Beams3D

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Elastic Beams in Three Dimensions

Lars Andersen and Sren R.K. Nielsen

ISSN 1901-7286

DCE Lecture Notes No. 23 Department of Civil Engineering

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Aalborg UniversityDepartment of Civil Engineering

Structural Mechanics

DCE Lecture Notes No. 23


by

Lars Andersen and Sren R.K. Nielsen

August 2008

c Aalborg University

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Scientific Publications at the Department of Civil Engineering

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Published 2008 by

Aalborg University

Department of Civil Engineering

Sohngaardsholmsvej 57,

DK-9000 Aalborg, Denmark

Printed in Denmark at Aalborg University

ISSN 1901-7286 DCE Lecture Notes No. 23

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Preface

This textbook has been written for the course Statics IV on spatial elastic beam structures

given at the 5th semester of the undergraduate programme in Civil Engineering at Aalborg Uni-versity. The book provides a theoretical basis for the understanding of the structural behaviour

of beams in three-dimensional structures. In the course, the text is supplemented with labora-

tory work and hands-on exercises in commercial structural finite-element programs as well as

MATLAB. The course presumes basic knowledge of ordinary differential equations and struc-

tural mechanics. A prior knowledge about plane frame structures is an advantage though not

mandatory. The authors would like to thank Mrs. Solveig Hesselvang for typing the manuscript.

Aalborg, August 2008 Lars Andersen and Sren R.K. Nielsen

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Contents

1 Beams in three dimensions 1

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Equations of equilibrium for spatial beams . . . . . . . . . . . . . . . . . . . . . 1

1.2.1 Section forces and stresses in a beam . . . . . . . . . . . . . . . . . . . 3

1.2.2 Kinematics and deformations of a beam . . . . . . . . . . . . . . . . . . 5

1.2.3 Constitutive relations for an elastic beam . . . . . . . . . . . . . . . . . 10

1.3 Differential equations of equilibrium for beams . . . . . . . . . . . . . . . . . . 12

1.3.1 Governing equations for a Timoshenko beam . . . . . . . . . . . . . . . 13

1.3.2 Governing equations for a Bernoulli-Euler beam . . . . . . . . . . . . . 14

1.4 Uncoupling of axial and bending deformations . . . . . . . . . . . . . . . . . . . 15

1.4.1 Determination of the bending centre . . . . . . . . . . . . . . . . . . . . 15

1.4.2 Determination of the principal axes . . . . . . . . . . . . . . . . . . . . 21

1.4.3 Equations of equilibrium in principal axes coordinates . . . . . . . . . . 25

1.5 Normal stresses in beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

1.6 The principle of virtual forces . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

1.7 Elastic beam elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

1.7.1 A plane Timoshenko beam element . . . . . . . . . . . . . . . . . . . . 34

1.7.2 A three-dimensional Timoshenko beam element . . . . . . . . . . . . . . 41

1.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

2 Shear stresses in beams due to torsion and bending 45

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452.2 Homogeneous torsion (St. Venant torsion) . . . . . . . . . . . . . . . . . . . . . 46

2.2.1 Basic assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

2.2.2 Solution of the homogeneous torsion problem . . . . . . . . . . . . . . . 48

2.2.3 Homogeneous torsion of open thin-walled cross-sections . . . . . . . . . 57

2.2.4 Homogeneous torsion of closed thin-walled cross-sections . . . . . . . . 59

2.3 Shear stresses from bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

2.3.1 Shear stresses in open thin-walled cross-sections . . . . . . . . . . . . . 69

2.3.2 Determination of the shear centre . . . . . . . . . . . . . . . . . . . . . 75

2.3.3 Shear stresses in closed thin-walled sections . . . . . . . . . . . . . . . . 82

2.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

References 93

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iv Contents


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CHAPTER 1

Beams in three dimensions

This chapter gives an introduction is given to elastic beams in three dimensions. Firstly, the

equations of equilibrium are presented and then the classical beam theories based on Bernoulli-Euler and Timoshenko beam kinematics are derived. The focus of the chapter is the flexural de-

formations of three-dimensional beams and their coupling with axial deformations. Only a short

introduction is given to torsional deformations, or twist, of beams in three dimensions. A full de-

scription of torsion and shear stresses is given in the next chapters. At the end of this chapter, a

stiffness matrix is formulated for a three-dimensional Timosheko beam element. This element can

be used for finite-element analysis of elastic spatial frame structures.

1.1 Introduction

In what follows, the theory of three-dimensional beams is outlined.

1.2 Equations of equilibrium for spatial beams

An initially straight beam is considered. When the beam is free of external loads, the beam

occupies a so-called referential state. In the referential state the beam is cylindrical with the

length l, i.e. the cross-sections are everywhere identical. The displacement and rotation of thebeam is described in a referential (x,y,z)-coordinate system with base unit vectors {i,j,k}, theorigin O placed on the left end-section, and the x-axis parallel with the cylinder and orientatedinto the beam, see Fig. 11. For the time being, the position of O and the orientation of the y-and z-axes may be chosen freely.

The beam is loaded by a distributed load per unit length of the referential scale defined by

the vector field q = q(x) and a distributed moment load vector per unit length m = m(x). Adifferential beam element of the length dx is then loaded by the external force vector qdx andexternal moment vector mdx as shown in Fig. 11. The length of the differential beam elementmay change during deformations due to axial strains. However, this does not affect the indicated

load vectors which have been defined per unit length of the referential state. Measured in the

(x,y,z)-coordinate system, q and m have the components

q =

qxqyqz

, m = mxmy

mz

. (11) 1

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2 Chapter 1 Beams in three dimensions

i

j

k

x

y

z

idx

qdx mdx

dx

l

M

M + dM

F

F + dF

Figure 11 Beam in referential state.

As a consequence of the external loads, the beam is deformed into the so-called current state

where the external loads are balanced by an internal section force vector F = F(x) and aninternal section moment vectorM = M(x). These vectors act on the cross-section with the baseunit vector i of the x-axis as outward directed normal vector. With reference to Fig. 12, thecomponents ofF and M in the (x,y,z)-coordinate system are:

F =

NQyQz

, M =

MxMyMz

(12)

Here, N = N(x) is the axial force, whereas the components Qy = Qy(x) and Qz = Qz(x) sig-nify the shear force components in the y- and z-directions. The axial component Mx = Mx(x) ofthe section moment vector is denoted the torsional moment. The components My = My(x) andMz = Mz(x) in the y- and z-directions represent the bending moments. The torsional momentis not included in two-dimensional beam theory. However, in the design of three-dimensional

frame structures, a good understanding of the torsional behaviour of beams is crucial.

Assuming that the displacements remain small, the equation of static equilibrium can be

established in the referential state. With reference to Fig. 11, the left end-section of the element

is loaded with the section force vector F and the section moment vector M. At the right

end-section, these vectors are changed differentially into F + dF and M + dM, respectively.Force equilibrium and moment equilibrium formulated at the point of attack of the section force

vector F at the left end-section then provides the following equations of force and moment


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1.2 Equations of equilibrium for spatial beams 3

x

y

z

Mx

My

Mz

N

Qy

Qz

Figure 12 Components of the section force vector and the section moment vector.

equilibrium of the differential beam element:

F + F+ dF + qdx = 0

dF

dx+ q = 0 (13a)

M + M + dM + idx (F + dF) + mdx = 0

dM

dx+ i F + m = 0 (13b)

From Eqs. (11) and (12) follows that Eqs. (13a) and (13b) are equivalent to the following

component relations:

dN

dx+ qx = 0,

dQydx

+ qy = 0,dQzdx

+ qz = 0, (14a)

dMx

dx + mx = 0,

dMy

dx Qz + my = 0,

dMz

dx + Qy + mz = 0. (14b)

At the derivation of Eq. (14b), it has been utilised that

i F = i (Ni + Qyj + Qzk) = Ni i + Qyi jQzi k = 0i Qzj + Qyk. (15)

Hence, iF has the components {0, Qz, Qy}. It is noted that a non-zero normal-force compo-nent is achieved when the moment equilibrium equations are formulated in the deformed state.

This may lead to coupled lateral-flexural instability as discussed in a later chapter.

1.2.1 Section forces and stresses in a beam

On the cross-section with the outward directed unit vector co-directional to the x-axis, the normalstress xx and the shear stresses xy and xz act as shown in Fig. 13. These stresses must be


+

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statically equivalent to the components of the force vector F and the section moment vector M

as indicated by the following relations:

N =A

xxdA, Qy =A

xydA, Qz =A

xzdA, (16a)

Mx =

A

(xzy xyz)dA, M y =

A

zxxdA, M z =

A

yxxdA. (16b)

x

y

z

dAMx

My

Mz

N

Qy

Qz

xx

xy

xz

Figure 13 Stresses and stress resultant on a cross-section of the beam.

x

y

z

dx

dy

dz

xx

yy

zz

xy

yx

xzzx

yz

zy

Figure 14 Components of the stress tensor.

On sections orthogonal to the y- and z-axes, the stresses {yy , yx, yz} and {zz , zx , zy}

act as shown in Fig. 14. The first index indicates the coordinate axis co-directional to theoutward normal vector of the section, whereas the second index specifies the direction of action

of the stress component. The stresses shown in Fig.14 form the components of the stress tensor


Momentet findes som vedsimpel moment tagning.For Mx indgr sigma forbde z og y retningen, dade begge bidrager tilmomentpvirkningomkring x-aksen.Afstanden z er negativ,derfor indgr minus iformlerne.

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in the (x ,y,z)-coordinate system given as

= xx yx zx

xy yy zyxz yz zz

. (17)

Moment equilibrium of the cube shown in Fig. 14 requires that

xy = yx, xz = zx, yz = zy . (18)

Hence, is a symmetric tensor.

1.2.2 Kinematics and deformations of a beam

The basic assumption in the classical beam theory is that a cross-section orthogonal to the x-axisat the coordinate x remains plane and keeps its shape during deformation. In other words, thecross-section translates and rotates as a rigid body. Especially, this means that Poisson contrac-

tions in the transverse direction due to axial strains are ignored. Hence, the deformed position

of the cross-section is uniquely described by a position vector w = w(x) and a rotation vector = (x) with the following components in the (x ,y,z)-coordinate system:

w =

wxwywz

, =

xyz

. (19)

Further, only linear beam theory will be considered. This means that the displacement compo-

nents wx, wy and wz in Eq. (19) all small compared to the beam length l. Further the rotationcomponents x, y and z are all small. Especially, this means that

sin tan , (110)

where represents any of the indicated rotation components measured in radians. The variousdisplacement and rotation components have been illustrated in Fig. 15. The rotation component

around the x-axis is known as the twist of the beam.Now, a material point on the cross-section with the coordinates (x ,y,z) in the referential

state achieves a displacement vector u = u(x ,y,z) with the components {ux, uy, uz} in the(x,y,z)-coordinate system given as (see Fig. 15):

ux(x,y,z) = wx(x) + zy(x) yz(x), (111a)

uy(x,y,z) = wy(x) zx(x), (111b)

uz(x ,y,z) = wz(x) + yx(x). (111c)

It follows that the displacement of any material point is determined if only the 6 components of

w(x) and (x) are known at the beam coordinate x. Hence, the indicated kinematic constraint

reduces the determination of the continuous displacement field u = u(x,y,z) to the determina-tion of the 6 deformation components wx = wx(x), wy = wy(x), wz = wz(x), x = x(x),y = y(x) and z = z(x) of a single spatial coordinate along the beam axis.


(Liner relation, se wiki http://en.wikipedia.org/wiki/Tensor)

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x

xx

y

y

y

z

z

z

dwy

dx

dwzdx

wxwx

wy wz

x

yz

Figure 15 Deformation components in beam theory.

The strains conjugated to xx, xy and xz are the axial strain xx and the angular strainsxy = 2xy and xz = 2xz . They are related to displacement components as follows:

xx =uxx

=dwxdx

+ zdydx

ydzdx

, (112a)

xy =

uxy +

uyx =

dwydx z

dxdx z(x), (112b)

xz =uxz

+uzx

=dwydx

+ ydxdx

+ y(x). (112c)

From Eq. (112) follows that xy = xy(x, z) is independent of y as a consequence of thepresumed plane deformation of the cross-section. Then, the shear stress xy = xy(x, z) mustalso be constant over the cross-section. Especially, xy = 0 at the upper and lower edge of thecross-section as illustrated in Fig. 16a. However, if the cylindrical surface is free of surface

shear tractions, then yx = 0 at the edge. Hence, xy = yx in contradiction to Eq. (18).In reality xy = 0 at the edges, corresponding to xy = 0. This means that the deformed

cross-section forms a right angle to the cylindrical surface as shown in Fig. 16b.The displacement fields Eq. (111) are only correct for beams with cross-sections which are

circular symmetric around the x-axis. In all other cross-sections, the torsional moment Mx will


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x x

y y

z z

dwydx

dwydx

wx wx

wy wy

xyxy

yxxy

(a) (b)

Figure 16 Shear stresses on deformed beam section: (a) Deformation of cross-section in beam theory and (b) real

deformation of cross-section.

induce an additional non-planar displacement in the x-axis, which generally can be written in theform ux(x,y,z) = (y, z)dx/dx. This is illustrated in Fig. 17. Hence, the final expressionfor the axial displacement reads

ux(x,y,z) = wx(x) + zy(x) yz(x) + (y, z)dxdx

. (113)

The expressions for uy and uz in Eq. (111) remain unchanged, and (y, z) is called the warp-ing function. Whereas y and z in Eq. (113) may be considered as shape functions for thedeformations caused by the rotations z(x) and y(x), the warping function is a shape functiondefining the axial deformation of the cross-section from the rotation component. The definition

and determination of the warping function is considered in a subsequent section.

Deformed bottom flange

Deformed top flange Undeformed stateSection AA

Section BBA

A

B

B

Figure 17 Warping deformations in an I-beam induced by homogeneous torsion. The cross sections AA and BB areshown with the top flange on the left and the bottom flange on the right.

As a consequence of the inclusion of the warping, the strain components in Eq. (112) are

modified as follows:

xx =uxx

=dwxdx

+ zdydx

ydzdx

+ d2xdx2

, (114a)

xy =ux

y

+uy

x

=dwy

dx

z + y

z dxdx

, (114b)

xz =uxz

+uzx

=dxzdx

+ y +

z+ y

dxdx

. (114c)


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xxy

y z

z

dwydx

dwzdx

wxwx

wy wz

yz

Figure 18 Kinematics of Bernoulli-Euler beam theory.

Bernoulli-Euler beam kinematics presumes that the rotated cross-section is always orthogo-

nal to the deformed beam axis. This involves the following additional kinematical constraints on

the deformation of the cross-section (see Fig. 18):

y = dwzdx

, z =dwydx

. (115)

Assuming temporarily that x 0 in bending deformations, i.e. disregarding the twist of thebeam, Eqs. (114) and (115) then provide:

xy = xz = 0. (116)

Equation (116) implies that the shear stresses are xy = xz = 0, and in turn that the shearforces become Qy = Qz = 0, cf. Eq. (16). However, non-zero shear forces are indeed presentin bending of Bernoulli-Euler beams. The apparent paradox is dissolved by noting that the shear

forces in Bernoulli-Euler beam theory cannot be derived from the kinematic condition, but has

to be determined from the static equations.

The development of the classical beam theory is associated with names like Galilei (15641642),Mariotte (16201684), Leibner (16461716), Jacob Bernoulli (16541705), Euler (17071783),

Coulomb (17361806) and Navier (17851836), leading to the mentioned Bernoulli-Euler beam

based on the indicated kinematic constraint. The inclusion of transverse shear deformation was

proposed in 1859 by Bresse (18221883) and extended to dynamics in 1921 by Timoshenko (1878

1972). Due to this contribution, the resulting beam theory based on the strain relations Eq. (112),

is referred to as Timoshenko beam theory (Timoshenko 1921).

The first correct analysis of torsion in beams was given by St. Venant (1855). The underlying

assumption was that dx/dx in Eq. (113) was constant, so the warping in all cross-sections

become identical. Then, the axial strain xx from torsion vanishes and the distribution of theshear strains xy and xz are identical in all sections. Because of this, St. Venant torsion is alsoreferred to as homogeneous torsion.


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Whenever the twist or the warping is prevented at one or more cross-sections, dx/dx is nolonger constant as a function of x. Hence, axial strains occur and, as a consequence of this,axial stresses arise and the shear strains and shear stresses are varying along the beam. These

phenomena were systematically analysed by Vlasov (1961) for thin-walled beams, for whichreason the resulting theory is referred to as Vlasov torsion or non-homogeneous torsion. Notice

that the shear stresses from Vlasov torsion have not been included in the present formulation.

These will be considered in a subsequent chapter.

Seen from an engineering point of view, the primary advantage of Vlasov torsion theory is that it

explains a basic feature of beams, namely that prevention of warping leads to a much stiffer struc-

tural elements than achieved in the case of homogeneous warping, i.e. a given torsional moment

will induce a smaller twist. Warping of the cross-section may, for example, be counteracted by the

inclusion of a thick plate orthogonal to the beam axis and welded to the flanges and the web. The

prevention of torsion in this manner is particularly useful in the case of slender beams with openthin-walled cross-sections that are prone to coupled flexuraltorsional buckling. Obviously, Vlasov

torsion theory must be applied for the analysis of such problems as discussed later in the book.

Next, the deformation of the cross-section may be decomposed into bending and shear com-

ponents. The bending components are caused by the bending moments My and Mz and deformas a Bernoulli-Euler beam. Hence, the bending components are causing the rotations y and zof the cross-section. The shear components are caused by the shear forces Qy and Qz . Thesecause the angular shear strains xy and xz without rotating the cross-section. Further, the dis-

placement of the beam axis in shear takes place without curvature. Hence, the curvature of thebeam axis is strictly related to the bending components, see Fig. 19.

With reference to Fig. 110, the radii of curvatures ry and rz are related to the rotationincrements dz and dy of the end-sections in the bending deformations of a differential beamelement of the length dx as follows

rydz = dx

rzdy = dx

y = 1/rz = dy/dx

z = 1/ry = dz/dx(117)

Here, y and z denote the components of the curvature vector of the x-axis. Especially, for a

Bernoulli-Euler beam the curvature components become, cf. Eq. (115),

y = d2wzdx2

, z =d2wydx2

. (118)

From Eqs. (114) and (117) follows that the axial strain may be written as

xx(x ,y,z) = (x) + zy(x) yz(x) + (y, z)d2xdx2

, (119)

where (x) denotes the axial strain of the beam along the x-axis given as

(x) =dwx

dx

. (120)

Here, (x), y(x) and y(x) define the axial strain and curvatures of the beam axis, i.e. the(x)-axis.


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=

+

x

x

x

y

y

y

z

z

z

Mz

Qy

xy

xy

Figure 19 Decomposition of cross-section deformation into bending and shear components.

xxy

y z

z

ds dx ds dx

ry rz

wy wz

dydz

Figure 110 Definition of curvature.

1.2.3 Constitutive relations for an elastic beam

In what follows we shall refer to N(x), Qy(x), Qz(x), Mx(x), My(x) and Mz(x) as generalisedstresses. These are stored in the column matrix

(x) =

N(x)Qy(x)Qz(x)

Mx(x)My(x)Mz(x)

. (121)


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The internal virtual work of these quantities per unit length of the beam is given as

= N + Qyxy + Qzxz + Mxx + Myy + Mzz = T, (122)

where

(x) =

(x)xy(x)xz(x)x(x)y(x)z(x)

. (123)

The components of(x) are referred to as the generalised strains. The components of(x) and(x) are said to be virtual work conjugated because these quantities define the internal virtualwork per unit length of the beam.

Let E and G denote the elasticity modulus and the shear modulus. Then, the normal stressxx and the shear stresses xy and xz may be calculated from Eq. (114) as follows:

xx = Exx = E

dwxdx

+ zdydx

ydzdx

+ (y, z)d2xdx2

, (124a)

xy = Gxy = G

dwydx

z +

y z

dxdx

, (124b)

xz = Gxz = Gdw2

dx + y +

z + y dx

dx

. (124c)

By integration over the cross-sectional area, it then follows that

N = E

A

dwxdx

+ Sydydx

Szd2dx

+ Sd2xdx2

, (125a)

Qy = G

Ay

dwydx

z

+ Ry

dxdx

, (125b)

Qz = GAz dwzdx

+ y+ Rz dxdx , (125c)

Mx = G

Sz

dwzdx

+ y

Sy

dwydx

z

+ K

dxdx

, (125d)

My = E

Sy

dwxdx

+ Iyydydx

Iyzdzdx

+ Izdxdx

, (125e)

Mz = E

Sz

dwxdx

Iyzdydx

+ Izzdzdx

Iydxdx

, (125f)

where Ay, Az , Ry, Rz , Sy, Sz, S, K, Iyy , Izz , Iyz = Izy , Iy and Iz are cross-sectional (or

geometrical) constants identified as:

A =

A

dA, Ay = yA, Az = zA, (126a)


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Ry =

A

y z

dA, Rz =

A

z+ y

dA, (126b)

Sy =A zdA, S

z =A ydA, S

=A dA, (126c)

Iyy =

A

z2dA, I zz =

A

y2dA, (126d)

Iyz =

A

yzdA, I y =

A

ydA, I z =

A

zdA, (126e)

K =

A

y2 + z2 + y

z z

y

dA. (126f)

Here, A is the cross-sectional area, whereas Ay and Az signify the so-called shear areas. Beamtheory presumes a constant variation of the shear stresses in bending, whereas the actual variationis at least quadratic. The constant variation results in an overestimation of the stiffness against

shear deformations, which is compensated by the indicated shear reduction factors y and z . Ifthe actual distribution of the shear stresses is parabolic, these factors become y = z = 5/6.For an I-profile, the shear area is approximately equal to the web area.

For GAy we have xy = Qy/(GAy) = 0. Bernoulli-Euler beam theory is charac-terised by xy = 0. Hence, Timoshenko theory must converge towards Bernoulli-Euler theoryfor the shear areas passing towards infinity. The magnitude of the shear deformations in propor-

tion to the bending deformations depends on the quantity (h/l)2, where h is the height and l is

the length of the beam. This relation is illustrated in Example 1-3.Ry and Rz are section constants which depend on the warping mode shape (y, z) as wellas the bending modes via y and z. Further, the section constants Sy and Sz are denoted the staticmoments around the y- and z-axes. S specifies a corresponding static moment of the warpingshape function.

Iyy and Izz signify the bending moments of inertia around the y- and z-axes, respectively.Iyz is denoted the centrifugal moment of inertia, whereas Iy and Iz are the correspondingcentrifugal moments of the warping shape function and the bending mode shapes.

K is the so-called torsion constant. This defines merely the torsional stiffness in St. Venanttorsion. As mentioned above, the additional contribution to Mx from Vlasov torsion will beconsidered in a subsequent section.

1.3 Differential equations of equilibrium for beams

In what follows, the governing differential equations for Timoshenko and Bernoulli-Euler beams

are derived. At this stage, the twist x and the torsional moment Mx are ignored. With no furtherassumptions and simplifications, Eq. (125) reduces to

N

My

MzQyQz

= EA ESy ESz 0 0ESy EIyy EIyz 0 0

ESz EIyz EIzz 0 00 0 0 GAy 00 0 0 0 GAz

dwx/dxdy/dx

dz/dxdwy/dx zdwz/dx + y

. (127)


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1.3 Differential equations of equilibrium for beams 13

The coefficient matrix of Eq. (127) is symmetric. When formulated in a similar matrix format,

the corresponding matrix in Eq. (125) is not symmetric. This is a consequence of the ignorance

of the Vlasov torsion in Mx.

1.3.1 Governing equations for a Timoshenko beam

Next, Eq. (127) is inserted into the equilibrium equations (14a) and (14b), which results in

the following system of coupled ordinary differential equations for the determination ofwx, wy ,wz , y and z:

dN/dx

dMy/dxdMz/dx

dQy/dxdQz/dx

=

0

QzQy

00

qx

mymz

qyqz

d

dx

EA ES y ESz 0 0ESy EIyy EIyz 0 0

ESz EIyz EIzz 0 00 0 0 GAy 00 0 0 0 GAz

dwx/dxdy/dxdz/dx

dwy/dx zdwz/dx + y

=

0 0 0 0 0

0 0 0 0 GAz0 0 0 GAy 00 0 0 0 00 0 0 0 0

dwx/dx

dy/dxdz/dx

dz/dx zdwz/dx + y

qx

mymzqyqz

. (128)

Equation (128) specifies the differential equations for Timoshenko beam theory. These should

be solved with proper boundary condition at the end-sections of the beam. Let x0 denote theabscissa of any of the two end-sections, i.e. x0 = 0 or x0 = l, where l is the length of the beam.At x = x0 either kinematical or mechanical boundary conditions may be prescribed.

Kinematical boundary conditions mean that values of wx, wy, wz, y and z are prescribed,

wx(x0) = wx,0wy(x0) = wy,0wz(x0) = wz,0y(x0) = y,0z(x0) = z,0

, x0 = 0, l, (129)

whereas mechanical boundary conditions imply the prescription ofN, Qy, Qz , My and Mz,

N(x0) = N0Qy(x0) = Qy,0

Qz(x0) = Qz,0My(x0) = My,0Mz(x0) = Mz,0

, x0 = 0, l. (130)


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In Eq. (130), the left-hand sides are expressed in kinematical quantities by means of Eq. (127).

Of the 10 possible boundary conditions at x = x0 specified by Eqs. (135) and (130), only 5can be specified. The 5 boundary conditions at x0 = 0 and x0 = l can be selected independently

from Eq. (135) and Eq. (130).With given boundary conditions Eq. (128) can be solved uniquely for the 5 kinematic quan-

tities wx, wy, wz , y, z, which make up the degrees of freedom of the cross-section. Althoughan analytical solution may be cumbersome, a numerical integration is always within reach.

1.3.2 Governing equations for a Bernoulli-Euler beam

Next, similar differential equations are specified for a Bernoulli-Euler beam. At first the shear

forces Qy and Qz in the equations of equilibrium for My and Mz in Eq. (14b) are eliminatedby means of the 2nd and 3rd equations in Eq. (14a):

d2My/dx2 dQz/dx + dmy/dx = 0

d2Mz/dx2 + dQy/dx + dmz/dx = 0

d2My/dx2 + qz + dmy/dx = 0

d2Mz/dx2 qy + dmz/dx = 0.

(131)

Using the Bernoulli-Euler kinematical constraint Eq. (115), the constitutive equations for the

resulting section forces may be written as

NMyMz

= EA ES y ESz

ESy EIyy EIyzESzz EIyz EIzz

dwx/dxd

2

wz/dx2

d2wy/dx2

. (132)

Then, the equations of equilibrium Eq. (14a) and Eq. (131) may be recasted as the following

system of coupled ordinary differential equations

d

dx

EA

dwxdx

ESyd2wzdx2

ESzd2wydx2

+ qx = 0, (133a)

d2

dx2ESy dwx

dx EIyy

d2wz

dx2 EIyz

d2wy

dx2+ qz + dmy

dx= 0, (133b)

d2

dx2

ESz

dwxdx

+ EIyzd2wzdx2

+ EIzzd2wydx2

+ qy +

dmzdx

= 0. (133c)

The governing equations (133) should be solved with 5 of the same boundary conditions as

indicated by Eqs. (135) and (130). The difference is that y(x0), z(x0), Qy(x0) and Qz(x0)are represented as, cf. Eqs. (14b) and (115),

dwz(x0)

dx= y,0,

dwz(x0)

dx= z,0, (134a)

dMz(x0)

dx mz(x0) = Qy,0,

dMy(x0)

dx+ my(x0) = Qz,0. (134b)


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1.4 Uncoupling of axial and bending deformations 15

With this in mind, the kinematic boundary conditions for Bernoulli-Euler beams are given in the

form

wx(x0) = wx,0wy(x0) = wy,0wz(x0) = wz,0

dwz(x0)/dx = y,0dwy(x0)/dx = z,0

, x0 = 0, l, (135)

whereas the mechanical boundary conditions defined in Eq. (130) are still valid.

1.4 Uncoupling of axial and bending deformations

Up to now the position of the origin O and the orientation of the y- and z-axes in the cross-section have been chosen arbitrarily. As a consequence of this, the deformations from the axial

force and the deformation from the bending moments My and Mz will generally be coupled.This means that the axial force N referred to the origin O will not merely induce a uniformdisplacement wx of the cross-section, but also non-zero displacements wy and wz of O as wellas rotations y and z . Similarly, the bending moment My will not merely cause a displacementwy and a rotation y of the cross-section, but also a non-zero displacement wy and a rotation zin the orthogonal direction in addition to an axial displacement wy of the origin. The indicatedmechanical couplings are the reason for the couplings in the differential equations (128) and

(133). The couplings may have a significant impact on the structural behaviour and stability of

an engineering structure and the position of the origin for a given beam element as well as theorientation of the coordinate axes must be implemented correctly in a computational model.

In this section, two coordinate transformations will be indicated, in which the axial force re-

ferred to the new origin B, called the bending centre, only induces a uniform axial displacementover the cross-section. Similarly, the bending moments My and Mz around the new rotated y-and z-axes, referred to as the principal axes, will only induce the non-zero deformation compo-nents (wz , y) and (wy , z), respectively. Especially, the moments will induce the displacementwx = 0 of the bending centre, B.

1.4.1 Determination of the bending centreThe position of the bending centre B is given by the position vector rB with the components{0, yB, zB} in the (x,y,z)-coordinate system. In order to determine the components yB and zB ,a translation of the (x,y,z)-coordinate system to a new (x, y, z)-coordinate system with originin the yet unknown bending centre is performed (see Fig. 111). The relations between the new

and the old coordinates read

x = x, y = y + yB, z = z + zB, (136)

In the new coordinate system, the displacement of B (the new origin) in the x-direction (the

new beam axis) becomes (see Fig. 111):

wx = wx + zBy yBz. (137)


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y , My

z , Mz

y , M

y

z , M

z

N

N

y

z

y

z

yB

zB

O

B

rB

Figure 111 Translation of coordinate system.

The axial strain of fibres placed on the new beam axis becomes

(x) =dw

dx=

dw

dx= + zBy yBz, (138)

where Eq. (117), Eq. (120) and Eq. (137) have been used. The components of the rotationvector of the cross-section are identical, i.e.

x = x,

y = y,

z = z. (139)

In turn, this means that the components of the curvature vector in the two coordinate systems

are identical as well

y =dydx

=dydx

= y,

z =dzdx

=dzdx

= z. (140)

Further, the components of the section force vector F in the two coordinate systems becomeidentical, i.e.

N = N, Qy = Qy Q

z = Qz. (141)

As a consequence of referring the axial force N = N to the new origin B, the componentsof the section vector in the (x, y, z)-coordinate system are related to the components in the(x,y,z)-coordinate as follows:

Mx = Mx, M

y = My zBN, M

z = Mz + yBN. (142)

Equations (138) and (140) provide the following relation for in terms of, y and z:

= zBy + yBz = zB

y + yB

z. (143)


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Then the relation between {N, My, Mz} and {N, My, M

z} and {, y, z} and {, y, z}

may be specified in the following matrix formulation:

= A

T

, (144a)

= A, (144b)

where

=

NMy

Mz

, =

y

z

, (145a)

=

N

MyMz

, =

yy , (145b)

A =

1 zB yB0 1 0

0 0 1

. (145c)

The components {N, My, Mz} and {, y, z} of and may be interpreted as work conjugatedgeneralised stresses and strains.

With reference to Eq. (132), the constitutive relation between and is given as

= C, (146)

where C denotes the constitutive matrix,

C = E

A Sy SzSy Iyy Iyz

Sz Iyz Izz

. (147)

Likewise, the constitutive relation in the (x, y, z)-coordinate system reads

= C (148)

where the constitutive matrix has the form

C = E

A Sy SzSy Iyy Iyz

Sz Iyz Izz

. (149)

Obviously, as given by Eqs. (147) and (149), the cross-sectional area A is invariant to a rotationof the cross-section about the x-axis and a translation in the y- and z-directions.

From Eqs. (144a), (144b) and (146) follows that

= ATC = ATCA

C = ATCA = E

1 0 0zB 1 0yB 0 1

A Sy SzSy Iyy IyzSz Iyz Izz

1 zB yB0 1 00 0 1


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C = E

A SyzBA (SzyBA)

SyzBA Iyy2zBSy+z2BA Iyz + yBSy+zB(SzyBA)(SzyBA) Iyz + yBSy+zB(SzyBA) Izz 2yBSz+y2BA

.

(150)

The idea is now to use the translational coordinate transformation to uncouple the axial defor-

mations from the bending deformations. This requires that Sy = Sz = 0. Upon comparisonof Eq. (149) and Eq. (150), this provides the following relations for the deformation of the

coordinates of the bending centre:

yB =SzA

, zB =SyA

. (151)

With yB and zB given by Eq. (151), the bending moments of inertia, Iyy and Izz , and the

centrifugal moment of inertia, Iyz , in the new coordinate system can be expressed in terms ofthe corresponding quantities in the old coordinate system as follows:

Iyy = Iyy 2zB(AzB) + z2BA = Iyy z

2BA, (152a)

Izz = Izz 2yB(AyB) + y2BA = Izz y

2BA, (152b)

Iyz = Iyz yB(AzB) zB(AyB) + yBzBA = Iyz yBzBA. (152c)

The final results in Eq. (152) are known as Knigs theorem.

x

y

z

yB

O

Figure 112 Single-symmetric cross-section.

If the cross-section is symmetric around a single line, and the y-axis is placed so that itcoincides with this line of symmetry, then the static moment Sy vanishes, i.e.

Sy =

A

zdA = 0 (153)

As a result of this, the bending centre B will always be located on the line of symmetry in asingle-symmetric cross-section, see Fig. 112. Obviously, if the cross-section is double symmet-

ric, then the position ofB is found at the intersection of the two lines of symmetry.


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Example 1.1 Determination of bending and centrifugal moments of inertia of non-symmetric

thin-walled cross-section

The position of the bending centre of the cross-section shown in Fig. A is determined along with the

bending moments of inertia Iyy and Izz and the centrifugal moment of inertia Iyz .

x

y

z

t

t

2t

a

2a

2aO

Figure A Thin-walled cross-section.

The (x,y ,z)-coordinate system is placed as shown in Fig. A. Then, the following cross-sectional con-stants are calculated:

A = 2a 2t + 2a t + a t = 7at, (a)

Sy = 2a 2t a + 2a t t2

+ a t a2

=1

2(2t + 9a) + ta, (b)

Sz = 2a 2t t + 2a t (2t + a) + a t

2t + 2a +t

2

=

1

2(21t + 8a)ta, (c)

Iyy =1

3 2t (2a)3 + 1

3 2a t3 + 1

3 t a3 = 1

3 2t2 + 17a2

ta, (d)

Izz = 13 2a 2a (2t)3 + 1

12(2a)3 t + 2a t (2t + a)2 + 1

12 a t3 + a t

2t + 2a + t

2

2

=1

3

59t2 + 54ta + 20a2

ta, (e)

Iyz = 2a 2t t a + 2a t (2t + a) 12

+ a t

2t + 2a +t

2

a

2=

1

4

8t2 + 25ta + 4a2

ta. (f)

Here, use has been made of Knigs theorem at the calculation of contributions to Iyy , Izz and Iyz fromthe three rectangles forming the cross-section.

The coordinates of the bending centre follow from Eq. (151) and Eqs. (a) to (c). Thus,

yB =SzA =

3

2 t +4

7 a, zB =

SyA =

1

7 t +9

14 a. (g)

(continued)


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x

y

zB

32

t + 47

a

17

t + 914

a

Figure B Position of bending centre in the thin-walled cross-section.

Subsequently, the moments of inertia around the axes of the (x, y, z)-coordinate system follow fromEq. (152), Eq. (153) and Eqs. (d) to (f):

Iyy =1

3(2t2 + 17a2)ta

1

7t +

9

14a

2 7ta = 1

84(44t2 108ta + 233a2)ta, (h)

Izz = 13 (59t2 + 54ta + 20a2)ta 32 t + 47 a

2

7ta = 184 (329t2 + 504ta + 368a2)ta, (i)

Iyz =1

4(8t2 + 25ta + 4a2)ta

1

7t +

9

14a)

3

2t +

4

7a

7ta = 1

14(7t2 15ta 22a2)ta. (j)

Now, for a thin-walled cross-section the thickness of the flanges and the web is much smaller than the

widths of the flanges and the height of the web. In the present case this means that t a. With this inmind, Eqs. (g) to (j) reduce to

yB 47

a, zB 914

a (k)

and

Iyy 23384

ta3, Izz 9221

ta3, Iyz 117

ta3. (l)

It is noted that the error on Izz estimated by Eq. (l) increases rapidly with increasing values of t/a.Thus, for t/a = 0.1, the error is about 13%. The errors related to the estimated values of Iyy and Iyzare somewhat smaller, i.e. about 5% and 7%, respectively.

From now on, the origin of the (x,y,z)-coordinate system is placed at the bending centre.Then, the constitutive matrix given by Eq. (147) takes the form

C = E

A 0 00 Iyy Iyz0 Iyz Izz

. (154)DCE Lecture Notes No. 23

Byttes om

zB

yB

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As a result of this, an axial force N no longer induces deformations in the y- and z-directions,and the bending moments My and Mz do not induce axial displacements. However, the bendingmoment My will still induce displacements in the y-direction in addition to the expected dis-

placements in the z-direction. Similarly, the bending moment Mz induces displacements in boththe y- and z-directions.

1.4.2 Determination of the principal axes

In order to uncouple the bending deformations, so that My will only induce deformations in thez-direction, and Mz only deformations in the y-direction, a new (x

, y, z)-coordinate system isintroduced with origin in B and rotated the angle around the x-axis as shown in Fig. 113.

y , My

z , Mz

y, M

y

z , M

z

N = N

y

z

yz

B

Figure 113 Rotation of coordinate system.

Let {N, My, Mz} and {N, My, M

z} denote the components of the generalised stresses inthe (x ,y,z)-coordinate system and the (x, y, z)-coordinate system, respectively. The two sets

of generalised stresses are related as

= B, (155a)

where

=

NMy

Mz

, =

NMy

Mz

, B =

1 0 00 cos sin

0 sin cos

. (155b)

Likewise, the components of the generalised strains in the two coordinate systems are denoted

as {, y, z} and {, y, z}, respectively. These are related as

= B, (156a)


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(a) (b)

y y

z z

y

yz

z

x = x

x = x

BB

Figure 114 Position of principal axes: (a) Iyz < 0 and (b) Iyz > 0.

The coordinate axes y and z are known as the principal axes of the cross-section, whereas Iyyand Izz are called the principal moments of inertia. It follows from Eqs. (163) and (167) thatthe choices of signs for sin(2) implies that Iyy becomes the larger of the principal momentsof inertia and Izz is the smaller principal moment of inertia. It is emphasised that this choiceis performed merely to have a unique determination of . Three other choices of are possibleobtained by additional rotations of the magnitudes

2, and 3

2 relative to the indicated.

If the cross-section has a symmetry line, and the y-axis is placed along this line, then Iyz = 0.Hence, a symmetry line is always a principal axis. Since the principal axes are orthogonal, the

z-axis is also a principal axiseven if the cross-section is not symmetric around the axis.

Example 1.2 Determination of principal axes coordinate system

The cross-section analysed in Example 1.1 is reconsidered. The thin-wall approximation is used, so the

moments of inertia are given by Eq. (l) in Example 1.1 and repeated here (without the primes):

Iyy 23384

ta3 2.7738 ta3, Izz 9221

ta3 4.3810 ta3, Iyz 117

ta3 1.5714 ta3.(a)

The position of the bending centre relatively to the top-left corner of the cross-section is provided inFig. A. From Eq. (a) and Eq. (165) follows that

J =

233

84ta3 92

21ta32

+ 4

11

7ta32

= ta3

9769

28, (b)

which by insertion into Eq. (164) provides:

sin(2) = 2 117 ta3 28ta3 9769 =

889769

0.8903, (c)

cos(2) = 23384 ta

3

92

21 ta

3 28ta3 9769 = 378084 9769 0.4553. (d)

(continued)


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y

z

y

z

x = x

t

t

2t

a

2a

2a

B

yB 47

a

zB 914

a

Figure A Position of principal axes coordinated system for the thin-walled cross-section.

From Eqs. (c) and (d) it is found that = 1.0217 radians corresponding to = 58.5418 . Hence, [0, 2 ] in agreement with Iyz = 117 ta3 < 0.

Finally, the moments of inertia in the principal axes coordinate system follow from Eq. (167), i.e.

Iyy

Izz

=

1

2

233

84+

92

21

1

2

233

84 92

21

2+ 4

11

7

2

ta3 =

5.3423 ta3,

1.8124 ta3.(e)

Clearly, Iyy is greater than any of Iyy or Izz , whereas Iyy is smaller than the bending moments ofinertia defined with respect to the original y- and z-axes.

1.4.3 Equations of equilibrium in principal axes coordinates

From now on it will be assumed that the (x ,y,z)-coordinate system forms a principal axes coor-dinate system with origin at the bending centre. In this case, the system of differential equations

(128) for a Timoshenko beam uncouples into three differential subsystems. Thus, the axial

deformation is governed by the equation

d

dx

EA

dwxdx

+ qx = 0, (169)

whereas bending deformation in the y-direction is defined by the coupled equations

d

dx

EIz

dzdx

+ GAz

dwydx

z

+ mz = 0, (170a)

d

dxGAy

dwy

dx

z+ qy = 0, (170b)where the double index yy on the bending moment of inertia has been replaces by a single indexy in order to indicate that the principal-axes coordinates are utilised.


y

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Similarly, the flexural deformations in the z-directions are determined by

d

dx EIydydx GAz

dwzdx

+ y+ my = 0, (171a)d

dx

GAz

dwzdx

+ y

+ qz = 0, (171b)

where again the double index on the bending moment of inertia has been replaced by a single

index. As seen from Eqs. (170) and (171), {wy, z} and {wz, y} are still determined bypairwise coupled ordinary differential equations of the second order.

For a Bernoulli-Euler beam, the system of ordinary differential equations (133) uncouples

completely into the following differential equations for the determination of wx, wy and wz :

d

dxEA

dwx

dx+ qx = 0, (172a)

d2

dx2

EIz

d2wydx2

qy +

dmzdx

= 0, (172b)

d2

dx2

EIy

d2wzdx2

qz

dmydx

= 0. (172c)

Example 1.3 Plane, fixed Timoshenko beam with constant load per unit length

Figure A shows a plane Timoshenko beam of the length l with constant bending stiffness EIz and shearstiffness GAy. The beam is fixed at both end-sections and is loaded with a constant load qy and a constantmoment load mz . The displacement wy(x), the rotation z(x), the shear force Qy(x) and the bendingmoment are to be determined.

x

y

z

l

mzqy

EIzz , GAy

Figure A Fixed beam with constant load per unit length.

The differential equations for determination ofwy (x) and z(x) follow from Eq. (170). Thus,

EIzd2zdx2

+ GAy

dwydx

z

+ mz = 0,d

dx

GAy

dwydx

z

+ qy = 0. (a)

According to Eq. (135), the boundary conditions are:

wy (0) = wy(l) = 0, z (0) = z(l) = 0. (b)

(continued)


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Integration of the second equation in Eq. (a) provides:

Qy = GAy dwydx

z = qyx + c1 (c)Then, the following solution is obtained for z (x) from the first equation in Eq. (a):

EIzd2zdx2

= qyx (c1 + mz) EIzz (x) = 16

qyx3 1

2(c1 + mz)x

2 + c2x + c3. (d)

Further, the boundary conditions z (0) = z(l) = 0 provide

c3 = 0, c2 = 16

qyl2 +

1

2(c1 + mz)l. (e)

Hence, the following reduced form is obtained for z (x):

z(x) = 16EIz

qy (x

3 xl2) 3(c1 + mz)(x2 xl)

. (f)

Next, Eq. (f) is inserted into Eq. (c) which is subsequently integrated with respect to x, leading to thefollowing solution for wy(x):

GAydwydx

= qyx + c1 + GAy6EIz

qy (x

3 xl2) 3(c1 + mz)x2 xl)

GAywy(x) = 12

qy x2 + c1x + c4 +

GAy6EIz

1

4qy (x

4 2x2l2) 12

(c1 + mz)(2x3 3x2l)

. (g)

The boundary conditions wy (0) = wy(l) = 0 provide the integration constants

c4 = 0, c1 =1

2qyl 1

y + 1mz, (h)

where

y = 12EIz

GAyl2(i)

Then, Eq. (a) and Eq. (f) provide the following solutions:

wy (x) =qy

2GAy(l

x)x +qy

24EIz(l

x)2x2

mz

GAy

x

y + 1 mz

12EIz

y

y + 1(2x3

3x2l)

wy (x) =

qy24EIz

(l x)2x2 + y l2(l x)x

mz12EIz

yy + 1

(2x3 3x2l + xl2), (j)

z(x) =qy

12EIz(2x3 3x2l + xl2) + mz

2EIz

yy + 1

(l x)x. (k)

The non-dimensional parameter y is a measure of the influence of the shear deformations. For arectangular cross-section with the height h we have Iz =

112h

2A and Ay =56A. Then y becomes

y =72

5 h

2

l2E

G. (l)

Hence, shear deformations are primarily of importance for short and high beams. On the other hand, for

long beams with a small height of the cross-section, shear deformations are of little importance, i.e. only

the bending deformation is significant. (continued)


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1.6 The principle of virtual forces 29

Eqs. (145a), (145b), (146) and (147) for Sy = Sz = Iyz = 0, i.e.

=N

EA

, y =My

EIy, z =

Mz

EIz. (174)

Insertion of Eq. (174) into Eqs. (119) and (124) provides the result for the axial stress in

terms of the generalised stresses,

xx =N

A

MzIz

y +MyIy

z. (175)

Equation (175) is due to Navier, and is therefore referred to as Naviers formula. It should be

noticed that Eq. (175) presumes that the stresses are formulated in a principal axes coordinate

system, so Iy and Iz indicate the principal moments of inertia. The relation is valid for both

Timoshenko and Bernoulli-Euler beams. This is so because only the relation (115), but notthe relation (118) has been utilised. Hence, Eq. (175) is based on the assumption that plane

cross-sections remain plane, but not that they remain orthogonal to the beam axis.

The so-called zero line specifies the line in the (y, z)-plane on which x = 0. The analyticalexpression for the zero line becomes

N

A

MzIz

y +MyIy

z = 0. (176)

It is finally noted that warping introduces displacements in the axial direction in addition to

those provided by bending. However, if the torsion is homogeneous, these displacements will

not introduce any normal strains and therefore no normal stresses. Hence, Naviers formula is

also valid in the case of St. Venant torsion, but in the case of Vlasov torsion, or inhomogeneous

torsion, additional terms must be included in Eq. (175).

1.6 The principle of virtual forces

In this section theprinciple of virtual forces is derived for a plane Timoshenko beam of the length

l. The deformation of the beam is taking place in the (x, y)-plane. In the referential state, theleft end-section is placed at the origin of the coordinate system and the x-axis is placed along the

bending centres of the cross-sections, see Fig. 115.The principle of virtual forces is the dual to the principle of virtual displacements. In the

principle of virtual displacements the actual sectional forces and sectional moments are assumed

to be in equilibrium with the loads and the reaction forces applied at the end sections. The virtual

displacements and rotations are considered as arbitrary increments to the actual displacements

and they only need to fulfil homogeneous kinematic boundary conditions, so that the combined

field made up by the actual and the virtual fields always fulfils the actual non-homogeneous

boundary conditions as given by Eq. (135). Further, the generalised virtual strains defining the

internal virtual work must be derived from the virtual displacement and rotation fields.

In contrast, the principle of virtual forces presumes that the displacements and rotations of the

beam are fulfilling the kinematic boundary conditions, and that the generalised internal strainsare compatible to these fields. The actual loads on the beam are superimposed with the virtual

incremental loads per unit length qx and qy, the virtual moment load per unit length mz ,


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1.6 The principle of virtual forces 31

Upon utilisation of Eq. (178), this is reduced to

N wx + Qy wy + Mz z

l0 +l0

qx wx + qy wy + mz z

dx

=

l0

N + Qy xy + Mz z

dx. (181)

The generalised strains on the right-hand side of Eq. (181) are now expressed in mechanical

quantities by means of Eq. (174). Further, N, Qy and Mz fulfil the following boundaryconditions at x = 0 and x = l, cf. Fig. 115,

N(0) = N1, Qy(0) = Qy,1, Mz(0) = Mz,1, (182a)

N( l) = N2, Qy( l) = Qy,2, Mz( l) = Mz,2. (182b)

Equation (181) then obtains the following final form:

2j=1

Nj wx,j + Qy,j wy,j + Mz,j z,j

+

l0

qx wx + qy wy + mz z

dx

=l0N N

EA +

Qy Qy

GAy +

Mz Mz

EIz

dx, (183)

where wx,j , wy,j and z,j denote the displacements in the x- and y-directions and the rotationin the z-direction at the end-sections, respectively. Equation (183) represents the principle ofvirtual forces. The left- and right-hand sides represent the external and internal virtual work,

respectively.

The use of Eq. (183) in determining the displacements and rotations of a Timoshenko beam

is demonstrated in Examples 1.4 and 1.5 below. Furthermore, the principle of virtual forces may

be used to derive a stiffness matrix for a Timoshenko beam element as shown later.

Example 1.4 End-displacement of cantilevered beam loaded with a force at the free end

Figure A shows a plane Timoshenko beam of the length l with constant axial stiffness EA, shear stiffnessGAy and bending stiffness EIz . The beam is fixed at the left end-section and free at the right end-section,where it is loaded with a concentrated force Qy,2 in the y-direction. The displacement wy,2 at the freeend is searched.

The principle of virtual forces Eq. (183) is applied with the following external virtual loads: qx =qy = mz = 0, N1 = Qy,1 = Mz,1 = N2 = Mz,2 = 0 and Qy,2 = 1. Further N(x) = 0.Then, Eq. (183) reduces to

1 wy,2 = l

0

Qy QyGAy + Mz MzEIz dx. (a)(continued)


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1

xx

yy

zz

l

ll

Mz

Qy

Mz

Qy

Qy,2

Qy,2

Qy,2 Qy,2 = 1

EA,GAy , EIz

Figure A Fixed plane Timoshenko beam loaded with a concentrated force at the free end: Actual force and section

forces (left) and virtual force and section forces (right).

The variation of the bending moment Mz(x) and the shear force Qy(x) from the actual load Qy,2 hasbeen shown in Fig. A on the left. The corresponding variational moment field Mz (x) and shear forceQy(x) from Qy,2 = 1 are shown in Fig. A on the right. Insertion of these distributions in Eq. (b)provides the solution

wy,2 =Qy,2l

GAy+

1

3

Qy,2l3

EIz=

1

12(4 + y) Qy,2l

3

EIz, (b)

where y is given by Eq. (i) in Example 1.3. The deformation contributions from shear and bendingare additive. This is a consequence of the additive nature of the flexibilities indicated by Eq. (183)

in contribution to the fact that the beam is statically determinate, which provides the fields Mz (x) andQy(x) as well as Mz(x) and Qy(x) directly.

Example 1.5 End-deformations of fixed beam loaded with a moment at the free end

The beam described in Example 1.4 is considered again. However, now the free end is loaded with a

concentrated moment Mz,2. The displacement wy,2 and the rotation z,2 of the end-section is to befound.

At the determination of wy,2 from Mz,2, the principle of virtual forces given by Eq. (183) is againapplied with Qy,2 = 1 and all other external variational loads equal to zero, leading to Eq. (a) inExample 1.4. However, Mz(x) and Qy(x) are now caused by Mz,2, and are given as shown in Fig. A,whereas Mz(x) and Qy(x) are as shown in Fig. A of Example 1.4. Then, wy,2 becomes

wy,2 =l0

Mz

MzEIz dx =

1

2

My,2l2

EIz . (a)

(continued)


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1.7 Elastic beam elements 33

1

xx

yy

zz

ll

Mz Mz

EA,GAy, EIz

Mz,2

Mz,2

Qy = 0 Qy = 0

Mz,2 = 1

Figure A Fixed plane Timoshenko beam loaded with a moment at the free end: Actual moment and section forces

(left) and virtual moment and section forces (right).

At the determination of z,2 the principle of virtual forces Eq. (183) is applied with the followingexternal virtual loads qx = qy = mz = 0, N1 = Qy,1 = Mz,1 = N2 = Qy,2 = 0 andMz,2 = 1. Then, Eq. (183) reduces to

1 z,2 = l0

Qy QyGAy + Mz MzEIz

dx. (b)

The variation ofQy(x) and Mz (x) from My,2 has been shown in Fig. A on the left, and the variation ofQy(x) and Mz(x) from Mz,2 = 1 is shown in Fig. A on the right. Then z,2 becomes

z,2 =

l0

1 Mz,2EIz

dx =Mz,2l

EIz. (c)

In the present load case, the shear force is given as Qy(x) = 0. Consequently it will not induce anycontributions in Eq. (a) and Eq. (c).

1.7 Elastic beam elements

When frame structures consisting of multiple beams are to be analysed, the establishment of

analytical solutions is not straightforward and instead a numerical solution must be carried out.

For this purpose, a discretization of the frame structure into a number of so-called beam ele-

ments is necessary, eventually leading to a finite-element model. The aim of the present section

is not to provide a full introduction to the finite-element method for the analysis of frame struc-

tures, e.g. tower blocks with a steel frame as the load-carrying structure. However, a formula-

tion is given for a single beam element to be applied in such analyses. Both the Timoshenkoand Bernoulli-Euler beam theories are discussed in this context, and plane as well as three-

dimensional beams are touched upon.


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1.7.1 A plane Timoshenko beam element

Firstly, the stiffness matrix and element load vector is derived for a plane Timoshenko beam ele-

ment with constant axial stiffness EA, shear stiffness GAy and bending stiffness EIz , cf. Fig. 116. The stiffness relation is described in an (x, y)-coordinate system with origin at the left end-section and the x-axis along the bending centres.

x

y

z

l

EA,GAy , EIz

Qy,1, wy,1

N1, wx,1

Mz,1, z,1

Qy,2, wy,2

N2, wx,2

Mz,2, z,2

1 2

Figure 116 Plane Timoshenko beam element with definition of degrees of freedom and nodal reaction forces.

At the end-nodes, nodal reaction forces Nj and Qy,j are acting along the x- and y-directions,respectively, and reaction moments Mz,j are applied around the z-axis. Here, j = 1 and j = 2stand for the left-end and right-end nodes of the beam element, respectively, and the reaction

forces and moments are in equilibrium with the remaining external loads on the element for

arbitrary deformations of the beam.

The element has 6 degrees of freedom defining the displacements and rotations of the end-

sections, cf. Fig. 116. These are organised in the column vector

we =

w

e1

we2

=

wx,1 wy,1 z,1 wx,2 wy,2 z,2T

(184)

The sub-vector wej defines the degrees of freedom related to element node j.Similarly, the reaction forces Nj, Qy,j and Mz,j , j = 1, 2, at the end-sections, work conju-

gated to wx,j , wy,j and z,j , are stored in the column vector

re = re1re2 = N1 Qy,1 Mz,1 N2 Qy,2 Mz,2

T(185)

x

y

z

mzpx

py

Qy,1

N1

Mz,1

Qy,2

N2

Mz,2

Figure 117 External loads and reaction forces from external loads on a plane beam element.

The equilibrium of the beam element relating the nodal reaction forces to the degrees of

freedom of the element may be derived by the principle of virtual displacements as demonstrated


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in a subsequent paper. The resulting equilibrium equations on matrix form may be written on the

form

re

= Ke

we

+ fe

. (186)

The vector fe in Eq. (186) represents the nodal reaction forces from the external element loads

when we = 0, i.e. when the beam is fixed at both ends as shown in Fig. 117. We shall merelyconsider constant element loads qx and qy per unit length in the x- and y-directions, and a con-stant moment load per unit length mz in the z-direction, see Fig. 117. The reaction forces andreaction moments follow from Eqs. (s) and (t) in Example 1.3:

fe =

12

qxl12

qyl + mz 112

qyl2 12

yy+1

mzl

12qxl12

qyl mz112

qyl2 12

yy+1

mzl

. (187)

The matrixKe in Eq. (186) denotes the stiffness matrix in the local (x,y,z)-coordinate system.Let wi denote the ith component ofwe. Then, the ith column inKe represents the nodal reactionforces for fe = 0, and with wi = 1 and wj = 0, j = i. These forces are obtained followingthe derivations in Example 1.3 from Eq. (a) to Eq. (t) with qy = mz = 0 and with the boundarycondition in Eq. (b) replaced by the indicated conditions. Because of the symmetry of the prob-

lem, only two such analyses need to be performed. Still, this is a rather tedious approach. Partly

because of this, and partly in order to demonstrate an alternative approach, the stiffness matrix

will be derived based on the principle of virtual forces.

x x

yy

wx

wyz

Undeformed stateUndeformed state

(a) (b)

Figure 118 Rigid-body modes of a plane beam element: (a) Translation and (b) rotation.

The beam element has 6 degrees of freedom, by which a total of 6 linear independent modes

of deformation may be defined. These consist of 3 linear independent rigid body modes and

3 linear independent elastic modes. The rigid modes may be chosen as a translation in the

x-direction, a translation in the y-direction and a rotation around the z-direction as shown inFig. 118. Any rigid body motion of the beam element may be obtained as a linear combination

of these component modes of deformation. Obviously, the rigid body motions do not introduce

stresses in the beam. Hence, the axial force N, the shear force Qy and the bending moment Mz

are all zero during such motions.Since, axial elongations are uncoupled from bending deformations, the elastic elongation

mode is uniquely defined as shown in Fig. 119a. The two bending deformation modes may be


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x

x

x

y

yy

Mz

Mz

(a)

(b) (c)

Qy = 0

N

N0

N0

N0

u02

u02

ss a

aMs Ms

Ms

Ma

Ma

Ma

Ma

2Mal

2Mal

2Ma

l Qy

Figure 119 Elastic modes and related section forces in a plane beam element: (a) Axial elongation; (b) symmetricbending and (c) antisymmetric bending.

chosen in arbitrarily many ways. Typically, these are chosen by prescribing an angle of rotation

at the other end-section. Following an idea by Krenk (2001), a more convenient formulation may

be obtained by choice of two bending modes symmetric and anti-symmetric around the mid-point

of the beam element as shown in Figs. 119b and 119c. It should be noticed that these modes

also apply if the material properties of the beam are not symmetrical around the mid-point.

The axial elongation and conjugated axial force related to the axial elongation mode are de-

noted u0 and N0, respectively. The symmetric and anti-symmetric bending modes are describedby the end-section rotations s and a defined in Figs. 119b and 119c, respectively. The con-

jugated moments are denoted Ms and Ma, respectively. The related distributions of the shearforce Qy(x) and the bending moment Mz(x) are shown in Figs. 119b and 119c.

The shear force is equal to Qy = 0 in symmetric bending, because the bending moment isconstant. Then, no shear deformations are related to this mode. In contrast, a constant shear

force appears in the anti-symmetric bending mode. Hence, the deformations occurring in this

mode are affected by bending as well as shear contributions.

At first the constitutive relations between the deformation measures and the conjugated gen-

eralised strains for the indicated elastic modes are found by means of the principle of virtualforces Eq. (183). In all cases, the beam element is unloaded, so qx = qy = mz = 0. For theaxial elongation mode N = N0, Qy = Mz = 0, and N = 1, N1 = 1, N2 = 1. Further,


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wx,1 = 12

u0, wx,2 =12

u0, wy,j = z,j = 0. Then, Eq. (183) reduces to

(1) (1

2

u0) + 1 1

2

u0 = l

0

1 N0

EA

dx u0 = N0 l

0

dx

EA

=l

EA

N0. (188)

The last statement holds for a beam element with constant axial stiffness EA. If EA varies, theintegral in the middlemost statement must be evaluated analytically or numerically.

For the symmetric bending mode N = Qz = 0, Mz = Ms, Mz = 1, Mz,1 = 1, andMz,2 = 1. Further, z,1 = s, z,2 = s and wx,j = wy,j = 0. Then, Eq. (183) provides

1 s + (1) (s) =

l0

(1)(Ms)

EIzdx 2s = Ms

l0

dx

EIz=

l

EIzMs. (189)

Again, the last statement only applies for a homogeneous beam, whereas the middlemost state-

ment applies for any variation of the bending stiffness EIz along the beam.

For the anti-symmetric bending mode N = 0, Qy = 2Mal , Mz(x) = (1 + 2x/l)Ma, andQy = 2Mz(x)/l = (1 + 2x/l). Further, z,1 = a, z,2 = a, wx,j = wy,j = 0. Then,

Eq. (183) provides

1 a + 1 a =

l0

2l

2Mal

GAy+

(1 + 2 xl )(1 + 2xl )Ma

EIz

dx

2a = Ma

4

l2

l0

dx

GAy+

l0

(1 + 2 xl )2

EIzdx

2a

= Ma4l 1

GAyl2+

1

3

l

EIy = 1

3

l

EIy(1 +

y)M

a,

y= 12

EIz

GAyl2. (190)

As discussed in Example 1.3, the non-dimensional parameter y defines the contribution of shearflexibility relatively to the bending flexibility.

The flexibility relations provided by Eqs. (188), (189) and (190) may be written in the

following equivalent stiffness matrix formulation:

r0 = K0w0, (191)

where

r0 = N0

MsMa

, w0 = u0

2s2a

, K0 =

EAl 0 00 EIz

l0

0 0 31+yEIzl

. (192)

The nodal reaction forces re and r0 in Eq. (185) and Eq. (192) are related via the transfor-

mation

re = Sr0 = SK0w0, (193)

where

S =

1 0 00 0 2/l0 1 1

1 0 00 0 2/l0 1 1

. (194)


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Similarly, the elastic deformation measures stored in w0 can be expressed by the degrees of

freedom of the element stored in we as follows (see Fig. 120):

u0 = wx,2 wx,1, (195)

z,1 = a + s +1l (wy,2 wy,1)

z,2 = a s +1l (wy,2 wy,1)

2a = z,1 + z,2 2l (wy,2 wy,1)

2s = z,1 z,2.(196)

wy,1

wy,2

a + s

a s

z,1z,2

Figure 120 Connection between elastic deformation measures and element degrees of freedom.

Equations (195) and (196) may be rewritten in the common matrix form

w0 = STwe, (197)

where S is given by Eq. (194). Insertion of Eq. (197) into Eq. (193) provides upon compari-

son with Eq. (186):

re = SK0STwe Ke = SK0S

T. (198)

Insertion ofK0 and S as given by Eq. (192) and Eq. (194) provides the following explicit

solution for Ke:

Ke =E

l3

Al2 0 0 Al2 0 00 121+y Iz

61+y

Izl 0 12

1+yIz

61+y

Izl

0 61+y Izl4+y1+y

Izl2 0 6

1+yIzl

2y1+y

Izl2

Al2 0 0 Al2 0 00 121+y Iz

61+y

Izl 012

1+yIz

61+y

Izl

0 61+y Izl2y1+y

Izl2 0 6

1+yIzl

4+y1+y

Izl2

. (199)

The corresponding result for a plane Bernoulli-Euler beam element is obtained simply bysetting y = 0. The equivalent element relations for a three-dimensional beam formulated in a(x,y,x) principal axes coordinate system are given in the next section.


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m

x

x

x

y

y

y

z

z

EA,GK

GAy , EIz

GAz , EIy

qx

qy

qz

mx

mz

wx,1 wx,2

N1 N2

x,1

Mx,1

wy,1, Qy,1

y,1, My,1

wz,1, Qz,1

z,1, Mz,1

x,2

Mx,2

wy,2, Qy,2

y,2, My,2

wz,2, Qz,2

z,2, Mz,2

Figure 121 Three-dimensional Timoshenko beam element with definition of degrees of freedom, nodal reaction forces,element loads and sectional properties.

The element equilibrium equations may be expressed on the matrix form, cf. Eq. (186),

re = Kewe + fe (1100)

Here, re and we are 12-dimensional column vectors storing the reaction forces and the element

degrees of freedom, respectively, cf. Eqs. (184) and (185),

re =

re1re2

=

N1Qy,1

Qz,1Mx,1My,1Mz,1

N2Qy,2Qz,2Mx,2My,2Mz,2

, we =

we1we2

=

wx,1wy,1

wz,1x,1y,1z,1wx,2wy,2wz,2x,2y,2z,2

. (1101)

Likewise, fe is 12-dimensional column vector storing the contributions to the reaction forces


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The element equilibrium relation provided by Eq. (1100) presumes that only St. Venant

torsion is taken into consideration, corresponding to the torsional equilibrium equations

Mx,1Mx,2

= GKl

1 11 1

x,1x,2

12 m

xl

11

. (1107)

The torsional constant K is determined in the next chapter. It will be shown that the inclusion

of Vlasov torsion requires the introduction of two extra degrees of freedomdx,1dx

anddx,2dx

.

The conjugated generalised stresses are the so-called bimoments. Hence, a Timoshenko beam

element, where both St. Venant and Vlasov torsion are taken into consideration, is described by

a total of 14 degrees of freedom.

1.8 Summary

In this chapter, the basic theory of Timoshenko and Bernoulli-Euler beams in three-dimensional

space has been presented. Some of the main topics covered are summarised below.

Beams are one-dimensional structures that may carry loads in three dimensions including axial

forces, shear forces in two orthogonal directions and moments around three directions.

Bernoulli-Euler beam kinematics assume that cross-sections remain orthogonal to the beam

axis during deformation. Hence, no shear deformation occurs.

Timoshenko beam kinematics include shear flexibility, but still a cross-section remains plane

during deformation. Hence, shear strains and stresses are homogeneous over the beam height.

The bending centre of a beam cross-section is defined as the point of attack of an axial force

not producing a bending moment.

The principal axes of a beam cross-section are defined as the axes around which a bending

moment will neither produce an axial force nor flexural displacements in the other direction.

The principle of virtual forces can by applied to the analysis of deformations in a beam. In the

case of Timoshenko beam theory, both shear and bending deformation occurs, whereas only

bending deformation is present in a Bernoulli-Euler beam.

Plane beam elements have six degrees of freedom with three at either end, i.e. two displace-

ments and one in-plane rotation. In the general case, the rotations and the axial displacements

are coupled, but in a principle-axes description, they become uncoupled.

Spatial beam elements have 12 degrees of freedom, that is three displacements and three rota-

tions at either end. Generally, the displacements and rotations are coupled, but an uncoupling

can be achieved by a proper choice of coordinate system.

Thus, a detailed description has been given of the lateral and flexural deformations in a beam.

However, in this chapter only a brief introduction has been given to twist and torsion of a beam.

A thorough explanation and analysis of these phenomena will be the focus of the next chapter.


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46 Chapter 2 Shear stresses in beams due to torsion and bending

BB

BB

(a) (b)

(c) (d)

xx

xx

yy

yy

zz

zz

Qy , qyQy, qy

Qy , qy

SS

SS

Mx, mx

Figure 21 Coupled and uncoupled bending and torsion. Coupling exists in cases (a) and (c), whereas cases (b) and (d)involve no coupling.

2.2 Homogeneous torsion (St. Venant torsion)

It is assumed that the torsional moment Mx and the incremental twist per unit length dx/dx andthe warping of the cross-sections remain unchanged along the beam. Then all cross-sections of

the beam are exposed to the same distribution of the shear stresses xy and xz. For this reason,this case is referred to as homogeneous torsion. Since the solution of the problem was given

by St. Venant (ref.), the case is also called St. Venant torsion. Inhomogeneous torsion refers to

the case, where either Mx or the material properties vary along the beam. Then dx/dx or thewarping will vary as well.

Figure 22a shows a cross-section of a cylindrical beam of the length l. The cross-sectionalarea is A. The curve along the outer periphery is denoted 0. The cross-section may have anumber N of holes determined by the boundary curves j , j = 1, 2, . . . , N . At the boundary

curves arc-length coordinates s0, s1, . . . , sN are defined. The arc-length coordinate s0 along0 is orientated in the anti clock-wise direction, whereas s1, s2, . . . , sN, related to the interiorboundaries 1, 2, . . . , N, are orientated in the clock-wise direction. The outward directed unit


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2.2 Homogeneous torsion (St. Venant torsion) 47

(a)

B

y

z

S

0N

yS

zS

Mxj

(b)0

j

n0

nj

s0

s0

sj

sj

n0s0

nj

sj

Figure 22 Cross-section with holes: (a) Interior and exterior edges; (b) definition of local (x, nj , sj )-coordinatesystems.

vector on a point of the exterior or interior boundaries j is denoted nj , j = 0, 1, . . . , N . Theunit tangential vector to a boundary curve is denoted sj and is co-directional to the arc-length

coordinate sj , see Fig. 22b. Thus, a local (x, nj , sj)-coordinate system may be defined withthe base unit vectors {i,nj , sj}. The indicated orientation of the exterior and interior arc-lengthcoordinates sj , j = 0, 1, . . . , N , insures that the related (x, nj , sj)-coordinate system forms aright-hand coordinate system.

The beam material is assumed to be homogeneous, isotropic linear elastic with the shear

modulus G. In homogeneous torsion, only shear stresses are present for which reason G is theonly needed elasticity constant.

2.2.1 Basic assumptions

For convenience the index x is omitted on the twist x (the rotation angle around the x-axis),i.e. x. Figure 23 shows a differential beam element of the length dx. Both end-sectionsof the element are exposed to the torsional moment Mx, so the element is automatically inequilibrium. On the left and right end-sections the twists are and + d, respectively. Theincrement d may be written as

d =d

dxdx. (21)

Since Mx and the material properties are the same in all cross-sections, d/dx must be constantalong the beam. Further, the warping must be the same in all cross-sections, i.e. ux = ux(y, z).This implies that the warping in homogeneous torsion does not induce normal strains, i.e.

xx =uxx

= 0. (22)

In turn this means that the normal stress becomes xx = Exx = 0. Hence, only the shearstresses xy and xz are present on a cross-section in homogeneous torsion.


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x

y

z

l

Mx

Mx

Mx

Mx

+ d

dx

dx

Figure 23 Beam and differential beam element subjected to homogeneous torsion.

The only deformation measure of the problem is the twist gradient d/dx. Then, due to thelinearity assumptions, the torsional moment Mx must depend linearly on d/dx. Further, xyand xz (and hence Mx) depend linearly on G. This implies the following relation:

Mx = GKd

dx. (23)

The proportionality constant K with the dimension [unit of length]4 is denoted the torsionalconstant. The determination of this constant is a part of the solution of the torsion problem.

2.2.2 Solution of the homogeneous torsion problem

As for all beam theories, the shape of the cross-section is assumed to be preserved during the

deformation. Then, the displacements in the (y, z)-plane are caused merely by the rotation around the shear centre S. The warping displacements ux must also be linearly dependent on thestrain measure d/dx, corresponding to the last term in Eq. (113). This implies the displacementcomponents

ux = ux(y, z) = (y, z)d

dx, uy = (z zs), uz = (y ys). (24)

Here (y, z) is the so-called warping function as discussed in Section 1.2.2, and its determinationconstitutes the basic part of the solution of the homogeneous torsion problem.

Similarly to Eq. (114), it follows from Eq. (24) that the components of the strain tensor

become:

xx =uxx

= 0, yy =uyy

= 0, zz =uzz

= 0, (25a)


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Equation (218) is a Poisson differential equation (inhomogeneous Laplace equation), and the

boundary conditions Eq. (221) are classified as the so-called Dirichlet boundary conditions.

In principle, the solution to the indicated boundary value problem is unique. The problem is

that the constant values Sj of the stress function along the boundary curves are unknown. Thedetermination of these is a part of the problem. For profiles with interior holes this is only

possible by the introduction of additional geometric conditions. The boundary value problem

for the Prandtl stress function has been summarised in Box 2.2. The formulation in terms of

Prandtls stress function is especially useful in relation to homogeneous torsion of thin-walled

profiles and will be utilised in a number of examples below.

Box 2.2 Boundary value problem for the Prandtl stress function

The equilibrium equation is automatically fulfilled. The compatibility condition is represented by the

following differential equation:

2S

y2+

2S

z2= 2G d

dx, (x, y) A. (222a)

The Dirichlet boundary condition, representing the Cauchy boundary condition, reads:

S = Sj , j = 0, 1, . . . , N , (x, y) 0 1 N. (222b)

The shear stresses xy and xz must be statically equivalent to the shear forces Qy = Qz =0 and the torsional moment Mx. Application of Gausss theorem on the vector field v

T =

[vy , vz] = [0, S] provides,

Qy =

A

xydA =

A

0

y+

S

z

dA =

Nj=0

j

(0 dz S dy) = Nj=0

Sj

j

dy, (223)

where the circulation is taken anticlockwise along 0 and clockwise along j , j = 1, , N.Further it has been used that Sj is constant along the boundary curve, and hence may be trans-ferred outside the circulation integral. Now,

j

dy = 0. Hence, it follows that any solution

to the boundary value problem defined by Eqs. (222a) and (222b) automatically provides a

solution fulfilling Qy = 0. Using vT

= [S, 0] it can in the same way be shown that

Qz =

A

xzdA =

A

S

y+

0

z

dA

Qz = Nj=0

j

(S dz 0 dy) = Nj=0

Sj

j

dz = 0. (224)

The torsional moment Mx must be statically equivalent to the moment of the shear stressesaround S, i.e.

Mx =

A

((y yS)xz (z zS)xy) dA =

A

(yxz zxy)dA, (225)


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2.2 H