,0 2
As illustrated in Fig. 3.38, in , axes this is a circle with its center at 2
and with radius
z 2 r x xz 2
Fig. 3.38 Mohr’s stress circle
80
3 THE THEORY OF STRESS AND STRAIN
Further geometric properties of the principal stresses are illustrated in Fig. 3.39. (It is from this diagram readily seen that Eq. 3.50 applies.)
Fig. 3.39 Special properties of Mohr’s circle
3.6 Decomposition of strain components, principal strains Similar to that of decomposing stress components into principal axis, it is of interest to perform the same transformations with strain, see Fig. 3.40 and 3.41. Before commencing on the task it is necessary to acknowledge that the shear angle has the following properties (see Fig. 3.41):
1 2 2 1 1 2 2 1 2
(3.51)
Let us first consider x rotated an arbitrary angle , as illustrated in Fig. 3.42.a. It is seen that this entails (all quantities defined on the illustration)
3.6 DECOMPOSITION OF STRAIN COMPONENTS, PRINCIPAL STRAINS
ds x dx cos dx sin 1 x
x cos2 ds 1 2 2 x sin cos 2 1 2 1
81
(3.52)
where it has been introduced that dx ds cos . Let us then consider z rotated an arbitrary angle , as illustrated in Fig. 3.42.b. It is seen that this entails
ds y dy sin
y sin 2 y dy cos 1 ds 1 2 2 y sin cos 2 1 2 1
(3.53)
where it has been introduced that dy ds sin . Finally, let us consider xz rotated an arbitrary angle , as illustrated in Fig. 3.42.c.
Fig. 3.40 Stress components and corresponding strain in x , z axes
Fig. 3.41 Strain components at angle with structural x axis
82
3 THE THEORY OF STRESS AND STRAIN
a)
x alone
b)
z alone
c)
xz alone
Fig. 3.42 Strain components x , z and xz rotated an arbitrary angle
3.6 DECOMPOSITION OF STRAIN COMPONENTS, PRINCIPAL STRAINS
83
It is seen that this entails
xz dy cos
ds xz sin cos dy sin 1 xz ds 1 2 xz cos2 sin 2 2 1 2 The system is linear, and thus, the superposition principle applies.
(3.54)
x , z and
xz all acting simultaneously will then render x cos2 z sin 2 xz sin cos 2 x z sin cos xz
cos2 sin 2
(3.55)
Again, introducing the trigonometric properties given in Eq. 3.41, the following is obtained x z x z cos 2 xz sin 2 2 2 2 (3.56) x z xz sin 2 cos 2 2 2 2 Extreme (maximum and minimum) values of and 2 may be obtained by setting their first derivatives equal to zero:
d x z sin 2 xz cos 2 0 d d 2 x z cos 2 xz sin 2 0 d
(3.57)
from which it is seen that the variation has an extreme value at tan 2 xz x z
(3.58)
while the 2 variation has an extreme value at cotan 2 xz x z
(3.59)
3 THE THEORY OF STRESS AND STRAIN
84
Again, since cotan tan 2 , it is seen that the extreme values of
and 2 are orthogonal.
Elaboration 3.4: Mohr’s strain circle Taking the square of both expressions in Eq. 3.56 2 2 x z x z xz cos 2 sin 2 2 2 2 and adding them, 2 2 xz xz x z sin 2 cos 2 2 2 2 2
the following is obtained:
2
2
x z xz x z xz 2 2 2 2
As illustrated in Fig. 3.43, in
, xz 2
2
axes this is a circle with its center at 2
x z and with radius r x z xz ,0 2 2 2
2
Fig. 3.43 Mohr’s strain circle
3.6 DECOMPOSITION OF STRAIN COMPONENTS, PRINCIPAL STRAINS
85
It is readily seen from Fig. 3.43 that the extreme values of and 2 follows the same pattern as the corresponding stress components. Thus, the extreme values are defined by
and
2 2 max x z x z xz min 2 2 2
(3.60)
2 2 2 max x z xz 2 min 2 2
(3.61)
It is also seen from Fig. 3.43 (and the equations above) that the extreme values of 2 may be calculated from the extreme values of :
2max 1 2min 2 max min
(3.62)
Fig. 3.44 Typical pattern of strain gauge filaments The load induced strain on structural systems may be measured by strain gauges, made up of an insulating flexible backing paper onto which is embedded a pattern of metal filament suitable for the recording of any stretching or compressive motion, see Fig. 3.44. The gauge may be glued to the surface of the structural system and thus, any stretching or compression of the system will also render corresponding deformations of the metal filament. Thus, the cross section of the metal filament will be changed and a low voltage signal through the filament will register a corresponding change of its electric resistance. As illustrated in Fig. 3.44, the orientation of the filament will decide which direction the strain is registered, e.g. parallel or at 45 to the structural displacement.
3 THE THEORY OF STRESS AND STRAIN
86
The relationship between the strain and corresponding change of electric resistance R is defined by the gauge factor , such that
1 R R
(3.63)
where R is the electric resistance in the un-stretched gauge. The change of electric resistance R is usually recorded in a Wheatstone bridge [7]. Obviously, a single filament is suitable for the recoding of (e.g. x or z ). Strain gauges with filaments at 45 and at 45 90 are suitable for the recording of the shear angle (e.g. xz ), because
x 2
xz
2 2 45 90 x 2 xz 2 2
45
xz 45 45 90
(3.64)
Elaboration 3.5: The connection between E and G Let us consider a two-dimensional infinitesimal element dx, dz subject to shear stress
xz alone. The strain is then shear alone, xz and according to Eqs. 3.47 and 3.60 max max xz and xz min min 2
and they occur at an angle of 45 . Then, according to Hooke’s law (see Eq. 3.2)
1 1 xz xz xz E E 1 xz xz . Hence is also equal to xz 2 we must have that 2 E E G xz xz 2 1
max
but , because max
Thus, it is seen that E , G and are interconnected material properties.
Chapter 4 Equation Chapter 4 Section 1
DESIGN CRITERIA 4.1 Basic material properties
Fig. 4.1
Possible sources of material failure or loss of structural stability
The safety against failure of structural systems is evaluated by checking extreme load induced stress components against material failure or against loss of stability. Under static (more or less time invariant) load, material failure for nonbrittle materials may occur as extensive normal or shear stress yielding, and ultimately, failure occurs as material fracture. In brittle materials yielding will be © Springer Nature Switzerland AG 2020 E. N. Strømmen, Structural Mechanics, https://doi.org/10.1007/978-3-030-44318-4_4
87
4 DESIGN CRITERIA
88
small or practically non-existent, and then failure will occur as a sudden fracture event. Under dynamic (cyclic) load, material failure may occur due to fatigue. Systems subject to compression may lose their stability and buckle at local or at global structural level. Plates may buckle and cause structural collapse.
Fig. 4.2
Typical material properties
Typical material properties are illustrated in Fig. 4.2. Some mechanical properties are given in Table 4.1 (though variations may occur). Brittle materials (illustrated in Fig. 4.1.a) should as far as possible be avoided in structural systems. The reason for this is that failure will occur as a sudden event and leaves no warning for people to escape. Failure in materials with stress strain properties as illustrated in Fig. 4.2.a will occur as a gradual event. Loss of stability may occur as a more or less sudden event, but elastic-plastic materials will in general render some remaining strength after considerable buckling deformations. In earthquake prone areas any use of brittle material should altogether be banned.
4.2 TRESCA AND VON MISES YIELD CRITERIA
Table 4.1
89
Basic properties of typical structural materials
Material
kg m3
Aluminium; 2700 normal, 2800 high-str. Concrete; 2300 normal 2400 high str. 2600 Glass 7200 Iron, wrought Steel; 7850 normal, 8030 stainless 7900 high-str. 700 Wood; fibres ┴ fibres *) Compressive strength
E
G
3
10 N mm 2
3
10 N mm 2
70
27
25 30 50-90 80-190
fy
10 C
fu
N mm 2
N mm 2
0.33
23
240 410
12.5 13 20-35 30-70
0.1-0.2 0.2 0.2-0.3 0.2-0.3
7-14
210
80
0.3
12
12 0.4
0.8
6
5-11 11
4-6 30-40
130
20-30* 40-80* 33 200
250 520 690
420 860 760 40
The safety mode of any normal or shear stress failure for most structural materials may be evaluated by some simple criteria, which are based on observations and model experiments, e.g. by the Tresca shear failure criterion or by Von Mises normal yield stress hypothesis. It should be noted that although these two hypotheses are widely in use, they are based on the assumption that the rate of loading is low and non-cyclic, and that possible high temperature effects are duly accounted for. E.g. high impacts, dynamic load as well as the event of fire need separate evaluation beyond the criteria presented below. For simplicity, only the condition of two-dimensional stress is included.
4.2 Tresca and von Mises yield criteria For a simple bar subject to axial load (Fig. 4.3) it is for some materials observed that failure occurs along yield lines at 45 degrees angle to the direction of axial stress. Since maximum shear coincides with a 45 degrees angle to the direction of uniaxial normal stress it is concluded that this is a shear failure mode, i.e. that failure occurs due to exceedance of the material shear strength.
4 DESIGN CRITERIA
90
Fig. 4.3 The Tresca shear yield criterion, [8]
Fig. 4.4
Shear failure in plate element
Consider the uniaxial stress condition x N A in Fig. 4.3. Then, according to Eq. 3.49 max x 2 , and thus, defining the yield limit at x f y , the following shear yield criterion is obtained
4.2 TRESCA AND VON MISES YIELD CRITERIA
max f y 2
91 (4.1)
This is Tresca shear yield criterion (named after the French mechanical engineer Henri Tresca, 1814-1885). In case of a biaxial stress condition (see Eq. 3.50)
max max min 2
(4.2)
and thus, the criterion in Eq. 4.1 may be replaced by
max min f y
(4.3)
The permissible combination of max and min according to Eq. 4.3 has been illustrated at right hand side in Fig. 4.3. If the relevant system consists of fairly thin-walled plates, it should be noted that it is necessary not only to consider inplane yield lines but also to include the possibility that shear yield lines may occur in a plane perpendicular to the direction of loading, see Fig. 4.4. In that case, it may be necessary to expand Eq. 4.3 into
max min max min
fy
(4.4)
Von Mises yield criterion [9] is based on the assumption that the effective stress 2 2 min max min does not exceed the yield stress f y , defined by j max
i.e. that 2 2 j max min max min f y
(4.5)
Von Mises yield criterion is illustrated in Fig. 4.5. Introducing Eq. 3.47 it is seen that 2 x z 2 x z 2 3 xz 2 2
2 2 min max min max
(4.6)
2 x2 z2 x z 3 xz
and thus, Eq. 4.5 may be replaced by 2 j x2 z2 x z 3 xz fy
(4.7)
4 DESIGN CRITERIA
92
Fig. 4.5
Von Mises yield criterion [9], (dotted lines: Tresca)
Example 4.1: Circular tube
Fig. 4.6
Thin walled tube subject to axial force N and torsion moment M x
4.3 SAFETY CONSIDERATIONS
93
A thin-walled tube (Fig. 4.6) has centre radius r 300 mm and thickness t 20 mm . It is subject to axial force N 3200 kN and torsion moment Mx 800 kNm (see Chapter 10.2).
A 2 rt 376.8 102 mm2 and I t 2 r 3t 33.9 108 mm 4 The material yield stress is f y 250 N mm2 . Thus:
x
N 85 N mm2 A
and
xt
Mx r 71 N mm 2 It
2 max x 124.25 x N mm 2 xt2 42.5 81.75 min 2 39.25 2
Safety factor
against shear yielding according to Tresca’ criterion is then given by
max min 124.25 39.25 163.5 f y The effective stress is given by
250 1.5 163.5
j x2 3 xt2 150 N mm2 and thus, the safety
factor against normal stress yielding according to von Mises criterion is given by:
250 150 1.67 I.e. the safety factor according to von Mises is slightly larger than the safety factor according to Tresca, which complies with that which may be concluded from Fig. 4.5.
4.3 Safety considerations It should be acknowledged that all loads on civil engineering structures vary with time, some slowly (e.g. the changing position of gravity loads), others more rapid (e.g. wind loads). Thus, so does the cross sectional forces in a structural system. Let us for instance consider an arbitrary bending moment M in a beam. Within a reasonable time window stationarity may be assumed, and thus, within this period it will have a mean value M and a fluctuating part M t , i.e. M M M t
(4.8)
94
4 DESIGN CRITERIA
which means that there is a corresponding maximum value M max as illustrated in Fig. 4.7. Over a normal life period of a structural system (e.g. 100 years), this maximum value will have a probability of occurrence p M max .
Fig. 4.7 The probability of failure
The ultimate bending strength M u of the system will also have certain probability of occurrence p M u , and, as illustrated in Fig. 4.7, the necessary safety consideration is then to make sure p M u is sufficiently larger than
p M max . This is usually obtained by applying a safety coefficient d to the load such that the design bending moment M max d F or d q is a nominal value at the upper tail of p M max , while another safety coefficient m is applied to the relevant material properties (e.g. the yield stress) such that a
nominal bending moment strength M u y m
is located at the lower tail of
p M u . These values are taken from research and stipulated by the appropriate national or international authority. Setting
4.3 SAFETY CONSIDERATIONS
M max d F or d q M u y m
95
(4.9)
will then render a nominal value of safety in the order of
d m
(4.10)
The value of m is material dependent, while the value of d is dependent on the type of load, the relevant environmental conditions as well as the expected life time of the system. In general the insecurity of material properties is considerably lower than the insecurity of environmental conditions. Therefore, m will most often be in the order of 1 1.5 while d will be in the order of 1.5 3.0 . The most prominent environmental loads on structural systems are gravity, wind and snow (in some cases also from earth or water pressure). Gravity loads come from the weight of the structure itself plus the weight of any people, furniture, machines or any other human activity. Density of typical structural materials is given in Table 4.1, while gravity loads from human activities are given in Table 4.2. The density of snow under various conditions is given in Table 4.3 (it should be noted that snow may build up in drifts). The load from wind, either given as a pressure or as a distributed force, is defined by
1 2 (with unit N m2 ) for surface pressure q p c p Vmax 2 1 2 (with unit N m) for distributed load qF CF D Vmax 2
(4.11)
where c p and CF are coefficients characteristic to structural shape, D is a representative structural dimension (usually perpendicular to the wind direction). c p and CF may be found in national or international standards as determined from wind tunnel tests. Vmax is gust wind velocity (with duration of about 3s). Table 4.2
Gravity loads from human activities
Type of human activity Areas of domestic and residential activities Office areas Areas where people congregate Shopping areas Areas for storage and industrial activities
Distributed load ( kN m 2 ) 1.5 – 4.0 2.0 – 3.0 2.0 – 7.5 4.0 – 5.0 5.0 - 7.5
4 DESIGN CRITERIA
96
Table 4.3
Density of snow
Type of snow condition Newly fallen Damp new snow Settled snow Depth hoar (also called sugar snow) Wind packed snow Firm snow (left from last season and recrystallized) Very vet snow Glacier ice
Density ( kg m 3 ) 50 – 70 100 – 200 200 – 300 100 - 300 350 – 400 400 – 850 700 – 800 800 - 950
Chapter 5 Equation Chapter 5 Section 1
DEFORMATIONS OF BEAMS, TRUSSES AND FRAMES 5.1 Deformations due to axial force
Fig. 5.1
Axial elongation
It was shown in Chapter 3.3 (see Fig. 3.12 and Eqs. 3.15 – 3.16) that an axial force N acting alone on a line-like member will render a constant axial stress distribution across the entire cross section x N A , where A is the area of the cross section, and that the corresponding axial strain is defined by
x u du u dx du dx , where u x is the elongation of the member. Assuming linear material behaviour, then x x E , and thus, the elongation at arbitrary position x of the cantilevered member in Fig. 5.1 will be x
x
x
u x du x dx 0
0
0
x E
x
dx
N N dx x EA 0 EA
(5.1)
where origo of x is located at the left hand side of the member (i.e. x 0 at the point of embedment, where the deformation is zero). It is seen that the total elongation at the tip of the member is
u x L
© Springer Nature Switzerland AG 2020 E. N. Strømmen, Structural Mechanics, https://doi.org/10.1007/978-3-030-44318-4_5
N L EA
(5.2)
97
5 DEFORMATION OF BEAMS, TRUSSES AND FRAMES
98
5.2 The differential equation of beam bending
Fig. 5.2
Simple examples of pure bending
Let us then consider the simple cases of pure bending of beams, e.g. cases similar to the cases of y axis bending shown in Fig. 5.2. It is taken for granted that the cross section of the beam is symmetric about the z axis (i.e. about the axis of load direction). For such cases the differential equations 3.26, 3.25 and 3.20 that was developed in Chapter 3.3 are applicable, i.e. EI y w =qz
M y EI y w
and
Vz M y EI y w
(5.3)
Successive integration of w= qz EI y , will then (assuming qz is constant) render the following beam displacements:
2 q x w z C1 x C2 EI y 2 3 2 qz x x w C1 C2 x C3 EI y 6 2 qz x 4 x3 x2 w C1 C2 C3 x C4 EI y 24 6 2 w
qz x C1 EI y
(5.4)
where C1 , C2 , C3 and C4 are unknown constants that may be determined from the boundary conditions of the system. A number of simple worked out cases are shown in Fig. 5.3 below. It is worth noting that
5.2 THE DIFFERENTIAL EQUATION OF BEAM BENDING
and
99
q x2 C1 x C2 M y EI y w EI y z EI y 2 q Vz M y EI y w EI y z x C1 EI y
(5.5) (5.6)
Example 5.1: Deformations of simply supported beam For the simply supported beam at left hand side of Fig. 5.2, the following boundary conditions apply:
w x 0 C4 0 C4 0
M y x 0 EI y C2 0 C2 0 q L2 qL M y x L EI y z C1L 0 C1 z EI y 2 2 EI y 4 3 q L L q L3 w x L z C1 C3 L 0 C3 z 6 24 EI y EI y 24 Thus:
w x
3 2 qz L4 x x x 2 1 24 EI y L L L
and
L 5 qz L4 w x 2 384 EI y
Example 5.2: Deformations of cantilevered beam For the cantilevered beam at right hand side of Fig. 5.2 the following boundary conditions apply:
w x 0 C4 0 C4 0
w x 0 C3 0 C3 0
Vz x L EI yC1 Fz C1
Fz EI y
M y x L EI y C1L C2 0 C2 C1L 2
Thus:
F L3 x x w x z 3 6 EI y L L
and
w x L
Fz L EI y
1 Fz L3 3 EI y
100
5 DEFORMATION OF BEAMS, TRUSSES AND FRAMES
Fig. 5.3
Some worked out cases of pure beam bending
5.3 THE USE OF VIRTUAL UNIT FORCE
101
5.3 The use of virtual unit force
Fig. 5.4
Displacements of cantilevered beam or truss and corresponding stress resultants
Most cases are more complex than those presented in Chapters 5.1 and 5.2 above, and then a more general method is required. This is provided by the principle of virtual forces, which by far is the most effective way of calculating structural displacements. The theory is developed in Appendix A.5. A most simple case is illustrated in Fig. 5.5, where the task is to determine the vertical displacement at midspan w xr L 2 of a simply supported beam subject to a concentrated load F , also at mid span. (For more generality the effects of axial force N as well as cross sectional bending M y are included in the development below although axial force does not occur in the example shown in Fig. 5.5.)
5 DEFORMATION OF BEAMS, TRUSSES AND FRAMES
102
Fig. 5.5
The use of virtual forces to determine structural displacement
The loading sequence is the following: First we apply a virtual load F at the point of, as well as in the direction of the displacement we wish to calculate (in this case at x L 2 ). Under this load the axial force distribution is N x and the bending moment
x . distribution is M y Then the displacements w x as caused by the real load is applied to the
system (whatever it may be but without the real load itself) while F is kept unchanged. The internal normal force and bending moment distributions compatible with w x are N x and M y x . (When using the principle of virtual work to establish equilibrium conditions the loading sequence is the opposite, see Chapter A.4.) Recalling that F is constant x , then the energy balance during this motion, and hence, so are N x and M during w x will require (see Chapter A.5)
5.3 THE USE OF VIRTUAL UNIT FORCE
103
x dA dx Fw r x x
(5.7)
LA
where x
N M y z A Iy
and
x
x
M x N y Fw r A Iy L A
E
1 N My E A I y
1 N My z E A Iy
z . Thus
z dAdx
Taking for granted that x , z are main axes, in which case
zdA 0 , dA A A
and
(5.8)
A
z dA I y , the following is obtained: 2
A
M M y y x NN dx Fw r EA EI y dx L L
(5.9)
Since the normal force within an arbitrary member j is constant, the spanwise EA L , integration of the first term may for each member be replaced by NN j j
and thus, Eq. 5.9 may be expanded into J M M y y x NN L Fw dx r EI y j 1 EA j L
(5.10)
where J is the number of members subject to axial force in the system. The virtual force F is virtual, i.e. we are free to choose its magnitude, and because it is the displacement at xr we aim at quantifying, it seems convenient to set F 1 , in which case J M M NN y y w xr L dx EI y j 1 EA j L
(5.11)
This is what in the literature is often called the virtual unit load method. Its background is in detail explained in Appendix A.1 – A.5. In more general, the structural displacement w , or the cross sectional rotation at position xr may be obtained by
5 DEFORMATION OF BEAMS, TRUSSES AND FRAMES
104
M M y y L dx j L EI y M M x x xr dx GI t L
J NN w xr j 1 EA
(5.12)
where
N , M y and M x are normal force, bending and torsion moments
compatible with all the external loads that are acting on the system, and are the normal forces, bending and torsion moments and M N , M x y
compatible with a virtual unit load that is acting at xr and in the direction of w or , in the case of displacement w xr the virtual load is F 1 , while in
case of cross sectional rotation xr the virtual load is M 1 . Some standard integrations
L
0 f1 x f 2 x dx abL L
Table 5.1 Standard integrations
are given in Table 5.1.
f1 x f 2 x dx abL 0
5.3 THE USE OF VIRTUAL UNIT FORCE
105
Example 5.3: Simply supported beam
Fig. 5.6
Simply supported beam with evenly distributed load q
A simply supported beam with bending stiffness EI y and evenly distributed load q is shown in Fig. 5.6. A free body diagram and the real bending moment diagram M x qx L x 2 are shown at the upper right hand side of the figure. We wish to determine the displacement at mid span as well as the rotation at the left hand support. Then, first we apply a virtual unit load F 1 at mid span of the beam. The
x is shown at the right hand side corresponding virtual bending moment diagram M in
the
middle
of
Fig.
M M Mdx dx M EI y L L
5.6.
The
EI y .
displacement
is
then
given
by
This may readily be obtained by a formal
5 DEFORMATION OF BEAMS, TRUSSES AND FRAMES
106
integration of the relevant functions, but a more effective way of obtaining the same is to use the ready integration formulas given in Table 5.1. Thus
5 qL2 L L 1 5 qL4 2 12 8 4 2 EI y 384 EI y
which is identical to that which has been given in Fig. 5.3. Next, we apply a virtual unit 1 at the left hand side support. The corresponding virtual bending moment M
1 is shown at the lower right hand side of Fig. 5.6. Then the support moment M rotation is given by
L
M M EI y
1 qL2 1 L 2 qL2 1 L 5 qL2 1 L 1 qL3 dx 4 8 2 2 3 8 2 2 12 8 2 2 EI y 24
Example 5.4: Simple truss
Fig. 5.7
Simple truss (gantry)
5.3 THE USE OF VIRTUAL UNIT FORCE
107
A simple truss (gantry) with concentrated load F at its outer tip is shown in Fig. 5.7. We wish to calculate the vertical displacement under the load. EA is identical for the two members. It is readily seen that the tensile forces in members 1 and 2 are given by (with tension defined positive):
N1 F L2 H 2 H
N2 FL H
and
To determine we apply a vertical unit load F 1 at point B. The corresponding virtual forces in members 1 and 2 are then
N1 L2 H 2 H
N 2 L H
and
From Eq. 5.11, the following is obtained:
L2 H 2 2 H NN L p 1 EA p
F L2 H 2 H EA
L FL L2 H 2 H H L EA
2 2 2 FL L H H 1 1 1 EA H L L
Example 5.5: Cable supported mast A mast BD is shown in Fig. 5.8. It is sideways supported at mid span C by a cable AC. The mast is subject to wind loading, idealised into a single horizontal concentrated force F at its top point D. The bending stiffness of the mast is EI y , its axial stiffness is EAm and the axial stiffness of the cable is EAc . A sketch of the displacement figure of the mast is illustrated by broken lines at upper left hand side. As can be seen, the displacement is a combination of bending and compression of the vertical mast and stretching of the cable. The axial force and bending moment diagrams are shown in the middle section of Fig. 5.8. We wish to calculate the horizontal displacement at the top D of the mast, and hence, we apply a virtual horizontal load F 1 at D. The corresponding normal force and bending moment diagrams are shown at the lower section of Fig. 5.8. Thus
J 2 2 2 2 F NN NN MM MM dx L dx 2L EA EI y EI y EAc j 1 EA j L L
2 2 F L 1 FL L L 2 8 EAm
3
EI y
2
FL 4 FL 2 FL3 EAc EAm 3 EI y
108
5 DEFORMATION OF BEAMS, TRUSSES AND FRAMES
Fig. 5.8
A cable supported mast
Example 5.6: Continuous beam A continuous beam across three spans is shown in Fig. 5.9. In its middle span it is
subject to a distributed load q x . The corresponding bending moment diagram is shown in Fig. 5.9. We wish to determine the vertical displacement the mid-span.
w at the middle of
5.3 THE USE OF VIRTUAL UNIT FORCE
109
Fig. 5.9 Three span continuous beam We do this by applying a virtual unit force F 1 at the point of the unknown displacement, on a system which is identical to the parent system, except the outer supports have been removed. This is convenient because it is a statically determinate system, whose bending moment diagram is most easily obtained, as shown in the lower section of Fig. 5.9. Thus
5 DEFORMATION OF BEAMS, TRUSSES AND FRAMES
110
2 3qL2 qL2 L qL L 4 40 20 L MM 1 4 20 L 5 13 qL4 w dx 2 2 EI y EI y EI y 2 12 2 1920 EI y L
As can be seen, integration will only take part in the middle span of the system. This does not mean that the outer spans are obsolete. The effects of the outer spans are
included in the real bending moment diagram M x in the mid span, reflecting that the deformation of the mid span beam is entirely defined by its cross sectional bending curvature.
There are three issues that are important to acknowledge:
During the second part of the loading sequence in Fig. 5.5 (page 102) it is the displacements compatible with all external loads that are applied, without the external loads present. I.e., it is the energy balance during this change of motion that is investigated, not the energy development as the external loads are initially applied to the system. It is taken for granted that the real displacement is small, such that it may be regarded as incremental. The virtual load is imaginary; one is free to choose its position and its size, but not its direction. Otherwise its contribution to the energy x . balance during the real displacements will be different from Fw r Investigation of energy balance during the real displacements may be performed on any statically permissible part of the system, and as shown in Example 5.6 above, it may not necessarily be identical to all of the original system. One is free to choose whatever is most convenient; the only requirement is that it is in a condition of static equilibrium and that it contains the part of the system at the location of the relevant unknown displacement w xr or cross sectional rotation xr .
Chapter 6 Equation Chapter 6 Section 1
STATICALLY INDETERMINATE SYSTEMS 6.1 Introduction We shall in the following take linearity for granted, i.e. that displacements are small and that material behaviour is elastic. We shall limit ourselves to twodimensional systems, i.e. to systems that may be described in x, z coordinates.
As shown in Chapter 2, a system is statically determinate if all support forces may be determined by the general equilibrium conditions:
Fx 0 Fz 0
and
MP 0
(6.1)
where P is an arbitrary point. In this chapter we shall assume that the system is statically indeterminate, i.e. that the three conditions in Eq. 6.1 are not enough to determine support forces. The solution to such systems may be obtained
by using deformation compatibilities of the system (Chapter 6.2), or by using the superposition principle (Chapter 6.3), or to employ a matrix method approach (Chapters 6.4 and 6.5).
The matrix method presented in Chapter 6.4 is a simplified version of the more general finite element method whose basic theory is presented in Chapter 6.5.
6.2 The use of deformation compatibility for simple trusses The method of using deformation compatibility is limited to simple systems where deformations are obvious from the geometry and distribution of stiffness.
© Springer Nature Switzerland AG 2020 E. N. Strømmen, Structural Mechanics, https://doi.org/10.1007/978-3-030-44318-4_6
111
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6 STATICALLY INDETERMINATE SYSTEMS
Fig. 6.1 Rigid bar suspended by two cables
The system in Fig. 6.1 is a typical representative of such systems. It consists of a rigid bar suspended by two identical cables at right angle to the bar. It is simply supported at its left hand support and free at its other end, where it is subject to a concentrated vertical force F . At the right hand side of Fig. 6.1 is shown a freebody-diagram of the system. As can be seen, there are four unknown support reaction forces, and since we only have three equilibrium conditions available (Eq. 6.1), it is necessary to provide one additional equation. This is done by considering the deformation compatibility of the system as illustrated by the broken line in Fig. 6.1, from which it is seen that 1 2 a 2a
1 1 2 2
(6.2)
Since N1 x1 A x1 EA 1 a EA 1 N1a EA 1 N 1 (6.3) N a EA 2 N2 N 2 x2 A x2 EA 2 a EA 2 2
it is seen that
N1 1 N 2 2 N1 N2 2
(6.4)
This, together with the equilibrium conditions
Fx 0 Fz 0 M A 0 renders the following solution:
Rx 0 Rz N1 N 2 F 0 N1a N 2 2a F 3a 0
(6.5)
6.2 THE USE OF DEFORMATION COMPATIBILITY FOR SIMPLE TRUSSES
Rx 0 ,
4 Rz F , 5
3 N1 F and 5
N2
6 F 5
113
(6.6)
It may be of some interest to find the vertical displacement under the load F at the tip of the bar: N a 9 Fa 31 3 1 (6.7) EA 5 EA Another typical case is shown in Fig. 6.2; a gantry supported by three cables. All cables have identical cross section A and elasticity modulus E . A free-body diagram is shown at the upper right hand side, and the pattern of deformation is shown at the lower left hand side.
Fig. 6.2 Gantry suspended by three cables
6 STATICALLY INDETERMINATE SYSTEMS
114
The equilibrium conditions are given by (no moment equilibrium is relevant as all forces N 1 , N 2 and N 3 merge at A ):
Fx 0 Fz 0
N1 sin N 3 sin 0
N1 N 3 (6.8) N1 cos N 2 N 3 cos F 0 2 N1 cos N 2 F
It is taken for granted that the deformation is small, such that the change of angle is negligible, i.e. that d . Thus, if is the vertical displacement of point A , then the stretching of the three cables is given by 2 1 3 cos
(6.9)
From deformation compatibility the following is obtained:
1 cos 2 EA EA EA cos L1 L cos L N 2 x 2 A x2 EA 2 EA EA L2 L
N1 x1 A x1 EA
(6.10)
which combined with the second part of Eq. 6.8 renders the following result 2
FL 2 EA cos cos EA F 1 2cos3 L L EA
1
(6.11)
and thus, F cos 2 2 EA cos L 1 2 cos3 F N 2 EA L 1 2 cos3
N1 N 3
(6.12)
It is seen that the basic idea behind this approach is to use deformation compatibility to establish a relationship between unknown quantities. It goes without saying, that this approach is limited to simple systems. The method also illustrates the general property of elastic systems that if the deformations are known, then so are the internal distribution of forces.
6.3 THE USE OF THE SUPERPOSITIONS PRINCIPLE FOR SIMPLE BEAMS
115
6.3 The use of the superposition principle for simple beams The second option is an approach where known basic cases of beam deformation properties are used together with the superposition principle. Some useful cases have previously been shown in Fig. 5.3. More cases are shown in Fig. 6.3.
Fig. 6.3
Basic cases of beam deformation properties
116
6 STATICALLY INDETERMINATE SYSTEMS
Fig. 6.4 Statically indeterminate beam
Fig. 6.5
Solution strategy
The method is most conveniently illustrated by some typical examples. A statically indeterminate beam is shown in Fig. 6.4. The free-body diagram is shown to the right. As can be seen, there are four unknown support reaction forces, rendering three available force equilibrium equations inadequate. The solution strategy is illustrated in Fig. 6.5. The idea is that if support B is taken away, then the support reaction force RBz is gone, and the system is identical to the one at the top of the diagram on the right hand side of Fig. 6.5, in which case point B will get a downward vertical displacement (see Fig. 5.3):
1 qL4 8EI y
(6.13)
Then, the system is subject to the missing support reaction force RBz , rendering a vertical upward displacement (see Fig. 6.3):
2 RBz L3 3EI y
(6.14)
6.3 THE USE OF THE SUPERPOSITIONS PRINCIPLE FOR SIMPLE BEAMS
117
Since the system is linear, then the superposition principle applies, and thus, deformation compatibility will require
1 2 =0
RB L3 qL4 z 8EI y 3EI y
RBz
3qL 8
(6.15)
The corresponding bending moment and shear force diagrams are shown in Fig. 6.6.
Fig. 6.6
Bending moment and shear force diagrams
Fig. 6.7
Alternative solution strategy
The same result will be obtained if the alternative strategy shown in Fig. 6.7 had been chosen. Then, according to the second upper diagram in Fig. 5.3,
6 STATICALLY INDETERMINATE SYSTEMS
118
1 qL3 24 EI y , while 2 M AL 3EI y (see Fig. 6.3). Setting 1 2 0 the following is obtained
M AL qL3 3EI y 24 EI y
MA
qL2 8
(6.16)
which complies with what was obtained above.
6.4 The matrix method for simple frames
Fig. 6.8 Single bay frame subject to distributed load
A single bay frame is shown in Fig. 6.8. It may be subject to any arbitrary loads, in this case a distributed load along the top beam. Using the superposition principle as illustrated in Fig. 6.9, it is seen that this problem may be solved by the sum of two conditions
one in which the system is subject to artificial joint forces in addition to all its external forces such that all joint displacements are kept at zero, plus one where these artificial forces alone are applied to the system with opposite sign.
The idea is that the first condition can always be taken from known basic cases (e.g. the situation shown to the left in Fig. 6.9 is identical to that which has been given as the fourth case in Fig. 5.3), while the second condition is a case of joint loads alone and then a formal matrix approach may be adopted.
6.4 THE MATRIX METHOD FOR SIMPLE FRAMES
119
Fig. 6.9 Superposition of displacement configurations
Fig. 6.10
Nodal loads and degrees of freedom
Since the solution to the fixed joint condition may be taken from known simple beam cases (as those shown in Figs. 5.3 and 6.3) no further considerations are needed. For the case with only nodal loads, we start off by giving each node and each member a numerical assignment, as shown in Fig. 6.10. According to the boundary conditions of the system, and dependent on which cross sectional displacement components are included, each node has a set of
degrees of freedom of displacements, ri , and
a corresponding set of load components, Ri .
In this case, disregarding axial displacements in the columns, there is two degrees of freedom in each node. Since the system is linear, then there must exist a linear connection between nodal degrees of freedom and corresponding nodal loads, i.e. that
6 STATICALLY INDETERMINATE SYSTEMS
120
K11 Ki1 K Nr 1
K1 j
Kij
K Nr j
K1N r r1 R1 KiN r ri Ri K N r N r rN r RN r
(6.17)
where N r is the total number of degrees of freedom in the system,
K11 K Ki1 K Nr 1
K1 j
Kij
K Nr j
K1N r KiN r K Nr Nr
(6.18)
is the stiffness matrix,
r r1 ri rN r
T
(6.19)
is the nodal displacement vector, and
R R1 Ri RN r
T
(6.20)
is the load vector, such that the system of linear equations is defined by
Kr R
(6.21)
r K1R
(6.22)
The idea is that if
is solved, then the solution to the problem is given by the superposition of the two cases r 0 and r K1R . The load vector R contains all the forces that are required to keep r 0 with opposite signs. It should be noted that the number of cross sectional displacement components may be subject to choice. In Fig. 6.10 each joint has two degrees of freedom, one displacement and one rotation, which implies that the effects of cross sectional axial displacements in the columns have been excluded, assuming that they are small as compared to pure bending. Thus, the system in Fig. 6.10 has four degrees of freedom, i.e.:
6.4 THE MATRIX METHOD FOR SIMPLE FRAMES
K11 K 21 K 31 K 41
K12 K 22
K13 K 23
K 32 K 42
K 33 K 43
K14 r1 R1 K 24 r2 R2 where K34 r3 R3 K 44 r4 R4
R1 o R M 2 0 R3 0 R4 M 0
121
(6.23)
If we also choose to exclude axial deformations in the beam, then r1 r3 and the number of degrees of freedom is reduced to three, i.e.: K11 K12 K K22 21 K41 K42
K14 r1 R1 K 24 r2 R2 K 44 r4 R4
where
R1 0 R M 2 0 R4 M 0
(6.24)
What then remains is to determine the stiffness matrix K . This may most readily be done by observing that Eq. 6.17 is applicable for any external loading to the system. If one arbitrary degree of freedom is set equal to unity ( ri 1 ) while all the others are kept at zero, and
R1 ri 1, all others at 0 Ri ri 1, all others at 0 RN ri 1, all others at 0 r
(6.25)
is the load vector that is necessary for this displacement condition to be in a condition of equilibrium, then K11 K i1 K Nr1
K1 j
K ij
K Nr j
K1N r 0 K1 j R r 1, all others 0 1 i KiN r 1 K ij Ri ri 1, all others 0 (6.26) K N r N r 0 K N r j RN r ri 1, all others 0
from which it is seen that an arbitrary column j i in the stiffness matrix may be determined by consecutively setting one degree of freedom after the other equal to unity while all the others are kept at zero. The method is best understood by a few examples. The solution to a set of linear equations Kr R larger than three by three may be obtained by the Gauss-Jordan elimination
122
6 STATICALLY INDETERMINATE SYSTEMS
method shown in Elaboration 6.1 below (or by a computer programming solution to r K1R ).
Example 6.1: Single bay frame with two degrees of freedom
Fig. 6.11 Simple frame with horizontal load
Fig. 6.12 Relevant nodal loads and degrees of freedom The most simple case of a frame with only two members at right angle with bending properties EI and subject to horizontal load F is illustrated in Fig. 6.11. It is taken for granted that axial deformations are small as compared to bending, and that bending occurs about the y axis, such that for simplicity, the index y on I may be omitted. An illustration of the displacements is illustrated at right hand side of Fig. 6.11, from which it is seen that the system has two degrees of freedom, see Fig. 6.12. It is subject to joint loads R1 F and R2 0 , and since there are no loads in the span of any of its members, a condition where r 0 as illustrated to the left in Fig. 6.9 is obsolete. Thus the problem is defined by:
6.4 THE MATRIX METHOD FOR SIMPLE FRAMES
K11 K 21
K12 r1 R1 K 22 r2 R2
where
123
R1 F R 0 2
Fig. 6.13 Unit displacements and corresponding nodal forces The content of the stiffness matrix may be obtained by (see Fig. 6.3) first setting r1 1 and r2 0 :
K 11 K 21
K12 1 K11 24 EI L3 K 22 0 K 21 12 EI L2
and then setting r1 0 and r2 1 :
K 11 K 21
K12 0 K12 24 EI L2 K 22 1 K 22 9 EI L2
Thus the load situation in Fig. 6.11 is defined by K r R 8 4 L r F 1 3EI . 3 2 r 0 L 4 L 3L 2
Since this is a two by two matrix, inversion is simple and the following is obtained: 1
K r R 3 2 3L 4 L F r1 L 1 FL2 3L r 3EI 2 8 3L2 4 L 4 L 8 0 24 EI 4 2
The displacement figure of the frame is illustrated in Fig. 6.14, together with the corresponding bending moment diagram, which has been obtained for the standard beam and column displacements given in Fig. 6.3. It is worth noting that the stiffness matrix
124
6 STATICALLY INDETERMINATE SYSTEMS
K
3EI 8 4 L L3 4 L 3L2
is symmetric. This is a general property of the stiffness matrix in linear systems.
Fig. 6.14 Displacements and bending moments
Example 6.2: Simple frame with two degrees of freedom
Fig. 6.15 Simple frame with three members
6.4 THE MATRIX METHOD FOR SIMPLE FRAMES
125
Fig. 6.16 Degrees of freedom and cases r1 1, r2 0 , and r1 0, r2 1 A simple frame with three members with bending properties EI and horizontal load F is shown in Fig. 6.15. It has two degrees of freedom r r1 load vector R R1
r2 and corresponding T
R2 F 0 . The problem is defined by Kr R where the T
T
K
four elements of the stiffness matrix K 11 K 21
K12 may be obtained first by setting K 22
r 1 0 , and next by setting r 0 1 as illustrated at right hand side of Fig. T
T
6.1.6. Thus, (see Fig. 6.3):
3EI 12 EI 15EI 3 3 , L3 L L 3EI 6 EI 3EI 2 2 2 , L L L
3EI 6 EI 3EI and 2 2 2 L L L 3EI 3EI 4 EI 10 EI L L L L
K11
K 21
K12
K 22
6 STATICALLY INDETERMINATE SYSTEMS
126
K r R EI 15 3L r1 F L3 3L 10L2 r2 0
rendering
from which the following is obtained: 1
r1 L3 15 3L F L3 1 r EI 2 2 2 3L 10 L 0 EI 141L
10 L2 3L
FL2 3L F 15 0 141EI
10 L 3
The corresponding bending moment diagram as well as support reactions are shown in Fig. 6.17.
Fig. 6.17 Bending moment diagram and support reactions
Example 6.3: Two bay single storey frame A more complex case is illustrated in Fig. 6.18. A two bay frame with four members and a point load F at mid span of one if its members. The system is held against horizontal displacements at support 1, and thus, it has only two degrees of freedom r r1 and corresponding load vector R R1 that
r 0 0
R M 0
T
r2
T
R2 as illustrated in Fig. 6.19. The condition T
is illustrated in Fig. 6.20 from which it is seen that
M 0 where M 0 FL 8 (see Fig. 5.3). T
6.4 THE MATRIX METHOD FOR SIMPLE FRAMES
Fig. 6.18 Two bay frame with four members and mid span point load F
Fig. 6.19 Degrees of freedom
Fig. 6.20 Basic case r 0
Fig. 6.21 Case r1 1 and r2 0
127
6 STATICALLY INDETERMINATE SYSTEMS
128
Fig. 6.22 Case r1 0 and r2 1 Hence, the solution is defined by the situation shown in Fig. 6.20 plus:
K11 K 21
K12 r1 R1 M 0 K 22 r2 R2 M 0
The stiffness matrix may be determined by first setting r 1 0 , see Fig. 6.21: T
K11 K 21
3EI 4 EI 3EI 13EI K12 1 K11 L L L 2 L K 22 0 K 21 1 4 EI 2 EI L 2 L
and then setting r 0 1 , see Fig. 6.22, from which T
K11 K 21
1 4 EI 2 EI L K12 0 K12 2 L K 22 1 K 22 4 EI 3EI 10 EI L L 2 L K
and thus:
EI 13 2 L 2 10
The total solution is given by the situation with the presence of F and r 0 0
T
(Fig. 6.23), plus the situation that R FL 8 1 1 (Fig. 6.24), and defined by T
EI L
13 2 r1 FL 1 2 10 r 8 1 2
r1 FL2 1 r 8 EI 13 10 22 2
10 2
whose solution is given by:
2 1 FL2 13 1 336 EI
4 5
Bending moment diagram and the support reaction are shown in Fig. 6.25 and 6.26.
6.4 THE MATRIX METHOD FOR SIMPLE FRAMES
Fig. 6.23 Bending moment diagram associated with basic case r 0
Fig. 6.24 Nodal forces equal to and with opposite sign of the basic case
Fig. 6.25
Final bending moment diagram
Fig. 6.26 Support reaction forces
129
6 STATICALLY INDETERMINATE SYSTEMS
130
Elaboration 6.1: Solving linear equations by Gauss-Jordan elimination Gauss-Jordan [10,11] elimination may be used directly to solve a set of N linear equations Ax y . The basic idea is that if the expanded matrix G A
y is
consecutively, row by row, reduced by elementary operations, it will after the N th operation come to contain the identity matrix I and the solution x , i.e.:
A y G i th row operation I x N th row operation
Let us for instance consider the system:
12
i.e. A 6
6
12 x1 6 x1 6x 1
6 x2 8 x2
6 x3 x3
1 2
x2
4 x3
3
6 6 1 8 1 and y 2 .We first establish the expanded matrix 1 4 3
12 6 6 1 G A y 6 8 1 2 on which we perform the following manipulations: 6 1 4 3 1.
Row one is divided by its first coefficient a11 12 to create a new matrix
1 0.5 0.5 0.0833 8 2 1 G 6 4 3 6 1 after which we take row two minus row one multiplied by the first coefficient of row two a21 6 , and row three minus row one multiplied by the first coefficient of row three a31 6 , to create a new matrix
6.4 THE MATRIX METHOD FOR SIMPLE FRAMES
131
1 0.5 0.5 0.0833 5 2 2.5 G 0 2 1 2.5 0 2. Then row two is divided by its second coefficient a22 5 to create a new matrix
1 0.5 0.5 0.0833 1 0.4 0.5 G 0 2 1 2.5 0 after which we the take row one minus row two multiplied by the second coefficient of row one a12 0.5 , and row three minus row two multiplied by the second coefficient of row three a32 2 , to create a new matrix
1 0 0.7 0.3333 0.5 G 0 1 0.4 1.5 0 0 0.2 3. Finally we focus on row three by dividing row three by its third coefficient a33 0.2 to create a new matrix
1 0 0.7 0.3333 0.5 G 0 1 0.4 7.5 0 0 1 after which we the take row one minus row three multiplied by the third coefficient of row one a13 0.7 ), and row two minus row three multiplied by the third coefficient of row two a23 0.4 to create a new matrix
x1 4.9167 1 0 0 4.9167 2.5 G 0 1 0 and the answer then x x2 2.5 7.5 0 0 1 x3 7.5
132
6 STATICALLY INDETERMINATE SYSTEMS
6.5 Background to the finite element method
Fig. 6.27 Global degrees of freedom and external loads
For systems with many degrees of freedom the direct approach shown above becomes unmanageable, and then a formal finite element method suitable for computer programming and a numerical solution is more appropriate. The basic idea is that any line-like type of structural system can be described as a collection of beams or beam-column type of N k members (elements) joined together at N p nodes, as illustrated in Fig. 6.27. We have seen that in order to solve a general statically indeterminate system we have used the characteristic structural stiffness properties as described by the connection between element end forces and displacements. We call these connections the element properties. In the following, the end forces and corresponding displacements for an arbitrary element n , whose end points are denoted 1 and 2 (Fig. 6.28), are defined by F Fn 1 F2 n
where
F F F F T 1 1 2 3 T F2 F4 F5 F6
(6.27)
6.5 BACKGROUND TO THE FINITE ELEMENT METHOD
and
d dn 1 d2 n
where
d d d d T 1 1 2 3 T d2 d 4 d 5 d 6
133
(6.28)
Fig. 6.28 Element forces and displacements
We say that this type of element has six degrees of freedom, d1 d 6 . In the following, linearity is taken for granted, and thus, the connection between Fn and dn will be linear. It is assumed that any shape of structural displacements may with sufficient accuracy be described by a combination of rigid body motion and polynomials. We shall take it for granted that the problem at hand is two-dimensional, and hence, any structural system may be described in overall (global) coordinates X , Z (i.e. Y is obsolete). Since the system will comprise elements with three degrees of freedom in each element end, then the number of degrees of freedom in each structural node p will also be three. Thus, as illustrated in Fig. 6.27, at an arbitrary structural node p the nodal displacement degrees of freedom in the overall system are defined by r p r1
r2
r3
T
(6.29)
while the corresponding external structural node forces are defined by: R p R1
R2
R3
T
As there are N p nodes in the system, the global load vector is defined by
(6.30)
6 STATICALLY INDETERMINATE SYSTEMS
134
R R 1 R p
RNp
T
(6.31)
and a corresponding structural (global) displacement vector is defined by r r1 r p
rN p
T
(6.32)
Thus, the total number of degrees of freedom in the system will in general be 3N p . The strategy is the following: 1. All the external gravity and environmental loading is applied to the system under the condition that r 0 . This will require a set of nodal forces R . 2. From this situation all cross sectional forces (bending moments M y , shear forces Vz and normal forces N ) are recorded, e.g. using all the standard cases of simple beams as developed in Figs. 5.3 and 6.3. 3. The connection between element degrees of freedom dn and global system
degrees of freedom r is established (as presented below). 4. The stiffness connection between nodal displacement degrees of freedom r and corresponding forces R is established ( Kr R ). 5. Then nodal force vector R alone is applied to the system, and corresponding nodal displacements r K 1R are determined. 6. From this displacement situation all cross sectional forces ( M y , Vz and N ) are recorded, using the general knowledge between individual element displacements dn and forces Fn . 7. The solution will then be the sum of all force effects from items 2 and 6.
(If all elements are short, then items 1 and 2 may be omitted, as distributed loads may be lumped directly to the nodes as concentrated forces and moments.) Let us start with the analysis at element level. An arbitrary beam element n , with local axis x , y and z is illustrated in Fig. 6.29. The element is subject to a set of forces at element ends 1 and 2, F F1 F2 where F1 F1 F2 F3 T and T
F2 F4
F5
F6 , which are compatible with end displacements d d1 d2 T
T
where d1 d1 d 2 d 3 T and d2 d 3 d 4 d5 T . The element is in a condition of static equilibrium.
6.5 BACKGROUND TO THE FINITE ELEMENT METHOD
Fig. 6.29
Fig. 6.30
135
Six degrees of freedom beam element
Shape functions i , i 1 6 , xˆ x L
It is assumed that the cross sectional displacements at position x along its span
rel x u w
(6.33)
rel x Ψ x d
(6.34)
T
complies with d , such that where
0 0 4 0 0 ψ x 1 0 2 3 0 5 6
(6.35)
where i x , i 1 6 are the shape functions defined in Fig. 6.30. Since they are polynomials and the problem contains an axial force to the element, it should be noted that they represent an approximate solution, as they will not fully satisfy the spanwise differential equation, which, as we shall show in Chapter 12, will require a solution containing a combination of harmonic and hyperbolic
6 STATICALLY INDETERMINATE SYSTEMS
136
functions. However, this will usually not render unduly erroneous results as long as the element length is short and axial forces are not in the vicinity of a stability limit.
Fig. 6.31
Displacements, real and virtual
Applying the principle of virtual work (see Appendix A.3) at an arbitrary position of external and internal equilibrium defined by rel x , the system is subject to an incremental virtual displacement (see Fig. 6.31)
rel x u w
T
(6.36)
compatible with d d 1 d2
where
d d d d T 1 1 2 3 T d2 d 4 d 5 d 6
rel x Ψ x d
such that
(6.37)
(6.38)
Assuming that any shear force displacements may be neglected, the following work balance applies (see Eq. A.36): 6
d F i uEAdx wEI y dx wNdx i 1
L
L
(6.39)
L
where EA and EI y are cross sectional axial and bending properties associated with element n , and where N is the axial force in the element (defined positive as tensile and assumed constant along its span). Eq. 6.39 may also be written
6.5 BACKGROUND TO THE FINITE ELEMENT METHOD
T u T EA 0 u u T 0 0 u d1 F1 d F w 0 EI w w 0 N w dx y 2 2 L
137
(6.40)
Defining
0 4 0 0 0 ψ x 1 0 2 3 0 5 6 and (6.41) 0 4 0 0 0 ψ x 1 0 2 3 0 5 6
EA 0 k0 0 EI y 0 0 n 0 N
such that
u w ψ x d u w ψ x d
u w ψ x d u w ψ x d
and
(6.42)
then Eq. 6.40 may be written
T
T
dT F Ψ d k 0 Ψd x Ψ d n Ψd dx L
(6.43)
L
Performing the matrix transpositions, this may be further developed into
dT F dT ΨT k 0Ψ dx ΨT nΨ dx d L
L
(6.44)
from which it is seen that the pre-multiplication by dT is obsolete and may be discarded. Defining the element material stiffness matrix k and the element geometric stiffness matrix k G
T Ψ k k 0Ψ k T dx G L Ψ nΨ
(6.45)
it is seen that the connection between element end forces and corresponding displacements is given by
k kG d F
(6.46)
138
6 STATICALLY INDETERMINATE SYSTEMS
Performing the matrix multiplications ΨT k 0Ψ and ΨT nΨ in Eq. 6.45:
12 EA 0 0 1 4 EA 0 0 22 EI y 2 3EI y 0 2 5EI y 2 6EI y 32 EI y 0 3 5EI y 3 6EI y k dx 2 EA 0 0 L 4 2 Sym. 5 EI y 5 6 EI y 62 EI y
0 0 0 2 2 2 3 32 kG N L Sym.
0 0 2 5 2 6 0 3 5 3 6 dx 0 0 0 52 5 6 62 0
(6.47)
0
(6.48)
and the ensuing integrations, then the following is obtained:
EA 0 L 12 EI y L3 k sym.
0 6 EI y L2 4 EI y L
EA 0 L 0 12 EI y L3 2
0
6 EI y L
EA L
0 12 EI y L3
0 6 EI y 2 EI y 0 6 EI y 4 EI y
2 L L L2 L
(6.49)
6.5 BACKGROUND TO THE FINITE ELEMENT METHOD
0 0 0 36 3L 4 L2 N kG 30 L Sym.
0 0 0 0 36 3L L2 0 3L 0 0 0 36 3L 4 L2
139
(6.50)
Elaboration 6.2: Beam element with simple support at remote end
Fig. 6.32 Beam element with simple support at remote end In the case of a beam element with a simple support at its remote end (end point 2), see Fig. 6.32, the shape function matrix is defined by
0 0 4 0 ψ x 1 0 7 8 0 9
0 0
7 xˆ 0 1 3 1 7 xˆ 1 0 and 7 1 xˆ 2 xˆ 3 whose properties are 2 2 7 xˆ 1 0 7 xˆ 0 0 8 xˆ 0 0 1 8 xˆ 1 0 3 8 x xˆ xˆ 2 1 whose properties are and 2 2 8 xˆ 0 1 8 xˆ 1 0 9 xˆ 0 0 1 9 xˆ 1 1 and 9 xˆ 2 3 xˆ whose properties are 2 9 xˆ 0 0 9 xˆ 1 0 where ( xˆ x L ). Thus, for such an element the structural stiffness matrix and corresponding geometric matrix are given by
140
6 STATICALLY INDETERMINATE SYSTEMS
12 EA 0 0 1 4 EA 0 2 7 EI y 7 8EI y 0 7 9EI y 82 EI y 0 8 9EI y kn 42 EA 0 L 2 9 EI y Sym.
0 EA L 3EI y L3 Sym.
EA L
0 2
0 0 0 dx 0 0 0
0 3
3EI y L
0
3EI y L
3EI y L
0
3EI y L2
EA L
0 3EI y L3
0 0 0 0 0 0
and
0 0 0 72 7 8 82 kG N L Sym.
0
0
0 7 9 0 8 9 0 0
92
0 0 0 0 0 24 4 L N 0 4 L2 dx 20 L 0 Sym. 0 0
0 0 0 0 24 0 0 4 L 0 0 0 0 24 0 0
Elaboration 6.3: The truss element
Fig. 6.33 The truss element The truss element is defined by a bar without bending, as illustrated in Fig. 6.33. Then the element stiffness matrix is simply obtained from rows one and four in Eq. 6.49:
141
6.5 BACKGROUND TO THE FINITE ELEMENT METHOD
k ndn Fn
where
d dn 1 , d2 n
F Fn 1 F2 n
and
kn
EA 1 1 L 1 1
Elaboration 6.4: Coordinate transformations
Fig. 6.34
Elements at arbitrary position
In some cases the direction of an element does not coincide with the direction of global system axis X , Z (see Fig. 6.34), in which case coordinate transformations may be necessary. From Fig. 6.35 it is readily seen that
d1 d1n cos d 2n sin d1 cos d 2 d1n sin d 2n cos d 2 sin
sin d1n and that d3 d3n cos d 2n
Thus, defining identical displacements in coordinates x, z by d d1 d 2 in coordinates xn , zn by dn d1 n
d Tdn
where
d 2n
T
d 3n , then
cos T sin 0
sin cos 0
0 0 1
d 3 and T
142
6 STATICALLY INDETERMINATE SYSTEMS
Fig. 6.35 Coordinate transformations
1
It is seen that dn T d where T
1
cos sin 0
sin cos 0
0 0 (i.e. T is orthogonal, 1
T1 TT ). Since work is independent of coordinate system, we must have that the T work FT d performed in coordinates x, z must be identical to the work Fn dn performed in coordinates xn , zn , and thus
FT d FnT dn FnT Td FT FnT T
and consequently:
F TT Fn .
Thus, it is seen that forces and displacements follow the same transformations.
Before proceeding to the overall structural (global) level it is necessary to define the six by 3N p connectivity matrix A ( 3N p is the number of degrees of freedom in the system), one for each of the elements in the system, describing the relationship between element degrees of freedom d and global degrees of freedom r , defined such that:
d Ar
(6.51)
143
6.5 BACKGROUND TO THE FINITE ELEMENT METHOD
Then, let us apply to the discrete system (the system as reduced to a discrete collection of nodes) a set of virtual displacements r consistent with r
d A r
(6.52)
Since the virtual work exerted by the external forces at global level must be equal to the sum of the virtual work of the internal stress resultants at element level, then
r T R dT F
(6.53)
Nk
where N k is the total number of elements in the system. Introducing
dT A r r T AT , then T
r T R A r F r T AT F T
Nk
(6.54)
Nk
Again, it is seen that pre-multiplication by r T is obsolete, and thus, the equilibrium condition at a global structural level is given by R AT F
(6.55)
Nk
Recalling from Eqs.6.46 and 6.51 that F k kG d k kG Ar it is seen that R AT k k G Ar
(6.56)
Nk
Thus, the connection between nodal load vector R and structural degrees of freedom r is given by Nk
Ktotr R where Ktot K KG
n (6.57) T A kG A n
K AT kA and
n 1 Nk
KG
n 1
Since k n and k Gn are symmetric, so is K and K G . Thus, the solution with respect to structural displacements and internal forces is given by:
144
6 STATICALLY INDETERMINATE SYSTEMS
1 r K tot R d Ar F k k G d
(6.58)
The solution strategy may either be based on a matrix inversion of K tot (e.g. according to a Gauss-Jordan elimination as shown in Appendix B.1), or on a solution strategy based on a Cholesky decomposition of K tot , as described in Appendix B.2, though it should be noted that there are also other numerical solution strategies available. There are two alternatives to the Cholesky decomposition; the original approach and the alternative version. In the original approach K tot is decomposed into the product of a lower triangular matrix U (having values only on and below the diagonal) and its transposed, such that
Ktot UUT
UUT r R
(6.59)
Introducing the substitute variable
x UT r it is seen that
Ux R
r UT
Thus,
(6.60)
1
x U1R
x
(6.61) (6.62)
The reason why this approach is widely in use is that the solution to a linear system of equations is particularly effective when the coefficient matrix is a lower diagonal matrix. In the alternative version of the Cholesky decomposition K tot is decomposed into the product of a lower triangular matrix L with values only below the diagonal and ones on its diagonal, a diagonal matrix D and the transposed of L such that
K LDLT
(6.63)
LDLT r R
(6.64)
x DLT r
(6.65)
rendering
Substituting it is seen that Lx R
x L1R
(6.66)
6.5 BACKGROUND TO THE FINITE ELEMENT METHOD
145
from which
D x
r LT
1
1
(6.67)
Again, because L is triangular and D is diagonal the matrix operations are simple and quick.
Elaboration 6.5: Condensation of obsolete degrees of freedom Any degrees of freedom rn that are equal to zero due to structural restrictions or boundary conditions may be discarded simply by deleting the row and column associated with rn . If two degrees of freedom rn and rn j are linearly dependent of each other, then the relevant rows and columns associated with rn and rn j may be added linearly. E.g., if Kr R and of freedom, then:
r Trred where rred contains a reduced number of degrees
T KT r T
TT R .
red
In some cases it may also be a convenient solution strategy to reduce the number of degrees of freedom in a system from N r N r1 N r2 to N r1 , if for instance the N r2 degrees of freedom are considered obsolete and may be ignored. The system reduction may then be obtained by rewriting the original equilibrium condition
K totr R
where
from which it follows that condition is given by
K tot K KG
into
K11 K12 r1 R 1 K 21 K 22 r2 R 2
1 1 r2 K22 R2 K22 K21r1 and thus the reduced equilibrium
K
11
1 1 K12K 22 K 21 r1 R1 K12K 22 R2
Example 6.4: Portal frame Let us set out to determine the stiffness matrix K of the frame in Fig. 6.36.a, assuming that we may ignore axial deformations in the columns. Each of the elements n 1,2,3 has six degrees of freedom with local stiffness matrix (see Eq. 6.49):
146
6 STATICALLY INDETERMINATE SYSTEMS
0 EA L 12 EI y 0 6 EI y 0 kn 0 EA L 0 12 EI y 0 6 EI y
EA L
0 2
0
0
3
3
L
6 EI y L
0
12 EI y L
L2
4 EI y L
0
6 EI y L2
0
EA L
0
L
2
6 EI y L
0
12 EI y L3
L2
2 EI y L3
0
6 EI y L2
3
6 EI y L 2 EI y L 0 6 EI y L2 4 EI y L n
a) Frame geometry and global degrees of freedom
b) Six degrees of freedom element Fig. 6.36
Single bay single storey frame
2
147
6.5 BACKGROUND TO THE FINITE ELEMENT METHOD
Then the total stiffness of the system is defined by: K
3
ATn k n An
where An is the
n1
connectivity matrix for each of the elements, as defined by: dn A nr where
dn d1 d 2 0 1 0 A1 0 0 0
0 0 1 0 0 0
d3
d4
0 0 0 0 0 0
0 0 0 0 0 0
d 6 and r r1 r2 T
d5
1 0 0 A2 0 0 0
0 0 1 0 0 0
r4 . Thus, it is seen that: T
r3
0 0 0 0 0 1
0 0 0 1 0 0
0 0 0 A3 0 0 0
0 0 0 0 0 0
0 1 0 0 0 0
T
Performing the matrix multiplications An k n An the following is obtained:
12 EI y H 3 6 EI y H 2 1 1 2 4 EI y H A1T k1A1 6 EI y H 1 1 0 0 0 0
EA L 2 0 T A2 k 2 A2 EA L 2 0
0 0 AT3 k 3A3 0 0 Thus:
0 0 0 0
EA L 2
0
4 EI y L 2
12EI 6 EI
0 0 y y
EA L 2
2 EI y L 2
0
H
H3
1
2
1
2 EI y L 2 0 4 EI y L 2 0
0
0
0 0 0 0 0 0 0 0
0 0
6 EI y
H2 1 H 1
4 EI y
0 0 1 0 0 0
6 STATICALLY INDETERMINATE SYSTEMS
148
6 EI EA 12 EI1 EA2 21 2 0 H 3 L L H 6 EI1 2 EI 2 4 EI1 4 EI 2 2 0 H L L H K 6 EI 3 EA2 12 EI 3 EA2 2 0 3 L L H H 6 EI 3 2 EI 2 4 EI 2 4 EI 3 2 0 L H H L In many cases it may render a sufficiently accurate solution if axial deformations in the beam are also ignored. In that case r1 r3 , and thus, the number of degrees of freedom may be reduced to three by defining:
rˆ rˆ1
rˆ2
rˆ3
T
such that
r C rˆ r1 1 0 0 r 0 1 0 rˆ1 2 rˆ r3 1 0 0 2 rˆ3 r4 0 0 1
ˆ ˆ Rˆ substituting r Crˆ into Then the problem Kr R may be replaced by Kr
Kr R and pre-multiplying by CT such that CT KCrˆ CT R , and thus it is seen that: 12 E I1 I 3 6 EI 6 EI 21 23 3 H H H 6 EI1 2 EI 2 4 EI1 4 EI 2 T ˆ K C KC 2 and H L L H 6 EI 3 2 EI 2 4 EI 2 4 EI 3 2 L H H L R1 R3 T ˆ R C R R2 R4
Chapter 7 Equation Chapter 7 Section 1
STRESSES IN COMPOSITE BEAMS 7.1 Introduction
Fig. 7.1 Composite beam cross section
A composite (or hybrid) beam is in the following defined as a beam with cross section made up of a layer of segments with different thickness and material properties, as illustrated in Fig. 7.1. It is taken for granted that layers of material segments are parallel to the y axis and that the cross section is symmetric about the z axis. It is also taken for granted that there is a fully functioning shear connection between all layers, such that the cross section behaves as a compact unit. In structural engineering there is a multitude of such design systems, e.g. reinforced concrete beams, plywood or multi-layered wooden beams, and a large variety of composite laminates made up of plastic or glass fibre materials.
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149
7 STRESSES IN COMPOSITE BEAMS
150
7.2 Normal stresses in composite beams Normal stresses may occur due to axial force or bending, in this case about the y axis (parallel to the direction of material segment), as bending about the z axis is identical to that which has been presented in Chapter 3. Based on Navier’s hypothesis, that a cross section in the y , z plane will remain plane during any axial or bending displacements, the development of normal stresses for composite beams follows the same calculus procedures as shown for beams with homogeneous cross section in Chapter 3.
Composite beam subject to pure axial force N
Fig. 7.2
Hence, a beam (Fig. 7.2) with length L subject to axial force N acting at the cross sectional centroid will render an axial strain x L L and corresponding axial stress x z E z x (7.1) where E z is the relevant modulus of elasticity at position z . Thus, it is seen that pure axial force will render a stress variation in the cross section due to variations in elastic properties. Since
N x z dA E z x dA= x E z dA A
and thus
A
x
(7.2)
A
N
E z dA A
(7.3)
7.2 NORMAL STRESSES IN COMPOSITE BEAMS
x z E z x
it follows that
E z N
E z dA
151 (7.4)
A
Fig. 7.3
Composite beam subject to pure bending M y
During pure bending it is assumed that the deformation w x is small. During its deformation from zero to an arbitrary position w x an infinitesimal element with length dx has gone through the displacement shown in Fig. 7.3, and thus, a material element dA at position z above the main y axis has been subject to strain
x x, z Thus,
w x z w x dw x z dx
w x z
x x, z E z x x, z E z w x z
(7.5)
(7.6)
where E z is the z dependent elastic modulus. The corresponding bending moment is given by:
7 STRESSES IN COMPOSITE BEAMS
152
M y x z x x, z dA z E z w x z dA E z z 2 dA w x A
A
(7.7)
A
By defining cross sectional stiffness with respect to bending about y axis
EI y E z z 2dA
(7.8)
w x M y x EI y
(7.9)
A
it is seen that
from which the following is obtained (see Eq. 7.6)
x x, z E z w x z M y x
E z z EI y
(7.10)
Since the cross section in this case is subject to pure bending, we must demand
N x x, z dA A
M y x
E z zdA 0 EI y A
(7.11)
from which it is seen that the position of the neutral y axis is defined by
E z zdA 0
(7.12)
A
7.3 Shear stresses in composite beams The development follows the procedure outlined in Chapter 3.4. Consider an infinitesimal beam element dx subject to bending and shear, see Fig. 7.4. On its left hand side the cross section is subject to a linear variation of bending induced normal stresses x z , on its right hand side the normal stresses are
x z d x z . Recalling that shear stresses occur in pairs (see Eq. 3.33), it is the shear stress z z at arbitrary position z that has to balance out the normal stress increase d x z , as illustrated in Fig. 7.4. The part of the cross section outside of z is As .
7.3 SHEAR STRESSES IN COMPOSITE BEAMS
Fig. 7.4
153
Shear stress at arbitrary position z
Considering the slice of element with length dx and cross sectional area As , it is seen that it is subject to normal stresses x and x d x as well as shear stresses z z on either side. It is acknowledged that on the outer (upper) surface there cannot be any shear stress, because shear stresses occur in orthogonal pairs they must vanish at the outer edges of the cross section. Thus, equilibrium in x direction
x dA As
x d x dA z z bdx 0
(7.13)
As
where b is the width of the cross section at z . Thus, it follows that
z z
1 d x dA b A dx
(7.14)
s
where x x, z is given in Eq. 7.10. Recalling (Eq. 3.20) that M y x Vz x , the following is obtained:
z z
E z 1 d M y x b A dx EI y s
where ES y
V x z dA z b EI y
E z zdA
As
E z zdA is the y axis static moment of
As
As .
Vz ES y b EI y
(7.15)
7 STRESSES IN COMPOSITE BEAMS
154
Example 7.1: Three layer rectangular cross section
Fig. 7.5
Composite cross section subject to bending M y and shear force V z
Let us consider the single-symmetric three layer rectangular cross section in Fig. 7.5 subject to M y and V z . The position of neutral axis origo is given by:
b E1t1 t1 2 E2t2 t1 a t2 2 Ea t1 a 2 c b E1t1 E2t2 Ea
1 Eˆ1tˆ1 2 tˆ1 1 tˆ1 tˆ2 2 Eˆ 2tˆ2 1 2 a
where
1 Eˆ1tˆ1 Eˆ 2tˆ2
EI y
b E1t13 E2t23 Ea 3 12
E bt c t 1 1
2
2
t2 a E2bt2 h c Eba c t1 2 2 2 1
The normal stress distribution (shown in Fig. 7.6):
x1
My
EI y
Eˆ1 E1 E Eˆ 2 E2 E tˆ1 t1 a tˆ t a 2 2
E2 h c , x2
My
EI y
E2 h c t2
2
7.3 SHEAR STRESSES IN COMPOSITE BEAMS
x3 x2 E E2 , x4
My
EI y
x6
Fig. 7.6
155
E c t1 , x5 x4 E1 E , and My
EI y
E1c
Normal stress distribution due to pure bending moment M y
Let us assume the cross section is subject to shear force V z , as illustrated in Fig. 7.5. Then, within upper layer, h c t2 z h c :
hc z b 2 E2 h c z 2 2 2 VE t V E 1 2 z1 z 2 t2 h c 2 z z 2 h c z 2 2 EI y EI y 2
ES y E2b h c z
Within middle layer, c t1 z h c t2 :
ES y E2bt2 h c
t2 h c t2 z Eb h c t2 z 2 2
t Eb E2bt2 h c 2 h c t2 2 z 2 2 2
z
Vz t2 E 2 2 E2t2 h c h c t2 z 2 2 EI y
7 STRESSES IN COMPOSITE BEAMS
156
z2 Vz z3 EI y
Vz EI y
t2 E 2 E 2 t2 h c 2 2 h c t 2
t2 E 2 2 E2t2 h c h c t2 c t1 2 2
The shear stress distribution due to shear force V z is illustrated in Fig. 7.7.
Fig. 7.7
Shear stress distribution due to shear force V z
Elaboration 7.1: Bi-axial bending In case of bi-axial bending M y and M z of a composite beam whose cross section is symmetric about y and
z axes, the position of the centroid is still defined as developed
in Eq. 7.12, i.e. by
E z zdA 0 A
and
E y ydA 0 .Due A
superposition principle will apply, and thus x y, z M y
to linearity, the
E y, z E y, z z Mz y EI y EI z
where E y , z is the symmetric variation of material properties across the cross section, while EI y E y , z z 2 dA and EI z E y , z y 2 dA
A
A
Chapter 8 Equation Chapter 8 Section 1
NON-SYMMETRIC BEAM CROSS SECTIONS 8.1 Main axis and normal stress calculations
Fig. 8.1 Main axis system x, y, z
The main axis system x, y, z , with x in direction of the beam span, is the axis system of pure bending according to a linear variation of the strain distribution across the cross section as illustrated in Fig. 8.1. We have seen in Chapter 3 (see Eqs. 3.29, 3.31 and Elaboration 3.2) that origo and orientation of main axes are then defined by y 0 z dA 0 (8.1) A yz 0
© Springer Nature Switzerland AG 2020 E. N. Strømmen, Structural Mechanics, https://doi.org/10.1007/978-3-030-44318-4_8
157
8 NON-SYMETRIC BEAM CROSS SECTIONS
158
Fig. 8.2
Normal stresses due to axial stretching and bi-axial bending
Due to linearity the superposition principle applies, and hence, the stress effects of axial force N and biaxial bending components M y and M z about main axes are additive (see Fig. 8.2). Thus, the main axis system is defined such that x
Fig. 8.3
My N Mz y z A Iz Iy
(8.2)
Axis x1, y1, z1 with origo at centroid but arbitrary main axis orientation
8.1 MAIN AXIS AND NORMAL STRESS CALCULATIONS
159
Let us assume that the cross section is subject to axial force N and biaxial bending M y1 and M z1 in an axis system x1, y1, z1 different from the main axis system, see Fig. 8.3. It is assumed that its x1 axis coincides with the direction of main x axis, i.e. that the
y1, z1
system has its origo in the cross sectional
centroid, but it may have an arbitrary orientation . The position of cross sectional centroid may be determined from another and conveniently chosen axis
system x * , y * , z * , with its axis parallel to x1, y1, z1 , by demanding y1
0
z1 dA 0
(8.3)
A
i.e. by setting (see Fig. 8.3) y * y0* y0* y* y1 0 dA dA dA z1 z* z* z* z* A 0 A A A 0 0
(8.4)
and thus, the origo of the x1, y1, z1 axis system is defined by:
Fig. 8.4
y0* 1 y * * * dA z0 A A z
(8.5)
Axial stretching and bending
Since axial and lateral displacements u, v, w are all referred to the centroid of the cross section, then (see Fig. 8.4, see also similar development in Chapter 3.3):
8 NON-SYMETRIC BEAM CROSS SECTIONS
160
x u
vy1 v dv y1 wz1 w dw z1 dx1 dx1
u vy1 wz1 1 y1
T z1 u v w
x E x E 1 y1 z1 u v w
T
and thus
(8.6)
(8.7)
from which the cross sectional stress resultants are obtained by integration
N 1 1 u M z1 E y1 x dA E y1 1 y1 z1 v dA A z A z w 1 1 M y1 y1 z1 u 1 E y1 y12 y1 z1 dA v A 2 w z y z z 1 1 1 1
(8.8)
Then
N A 0 I z1 M z1 E 0 M y1 0 I y1z1
A 1 0 y 1 0 u 0 z1 I y1z1 v where 2 dA I y z1 A 1 I y1 w I 2 y1 z1 I y z y1z1 1 1
(8.9)
and thus
A 0 u v 1 0 I z1 E w 0 I y1z1
0 I y1x1 I y1
1
N M z1 M y1
(8.10)
The normal stress is then given by (see Eq. 8.7):
and thus:
x 1 y1
A 0 z1 0 I z1 0 I y1z1
0 I y1x1 I y1
1
N M z1 M y1
(8.11)
8.1 MAIN AXIS AND NORMAL STRESS CALCULATIONS
161
1 Iz I y1z1 M z1 1 z1 I y1z1 I y1 M y1 I z M y I y1z1 M z1 N I y M z I y1z1 M y1 1 1 y1 1 1 z1 2 A I y1 I z1 I y1z1 I y1 I z1 I 2y1z1
N x y1 A
Fig. 8.5
(8.12)
The orientation of the main axis system
The orientation of main axis system y, z is defined by I yz yzdA 0 . This A
defines the angle between
y1, z1
and the main axis
y, z
systems. This
transformation is illustrated in Fig. 8.5, from which it is seen that
y1 cos z sin 1
sin y cos z
Thus
y cos sin y1 z sin cos z 1
I z y 2 dA y12 cos2 y1z1 sin 2 z12 sin 2 dA A
A
I z1 cos2 I y1z1 sin 2 I y1 sin 2
I y z 2 dA y12 sin 2 y1 z1 sin 2 z12 cos2 dA A
A
I z1 sin 2 I y1z1 sin 2 I y1 cos2
(8.13)
(8.14)
(8.15)
8 NON-SYMETRIC BEAM CROSS SECTIONS
162
I yz yzdA y12 z12 sin cos y1 z1 cos2 sin 2 dA A
A
1 I z1 I y1 sin 2 I y1z1 cos 2 2
(8.16)
Since I yz 0 in main axes (see Eq. 8.1), it is seen that the angle between the
y1, z1
system and the y, z system is defined by tan 2
2 I y1z1
(8.17)
I z1 I y1
From the sum of Eqs. 8.14 and 8.15 it is also seen that
I z I y I z1 I y1
(8.18)
dI z I z1 I y1 sin 2 2 I y1z1 cos 2 d dI y I z1 I y1 sin 2 2 I y1z1 cos2 d
(8.19)
Since
are both equal to zero for the main axis condition in Eq. 8.17, it may be concluded that I z and I y are extreme values of the second area moments in the cross section, one a maximum value and the other a minimum. Cross sectional stress may either be calculated in the y1, z1 system according to Eq. 8.12, or in main axes according to Eq. 8.2, whichever is most convenient. The connection between bending moment components in the two systems are given by: M y cos M z sin
sin M y1 cos M z1
M y1 cos M z1 sin
sin M y cos M z
(8.20)
Example 8.1: Properties of equal legged L shaped cross section Given an L shaped cross section with equal length legs a and thickness t . We shall set out to determine the location and orientation of its main axis system.
8.1 MAIN AXIS AND NORMAL STRESS CALCULATIONS
Fig. 8.6
163
Angular cross section
We start out with choosing a convenient axis system y * , z *
shown at upper right hand
side illustration. The static moment of the cross section about these axes are given by a
* * * y dA y tdy A
0
a
1 2 a t 2
and
* * * z dA z tdz A
0
1 2 a t 2
from which the following position of the main axis origo is obtained ( see Eq. 8.5):
y0*
a 2t 2 a 2at 4
z0*
and
a 2t 2 a 2 at 4
Then we define a new axis system y1, z1 , parallel with y * , z * , but with its origo coinciding to the newly obtained origo of the main axis system (see lower left hand side illustration of Fig. 8.6). Thus,
I y1 z12dA A
3a 4
a 4
3a 4
z12tdz1
2
5 3 a 4 tdy1 24 a t a 4
8 NON-SYMETRIC BEAM CROSS SECTIONS
164
I z1 y12dA A
3a 4
3a 4
2
5 3 a 2 4 tdz1 y1 tdz1 24 a t a 4 a 4
I y1z1 y1z1dA A
3a 4
3a 4
1 3 a a 4 z1tdz1 y1 4 tdy1 8 a t a 4 a 4
The orientation of the main axis system is then defined by , such that (Eq. 8.17):
tan 2
2 I y1z1
I z1 I y1
2 90
45 (see Fig. 8.6)
The corresponding cross sectional bending properties are given by
I y z 2dA I z1 sin2 I y1z1 sin 2 I y1 cos2 a 3t 12 A
I z y 2dA I z1 cos2 I y1z1 sin 2 I y1 sin2 a 3t 3 A
I yz yzdA A
1 I z I y1 sin 2 I y1z1 cos2 0 2 1
8.2 The calculation of shear stresses
Fig. 8.7
Shear stress at arbitrary position z
8.2 THE CALCULATION OF SHEAR STRESSES
165
For non-symmetric cross sections the shear stresses du to shear forces may be calculated in main axes according to the theory outlined in Chapter 3.4. However, in most cases it is more convenient to perform the calculations in axes
y1, z1
that are parallel to the plate elements within the cross section, e.g. as
shown in Example 8.1 above and also illustrated in Fig. 8.7. The equilibrium condition in Eq. 3.34 still applies (see Fig. 8.7), i.e.
x dA As
x d x dA z z1 bdx 0
(8.21)
As
and thus Eq. 3.35 also holds, i.e.
z z1
1 d x dA b A dx
(8.22)
s
where z is defined positive parallel to axes y1, z1 . The only difference is that
x must be taken from Eq. 8.12, and thus I y M z I y1z1 M y1 I z M y I y1z1 M z1 d x 1 1 y1 1 1 z1 2 dx I y1 I z1 I y1z1 I y1 I z1 I y21z1
I y1V y1 I y1z1Vz1 I y1 I z1 I y21z1
y1
I z1Vz1 I y1z1V y1 I y1 I z1 I y21z1
(8.23)
z1
where it has been introduced that M y1 Vz1 and M z1 Vy1 (Fig. 3.13 and Eq. 3.20). Thus
z z1
I z Vz I y1z1V y1 1 d x 1 I y V y I y1z1Vz1 dA 1 1 y1 1 1 z1 dA 2 b A dx b A I y Iz I y z I y1 I z1 I y21z1 1 1 11 s s
I z1Vz1 I y1z1V y1 1 I y V y I y1z1Vz1 S S 1 1 z y 1 1 b I y I z I y2 z I y1 I z1 I y21z1 1 1 11
(8.24)
where b is the width of the cross section at position y1, z1 and where S y1 S z1
z1
y1 dA
As
(8.25)
8 NON-SYMETRIC BEAM CROSS SECTIONS
166
As shown in Chapter 8.1 above, the main axis system is defined by I yz 0 . I.e., if I y1z1 0 then axis system y1, z1 coincides with main axes y, z , and it is seen from Eq. 8.24 that
V 1 Vy Sz z S y b Iz I y
z z
where
S y Sz
z
y dA
(8.26)
As
which is identical to that of a cross section which is symmetric about y and z axes and is subject to bi-axial shear forces V y and V z .
Example 8.2: Bending and shear stresses in angular cross section
Fig. 8.8
Angular cross section subject to bending M y1 and shear Vz1
Let us consider the same equal legged L shaped cross section that was dealt with in Example 8.1 above, and assume it is subject to bending moment M y1 and shear force
Vz1 , i.e. that M z1 0 and V y1 0 . From Example 8.1: I y1 I z1 5a3t 24 and I y1z1 a3t 8 . Then the normal stress component is given by (see Eq. 8.12):
8.2 THE CALCULATION OF SHEAR STRESSES
x
I y1z1 M y1 I y1 I z1
I y21z1
y
I z1 M y1 I y1 I z1
I y21z1
from which the following is obtained:
z
x
M y1 I y1 1
1 I y21z1
167
I y1z1 y1 z1 I y1 I z1 I z1
15 M y1 3 y1 z1 3 2 a t 5
Fig. 8.9 Normal stress distribution x
M
y1
(see Fig. 8.9)
a 2t
The shear stress component is given by (see Eq. 8.24):
Vz I z1Vz1 1 I y1z1Vz1 1 1 S S z y 2 2 1 1 t I y Iz I y z tI y1 I y1 I z1 I y1z1 I y2 z 11 1 1 1 1 1 I y1 I z1
z
3 4
1 3 24
Iy z S y1 1 1 S z1 I z1
In vertical leg S z1 t a z1 a z1 and S y t a z1 1 Then
3 4
a 4
8 NON-SYMETRIC BEAM CROSS SECTIONS
168
z
Vz1 2 I y1
1 1
I 2y1z1
I y1 I z1
15 Vz1 3a 3a aI y1z1 3a 9a z1 z1 z1 z1 3 4 4 4 20 a t 4 2I z
a 4
Similarly, in the horizontal leg S z1 at t y1
a1 a y1 42 4
a a a t y1 , and then 4 4 4
and S y at 1
z
2 I y1z1 2 9a 2 a 3a ay 3a 2 1 15 Vz1 3 y y y 1 1 3 2 I y1 16 2 4 2 80 I y21z1 I z1 4 a t 5 1 I y1 I z1
Vz1
The shear stress distribution is shown in Fig. 8.10. The shear stress distribution has its apex values at the positions on the cross section where the normal stress is zero.
Fig. 8.10
Shear stress distribution x Vz1 at
Chapter 9 Equation Chapter 9 Section 1
SOME SPECIAL STRUCTURAL SYSTEMS 9.1
The theory of simple arches
Fig. 9.1
Triple joint circular arch subject to concentrated force at mid-joint
In spite of containing four support reaction forces, a symmetric triple joint arch (circular as well as parabolic) like the one illustrated in Fig. 9.1.a is a statically determinate system, because all reaction forces may be determined from the equilibrium conditions of forces as shown in Fig. 9.1.b, i.e.
Fx 0 Ax Bx 0 A B F 2 Fz 0 Az Bz F 0 Ax B x F 2 z z Az L 2 Ax L 2 0 M C 0 B L 2 B L 2 0 x z
© Springer Nature Switzerland AG 2020 E. N. Strømmen, Structural Mechanics, https://doi.org/10.1007/978-3-030-44318-4_9
(9.1)
169
9 SOME SPECIAL STRUCTURAL SYSTEMS
170
Equivalent equilibrium conditions apply if the arch is subject to a constant and vertically directed force qz x that is distributed along the x axis, in which case:
Ax Bx qz L 2
and Az Bz qz L 2
(9.2)
The circular triple joint arch in Fig. 9.2.a is subject to concentrated force Fz at position a from left hand side support and self-weight q z s mg (where m is mass per unit length of the arch). Since self-weight is distributed along the arch itself, its effect needs special considerations. A free-body diagram is shown in Fig. 9.2.b.
Fig. 9.2
Circular arch subject to self-weight qz s mg and concentrated force Fz at arbitrary position a from support A
First, let us consider the concentrated force Fz alone. Then:
Fx 0 Fz 0 M A 0 MC 0
Ax Bx 0 Ax Fz a L Az Bz Fz 0 Az Fz 1 a L Fz a Bz L 0 Bx Fz a L Bz L 2 Bx L 2 0 Bz Fz a L
(9.3)
9.1 THE THEORY OF SIMPLE ARCHES
171
The corresponding cross sectional normal force, shear force and bending moment may be determined from the equilibrium conditions in tangential and arch normal coordinates of a segment as shown in Fig. 9.3.a.
Fig. 9.3
Equilibrium conditions in tangential and arch normal coordinates
Then, in the interval 0 2 :
N Bx sin Bz cos V Bx cos Bz sin 0 V Bx cos Bz sin (9.4) L M Bx r sin Bz r 1 cos 0 M Bx sin Bz 1 cos 2 N Bx sin Bz cos 0
Similarly, in the interval 2 F , where cos F 1 a r and r L 2
N Bx sin Bz cos 0
N Bx sin Bz cos V Bx cos Bz sin 0 V Bx cos Bz sin M Bx r sin L Bz r 1 cos 0 M Bx sin Bz 1 cos 2 while, in the interval F :
(9.5)
172
9 SOME SPECIAL STRUCTURAL SYSTEMS
N Ax sin Az cos 0 N Ax sin Az cos V Ax cos Az sin 0 V Ax cos Az sin M Ax r sin L Az r 1 cos 0 M Ax sin Az 1 cos 2
a) Concentrated force Fz
(9.6)
b) Self-weight
Fig. 9.4 Normal force, shear force and bending moment of circular arch; left subject to concentrated force Fz at a L 0.25 , right hand side distributed load q z s
The result is shown in Fig. 9.4.a for the case that a L 4 , as functions of the
non-dimensional angular arch coordinate (as defined in Fig. 9.3.a). For the situation of self-weight alone (see Fig. 9.3.b), the system is symmetric, and acknowledging that ds rd and assuming for simplicity that qz s mg is constant, then, from equilibrium requirement in vertical direction
9.1 THE THEORY OF SIMPLE ARCHES
2
Az Bz
qz rd
0
2
qz r
4
173
qz L
(9.7)
while a moment equilibrium about point C renders 2
Bz r Bx r
qz r cos rd 0
(9.8)
0
from which the following is obtained:
Bx
qz L 1 2 2
(9.9)
Finally, from simple horizontal equilibrium: Ax Bx . Again, the cross sectional normal force, shear force and bending moment may be determined from the equilibrium conditions in tangential and arch normal coordinates of a segment as shown in Fig. 9.3.b. Then
0 V Bx cos Bz sin qz rd sin 0 (9.10) 0 M Bx r sin Bz r 1 cos r cos cos qz rd 0 0 N Bx sin Bz cos qz rd cos 0
from which the following is obtained:
q L V z 1 cos sin 2 2 2 2 qz L M sin cos 1 cos 1 sin 4 2 2 N
qz L 1 sin cos 2 2 2
(9.11)
These diagrams are shown in Fig.9.4.b as functions of non-dimensional angular arch coordinate (as defined in Fig. 9.3.b).
9 SOME SPECIAL STRUCTURAL SYSTEMS
174
Fig. 9.5
Parabolic arch subject to self-weight
Similarly, the support reaction forces and cross sectional normal force, shear force and bending moment as those developed above, may also be developed for triple joint arches with a parabolic geometry za 4 f x L 1 x L
(9.12)
as illustrated in Fig. 9.5 above. Let us for simplicity assume that the arch is subject to self-weight qz s alone. Then, vertical equilibrium requires Az Bz
1 qs ds 2
(9.13)
L
where
2 1 dz 2 dz ds dx 2 dza2 1 a dx 1 a dx 2 dx dx
(9.14)
Introducing dz a dx 4 f L 1 2 x L , and then
Az Bz
L 1 4 f 1 q z 1 2 2 L 0
x 1 2 L
2
2 qz L 8 f 1 dx 2 3 L
Similarly, due to symmetry and horizontal equilibrium: Ax Bx . Thus,
(9.15)
175
9.2 THE SHALLOW CABLE THEORY
moment equilibrium of the left hand side half about point C will require L Ax f Az 2
L2
0
L x qz ds 2
L L 1 4 f Ax f Az x qz 1 2 2 2 L 0 L
x 1 2 L
2
(9.16)
dx 0
from which the following is obtained:
Ax Bx
2 qz L2 4 f 1 8 f 3 L
(9.17)
It may be favourable to optimize the shape of an arch such that the bending moment along its span is as small as possible.
9.2
The shallow cable theory
While stiffness properties of beams or beam-columns mainly rely on cross sectional bending and torsion ( EI y , EI z and GIt ), a cable relies almost entirely on its axial stiffness property EA and the presence of an axial force N . It is in the following assumed that we are dealing with a cable whose bending and torsion properties are negligible. It is taken for granted that the cable is suspended in the gravity field and that it is fairly shallow, i.e. that its sag is less that about a tenth of its suspended length ( f L 0.1 ). To start off with, we shall develop a solution based on the assumption that the cable geometry can be approximated by a parabola, first for the situation that the cable supports are at identical levels, then the more general case that cable supports are at different levels. In Elaboration 9.1, the more accurate and classical catenary solution is presented (catenary as derived from the Latin word for chain and commonly used for the geometry of an idealised cable). It is convenient to choose coordinate axes as shown in Fig. 9.6.a. Thus, let us consider the situation of a shallow cable in the gravity field with the aim of determining the support forces Vz and H due to the distributed gravity force qz s mg , where m is the mass per unit length of the cable.
9 SOME SPECIAL STRUCTURAL SYSTEMS
176
Fig. 9.6
The shallow cable in the gravity field
Due to symmetry, vertical equilibrium requires
Vz x 0 Vz x L qz 2
(9.18)
where is the total length of the cable
ds
where
ds dx 2 dzc2
(9.19)
L
and where zc is the cable geometry (its position in z direction). Similarly, due to symmetry, moment equilibrium of the left hand side half about its support point will require (see Fig. 9.6.c) L2
Hf
0
q x qz ds 0 H z f
L2
xds
0
where f is the cable sag. Adopting a parabolic cable geometry
(9.20)
9.2 THE SHALLOW CABLE THEORY
zc
4f x x 1 L L
177 (9.21)
a solution may be obtained by introducing the approximation 12
ds dx
2
dzc2
dz 2 1 c dx dzc 4 f dx L
Then
1 dz 2 dx 1 c dx 2 dx
x 1 2 L
(9.22)
(9.23)
and thus L 1 4 f ds 1 2 L L 0
x 1 2 L
2
8 f 2 dx L 1 3 L
(9.24)
The vertical component of the support reaction forces is then given by (see Eq. 9.18) 2 1 q L 8 f Vz x 0 Vz x L qz z 1 (9.25) 2 2 3 L The corresponding horizontal component of the support reaction force is obtained from the integration of Eq. 9.20, and thus H
qz f
L2
0
xds
qz f
L2
0
1 4 f x 1 2 L
x 1 2 L
2
2 qz L2 4 f dx 1 (9.26) 8f 3 L
(This is equivalent to that which was obtained for the parabolic arch subject to self-weight in Chapter 9.1 above.) In the case of non-symmetric cable geometry (see Fig. 9.7), the same incremental equilibrium conditions apply, i.e. horizontal, vertical and moment equilibrium of an element with length
ds is defined by (see Fig. 9.7.c):
dH 0 dVz qz ds 0 (9.27) dz dz dx dx H c Vz H dH c Vz dVz Hdzc Vz dx 0 2 2 2 2
178
9 SOME SPECIAL STRUCTURAL SYSTEMS
Fig. 9.7
Non-symmetric cable geometry
Thus, it is seen that H is constant and
dVz qz ds 0 Vz dx Hdzc 0
(9.28)
If the cable is sufficiently shallow (e.g. f L less than about 1 10 ), then ds dx 1 dzc dx dx
(9.29)
dVz qz dx
(9.30)
2
in which case
H
and thus
d 2 zc qz dx2
(9.31)
From simple integration, it is then seen that
zc
qz 2 x C1 x C2 2H
(9.32)
9.2 THE SHALLOW CABLE THEORY
179
where C1 and C 2 are integration constants that may be determined from the boundary conditions zc x 0 0 and zc x L h . Thus
zc
qz L2 x x x 1 h 2H L L L
L h q L2 f zc x z 2 2 8H
The cable sag is given by:
zc
and thus
8f L2
and
H
qz L2 8f
(9.33)
(9.34)
(9.35)
Since H is constant, the equilibrium condition requires Ax Bx H and Az Bz qz qz L (see Fig. 9.7.b). Moment equilibrium about point A
requires qz L L 2 Hh Bz L 0 and thus
Bz
qz L h 1 2 4 f
and Az
qz L h 1 2 4 f
(9.36)
The axial forces N A and N B at supports A and B may then be obtained by: 2 qz L2 4f h 1 8f L L 2 qz L2 4f h 2 2 N B B x Bz 1 8f L L
NA
Ax2 Az2
(9.37)
Elaboration 9.1: The general shallow cable theory The more accurate classical solution may be developed by using a calculus approach. The force equilibrium requirements for an element ds subject to vertical gravity qz ds (see Fig. 9.6.d) are given by:
H x H dVz x dVz x qz ds 0 qz ds dH x 0
9 SOME SPECIAL STRUCTURAL SYSTEMS
180
Thus, it is seen that H will in any approach be constant along the entire span of the cable. Moment equilibrium about the finite element mid span is given by
H
dzc dx dz dx Vz H dH c Vz dVz Hdzc Vz dx 0 2 2 2 2
Vz x H
rendering
dzc from which it is seen that dx
H
d 2 zc dx
2
qz
ds 0. dx
H qz and acknowledging that ds dx 1 dzc dx 2 , it is seen that
Defining
zc 1 1 zc2 . Introducing zc dzc dx tan (see Fig. 9.6.d), then: zc
dx
d 1 d 1 1 1 1 tan 2 from which it is seen that tan 2 cos dx cos dx
cos 2
cos
d
cos 1 sin2
d . The integral of the left hand side of this equation is
dx x C1 , where C1 is an unknown integration constant. The integral on the right hand side may be obtained by substituting rendering
cos
sin (hence cos d d ),
1
1 sin2 d 1 2 d Arctanh ,
i.e.
x C1 Arc tanh ,
and thus sin tanh x C1 . Furthermore, since
cos 2 1 sin 2 1 2 1 tanh 2 x C1 cosh 2 x C1 it is seen that cos cosh 1 x C1 , and since
zc tan
tanh x C1 sin sinh x C1 cos co cosh x C 1 1
then the following cable geometry is obtained
zc sinh x C1 dx cosh x C1 C2
H q cosh z x C1 C2 qz H
where C1 and C2 are integration constants determined by the relevant geometric
boundary conditions. If the boundary conditions are defined by zc x 0 0 and
zc x L 0 , then
H H q L cosh C1 C2 0 and cosh z C1 C2 0 qz qz H
181
9.2 THE SHALLOW CABLE THEORY
q L which is satisfied if C1 z 2H
and C2
H q L cosh z . Thus, the cable qz 2H
geometry for a cable whose supports are at identical levels is given by:
H qz
zc
qz x qz L qz L cosh H 2 H cosh 2 H
qz x qz L qz L qz x It is seen that zc sinh sinh is zero at mid span, i.e. at H
2H
2H
H
x L 2 . Hence, the cable sag is given by: f zc x L 2
H qz
qz L cosh 2 H 1
and thus:
H
qz f q L cosh z 1 2H
The well-known parabola solution already presented above can be obtained by using a series expansion cosh 1 2 2! 4 4! , where it for small arguments of will suffice to include only the first two terms. Thus
zc
H qz
1 q L 2 1 q x q L 2 q L2 x x z z z z 1 1 1 2 2 H 2 H 2 H 2 H L L
2 2 N H 2 Vz2 H 1 Vz H 1 dzc H dx The tensile force is given by: qz x qz L 2 qz x qz L H cosh H 1 sinh H H 2 2H H qz L N x 0 N x L H cosh rendering: 2H N x L 2 H
The cable length may be obtained by integration, i.e. 2 L dzc 2 qz L qz x ds dx 1 dx 1 sinh 2 H H dx 0 0 L
L
L
H 2H q L q x q L q x q L sinh z cosh z z dx sinh z z H H q H H q 2 2 0 2H z z 0
9 SOME SPECIAL STRUCTURAL SYSTEMS
182 Since sinh
3 3!
, then:
2H qz
2 q L 1 q L 3 1 qz L z z L 1 2 H 3! 2 H 24 H
Assuming that the cable stiffness is constant along its span, then the cable elongation due to the gravity field is given by
ds
E L
L
L H 1 dzc dx N ds EA EA L 0
ds
2
1 dzc dx dx 2
2 L L H dzc H q x q L cosh 2 z z dx 1 dx EA 0 dx EA 0 H 2H
Substituting qz x H qz L 2 H then the following is obtained: qz L
H2 qz EA
2H
cosh
qz L
qz L
2
H 2 sinh 2 2 H HL H qL d sinh z 1 q L qz EA 4 2 z 2 EA qz L H
2H
2H
which may be expanded into: 3 2 HL H qz L 1 qz L HL 1 qz L 1 1 2 EA qz L H 3! H EA 12 H
2 8 f 2 HL 16 f q z L2 L 1 1 then: Introducing f and EA 3 L 8H 3 L
9.3 Beams on elastic foundation A beam resting on an elastic foundation along its entire length is shown in Fig. 9.8.a. As usual, we set out to determine the equilibrium conditions in a calculus approach, see Fig. 9.8.b. It is assumed that deformations are small and that the beam and its foundation are elastic. With the definition in Fig. 9.8, equilibrium requirements are identical to those developed in Eqs. 3.17 – 3.19, except there is now an additional resistance force from the foundation equal to k w x , where
k is the elastic stiffness coefficient of the foundation (with unit N m2 ). For simplicity, it is assumed that k is constant along the span of the beam.
9.3 BEAMS ON ELASTIC FOUNDATION
183
a) Line like beam on elastic foundation
b) Equilibrium conditions Fig. 9.8
Beam on elastic foundation
Thus,
Fz 0 :
Vz Vz dVz qz dx kw dx 0 Vz qz kw 0 (9.38) dx dx M y 0 : Vz 2 Vz dVz 2 M y M y dM y 0 Vz M y
With sign definitions in Fig. 9.8, Eq. 3.25 ( M y EI y w ) is still applicable, and thus
9 SOME SPECIAL STRUCTURAL SYSTEMS
184
w x
q x k wx z EI y EI y
(9.39)
This is the differential equation of a beam on elastic foundation. Its solution to this differential equation may be obtained as the sum of the particular solution
w p x qz x k
(9.40)
and the general solution to the homogeneous part
wh x
k wh x 0 EI y
(9.41)
which is given by wh x exp xˆ C1 cos xˆ C2 sin xˆ exp xˆ C3 cos xˆ C4 sin xˆ
where
xˆ x Le
(9.42)
and
Le 4 EI y k
14
(9.43)
is the “elastic length”, and C1,, C4 are constants dependent on the system load and geometry. Thus, a general solution is given by
w x w p wh
(9.44)
Example 9.1: Beam on elastic foundation subject to concentrated force Let us consider an infinitely long beam on an elastic foundation which is subject to a concentrated force F . It is convenient to choose an axis system where origo coincides with the position of F . Since qz 0 then w p x 0 . To obtain a converging solution at the boundary conditions lim w x 0 it is a necessary requirement that x
C3 C4 0 . Thus: w x exp xˆ C1 cos xˆ C2 sin xˆ It is readily seen that:
where
xˆ
x Le
185
9.3 BEAMS ON ELASTIC FOUNDATION
w
1 exp xˆ C2 C1 cos xˆ C2 C1 sin xˆ Le
M y EI y w EI y Vz M y EI y
2 exp xˆ C1 sin xˆ C2 cos xˆ L2e
2 exp xˆ C2 C1 cos xˆ C2 C1 sin xˆ L3e
Due to symmetry and vertical equilibrium at the point of the concentrated load, the boundary conditions are the following:
1 C1 C2 0 Le
C1 C2
2 F C C2 3 1 2 Le
C1 C2
w x 0 Vz x 0 EI y
FL3e 8EI y
Fig. 9.9 Beam with infinite length subject to concentrated force F
9 SOME SPECIAL STRUCTURAL SYSTEMS
186 Thus:
FLe x exp xˆ cos xˆ sin xˆ at xˆ M y x EI y w x 0 4 Le F Vz x exp xˆ cos xˆ 2 w x
FL3e exp xˆ cos xˆ sin xˆ 8 EI y
These are illustrated in Fig. 9.9 above.
9.4 Cylindrical shells As illustrated in Fig. 9.10, it is in the following assumed that the shell is cylindrical and that it has a constant radius r , which is much larger than its thickness t , i.e. the shell is thin walled such that t r 1 . It is not a requirement that t is constant, but a non-constant t will require a numerical approach not included below. It is assumed that the load q x is radial. A free body diagram of an infinitesimal element dx , ds where ds rd is shown at the upper right hand side of Fig. 9.10. Radial equilibrium requires:
Vx Vx dVx rd N ddx qdxrd 0
(9.45)
where Vx is the shear force and N is the normal (membrane) shell force, both per unit length (i.e. with unit N m ). Thus, radial equilibrium requires Vx N r q 0
(9.46)
Similarly, moment equilibrium about x axis requires
M dx M dM dx 0
dM 0
(9.47)
Moment equilibrium in tangential direction will require:
M x rd M dM rd Vx rd
dx dx Vx dVx rd 0 2 2
(9.48)
9.4 CYLINDRICAL SHELLS
Fig. 9.10
187
Cylindrical shell
It is seen (Eq. 9.47) that M must be constant. Thus, ignoring higher order terms, it is seen that
Vx M x
(9.49)
Combining Eqs. 9.46 and 9.49, the following is obtained: 1 M x N q r
(9.50)
Based on the assumption that r t it is taken for granted that the radial deformations are purely radial, as illustrated at the lower right hand side of Fig. 9.10, while the longitudinal of deformations (in the x direction) follows Navier’s theory of pure bending (see Eq. 3.22). Thus,
9 SOME SPECIAL STRUCTURAL SYSTEMS
188
s r w d rd w x ds rd r
w dw z wz w x z r x x, z dx
x
E w x r x x, z E z r w x
x
(9.51)
(9.52)
from which it is seen that N x t x tE x
Et w x r
(9.53)
and (see Eq. 3.23): 2 r t 2 w z r z r x Mx Et dz D w M z r dA 0 2 w 1 r t 2 z r A r
where D
Et 3
12 1 2
(9.54)
is the plate stiffness of the shell wall. [That M 0 is
obvious since we already have established that E Ew x r is constant across the thickness of the shell wall.] From Eqs. 9.50, 9.53 and 9.54, the following differential equation for a cylindrical shell is then obtained
d 2 d 2 w Et D w q dx 2 dx 2 r 2
(9.55)
If t is constant this simplifies into: w
Et q w 2 D Dr
(9.56)
The solution is a sum of particular and homogeneous solutions, w p and wh : w w p wh
(9.57)
9.4 CYLINDRICAL SHELLS
189
Et q w 2 p D Dr
(9.58)
where
wh
and
4 wh 0 L4e
14
and where
4 Dr 2 Le Et
rt
12
3 1 2
(9.59)
1 4
(9.60)
is the elastic length. Thus
wp qr2 Et and
wh e xˆ C1 cos xˆ C2 sin xˆ e xˆ C3 cos xˆ C4 sin xˆ
(9.61) (9.62)
where xˆ x Le and C1 ,, C4 are constants to be determined from the relevant boundary conditions. It is readily seen that Eq. 9.59 may be replaced by
d 4 wh 4wh 0 dxˆ 4
(9.63)
Example 9.2: Cylindrical shell subject to inner pressure A thin walled cylinder shell (Fig. 9.11), whose length is much larger that its radius,
r L 1 , is subject to inner pressure p (with unit N m 2 ). At one end the shell is clamped to a wall, i.e. at x 0 it is fixed against any deformations. As illustrated in Fig. 9.11, the solution may then be based on the following superposition procedure:
First we apply the inner pressure p and let the shell be free to move at x 0 , a situation in which the shell displacement will be a constant expansion:
wp x pr2 Et .
Then, with no inner pressure we apply a set of unknown forces M 0 and V0 at
x 0 , such that the system is brought back to its original position where w x 0 0 and w x 0 0 , a situation in which the solution is wh e xˆ C1 cos xˆ C2 sin xˆ e xˆ C3 cos xˆ C4 sin xˆ .
9 SOME SPECIAL STRUCTURAL SYSTEMS
190
Fig. 9.11
Cylindrical shell fixed at one end and subject to inner pressure
To obtain a converging solution it is a requirement that C3 C4 0 . Thus:
w wp wh pr2 Et e xˆ C1 cos xˆ C2 sin xˆ from which
w
1 xˆ e C1 cos xˆ sin xˆ C2 cos xˆ sin xˆ Le
w
2 xˆ e C1 sin xˆ C2 cos xˆ Le
w
2 xˆ e C1 cos xˆ sin xˆ C2 cos xˆ sin xˆ L3e
From boundary conditions
9.4 CYLINDRICAL SHELLS
w x 0
191
C C1 pr 2 0 C1 0 and w x 0 2 Le Et
the following is obtained:
C1 C2
pr 2 Et
Thus:
w
pr 2 1 e xˆ cos xˆ sin xˆ Et
M x Dw D
2 pr 2 xˆ pL2e xˆ ˆ ˆ e x x e cos xˆ sin xˆ cos sin 2 L2e Et
Vx M x pLe e xˆ cos xˆ N
Et w pr 1 e xˆ cos xˆ sin xˆ r
Fig. 9.12 Radial displacement w and forces ( M x , Vx , N ) in cylindrical shell fixed at one end and subject to constant inner pressure p
192
9 SOME SPECIAL STRUCTURAL SYSTEMS
A plot of these quantities is shown in Fig. 9.12 above. (Please note that M x has the unit
Nm m , while Vx and N have units N m .)
As can be seen:
pr 2 w Et 0 M lim x x Vx 0 N pr
Chapter 10 Equation Chapter 10 Section 1
THE THEORY OF TORSION 10.1 Introduction
Fig. 10.1 Cantilevered beam with crossbar subject to transverse forces F Consider a cantilevered beam with a stiff crossbar at its outer tip, as illustrated in Fig. 10.1. At either end the crossbar is subject to a transverse force F . This will cause a twisting x along the entire length of the beam. The force effect that causes this twisting is the torsion moment M x , which in this particular case is equal to 2Fa . The way a line-like beam type of system will carry a torsion moment depend on the type of cross section, as illustrated in Fig. 10.2. In a circular or rectangular tube, the torsion moment will cause a constant shear stress flow around the entire cross section, as illustrated in Fig. 10.2.a. In a thin walled open cross section, the torsion moment will cause two shear stress effects, as illustrated in Fig. 10.2.b. First, there will be a shear stress flow in each of the plate elements within the cross section. This shear flow varies linearly across the thickness of each of the plate elements. This effect is called St. Venant torsion. Secondly, there will be a constant shear flow across each of the outer flange elements, an effect which will cause the flanges to bend in opposite directions as illustrated in Fig. 10.3. This effect is called warping torsion. As can be seen, the effect of warping is not only shear, but also normal stresses due to bending in the flanges. In general, the torsion moment stress effects will be a combination of St.
© Springer Nature Switzerland AG 2020 E. N. Strømmen, Structural Mechanics, https://doi.org/10.1007/978-3-030-44318-4_10
193
194
10 THE THEORY OF TORSION
Venant torsion and warping. We shall in the following assume elastic, homogeneous and isotropic material behaviour. We shall take it for granted that any cross section is thin walled or else compact.
a) Circular or rectangular cross section
b) Thin-walled cross sections Fig. 10.2 Torsion induced shear stresses
Fig. 10.3 Torsion induced cross sectional warping
10.2 TORSION OF THIN-WALLED TUBE OR COMPACT CIRCULAR SECTION
195
10.2 Torsion of thin-walled tube or compact circular section
Fig. 10.4 Circular thin walled tube subject to torsion moment M x Let us first consider the simple cases of circular cross sections, starting with the tube shown in Fig. 10.4. The tube is thin-walled, such that t r 1 (e.g.
t r 0.1 ), where r is the mean radius of the tube. It is a fair assumption that the shear stress is constant across the wall thickness t , and that the shear flow t is constant around the entire cross section, i.e. that t is independent of its position . Then the torsion moment M x is given by M x r dA
(10.1)
A
As shown in Fig. 10.4, the connection between the shear angle caused by at an arbitrary position around the tube wall and the cross sectional twist is that
and thus
dx rd r
(10.2)
G G r
(10.3)
Introducing Eq. 10.3 into Eq. 10.1 the following is obtained:
196
10 THE THEORY OF TORSION
M x r G r dA
2
A
3 r G r t rd G 2 r t
(10.4)
0
I t 2 r 3t
Identifying the torsion constant it is seen that
x M x GI t
and thus
x
(10.5) (10.6)
Mx r It
(10.7)
The corresponding cross sectional twist along the span of a cantilevered beam like the one in Fig. 10.1, is then given by x
x
Mx M dx x x GI t GI t 0
x d 0
x L
Mx L GI t
(10.8)
Fig. 10.5 Compact circular cross section with diameter 2 R
Similarly, for a compact circular cross section with diameter 2 R , see Fig. 10.5, the shear angle of an element dx, rd at arbitrary position , r is dx rd r and G Gr . Thus
M x r dA A
2 R
2 4 R 4
r Gr rd dr G 0 0
(10.9)
10.3 ST. VENANT TORSION OF COMPACT CROSS SECTION
Identifying the torsion constant
It
2 4 R 4
(10.10)
Mx GI t
(10.11)
Mx r It
(10.12)
x
it is seen that
x, r
and thus
197
I.e., the shear stress x, r is linearly distributed, as illustrated in Fig. 10.5. For any other case than those presented above, a more comprehensive approach will be required, as presented below.
10.3 St. Venant torsion of compact cross section For any compact cross section (other than the circular cross section presented in Chapter 10.2 above) the semi inverse theory of St. Venant is adopted. The situation is illustrated in Fig. 10.6. Cross sectional rotation due to a constant torsion moment M x is x , and the displacements at an arbitrary point on the cross section are u , v and w . The basic assumption is that u , v and w are proportional to d and that u (the warping component) is independent of spanwise position x . It is assumed that the member is line like such that is constant (see Eqs. 10.6 and 10.11) and that x 0 0 , i.e. that
x x 0 x .
Fig. 10.6
Torsion of compact cross section
198
10 THE THEORY OF TORSION
u y , z v xz w xy
Thus
(10.13)
where y , z is an unknown St. Venant warping function (which for a compact circular cross section is zero). From elementary theory of elasticity (see Chapter 3) the following applies (please recall that is constant, see Eqs 10.6 and 10.11): x du dx 0 x 0 (10.14) y dv dy 0 y 0 z dw dz 0 z 0
and (see Fig. 10.7 and 10.8) d du dv d z xy G xy G z dy dx dy dy dv dw yz v, w yz G yz 0 0 dz dy du dw d d xz u, w y xz G xz G y dz dx dz dz
xy u, v
(10.15)
As could be expected in a case of pure torsion, all normal stresses are zero, and that the shear stress on the y plane in the direction of z , i.e. yz , is also zero. The other shear stress components are illustrated in Fig. 10.7. The equilibrium conditions of the element dx, dy , dz in Fig. 10.7 are as follows
Fx zx d zx dxdy zx dxdy yx d yx dxdz yx dxdz 0 Fy xy d xy dydz xy dydz 0 Fz xz d xz dydz xy dydz 0
(10.16)
(No contributions) dx dz M d dydz d dxdy 0 (10.17) y xz xz xz zx zx zx 2 2 dx dy M z xy d xy xy dydz 2 yx d yx yx dxdz 2 0
Mx 0
10.3 ST. VENANT TORSION OF COMPACT CROSS SECTION
199
Fig. 10.7 Shear stress components
Fig. 10.8
The connection between displacements and shear angles
Ignoring higher order terms, the following is obtained: d zx d yx 0 dz dy
(10.18)
d xy 0 xy is constant zx xz and that d xz 0 xz is constant xy yx
(10.19)
From Eqs. 10.15, 18 and 19 it is seen that
d 2 d 2 d zx d yx d xz d xy G 2 2 0 dz dy dz dy dy dz and thus
(10.20)
200
10 THE THEORY OF TORSION
2 y , z 0
(10.21)
d d 2 is the Laplace operator. For practical use St. Venant 2 dy dz warping function is cumbersome. It is far more convenient to introduce Prandtl stress function y , z , defined by
where 2
d xy dz
and
d xz dy
(10.22)
With this definition the equilibrium condition in Eq. 10.18 is fulfilled, and from Eq. 10.15 it is seen that d d G z dz dy
d d G y dy dz
xy xz
d 2 d 2 G 1 2 dz dydz d 2 d 2 G 1 dy 2 dydz
(10.23)
Thus,
d 2 d 2 2 2 y, z 2G 2 dy dz
(10.24)
This is then the differential equation of Prandtl stress function. The boundary condition of torsion induced stresses is that any shear stress component perpendicular to the surface of the cross sectional must be zero (Fig. 10.9), i.e.:
xy
dz dy xz 0 ds ds
d dz d dy 0 dz ds dy ds
d 0 ds
(10.25)
which implies that y , z is constant at the edges of the cross section. Since this is the only boundary condition of y , z it is usually convenient to choose
y, z equal to zero at the outer edges of the cross section. The torsion moment is then given by
d d M x xz y xy z dA y z dA dy dz A A d d ydz dy zdy dz dy dz A A
(10.26)
10.3 ST. VENANT TORSION OF COMPACT CROSS SECTION
201
Fig. 10.9 Boundary conditions Noting that is constant (or zero) at outer edges of the cross section, and integrating the two terms by parts, the following is obtained:
M x ydz y dzdy zdy z dydz 2 dA z
A
y
A
(10.27)
A
In general we define the St. Venant torsion constant It by (see Eq. 10.11)
Mx GI t
(10.28)
similar to that of cross sectional bending moment resistance. Thus (see Eqs. 10.24 and 10.27) M 4 (10.29) It x dA G 2 A The bending moments M y and M z are zero, because x is zero.
Example 10.1: St. Venant torsion of rectangular cross section Let us consider the massive rectangular cross section shown in Fig. 10.10.a. Let us for 2 simplicity adopt the parabolic Prandtl stress function c 1 2 z t .
202
10 THE THEORY OF TORSION
Fig. 10.10
Rectangular massive cross section
It is illustrated in Fig. 10.10.b, it satisfies the boundary conditions at z t 2 , but not at the shorter edges of the cross section, i.e. at y b 2 . However, we shall assume that
t b , such that the error is sufficiently small (e.g. t b 0.1 ). Then
M x 2 dA 2b A
2
2
t2
t 2
4 2 c 1 2 z t dz btc 3
2
d d 8c 2 2 2G 2 dy dz t and since
Mx it is seen that: GI t
c
3 Mx 4 bt
4c 3M x 2 Gt Gbt 3
1 I t bt 3 3
d 0 xz dy The shear stress is defined by: xy d 8 cz 6 M x z 2 M x z dz It t2 bt 3 M (see Fig. 10.10.c) from which it is seen that xy x t max It
203
10.4 SINGLE CELL THIN WALLED HOLLOW CROSS SECTION
Example 10.2: St. Venant torsion of a thin walled open cross section As shown in Example 10.1 above, for a compact rectangular cross section I t bt 3 3 and the shear stress varies linearly across its thickness. For a thin walled open cross section containing several rectangular plate elements, each denoted i , and each following
the total rotation of the cross section, such that i , where i is the rotation of an arbitrary plate element and is the rotation of the entire cross section, then
M x M xi GI t i i GI t i i
such that
i
i
Mx M x and thus GI t i GI t
GI t GI t i i
i
The torsion moment carried by each plate is M xi GI t i i GI t i
Thus: xyi 2
M xi I ti
zi 2
GI t i GI t
Mx
1 Gbt 3 3 i
GI t i GI t
i
Mx
M G 1 M G zi 2 x i zi and xyi x i ti max I ti GI t GI i
If the cross section is homogeneous and all parts have same G then: I t
1 bt 3 3 i
i
10.4 Single cell thin walled hollow cross section The circular thin walled tube has been dealt with in Chapter 10.2 above. Let us now turn to the more general case of a single cell hollow cross section of any shape. Since torsion induced cross sectional displacements take place in the direction of its surface coordinates u and vs (see Fig. 10.11), the shear angle is given by: dv dv du xs (10.30) xs s du xs s ds dx dx G dx G and thus, the cross sectional continuity condition is defined by dv u Gxs dxs ds 0 C
dv
Gxs ds dxs ds C
C
(10.31)
204 where
10 THE THEORY OF TORSION
indicates integration around any closed curve C within the solid area
C
of the cross section (where there are torsion induced stresses), see Fig. 10.11. If we consider an arbitrary point of rotation P , then
dvs r dx where As
Gxs ds rds 2 As
C
(10.32)
C
1
2 rds is the area within C .
Fig. 10.11
Fig. 10.12
Single cell hollow cross section
Prandtl stress function and torsion moment integration
10.4 SINGLE CELL THIN WALLED HOLLOW CROSS SECTION
205
Since we define the torsion constant M x GI t it follows from Eq. 10.32:
M
Gxs ds GIxt 2 As
xs ds =2
C
C
Mx As It
(10.33)
As illustrated in Fig. 10.12, it is a permissible assumption that Prandtl stress function is zero at the outer perimeter R y and constant at the inner perimeter Ri . In most cases a parabolic shape will suffice, i.e. that
1
r
2 r
c1 c2 1 2 t t 2
(10.34)
where r is radial distance from center of rotation and t is wall thickness, both functions of the perimeter coordinate s , and where c1 and c2 are unknown constants. However, for simplicity we shall assume that the cell wall is thin, such that a linear variation ( c2 0 ) is considered sufficient, i.e.
c1 1 2 r t
(10.35)
Thus, the shear stress components are defined by
xr
d 0 ds
and
xs
d c1 dr t
(10.36)
Defining the shear flow q xst it is seen that this is a quantity that is constant
(equal to c1 ), although both xs and t may vary around the circumference of the cross section. Performing the integration in Eq. 10.33 with respect to the curve Rm through the mid thickness of the cell wall (see Fig. 10.12 and Eq. 10.33), then ds M M 2A (10.37) xs ds c1 t 2 I tx Am c1 I tx dsm Rm Rm t R m
From Fig. 10.13 it is also seen that
Mx
1
xst rds 2q 2 rds 2q Am 2c1 Am
Rm
Rm
(10.38)
206
10 THE THEORY OF TORSION
Shear flow q and torsion moment M x
Fig. 10.13
Thus, from Eqs. 10.37 and 10.38, Bredt’s formula is obtained It
4 Am2
(10.39)
ds t
Rm
10.5 Multiple cells thin walled hollow cross section For a multiple cells cross section (see Fig. 10.14) we shall assume the same Prandtl stress function that we adopted for the single cell case presented above, i.e. it is assumed thin walled. Thus, for an arbitrary cell i
i ci 1 2 ri ti
xsi
d i ci dri ti
q i xsi ti ci
(10.40)
and the shear flow around each cell is constant such that (see Eq. 10.37) 2
Mx ds ds Ami xs i ds ci q i It t i t i Rm Rm Rm i
i
(10.41)
i
where Ami is the area within the cell center wall thickness periphery Rmi , see Fig. 10.16. Let us for instance consider a three cell cross section, see Figs. 10.14, 10.15 and 10.16. Then the following applies:
10.5 MULTIPLE CELLS THIN WALLED HOLLOW CROSS SECTION
Fig. 10.14
Fig. 10.15
Fig. 10.16
Multiple cells thin walled cross section
Shear flow at cell wall intersections A and B
Definition of cell area within mid thickness periphery
207
208
10 THE THEORY OF TORSION
Cell 2:
Rm 2
Cell 3:
where
Mx A m1 t t It 12 Rm 1 q 3 q 2 q1 Mx ds ds ds 2 Am2 t t t It 12 23 q 3 q 2 Mx t ds t ds 2 I t Am3 32 Rm 3 q1
Cell 1:
ds
q 2
ds 2
(10.42)
means integration along joint walls between cells i and j , and
ij
q i xs t i is the shear flow in cell i . Since q i is constant, this may also be written
a11 a12 a a 21 22 0 a32
0 q1 2M x a23 q 2 It a33 q 3
ds Am aii t i 1 Rm i Am2 where a ds ij t ij Am3 ij
(10.43)
Defining
I t q q 2M x a Am1
Am2
where
Am3
T
q q1 q q 1
q 2 q 2
a11 and A a21 0
T q 3 a12 0 a22 a23 a32 a33 q 3
T
(10.44)
then the equation above and its solution is given by
a Aq Since M xi 2 q i Ami and q i
A 1a q
2M x q i then It
M x M xi 2 q i Ami 4 i
and thus
(10.45)
i
Mx It
q i Ami i
(10.46)
10.5 MULTIPLE CELLS THIN WALLED HOLLOW CROSS SECTION
T a 4 A 1a I t 4 q i Ami 4q i
T
a 4aT A 1
T
a
209 (10.47)
T
and since xsi q i ti the shear stress vector τ xs1 xs2 xs3 is given by
τ
2M x Cq It
where
1 C diag ti
(10.48)
Example 10.3: Three cells rectangular cross section
Fig. 10.17
Three cells rectangular cross section
For the cross section in Fig. 10.17 the following is obtained:
b b b b b 2 t 4t 4t 2t t A1 A3 b2 2b b 2b b 7 b a22 and 2 4t t 2t t 2 t A2 2b b a12 a21 a23 a32 t q 2 1 0 q1 1 6 2 1 1 1 1 t b 2 q 2 4 2 b2 2 1 7 2 1 q 2 b 2 and q 2 t 10 b 1 2 6 1 q 0 1 2 q 3 1 3 a11 a33
210
10 THE THEORY OF TORSION
rendering:
q 3 1 bt 4 q q 2 10 3 q 3
T
and
3 1 bt 2 28 3 T I t 4q a 4 4 b 2 bt 5 10 1 3
The shear stress flow in each cell is then:
xs 1 2M x τ xs2 It xs3
0 0 3 1 t1 s Mx bt 1 t2 s 0 4 0 10 28b2 0 0 1 t3 s 3
3 t1 s 4 t2 s 3 t3 s
10.6 Warping torsion We have seen in Chapter 10.3 above that torsion will render deformations u in the x direction which we call warping deformations (see Eq. 10.13). In St. Venant torsion we avoided the effects of these deformations by adopting Prandtl stress function. Thus, St. Venant torsion and warping torsion are independent, and therefore additive. I.e. torsion may be regarded as a combined effect of St. Venant torsion and warping. In this chapter we shall consider the torsion effects of warping for thin walled open cross sections and for a single cell closed cross section, as first developed by V.Z. Vlasov [16].
Elaboration 10.1: A simplified approach to warping torsion The torsion effect of warping may best be illustrated by the case shown in Fig. 10.18. A cantilevered beam with H or I type of cross section is subject to a torsion moment M x . The torsion moment will rotate the cross section an angle (see Fig. 10.18.a), implying that the upper flange is bent to the left and the lower flange to the right, rendering bending of the flanges while the web remains largely only twisted. This flange bending is what we call warping torsion. In this particular case, the torsion moment may be replaced by shear forces V f M x h in opposite directions, one in upper flange and one in lower flange, see Fig. 10.18.b, where h is the distance between the two flanges. The bending moment in the flanges is then M f V f L x where L is the length of the beam.
10.6 WARPING TORSION
211
Fig. 10.18 Cantilevered beam with H or I type of cross section subject to torsion This bending moment creates deformations
u in x direction and corresponding normal
stresses x M f I f y where I f tb3 12 (see Chapter 3.3), t is the plate thickness of the flange and b is its width. We define the cross sectional warping parameter h 2 y , the bi-moment B M f h and the warping normal stresses
x , and thus:
Mf If
y
B 1 B h I f h 2 I
where
I
h2 th 2b3 If 2 24
I is the cross sectional warping constant (similar to that of I y , I z and I t , only that I will have the dimension m 6 ). Then:
max
B B h b 6M f 6M x 3 Vf 3 Mx max L x and max 2 2 I I 2 2 2 A f 2 thb tb thb
(see Example 3.4). It is seen from Fig. 10.18.c that the warping stresses make no contribution to the stress resultants N , M y and M z .
212
10 THE THEORY OF TORSION
a) Definition of axes and cross sectional stress resultants
b) Displacements and stress components at arbitrary point P Fig. 10.19
Arbitrary open and thin walled cross section
10.6 WARPING TORSION
213
Let us consider an arbitrary open and thin walled cross section, as illustrated in Fig. 10.19. The displacements and twisting of the shear center is uD in x direction vD in y direction and is the rotation wD in z direction
(10.49)
Vlasov’s theory [16] is based on the assumptions that:
the cross section maintains its shape, shear strain in the middle of any plate element in an open section is zero, stress component perpendicular to the x axis ( s ) is zero, wall thickness is small as compared to width or height of cross section (and that shear flow in a closed section is constant).
Then the torsion induced contributions to the y and z direction displacements of an arbitrary point P on the cross section is given by (see Fig. 10.19)
v vD z ez w wD y e y
(10.50)
where is the torsion induced cross sectional rotation and
e y , ez
are the
coordinates of the shear center (the center of cross sectional rotation). In the case of pure torsion, it is taken for granted that the shear center is defined such that its displacement is zero. The tangential displacement of an arbitrary plate element dx, ds at mid thickness sector coordinate s , see Fig. 10.20.a, is given by
vs vD cos wD sin r
(10.51)
where r is radial distance, is the angle between the y axis and the tangent of the surface element. (The s coordinate is also defined in Fig. 10.19.b.) The corresponding shear angle (see Fig. 10.20.b) is then
xs
du dvs ds dx
(10.52)
Since shear strain xs in the middle of a plate element is assumed zero, then
du
dvs ds dx
(10.53)
214
10 THE THEORY OF TORSION
Fig. 10.20
Contour displacement and corresponding shear angle
Integrating u
dv s dw d dv ds D cos D sin dx dx dx dx
r ds
(10.54)
and noting that cos dy ds and sin dz ds , the following is obtained
dz rds uD vD y wD z u x, y, z vD dy wD
(10.55)
where uD is the integration constant that represents a constant (unanimous) axial stretching of the cross section, and is the sector coordinate defined by
d where d rds
(10.56)
In Vlasov’s theory it is assumed that due to torsion alone the only stress components are normal stress x (i.e. s 0 ) and tangential shear stress xs (index s to indicate dependency of surface direction). Assuming elastic behaviour du x E x E (10.57) dx the following is obtained
x E uD vD y wD z
(10.58)
10.6 WARPING TORSION
Fig. 10.21
Fig. 10.22
215
Stress components
Stress components and cross sectional stress resultants
Recalling that x , y , z are main axes, rendering
ydA zdA yzdA 0 , then A
A
A
the stress resultants (see Fig. 10.22) are given by
M y x zdA E wD I y zdA A A I z ydA M z x ydA E v D A A N x dA E uD A dA A A
where
1 A 2 I y z dA I A y2 z
(10.59)
216
10 THE THEORY OF TORSION
Since pure torsion (i.e. in the case that uD , vD , wD are all zero) will render no normal force N nor bending moments M y or M z ,
1 0 the shear centre must be defined by y dA 0 A z 0 N EAuD M y EI y wD and then M z EI z v D
x
Thus, it is seen that
We define the bi-moment
(10.61)
My N Mz y z E A Iz Iy
(10.62)
B xdA E 2 dA
(10.63)
A
and the warping constant
(10.60)
A
I 2dA
(10.64)
B EI
(10.65)
A
such that
x
and then
My B N Mz y z A Iz Iy I
(10.66)
As illustrated by the right hand side illustration in Fig. 10.23 (see also Fig. 10.21), the corresponding shear stress may be obtained from the x direction equilibrium condition of an infinitesimal element dx , ds :
x d x tds x tds xs d xs tdx xstdx 0 td xs t
d x ds dx
(10.67)
and thus s My d x d N M B ds t z y z ds dx dx A I z Iy I 0 0 s s s s M y N M B ztds tds tds z ytds A 0 Iz 0 Iy 0 I 0 s
t xs t
(10.68)
10.6 WARPING TORSION
Fig. 10.23
217
Shear stresses
As s 1 s S z s t y ds S y s z 0 S s
By defining
(10.69)
then the following is obtained: t xs q
M y N M B As z S z S y S A Iz Iy I
(10.70)
What remains is to establish the equilibrium requirements for an element dx , similar to that which we have done previously , e.g. in Chapter 3.3 (see Fig. 3.14). Force equilibrium requirements in x , y and z direction:
Fx 0 N dN dN qx dx 0 Fy 0 V y dV y dV y q y dx 0 Fz 0 Vz dVz dVz qz dx 0
N qx V y q y (10.71) Vz qz
where qx , q y and qz are distributed loads in x , y and z direction. The corresponding moment equilibrium about x , y and z axis are the following:
M x q M y 0: M y dM y Vz dVz dx 2 Vz dx 2 0 M y Vz (10.72) M z 0: M z dM z Vy dVy dx 2 Vy dx 2 0 M z Vy
M x 0 : M x dM x M x q 0
218
10 THE THEORY OF TORSION
where q is distributed torsion load. Thus Vy B 1 qx V As S z z S y S t A Iz Iy I
xs
(10.73)
s
The sector coordinate is defined by rds (see Eq. 10.56), where r is the 0
sector radius from the shear center to an arbitrary point s , with the starting point s 0 taken at a free edge of the cross section, and integration is taken anticlockwise around the surface of the cross section (because this coincides with positive direction of M x ). The sector area in Cartesian coordinates is defined by 1 1 1 1 rds d y e y dz z ez dy 2 2 2 2
where e y , ez
are the shear center coordinates. Let us consider another arbitrary
(but conveniently chosen) point P with main axes coordinates Defining p
(10.74)
yp, zp .
s
rp ds , where
rp is the radial distance from P to point s , it is
s 0
seen that the sector area as developed from P is 1 1 1 1 rp ds d p y y p dz z z p 2 2 2 2
dy
(10.75)
Taking the difference d d p y p e y dz z p ez dy and integrating
d d p p y p e y z z p ez y C
(10.76)
where C is an integration constant. Then:
p C y p e y z z p ez y
(10.77)
The shear centre position is defined in Eq. 10.60, and thus
p 1 1 y dA p y C y y p e y A z A z p z
z yz z p ez z2
y 0 2 y dA 0 (10.78) yz 0
10.6 WARPING TORSION
1 A 2 Introducing z dA I y and A 2 I z y
219
y 0 z dA 0 the following is obtained: A yz 0
1 e y y p I p zdA y A 1 C p dA and AA ez z p 1 p ydA I z A
(10.79)
Example 10.4: Warping torsion of thin walled I or H section A simple example is the I or H section shown in Fig. 10.24 (previously also presented in a more physical approach in Elaboration 10.1 above). Since the cross section is symmetric about both axes y and z the shear centre coincides with the centroid.
Fig. 10.24
Thin walled I or H section, and S diagrams
Then it is readily seen that in its web r 0 and thus: y 0, z 0 , while in the upper right hand side flange r h 2 and ds dy (because s b 2 y ) and thus:
220
10 THE THEORY OF TORSION
s
rds 0
h 2
b 2 y
h hb dy 2 y 2 2 s
b2
Similarly, on the other flange may readily be obtained, as shown in Fig. 10.24. Then 2
b2 t f h 2 b3 h b I 2 dA 4 s t f ds 2 2 24 0 A
which is identical to that which has previously been obtained in Elaboration 10.1. s
s
0
0
S t f ds t f
tf h hb s b s s ds 22 4
Thus, the normal and shear stress components are given by:
x
B I
and
xs
B S I
where
B EI
Example 10.5: Warping torsion of thin walled channel cross section
a)
Cross section
Fig. 10.25
b) p , P a 4,0
Thin walled channel cross section
c) p C
10.6 WARPING TORSION
221
Let us consider the more complex case of the channel section shown in Fig. 10.25.a. Then: A 4 at and the distance from the web to the centroid is y0 (Because of symmetry about y axis z0 0 .)
2at a 2 a 4at 4
I y 2 at a 2 t 2a 12 8a 3t 3 and 3
2 2 ta 3 a a a Iz 2 at 2 at 5a 3t 12 2 4 4 12 Choosing P at y p , z p a 4,0 , i.e. at the point where the y axis intersects with
the web, and choosing the starting point of
s at the outer tip of the lower flange, then
In lower flange: p as, y s 3a 4, z a p rp ds In web: p a 2 , y a 4, z s 2a 0 2 Upper fl.: p as 2a , y 4a s 3a 4 s 13a 4, z a s
Then: a
3a
2 pdA astds a tds
A
a
0
a
p ydA as s A
0
4a
2 3 as 2a tds 4a t
3a
3a
4a
1 p dA a 2 AA
3a 2a 2 13a s tds 0 tds a tds as 2a 4 4 4 a 3a
a
3a
4a
0
a
3a
2 p zdA as a tds a s 2a tds
A
C
2 4 as 2a atds a t
1 a a 4t 5a e y y p p zdA 3 I 4 8a t 3 8 y A Thus: 1 ez z p p ydA 0 I z A The sector coordinate p C y p e y z z p ez y p a 2 3a 8 z :
In lower flange: as a 2 3a 8 z, z a as 5a 2 8 In web: 3a 8 z, z s 2a 3a 8 s 2a 2 2 In upper flange: p as 3a 3a 8 z, z a as 27a 8 as shown in Fig. 10.26. The torsion stiffness constant is then given by
222
10 THE THEORY OF TORSION
I 2 dA A
2
2
2 3a 4a 5a 2 27a 2 7a 5t 3a as tds s 2a tds as tds 8 8 8 24 0 3a a a
Fig. 10.26
diagram
Fig. 10.27
S diagram
The static torsion moment (shown in Fig. 10.27) is given by
10.6 WARPING TORSION
223
s s 2 5a 2 5a 2 Lower flange: S as tds st a 8 8 2 0 s s 3a a 3t 3a 2 11a 3t 3at s2 S tds Web: S tds as s 8 a 8 4 16 4 4 0 s 3 2 3 2 2 Upper fl.: S a t as 27a tds 11a t ats 27a ts 8 3a 8 2 2 8
For the case of a single cell thin walled cross section, the basic assumption is that the shear flow will be the same as that which was obtained from the St. Venant theory, i.e. q xs t
Mx 2 Am
where M x GI t and I t
4 Am2 ds t
(10.80)
Rm
and where Am is the area within the mid wall thickness curve, Rm (see Chapter 10.4). Thus, Eq. 10.30 still applies:
xs
du dvs ds dx
(10.81)
and so does Eq. 10.51:
vs vD cos wD sin r
(10.82)
where cos dy ds and sin dz ds (see Fig. 10.20.a). Assuming elastic behaviour Mx I xs xs t (10.83) G 2 Am t 2 Amt
and then the following is obtained: du dv dy dz I s xs v D wD r t 2 Am t ds dx G ds ds
(10.84)
Defining the sector coordinate: s
r r ds 0
where r
It 2 Amt
(10.85)
224
10 THE THEORY OF TORSION
and integrating
it is seen that
I du vD dy wD dz r t ds 2 Amt
(10.86)
u uD vD wD
(10.87)
which is identical to that which was obtained for an open cross section in Eq. 10.55, except with now replaced by . I.e., apart from that, all the procedures developed above for the warping torsion of an open cross section may also be used for a single cell cross section. Thus, for a single cell cross section
M N M B x z y y z A Iz Iy I
where
B EI I 2 dA A
(10.88)
(Multiple cell cross sections or partly closed partly open cross section will require a twostep procedure, where the cross section is first treated as open, rendering warping displacements, then cell continuity is established by an unknown shear flow providing cross sectional continuity.)
10.7 The differential equations of beam-columns What then remains is to develop the differential equations that govern the combined problem of axial force, bi-axial bending and torsion. We have previously shown (see e.g. Eqs. 10.11 and 10.28) that the St. Venant contribution to the torsion moment is: M x GI t . The warping contribution to the torsion moment may readily be developed by
M x t xs rds t xs d A
(10.89)
A
which integrated by parts renders ( q t xs 0 at outer edges of cross section)
M x t xs d t xs td xs A
A
A
(10.90)
10.7 THE DIFFERENTIAL EQUATION OF BEAM-COLUMNS
Then, by introducing td xs t
225
d x ds from Eq. 10.67 and B xdA dx A
from Eq. 10.63, the following is obtained:
M x t A
d x dB d ds x dA dx dx A dx
(10.91)
The warping contribution to the torsion moment is then defined by (see Eq. 10.65) M x B EI (10.92) Thus, the total torsion moment from St. Venant torsion as well as warping is given by M x GIt EI (10.93) which is a third order differential equilibrium equation for pure torsion. Then from Eqs. 10.61 and 10.72 we see that
EAuD qx 0 EI z vD q y 0
EI y wD qz 0 EI GI t q 0
(10.94)
These are the differential equations for axial force, bi-axial bending and torsion of a thin walled open cross section (or for a single cell cross section, only replacing by , see Eqs. 10.56 and 10.85). The solution to the equation of pure torsion
EI GI t q 0
(10.95)
M x GIt EI
(10.96)
or, if q 0 and M x 0 ,
is given by:
p h
(10.97)
p is a particular solution where h is the solution to the homogeneous problem: EI GI t 0
(10.98)
226
10 THE THEORY OF TORSION
In addition to the ordinary boundary conditions that: 0 wherever the cross section is prevented from rotation (e.g. at a fork bearing or where the beam is embedded to a fixed connection), 0 wherever the beam is embedded to a fixed connection the warping effects must render no normal stresses at a free end, and because x B I and B EI then 0 at a free end of the beam. Solution to the differential equations of cross section normal force or bending
EAuD qx 0 EI z vD q y 0 EI y wD qz 0
(10.99)
have been presented in Chapter 3.
Elaboration 10.3: The connection between shear centre and centroid displacements
Fig. 10.28
Centroid displacements
Although bending occurs about main y and z axis, it should be noted that the differential equations of bending are described by the displacements of the shear centre, as it has been one of the basic assumptions that pure torsion will not render contributions
10.7 THE DIFFERENTIAL EQUATION OF BEAM-COLUMNS
to bending. The connection between shear centre (SC) displacements uD and bending centroid (CC) displacements u0
0 u0 uD v v e z 0 D e y w0 wD
v0
where e y , ez
227 vD
wD
T
w0 is defined by: T
are shear centre coordinates.
Example 10.6: Cantilevered beam with torsion moment at outer end Let us for instance consider the cantilevered beam with a torsion moment M x at its outer end in Fig. 10.29. In this case it suffices to solve the third order differential Eq. M x GIt EI 10.81:
Fig. 10.29
Cantilevered beam with torsion moment M x at outer end
The particular solution is:
p
Mx M , i.e.: p p dx x x . The solution to the GI t GI t
homogeneous equation GI th EIh 0 is h c1 c2 exp kx c3 exp kx where k GI t EI
and c1 , c2 and c3 are constants to be determined from
boundary conditions. Thus:
c1 c2 exp kx c3 exp kx
Its derivatives are given by c2 ke kx c3ke kx Boundary conditions are defined by:
Mx x GI t
Mx and c2 k 2 e kx c3k 2 e kx GI t
10 THE THEORY OF TORSION
228
c c c 0
1 2 3 x 0 0 Mx 0 x 0 0 c2 k c3k GI t x L 0
c k 2 e kl c k 2e kl 0 3 2
Thus:
c1 Mx 1 c2 kGI 1 e 2 kL t c3
1 2 kLkx 2 kL kx kx 1 e2 kL 1 e e e e kx M x e2 kL e 2 kx 1 2 kL kGI 1 e t k 2 kL kx e ekx 2 kL e 1
1 e 2 kL 1 e 2 kL
In a non-dimensional format and B are plotted in Fig. 10.30 for three different values of kL L GI t EI .
Fig. 10.30
Cantilevered beam; upper diagram: cross sectional rotation, lower diagram: bi-moment
Chapter 11 Equation Chapter 11 Section 1
BENDING OF PLATES 11.1
Introduction to the theory of plate bending
The development below is limited to cover the theory of rectangular or circular thin and plane plates, see Fig. 11.1. It is often referred to as the Kirchhoff – Love theory, as it was first presented by A.E.H. Love [7] based on basic assumptions outlined by G.R. Kirchhoff (1824 – 1887). Perpendicular to the plate it is subject to a distributed load q z x , y (with unit N m 2 ). In the plane of the plate it
may also be subject to evenly distributed axial loads N x and N y (defined positive tensile and with unit N m ). These are included in the development of the differential equation of the system as they will play an important part in the next chapter about buckling. However, in this chapter we will primarily focus on pure bending. It is taken for granted that the plate is homogeneous and isotropic, that it is linear elastic, and, although this is not a general requirement, it is assumed that the plate thickness t is constant. Since the plate is thin as compared to its overall dimension it is taken for granted that t Lx and L y ,
and that stresses perpendicular to the plate plane ( z ) may be ignored. For the same reason, Navier’s hypothesis [4] is adopted, implying that a straight section which is normal to the middle surface before any plate deformations will remain straight after deformations, i.e. there are no cross sectional distortion. Finally, the theory is limited to plate deformations w x , y which are small, such that for any cross sectional rotation, , it will be sufficiently accurate to assume that cos 1 and that sin tan w j , j x or y . Thus, it is seen that the Kirchhoff – Love theory is an extension of the beam theory first presented in Chapter 3.3. The relevant normal and shear stress components x , y , xy ,
yx , xz and yz are shown in Fig. 11.2. For an infinitesimal element
dx, dy, dz , force equilibrium in x and
© Springer Nature Switzerland AG 2020 E. N. Strømmen, Structural Mechanics, https://doi.org/10.1007/978-3-030-44318-4_11
y directions
229
11 BENDING OF PLATES
230
a) Overall geometry and transverse load
b) Stress resultant on infinitesimal element dx, dy Fig. 11.1
Rectangular, plane and isotropic plate
x d x dydz x dydz yx d yx dxdz yx dxdz 0
y d y dxdz y dxdz xy d xy dydz xy dydz 0
(11.1)
together with moment equilibrium about a vertical axis through the middle of the element
11.1 INTRODUCTION TO THE THEORY OF PLATE BENDING
231
xy d xy dydz dx2 xydydz dx2 yx d yx dxdz dy2 yxdxdz dy2 0 (11.2) will require that
d x d yx 0 , dx dy
d y dy
d xy dx
0 and xy yx
Fig. 11.2
Stress components due to bending and shear
Fig. 11.3
Superposition of in-plane strain components
(11.3)
232
11 BENDING OF PLATES
The relationship between stresses x , y , xy and corresponding strain components x , y and xy may, as illustrated in Fig. 11.3, be obtained by using the principle of superposition (see also Chapter 3.3), i.e. by adding the effects of each strain component separately: x x E x alone y x
y x x E E y y E y x (11.4) y alone y E E x y xy xy alone xy xy G xy G where E is the elastic normal stress modulus, G is the corresponding shear stress modulus and is the Poisson ratio. For an elastic homogeneous and isotropic material the shear stress modulus is given by G E 2 1 , as shown in Elaboration 3.5. Thus
1 E E 0 x x 0 y y E 1 E 0 xy 0 1 G xy
(11.5)
from which the following in-plane stress-strain relationship is obtained
x E y 2 1 xy
1 1 0 0
0 0
1
x y 2 xy
(11.6)
It has been assumed that displacement w x , y is small such that only insignificant stretching will occur, i.e. that ds x dx and ds y dy , and thus, the normal stresses x and y that are associated with the deformation w x , y will largely be pure plate bending.
11.1 INTRODUCTION TO THE THEORY OF PLATE BENDING
Fig. 11.4
233
Pure plate bending
Bi-axial pure bending and shear deformations are illustrated in Fig. 11.4. It is seen that for an infinitesimal element dx, dy, dz
11 BENDING OF PLATES
234
2 w v w v z see Fig. 11.4.b y 2 z y y y 2 u v w w w z Fig. 11.4.c xy z z 2 y x y x x y xy u
w z x
see Fig. 11.4.a
x
u 2w 2 z x x
2 w x 2 x 2 2 y z w y 2 2 w xy xy
and thus
(11.7)
(11.8)
From this the following connection between displacement w x , y and the cross sectional stress components is obtained (see Eq. 11.6)
x Ez y 1 2 xy
1 1 0 0
2 2 0 w x 0 2 w y 2 1 2 w xy
(11.9)
and thus, the stress resultants M x , M y and M xy (moments per unit length, see Figs. 11.1.b and 11.2.b) is obtained:
Mx z y 1 M h2 1 y z x z yx dx D 0 0 M xy h 2 M yx z xy 0 0
0 2 w x 2 0 2 w y 2 (11.10) 1 2 w xy 1
where the plate stiffness
D
E
t2
1 2 t 2
z2dz
Et 3
12 1 2
and from which it is seen that M yx M xy .
(11.11)
11.2 THE DIFFERENTIAL EQUATION OF RECTANGULAR PLATES
235
11.2 The differential equation of rectangular plates
Fig. 11.5
Fig. 11.6
Definition of stress resultants
Cross sectional stress resultant variation in x and y directions
What then remains is to establish the equilibrium conditions for an infinitesimal plate element
dx, dy
(with thickness t ). For such an element the stress
resultants (per unit length) are defined in Fig. 11.5, where external load qz x, y
11 BENDING OF PLATES
236
as well as normal and shear forces N x , N y , N xy and N yx are assumed constants. The relevant plate deformations and variation of stress resultants in x and y directions are illustrated in Fig. 11.6. Thus, the following equilibrium requirements are obtained: 1) Force equilibrium in x -direction:
Vyx dVyx dx Vyxdx 0
dV yx 0 V yx N yx
(11.12)
2) Force equilibrium in y -direction:
Vxy dVxy dy Vxydy 0
dVxy 0 Vxy N xy
(11.13)
3) Force equilibrium in z -direction: w 2 w w 2 w w w N x dy dx N x dy dy N y dx N y dx x x 2 y y 2 x x 2 w 2 w w w w dx N xy dy dy N xy dy N yx dx y xy x xy y w N yx dx 0 x Vxz V yz 2w 2w 2w 2w N x 2 N y 2 N xy N yx 0 qz y xy xy x x y
qz dxdy Vxz dVxz dy Vxz dy V yz dV yz dx V yz dx
(11.14)
4) Moment equilibrium about an axis through the mid-point p and parallel to x:
dydx V yz M x dM x dx V yz dV yz dxdy 2 2 M x dx M yx dM yx dy M yx dy 0 V yz
M x M yx y x
(11.15)
5) Moment equilibrium about an axis through the mid-point p and parallel to y:
11.2 THE DIFFERENTIAL EQUATION OF RECTANGULAR PLATES
237
dxdy dydx Vxz M y dM y dy 2 2 M y dy M xy dM xy dx M xy dx 0 M y M xy Vxz x y
Vxz dVxz
(11.16)
6) Moment equilibrium about an axis through the mid-point p and parallel to z:
N xy dN xy dy dx2 N xy dy dx2 N yx dN yx dx dy2 N yx dx dy2 0
N xy N yx
(11.17)
Introducing Eqs. 11.12, 11.13, 11.15, 11.16 1 and 11.7 into Eq. 11.14
qz
M y M xy M x M yx w w w N x 2 N y 2 2 N xy 2 2 x y x y xy x y x y 2 2w 2w 2 2w 2w (11.18) q z D 2 2 2 D 2 2 2 x x y y x y 2 2w 2w 2w 2w 2 D 1 0 N x 2 N y 2 N yx xy xy xy x y 2
2
2
2
2
2
2
and substituting M x , M y and M xy M yx from Eq. 11.10, the following differential equation is obtained:
4w 4w 4w 2 w 2 w 2w D 4 2 2 2 4 N x 2 N y 2 2 N xy qz xy x y y x y x
(11.19)
If N x N y N xy 0 , this reduces into
4 w 4 w 4 w qz 2 x 4 x 2y 2 y 4 D
(11.20)
which may be written in the following more compact way w q z D
(11.21)
11 BENDING OF PLATES
238
where is the Laplace operator: f x1 ,, xm ,, x M
2 f 2. m 1 xm M
Example 11.1: Rectangular plate subject to distributed load
Fig. 11.7
Simply supported rectangular plate
Let us consider a simply supported rectangular plate with side lengths Lx and L y that is subject to a distributed load q z x , y across its entire surface. For this problem it is convenient to resort to Navier’s solution which is based on a Fourier transform of the load, such that
qz x, y
M
N
amn sin
m 1 n 1
m x n y sin Lx Ly
where amn are the Fourier coefficients, and where the number of elements M and N are large enough to obtain a sufficiently accurate solution. Noting that L
sin 0
0 when p k p k sin d L L L 2 when p k
11.2 THE DIFFERENTIAL EQUATION OF RECTANGULAR PLATES
the coefficients amn may be obtained by first multiplying either side by sin
239
p y , Ly
p 1,, N , and integrating across entire y direction side length Ly
0
M N p y m x qz sin dy amn sin Ly Lx m1 n1
Ly
sin 0
Ly M p y n y m x dy amn sin sin Ly Ly Lx 2 m1
for p n (otherwise zero), followed by multiplication of either side with sin
k 1,, M , and integrating across entire Lx Ly
qz sin
0 0
x
k x , Lx
direction side length
Lx Ly M Lx Ly k x n y m x k x dxdy a dx amn sin sin mn sin Lx Ly L L 2 m1 4 x x 0
for k m . Thus it is seen that: amn
4 Lx Ly
Lx Ly
qz x, y sin 0 0
m x n y sin dxdy Lx Ly
The solution to the differential equation (Eq. 11.20) is then a corresponding Fourier series:
w x, y
M
N
cmn sin
m 1 n 1
m x n y sin Lx Ly
It is seen that the solution fulfils the boundary conditions of simply supported edges (see Eq. 11.10):
w x, y
M
N
cmn sin
m 1 n 1
x 0 and Lx m x n y sin 0 at Lx Ly y 0 and Ly
2w 2w M x D 2 2 0 at y 0 and Ly y x 2w 2w M y D 2 2 0 at x 0 and Lx y x Introducing this into the differential equation Eq. 11.20, then the following is obtained: 2
2 2 m n m x n y M N m x n y D L L cmn sin L sin L amn sin L sin L x y x y m 1 n 1 m 1 n 1 x y M
N
11 BENDING OF PLATES
240
cmn
Thus, it is seen that:
amn 2 2 4 D m Lx n Ly
If qz is constant, then amn
w x, y
M
16qz 6 D m 1 n 1
16qz where 2 mn
sin
N
2
m and n
are odd numbers, and then
m x n y sin Lx Ly
mn m Lx n L y 2
where
2 2
m 1,3,5, . n
Convergence should be checked, but M and N 4 6 will usually suffice. Cross sectional resultants are given by 2
2w 2w 16q M x D 2 2 4 z y x
M
N
m 1 n 1
m n Lx Ly
2 m n mn Lx L y 2
2w 2 w 16q M y D 2 2 4z y x
M
N
m 1 n 1
2
n m Lx Ly 2 m n mn Lx L y
sin
m x n y sin Lx Ly
sin
m x n y sin Lx Ly
2 2
2
2 2
M xy M yx D 1
Vxz
M y x
2
16q w 4z xy
M xy y
16qz
3
M
1
N
m 1 n 1
1 1 Lx L y 2 2
2
cos
m n Lx Ly
M N
1 Lx
m1 n 1 n
m Lx
2
n Ly
2
m x n y cos Lx Ly
cos
m x n y sin Lx Ly
11.2 THE DIFFERENTIAL EQUATION OF RECTANGULAR PLATES
V yz
241
1 Ly M x M yx 16qz M N m x n y 3 sin cos y x Lx Ly m 1 n 1 m 2 n 2 m Lx Ly
The distribution of M x , M y , M xy , Vxz and V yz are illustrated in Fig. 11.8.
Fig. 11.8
Rectangular plate with constant transverse load qz
Example 11.2: Rectangular plate subject to concentrated load Let us consider a simply supported rectangular plate with side lengths Lx and L y (see Fig. 11.7) that is subject to a concentrated force F at arbitrary position
xF , y F
within its span, and where in-plane loads N x N y N xy 0 . Then the solution may be obtained from Example 11.1 above by letting
11 BENDING OF PLATES
242
lim lim qz xF dx, y F dy dxdy Fz xF , y F
dx 0 dy 0
amn
and thus:
where
w x, y
4 Fz m xF n yF sin sin Lx Ly Lx Ly
M
N
sin
m x F n y F sin Lx Ly
4 Fz Lx L y D m 1 n 1 m Lx 2 n Ly 4
2 2
sin
m x n y sin Lx Ly
m Ly L , then 1,3,5, . If Fz is located at mid span, i.e. x F x and y F n 2 2
w x, y
M
N
sin
m x n y sin Lx Ly
4 Fz Lx Ly D m 1 n 1 m Lx 2 n L y 4
2 2
where
11.3 The differential equation of circular plates
Fig. 11.9
Polar coordinates
m 1,3,5, n
11.3 THE DIFFERENTIAL EQUATION OF CIRCULAR PLATES
243
We shall in the following consider circular plate bending, and assume that the load is either evenly distributed qz or a concentrated load Fz at mid span. This ensures symmetry, and thus, the problem is greatly simplified. As illustrated on Fig. 11.9, it is convenient to introduce polar coordinates r and with origo at the centre of the plate. It is readily seen that:
r x2 y2
12
and arctan y x
(11.22)
From the chain and sum rules of derivation it follows that: r 1 2 x 2 y 2 x r 1 2 x 2 y 2 y yx 2 x x 1 y
1 2 2 y y r sin 1 y x 2 y r 2 sin r 1 y x 2 x r 2 cos r 1 2
and thus: w w r w x r x w w r w y r y
2 x x r cos
w 1 w cos sin r x r 1 w w sin cos r y r
(11.23)
(11.24)
from which it is seen that:
(11.25) 2 2 2 1 w 1 w 2 1 w 1 w 2 w 2 cos sin 2 sin 2 sin 2 2 sin r r r r r r 2 r 2 w 1 1 w 2w sin cos sin cos 2 r r r r x
(11.26) 2 2 2 1 w 1 w 1 w 1 w w 2 2 2 sin sin 2 cos 2 sin 2 2 cos r r r r r r 2 r 2 w 1 1 w 2w cos sin cos sin 2 r r y r r
and
11 BENDING OF PLATES
244
w 1 2w 1 1 w 2w sin sin cos sin cos 2 r r xy r r 2 r (11.27) 2 2 1 w 1 1 w 1 w 1 1 w cos 2 sin 2 2 cos 2 2 sin 2 2 r r 2 r r 2 r r Thus, the transformation from Cartesian to polar coordinates is given by:
w x, y
2w 2w x 2 y 2
w r,
2 w 1 w 1 2 w r 2 r r r 2 2
(11.28)
and
2 1 1 2 2 w 1 w 1 2 w qz w 2 r r r 2 2 r 2 r r r 2 2 D r
(11.29)
This equation is generally applicable to any problem suitable for a formulation in polar coordinates, but, since we have restricted our problem to symmetric cases only, then the deformation w r, is independent of , and thus, the differential equation simplifies into
2 1 2 w 1 w qz w 2 r r r 2 r r D r
(11.30)
which may also be written in the following more compact way: 1 1 w q z r r r r r r r r D
(11.31)
Recalling that due to symmetry, w is independent on , the bending moment and shear forces are given by:
2 w 1 w M D 2 r r r 2w 2w 1 w 2w M r D 2 M y D 2 2 r r y x r 2w 1 2 w 1 w M r 2 0 M xy M yx D 1 xy r r r 2w 2w M x D 2 2 y x
(11.32)
11.3 THE DIFFERENTIAL EQUATION OF CIRCULAR PLATES
245
M x M yx 2w 2w 1 D 2 2 V r D w 0 y x y x r y (11.33) 2 2 M y M yx w w D 2 2 Vr D w Vxz x y x x r y
V yz
w r
where
2 w 1 w 1 w r r 2 r r r r r
(11.34)
Example 11.3: Simply supported circular plate Let us consider a simply supported circular plate with radius R and plate thickness t . The plate is subject to a constant evenly distributed load qz . Since the vertical shear force Vr is independent of (due to symmetry), it is seen that the vertical equilibrium condition for a disc with arbitrary radius r is that
qz r2 Vr 2 r 0
and thus
Vr D
q w z r r 2
1 w 1 w qz r which can be r it is seen that r r r r r r r r 2 D q c w z r 4 1 r 2 c2 ln r c3 integrated into the following solution: 64 D 4
Since w
where c1 , c2 and c3 are integration constants to match the boundary condition :
w r 0 0 r
c2 0
M r r R 0 wr R 0
(due to symmetry), and due to simple supports:
qz 2 c 3 qz R 2 R 1 1 0 c1 16 D 2 1 8D qz 4 c1 2 5 qz R 4 R R c3 0 c3 64 D 4 1 64 D
3
qz R 4 r 3 2 64 D R 1 4
and thus
w
2 r 5 R 1
11 BENDING OF PLATES
246
2 2w 1 w qz R 2 r M r D 2 3 1 r r 16 R r 2 2 w 1 w qz R 2 r M D 2 1 3 3 r r 16 R r
M r and M are illustrated in Fig. 11.10, ( M r 0 ). It is readily seen that w r R 0 and that wmax
Fig. 11.10
qz R 4 5 occurs at r 0 . 64 D 1
Simply supported circular plate subject to evenly distributed load
Chapter 12 Equation Chapter 12 Section 1
ELASTIC BUCKLING 12.1 Columns
Fig. 12.1
Buckling of simply supported column
It is taken for granted that material behaviour is linear elastic and that axial compression u NL EA is negligible. Let us first consider the case of an axially loaded and simply supported column, as illustrated in Fig. 12.1. The column is perfectly straight and the axial load is perfectly aligned with the centre axis of the column. In this introductory part it is assumed that the column is free to
© Springer Nature Switzerland AG 2020 E. N. Strømmen, Structural Mechanics, https://doi.org/10.1007/978-3-030-44318-4_12
247
12 ELASTIC BUCKLING
248
lateral displacements in the z direction, see Fig. 12.1.a, but held against any out of plane displacements in the y direction, as well as against any cross sectional twist. Let us imagine that N is gradually increased to higher and higher values, and that we set out to monitor the lateral displacement at mid-span of the column. The most obvious path of system equilibrium is that there is no lateral deformation at all (Alt. 1 in Fig. 12.1.b). H owever, let us investigate the possibility that at some unknown value of N N E there is an alternative path of equilibrium (Alt. 2 in Fig. 12.1.b) which involves an unknown lateral displacement w x . For this to be possible it is an inescapable requirement that the system is still in a state of equilibrium.
Fig. 12.2
Equilibrium requirement and column displacements
Assuming that w x is small, then the equilibrium requirement at N N E is defined by (Fig. 12.2.a) M y x NEw x 0
(12.1)
where, from theory of elasticity, we know that M y x EI y w x , and thus, the following differential equation is obtained
12.1 COLUMNS
EI y w x N E w x 0
249 (12.2)
It is seen that the properties of w x must be such that its second derivative is negative and congruent with itself, and thus, the only possible solution is that
w x C1 sin x C2 cos x
(12.3)
where C1 and C 2 are unknown coefficients and is the unknown wave length of the displacement. Introducing the boundary conditions at x 0 and x L , it is seen that w x 0 C2 0 (12.4) w x L C1 sin L 0 This may be fulfilled if C1 0 , which is equivalent to path alternative 1 in Fig. 12.1.b, or if L n , n 1, 2,3, , which is equivalent to path alternative 2. Since we are interested in the lowest possible axial load that causes the column to buckle sideways, we choose n 1 , in which case x w x C1 sin L
(12.5)
L 2 EI N w x 0 y E
(12.6)
and thus, Eq. 12.2 becomes
which is an eigenvalue problem, where N E is the eigenvalue and w x is the eigenfunction. From Eq. 12.6 it may be concluded that the elastic buckling load of a simply supported column is
NE
2 EI y L2
(12.7)
For columns with other support conditions (see Fig. 12.3) NE
2 EI y L2k
(12.8)
where Lk is the buckling length, dependent of system boundary conditions.
12 ELASTIC BUCKLING
250
Fig. 12.3
Fig. 12.4
Buckling length Lk of simple columns
Simply supported column with geometric imperfection
12.1 COLUMNS
251
Obviously, the case with a perfectly straight column is an idealised situation. Let us now turn to the more realistic situation of a column with a geometric imperfection, as illustrated in Fig. 12.4. We still assume elastic behaviour, small deformations and that the column is simply supported. Then the equilibrium condition is defined by (see Fig. 12.4.b) M y x N w x w0 x 0
(12.9)
where w0 x is the geometric misalignment prior to any loading (at N 0 ), and w x is the additional displacement at an arbitrary value of N . Again, from theory of elasticity we have that M y x EI y w x , and thus Eq. 12.9 becomes EI y w x N w x w0 x 0
(12.10)
which is similar to the differential equation Eq. 12.2, only with the initial imperfection w0 x added to the moment effect of a displaced N . If, for
simplicity, w0 x 0 sin x L , then the solution to Eq. 12.10 is identical to the solution of Eq. 12.2, i.e. w x C1 sin x L . For convenience, setting the buckling induced lateral displacement at mid-span C1 , and introducing
w x and w0 x back into Eq.12.10, then the following is obtained 2 EI y N 0 sin x 0 L L
where
0 2
0
1 N NE
(12.11)
is the total lateral displacement at mid-span, and
2
N E EI y L is the elastic buckling load. Thus, as shown on the left hand
side illustration in Fig. 12.5, the effect of an increasing axial load N is to gradually increase the lateral displacement of the column, which in the limit N N E
(12.12)
Thus, it is seen that the physical phenomenon of buckling is a load-displacement problem which may take place more or less gradually. If the initial deformation 0 is small, then the buckling effect will occur rapidly at values of N in close vicinity to N E , while if the initial deformation 0 is large, then the buckling
12 ELASTIC BUCKLING
252
effect will occur more slowly (but with larger lateral displacements). Under more realistic conditions there are several initial imperfections that will contribute to the shape of the load-deformation curve, e.g. load misalignments, material and structural imperfection or minor lateral loads (e.g. gravity). If the slope of the load-displacement curve, N , is perceived as stiffness, then the buckling problem may also be described by loss of stiffness.
Fig. 12.5
Elastic (left) and inelastic buckling (right)
Elaboration 12.1: The moment amplification effect of buckling As shown on the right hand side illustration in Fig. 12.5, if we take inelastic behaviour into account, then at some value of N the largest bending moment (in this case at midspan)
M y N N N 0 1 N N E
will sooner or later cause the outer fibres of the cross section to yield at max f y , and from this point onwards yielding will progress and the effective bending stiffness of the column cross section in the vicinity of the mid span will decrease with increasing values of N . Ultimately the cross section at mid span will be fully plasticised; the system will be unable to sustain the loading and collapse. Thus, it is seen that the real buckling load N k will always be lower than N E . While N E is the elastic buckling load, N k is the inelastic (plastic) buckling load. A typical variation of the buckling stress k N k A versus the slenderness parameter k Lk ik , where Lk is the buckling length and
12.1 COLUMNS
253
ik I y A , is shown in Fig. 12.6. Obviously, the upper boundary of the buckling curve is defined by the yield stress
E N E A 2 EI y
L A 2 k
2
f y and the elastic buckling stress
E k2 . The real buckling stress k will
consistently follow a lower curve. A simplified curve may be obtained by regarding the inelastic buckling to be an interaction between yielding and elastic buckling, mathematically represented by 1 N k 1 N y 1 N E . Thus Nk N y
1 N y
N E , from which k N k A f y 1 f y 2 E k2 .
This may then be written k f y
1 2 k
12 where k k f y E .
In any case, the evaluation of safety against buckling starts with the calculation of the elastic buckling load.
Fig. 12.6
Typical variation of buckling stress k N k A versus the slenderness parameter k Lk i k
While we above restricted our presentation to the case of a column free to move in the lateral z direction and held against any out of plane motion, we shall now turn to the more general case of a column which is free standing, i.e. it is free to move in both y and z direction as well as cross sectional twisting, see Fig. 12.7. We shall still assume elasticity and that the column is simply supported with fork bearings at both ends. We shall also assume that the cross section of the column is symmetric about its main y and z axes.
12 ELASTIC BUCKLING
254
Fig. 12.7
Fig. 12.8
Buckling motion of free standing columns
Lateral displacement of infinitesimal element dA dx
This is a more complex case than that which was handled above, and it is convenient to resort to a Rayleigh – Ritz approach, see Appendix A.6. The strain energy in a column subject to a buckling motion of bi-axial bending and torsion has been developed in Appendix A.2 (see Eq. A.13). Due to symmetry the shear centre coincides with the centroid ( e y ez 0 ), and thus displacements vD v0 and wD w0 rendering
U
1 2 2 2 2 EI z v0 EI y w0 GI t EI w dx 2 L
(12.13)
As illustrated in Fig. 12.7, the energy loss P of the external axial load N :
12.1 COLUMNS
P N N
L A
255
x dAdu
(12.14)
where x N A is axial stress and du is the axial displacement of an infinitesimal element dx due to sideways motion v and w as well as cross sectional twisting . It is seen from Fig. 12.8 that
2 12 2 1 2 2 v 2 w du dx dx dx dx dx x x 1 2 1 2 1 2 2 1 1 v x 1 w x dx v2 w2 dx 2
(12.15)
where the Binomial series expansion 1 a 1 na n n 1 a 2! has been used to obtain an approximate solution. As shown at right hand side illustration in Fig. 12.7, the motion of an infinitesimal cross section element dA is given by v v0 z (12.16) w w0 y n
du
Then
2
1 2 1 2 2 v w2 dx v0 z w0 y dx 2 2
(12.17)
and thus N 1 v z 2 w0 y 2 dAdx 0 A 2 LA
P x dAdu LA
Recalling that
ydA zdA yzdA 0 , A
P
N v02 w02 y 2 z 2 2 2 zv0 yw0 dAdx 2 A L A
A
A
N v02 w02 i 2p 2 dx 2 L
(12.18)
I y z 2 dA and I z y 2 dA , then: A
A
i 2p
where
I y Iz
(12.19)
A
Thus L
1 EI z v02 EI y w02 GI t 2 EI w 2 N v02 w02 i 2p 2 dx (12.20) 2 0
12 ELASTIC BUCKLING
256
Based on the solution in our first approach (see Eq. 12.5) it is a reasonable assumption that v0 c1 w c sin x (12.21) 0 2 L c3 L
L
L x 2 x Observing that sin 2 dx cos dx , then: 2 L L 0 0 2 2 2 2 EI y L 2 EI z 2 EI z 2 2 N c GI 2 N c1 i p N c3 2 t L2 4 L L L2 (12.22) By introducing the individual buckling loads associated with bending about y axis, bending about z axis and cross sectional twisting
NEy
2 EI y 2
L
, N Ez
2 EI z
and N E
2
L
GI
t
2 EI w L2 i 2p
(12.23)
this may be written: 2
L 2 2 2 2 N Ez N c1 N E y N c2 N E N i p c3 4 L
(12.24)
The Rayleigh – Ritz idea is that since we have chosen displacement functions that are not necessarily consistent with the solution to the differential equation of the system, it follows that is a function in the space of the unknown coefficients c1 ,c2 ,c3 , and then the best possible solution is obtained if is at a minimum, i.e. 2 L N Ez N c1 0 c1 2 L 2 L N E y N c2 0 (12.25) c2 2 L 2 L N E N i 2p c3 0 c3 2 L
which may be written
12.1 COLUMNS
NE N z 0 0
0
N
Ey
257
c1 0 c 0 0 2 c 0 N E N 3 0
N
0
(12.26)
or alternatively
NE N I c 0
where
N diag N N E y N E Ez E I is a 3 by 3 identiy matrix T c c1 c2 c3
(12.27)
Again, this shows that elastic buckling is an eigenvalue problem, where N is the eigenvalue and c is the eigenvector, containing the coefficients of the buckling shape of the system, i.e. they represent the buckling mode shape of the system. As can be seen, Eq. 12.26 can be fulfilled for values of c unequal to zero only if the determinant to the coefficient matrix is zero, i.e. if
N Ez N N E y N N E N 0
(12.28)
Thus, it is seen that there are three possibilities:
If
N N Ez
then
c1 0
while
c2 c3 0 , which means that
v c1 sin x L and w 0 , i.e. the buckling displacements occur in the y direction alone (with pure bending about the z axis).
If
N NEy
then
c2 0 while c1 c3 0 , which means that
w c2 sin x L and v 0 , i.e. the buckling displacements occur in the z direction alone (with pure bending about the y axis).
If
N N E
then
c3 0 while c1 c2 0 , which means that
c3 sin x L and v w 0 , i.e. buckling occurs in the shape of a cross sectional twisting along the span of the column with pure torsion about the x axis (called torsion buckling load). Obviously, the critical buckling load is the lowest of N E y , N E z and N E .
12 ELASTIC BUCKLING
258
Elaboration 12.2: Column with non-symmetric cross section
Fig. 12.9
Non-symmetric cross section
If the cross section is non-symmetric, with shear centre coordinates axes ey , ez , then, as shown in Fig. 12.9, twisting occurs about the shear centre (see Eq. A.13):
U
1 2 2 2 2 EI y wD GI t EI w dx EI z vD 2L
where v D and wD are shear centre displacements. The displacements of an arbitrary
element dA is now defined by v v D z ez and w wD y ey
du
. Thus
1 2 1 v w2 dx v D2 wD2 2 z ez vD y e y wD 2 2
y 2 z 2 2 e y y ez z e 2y ez2 2 dx Since we still have that
ydA zdA 0 , I y z dA and I z y dA , then 2
A
P
A
A
A
1 N x dAdu vD2 wD2 2 z ez vD y e y wD 2 L A 2A L A
y 2 z 2 2 e y y ez z e2y ez2 2 dAdx
2
N 2 e y wD i 2 2 dx vD wD2 2 ez vD 2L
12.1 COLUMNS
where i 2
I y Iz A
259
e 2y ez2 . Then the energy function U P is given by 1 2 2 2 EI y wD 2 GI t EI w dx EI z v D 2L N 2 wD2 2 ez v D e y wD i 2 2 dx vD 2 L
Assuming simple bending supports and torsion fork bearings at either end,
vD c1 w c sin x will render a sufficiently accurate solution. Thus: D 2 L c3
then
2
L 2 2 2 2 N E z N c1 N E y N c2 N E N i c3 2 N ez c1 e y c2 c3 4 L where
N E
obtained:
GI t 2 EI w L2 . Setting 0, c j i2
NE N z 0 Nez
Ey
N
Ne y
j 1, 2,3 the following is
c1 0 c 0 Ne y 2 c 0 N E N e02 3 Nez
0
N
The determinant to this coefficient matrix is zero for the condition that:
N Ez N N E y N N E N e02 Ne y N E y N Nez 2 0 2
which may also be written
N
Ey
N
N
Ez
N
N E N N E y N Neˆz 2 N Ez N Neˆ y
2
0
where: eˆ y e y i and eˆz ez i . This is a third degree polynomial which can only be solved numerically. However, if for instance e y 0 then the cross section is symmetric about the
z
axis, and the third order equation above is reduced into
N Ez N N Ey N N E N Neˆz 2 0
12 ELASTIC BUCKLING
260
In this case there are two alternative solutions:
N N E z which is equivalent to the case of pure buckling in the y direction, with pure bending about the
N
Ey
N
N
2
N
1
1 eˆz2
E
z
axis, or
N Neˆz 0 from which the following is obtained:
N E y N E
2
N
Ey
N E 4 1 eˆz2 N E y N E
which represents buckling displacements with contribution from a lateral motion
w c2 sin x L in the z direction as well as cross section twisting
c3 sin x L .
12.2
Beams
Let us consider a beam with length L and a cross section that is symmetric about y and z axes, see Fig. 12.10. It is simply supported with fork bearings at both ends. Let us assume that I y I z , such that z direction displacements ( w ) may be ignored as compared to those in the y direction ( v ) and cross sectional rotation ( ). The beam is subject to external y axis bending moments M at its supports, acting about the y axis with positive direction at x 0 and with negative direction at x L , such that they both create compression in the upper parts of the cross section. Under these circumstances there is a danger that the upper parts will act similarly to that of a column, and a buckling mode comprising lateral displacements v x and cross sectional rotation x will occur, as illustrated in Fig. 12.10. This is what is called lateral torsional buckling. As shown in Fig. 12.11 the lateral displacement v creates a torsion moment M x Mv , while the cross sectional rotation creates a bending moment M z M . Recalling the differential equations for bending and torsion from Chapters 3 and 10 (see Eqs. 3.25 and 10.96), the following applies:
M z M EI z v v M EI z M x Mv GI t EI w Mv GI t EI w
(12.29)
12.2 BEAMS
Fig. 12.10
Fig. 12.11
261
Beam subject to external y axis bending moments M
Bending and torsion effects due to v and displacements
Thus, combining first and second expression in Eq. 12.29, the following fourth order differential equation is obtained for the problem of lateral torsional buckling of a beam with double symmetric cross section:
EI w GI t
M2 0 EI z
(12.30)
It is seen that the second and fourth derivatives of must be congruent with itself, and thus the general solution is given by:
c1 sinh 1 x c2 cosh 1 x c3 sin 2 x c4 cos 2 x
(12.31)
12 ELASTIC BUCKLING
262
where c1 ,, c4 are unknown coefficients determined by the boundary conditions of the system, 1 and 2 are wave length constants of the harmonic hyperbolic and trigonometric shape functions. Introducing Eq. 12.31 and
c112 sinh 1 x c212 cosh 1 x c322 sin 2 x c422 cos 2 x
c114 sinh 1 x c214 cosh 1 x c324 sin 2 x c424 cos 2 x
(12.32)
into Eq. 12.30, and solving for 1 and 2 , it is seen that 12 1 2 2 2 GI t M GI t 2 EI z EI w 2 EI w 2 EI w 12
12 GI 2 M2 GI t t 1 EI z EI w 2 EI w 2 EI w
(12.33)
From the boundary conditions: c2 c4 0 0 x0 c2 c4 0 2 2 0 c21 c42 0
(12.34)
c1 sinh 1L c3 sin 2 L 0 0 xL 2 2 0 c11 sinh 1L c32 sin 2 L 0
(12.35)
then the following conditions are obtained:
c1 12 22 sinh 1L 0 c1 0 2 2 c3 1 2 sin 2 L 0 c3 sin 2 L 0
(12.36)
from which it is seen that if buckling occurs then c3 0 , and hence, we must demand 2 L , i.e. that (see Eq. 12.33) 12
12 2 GI 2 M GI t t 2 2 EI w 2 EI w EI z EI w
L
(12.37)
263
12.2 BEAMS
Solving with respect to M , the following elastic buckling moment M E M is obtained:
ME
12
2 EI w EI z GI t L L2
(12.38)
In general, the use of the differential equation to solve this type of problem is rather cumbersome (unless a Galerkin’s approach is adopted). We shall in the following turn to a Rayleigh-Ritz approach, which in most cases will render a manageable solution.
Fig. 12.12
Simply supported beam subject to transverse loads
12 ELASTIC BUCKLING
264
Let us consider the case shown in Fig. 12.12. The beam is still subject to external bending moments M at both supports, but the beam is now also subject to a concentrated load Fz at arbitrary position x F along the span and an evenly distributed load qz . The strain energy in the system is given by (see Appendix A.2, Eq. A.14):
U
1 2 2 2 EI z v GI t EI w dx 2 L
(12.39)
and, as illustrated in Fig. 12.12.c, the energy loss P from the motion of M , Fz and qz is given by: P M x 0 M x L F xF qz x dx
(12.40)
L
where x 0 and x L are cross sectional rotations at either end of the beam and is vertical motion of the beam centroid. At arbitrary position, x may be determined by observing (see Fig. 12.12.c) that r sin v , and thus v v 2 1 (12.41) x r 1 cos 1 cos 1 1 v sin 2 2 where the series approximation cos 1 2 2! 4 4! 1 2 2 has been introduced. Similarly, the cross sectional rotations at either end of the beam, x 0 and x L , may be obtained by observing that
tot x 0 x L u l h
(12.42)
where u and l are the displacements in x axis direction of upper and lower edges of the cross section, and h is the distance between upper and lower edges. It is seen from Fig. 12.12.c that vu v h 2 and vl v h 2 . Thus 2 1 h v dx 2 2 2 12 h 1 1 d l dx 1 vl2 dx vl2 dx v dx 2 2 2
d u dx 1 vu2
12
1 dx vu2 dx 2
(12.43)
12.2 BEAMS
265
2 2 u l 1 1 h h v v dx v dx h h 2 L 2 2 L
(12.44)
Hence
tot
The energy balance in the system is then defined by: U P
M v dx L
1 EI z v 2 GI t 2 EI w 2 dx 2 L Fz q v x F x F z v dx 2 2 L
(12.45)
Let us assume that xF L 2 and that the beam is simply supported with fork bearings at either ends, rendering the boundary conditions v 0, v 0 and 0 at x 0 and at x L
(12.46)
which are all fulfilled by the following displacement functions:
v c1 c sin L x 2
(12.47)
Introducing Eq. 12.47 into 12.45 and performing the necessary integrations, L
L
L x 2 x recalling that sin 2 dx cos dx , the following is obtained: 2 L L 0 0
2 1 2 EI z 2 L 2 L2
2 EI w 2 Fz L qz L2 c12 GI t c M c c (12.48) 2 1 2 L2 2 2 2
The function c1 , c2 is then minimised by demanding 2 EI w Fz L qz L2 2 c2 0 M 2 c1 GI t c2 2 L L2 2 2 2 2 EI z Fz L qz L2 c M c2 0 1 c1 2 L L2 2 2 2
(12.49)
12 ELASTIC BUCKLING
266
2 EI z F L q L2 M z2 z 2 2 L 2 c1 0 2 2 EI w c2 0 M Fz L qz L GI t 2 2 2 L2
(12.50)
This can only be fulfilled with non-zero coefficients if the determinant to the coefficient matrix is zero. Thus, the buckling condition becomes:
2 EI z
GI t
L2
2
2 EI w L2
Fz L qz L2 M 0 2 2 2
(12.51)
Case 1: Let Fz 0 and qz 0 , then the elastic buckling condition is defined by
12
2 EI w M E M EI z GI t L L2
(12.52)
which is identical to what was obtained in Eq. 12.38 above. Case 2: Let M 0 and qz 0 , then the elastic buckling condition is defined by: 12
FzE
2 2 EI w Fz EI z GI t L2 L
(12.53)
Case 3: Let M 0 and Fz 0 , then the elastic buckling condition is defined by:
qz E
2 qz L
12
2 2 EI w EI GI z t L2 L
(12.54)
Example 12.1: Lateral torsional buckling of a cantilevered beam Let us consider the cantilevered beam in Fig. 12.13. At its tip it is subject to a concentrated force Fz elevated a distance e above the centroid. It is for simplicity assumed that the cross section is symmetric about both y and
z
axes.
267
12.2 BEAMS
Fig. 12.13
Thus:
Cantilevered beam with elevated force at its tip
U P
1 EI z v2 GI t 2 EI w 2 dx Fz F 2L
where F is the vertical displacement of Fz , which may be calculated according to Eq. 12.41, only with v replaced by v e (see right hand side illustration in Fig. 12.13),
i.e. F 1 2 v e .
U P
1 F EI z v2 GI t 2 EI w 2 dx z v L e L L 2L 2
Assuming v x c1 f x and x c2 f x where
f x 1 cos x 2 L ,
then the following is obtained:
2 EI c 2 c2 4 2 EI w 8 z 1 2 GI LeF 2 LFz c1c2 t z 2 2 2 2 2L 2 L 2 Minimising by demanding c1 0 and c2 0 the following is obtained: c1 , c2
L 2L
2
2 EI z 2 2L 4 2 LFz
c1 0 2 c2 0 EI w 8 GI t 2 LeFz 2 2 L
4
2
LFz
12 ELASTIC BUCKLING
268
This condition can only be obtained if the determinant to the coefficient matrix is zero, and thus, the elastic critical buckling load is given by:
2 2 2 EI z 2 L GI t I w 2 Fz e 4 L 2 L 2 EI z I z
12
FzE
It is readily seen that if e 0 then: FzE
2 2 EI z
e 12
2 EI w
GI t 2 4 L 2 L 2 2 L
Elaboration 12.3: Beam with non-symmetric cross section Let us consider the beam in Fig. 12.12.a subject to constant bending moment M alone (i.e. Fz and qz are zero while end moments M 0 ). Let us assume that the cross section is single symmetric, with symmetry about weak y axis, i.e. the shear centre coordinates are 0, ez . Then the external load energy is
P x dA du
x
where
LA
M z Iy
1 2 v dx and v vD z ez . Thus: 2 1 2 M 1 M z ez 2vD z 3dA 2ez 2 dx P z vD I 2 2 L LA y Iy A
and as shown in Elaboration 12.2: du
im 2 M vD dx 2 L
where
im
1 3 z dA 2ez I y A
The energy balance is then defined by:
U P
1 i EI z v2 GI t 2 EI w 2 dx M vD m 2 dx 2L 2 L
vD
c1
x
Assuming that the beam is simply supported with fork bearings sin L c2 will satisfy boundary conditions. Thus:
269
12.3 BEAM-COLUMNS
from which:
2 1 2 EI z 2 L 2 L2
2 EI 2 im 2 c12 GI t c M c c c2 2 1 2 2 L2
2 2 EI z 2 c1 Mc2 0 c1 2 L L 2 2 EI c 2 L GI t L2 c2 M c1 im c2 0 2
2 EI z 2 L M
c1 0 c2 0 2 EI GI i M t m L2 M
Setting the determinant to the coefficient matrix equal to zero, the following is obtained:
M2
2 EI z L2
im M
2 EI z L2
GI t
2 EI L2
0
Thus, the elastic buckling moment is given by: 2 2 L 2 GI I i i t M E M EI z m m 2 EI z I z 2 L
12.3
Beam-columns
In this chapter we shall consider the combined buckling effects that my occur for a structural member subject to axial load N as well as transverse loads q y and
qz , and torsion load q . For simplicity, qx and any concentrated transverse forces are not included, see Fig. 12.14. Axis system and sign conventions are shown in Fig. 12.15. As illustrated in Fig. 12.16.a, the load components are referred to the shear center of the cross section. The relevant displacements and stress resultants are shown in Fig. 12.16.b and c. The buckling deformations may then be a combination of sideway displacements in y and z directions as well as cross sectional twisting, see Fig. 12.17.
12 ELASTIC BUCKLING
270
Fig. 12.14
Fig. 12.15
The beam column
Axis system and sign conventions
12.3 BEAM-COLUMNS
Fig. 12.16
Fig. 12.17
External loads and load effects
Bending and torsion buckling deformations
271
12 ELASTIC BUCKLING
272
Shear center and centroid displacements are defined by:
vD
wD
T
v0 vD ez w0 wD e y
and
where ey , ez
(12.55) (12.56)
are the shear center coordinates. Since we are dealing with a
beam-column stability problem the axial displacement component u0 uD is ignored.
a) Normal force, N
b) Shear forces, V y and Vz
c) Bending moments, M y and M z Fig. 12.18
Stress resultants in displaced positions on element dx
12.3 BEAM-COLUMNS
273
The stress resultants N , V y , Vz , M y and M z are illustrated in displaced positions in a calculus equilibrium type of approach in Fig. 12.18. Torsion will be handled separately. (Please note that in spite of presently handling a buckling problem, the equilibrium conditions are developed with N defined in accordance with a vector description, i.e. positive tensile.) The force equilibrium conditions in x , y and z directions of an infinitesimal element dx are then:
F 0 d V Nv V q dx 0 V q y 0 y z y y y F 0 d V Nw V q dx 0 V q z 0 z y z z z
Fx 0
dN qx dx 0 N qx
(12.57)
where higher order terms have been ignored (e.g. Nv0 Vz as compared to Vy , and similarly in z direction). Since qx 0 , then N is constant. The moment equilibrium conditions about y and z axis of an infinitesimal element dx are as follows:
M y 0 Mz 0
Fig. 12.19
d M y M z Vz dx Ndw0 0 d M z M y V y dx Ndw0 0
Normal stress component at arbitrary position y , z
(12.58)
12 ELASTIC BUCKLING
274
Ignoring higher order terms, the following is obtained: M y M z Nw0 Vz M z M y Nv0 V y
(12.59)
The moment equilibrium condition about an axis through the shear center consists of two effects; one by the variation of the torsion moment from M x at
x increasing to M x dM x at x dx plus the effects of normal stress x y , z inclinations at a displaced cross sectional position as illustrated in Fig. 12.19:
x v x vD z ez
and x w x wD y e y
(12.60)
Then torsion equilibrium of an infinitesimal element dx is defined by:
Mx 0
dM x d x vD z ez z ez dA A d x wD y e y y e y dA q dx 0 A
M x
(12.61)
d d vD x z ez dA wD x y e y dA dx A dx A 2 d 2 x z ez y e y dA q 0 dx A
(12.62)
Introducing x from Eq. 10.66 and integrating, acknowledging that x, y , z are main axes, such that 1 A 1 y 2 I y z z dA 0 (12.63) dA y dA 0 I z y2 A yz A z A 2 I then torsion equilibrium is defined by:
M x
d d vD M y Nez wD M z Ne y dx dx
d Ni 2 M y i y 2ez M z iz 2e y Bi q 0 dx
(12.64)
275
12.3 BEAM-COLUMNS
i2
where:
I y Iz A
e2y ez2
1 2 2 I z y z y iy 1 2 2 iz I y y z dA i A z 1 2 2 y z I
and
(12.65)
(Please note that i , i y and iz have dimension m , while i is non-dimensional). The differential equilibrium conditions for bi-axial bending and torsion are then obtained by introducing (see Eq. 10.61):
M y EI y wD
M z EI z vD M x EI GI t
(12.66)
into Eqs. 12.59 and 12.64, from which the following fourth order differential equations are obtained:
EI y wD M z N wD ey qz 0
(12.67)
M y N vD ez q y 0 EI z vD
(12.68)
i y 2ez M z wD iz 2e y EI GI t M y vD
e y wD i 2 Bi q 0 N ez v D
(12.69)
(Please note that N is defined positive as tensile, see Fig. 12.14.)
Example 12.2 Beam-column with symmetric cross section Let us for simplicity consider a beam-column with a cross section that is symmetric about both y and z axis, in which case vD v0 , wD w0 and i i p where
i 2p I y I z
A . Let us assume it is subject to a normal compressive force N and a
bending moment M at either end. Let us also assume the beam-column is simply
12 ELASTIC BUCKLING
276
supported with a fork bearing at its supports (i.e. the differential equations reduce into:
0 at x 0 and at x L ). Then
EI y w0 Nw0 0 EI z v0 M Nv0 0 EI GI t Ni 2p Mv0 0 As can be seen, any motion in w direction may occur alone as w is independent of v and , while any v or motion will always involve both, as they are interconnected by the second and the third equilibrium conditions. Thus, the solution to the first equation alone is what we previously have dealt with in Chapter 12.1 above, i.e. that
NEy N
2 EI y L2
For v0 and the following displacement functions will satisfy boundary conditions:
v0 c1 x c sin L 2
2 4 v0 v0 v0 v0 and L L
Then the second and third differential equations above become:
2 EI z M 2 N L c1 0 c 0 2 2 EI 2 GI t Ni p M 2 L The stability condition is then defined by a zero coefficient matrix determinant:
2 EI z 2 EI GI t Ni 2p M 2 0 2 N 2 L L 2 EI z N Ez L2 By defining: 2 N 1 EI GI E t i 2p L2
and
M yE
L
EI zGI t 1
2 EI L2GI t
12.4 FRAMES
then this stability condition may be written:
M My E
277 2
N N 0 1 1 NE N E z
If N 0 then the buckling mode is lateral torsional buckling moment, M M y E .
either buckling about the z axis: N N Ez or else torsional buckling: N N E
If M 0 then the buckling mode is
The interaction effects between sideway buckling about z axis, torsional buckling and lateral torsional buckling is illustrated in Fig. 12.20 below. (For these types of problems Galerkin’s approach will usually be the most effective strategy.)
Fig. 12.20
12.4
Stability interaction between N and M , N Ez N E
Frames
Based on the differential equation, analytical continuous solutions to the problem of elastic buckling of some simple frames may be found in the literature, but these are often cumbersome.
12 ELASTIC BUCKLING
278
Fig. 12.21
Four degree of freedom element with constant axial load
In general it is more convenient to resort to a finite element approach, rendering an eigenvalue problem which for simple frames most often involves a manageable numerical solution with a limited number of degrees of freedom. The finite element in a general format for line-like systems was presented in Chapter 6.5. Six degrees of freedom element stiffness and geometric matrices are given in Eqs. 6.49 and 6.50, while global equilibrium condition is given in Eq. 6.57. H ere we shall ignore axial deformations u0 NL EA and limit our analysis to pure bending, and thus, four degrees of freedom elements will suffice, see Fig. 12.21. The general equilibrium condition is then defined by
K K G r R
Nk K K n where K n AT kA n n 1 Nk T K G K Gn where K Gn A k G A n 1
where
(12.70) n
where N k is the number of elements in the system, An is the connectivity matrices (defined in Eq. 6.51) and the bending and geometric stiffness matrices kn and kGn are given by (see Eqs. 6.49 and 6.50) 12 6 L 12 6 L 2 6 L 2 L2 EI y 6 L 4 L kn 3 6L L n 12 6 L 12 6 L 2 L2 6 L 4 L2 n and
(12.71)
12.4 FRAMES
k Gn
36 3L 36 3L 2 3L L2 N 3L 4 L 36 3L 30 L n 36 3L 2 3L 4 L2 n 3L L
279
(12.72)
It should be noted that the geometric stiffness matrix above was developed with the axial force N n defined positive tensile, and that any compressive forces must then be introduced into Eq. 12.72 as negative quantities. The solution strategy is the following:
First the external loading on the system ( R ) is applied without taking any buckling effects into account, i.e. the following equation
Kr0 = R
(12.73)
is solved, rendering elastic displacements r0 and a set of corresponding element axial forces N n .
Then all external loads are removed and elastic displacements are reset to zero, while axial element forces N n are upheld constants, rendering the following equilibrium condition K K G N n r 0
(12.74)
which is a classic eigenvalue problem from which the critical elastic axial force condition may be obtained.
The method is best understood by an example. Let us consider the single bay portal frame shown in Fig. 12.22 (similar to that which was solved for pure bending in Example 6.4). The horizontal beam is subject to a concentrated force F at midspan. Disregarding axial deformations, the system has three degrees of freedom: r r1 r2
r3 , illustrated at the right hand side in Fig. 12.22. The T
solution strategy is illustrated in Fig. 12.23: First the external load is applied (ignoring any normal force effects). Due to symmetry it is readily seen that the axial forces from this analysis are N 2 N 3 F 2 and N1 H , where H is the horizontal support reaction force at base 1 and 4 (due to horizontal equilibrium they are equal but with opposite signs). Then these axial forces are applied to the system (see right hand side illustration in Fig. 12.23), rendering an eigenvalue
12 ELASTIC BUCKLING
280
problem from which the stability condition may be obtained. The connectivity matrices for the three elements are 0 0 A1 0 0
0 1 0 0
0 0 , 0 1
Fig. 12.22
Fig. 12.23
1 0 A2 0 0
0 1 0 0
0 1 0 0 and A3 0 0 0 0
0 0 0 0
0 1 0 0
Simple frame subject to spanwise load
Reaction forces and element axial forces
(12.75)
12.4 FRAMES
281
Performing the matrix multiplications k n AT kA
n
and k Gn AT k G A
n
the
following is obtained (please note that for simplicity the index y on the bending stiffness is in the following omitted, as it is to be taken for granted that all bending deformations takes pace about the y axis): 0 0 0 k1 0 4 EI1 L1 2 EI1 L1 , 0 2 EI1 L1 4 EI1 L1
12 EI 2 L32 k 2 6 EI 2 L22 0
6 EI 2 L22
12 EI 2 L32 k3 0 6 EI L2 2 2
0 6 EI 2 L22 0 0 , 0 4 EI 2 L2
4 EI 2 L2 0
0 0 , 0
k G1
0 0 N1 0 4 L12 30 L1 0 L2 1
3L2
k G2
36 F 2 3L2 30 L2 0 36 F 2 0 30 L2 3L 2
0 3L2 0 0 (12.78) 0 4 L22
k G3
0 L12 4 L12
4 L22 0
(12.76)
0 0 (12.77) 0
Thus,
3L2 12 2 EI 2 K k n 3 3L2 2 1 L22 L2 n 1 L22 3L2 and 3
3
KG kGn n 1
22 F 3L2 60L2 3L2
3L2
EI1 L2 (12.79) where L1 EI 2 2 2 1 L2
L22
L1 L 2 L22 where N 1 2 1 L22 F (12.80)
3L2
3L2
4 2 1 L22
L22
What remains before the frame stability condition can be obtained is to determine the axial force N 1 from a first order static problem with F in its original position at midspan of the beam. This may readily be obtained by superposition of the two static systems illustrated in Fig. 12.24, one where r0 0 , and the other where and F 0 F 0
R R1 R2 R3 0 FL1 8 FL1 8 . T
T
12 ELASTIC BUCKLING
282
Fig. 12.24
Load cases for the determination of N 1
Since these systems are symmetric ( r1 and R1 are equal to zero), the first degree of freedom may be ignored, and thus
r2 FL1 1 r2 1 2 EI 2 2 1 FL1L2 (12.81) 2 1 r3 L2 8 1 r3 16 EI 2 2 1 from which (see Fig. 6.3)
N1 H
6EI 2 3 r2 F 2 8 2 L2
(12.82)
where and are defined in Eqs. 12.79 and 12.80 above. Thus, the eigenvalue problem defining the elastic stability properties of the system is given by: 12 3L2 3L2 2 L22 3L2 2 1 L2 L22 2 1 L22 3L2 (12.83) 3L2 3L2 22 r1 0 3L2 4 2 1 L22 L22 r2 0 2 2 r 3 2 1 L L L 3 2 2 2 where
FL22 . As illustrated in Fig. 12.25, this system has two primary 120 EI 2
buckling eigenmodes, one asymmetric and one symmetric.
12.4 FRAMES
283
Since the system has three degrees of freedom there is also a third eigenmode, which is a higher order inverse version of one of the other two, and therefore of no interest to the stability limit of the system, which will be determined by the first or the second, whichever is the lower. As can be seen, we are free to scale r by any linear transformation without changing the eigenvalues of the system. This is desirable, because from a computational point of view it is convenient to
r r1 r2 introduce a set of degrees of freedom
r3
T
which is defined such
that they all have the same dimension.
a) Asymmetric mode Fig. 12.25
b)
Symmetric mode
Buckling modes
In this case it will render a non-dimensional eigenvalue problem if we introduce:
0 1 0 r Cr where C 0 1 L2 0 0 0 1 L2
(12.84)
T into Eq. 12.83, followed by pre-multiplication by C . Then the following is obtained
12 3 3 3 3 r1 72 0 3 4 2 1 2 r2 (12.85) 3 2 1 0 3 3 2 1 2 2 1 r3
12 ELASTIC BUCKLING
284
For systems with two degrees of freedom it is fairly easy to obtain a solution by demanding zero determinant of the coefficient matrix, but already at three degrees of freedom such an approach is cumbersome. Usually, it is far more convenient to adopt a mathematical eigenvalue approach, which will not only render eigenvalues but also the eigenmodes, making it easy to identify critical buckling shapes. An introduction to the numerical solution of eigenvalue problems is presented in Chapter 12.6. Thus, Eq. 12.85 may be written:
A B r 0
(12.86)
where 3 3 3 3 12 72 A 3 2 1 2 (12.87) and B 3 4 2 1 3 3 2 1 2 4 1
Fig. 12.26
Asymmetric and symmetric frame buckling length Lk
For instance, introducing 1 and 1 , then a solution based on MATLAB software [21] renders the following eigenvalues and corresponding eigenmodes:
285
12.5 RECTANGULAR PLATES
0 0 0 0.0015 0.1225 0.1208 λ 0 0.6316 0 and Φ 0.073 0.3244 0.3077 (12.88) 0 0.073 0.3244 0.3077 0 0.9525 contains three eigenvalues where λ contains eigenvalues on its diagonal and Φ on each of its columns. Defining the critical elastic buckling length Lk such that FE 2 EI 2 L2k for asymmetric buckling and N1E 2 EI1 L2k for symmetric buckling, the results of a MATLAB [21] calculation is shown in Fig. 12.26.
12.5
Rectangular plates
The relevant types of distributed edge loading, normal forces N x or N y and shear force N xy N yx (all with unit N m ), are shown in Fig. 12.27.a. The corresponding stress resultants are shown in Fig. 12.28. It is taken for granted that there is no transverse load ( qz ) on the plate. A collection of typical buckling modes are illustrated in Fig. 12.29. Apart from purely numerical solution strategies (e.g. finite elements or finite differences), there are three alternative methods of solving the plate buckling problems: the use of the differential equation, an energy method (e.g. a Rayleigh-Ritz approach), or Galerkin’s method of weighted residuals.
1) Solution by the differential equation:
The differential equation was developed in Chapter 11 (see Eq. 11.19) with the sign convention that all forces are vectors as defined in Fig. 11.5. Here we shall assume that the external forces are in the opposite direction, i.e. that N x and N y are compressive forces while N xy gives a positive displacements in the
x, y
plane, see Fig. 12.27.a. Hence, the differential equation is given by
4w 4w 4w 2w 2w 2w D 4 2 2 2 4 N x 2 N y 2 2 N xy 0 xy x y y x y x
(12.89)
286
12 ELASTIC BUCKLING
Fig. 12.27
Rectangular plane and isotropic plate
The use of the differential equation is usually the most cumbersome approach, so let us consider the case of a simply supported plate as shown in Fig. 12.30, which is subject to normal force N x alone. It is taken for granted that Lx Ly .
287
12.5 RECTANGULAR PLATES
The following displacement function will satisfy the boundary conditions that w and w are zero along all its outer edges: M m w x , y sin y c sin x L y m Lx m 1
(12.90)
where cm are unknown coefficients and M is a sufficiently large number to obtain a satisfactory result. Introducing m L x 2 w x 2 4 2 m Lx 4 M w x sin y c 4 2 2 L y m m L 2 L x y m1 w x y 4 w y 4 4 Ly 2
m 2 sin Lx
x
(12.91)
into the differential equation (Eq. 12.89 above), the following is obtained:
2 2 2 2 m N x m m sin y cm x0 sin Ly m1 Lx Ly D Lx Lx
M
Fig. 12.28
Stress resultants
(12.92)
288
12 ELASTIC BUCKLING
a) Constant normal load in x direction alone
b)
z axis pure bending at either ends
c) Shear force loading Fig. 12.29
Typical buckling modes of rectangular plates
12.5 RECTANGULAR PLATES
Fig. 12.30
Fig. 12.31
289
Rectangular plate with axial load N x
Buckling coefficients for a rectangular plate; upper diagram: normal load N x , lower diagram: shear loading N xy
This is an eigenvalue problem whose only non-trivial solution is defined by 2
2 2 2 m N x m 0 Lx L y D Lx
Thus, the elastic buckling load is given by:
(12.93)
290
12 ELASTIC BUCKLING
N xE N x k x
2D L2y
m where k x m
2
and = Lx Ly
(12.94)
The buckling coefficient k x is illustrated in the upper diagram in Fig. 12.31. It is worth noting that k x 1 4 and that k x 4.
2) Solution by Rayleigh-Ritz’s energy method: The strain energy of plate bending was developed in appendix A.3 (Eq. A.24): 2 2 2 ab 2w 2w 2w D 2 w 2 w 2 1 U 2 2 2 2 dxdy (12.95) 2 0 0 x y x y x y 2
As illustrated in Fig. 12.27.b, c and d, the loss of load energy P follows the same development as that which was shown for columns (see Fig. 12.8 and Eq. 12.15). Thus 2 2 1 w w dPx N x dy d x where d x dx dx dx dx 2 x x 2 2 1 w 2 w dPy N y dx d y where d y dx dx dy dy 2 y y (12.96) w w w w dPxy N xy dx d xy where d xy dy dy x y x y w w w w dx dx dPyx N yx dy d xy where d yx y x x y 2
from which the following is obtained: 1 P 2
Lx L y
0 0
2 w 2 w w w Nx dxdy 2 N xy Ny x y x y
Hence, the energy balance function U P is given by:
(12.97)
291
12.5 RECTANGULAR PLATES
D 2
1 2
Lx L y
0 0
Lx L y
0 0
2 2 w 2 2 w 2 2w 2w 2w 2 1 2 2 dxdy xy x 2 y 2 x y 2 2 w 2 w w w Nx dxdy N 2 N xy y x y x y
(12.98)
Let us again consider a simply supported rectangular plate with dimensions Lx and Ly , assuming Lx Ly , now subject to bi-axial loads N x and N y . The following trigonometric series solution, which satisfies all boundary conditions, is adopted: w x, y
M
m n x sin y x L y
N
cmn sin L
m 1 n 1
(12.99)
It is readily seen that its derivatives are given by m n w x sin y m Lx cos x M N Lx L y cmn n w m 1 n 1 m y n L y sin L x cos L y x y 2 w 2 m L 2 M N x m n x cmn x sin y 2 sin 2 w n L Lx Ly m 1 n 1 y 2 y
(12.100)
(12.101)
and that Lx L y
0 0
m L 2 L L M N w x 2 x 2 x y dxdy cmn 2 2 4 w y m 1 n 1 n Ly
2 w x 2 2 m Lx 4 L Lx y M N 2 4 2 2 2 w y dxdy n Ly cmn m 1 n 1 0 0 2 2 w xy m Lx n Ly
Lx Ly 4 2
(12.102)
(12.103)
292
12 ELASTIC BUCKLING
Introducing this into Eq. 12.98, the following is obtained:
4 DLx Ly 8
2 2 n 2 N x m2 N y n 2 2 m c L2 L2 2 2 2 2 mn x D Lx D Ly m 1 n 1 y M
N
(12.104)
Minimising the energy balance function by demanding cmn 0 for all values of m 1,, M and n 1,, N renders the eigenvalue problem: 4 DLx Ly 4 cmn
2 2 n2 N x m2 N y n2 m L2 L2 2 D L2 2 D L2 cmn 0 (12.105) m 1 n 1 y x y x M
N
whose solution is given by N x m2 N y n2 m2 n2 2 D L2x 2 D L2y L2x L2y
2
(12.106)
Thus, the elastic buckling load is defined by
N x L2y
2D
m2
N y L2x
m2 2 n2 n 2 D
2
where
Lx Ly
(12.107)
E.g., if N y 0 , then n 1 and the elastic buckling load is identical to that which was obtained by the differential equation and given in Eq. 12.94.
3) Solution by Galerkin’s principle of weighted residuals:
For more complex plate buckling problems than those handled above, Galerkin’s approach is by far the most effective. It renders an eigenvalue problem which is easily manageable from a computer programming point of view. It is explained in appendix A.7. Let us for simplicity consider a simply supported rectangular plate with dimensions Lx and Ly , assuming Lx Ly . A linear solution M
N
w x, y cmn mn x, y m 1 n 1
(12.108)
293
12.5 RECTANGULAR PLATES
with M N unknown coefficients cmn , and shape functions m n x sin y Lx L y
mn x, y sin
(12.109)
will comply with the boundary conditions, which is hugely advantageous for the accuracy of the solution. If this solution is introduced into the differential function 4w 4w 4w 2w 2w 2w 0 f w D 4 2 2 2 4 N x 2 N y 2 2 N xy xy x y y x y x (12.110)
(see Eq. 12.89) then the following is obtained: 2 2 2 m n m n f w cmn D x sin y sin Lx Ly m 1 n 1 Lx Ly M
N
2 n m Nx N y Lx Ly
2 N xy
2
sin m x sin n y Lx Ly
m n m n cos x cos Lx Ly Lx Ly
(12.111)
y 0
Now let f w be successively multiplied, first by 11 and integrated across the space 0 Lx and 0 Ly , then by 12 and integrated, and so on throughout all shape functions, i.e. f w is successively weighed by the shape functions p Lx
pk x, y sin one after the other, where This will render
M N
k x sin y L y
(12.112)
p 1,, P and k 1, , K , P K M N . equations, all equal to zero, and with unknown
coefficients cmn . Thus, the following eigenvalue problem is obtained:
K0 N xGx N yGy N xyGxy c 0
(12.113)
294
12 ELASTIC BUCKLING
where c c11 c12 c1N
c21 c22 cmn cMN and where K 0 ,
G x , G y and Gxy are square matrices with size MN by MN . K0 K0 pk ,mn ,
Gx Gx pk ,mn ,
Gy G y pk ,mn ,
G xy
Gxy pk ,mn
(12.114)
whose content is defined by: K 0 pk ,mn
2 2 n m D Lx Ly
2
Ly
Lx
sin
0
p m k n x sin dx sin x sin dy Lx Lx L Ly y 0 (12.115)
m Gx pk ,mn Lx
2 Lx
p
sin Lx 0
2L x
G y pk ,mn
n Ly
Gxy pk ,mn
m n 2 Lx Ly
Ly
x sin
m k n dx sin x sin dy Lx Ly Ly 0
(12.116)
Ly
p m k n sin Lx x sin Lx dx sin Ly x sin Ly dy 0 0 Lx
(12.117)
Ly
p m k n sin Lx x cos Lx dx sin Ly x cos Ly dy 0 0
(12.118)
0 0 if i j is an even number L i j sin x cos xdx 2 L i L if i j is an even numbe r L i2 j2 0
(12.119)
Since L
i
sin L x sin
it is readily seen that
0 if i j j xdx L L 2 if i j
295
12.5 RECTANGULAR PLATES
2 4D if p m and k n K 0 pk ,mn m2 n2 4 Lx Ly Gx pk ,mn 2 4 m 2 if p m and k n (12.120) G y pk ,mn 2 4 n 2 if p m and k n m p 2 2 2 2 are odd numbers Gxy pk ,mn 8mnpk p m k n if nk
where Lx L y . It is convenient to re-write the eigenvalue problem into a non-dimensional format by defining: Nˆ x N x
Lx L y 2
D
,
Nˆ y N y
Lx L y 2
D
and
Nˆ xy N xy
32 Lx L y
(12.121)
4D
in which case the solution is defined by the following eigenvalue problem
Kˆ
0
ˆ Nˆ G ˆ ˆ ˆ Nˆ x G x y y N xy G xy c 0
(12.122)
where 2 m 2 2 2 2 ˆ ˆ ˆ K 0 diag n , G x diag m , G y diag n mnpk G pk ,mn p 2 m 2 k 2 n 2 (12.123) ˆ G pk ,mn G xy where m p if n k are odd number
Example 12.3: Shear buckling of rectangular plate Let us still consider a rectangular plate and let us for simplicity chose M N 3 , i.e.
w x, y
3
3
cmn mn x, y
m 1 n 1
where
m n x sin y Lx L y
mn x, y sin
296
12 ELASTIC BUCKLING
Then
ˆ diag 1 2 K 0
1 4 2 1 9 2 4 2
4 4 2 4 9 2 9 2 9 4 2 9 9 2 ˆ diag 1 G x
1
1
ˆ diag 4 9 G y
4
4
4
9
9
9
4 9 4 9
and
ˆ G xy
0 0 0 49 0 0 0 0 0 0 4 9 0 0 0 45 0 0 0 0 4 5 0 0 0 0 0 0 0 0 0 0 0 0 45 0 0 4 9 4 9 0 0 0 0 0 36 25 4 5 4 5 45 0 0 0 0 0 0 36 25 0 0 4 5 0 0 0 0 0 0 0 36 25 0 0 45 0 0 0 0 0 0 0 0 0 36 25 0 0 0 0
Let us assume that the loading is shear alone, i.e. N x N y 0 . Then the eigenvalue problem is defined by
Kˆ
0
ˆ Nˆ xy G xy c 0
The eigenvalues Nˆ xy for a variation of
between 1 and 2.2 is shown in Table 12.1
below. Defining elastic buckling load N xy E k xy 2 D L2y , then the relationship
2 ˆ N xy . 32 presented in Table 12.1 have been plotted versus in
between k xy and the eigenvalues Nˆ xy are given in Eq, 12.121, i.e. k xy The buckling coefficients k xy
the lower diagram in Fig. 12.31. Larger values of
will require larger values of M
and N . As mentioned above, one of the advantages with Galerkin’s approach is that the
297
12.5 RECTANGULAR PLATES
buckling mode shape is the eigenvector to the solution. For instance, at
1
the
eigenvalue is Nˆ xy 30.55 and the corresponding eigenvector is
c c11 c12
c13 c21 c22
c23
c31 c32
c33
T
1 0 0.072 0 0.2964 0 0.072 0 0.04
T
The buckling mode shape w x , y
3
m n x sin y is shown in x L y
3
cmn sin L
m 1 n 1
Fig. 12.32. Table 12.1
Shear force load 1.0 1.2
1.4
1.6
1.8
2.0
2.2
Nˆ xy
30.55
31.39
33.44
36.28
39.7
43.55
47.78
k xy
9.42
8.07
7.37
7
6.8
6.72
6.7
Fig. 12.32
Buckling mode shape, shear loading, 1 , Nˆ xy 30.55
298
12.6 The numeric eigenvalue problem The solution to an eigenvalue problem plays a significant role in mathematics. It is beyond the scope of this book to cover the subject in great detail. But, for the sake of completeness a brief coverage of the simplest routines available is given below. In general the eigenvalue problem is defined by
A B r 0
(12.124)
where A and B are N by N square matrices, is the eigenvalue and r is the eigenvctor. The solution to this equation is in mathematics known as the general 1 eigenvalue problem. By pre-multiplication with B then the special eigenvalue problem is obtained (12.125) C I r 0 1 where I is the identity matrix and C B A . Similarly, by pre-multiplication 1
with A obtained, i.e.
then the inverse version of the special eigenvalue problem is
D I r 0 1
(12.126)
where 1 and D A B . It is readily seen that any of these equations may be chosen for the determination of and r .
Fig. 12.33
N dimensional polynomial f
12.6 THE NUMERIC EIGENVALUE PROBLEM
299
In general there will be N possible solutions, as a nontrivial r will require that the determinant to C I is equal to zero, which may be expressed by expanding det C I into an N dimensional polynomial
f det C I 1 2 i N r
(12.127)
whose roots 1 , 2 ,, N (see Fig. 12.33) all represents a possible solution to the eigenvalue problem. It is customary to arrange eigenvalues in ascending order, i.e. 1 2 i N , because it is the lowest that requires the least amount of system energy to occur. The corresponding set of eigenmodes ri , i 1, 2,, N , may then be obtained by re-introducing any of the solutions i back into
C iI ri 0
(12.128)
Thus, it is seen that ri is scalable, i.e. that it may be multiplied or divided by any constant number. It only represents a characteristic shape of structural displacements. The actual displacements cannot be quantified unless an external loading is included. Re-writing the equation above into
Cri iri
(12.129)
it is seen that an eigenmode ri of a matrix C is a non-zero vector which multiplied by C is the vector itself multiplied by a constant i called the eigenvalue of C . Let a a C 11 12 (12.130) a21 a22 Then the solution to the eigenvalue problem a11 a12 1 0 r1 a11 0 1 r2 a21 a21 a22
C I r
a12 r1 0 a22 r2 0
(12.131) is that a det 11 a21
a12 a11 a22 a12 a21 a22
a11 a22 a11a22 a12 a21 0 2
rendering
(12.132)
300
1,2 and
2
a11 a22 a a22 11 a11a22 a12 a21 2 2
(12.133)
a11 1,2 r1 a12 r2 0
(12.134)
r2 1,2 a11 r1 a12
(12.135)
from which it is seen that
2 2 Let for instance C . Then 2 5
1 6 r1 r1 while r1 and r2 . 2r1 0.5r1 2 1
Let i and j , and corresponding ri and r j be two arbitrary but independent and non-trivial solutions to the eigenvalue problem A B r 0 , i.e.
A iB ri 0
A jB r j 0
(12.136)
Pre-multiplying the first with r Tj and the second by riT r j Ari ir jBri 0 ri Ar j jriBr j 0
(12.137)
then transpose throughout the second (recalling that A and B are symmetric) r Tj Ari jr Tj Bri 0
(12.138)
and subtracting the first, then the following is obtained:
i j rTj Bri 0
(12.139)
Thus, since i j we must conclude that r Tj Bri 0
(12.140)
which means that the eigenmodes ri and r j are orthogonal with respect to B . Pre-multiplying A iB ri 0 by riT it is seen that
12.6 THE NUMERIC EIGENVALUE PROBLEM
i
riT Ari riT Bri
301 (12.141)
which is called the Rayleigh quotient. Since the eigenvectors are arbitrarily scalable, it is often convenient to scale them such that riT Bri 1 , which is obtained by setting
riT Bri
rˆi ri
(12.142)
i rˆiT Arˆi
Then the Rayleigh quotient is given by
(12.143)
1) Direct (power) vector iteration:
Starting with a sound engineering guess ψ0 as an initial solution to the 1 eigenvalue problem C k I ψk 0 , where C B A , may iteratively be expanded into an improved solution by
ψk
Cψk 1 Cψk 1
where k 1,2,, N
(12.144)
and where
Cψk 1
Cψk 1 T Cψk 1
(12.145)
is the norm of the previous solution. The corresponding eigenvalue may be determined by using the Rayleigh quotient
k
ψTk Cψk ψTk ψk
(12.146)
ˆ k ψk ψTk ψk is optional.) (Normalisation of ψk such that after each step ψ The method is not widely in use because the solution converges towards a larger dominant eigenvalue, provided ψ0 has a non-zero component in the direction of the corresponding eigenmode. 2) Inverse vector iteration:
The inverse vector iteration method is in principal equivalent to the direct method presented above, only it operates on the inverse eigenvalue problem
302
D k I ψk 0 ˆ1 Let ψ
ψ0 ψT0 ψ0
where D A1B and k k1
be the normalised vector of an initial guess ψ0 . Then
ˆk ψk Dψ Because
(12.147)
where
ˆk ψ
ψk 1 ψTk1ψk 1
(12.148)
ˆ Tk D k I ψ ˆk ψ ˆ Tk Dψ ˆ k kψ ˆ Tk ψ ˆk ψ ˆ Tk ψk k 0 ψ
(12.149)
ˆ Tk ψk k ψ
(12.150)
then
The main advantage with this method is that with a choice of ψ0 that is not unduly off target it will always converge towards the solution with numerically lowest eigenvalue 1 and corresponding eigenmode rˆ1 , i.e.
k 11 and k
Fig. 12.34
ˆ k rˆ1 ψ k
Eigenvalue polynomial f with shift
12.6 THE NUMERIC EIGENVALUE PROBLEM
303
Second or any higher eigenvalues may be determined by introducing a shift (see Fig. 12.34) and rewrite the original eigenvalue problem into A B B r 0
(12.151)
1
Defining and pre-multiplication by A B render the following equivalent eigenvalue problem:
D I r 0
1
1
where D A B B
will then
(12.152)
Thus, iterations may be performed in an equivalent procedure to that which has been shown above, and the solution will converge towards the eigenvalue immediately above , while k 1 k (12.153) There are numerous far more effective routines for a numerical solution to the eigenvalue problem than those included above (for an overview, see e.g. Bell [19]). MATLAB [21] and similar programming tools are obvious sources for a quick solution to problems with a limited number of degrees of freedom.
Example 12.4: Inverse vector iteration 5 Let us consider matrices A 6 10 2 0 1 30 1 D 0 A B 1 30 Then 1 30 4
2
0 2 0 0 3 3 1 and B 3 10 0 2 0 1 1 0 0 1 1 30 1 60 1 12 1 24 1 12 11 120
The successive inverse iterations are shown in Table 12.2 below, starting with
ψ 0 1 1 1
T
ˆ 1 ψ0 ψ
ψT0 ψ0 0.5774 0.5774 0.5774
T
As can be seen the convergence is approaching 1 6.278 from below. It represents the first (lowest) eigenvalue of the system. The corresponding normalised eigenmode is given by
ˆ 6 ψ6 r1 ψ
ψT6 ψ6 0.2504 0.5478 0.7982
T
304 To obtain the second eigenvalue
24
D 24 A B
1
is a suitable choice. Then
5.303 1.894 4.735 B 10 1.894 2.462 6.155 9.47 12.31 5.777 2
ψ 0 1 1 1 , T
The successive inverse iterations (Table 12.3), starting with rendering
ˆ 1 ψ0 ψ
ψT0 ψ0 0.5774 0.5774 0.5774 and so on. T
ˆ 6 ψ6 Thus, 2 30 and r2 ψ Table 12.2 k
ψT6 ψ6 0.4099 0.4111 0.8143
T
1
2
3
D 0 0.0333 0.0333 0.0167 0.0333 0.0833 0.0417 0.0333 0.0833 0.0914
ˆk ψk Dψ
ˆ Tk ψk k ψ k 1 k Table 12.3 k
ˆ k ψk 1 ψ
ˆ Tk ψk k ψ k 1 k
6
0.3034 0.5766 0.7586 0.0420 0.0898 0.1277
0.2597 0.5554 0.7900 0.0403 0.0879 0.1274
0.2523 0.5495 0.7965 0.0400 0.0874 0.1272
0.2509 0.5481 0.7979 0.0399 0.0873 0.1272
0.2506 0.5478 0.7982 0.0399 0.0873 0.1272
0.15
0.1614
0.1599
0.l594
0.1593
0.1593
6.667
6.196
6.254
6.274
6.278
6.278
1
2
3
4
5
6
10 D 24
ˆk ψk Dψ
5
ψTk 1ψk 1
0.5774 0.5774 0.5774 0.0481 0.0914 0.1203
2
5.303 1.894 -4.735 1.894 2.462 -6.155 -9.47 -12.31 5.777
4
ˆ k ψk 1 ψ 0.5774 0.5774 0.5774 0.0142 -0.0104 -0.0924
ψTk 1ψk 1
0.1511 -0.1104 -0.9823 0.0524 0.0606 -0.0575
0.5317 0.6146 -0.5827 0.0674 0.0611 -0.1597
0.3669 0.3323 -0.8688 0.0689 0.0689 -1.1258
0.42299 0.4338 -0.7956 0.06831 0.0677 -0.1394
0.4034 0.3995 -0.8232 0.0679 0.0681 -0.1349
0.0577
0.1664
0.1567
0.1692
0.1657
41.33
30.01
30.38
29.91
30.04
Appendix A Equation Chapter 1 Section 1
BACKGROUND TO THE ENERGY METHODS A.1 The concept of energy conservation The law of energy conservation states that for a system in equilibrium the total energy remains constant; energy can neither be increased nor lost; rather, it transforms from one type to another. In the following it is this principle we are going to take advantage of. It is assumed that material behaviour is linear elastic, as indicated in Fig. A.1.a. It is a general requirement that any force, concentrated or distributed, is conservative, i.e. during a motion from position A to another position B through any path s the size and direction of the force remains unchanged, see Fig. A.1.b. It is taken for granted that the system under all conditions is at rest in a state of static equilibrium, and that any deformations or rigid body displacements are small. Kinetic or thermal contributions are disregarded.
Fig. A.1
Linear elasticity and conservative forces
© Springer Nature Switzerland AG 2020 E. N. Strømmen, Structural Mechanics, https://doi.org/10.1007/978-3-030-44318-4
305
A BACKGROUND TO THE ENERGY METHODS
306
The instantaneous energy in the system at any position is then defined by
the energy P held by external forces acting on the system, the strain energy U stored in the material fibres of the system.
The principle of energy conservation states that the energy function
U P
(A.1)
is constant.
Fig. A.2
Energy considerations in a simple mass-spring system
Energy considerations in a simple mass-spring system are illustrated in Fig. A.2. Let us assume that the external force has been slowly increased from zero to F by adding small weights to a dish suspended below the body. During this loading process the displacement has increased from zero to r . If a change of
A.1 THE CONCEPT OF ENERGY CONSERVATION
307
force dF is moving the load a displacement dr in its own direction (i.e. in the direction of F itself), then the force has perform the work dFdr . Energy is ability to work. Hence, work done is energy lost. I.e., the load has had its energy level reduced by dFdr , and since it had no energy prior to the loading process, it has the energy
P
r
F
dFdr Fr
(A.2)
0 0
at the end of the loading process. During the same loading procedure, the spring (the structural system) has gained energy U in the form of elastic strain: r
r
0
0
U Fs r dr Kr dr Kr 2 2
(A.3)
Thus, at the end of the loading process the total energy in the system is: U P Kr 2 2 Fr
(A.4)
All the energy methods applied herein are based on the principle of energy conservation, i.e. that any incremental variation of with respect to its displacement variables is equal to zero, i.e. that
0
(A.5)
For the system in Fig. A.2, the only variable is r , and thus:
d dr r Kr F r 0
Kr F 0
(A.6)
which is identical to the equilibrium condition of the system. In general, if a structural system is subject to external concentrated or distributed forces Fx
Fy
T
Fz , M x j
M z
My
T j
or q y
T
qz , applied during structural
displacements u v w , then the energy held by these forces is given by: T
T
Fx P Fy j Fz j
u xN x v x Fy w xFz
T
M x M y j M z j
x M x v x My w xM z
T
q y v x dx q w x L z
(A.7)
A BACKGROUND TO THE ENERGY METHODS
308
P is negative because positive force and motion have same directions, rendering loss of ability to perform the work it has done through these displacements. Please note that it is the change of with respect to its displacement variables that is zero, not itself. Relevant expressions for the build-up of strain energy U will be developed in the following (for beam-columns and plates), while the loss of external load energy P will be developed wherever applicable.
A.2 Strain energy in beam-columns and plates
Fig. A.3
Beam-column axis, external forces and stress resultants
Strain energy is the energy stored in the system because material fibres have been stretched or compressed and because they have been distorted by shearing forces or torsion. Let us first turn to the general case of a line-like beam or beam-column type of system. Cross sectional loads, displacements and stress resultants are defined in Fig. A.3. From linear elasticity
E 0 x σ x Eε 0 G
(A.8)
where x is normal stress due to axial force and bending, x is corresponding strain, xs is shear stress and xs is corresponding shear angle (see Chapter 10).
309
A.2 STRAIN ENERGY IN BEAM-COLUMNS AND PLATES
Then strain energy for an incremental material element dAdx is defined by ε
ε
ε
dU σT d ε E ε d ε εT E d ε 0
x
0
E xd x G 0
T
0
yz
xs d xs E 0
x2 2
G
2 xs
2
x2 2E
(A.9)
2 xs
2G
Thus, the total strain energy in the system is given by U
2 1 x2 xs dA dx 2 L A E G
(A.10)
In Chapter 10 we have shown that (see Eq. 10.58) that
x E uD vD y wD z
(A.11)
where the connection between centroid and shear centre displacements are:
0 u0 uD v v e z 0 D e y w0 wD
(A.12)
Recalling that main axes x , y , z and shear centre position are defined by y 0 z dA 0 , A yz 0
1 0 y dA 0 A z 0
and
1 A 2 I y z dA z2 Iy A 2 I
(A.13)
then the normal stress contribution to the strain energy is 2
z 1 x2 1 E uD vD y wD dA dA 2A E 2A E
(A.14)
1 2 2 2 2 EI z vD EI y wD EI EA uD 2 The effects of shear strain due to shear forces and warping are in the following assumed negligible. Then, we have shown in Chapters 10.3 and 10.4 that the
A BACKGROUND TO THE ENERGY METHODS
310
connection between torsion moment and cross sectional rotation is defined by M x GI t , and that 1 Mx 1 1 for a single cell cross section GI t t 2 Am t 2 Am
xs
xs
Mx 2 z G 2 z It
for an open section
(A.15)
Thus: bi ti 2 G 2 z 2 b t3 2 dA G 2 2 zi dzi dyi G 2 i i GI t 2 G 3 2 A i 0 ti 2 i xs dA G 2 2 A 1 1 1 ds 2 It GI dA G ds GI t 2 G t t 2 Am 2 t 4 A m Rm A (A.16) from which it is seen that: 2 1 xs 1 2 (A.17) dA GI t 2A G 2
Thus, the total strain energy in a continuous line-like beam-column type of system subject to axial force, bi-axial bending and torsion is the sum of the expressions in Eqs. A.14 and A.17, i.e.
U
1 2 2 2 2 2 EI z vD EI y wD GI t EI dx (A.18) EA uD 2 L
Let us then turn to the corresponding development for a rectangular plate, where the relevant stress components are shown in Fig. A.6. Still, x x E and
xy xy G applies. Due to the complexity of the problem it is convenient to use the superposition principle and apply one stress component after the other. First, x is increased from zero to an arbitrary value, rendering the following contribution to the strain energy of an incremental element dx , dy , dz :
1 12 x dydz x dx x dxdydz 2 2 E
(A.19)
A.2 STRAIN ENERGY IN BEAM-COLUMNS AND PLATES
Fig. A.6
311
Stress components in a plate
Then, while x is constant, y is increased from zero to an arbitrary value, rendering the following energy contribution:
1 11 2 y dxdz y dy x dydz y dx y 2 x y dxdydz 2 2E
(A.20)
where is Poisson’s ratio (and no half in front of the y contribution because
x is constant). Finally, while x and y are constant, xy yx is increased from zero to an arbitrary value, rendering the following energy contribution: 2
2
1 1 xy dydz xy dx xy dxdydz 1 xy dxdydz 2 2 G E
(A.21)
where it has been introduced that G E 2 1 , see Elaboration 3.5. Thus, the strain energy due to plate bending and shear is given by: U
1 2E
Lx L y t 2
0 0 t 2
2 x2 2y 2 x y 2 1 xy dxdydz
(A.22)
where Lx and Ly are the length of the plate in x and y directions and t is its thickness. From the elastic theory of plates (see Chapter 11, Eq. 11.9)
A BACKGROUND TO THE ENERGY METHODS
312
x Ez y 1 2 xy
1 1 0 0
2 2 w x 0 2 w x 2 1 2 w xy
0
(A.23)
where w x , y is the transverse bending displacement of the plate. Introducing this into Eq. A.22, then the general expression of strain energy due to plate bending and shear is given by: D U 2
Lx L y
0 0
where D
2 2 w 2 2 w 2 2w 2w 2w 2 1 2 2 dxdy (A.24) xy x 2 y 2 x y 2
Et 3
12 1 2
is the plate bending stiffness and is Poisson’s ratio.
Before proceeding, Maxwell-Betti’s and Castigliano’s theorems will be presented in Elaborations A.1 and A.2 below, not because they are widely in use, but because they will add to the understanding of the principle of virtual work.
Elaboration A.1: Maxwell-Betti’s theorem We have seen that according to the principle of energy conservation the total energy of a system in equilibrium is constant, it cannot be changed, i.e.
0 where
U P
and
U is the material strain energy P is the external load energy
An elastic system is subject to a set of external forces at positions xi , F xi , rendering deformations wF x , corresponding strains F and stresses F . The energy function from these forces is F . The same system is also subject to another set of external
forces at positions x j , Q x j , rendering deformations wQ x , corresponding strains
Q and stresses Q . The energy function from these forces is Q . 1. Let the system first be subject to forces F xi , and then, keeping F xi constant, let the system be subject to additional deformation wQ x . The change to the energy function of the system is then given by
313
A.2 STRAIN ENERGY IN BEAM-COLUMNS AND PLATES
F F Q dV F xi wQ xi V
and where
V
where
F E F
i
means integration over the entire volume of the system. Thus:
F E F Q dV F xi wQ xi 0 i
V
constant,
2. Let the system first be subject to forces Q x j , and then, keeping Q x j
let the system be subject to additional deformation wF x . The change to the energy function of the system is then given by
Q Q F dV Q x j wF x j
where
Q E Q
j
V
Q E Q F dV Q x j wF x j 0 j
V
Taking F Q , it is seen that
F xi wQ xi Q x j wF x j i
j
which is what is called Maxwell-Betti’s theorem. Let for instance F at position x F give deformation wF at position x j and Q at position x j give deformation wQ at
position x F , then, according to Betti’s theorem: F xF wQ xF Q xQ wF xQ .
Elaboration A.2: Castigliano’s first and second theorems A linear elastic system is subject to a force F at position x F , rendering the deformation w x , corresponding strains F and stresses F . The total energy in the system is (see Eqs. A.7 and A.10)
U P
1 F2 dV F x F w x F 2 V E
Let the system first be subject to the force F x F alone, and then, keeping F x F constant, let the system be subject to an additional incremental deformation w x . The change to the total energy in the system is then given by
A BACKGROUND TO THE ENERGY METHODS
314
U w x F F xF w xF 0 w xF
U F xF w xF
This is Castigliano’s first theorem. Similarly, let the system first be subject to the incremental force F x F alone, rendering the deformation w x . Then:
U P
1 F2 dV F x F w x F 2 V E
Keeping F x F constant, let the system be subject to an additional small deformation
w x . Then the change to the total energy in the system is given by
U F xF F x F w x F 0 F xF
U w xF F xF
This is Castigliano’s second theorem.
Fig. A.7
Cantilevered beam
Let us for instance consider the cantilevered beam shown in Fig. A.7, and include only the effects of bending (see bending moment diagram at right hand side of Fig. A.7):
M 1 U dV 2E 2 E L A I y V
x2
2
L 1 2 z dAdx z dA M 2 dx 2 EI 2 y A 0
L
2 F 2 L3 1 F L x dx 2 EI y 0 6 EI y
U FL3 w L F 3EI y
which is identical to that which was obtained in Example 5.2.
A.3 THE PRINCIPLE OF VIRTUAL WORK
315
A.3 The principle of virtual work The principle of virtual work is usually attributed to d’Alambert (1717-1783), but it was first presented in a variation format by Lagrange [20]. The basic assumption is that for an observer at rest outside of any system that is in an arbitrary position of equilibrium r the total energy level will not change for any infinitesimal change r of the position of the system. The only restriction imposed on r is that it is small and time invariant. Otherwise, r and r are both arbitrary, one of them may be virtual. H ence, it has been labelled the principle of virtual work. It is important to realise that the principle is a balance of energy conservation from an independent initial condition of the system. I.e., one is free to choose which one of r or r is virtual and which one is real, and thus, there are two alternatives:
the original position may be an equilibrium condition caused by real forces acting on the system while the infinitesimal change is virtual, this is the principle of virtual displacements (which may be used to establish equilibrium conditions), or, the original position may be caused by virtual forces acting on the system while the infinitesimal change is real, this is the principle of virtual forces (which may be used to determine displacements).
At the upper left hand side of Fig. A.7 is shown the free-body-diagram of a linelike continuous type of system. For simplicity, its displacement is in the z direction alone and with corresponding bending moment (about y axis). At arbitrary position w x with corresponding support reaction forces Rn or support moments M n , n 1,2,, N R , it is subject to external loads Fz , M y
and qz x . The lower left hand side illustration shows the same system, but now with an additional incremental displacement w x . [It should be noted that it is not a requirement that points of support forces are exempt, i.e. the displacement w x may also contain motion of support forces Rn or support moments M n ]. During the motion w x , external forces and reaction forces have had their energy level changed:
if force and motion are in the same direction then their energy level has been reduced (they have lost their ability to perform this work),
A BACKGROUND TO THE ENERGY METHODS
316
if force and motion are in opposite directions then their energy level has been increased (they have gained ability to perform work).
For simplicity, we shall in the following limit our development to the case of line-like two-dimensional systems, i.e. the displacements are vertical w x or cross sectional rotation x .
Fig. A.7 The principle of virtual work applied to a line-like continuous system
Thus, during an incremental motion from w x to w x w x or
x x the contribution from external forces to changes in the energy function is defined by
317
A.3 THE PRINCIPLE OF VIRTUAL WORK
F dw
w xF w xF z z w xF z
PF
z
w x w x
w x
qz x dxdw
y w xM y M dw xF xF M d y x xF w xM y w x M
(A.25)
where Fz is concentrated force, qz is distributed force, M y and M x are external bending and torsion moments. Thus
PF w xF Fz w xM M y xM x M x w x qz x dx
(A.26)
L
Similarly, the corresponding changes to the energy function for arbitrary support forces Rn or support moments M n is given by
PR w x Rn Rn w x M n M n n
(A.27)
Assuming that shear stresses due to shear forces and warping may be neglected, the normal stresses x and shear stresses xs (due to St Venant torsion) in the system have also had their energy level changed [where coordinate s is in the direction of the contour of the cross section (i.e. s y or z for cross section with plate elements parallel to y or z axes), see Chapter 10]. During incremental motion from w x to w x w x the strain in the system has changed from
x x to x x x x as shown on the upper right hand side illustration in Fig. A.7. Similarly, during the incremental rotation from x to x x the shear angle in the system has changed from xs x xs x . The corresponding change of the energy function of the system is then:
x x xs xs U x d x dAdx xs d xs dAdx L A L A x xs
(A.28)
x x dAdx xs xs dAdx LA
LA
The basic idea is that during the motion w and the total change of the energy function is zero, i.e. that PF PR U 0 , and thus
A BACKGROUND TO THE ENERGY METHODS
318
w x F Fz w xM M y xM x M x w x qz x dx
L
w x Rn Rn w xM n M n x x dAdx xs xs dAdx n
LA
(A.29)
LA
Since we have restricted ourselves to in-plane motion or cross sectional rotation, then (see Eq. 10.56 and Chapters 10.3 and 10.4, see also Eqs. A.11 and A.15):
x E uD wD z 1 Mx 1 1 GI t for a single cell cross section t 2 Am t 2 Am
xs
xs
Mx 2 z G 2 z It
for an open section
x uD wD z xs 2 z
xs
(A.30)
1 1 for a single cell cross section I t t 2 Am
(A.31)
(A.32)
for an open section
(A.33)
Thus U x x dAdx xs xs dAdx uD wD z E uD wD z dAdx LA
LA
LA
2 z G 2 z dAdx for an open section LA I t 1 1 I t 1 1 dAdx for a single cell cross section t 2 Am t 2 Am LA (A.34) Performing cross sectional integration, considering the properties in Eq. A.13, then U uD EAuD wD EI y wD GI t dx (A.35) L
and thus, the energy balance in Eq. A.29 becomes:
A.4 THE PRINCIPLE OF VIRTUAL DISPLACEMENTS
319
w xF Fz w xM M y xM x M x w x qz x dx L
w xRn Rn w xM n M n uD EAuD wD EI y wD GI t dx n
L
(A.36) This is the equation that ensures the energy balance of a line-like body in a condition of equilibrium. Since one is free to choose whether the incremental motion is real or virtual, Eq. A.36 is called the equation of virtual work. It may be used to establish equilibrium, as shown in Chapter A4, or to determine structural displacements, as shown in Chapter A.5.
A.4 The principle of virtual displacements Let us first show how this may be used to establish equilibrium conditions. Since we are free to choose, let the initial position u, w, be real while the incremental change u, w, is a rigid body virtual motion
u, w, u, w,
(A.37)
In a rigid body motion there are no internal strain contributions, and thus energy balance is:
x F Fz w x M M y x M x M x w x qz x dx w
w xRn Rn w xM n M n 0 n
L
(A.38)
Example A.1: Cantilevered beam Let us consider the simple example of a cantilevered beam subject to a force F as illustrated in Fig. A.8.a. The free body diagram of the system is shown in Fig. A.8.b. As can be seen, the equilibrium condition is determined by the two reaction forces R1 and R2 . As illustrated in Fig. A.8.c, let us give the entire system a rigid virtual motion
F . w w
A BACKGROUND TO THE ENERGY METHODS
320
Fig. A.8
Fig. A.9
Simple cantilevered beam
Continuous beam subject to distributed force q
A.4 THE PRINCIPLE OF VIRTUAL DISPLACEMENTS
a) Structural system
b) Structural system with member
n replaced by unknown forces Sn
F c) Virtual motion w w Fig. A.10
Three bay cantilevered truss
321
A BACKGROUND TO THE ENERGY METHODS
322
Then the energy balance is given by:
F Fw F 0 R1w
R1 F
has been used to determine the Thus, an arbitrary virtual (non-existing) motion w reaction force R1 . But, although the size of the motion is arbitrary, the clue is that its shape has been deliberately chosen to provide energy balance where R1 is the only
F L unknown quantity. This may be further illustrated by the virtual rotation w w shown in Fig. A.8.d, where it is seen that the energy balance is defined by F L Fw F 0 R2 w
R2 FL
Example A.2: Continuous beam A more demanding example is shown in Fig. A.9. The free body diagram shows that there are three unknown support forces: R A , RB and RC . To quantify RC we simply
C , and since the virtual displacement is give the support C a virtual displacement w C 1 . arbitrary we chose w x dx 0 1 RC qw
Thus:
L
from which:
1 1 L 1 qL x dx q L RC q w q L 1 2 L 2 2 4 L
Example A.3: Cantilevered truss The example of a cantilevered truss subject to the outer end force F is shown in Fig. A.10. We shall set out to determine the axial force in member n . We do this by isolating the unknown force Sn (positive tensile) in a free body type of situation as
F L of shown in Fig. A.10.b. In this position we introduce a virtual rotation w w
F and Sn is moved a distance the outer bay, such that F is moved a distance w
w F
L H . Then the energy balance is given by
L H Fw F 0 Sn w from which it is seen that
Sn FL H
A.4 THE PRINCIPLE OF VIRTUAL FORCES
323
A.5 The principle of virtual forces The principle of virtual forces is widely used for the calculation of displacements. Since we are free to choose, let the initial position of equilibrium be virtual (A.39) u, w, u, w,
including all its external forces
Fz , qz , M y , M x Fz , qz , M y , M x
(A.40)
while the incremental change is real
u, w, u, w,
(A.41)
in which case there can be no incremental move of supports (because it is real), and thus, Eq. A. 36 becomes
x z x dx w xF Fz w xM M y Mx M x w xq L
wD EI y w D GI t dx uD EAuD
(A.42)
L
Since we are now only interested in determining one of the displacements w at position xr x F or w at position xr xM y or at position xr x M x
(A.43)
alone, or M let us for convenience choose qz 0 and apply any of Fz , M x y
then Eq. A.42 may be written Fz w xr w x u EAu w M y r D D D EI y wD GI t dx x L M x r
Acknowledging that (see Eqs. 3.16, 3.25 and 10.28)
(A.44)
A BACKGROUND TO THE ENERGY METHODS
324
x M y EI y M y x wD EI y
N x uD EA N x uD EA
D w
x M x GI t M x x GI t
(A.45)
where
x are the normal forces, bending moments and x and M N x , M x y
torsion moment in the system due to the virtual force, and
N x , M z x , M y x and M x x are normal forces, bending moments and torsion moment due to real external forces on the system. Thus Fz w xr M M w x NN y y M x M x M y r EA EI y GI t x L M x r
dx
(A.46)
As mentioned above, this procedure may be used to determine the displacement w , slope w or cross sectional rotation at any position xr in a system by 1 or a applying either a virtual force Fz 1 , a virtual bending moment M y virtual torsion moment M 1 to the system. The only requirement to the x
virtual forces is that they are applied at xr and in the direction of the displacement or rotations we wish to determine. For instance, if we want to determine the displacement w xr , then we apply a virtual force Fz 1 at xr and in the direction of w , and then: M NN M y y w xr EA EI y L
dx
(A.47)
Or, for instance, if we want to determine the cross sectional rotation xr , 1 at x , and then: then we apply a virtual torsion moment M x r M M x x dx GI t L
xr
(A.48)
A.4 THE PRINCIPLE OF VIRTUAL FORCES
325
Example A.3: Simply supported beam
Fig. A.11 Simple example on the use of the principle of virtual forces This may be further illuminated by considering the example illustrated in Fig. A.11, where a simply supported beam is subject to a concentrated force F at midspan, and where the problem is to calculate the vertical displacement at the position of the load (for simplicity, we shall assume that the cross section is symmetric about y and z axes). To do this we first apply a virtual load F at the position of and in the direction of the unknown displacement w x L 2 . Most conveniently we choose F 1 . In the presence of this load the real structural displacements w w x due to external load
F is applied to the system, without F itself. In this case the only stress-strain effect is bending. Then Eq. A.47 becomes: L
w xr L 2 0
x M y x M y EI y
dx
A BACKGROUND TO THE ENERGY METHODS
326
x and M x are the bending moments due to F and F , respectively. where M y y It is readily seen that
F L 1 L x at 0 x M y x at 0 x 2 2 2 2 and F L 1 L L x at x L M M y L x at x L y 2 2 2 2 My
Then the displacement at mid-span is given by:
L w x 2
L2
0
1 F EI y 2
L
1 1 x x dx EI y 2 L2
F 1 2 L x 2 L x dx
Thus, assuming for simplicity that EI y is constant, the following is obtained:
L w xr 2
L2
0
L F L x Fx 2 FL3 dx dx 4 EI y 4 EI y 48EI y L2 2
A.6 Rayleigh-Ritz method As shown above, at any displaced position of equilibrium the total energy U P of a structural system is constant, and thus the variation of with respect to its deformation r is equal to zero, i.e. r 0 . Obviously, if the same variation is performed with respect to displacements r that are not rendering the system in perfect equilibrium, then r may not necessarily be identical to zero. The Rayleigh-Ritz method is based on the assumption that if r is close to the exact shape of r , then minimizing r with respect to the inferior solution r by setting
r 0
(A.49)
will render a solution where the variation of is as close to zero as one can get within the limits of the chosen solution r . It may be used in a continuous as well as on a discrete system. The great advantage is that sound engineering choice of displacement functions may render a solution that is accurate enough
327
A.6 RAYLEIGH-RITZ METHOD
to problems which otherwise would be demanding to obtain. The total energy function for a line-like system is then given by (see Eqs A.2, A.3 and A.16):
2 2 1 2 2 2 EA u EI z v EI y w GI t EI w dx 2 L Fyi v xi Fzi w xi M j wj q y v x dx qz w x dx
i
i
j
L
(A.50)
L
while the total energy function for a rectangular plate is given by (see Eqs A7, A.18 and A.24): 2 ab 2w D 2 w 2 w 2w 2w 2 2 2 2 2 1 dxdy 2 0 0 x y xy x y 2 Fi w xi i
Lx L y
(A.51)
qz w x, y dxdy 0 0
Let us adopt a series solution to the relevant displacement functions, e.g. in the case of a line-like system, let
T u x v x w x x auk k
avk
awk
T
ak k x
(A.52)
or, in the case of a rectangular plate, let w x, y awk k x, y
(A.53)
k
where k are chosen (known) functions, based on good engineering judgement of what can be expected of system behaviour. It is a point in itself that they are approximations, but it is hugely advantageous that they satisfy the boundary (support) conditions. Polynomials or harmonic functions are often a convenient choice. The total energy function will then be a function of the coefficients in the series solution, i.e.:
auk , avk , awk , ak for a line-like system for a plate awk
(A.54)
328
A BACKGROUND TO THE ENERGY METHODS
As mentioned above, the idea is that an approximate solution is obtained if is minimised in the space of its relevant variables
a avk uk =0 awk
awk
ak
T 0 for a line-like system for a plate
(A.55)
from which a set of linear equations with zero on the right hand side will be obtained: a uk av kp k 0 (A.56) awk ak and thus, the solution has been turned into an eigen-value problem where the final requirement is defined by a zero determinant of the coefficient matrix.
A.7 Galerkin’s principle of weighted residuals Galerkin’s method is based on the equilibrium requirement in the form of one or several differential equations (or an interconnected group of differential equations)
f r1,, rj ,rJ 0
(A.57)
where rj are the relevant displacement components, for instance qx 0 EAuD EI z vD q y 0 f uD , vD , wD , EI y wD qz 0 EI GI q 0 t for a line-like system, and
(A.58)
329
A.7 GALERKIN’S PRINCIPLE OF WEIGHTED RESIDUALS
4 w 4w 4w 2 w 2w 2w f w D 4 2 2 2 4 N x 2 N y 2 2 N xy qz (A.59) xy x y y x y x for rectangular plate, or
f w
1 1 w qz r r r r r r r r D
(A.60)
for a circular plate. Again, the idea is that these equations are converted into an eigen-value problem (A.61) Aa 0 T
with unknown coefficients a a1 ak aN , by imposing an approximate solution comprising a linear combination of N unknown coefficients ak and a corresponding set of known shape functions k x : N
rj ak k x
(A.62)
i 1
and then applying the method of weighted residuals in its functional space, i.e. in
its length L or across its surface Lx , Ly . Thus, the approximate solution N f ak k 0 i 1
(A.63)
is successively weighed with the same functions k x [or k x, y for a plate], k 1,2,......, N , and integrated over L [or
Lx , Ly
for a plate],
rendering a set of N equations, which fully written is given by:
A11 A k1 AN 1
A1 j
Akj
AN j
A1N a 1 AkN ak 0 AN N a N
(A.64)
A BACKGROUND TO THE ENERGY METHODS
330
where
Akj k f j dx for a line-like system L L y Lx A kj k f j dxdy for a plate 0 0
(A.65)
Again, this is an eigen-value problem whose solution is defined by
A11 Det Ak 1 AN 1
A1 j
Akj
AN j
A1N AkN 0 AN N
(A.66)
This is a general method which may offer an approximate solution to particularly complex systems. It is a requirement to the accuracy of the solution that k x or k x, y fulfils (more or less) the geometric boundary (support) conditions of the system. It is advantageous that they are as close to orthogonal as possible (as this will render a predominantly diagonal structure in the coefficient matrix).
Appendix B Equation Chapter 2 Section 1
SOLUTION TO SYSTEMS OF LINEAR EQUATIONS B.1
Matrix inversion by Gauss-Jordan elimination
Given a square N by N matrix A . The first step is to establish the expanded matrix G A I , where I is the N by N unity matrix. The basic idea is that a successive reduction of row entries aij in G by linear operations, one row after another, will render G I A 1 . The method is best illustrated by an example. Given
Fist we establish
12 6 6 A 6 8 1 6 1 4
(B.1)
12 6 6 1 0 0 G 6 8 1 0 1 0 6 1 4 0 0 1
(B.2)
Then we start our linear reductions with focusing on row one by dividing it by its first coefficient a11 12 to create a new matrix 1 0.5 0.5 0.0833 0 0 8 0 1 0 1 G 6 4 0 0 1 6 1
(B.3)
after which we take row two minus row one multiplied by the first coefficient of row two a 21 6 , and row three minus row one multiplied by the first coefficient of row three a31 6 to create a new matrix
© Springer Nature Switzerland AG 2020 E. N. Strømmen, Structural Mechanics, https://doi.org/10.1007/978-3-030-44318-4
331
332
B SOLUTION TO SYSTEMS OF LINEAR EQUATIONS
1 0.5 0.5 0.0833 0 0 5 2 0.5 1 0 G 0 2 1 0.5 0 1 0
(B.4)
Then we focus on row two by dividing it by its second coefficient a22 5 to create a new matrix 1 0.5 0.5 0.0833 0 0 G 0 1 0.4 0.1 0.2 0 2 1 0 1 0.5 0
(B.5)
after which we take row one minus row two multiplied by the second coefficient of row one a12 0.5 , and row three minus row two multiplied by the second coefficient of row three a31 2 to create a new matrix 1 0 0.7 0.1333 0.1 0 G 0 1 0.4 0.1 0.2 0 0 0 0.2 0.7 0.4 1
(B.6)
Finally we focus on row three by dividing it by its third coefficient a 33 0.2 to create a new matrix 1 0 0.7 0.1333 0.1 0 G 0 1 0.4 0.1 0.2 0 3.5 2 5 0 0 1
(B.7)
after which we the take row one minus row three multiplied by the third coefficient of row one a13 0.7 ), and row two minus row three multiplied by the third coefficient of row two a 23 0.4 to create a final matrix 1 0 0 2.5833 1.5 3.5 1.5 1 G 0 1 0 2 I A 1 5 0 0 1 3.5 2
Thus:
A
1
2.5833 1.5 3.5 1 1.5 2 5 3.5 2
(B.8)
(B.9)
B.2 SOLUTION BY CHOLESKY DECOMPOSITION
B.2
333
Solution by Cholesky decomposition
There are two versions of the Cholesky decomposition [12]. First, the original definition is presented. Given a positive definite and symmetric matrix X , the Cholesky decomposition of X is defined by a lower triangular matrix Y of the same size that satisfies the following:
X YYT
(B.10)
Expanding this equation
x11 x1i x1N y11 xi1 xii xiN yi1 x N 1 x Ni x NN y N 1
0 y11 0 0 y NN 0
yN1 yii yii y Ni y Ni 0 y NN (B.11) and developing the matrix multiplication column by column, will render first column entries y11 x11 x11 y11 y11 x y y 21 y21 x21 / y11 21 11 (B.12) x N 1 y N 1 y11 y N 1 x N 1 / y11 0
yi1
while from second column x22 y21 y21 y22 y22 x y y y y 32 31 21 32 22 x N 2 y N 1 y21 y N 2 y22
y22 x22 y21 y21 y32 x32 y31 y21 / y22 y N 2 x N 2 y N 1 y21 / y22
(B.13)
and so on. This can be summarized as follows:
y11 x11
1/2
1/2
i 1 yii xii yik2 k 1
yij
for i 2,....., N 1
j 1 1 x ij yik ykj y jj k 1
for all i j
(B.14)
334
B SOLUTION TO SYSTEMS OF LINEAR EQUATIONS
For instance 0 0 3.464 1.732 1.732 12 6 6 3.464 X 6 8 1 1.732 2.236 0 0 2.236 0.894 (B.15) 6 1 4 1.732 0.894 0.447 0 0 0.447 The following is the alternative definition (which eliminates the necessity of taking the square root). Given a positive definite and symmetric matrix X , the alternative Cholesky decomposition of X is defined by a lower triangular matrix Z of the same size with unit entries on its diagonal, and a diagonal matrix D , also of the same size, that satisfies the following: X ZDZ T
(B.16)
i.e. X 1 zi1 z N 1
0 1
z Ni
0 d11 0 0 1 0
0 d ii
0
0 1 0 0 d NN 0
zi1 z N 1 (B.17) z Ni 1 0 1
Performing the matrix multiplications column by column similar to that of the classical case above, it is readily seen that i 1
d ii xii zik2 d kk for i 1,....., N k 1
j 1 1 zij x ij zik zkj d kk for all i j d jj k 1
(B.18)
For instance 0 0 12 0 0 1 0.5 0.5 12 6 6 1 X 6 8 1 0.5 1 0 0 5 0 0 1 0.4 (B.19) 0 1 6 1 4 0.5 0.4 1 0 0 0.2 0 The connection between the two is easily seen by expanding X ZDZ T into
X ZDZT ZD1 2D1 2ZT ZD1 2 ZD1 2
T
Y ZD1 2 Z YD1 2 (B.20)
Appendix C COLLECTION OF IMPORTANT FORMULAS A collection of structural engineering formulas derived in this book are presented below, as well as the more advanced mathematical formulas that have been employed in their development.
C.1 Definitions and formulas in structural mechanics Basic units (as defined by ISO) are as follows:
displacement: meter (m ) time: second ( s ) mass: kilogram ( kg )
force:
Newton ( N kgm s 2 ).
Fx 0 The equilibrium requirements: Fz 0 where P is an arbitrary point. M P 0 The superposition principle:
the force effects due to external forces F1 and F2 together is equal to the force effects of F1 alone plus the force effect of F2 alone,
the structural displacements due to external forces F1 and F2 together is equal to the deformations due to F1 alone plus the deformations due to F2 alone.
© Springer Nature Switzerland AG 2020 E. N. Strømmen, Structural Mechanics, https://doi.org/10.1007/978-3-030-44318-4
335
336
C. COLLECTION OF IMPORTANT FORMULAS
Fig. C.1
Fig. C.2
Stress sign conventions
Axis system and sign conventions
337
C.1 DEFINITIONS AND FORMULAS IN STRUCTURAL MECHANICS
Cross sectional properties:
y 0 Main axes and shear center position:: z dA 0 and A yz 0
1 0 y dA 0 A z 0
s
rds where r is distance from shear center, with coordinates e y , ez 0
1 A 2 I z y dA 2 Iz A y 2 I
i 2p
I y Iz A
As S y Sz S
i 2 i 2p e2y ez2
im
1 2 2 I z y z y iy 1 2 2 iz I y y z dA i A z 1 2 2 y z I
1 z y dA As
1 z 3dA 2ez I y A
ik I y A
x E 1 x In two dimensions: 2 y 1 1 y Three-dimensional version of Hooke’s law :
E x Hooke’s law: x xz G xz
x 1 E y 1 1 2 z xy 1 0 0 xy yz G 0 1 0 yz 0 0 1 zx zx
x y 1 1 z
xy yx yz zy zx xz
G
E 2 1
Normal stresses and shear stresses:
x
N My M B z z y A Iy Iz I
Vy 1 q V B and xs x As z S y S z S t A Iy Iz I
Decomposition of stress components, principal stresses :
C. COLLECTION OF IMPORTANT FORMULAS
338
x z 2
x z
x z
2
2
cos 2 xz sin 2
sin 2 xz cos 2
2 max x z x z 2 xz min 2 2
for
2 max x z 2 xz min 2
cotan 2
for
tan 2
2 xz x z
2 xz x z
Decomposition of strain components, principal strains:
x z
2
2
x z 2
x z 2
cos 2
sin 2
2 2 max x z x z xz min 2 2 2 2 2 2 max x z xz 2 min 2 2
xz 2
xz 2
cos 2
for for
sin 2
tan 2
cotan 2
xz x z
2 xz x z
Tresca yield criteria:
f y 2 or alternatively: max min f y
Von Mises yield criterion:
2 j x2 z2 x z 3 xz fy
Axial deformation:
u x
N EA
Bending of beam with cross section symmetric about z axis ( wD w0 w ): M y EI y wD q x4 x3 x2 M y Vz w z C1 C2 C3 x C4 EI y 24 6 2 Vz qz
C.1 DEFINITIONS AND FORMULAS IN STRUCTURAL MECHANICS
339
The differential equation of beam bending and twisting:
EAuD qx 0 q y 0 EI z vD
EI y wD qz 0 EI GI t q 0
M x GI t EI M y EI y wD M z EI z vD
and
where v D and wD are shear centre displacements Virtual unit force method:
v xr M M M NN M M B B M y y w xr z z x x dx EA EI z EI y GI t EI xr L
N , M y , M z , M x and B are normal force, bending moments, torsion
moment and bi-moment from external loads acting on the system, , M and B are the normal force, bending moments, , M N , M z x y torsion moment and bi-moment from a virtual unit load F 1 at x r and
in the direction of the displacement v xr or w xr , or from a virtual 1 about y , z or x axes at x and in the unit load moment M r
direction of xr .
Normal stresses and shear stresses in composite beams:
x N z
E z E z My z EA EI y
Vz x ES y b EI y
where
EA 1 E z 2 dA EI y A z
where
ES y E z zdA
E z zdA 0
As
A
Normal stresses and shear stresses in non-symmetric beams:
x
I z M y I y1z1 M z1 N I y1 M z1 I y1z1 M y1 y1 1 1 z1 2 A I y1 I z1 I y1z1 I y1 I z1 I y21z1
where
y1
0
z1 dA 0 A
C. COLLECTION OF IMPORTANT FORMULAS
340
A 1 I 2 z1 y1 I y 2 dA 1 A z1 I y1z1 y1z1
M y cos and 1 M z1 sin
I z Vz I y1z1V y1 1 I y1V y1 I y1z1Vz1 S z1 1 1 S y1 2 2 b I y Iz I y z I y1 I z1 I y1z1 1 1 11
sin M y cos M z
S y1 S z1
z z1
y cos Property transformations in non-symmetric beams: z sin
I z y dA I z1 cos I y1z1 sin 2 I y1 sin A I y z 2 dA I z1 sin 2 I y1z1 sin 2 I y1 cos2 A 1 I yz yzdA I z1 I y1 sin 2 I y1z1 cos 2 2 A 2
2
H
qz L2 8f
4 f 2 1 3 L
As
sin y1 cos z1
2
The shallow cable: zc
z1
y1 dA
tan 2
2 I y1z1
I z1 I y1
H qz
x qz x qz L qz L 4 f cosh H 2 H cosh 2 H L x 1 L
Vz x 0
2 1 q L L 1 z 24 H
Beam on elastic foundation: w
Et q Cylindrical shell: w 2 w D Dr
2 qz L 8 f 1 2 3 L
H 2 Vz2
2 HL 16 f 1 EA 3 L
k q w z EI y EI y
D
N
w w p wh
Et 3
12 1 2
w w p wh N E t r w M x Dw
341
C.1 DEFINITIONS AND FORMULAS IN STRUCTURAL MECHANICS
St. Venant torsion, compact cross section: M x GI t
Single cell cross section: M x GI t , I t
4 Am2
ds t
1 3 I t 3 bt i i 2 M x z i xyi It
, q xst
Mx 2 Am
Rm
Warping torsion: x
B I
q t xs
B S I
B EI
Differential equation of torsion: q EI GI t or M x GI t EI Bending of rectangular plate: 4w 4w 4w 2w 2w 2w qz D 4 2 2 2 4 N x 2 N y 2 2 N xy xy x y y x y x
x Ez y 1 2 xy
1 1 0 0
2 2 0 w x 0 2 w y 2 1 2 w xy
Mx z y 1 M h2 1 y z x z yx dx D 0 0 M xy h 2 0 M yx z xy 0
Bending of circular plate:
2 EI y 2
L
, N Ez
Eh3
12 1 2
2 w x 2 2 w y 2 1 2 1 w xy 0 0
1 1 w qz r r r r r r r r D
Elastic buckling of column: N E z N NEy
D
2 EI z 2
L
,
N E y N N E N 0
N E
GI t 2 EI w L2 and i 2p
where
i 2p
I y Iz A
C. COLLECTION OF IMPORTANT FORMULAS
342
Elastic buckling of rectangular plate: 1) by the differential equation (or by Galerkin’s principle of weighted residuals): 4w 4w 4w 2w 2w 2w D 4 2 2 2 4 N x 2 N y 2 2 N xy 0 xy x y y x y x 2) or by Rayleigh-Ritz:
D 2 1 2
Lx L y
0 0
Lx L y
0 0
N y 0 N xy 0
2 2 w 2 2 w 2 2w 2w 2w 2 1 2 2 dxdy x y x 2 y 2 x y 2 2 w 2 w w w Nx dxdy 2 N xy N y x y x y
N xE k x
2D L2y
where
m kx m
2
and = Lx Ly
Lateral torsional buckling of simply supported beam: M 1 E 2 2 EI w EI z GI t FzE L L L2 2 qzE 2 L The differential equation of beam-column:
EI y w M z N w e y qz 0 EI z v M y N v ez q y 0
EI GI t M y v i y 2ez
M z w iz 2e y
N ez v e y w i 2 Bi q 0
I y Iz 2 i A 1 I z i y y i 1 y z A I z i 1 I
e 2y ez2
y
z2 2 2 y z dA 2 2 y z 2
343
C.2 SOME USEFUL MATHEMATICAL FORMULAS
Elastic buckling of simple frame:
K AT kA K N k K n n n K K G r 0 where K and T K K Gn A k G A G n 1 Gn n
6 3 L 6 3 L 36 3L 36 3L 2 2 2 3L L2 N 3L 4 L EI y 3L 2 L 3L L , k Gn kn 2 3 30 L n 36 3L 36 3L L n 6 3 L 6 3 L 2 2 3L 2 L2 n 3L 4 L2 n 3L L 3L L ( N positive tensile) Strain energy in beams and beam-columns: U
1 2 2 2 2 2 EI y wD GI t EI w dx EA u D EI z v D 2L
where u D and wD are shear center displacements The connection between centroid (CC) and shear center (SC) displacements: ez v0 vD w w e 0 D y
where e y , ez
are shear centre coordinates
C.2 Some useful mathematical formulas a b 2 a 2 2ab b 2 Algebra: 2 2 2 a b a 2ab b
x 2 bx c 0 ax 2 bx c 0 x1 x2 c a x1 x2 b a
a b a b a 2 b2 a b 3 a 3 3a 2b 3ab2 b3
2 x1 b b c x2 2 2 2 x1 b b2 4ac b b c x2 2a 2a 2a a
C. COLLECTION OF IMPORTANT FORMULAS
344 Series:
1 x
1 x
f a b f a
sin x x ex 1
Trigonometry:
1 2!
x2
1 2 3!
x3
2
b b f a f a 1! 2!
x3 x5 x7 , 3! 5! 7!
(Taylor)
cos x 1
x2 x4 x6 2! 4! 6!
x x2 x3 1! 2! 3!
sin sin cos cos sin cos cos cos sin sin tan tan tan 1 tan tan sin 2 2sin cos cos 2 cos 2 sin 2 1 2 sin 2 2 cos 2 1
tan 2 2 tan 1 tan 2
Calculus:
d df dg f g dx dx dx d df f n nf n1 dx dx
Differentials:
d dg df f g f g dx dx dx d f df dg 2 g f g dx dx g dx
dy dy dz dx dz dx d df dz f z dx dz dx
d d f ax a f x where x ax dx dx
f x
f x
f x
f x
sin x
cos x
tan x
cos 2 x
cos x
sin x
cot x
sin 2 x
ex
ex
sinh x
cosh x
cosh x
sinh x
ln x
x
1
345
C.2 SOME USEFUL MATHEMATICAL FORMULAS
Integrals*:
1
f ax dx a f x dx
where x ax
f x
f x dx
f x
f x dx
sin x
cos x
tan x
ln cos x
cos x
sin x
cot x
ln sin x
sinh x
cosh x
cosh x
sinh x
e
x
e
ln x
x
x ln x 1
*) Integration constants omitted)
f x
f x dx
f x
f x dx
sin 2 x
x 2 sin 2 x 4
sin x cos x
sin 2 x 2
cos2 x
x 2 sin 2 x 4
sin 2 x cos 2 x
x 8 sin 4 x 4
xn
x n 1 n 1
f x
f x dx
sin mx sin nx
f x sin 2 mx cos2 mx
sin m n x
for m n
sin m n x
sin m n x
for m n
2 m n
cos mx cos nx
sin m n x
f x dx
0
2 2
2m n
2 m n
2 m n
f x
sin mx sin nx cos mx cos nx sin mx cos mx
f x dx 0
0 0 0
C. COLLECTION OF IMPORTANT FORMULAS
346
f x
f x dx
sin mx cos nx ,
2m if m n is odd, 0 if m n is even 2 m n2
0
mn
udv uv vdu
Integration by parts:
du
uvdx uw dx wdx
where w vdx
d2 f d2 f d2 f 2 In cartesian coordinates: , , f x y z dx 2 dy 2 dz 2 Laplace operator: 2 2 Polar coordinates: 2 f r, , z 1 r df 1 d f d f r dr r 2 d 2 dz 2
C.3 Basic matrix operations a11 a1k A a j1 a jk a m1 amk E.g., let
a A 11 a12 T
a11 a j1 am1 AT a1k akj amk a 1n a jn amn
a1n a jn amn
a12 a A 11 a21 a22
and
b11 b21 B b12 b22 b13 b23
a21 a22
T
b b B 11 12 b21 b22
b13 b23
then
AB T BT AT In general: T T T T ABC C B A
Let A be size m n and B size n p then AB C is size m p where: n
c jk a ji bik i 1
j 1,2,, m and k 1,2,, p
347
C.3 BASIC MATRIX OPERERATIONS
In general: AB BA , while AB C A BC ABC E.g.: a b b a AB 11 12 11 12 a21 a22 b21 b22
b13 c11 c12 b23 c21 c22
c13 c23
c a11b11 a12b21 c12 a11b12 a12b22 c13 a11b13 a12b23 11 c21 a21b11 a22b21 c22 a21b12 a22b22 c23 a21b13 a22b23
Matrix multiplication by Falk’s scheme: AB C
a a12 A 11 a21 a22
b b12 B 11 b21 b22
c C 11 c21
c12 c22
b13 b23
c13 c23
A B D where d jk a jk b jk For equal size matrices A and B : A B C AC BC C A B CA CB
For square matrices: 1 B AB BB 1A 1 A 1 AB 1 B1A1, det AB det A det B
AA 1 I ,
a11 a1k a1m The determinant of square matrices: A a j1 a jk a jm a m1 amk amm m
det A mm 1
k 1
jk
a jk det A jk with j fixed between 1 and m (e.g. j 1)
and det A jk is the determinant to A reduced by line j and column k
C. COLLECTION OF IMPORTANT FORMULAS
348 E.g.:
a12 a A 11 a21 a22
a11 a12 A a21 a22 a31 a32
det A a11a22 a12 a21
a13 a23 a33
det A a22 a32
111 a11 det
a23 a a a a 1 2 1 3 1 a12 det 21 23 1 a13 det 21 22 a33 a31 a33 a31 a32
a11 a22 a33 a23a32 a12 a21a33 a23a31 a13 a21a32 a22 a31 For square and symmetric matrices: AT A a a 1 a22 a12 The inverse of 2 by 2 matrix A 11 12 A 1 det A a21 a11 a21 a22
a11 The inverse of a symmetric 3 by 3 matrix: A a12 a13
a C11 det 22 a23
a23 a33
a C12 det 12 a13
a a C22 det 11 13 a13 a33
a12 a22 a23
a23 a33
a C13 det 12 a13
a22 a23
a a C23 det 11 12 a13 a23
a C33 det 11 a12
a12 a22
det A a11C11 a12C12 a13C13
A
1
a13 a23 a33
C11 1 C12 det A C13
C12 C22 C23
C13 C23 C33
References [1]
Timoshenko, S.P. & Goodier, J.N., Theory of elasticity, McGraw-H ill, 1951.
[2]
Newton, Isaac, Philosophiae Naturalis Principa Mathmatica, July 5 1687, first published on July 5, 1687, Universtity of Cambridge, UK.
[3]
Hooke, Robert, De Potentia Restitutiva, or of Spring, Explaining the Power of Springong Bodies, Londo, 1678.
[4]
Leibniz, Gottfried Wilhelm, The Early Mathematical Manuscripts of Leibniz, Cosimo Inc., 2008. Page 228.
[5]
Navier, Claude-Louis, École Nationale des Ponts et Chaussées, Paris, Navier’s hypothesis in theory of elasticity first published in 1821.
[6]
Mohr, Christian Otto, Dresden Polytechnic, 1882.
[7]
Wheatstone, Charles, An Account of Several New Processes for Determining the Constants of a Voltaic Circuit, The Royal Society, 1843
[8]
Tresca, Henri, Mémoire sur l'écoulement des corps solides soumis à de fortes pressions. C. R. Acad. Sci. Paris, vol. 59, p. 754, 1864.
[9]
von Mises, Richard, Mechanik der Festen Korper im plastisch deformablen Zustand. Göttingen, Nachr. Math. Phys., vol. 1, pp. 582–592, 1913.
[10] Gauss, Carl Friedrich, Disquisitiones Arithmeticae, 1801. English translation by Arthur A. Clarke, Disquisitiones Arithmeticae (Second, corrected edition). New York: Springer. 1986. ISBN 0-387-96254-9. [11] Jordan, Camille, Traité des substitutions et des équations algébriques, Paris: Gauthier-Villars, 1870. [12] Cholesky, André-Louis, Posthumously published by Commandant Benoit, Note sur une méthode de résolution des équations normales provenant de l'application de la méthode des moindres carrés à un système d'équations linéaires en nombre inférieur à celui des inconnues (Procédé du Commandant Cholesky), Bulletin Géodésique 2 (1924), 6777.
© Springer Nature Switzerland AG 2020 E. N. Strømmen, Structural Mechanics, https://doi.org/10.1007/978-3-030-44318-4
349
350
REFERENCES
[13] Love, A.E.H., A treatise on the mathematical theory of elasticity, Cambridge University Press, Vol. 1 (1906). First published in 1892. [14] Timoshenko, S.P. & Woinowsky-Krieger, S., Theory of plates and shells, McGraw-Hill, 2nd ed., 1959. [15] Timoshenko, S.P. & Gere, G.M., Theory of elastic stability, McGrawHill, 1961. [16] Vlasov, V.S., Thin-walled elastic beams, Israel Program for Scientific Translations, 1961. [17] Kollbrunner, C.F. & Basler, K., Torsion, Springer Verlag, 1966. [18] Strømmen, Einar N., Structural Dynamics, Springer Verlag, 2014. [19] Bell, K., Eigensolvers for structural problems, Delft Univ. Press, 1998. [20] Lagrange, Joseph-Louis, (1736-1813). Mecanique Analytique, Courcier, Paris 1811. (Reissued by Cambridge University Press, 2009, ISBN 9781108001748.) [21] MATLAB, Mathematical and computing software for engineers and scientists, The MathWorks Inc., Natick, Massachusetts, USA. [22] ISO, International Standards Organisation. Geneva, Switzerland.
Index of beam-columns 270, 276 of columns 249, 253 of frames 278 of plates 285 length 251 stress 255
A Amplification effect 252 Angle section 162, 166 Approximate calculations by Galerkin’s method 292, 328 by Rayleigh-Ritz 290, 326 Arch circular 171 parabolic 176 B Beam bending 98, 100 cantilevered 37, 99 composite 149 differential equation 98 element 135, 139 non-symmetric 157 on elastic foundation 184, 186 simply supported 32, 38, 99, 105 Beam-columns differential equation 226, 227 Bending 54, 56, 58 bi-axial 60, 156 moment 36, 37, 39, 41 of beams 54, 56 of plates 231 Bi-axial bending 60, 156 Bi-moment 218 Boundary conditions 16 Buckling 87 inelastic 254 of beams 262
© Springer Nature Switzerland AG 2020 E. N. Strømmen, Structural Mechanics, https://doi.org/10.1007/978-3-030-44318-4
C Cable geometry 178, 182 non-symmetric 179 theory 175, 177, 181 Cantilevered beam 37, 99 beam buckling 266 torsion 229 Castigliano’s theorems 315 Channel section 222 Cholesky decomposition 144, 335 Circular arch 171 tube 52, 71 Circular plates 244, 247 differential equation 246 Columns buckling 249, 253 Connectivity matrix 143 Conservative force 307 Continuous beam 108, 322 Coordinate transformations 141 Critical load 252 stress 253 Cross sectional forces 35 Cylindrical shell 188, 191
351
352 D Degree of freedom 119, 132, 135 Design criteria 87 Determinate system 27 Differential equation of beams 59 of beam-columns 226, 227 of plates 237, 244 Displacement calculations 101, 102, 103 cross sectional 35 of contour 216 shear 216 vector 120, 134 Deformations axial 97 bending 98 E Eigenmodes orthogonality 301 Eigenvalue problem general 299 special 299 Elastic foundation 184, 186 modulus 46, 86 Element displacements 133 forces 133 Energy conservation 307 methods 305 Equilibrium 5, 8, 20, 30, 41
INDEX
Finite element method 132 Formulas 337 Force 1, 2 axial 54, 55, 150 collinear 3, 4 cross sectional 35, 36 vector 4 Fracture 46 Frame 33, 118, 122, 124, 126, 146 buckling 278 Free body diagram 8, 37 Friction 12 coefficient 12 G Galerkin’s method 293, 330 Gantry 30, 113 Gauge factor 86 Gauss-Jordan 130, 144, 333 Geometric stiffness 137 Gravity loads 95 H Hollow cross section single cell 205 multiple cell 208 Hooke’s law 46, 49, 50 I Initial imperfection 251 I-section 69 Inelastic buckling 254 Indeterminate system 28, 111
F
L
Failure 87 Fatigue 87
Laplace operator 202, 240, 348 Lateral torsional buckling 262, 268
INDEX
Linear equations 130 solution 333 Load vector 120, 134 L-section 162 M Main axes 59, 157, 161 Mass centre 2 Material properties 87, 88, 89 Matrix inversion 333, 348 method 118 Maxwell-Betti’s theorem 314 Mohr’s circle of stresses 79 of strains 84 Modulus of elasticity 46, 86 Moment amplification 254 bending 35, 56 equilibrium 7, 11 torsion 35, 193
353 buckling 285 circular 242 differential equation 237, 239 rectangular 240, 243 stiffness 234 Poisson’s ratio 47, 86 Polar coordinates 244, 246 Prandtl stress function 202, 203 Principal strain 80, 85 stresses 75 Principle of superposition 17 Probability of failure 94 R Rayleigh-Ritz 256, 265, 290, 328 Rayleigh quotient 301 Rectangular cross section 60, 66, 154 plate 238 Rotation cross sectional 193
N
S
Navier’s hypothesis 57 solution 241 Newton 1, 2 Nodal load 119 Normal force 36, 37, 43 stress 55, 59 stress failure 87, 91
Safety coefficient 93, 94, 95 Sector coordinate 216, 226 Shallow cable theory 175 Shape function 136 Shear buckling 296 centre 216 displacement 216 flow 207, 208, 209 force 36, 37, 39, 41, 63 modulus 47, 86 stress 47, 64, 66, 152, 153, 164, 165, 166, 220 stress failure 87, 90
P Parabolic arch 176 Plate bending 235
354 strain 46, 80 Shells cylindrical 186 Slenderness parameter 255 Snow density 96 Standard integrations 104 Steiner’s theorem 61 Stiffness matrix 120 Strain 45, 55 components 83, 84 elastic 46 energy 308, 310 gauge 85 in beam-columns 312 in-elastic 46 in plates 314 Stress axial 54, 56 bending 54, 59 components 74 in beams 60, 61 function 200 in plates 234 normal 45, 54, 55, 57, 59, 150, 158, 161, 218, 226 principal 75 thermal 54 Structural system 13, 27 Superposition principle 17, 115, 119 Support conditions 16 T Thermal variation 52 Torsion buckling 257 constant 196, 197 moment 36, 37, 195, 197 of tube 197 of circular section 197
INDEX
of rectangular section 199, 203 St. Venant 196 stress function 200 theory 195 warping 196, 212 Tresca yield criterion 89, 90, 92 Truss 41, 106, 111 element 141 T-section 61, 67 Twisting 193 U Units 2 V Vector iterations 302 Virtual displacement 136, 321 force 101, 325 unit force method 103, 341 work 317, 321 von Mises yield criterion 89, 91, 92 W Warping constant 213, 218 normal stress 213 Wheatstone bridge 86 Wind load 95 Y Yield stress 47