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Eindhoven University of Technology
MASTER
Finite element methods for geometrically linear curved beams
Kinyanjui, Tabitha Wangari
Award date:2012
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TECHNISCHE UNIVERSITAT KAISERLSLAUTERNDEPARTMENT OF MATHEMATICS
MASTER’S THESIS
Finite element methods forgeometrically linear curved beams
by:
Tabitha Wangari Kinyanjui
Supervisors:
Dr - ing. Joachim Linn (ITWM)
Dr. Joseph M. Maubach (TU/e)
Kaiserslautern, Germany
November 22, 2007
The Faculty members appointed to examine the thesis by
Tabitha W. Kinyanjui find it satisfactory and recommend that it be
accepted.
TU-Kaiserslautern Date:
TU-Eindhoven Date:
Contents
1 Introduction 8
2 Straight Beam Theory 11
2.0.1 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.0.2 Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.1 Euler - Bernoulli Beam Theory . . . . . . . . . . . . . . . . . . . . . 12
2.1.1 Total potential energy . . . . . . . . . . . . . . . . . . . . . . 13
2.2 Timoshenko Beam Theory . . . . . . . . . . . . . . . . . . . . . . . 14
2.2.1 Total potential energy . . . . . . . . . . . . . . . . . . . . . . 15
3 Finite Element Method 16
3.0.2 Formulation of the Finite Element Method . . . . . . . . . . 17
3.1 Bubnov-Galerkin’s Approximation Method . . . . . . . . . . . . . . 17
3.1.1 The Stiffness Matrix and the Load vector . . . . . . . . . . . 19
3.2 Assembly of global stiffness matrix and force vector . . . . . . . . . 21
4 Linear Curved Beams 22
4.1 The Equilibrium Equations . . . . . . . . . . . . . . . . . . . . . . . 24
4.2 Normal stresses in Curved Beams . . . . . . . . . . . . . . . . . . . 25
4.3 Deflection of a Curved Beam . . . . . . . . . . . . . . . . . . . . . . 28
4.4 Shear Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
4.5 Finite Element Implementation . . . . . . . . . . . . . . . . . . . . . 32
4.5.1 Stiffness Matrix and Load Vector . . . . . . . . . . . . . . . 34
1
Contents4.5.2 Timoshenko beam stiffness matrix . . . . . . . . . . . . . . . 34
4.5.3 Euler-bernoulli beam stiffness matrix . . . . . . . . . . . . . 36
4.6 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . 39
4.7 Displacement Transformation Matrix . . . . . . . . . . . . . . . . . . 39
4.8 Common Boundary Conditions . . . . . . . . . . . . . . . . . . . . . 42
4.8.1 Numerical results of beams under different boundary conditions 44
4.9 Shear and Membrane Locking . . . . . . . . . . . . . . . . . . . . . 49
4.10 Numerical Test Examples . . . . . . . . . . . . . . . . . . . . . . . . 50
4.10.1 Example 1: An arc for various subtended angles. . . . . . . . 51
4.10.2 Castigliano’s Energy Method . . . . . . . . . . . . . . . . . . 53
4.10.3 Example 2 :A quarter-Circular Cantilever Beam . . . . . . . . 54
4.10.4 Results for Timoshenko curved Beam . . . . . . . . . . . . . 57
4.10.5 Results for the Euler-Bernoulli curved beam . . . . . . . . . . 58
4.11 Points to Note: Timoshenko curved beam strains definition . . . . . . 59
4.12 Summary and Conclusion . . . . . . . . . . . . . . . . . . . . . . . . 62
Appendix 64
.1 Matlab Programmes . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
.1.1 Common programmes and Parameters for both beam theories 64
.1.2 Curved Beam - Euler-Bernoulli beam theory . . . . . . . . . 66
.1.3 Curved Beam - Timoshenko Beam Theory . . . . . . . . . . 71
Declaration 79
2
List of Figures
2.1 Geometry of a simple Beam . . . . . . . . . . . . . . . . . . . . . . 11
2.2 Deformation of cross-section of an Euler - Bernoulli beam . . . . . . 13
2.3 Deformation of the cross-section of a Timoshenko beam . . . . . . . 15
4.1 Bending stress in curved beams . . . . . . . . . . . . . . . . . . . . . 23
4.2 Forces and displacements on an infinitesimal curved beam element . . 24
4.3 Geometry of a beam’s element vertical displacement of magnitude y . 25
4.4 (a.)Tangential displacement (b.)Radial displacement(c.)cross-sectional
displacements and rotations of the cross-sectional plane around the y-
axis (d)About the z-axis. . . . . . . . . . . . . . . . . . . . . . . . . 29
4.5 Geometry of a curved beam element with three degrees of freedom at
each node. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
4.6 curved beam element related to local and global displacements . . . . 40
4.7 straight beam element displacement . . . . . . . . . . . . . . . . . . 40
4.8 Relation of curved beam element in local and global coordinates system 41
4.9 Different beam supports and their reactions . . . . . . . . . . . . . . 44
4.10 Tangential displacement and Radial displacements for Euler -Bernoulli
and Timoshenko beams fixed at both ends . . . . . . . . . . . . . . . 45
4.11 Cross sectional Rotation for Euler -Bernoulli and Timoshenko beams
fixed at both ends . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
4.12 Tangential displacement and Radial displacements for cantilever Euler
-Bernoulli and Timoshenko beams subjected to external load of 500 . 46
4.13 Cross sectional Rotation for cantilever beam Euler -Bernoulli and Tim-
oshenko beams,subjected to external load of 500 . . . . . . . . . . . 47
3
List of figures
4.14 Tangential Displacement and Rotational Displacement for increased
external load from 100 - 500 of a beam clamped at both ends . . . . . 47
4.15 Cross sectional Rotation for increased external load from 100 - 500 of
a beam clamped at both ends . . . . . . . . . . . . . . . . . . . . . . 48
4.16 Displacement of an Euler- Bernoulli Beam and a Timoshenko Beam
that are both simply supported at both ends,subjected to external load
from 100 - 500. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
4.17 Cross sectional Rotation of an Euler- Bernoulli Beam and a Timo-
shenko Beam that are both simply supported at both ends, subjected
to external load from 100 - 500. . . . . . . . . . . . . . . . . . . . . 49
4.18 Effects of reduced membrane energy term integration for Euler-Bernoulli
beam, and reduced shear and membrane energy terms integration for
Timoshenko beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
4.19 An arc fixed on one end and supported on rollers and subjected to point
load on the other end . . . . . . . . . . . . . . . . . . . . . . . . . . 51
4.20 Radial displacement (a)With reduced integration, (b) Without reduced
integration for an arc fixed on one end and supported on rollers and
subjected to point load on the other end . . . . . . . . . . . . . . . . 52
4.21 A quarter-circular cantilever beam subjected to radial point load Q, at
the free end. (Lee, Sin [22]) . . . . . . . . . . . . . . . . . . . . . . 54
4.22 Free body diagram of a quarter-circular cantilever beam subjected to
radial point load Q, and we introduce axial dummy force Θ and a
dummy moment - m at the free end. . . . . . . . . . . . . . . . . . . 55
4.23 Maximum Absolute Error for Tangential, Radial displacements and
Cross sectional Rotation when the conventional strains of a Timoshenko
curved beam are used . . . . . . . . . . . . . . . . . . . . . . . . . . 61
4
List of Tables
4.1 Parameters used for a curved beam subtended by angle π . . . . . . . 44
4.2 Comparison of the finite element radial displacement with reduced in-
tegration for the Timoshenko beam and Euler-Bernoulli beam with, the
method by Saffari and Tabatabaei [2] . . . . . . . . . . . . . . . . . 53
4.3 Comparison of the two different finite element solution models loaded
with tip radial force, of a quarter circular cantilever beam with Cas-
tigliano’s energy solutions for Timoshenko beam. . . . . . . . . . . . 57
4.4 Comparison of the two different finite element solution models loaded
with tip radial force, of a quarter circular cantilever beam with Cas-
tigliano’s energy solutions for Euler-Bernoulli Beam. . . . . . . . . . 58
4.5 Comparison of the two different strain definition for fem solution mod-
els loaded with tip radial force, of a quarter circular cantilever beam
with Castigliano’s energy solutions for Timoshenko Beam. . . . . . . 60
5
6
Preface
I would like to express my sincere thanks and appreciations to my supervisor
Dr - ing. Joachim Linn for his great effort and a lot of time spent in introducing me into
this area of study and guiding me through. Without his undoubted advice and guidance,
it would have been impossible to get through this work.
I would like to thank my lecturers at TU - Kaiserslautern and TU - Eindhoven.
They kept us very busy, we barely noticed days passing by, but they tireless helped and
assisted us all the way.
I would like to thank the consortium for industrial mathematics for giving me
this chance to participate in the Erasmus Mundus programme. Indeed it is a privilege
to be a participant and the knowledge, and skills acquired during my two years of study
are without doubt enormous.
Many thanks to my friends in Eindhoven, and in Kaiserslautern. I wish to men-
tion especially Changgih, Henry and Vincent. Thanks for keeping the long hours in the
discussion groups.
Lastly, i thank my parents and my siblings, for their tireless and relentless love,
continuous support, and countless sacrifices they’ve had to make on my behalf. Thank
you for believing in me.
Tabitha W. Kinyanjui
November 22, 2007
Kaiserslautern
Chapter 1
Introduction
Beams are the most common structural components found in most mechanical struc-
tures. They can be defined as thin long structures that are capable of carrying loads in
flexure, and are characterized by
(a) The longitudinal dimension, which defines the axial direction of the beam is
considerably larger than the transverse dimensions, the dimension defining the
directions normal to the axial directions.
(b) If the beam is restricted to deform in plane, then it resists primarily loading
applied in one plane and has a cross section that is symmetric with respect to
that plane, hence it can be analyzed with two-dimensional idealizations [1].
A variety of practical engineering problems for example some of the structures such
as tyres, pipelines, rings, arc-like structures, just to mention a few however are mod-
eled by curved finite elements. A lot can be learned by studying the performance of
simple (essentially one-dimensional) models so as to gain insight into the more gen-
eral behavior of the overall structures. The circular ring serves as a simple yet realistic
model in addition for analyzing the effects of symmetry in structural systems elements.
Since the deformations are symmetric, a closed ring can be regarded as an example
of a curved beam with restrained sections or ends. From conditions of symmetry, the
distribution of stress in one quadrant is know to be the same as in another. Thus one
quadrant may be considered to be the curved beam in which the behavior of the ring
under load, at any section is to be found. Furthermore, curved beams are known to be
more efficient in transfer of loads than straight beams because the transfer is affected
by bending, shear, and membrane action (Saffari, Tabatabaei [2]).
Therefore the main objective in this thesis is
8
(a) Give a detailed derivation of equations of a curved beam in three dimesions, with
reference to the quoted literature, having first introduced a straight beam model
as a starting ground.
(b) To investigate the behavior of a plane curved beam, resulting from loading ap-
plied to the beam with cross-sections symmetric about that plane.
(c) The behavior of the beam under the assumption that there is a linear variation
of displacement over the cross-section, the Euler-Bernoulli beam theory. This
implies that the cross-sections which are normal to the centroid axis of the beam
before deformation remain plane, undeformed and perpendicular to the deformed
axis.This model accounts for bending moment effects on stresses and deforma-
tions. This implies that transverse shear forces are recovered from equilibrium
but their effect on beam deformations is neglected.
(d) Account for shear deformation by constraining the normal to remain straight,
but not necessarily normal, to the centroidal axis after shearing and bending;
Timoshenko beam theory.
(e) Implement the finite element method to model the numerical analysis of the be-
havior of the displacement of curved beams.
(f) To check how accurate the finite element we are going to use here, we com-
pare the finite element approximation solution of the analytical solution, and this
comparison is carried out with already existing finite element results in literature
which have been proven to be very accurate in their approximation. The exam-
ples we are going to use are the typical examples of the shear and membrane
locking. They are frequently dealt with to check the capability of analysis of the
elements developed and to see if the locking phenomena is property overcome.
Assumptions
The main assumptions are that
(a) The beam deformation is geometrically linear. Therefore the displacement gra-
dients are small compared to unity1, implying that the difference between the
deformed and the undeformed beam is very small.1
||G ||=‖ ∂U
∂ X‖=: ε ¿ 1
where U is the displacement vector, and X is position of material point p in a reference configuration with
respect to a given origin O.
9
(b) The beam is prismatic with constant circular cross section and constant radius of
curvature.
(c) We are neglecting changes in dimensions of the cross sections as the beam is
displaced.
10
Chapter 2
Straight Beam Theory
To set ground for curved beams, we first give an overview of the deformation of a
straight beam with a square cross section, while adopting definitions according to [3].
2.0.1 Geometry
h
b
q(x)
Cross − Section
Z
Y
X
L
Y
Figure 2.1: Geometry of a simple Beam
(a) The beam is long and thin, implying that L >> b and h.
(b) Loading is in the y− direction to obtain the bending moment.
(c) Constant cross section.
(d) The beam does not experience torsion.
2.0.2 Stresses
(a) No load applied in the z- direction implying the stresses are zero in that direction.
(b) The only significant stress are σxx and τxy.
11
2.1 Euler - Bernoulli Beam Theory
We define v the axial deflection and u as the transverse deflection of midplane.
After deformation,
(a) Plane sections remain plane and perpendicular to the midplane. For this to be
true, we require that
(i). The beam should be bent only with bending moments, therefore there
are no shear on transverse planes, as we will explain below;
(ii). The applied loads are such that no twisting occurs.
(b) In addition to this, we use the assumptions that the axis of the rod is inextensible.
(c) The Euler-Bernoulli beam theory assumes that the internal energy of a beam is
entirely due to bending strains and stresses, therefore the predominant stresses
and strains are σxx and εxx.
Therefore
εyy =∂u∂y
= 0 (2.1)
εxy =12
(∂u∂x
+∂v∂y
)= 0 (2.2)
Thus from (2.1) u is not a function of y, thus
u = u(x) (2.3)
This implies that every point of the cross section at a given location along the x−axis
has the same vertical displacement.(Timoshenko, Goodier [4])
Therefore
εxy = 0⇒ ∂v∂y
=−∂u∂x
=−dudx
(2.4)
Thus
v =−ydudx
=−yθ (2.5)
implying that the axial displacement vary linearly with the y distance from the neutral
axis. We also note that the equation (2.5) implies that the cross-sections’ rotational
(bending) angle directly depend on the gradient of deformation, i.e
θ =dudx
(2.6)
12
which along with (2.3) confirms the assumption (b) above that any cross-sectional
plane stays a plane and normal to the deformed longitudinal axis, though it displaces
and rotates a bit from its original position.
deformed state
undeformed state
θ
Normal to deformed beam axis, which is also
the direction of the deformed crosssections.
Normal to the reference beam
du/dx
y,u
x, v
u
Figure 2.2: Deformation of cross-section of an Euler - Bernoulli beam
2.1.1 Total potential energy
For a given loaded elastic body under geometrical constrains or boundary conditions,
the potential energy of the deformed body assumes a stationary value, and it attains
an absolute minimum when the displacement of the body are those of equilibrium
configuration. From (2.5), the axial strain distribution is
εxx =∂v∂x
=−y∂ 2u∂x2 =−y
d2udx2 (2.7)
Using the Hooke’s law, (Timoshenko,Goodier, [4]), the constitutive equation is thus
σxx = Eεxx =−Eyd2udx2 (2.8)
the stress distribution in terms of displacement field, where E is the Young’s modulus
of elasticity.
The resulting bending moment M is defined as
M =∫
A−yσxxdx = E
d2udx2
∫
Ay2dA = EIκ for κ =
d2udx2 (2.9)
Where I is the area moment of inertia∫
A y2dA a property of the beam that is used to
predict its resistance to bending and deflection, and κ is the curvature of the beam.
Since the internal energy for the Euler-Bernoulli beam accounts only for the bending
13
moment deformation, the total potential energy of the beam, subjected to force q acting
in the y− direction is
Wi =12
∫
VσxxεxxdV =
12
∫ L
0EI
(d2udx2
)2
dx (2.10)
We = −∫ L
0qudx
⇒Π = Wi +We =12
∫ L
0EI
(d2udx2
)2
−∫ L
0qudx
Where Wi and We are internal and external virtual work respectively and L is the length
of the beam.(Oden, [5])
2.2 Timoshenko Beam Theory
As cited by Felippa [1] the Timoshenko beam corrects the classical beam theory with
first-order shear deformation effects. Therefore the effect of the shear stresses on the
deformation are taken into account.
After deformation,
(a) Plane sections are assumed to remain plane and rotate about the same neutral
axis, with the assumption under (a) - (ii) in section 2.1 also holding for this case
too.
(b) Unlike the euler-Bernoulli beam, the cross sections do not remain perpendicular
to the deformed longitudinal axis.
Therefore
εxx =∂v∂x
=−ydθdx
(2.11)
εxy =12
(∂v∂x
+∂u∂x
)=
12
(−θ +
dudx
)=
12
ϒ (2.12)
Thus the deviation from the normal direction which is the difference between the nor-
mal to the longitudinal axis and the plane section rotation is the shear angle ϒ. It
consists of the contributions from the gradient of the deflection (the rotation of the
longitudinal axis), and the cross sectional rotation bending angle θ . This is further
illustrated in figure (2.3). For Timoshenko beam the cross section deformation is the
sum of two contributions that is one is due to bending, and the other is the shear defor-
mation.
14
du/dx
θ
ucross sectionDirection of deformed
x, v
y, u
Shear angleγ −
du / dx
Normal to defromed beam axis
Normal to the reference beam
Figure 2.3: Deformation of the cross-section of a Timoshenko beam
2.2.1 Total potential energy
From equations (2.11) and (2.12), the axial and shear strains for a Timoshenko beam
are
ε = −ydθdx
(2.13)
ϒ = −θ +dudx
(2.14)
Therefore
Π =12
∫
V(σxxεxx +σxyεxy)dV −
∫ L
0qudx (2.15)
=EI2
∫ L
0
(dθdx
)2
dx+κGA
2
∫ L
0
(−θ +
dudx
)2
dx−∫ L
0qudx
(2.16)
where κ is the shear correction factor, G the shear modulus,A is the cross sectional
area, and q is the external load applied to the beam, as has been explained by (Kwon,
Bang [6]).
15
Chapter 3
Finite Element Method
In this chapter we introduce the finite element numerical methods, that will be adopted
for the problems to follow.
Most numerical methods involve an approximation to an unknown function U by a
new function U , which is a linear combination of the basis functions
U =n
∑i=1
UiWi
where U are the unknown coefficients, Wi are the chosen weight functions. While
classical methods employ global basis functions of various frequencies, approximation
of this form will work adequately for simple geometries but yield poor approximations
for more complex geometries, as global basis functions cannot reflect properties of the
solution that are induced by the geometry.
Instead of employing global basis functions, the finite element method approximates a
solution for problems which are described by partial differential equations or formu-
lated as functional minimization, based on the concept of discretization of the solution
domain into sub-domains called finite elements. As summarized by (Nikishkov, [7]),
this can be achieved with the following general steps.
(a) Discretize the domain, that is to divide the solution region into finite elements.
The solution domain is divided into several simpler finite elements, where each
element has a simple geometry, so appropriate assumed solutions can easily be
written for the element.
(b) Selection of interpolation functions to describe solution over an element is the
next step.
16
If we use a polynomial function to interpolate U within elements from nodal
values of U , then for U to approach the exact values as the finite element mesh
is repeatedly refined and if α is the highest derivative of U , then
(i) Within each element, the assumed field for U must contain a complete
polynomial of degree α or higher,
(ii) Across the elements’ boundaries, there must be a continuity of order
α−1 for U and its derivatives.
(c) Establish the matrix equation for the finite element which relates the nodal values
of the unknown function to other parameters.Different approaches used are for
example the most the variational approach and the Galerkin method.
(d) To find the global equation system for the whole solution, all element equations
must be assembled. This involves the combination local element equations for
all elements used for discretization. Element connectivities are used for the as-
sembly process. Before solution, boundary conditions (which are not accounted
in element equations) should be imposed.
(e) Solving the global equation system. Since the finite element global equation
system is typically sparse, symmetric and positive definite, direct and iterative
methods can be used for solution. The nodal values of the sought function are
produced as a result of the solution,since approximating functions are determined
in terms of nodal values of as physical field which is sought.
3.0.2 Formulation of the Finite Element Method
As we have mentioned in (c) above, several approaches can be used to transform
the physical formulation of the problem to its finite element discrete state. If the
physical formulation of the problem is described by a differential equation then
the most popular method of its finite element formulation is the Galerkin method.
If the physical problem can be formulated as minimization of a functional then
variational formulation of the finite element equations is usually used.
3.1 Bubnov-Galerkin’s Approximation Method
The formulation given here is standard, and can be found in many finite element
resources. The construct of the finite element given below is in reference mainly
from (Hughes, [8]); Also we refereed to approach used by (Nikishkov, [7]); (
17
Strang et al., [9]); With a general example, let the problem to be solved in one
dimensional be defined by the equation
L u = f in Ω (3.1)
u|Γ1= uρ on Γ1
upα p|Γ2= uβ on Γ2
for L a given operator acting on functions u ∈ Lp(Ω), such that the functions
satisfy in some sense the boundary conditions. The fundamental question is then:
To match such a space of functions u with a class of inhomogeneous terms f , in
such a way that to each f , there corresponds one and only one solution u.
Let the space
δ = u | u ∈ Hs,u|Γ1= uρ (3.2)
be defined as a function space for trail solutions u.
Define a second collection of functions know as test functions such that we re-
quire that they vanish at boundaries corresponding to the essential boundary con-
dition uρ . That is
V = ω | ω ∈ Hs,ω|Γ1= 0 (3.3)
and
L : H s → (H s)∗, and f ∈ (H s)∗ the dual space of H s. (3.4)
Then the weak formulation of the problem is to find u ∈ δ , such that for all
ω ∈ V ,
a(u,ω) = ωT L u = (ω, f ) (3.5)
Where a : δ × V → R is a bilinear operator linear in its second argument. The
∂upα p for any α with p α p ≤ s exist in the weak sense, and derivatives of small
order needn’t exist, like in the classical sense.
The Galerkin’s method with piecewise polynomial subspaces δ h and V h is the
finite element method.
Let the approximations of sets δ and V , denoted by δ h and V h respectively, be
such that
δ h ⊂ in δ (i.e.; if u ∈ δ h, then uh ∈ δ ) (3.6a)
V h ⊂ in V (i.e.; if ω ∈ V h, then ωh ∈ V ) (3.6b)
18
Consequently from (3.1)
uh|Γ1
= uξ and ωh|Γ1
= 0 (3.6c)
since this is an essential boundary condition then it must be satisfied by every
function ωh Implying that the functions ωh, should be zero at that boundary.
Given the collection V h, then to each member νh ∈ V h, we construct a function
uh ∈ δ h, by
uh = νh +uhρ (3.7)
where uhρ = uρ . Therefore δ h is all functions of the form (3.6a). The problem is
now defined as:
Given f ,uρ , and uβ , find uh = νh +uhρ , where νh ∈ V such that for all ωh ∈ V h,
a(ωh,νh) = (ωh, f )+ωh|Γ2
uβ −a(ωh,uhρ) (3.8)
uhρ and V h have to be explicitly defined. Note that the approximate solution is
required to fulfil the essential boundary conditions, but not the natural boundary
condition, which doesn’t prevent the approximations form converging in the Hs
norm, to a solution u, which does satisfy upα p|Γ2= uβ .
3.1.1 The Stiffness Matrix and the Load vector
The Galerkin method leads to a coupled system of linear algebraic equations.
Every test function νh is determined by its nodal parameters, which are the un-
knowns ci of the discrete problem. Each of these nodal parameters is the value
at a given node, of either the function or its derivatives.
Let V h consist of all linear combinations of given functions which are denoted
by φi : = Ω → R, for i = 1,2, ...,n. Then if ωh ∈ V h, then there exists constants
ci such that
ωh(x) =n
∑i=1
ciφi(x) (3.9)
for ci = wh(xi)
for ci = wh(xi), the nodal values for wh(x),
φ j(xi) =
1 if i = j
0 if i 6= j i, j = 1, ...,n
19
and
(φi)|Γ1= 0, i = 1,2, ...,n
Thus V h is a linear space of dimensions n with the basis φini=1.
To define members of δ , we need to specify uξ .
Defining another shape function as φn+1 : = Ω → R, with the property that
(φn+1)|Γ1= 1, φn+1 /∈ V h, then
uhρ = uφn+1
and thus
uhρ|Γ1
= uρ
For uh ∈ δ h, and constanst di, i = 1,2, ...,n
uh = νh +uρ (3.10)
=n
∑i=1
diφi +uρ φn+1
Substituting (3.9) and (3.10), into (3.8), then
a
(n
∑i=1
ciφi,n
∑j=1
d jφ j
)=
(n
∑i=1
ciφi, f
)+
[n
∑i=1
ciφi|Γ1
]uβ −a
(n
∑i=1
ciφi,uρ φn+1
)
(3.11)
which simplifies to
n
∑i=1
ciDi = 0 (3.12)
For
Di =n
∑j=1
a(φi,φ j)d j− (φi, f )−φi|Γ1uβ +a(φi,φn+1)uρ (3.13)
Since the solution hold ∀ ωh ∈ V h, and ∀ cini , i = 1, ...,n, arbitrary in (3.12),
thenn
∑j=1
a(φi,φ j)d j = (φi, f )+φi|Γ1uβ −a(φi,φn+1)uρ (3.14)
for each Di, i = 1, ..,n.
Since everything is know besides d j, then (3.14) constitute a system of n equa-
tions, in n unknowns. Let
Ki j = a(φi,φ j) (3.15a)
20
Fi = (φi, f )+φi|Γ1uβ −a(φi,φn+1)uρ (3.15b)
Then (3.14) while adopting a matrix notation can compactly be restated as solv-
ing for d in the algebraic equations defined as
Kd = F (3.16)
3.2 Assembly of global stiffness matrix and force
vector
The elements stiffness must now be assembled into the beam’s stiffness, imply-
ing now that to add values of element stiffnesses in global axes, we need to adjust
the numbering into a numbering system of the beam as a whole. This particular
task is accomplished by the location matrix.The location matrix (if we denote it
by loc) dimensions are the number of element nodes by the number of elements.
Thus give a particular degree of freedom, and an element number, for example
i and j respectively, then the value returned by the location matrix array is the
corresponding global equation number A such that
A = loc(i, j) =
j if i = 1
j +1 if i = 2
j +2 if i = 3...
j +nel −1 if i = nel
for j = 1,2, ...,nel
Therefore the contribution of an element to the beams’ global stiffness is ob-
tained by adding to the beams’ global stiffness matrix term in row loc(i), and
column loc( j), the term Kloc(i),Kloc( j), the element stiffness term K(e)i, j . This ap-
plies to the force vector too, where the representation of the global force vector
is now
Floc(i) = Floc(i) +F(e)loc(i) (3.17)
In the next two subsections, we are going to discuss a plane two-dimensional, lin-
ear Euler-Bernoulli beam finite element, and Timoshenko beam element. From
the definitions of the potential energy for each beam,we derive the displacement
formulation on the basis of the galerkin method already described.
21
Chapter 4
Linear Curved Beams
In our discussion in the previous chapters, the deformation of the beam subjected
to a load, has been restricted to beams with the assumption that the longitudinal
elements have the same length. The xy- longitudinal plane has been assumed to
be the plane of symmetry with the load applied in this plane. This has restricted
the beam theory formulated above to initially straight beams of constant cross
section.
Although considerable deviations from this restriction can be tolerated in real
problems, when the initial curvature of the beams becomes significant, the linear
variations of strain over the cross section is no longer valid, even though the
assumption of plane cross sections remaining plane is valid.
The cross section of part of an initially curved beam is described by the figure
(4.1), with xy plane the plane of symmetry.
The radii R references the location of the centroid of the cross sectional area; Rn
references the location of the neutral axis; and r references some arbitrary point
p of area element dA on the cross section.
For curved beams, the longitudinal deformation of any element will be propor-
tional to the distance of the element from the neutral surface, implying that the
total deformations are proportional to the distance from the neutral axis. This
implies then that the linear variation of the strain over the cross section is no
longer valid as the beam’s elements are not of equal length, even though the as-
sumptions of the plane cross section remaining plane after deformation is valid,
as we assumed for the straight beam case. This citation can be found at univer-
sity of Washington’s department of mechanical engineering website [13].
Since the stress during deformation is proportional to the strain of the beam, then
22
Cross sectional rotations∆θ =
M = Bending moments
r = Radius of general fiber in the beam
= Radius of Neutral axis n
R
R = Radius of curvature
− r n R
∆θ
∆θ
p
A’
A
A’
p
A
M
r
n R R
Stress distribution
Centroidal Axis
Neutral Axis
Figure 4.1: Bending stress in curved beams
the elastic stress of a curved beam is not promotional to the distance of the el-
ement from the neutral axis. For the same reason, the neutral axis in a curved
beam does not pass through the centroid section.
We define the geometry of the curved beam with reference to (Oden, [5]). Con-
sider a beam, having a constant cross section and a constant radius of curvature
in the plane of bending.
The geometry of the curved beam is defined by establishing two orthogonal co-
ordinates systems: one fixed coordinate system (x,y,z), with its origin located
on a convenient cross-section, and another curvilinear system (s,y,z) in which s
is tangent to the curve, y is a radial coordinate directed toward the center of the
geometric axis, and z is directed normal to the plane of the beam.
The assumptions taken are that,
(a) The beam material is homogeneous, isotropic and linearly elastic.
(b) The beam axis always lies in the plane of bending.
(c) The Radius of curvature is constant and the cross section is constant and has
symmetry about the plane of curvature, that is xy-plane.
23
(d) External loads act in the plane of curvature, therefore the plane of bending and
the plane of curvature are the same.
(e) Plane sections remain plane, and the cross section does not deform.
We note that these assumptions are almost the same as the ones used for straight
beams only that the beam is initially curved in this case.
y
y
s s
z
z
s M + s
s + R
y V
M
N M
y V ∆ + y V
M ∆ + y M
∆ N
s M ∆ M
s∆ s
y
z
s
z V ∆ + z V
z M ∆ + z M
M
V
Figure 4.2: Forces and displacements on an infinitesimal curved beam element
4.1 The Equilibrium Equations
With reference to figure (4.2), the equilibrium equations at a point p in the de-
formed beam element have been derived and found to be
dNs
ds− Vy
R+q1 = 0 (4.1)
dVy
ds+
Ns
R+q2 = 0
dVz
ds+q3 = 0
For the equilibrium of the moments we get
dMs
ds− My
R+m1 = 0 (4.2)
dMy
ds+
Ms
R−Vz +m2 = 0
dMz
ds+Vy +m3 = 0
24
Where Ns,Vy,Vz are Normal force along s, and shear forces in in y and z direction,
while My,Mz are bending moments around y and z axis, and Ms is the twisting
moment around the longitudinal axis, and qi,mi, i = 1, ...,3 are external
loads and moments applied in the tangential, radial direction and z direction
respectively.
Detailed derivation are derived by (Oden, [5]), (Blake [14], pg. 575 ) and (
Lebeck, Knowlton, [15]) derives the equilibrium equations in polar coordinates.
4.2 Normal stresses in Curved Beams
We follow the derivation given by (Oden, [5], pg. 64). Following similar deriva-
tion is (Wallerstein, [16], pg. 99).
To be able to find the displacement of the beam, we find the stress-strain rela-
tions,assuming that the material is elastic and obeys the Hooke’s law.
First the derivation of the pure bending of the curved beam is presented, implying
that we use the assumption that plane sections normal to the beam’s axis before
deformation remain plane and normal to this axis after deformation.
With this assumption,the components of displacement in a direction normal to a
cross section must satisfy the equation of a plane
u = c0 + c1y+ c2z (4.3)
where c0, c1, c2 are constants.
Let the geometry of element between the cross-sections of the rod be as in the
diagram below.
R
y
s − y
s
∆
∆
Figure 4.3: Geometry of a beam’s element vertical displacement of magnitude y
Materials located at a radial distance y from the axis are of different length ∆sy.
Thus taking limit as ∆s approaches zeros,dsdsy
=1
1− y/R
25
From strain-displacement formula,
εs =∂v∂ sy
(4.4)
Substituting this to (4.3), then
εs = c0dsdsy
+ c1dsdsy
y+ c2dsdsy
z
=dsdsy
(c0 + c1y+ c2z)
for
c0 =dc0
ds, c1 =
dc1
ds, c2 =
dc2
ds
Implying that
εs =1
1− yR
(c0 + c1y+ c2z) (4.5)
Thus the normal stress is
σs = Eεs =E
1− yR
(c0 + c1y+ c2z) (4.6)
To determine c0,c1,c2 then:
Ns =∫
AσsdA = c0E
∫
A
11− y
RdA+ c1E
∫
A
y1− y
RdA+ c2E
∫
A
z1− y
RdA(4.7)
Mz =∫
Aσs ydA = c0E
∫
A
y1− y
RdA+ c1E
∫
A
y2
1− yR
dA+ c2E∫
A
yz1− y
RdA
My =∫
AσszdA = c0E
∫
A
z1− y
RdA+ c1E
∫
A
yz1− y
RdA+ c2E
∫
A
z2
1− yR
dA
The inertia terms, also derived by (A.O.Lebeck, J.S.Knowlton, [15]) can be de-
fined as
Iy =∫
A
z2
1− yR
dA (4.8)
Iyz =∫
A
yz1− y
RdA
Iz =∫
A
y2
1− yR
dA
We note here when the dimensions of the cross section are small compared to
the radius of curvature of the longitudinal axis, that is y << R, then the term y/R
is negligibly small, and therefore the inertia terms are the same as for a straight
26
beam.
Simplifying further,
∫
A
11− y
RdA =
∫
AdA+
1R
∫
AydA+
1R2
∫
A
y2
1− yR
dA (4.9)
= A+1R
∫
AydA+
1R2 Iz
∫
A
z1− y
RdA =
∫
AzdA+
1R
∫
A
yz1− y
RdA =
∫
AzdA+
1R
Iyz
∫
A
y1− y
RdA =
∫
AydA+
1R
∫
A
y2
1− yR
dA =∫
AydA+
1R
Iy
Since the origin of the coordinate system is at the centroid of the section,∫
AydA =
∫
AzdA = 0
Therefore equations (4.7) reduces to
Ns
E=
(A+
Iz
R2
)c0 +
Iz
Rc1 +
Iyz
Rc2 (4.10)
Mz
E=
Iyz
Rc0 + Iyzc1 + Iyc2
My
E=
Iz
R− c0 + Izc1 + Iyzc2
Solving the equations,
Ec0 =Ns
A− Mz
AR(4.11)
Ec1 =MzIy−MyIyz
IyIz− I2yz
− Ns
RA+
Mz
AR2
Ec2 =MzIy−MyIyz
IyIz− I2yz
Therefore we get the normal stress as
σs =Ns
A− Mz
RA+
MzIy−MyIyz
IyIz− I2yz
y1− y
R+
MzIy−MzIyz
IyIz− I2yz
z1− y
R(4.12)
The term
Ns
A− Mz
RA
represents a uniform normal stress over the section, while the rest of the terms
represent a non uniform distribution of stress. Lack of symmetry is represented
by the terms Iyz, which we will equate to zero in this case. We refereed to
(Oden, [5], Wallerstein, [16]).
27
4.3 Deflection of a Curved Beam
Having noted down the formulation of the normal stress, the corresponding
strains, are formulated, so as to calculate the displacements of the beam after
deformation through the strain-displacement relations.
Let any point on the cross-section of the beam be displaced from its unstrained
position through small components of displacement v and u representing the tan-
gential and radial displacements of points on the cross section, and w represent
the displacement in a direction normal to the plane of the rod, with the previous
assumptions that plane cross sections remain plane after deformation still hold-
ing.
From (4.4),
εs =∂v∂ sy
(4.13)
The displacements can be described as in the following figure (4.4) described as
in (Oden, [5], pg 132)
(a.)Tangential displacement
(b)Radial displacement
(c.)cross-sectional displacements and rotations of the cross-sectional plane
around the y-axis
(d)About the z-axis.
Using the principle of superposition we can get the overall tangential strain
(4.13) as follows.
From (a), we can see that the strain as a result of tangential displacement of the
centroidal axis, while ignoring higher order terms is just the change of v, along s
dvds
(4.14)
The radial strain is as a result of the difference between the original length of ma-
terial at the neutral axis to a distance y below the axis. see (Timoshenko,Goodier,
[4]). Therefore the radial strain is
∆s[1− y/(R−u)]−∆s(1− y/R)∆s(1− y/R)
(4.15)
=−y
1− y/R
(1
R−u− 1
R
)− u
R
28
(d)(c)
(b)(a)
s ∆ s 2
w/d2
dw/ds + d dw/ds
s∆
z
z dw/ds
R − u
s ∆ 2
u/ds2
du/ds + ddu/ds
y du/ds
s∆
y
u
s − u∆
s∆ s
s ∆
∆
v + dv/ds v
Figure 4.4: (a.)Tangential displacement (b.)Radial displacement(c.)cross-sectional
displacements and rotations of the cross-sectional plane around the y-axis (d)About
the z-axis.
The strains from (c) and (d) are as a result of the rotation of sections owing to
change in u and w, thus from the diagrams (c)and(d), the rotations of the cross-
sections is
−yduds− z
dwds
Where the strain is calculated to be
∂∂ s
(−du
ds− z
dwds
)dsdsy
=1
1− y/R
(−y
d2uds2 − z
d2wds2
)(4.16)
Thus adding (4.14), (4.15) and (4.16), then we get
dvds
− uR−
(d2uds2 +
uR2
)y
1− yR− d2w
ds2z
1− yR
(4.17)
=σs
E
=Ns
AE− Mz
RAE+
MzIy−MyIyz
E(IyIz− I2
yz) y
1− yR
+MyIz−MzIyz
E(IyIz− I2
yz) z
1− yR
Equating the like y/(1− y/R) and (z/(1− y/R)) on each side we get the equa-
tions of the elastic curve
dvds− u
R=
Ns
AE− Mz
RAE(4.18)
29
−d2uds2 −
uR2 =
MzIy−MyIyz
E(IyIz− I2
yz) (4.19)
−d2wds2 =
MyIz−MzIyz
E(IyIz− I2
yz) (4.20)
Note The last equation describing the out of plane deformation is not considered
here as we are dealing with plane deformation, as we will see below.
4.4 Shear Deformation
Shearing stress for a symmetric circular curved bar is given in (Oden, [5]) as
τys =Vy
b(1− yR )
(Qz
Iz− A′
AR
)(4.21)
From Hooke’s law, we have the relation
γys =τys
G=
Vy
b(1− yR )G
(Qz
Iz− A′
AR
)(4.22)
Where
Q =∫
A
y1− y/R
dA
A′, element cross sectional area, with dimensions b of A′ parallel to the z- axis.
With reference to figure (4.4) the normal displacement due to the rotations of
the cross-sections, owing to change in u and w, about the y axis, as a result of
change in curvature, produces the displacement
−yduds− z
dwds
Since we assumed that there are no out-of -plane displacement (in the z - direc-
tion), then w = 0. Therefore the displacement is
v =−ydub
ds
for ub the deflection due to bending.
Thus
Vy
b(1− yR )G
(Qz
Iz− A′
AR
)=−y
∂ 2ub
∂ s∂y− ∂ub
∂ s+
∂u∂ s
(4.23)
30
The first term on the right side represent a slight warping of the cross section due
to shear deformation. On the assumption that planes remain plane, for which τys
was derived, then this term is small and can be neglected.
Vy
b(1− yR )G
(Qz
Iz− A′
AR
)=−dub
ds+
duds
(4.24)
But the left side of the equation is a function of both s and y. This contradicts the
assumption that planes remain normal after deformation, as ub is a function of s
only.
To remedy this we replace τsy by uniform stress distribution and account for its
variation with y by introducing a shear correction factor κ = β − β . Thus (4.24)
gives the shear deformation. That is
−dub
ds+
duds
= κVy
AG(1− yR )
(4.25)
Where
β =QAIzb
and β =ARb
−θ +duds
=κ
AG
(Vy
1− yR
)(4.26)
If we assume that the cross section is symmetric with respect to y, then Iyz. My,
and w are equal to zero, because of our choice of external loading to be in the
xy plane. The strain-displacement equations describing the deformation of a
shearable curved beam reduces to
EA(
dvds− u
R
)= Ns− Mz
R(4.27a)
EIz
(−d2u
ds2 −u
R2
)= Mz (4.27b)
AGκ
(−θ +
duds
)=
(Vy
1− yR
)(4.27c)
Where EIz the bending rigidity, G is the shear modulus of elasticity, A is the
cross sectional area of the curved element, and κ the shear correction factor.
The correction factor arise from the fact that by admitting a nonzero shearing
strain, and with the assumption of plane sections remaining plane leads to a
31
constant shear stress through the curved beam thickness. Therefore a shear cor-
rection factor is used to account for the difference in the constant state of shear
stress and the more realistic parabolic variation of shear stress. We notice the
difference between the displacement for a straight beam, and a curved beam is
that
(a) Due to the initial curvature of the beam, the terms are multiplied by 1/R,
where R is the radius of curvature.
(b) The tangential deformation, the bending and the rotation are intrinsically
coupled, unlike the case for straight beams, hence efficient transfer of loads
than for a straight beam because the transfer is affected by bending, shear,
and membrane action.
(a) However, since we are considering beams, whose longitudinal direction is
much larger compare to transverse directions, then the term y/R found in
the inertia terms is negligible, thus the inertia terms are the same as for the
straight beams.
4.5 Finite Element Implementation
Once again, the element kinematics of a plane beam is completely defined if
the following functions are given: the axial displacement v1, the transverse dis-
placement u2 and the cross section rotation θ . The geometry of the curved beam
element in figure (4.5) is as described by (Friedman, Kosmatka, [17])
At a typical node, three variables di = vi,ui,θi are used to define the element
behavior. The corresponding tangential ε , bending χ and shear strains γ respec-
tively are in the form:
ε =dvds− u
R(4.28a)
χ = −d2uds2 −
dvRds
(4.28b)
γ = −θ +duds
(4.28c)
1The use of axial and tangential displacements are used interchangeably, and is assumed to mean the
same displacement2The use of the transverse and Radial displacements are used interchangeably, and is assumed to refer to
the same displacement
32
u y,
v s,
R 1 u
1v
2 θ
2 u 2 v
1θ
Figure 4.5: Geometry of a curved beam element with three degrees of freedom at each
node.
For a given loaded elastic body under geometrical constrains or boundary condi-
tions, the potential energy of the deformed body assumes a stationary value, and
it attains an absolute minimum when the displacement of the body are those of
equilibrium configuration. We can find the total potential energy of the element
by thus
Π = UM +UB +US−∫ L(e)
0(q1v+q2u+mθ)ds (4.29)
where UM is the membrane energy, UB the bending strain energy, and US the
shear strain energy, and q1, q2 and m the external tangential and radial loads
respectively, and the distributed moment, while L(e) is the length of the element.
see (Seok-Soon Lee, [18]).
UM,UB, and US are expressed by
UM =∫ L(e)
0
12
EA(
dvds− u
R
)2
ds (4.30a)
UB =∫ L(e)
0
12
EIz
(−d2u
ds2 −dv
Rds
)2
ds (4.30b)
US =∫ L(e)
0
12
κGA(−θ +
duds
)2
ds (4.30c)
33
4.5.1 Stiffness Matrix and Load Vector
If we let the displacement vector be U , then following the Galerkin method
already introduced in section one, then we can define the finite element approxi-
mation of the displacements and the rotation of the cross sections with the finite
element shape functions φ j as follows
U hi =
nnd
∑j=1
φ jUh
i j
θ hi =
nnd
∑j=1
φ jθ hi j
where nnd is the number of nodes in the discretization.
4.5.2 Timoshenko beam stiffness matrix
With reference to (2.13), we saw that the cross sectional rotation for a Timo-
shenko beam are independent of the vertical displacement u, therefore they can
be interpolated independently,implying that the degrees of freedom are v, u and
θ . Thus the axial, bending and shear strains are
ε =dvds− u
R(4.31a)
χ = −dθds− dv
Rds(4.31b)
γ = −θ +duds
(4.31c)
The highest derivative in the displacements for the total potential energy func-
tional in (4.29) with the strains as defined in (4.31) is C(1) continuous, therefore
we only need C(0) continuous interpolation functions of the Timoshenko curved
beam, Thus for the shape functions φ = φ ,
v
u
θ
=
φ1 0 0 φ2 0 0
0 φ1 0 0 φ2 0
0 0 φ1 0 0 φ2
v1
u1
θ1
v2
u2
θ2
34
Introducing the non-dimensional coordinate system formulation
−1≤ ξ ≤ 1 3 such that
ξ =xL
(4.32)
then
φ1 =12
(1−ξ ) (4.33)
φ2 =12
(1+ξ )
Defining the vector of the unknown values as d, and since we have three degrees
of freedom per node, that is v,u and θ then we can write the axial, bending, shear
strains as
ε = Bad
χ = Bbd
γ = Bsd
respectively, for the axial, bending and shear - displacement matrices
Ba =[
φ ′1 − φ1R 0 φ ′2 − φ2
R 0]
Bb =[
φ ′1R 0 φ ′1 0 φ ′2
R φ ′1]
Bs =[
0 φ ′1 −φ1 0 φ ′2 −φ2
]
Therefore the element stiffness matrix is a contribution from the membrane,
bending, and the shearing stiffness such that
[Ke] = [Ka]+ [Kb]+ [Ks] (4.34)
where
[Ka] =dT
2
∫ +1
−1BaT
(EA)BadL(e)
2dξ
[Kb
]=
dT
2
∫ +1
−1BbT
(EI)BbdL(e)
2dξ
[Ks] =dT
2
∫ +1
−1BsT
(κGA)BsdL(e)
2dξ
3The global and local domains description can be related by the mapping ξ : [xi,xi+1]→ [ξ1,ξ2] : ξ (xi) =
ξ1, and ξ (xi+1) = ξ2, for example for a linear two noded element. Thus in general, for −ξ ≤ 1≤ ξ , and an
n-noded element, the lagrangian shape functions of n−1 degree is
φi(ξ )(e) =n
∏j=1, j 6=i
(ξ −ξ j
ξi−ξ j
)
35
The load vector is
F(e) = ft =
dT ∫ +1−1 qi φ j
L(e)
2 dξ t = 3 j−3+ i
dT ∫ +1−1 m φ j
L(e)
2 dξ t = 3 j for 1≤ j
and d is
d =
v1
u1
θ1
v2
u2
θ2
where qi represents external tangential and radial loads q1,q2, and m the external
moment
The stiffness matrix (4.34) represents the Timoshenko curved beam, with the
inclusion of shear strain.
4.5.3 Euler-bernoulli beam stiffness matrix
For the Euler-Bernoulli curved beam, shear strain is neglected, and therefore,
since we are assuming the beam is inextensional, the relevant strains in this case
are the axial strain ε , and the bending strain, χ , where the bending strain now
depends only on the radial displacement, u.
ε =dvds− u
R(4.35a)
χ = −d2uds2 −
uR2 (4.35b)
In choosing the shape for the distribution of the unknown displacement, the
geometry of the element is defined using the nodal coordinates and the same
shape functions which are used to interpolate the displacement. Since in Euler-
Bernoulli theory we have the normality assumption, cross-sectional rotations di-
rectly depend on the gradient of the deflection, the vertical displacement of a
beam must have continuous slope as well as continues deflection at any neigh-
boring beam elements. The bending of the beam has a second derivative term
in the bending strain, thus we must have C(1) continuity of u and its derivative,
hence C(1) continuous elements.
For the tangential displacement v, the highest derivative is first order, so we only
36
need C(0) continuous interpolation functions. Therefore the simplest approxi-
mation of u and v is
v(s) = b0 +b1s (4.36)
u(s) = c0 + c1s+ c2s2 + c3s3
duds
(s) = c1 +2c2s+3c3s2
for constants b0,b1,c0,c1,c2,c3.
The Euler-bernoulli beam element thus has v, and u as degrees of freedom. Cross
sectional rotations depend on the gradient of deflection, therefore θ is not a free
parameter, and hence we evaluate the deflection u and its slope du/ds at both
nodes,
u(0) = c0 = u1 (4.37)duds
(0) = c1 =du1
dsu(L) = c0 + c1L+ c2L2 + c3L3 = u2
duds
(L) = c1 +2c2L+3c3L2 =du2
ds
Solving (4.37) for the constants ci in terms of ui and dui/ds and substituting into
(4.36), and for shape functions φi = φi we get
u(s) = φ1u1 + φ2du1
ds+ φ3u2 + φ4
du2
ds(4.38)
Re-writing in non-dimensional coordinate system formulation, then we have
u(ξ )(e) = φ1(ξ )u(e)1 + φ2(ξ )
L(e)
2
(duds
)(e)
1+ φ3(ξ )u(e)
3 + φ4(ξ )L(e)
2
(duds
)(e)
2(4.39)
where
4
∑i=1
φi
are cubic hermitian shape functions, as below:
φ1 =14
(2−3ξ +ξ 3) (4.40a)
φ2 =14
(1−ξ −ξ 2−ξ 3) (4.40b)
φ3 =14
(2+3ξ −ξ 3) (4.40c)
φ4 =14
(−1−ξ +ξ 2 +ξ 3) (4.40d)
37
Thus for functions φi = φi defined in (4.33), and hermitian shape functions φi =
φi defined in (4.40) respectively, we can represent v,u and du/dx as
v
u
du/ds
=
φ1 0 0 φ2 0 0
0 φ1 φ2L2 0 φ3 φ4
L2
0 φ ′1 φ ′2L2 0 φ ′3 φ ′4
L2
v1
u1
(du/ds)1
v2
u2
(du/ds)2
Therefore the axial strain displacement matrix is
Ba =[
φ ′1 − φ1R − φ2
R φ ′2 − φ3R − φ4
R
]
⇒ [K(e)a ] =
dT
2
∫ +1
−1BT
a (EA)Ba dL(e)
2dξ (4.41)
and the bending strain displacement matrix is
Bb =[− φ ′1
R −φ ′′1 −φ ′′2 − φ ′2R −φ ′′3 −φ ′′4
]
⇒ [K(e)b ] =
dT
2
∫ +1
−1BT
b (EI)Bb dL(e)
2dξ (4.42)
Hence the element stiffness matrix is thus
[Ke] = [Ka]+ [Kb] (4.43)
=dT
2
∫ +1
−1BT
a (EA)BadL(e)
2dξ +
dT
2
∫ +1
−1BT
a (EI)BadL(e)
2dξ
While the load vector is now
[Fe] = dT∫ +1
−1q1
φ1
0
0
φ2
0
0
L(e)
2dξ +dT
∫ +1
−1q2
0
φ1
φ2
0
φ3
φ4
L(e)
2dξ
where d in this case is
d =
v1
u1
du1/ds
v2
u2
du2/ds
38
4.6 Numerical Integration
Evaluation of the integrals needed to determine the element characteristics can
be tedious and possibility of having so many errors is high, therefore we have
adopted Gauss-Legendre numerical integration rules through out.(Owen, [10])
This can be in short defined as follows:
If we have a function f (ξ ), and we wish to calculate the integral
I =∫ +1
−1f (ξ )dξ (4.44)
for pre-determined sampling position ξ = ξ , we sample the function f (ξ ), and
multiply the results by some predetermined weights Wi. Therefore for a p - point
rule,
Ip = W1 f (ξ )+W2 f (ξ )+ ...+Wp f (ξ ) (4.45)
We note that an n- point rule can integrate polynomial functions of degree 2n−1
or less exactly. Tables of sampling positions and weights can be found in many
literatures. see for example (Stewart, pg. 169 [11]).
4.7 Displacement Transformation Matrix
Since we have used a rectangular cartesian coordinates system, it is necessary to
relate the displacements defined along the local element axes to a global coor-
dinate system common to every element in the beam. 4 Let the displacements
of the beam in the local and global coordinate systems be v, u, θ and X , Y, θrespectively, as shown in the figure (4.6)
Then to determine the transformation matrix for the curved beam element, we
first consider a transformation matrix in two dimensions of straight beam ele-
ment, of length L. If the lateral and normal displacements are ν and υ , and if the
angle that the element makes with the X direction is α , as shown in figure (4.7),
then the relationship between the displacements ν ,υ , θ and X , Y, θ is known
4If elements matrices are derived based on the local polar coordinate system instead of the local Carte-
sian one, then the matrices’ coefficients are invariant for any curved beam element with constant radius of
curvature and subtended angle. Therefore there is no need for transformation.
39
x
y
θ
u
v
Figure 4.6: curved beam element related to local and global displacements
to be expressed by the well known equations see (A.H. Love, [21], §253) as
ν1
υ1
θ1
ν2
υ2
θ2
=
cosα sinα−sinα cosα
1
cosα sinα−sinα cosα
1
X1
Y1
θ1
X2
Y2
θ2
or more compactly
ds = S dg (4.46)
where ds, dg are the local straight beam displacement vector, and global coordi-
nate displacement vector, and S is the transformation matrix above. Now if we
x
y
ν υ
θ
α
Figure 4.7: straight beam element displacement
consider the relationship of the displacement of the curved beam element v, u, θ
40
with the displacements of the straight beam elements ν , υ , θ , then as described
in figure (4.8), see (Kikuchi, [22]), (Just, [23])
v u
ν
curved beam element
θ
υ
ψ / 2
Figure 4.8: Relation of curved beam element in local and global coordinates system
we can therefore relate the displacements of the curved beam element to those
of the straight beam element through the equations
v1
u1
θ1
v2
u2
θ2
=
cos(ψ) −sin(ψ)
−sin(ψ) cos(ψ)
1
cos(ψ) sin(ψ)
−sin(ψ) cos(ψ)
1
ν1
υ1
θ1
ν2
υ2
θ2
That is
dc = C ds (4.47)
where now dc is the displacement vector for a curved beam in the local coordi-
nate system, and C is the transformation matrix above.
Thus, from (4.46) and (4.47), the relation of the nodal displacement of the
curved beam in the local system dc and those in the global system dg is
dc = C S dg = T dg (4.48)
for the transformation matrix
T = C S (4.49)
41
Since for the nodal forces we also have to transform from local to global coordi-
nate system, by identifying the global resultant force component when a force is
applied in local coordinates, thus the force vector is transformed as below
Fc = T Fg (4.50)
Therefore from equations (3.16), (4.48) and (4.50) we get the well known rela-
tions as follows
Fc = K(e)dc (4.51)
T Fg = K(e)T dg
Fg = T T K(e)T dg
⇒ Kg = T T K(e)T
and
Fg = Kgdg
where the subscripts c,g refers to matrices defined in local and global coordinate
system respectively. We note that determination of the angles α and ψ depends
on the geometry of a particular problem.
4.8 Common Boundary Conditions
The type of beam support determines the type of load that the support can resist,
and how much load the beam can carry. Thus to get a unique solution, we have
to assign relevant boundary conditions and or geometric constrains, based on the
beam supports.
Boundary conditions that arise from the enforcement of prescribed displace-
ments or rotations are the geometric boundary conditions, while the ones from
arising from the enforcement of prescribed forces or moments are the natural
boundary conditions. See (Landau, Lifshitz [12]), and class notes by Carlos
Fellipa [1] chapter 8.
(a) Fully clamped beam
(i) If the beam is fixed at both ends, then it resists vertical and horizon-
tal forces as well as a moment. Therefore both rotation and translation are re-
strained. The boundary conditions are thus
u(0) = u(L) = u′(0) = u′(L) = 0 (4.52)
42
Note here that the reaction moment, and the shear force are unknown, and have
to be determined.
(b)Cantilever beam
If one end is fixed (for example at s = 0) while the other end is free (at s = L),
the beam is called a cantilever beam. The boundary conditions are such that
u(0) = u′(0) = u′′(L) = u′′′(L) = 0 (4.53)
This implies therefore that the moment M and force V at s = L is zero. Note that
If a concentrated force q is applied to the free end of the beam, then this induces
a shear on the end of the beam. Consequently, the boundary condition u′′′(L) = 0
is no longer valid, and is typically replaced by the condition u′′′(L) =−q
(c) Simply supported beam
This is a beam that is supported on a smooth surface at the ends. Since the beam
develops a reaction force normal to the beam, but will not produce a moment
at the reaction. However, the beam does not experience any bending moments,
and is free to rotate, therefore the bending moments u′′ are zero at both ends,
assuming it is supported at both ends. Thus
u(0) = u(L) = u′′(0) = u′′(L) = 0 (4.54)
Note that a simple support cannot resist lateral loads of any magnitude.
(d) Pinned support
A pinned support can resist both vertical and horizontal forces but not a moment.
Therefore the beam can rotate, but cannot translate in any direction, therefore u
is zero at the boundaries, while θ is unknown and has to be calculated. That is
u(0) = u(L) = u′′(0) = u′′(L) = 0 (4.55)
(e)Roller support
Roller supports are free to rotate and translate along the surface upon which
the roller rests. The surface can be horizontal, vertical, or sloped at any angle,
therefore this kind of support does not provide resistance to a lateral force. This
43
supports are similar to simple support in c, only that the simple supports cannot
resist lateral loads , while this support can.
The supports above and their reactions are as in figure (4.9)
Roller support
−
frictionless surface
Simple supportPinned support
Fixed supportOO
Figure 4.9: Different beam supports and their reactions
4.8.1 Numerical results of beams under different boundaryconditions
We consider several examples with different boundary conditions, and different
supports for both Euler-Bernoulli and Timoshenko beams.
We used the following parameters;
Parameters Value
Number of elements 40
Modulus of elasticity (E) 27.6×106
Shear Modulus of elasticity (G) 10.6×106
Beam Span (L) 40
Width (b) 1
Cross sectional Area (A) π(b/2)2
Subtended Angle (ψ) πRadius of curvature (R) L/ψSecond moment of inertia (I) πb4/64
Shear correction factor (κ) 0.85
Table 4.1: Parameters used for a curved beam subtended by angle π
We apply distributed tangential and radial loads of 500lbs, acting in the positive
direction.
44
(a) Clamped Beam
The boundary condition describing a fixed beam on both ends is
v(0) = v(L) = 0,u(0) = u(L) = 0,u′(0) = u′(L) = 0 (4.56)
This boundary condition implies that the beam cannot translate horizontally or
vertically, therefore there is no displacement at s = 0, and at s = L. since u′(0) =
θ(0) is also zero, the derivative of the deflection is zero at this point and therefore
the direction cannot change.
We compare the displacements, and the rotation of both Euler-Bernoulli and
Timoshenko beams.
(i) Tangential and Radial displacement, figure(4.10)
0 5 10 15 20 25 30 35 40−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4Tangential Displacement
BEAM LENGTH
DIS
PLA
CE
ME
NT
Eler−Bernoulli BeamTimoshenko Beam
0 5 10 15 20 25 30 35 40−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4Radial Displacement
BEAM LENGTH
DIS
PLA
CE
ME
NT
Eler−Bernoulli BeamTimoshenko Beam
Figure 4.10: Tangential displacement and Radial displacements for Euler -Bernoulli
and Timoshenko beams fixed at both ends
(ii) Cross sectional Rotation, figure (4.11)
45
0 5 10 15 20 25 30 35 40
−0.08
−0.06
−0.04
−0.02
0
0.02
0.04
0.06
0.08
Cross sectional rotation
BEAM LENGTH
RO
TA
TIO
N
Eler−Bernoulli BeamTimoshenko Beam
Figure 4.11: Cross sectional Rotation for Euler -Bernoulli and Timoshenko beams
fixed at both ends
(b) cantilever beam
If one end is fixed (for example at s = 0) while the other end is free (at s = L),
then we have that
u(0) = u′(0) = v(0) = 0,M = V = 0 at s = Ł; (4.57)
(i) Tangential and Radial displacement, figure(4.12)
0 5 10 15 20 25 30 35 40−120
−100
−80
−60
−40
−20
0
20Tangential displacement for a cantilever beam
BEAM LENGTH
DIS
PLA
CE
ME
NT
Euler−Bernoulli beamTimoshenko beam
0 5 10 15 20 25 30 35 40−60
−50
−40
−30
−20
−10
0Radial displacement for a cantilever beam
BEAM LENGTH
DIS
PLA
CE
ME
NT
Euler−Bernoulli beamTimoshenko beam
Figure 4.12: Tangential displacement and Radial displacements for cantilever Euler
-Bernoulli and Timoshenko beams subjected to external load of 500
(ii) Cross sectional Rotation, figure (4.13)
46
0 5 10 15 20 25 30 35 40−4
−3
−2
−1
0
1
2
3
4Cross sectional rotation for a cantilever beam
BEAM LENGTH
RO
TA
TIO
N
Euler−Bernoulli beamTimoshenko beam
Figure 4.13: Cross sectional Rotation for cantilever beam Euler -Bernoulli and Tim-
oshenko beams,subjected to external load of 500
Moment M and force V at s = L is zero. Therefore the beam is free to rotate.
There are no conditions imposed on u,v,θ at the free end, for which therefore
we have to calculate.
With a gradual increase in external load and moment from 100 - 500, the defor-
mation is as follows
0 5 10 15 20 25 30 35 40−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4Tangential Displacement for external load 100 − 500
BEAM LENGTH
DIS
PLA
CE
ME
NT
500
400
300
200
100
0 5 10 15 20 25 30 35 40−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4Radial Displacement for external load 100 − 500
BEAM LENGTH
DIS
PLA
CE
ME
NT
500
400
300
200
100
Figure 4.14: Tangential Displacement and Rotational Displacement for increased
external load from 100 - 500 of a beam clamped at both ends
47
0 5 10 15 20 25 30 35 40
−0.06
−0.04
−0.02
0
0.02
0.04
0.06
Cross sectional rotation for external load 100 − 500
BEAM LENGTH
Rot
atio
n
500
400
300
200
100
Figure 4.15: Cross sectional Rotation for increased external load from 100 - 500 of a
beam clamped at both ends
(c) Simply Supported Beam
If we apply an increasing external load and moment, from 300 - 500, as shown
in (b) above, while parameters defined in table (4.1) still holding, then the de-
formation of the curved beam is as below
0 5 10 15 20 25 30 35 40
−1
−0.5
0
0.5
1
Tangential displacement for simply supported beam
BEAM LENGTH
DIS
PLA
CE
ME
NT
0 5 10 15 20 25 30 35 40−1.5
−1
−0.5
0
0.5
1
1.5Radial displacement for simply supported beam
BEAM LENGTH
DIS
PLA
CE
ME
NT
Figure 4.16: Displacement of an Euler- Bernoulli Beam and a Timoshenko Beam that
are both simply supported at both ends,subjected to external load from 100 - 500.
The boundary conditions for a beam that is simply supported is
v(0) = v(L) = 0,u(0) = u(L) = 0,M(0) = M(L) = 0
This mean that when we apply external load to a beam that is simply sup-
ported,then it cannot translate hence tangential and radial displacement at both
ends is zero.
However, the beam does not experience any bending moments, and is free to
48
0 5 10 15 20 25 30 35 40−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4Cross sectional rotation for simply supported beam
BEAM LENGTH
RO
TA
TIO
N
Figure 4.17: Cross sectional Rotation of an Euler- Bernoulli Beam and a Timoshenko
Beam that are both simply supported at both ends, subjected to external load from 100
- 500.
rotate, therefore the bending moments M are zero at both ends, as can be seen
from the graphs in figure (4.17).
4.9 Shear and Membrane Locking
Shear and membrane locking is a common phenomenal in finite elements based
upon the virtual work principle for curved beam elements. This has been a topic
of intense research and according to [19], In a curved beam elements, shear lock-
ing occurs C(0) curved displacement elements with low-order (e.g. linear) poly-
nomial approximations, as a result of overestimating the transverse shear stiffens
because the lower-order conventional elements cannot represent the condition
of zero radial shear strains when the elements are subjected to bending moment
only , while membrane locking results mainly due to the inability of the element
to model inextensional (or nearly inextensional) deformation modes.see [20]. To
alleviate these problems, there are several procedures that have been proposed,
[24]. One among the several procedures is reduced or selective integration.
In using reduced integration, the stiffness matrix is evaluated with a lower order
of integration than the one that exactly integrates the finite element polynomial
approximations.
In our case, we have used 1− point gauss integration (Stewart, pg. 169 [11]) for
the shear,and membrane energy terms in Timoshenko curved beam, while for the
Euler bernoulli beam, a 1− point integration for the membrane energy term has
been adopted.
For example, if we approximate the tangential displacement, using parameters in
49
table (4.1), we can see from figures (4.18) that indeed reduced integration does
improve the results of the approximated solution.
0 5 10 15 20 25 30 35 40−0.25
−0.2
−0.15
−0.1
−0.05
0
0.05
0.1Tangential displacement for Euler bernoulli clamped beam
BEAM LENGTH
DIS
PLA
CE
ME
NT
1−point integration of mebrane energy termfull integration
0 5 10 15 20 25 30 35 40−0.25
−0.2
−0.15
−0.1
−0.05
0
0.05
0.1Tangential displacement for a clamped Timoshenko clamped beam
BEAM LENGTH
DIS
PLA
CE
ME
NT
1−point integration of shear and membrane termsfull integration
Figure 4.18: Effects of reduced membrane energy term integration for Euler-Bernoulli
beam, and reduced shear and membrane energy terms integration for Timoshenko
beam
4.10 Numerical Test Examples
We are going to use Castigliano’s energy theorem to derive the analytical solu-
tion of a beam subjected to point load in the examples that will follow.
With reference to the example given by Lee and Sin [24], and thereafter adopted
by [2] and [25], we investigate the accuracy of the current curved beam ele-
ment formulation, by comparing the finite element solution of the given exam-
ples, with those already in literature which have been proven to give accurate
results.The problems are the typical examples of the shear/membrane locking
which are frequently used to investigate the capability of analysis of the ele-
ments developed and to see whether the shear locking is properly overcome. We
use in this case the same number of elements as used in the examples cited from
literature.
In example one, we vary the size of the subtended angle, and observe the con-
vergence of the radial displacement to the analytical solution see([2]).
In example two, we vary slenderness ratio of the beam defined by R/b between
4 and 1000, that is from a thick beam to a thin beam limits, and observe the con-
vergence of the ratio of the normalized displacement, that is (numerical solution
dived by the analytical solution) to 1.
50
4.10.1 Example 1: An arc for various subtended angles.
In this example, with reference to [2], [24], and [25], we subject an arc of a ring
with angle ψ , fixed on one end and on the other end it is on rollers, and subjected
to a point radial force Q as shown in figure (4.19).
= 1 = u = 0 1 1
= 0 2 θ = 2 v
v θψ
b
2 u
Q
R
OO
OOO
Figure 4.19: An arc fixed on one end and supported on rollers and subjected to point
load on the other end
Therefore the boundary conditions are that at the fixed end, there is no transla-
tion nor rotation, therefore v1 = u1 = θ1 = 0, while at the end, the roller sup-
ports are free to translate along the surface upon which the roller rests, therefore
v2 = θ2 = 0, while the vertical displacement u2 is unknown, and has to be deter-
mined.
When we compare the present element by comparing results with an already ex-
isting element by Saffari and Tabatabaei see [2], we obtain the approximation of
radial displacement u2 to be the following, when the arc in (4.19) is subjected to
point radial force Q = 100, R = 100,E = 10.5×106,G = 4×106,b = 2.
From figure (4.20), (a), and the results from table (4.2), we can see that the re-
sults are not exactly precise. For the present element, and for an angle subtended
at 10o we get the approximate radial displacement to be 0.0069, a solution that
is achieved after using four elements, while the the element by [2], only uses one
element and gives very accurate results. The displacement considered is for a
Timoshenko beam, therefore we see that the Timoshenko beam approximation
51
0 20 40 60 80 100 120 140 160 1800
1
2
3
4
5
6
Angle (degrees)
Rad
ial d
ispl
acem
ent
Radial dispalcacement of a ring arc subjected to point load, for various angles
1−element solution Saffari & Tabatabei4−elements Timoshenko beam for present model 4−elements Euler beam for present model
0 20 40 60 80 100 120 140 160 1800.002
0.004
0.006
0.008
0.01
0.012
0.014
0.016
0.018
Angle (in degrees)
Rad
ial d
ispl
acem
ent
shear and membrane locking for an arc of varying subtended angles
4−element Euler Bernoulli beam4−elements Timoshenko beam
Figure 4.20: Radial displacement (a)With reduced integration, (b) Without reduced
integration for an arc fixed on one end and supported on rollers and subjected to point
load on the other end
for the present curved element are relatively a close approximate to the compar-
ison element, while the Euler-bernoulli beam, which assumes no shearing effect
overestimates the solution.
Reduced integration
To get these results, we have used reduce integration. We note that without
reduce integration, the results are clearly wrong, as shown in figure (4.20)(b).
The first two results for Euler bernoulli beam, for subtended angles 10,20 gives
the same results as the beam with reduced integration. Thus for a very thick
shallow beam, the Euler-bernoulli beam gives notable approximation. But as the
subtended angle becomes big, the results do not even converge to the same order
as the true solution.
For Timoshenko beam,the results are overly very stiff, and are not even correct
from onset.
52
Angle Saffari and Tabatabaei Timoshenko Beam Euler-Bernoulli Beam
10 0.00419 0.0063 0.0063
20 0.0077 0.0101 0.0109
30 0.01423 0.0176 0.0209
40 0.02876 0.0331 0.0416
50 0.05400 0.0573 0.0743
60 0.09316 0.0908 0.1197
70 0.14995 0.1339 0.1786
80 0.2288 0.1878 0.2522
90 0.3351 0.2544 0.3432
100 0.4752 0.3373 0.4564
120 0.8898 0.5789 0.7855
140 1.5589 1.0143 1.3783
160 2.6138 1.9151 2.6059
180 4.2539 4.1265 5.6205
Table 4.2: Comparison of the finite element radial displacement with reduced integra-
tion for the Timoshenko beam and Euler-Bernoulli beam with, the method by Saffari
and Tabatabaei [2]
4.10.2 Castigliano’s Energy Method
To solve the analytical solution of the next example in figure(4.21), we use Cas-
tigliano’s theorem. Adopting the explanation given in [29], when external forces
are applied to a beam or structure the structure deforms. The external forces
perform work and the energy is stored in the structure in the form stress and
elastic deformation. To obey the law of conservation of energy the work done
in the infinitesimal movements of the external forces and moments we denote
by F must be equal to the potential energy Π stored in the structure.When the
external forces are removed the beam or structure rebounds elastically, and the
stored potential energy returns to zero. Hence one can be able to investigate the
displacements of structures under external load. The Castigliano’s theorem is
therefore formally stated as:
When forces generate on elastic structures, the displacement corresponding to
any force may be found by obtaining the partial derivative of the total strain en-
ergy with respect to that force.
Therefore, for Π the total strain energy , an application point of the force Fi in
53
the direction of Fi , and the displacement Ui ,
Ui =∂Π∂Fi
(4.58)
Procedure
To determine a displacement Ui in the direction of a real or fictitious force Fi
(a) Obtain an expression for the total strain energy including the the loads
Mi,Ni,Vi and a dummy forces and moments if required.5
(b) Calculate the linear displacement Ui from the relationship
Ui =∂Π∂Fi
(c) If the force is fictitious set Fi = 0 and solve the resulting equation.
4.10.3 Example 2 :A quarter-Circular Cantilever Beam
We now take a look at how efficient the element derived is with respect to how
accurate it approximates the analytical solution.
With reference to [2], we investigate a quarter circular beam fixed at one end, and
subjected to a radial point force Q = 1 at the free end,R = 10 , b = 1, E = 10.56,
G = 4.0×106, as shown in the figure (4.21).
Q
2 θ
2 u
2v
R
b
Figure 4.21: A quarter-circular cantilever beam subjected to radial point load Q, at
the free end. (Lee, Sin [22])
5If displacement is required where there was no corresponding load, then a dummy load corresponding to
the desired displacement is added to the applied loading condition, and after calculating the value of dummy
load is set to zero.
54
In this case, the boundary conditions at the fixed end is as in figure (4.19), while
at the free are that the beam cannot translate, and rotate, therefore the moments
M, and the forces are zeros at that end, while the displacements are unknown and
have to be calculated.
The free body diagram for figure (4.21) is figure (4.22).
Dummy External Force Θ
Dummy External Momentsm
ψ
V y
M z
N s
Q
Extrnal force applied
Figure 4.22: Free body diagram of a quarter-circular cantilever beam subjected to
radial point load Q, and we introduce axial dummy force Θ and a dummy moment - mat the free end.
Using castigliano’s energy theorem, we calculate the analytical displacements
v,u and cross sectional rotations θ as follows
Radial Displacement
Thus refereing to (4.22)
Ns = −Qsin(ψ) (4.59)
Mz = −QRsin(ψ) (4.60)
Vy = −Qcos(ψ) (4.61)
where ψ is the angular coordinate indicated in the figure (4.22) defined as
ψ =sR
From (4.30a),(4.30b) , (4.30c), and the total axial, bending and shearing energy
of (4.21) for the radial displacement u is
Π =∫ π
2
0
(N2
s
2EA+
M2z
2EI+
V 2y
κGA
)Rdψ (4.62)
=πQ2R8EA
+πQ2R3
8EI+
πQ2R8GAκ
(4.63)
55
Thus for external force Q, the radial tip displacement
u =∂Π∂Q
=πQR4EA
+πQR3
4EI+
πQR4GAκ
(4.64)
Axial displacement
To calculate the axial displacement,we introduce a dummy axial external force
Θ such that forces in the axial direction are
Ns = −Qsin(ψ)+Θcos(ψ) (4.65)
Mz = −QRsin(ψ)−ΘRcos(ψ) (4.66)
Vy = −Qcos(ψ)+Θsin(ψ) (4.67)
Therefore total axial, bending and shearing energy with regard to the axial dis-
placement is
Π =∫ π
2
0
(N2
s
2EA+
M2z
2EI+
V 2y
κGA
)Rdψ (4.68)
=R
2EA
(−QΘ+
14
Θ2π +14
Q2π)
+R
2EI
(R2ΘQ+
R2Θ2π4
+R2Q2π
4
)
+R
2κGA
(−QΘ+
14
Θ2π +14
Q2π)
Therefore the axial tip displacement v is
v =∂Π∂Θ
=R
2EA
(−Q+
12
Θπ)
+R
2EI
(R2Q+
R2Θπ2
)+
R2κGA
(−Q+
12
Θπ)
(4.69)
Equating the dummy force Θ to zero, the axial displacement is thus
v =− QR2EA
+QR3
2EI− QR
2GAκ(4.70)
Cross sectional rotation
To calculate the cross sectional rotation θ , we introduce at the free end a dummy
external moment m, in the tangential direction. Therefore
Π =1
2EI
∫ π2
0(−QRsin(ψ)−m)2Rdψ (4.71)
=R
2EI
(2mQR+
m2π2
+R2Q2π
4
)
⇒ θ =∂U∂m
=R
2EI(2QR+mπ)
56
Equating m to zero,
θ =QR2
EI(4.72)
Lee and Sin [24], have solved the same problem but using a more involving
approach which still leads to the same answer, as we have found above. Hav-
ing derived the analytical solution, we now compare how accurate our model
approximates the particular example that we have put forth.
4.10.4 Results for Timoshenko curved Beam
Table (4.3) gives the results obtained for comparison of the finite element solu-
tion for the quarter - circular cantilever curved beam, for a cantilever beam.
The subscripts f and c denote the finite element solution and analytic solution
R/b 1 Element - Saffari and Tabatabaei 1 Element - Present Model
v f /vc u f /uc θ f /θc v f /vc u f /uc θ f /θc
4 1.0335 1.0000 1.0000 1.2386 0.6110 0.7117
10 1.0053 1.0000 1.0000 1.1914 0.5945 0.7306
20 1.0013 1.0000 1.0000 1.1775 0.5885 0.7361
50 1.0002 1.0000 1.0000 1.1696 0.5847 0.7393
100 1.0000 1.0000 1.0000 1.1670 0.5835 0.7403
200 1.0000 1.0000 1.0000 1.1657 0.5828 0.7408
500 1.0000 1.0000 1.0000 1.1649 0.5825 0.7411
1000 1.0000 1.0000 1.0000 1.1647 0.5823 0.7412
Table 4.3: Comparison of the two different finite element solution models loaded with
tip radial force, of a quarter circular cantilever beam with Castigliano’s energy solu-
tions for Timoshenko beam.
by Castigliano’s theorem respectively, while v,u are the tangential, radial dis-
placements respectively, and θ is the cross sectional rotation.
The results from the above table highlights one major issue; that this model is not
as accurate as we would want the results to be, as we would expect the normal-
ized displacement to converge to unity, when the ratio R/b >> 1. Further, there
is deterioration of radial deflection convergence, as the values are not converging
towards unity but away from unity.
Since we have used reduced integration, there is generally an improvement of
convergence, as seen in figures (4.18), and (4.20) but as from the results in table
57
(4.3), even with reduced integration for the choice of the interpolation functions
used, the reduced integration approach is not trouble free.
4.10.5 Results for the Euler-Bernoulli curved beam
If we again compare the analytical solution with the numerical solution for the
example give in figure(4.21) for an Euler-Bernoulli beam, the analytical solution
is without the shearing term, that is
vc =QR3
2EI− QR
2EA(4.73a)
uc =πQR3
4EI+
πQR4EA
(4.73b)
duc/dx =QR2
EI(4.73c)
Comparing to the results to an already existing model by Saffari and Tabatabeai
[2] , we get the results in (4.4) for the displacement of the quarter circular Euler-
Bernoulli beam fixed at one end and subjected to a radial point force at the other
free end. We note that first of all the tangential displacement in this case is not as
R/b 1 Element - Saffari and Tabatabaei 1 Element - Present Element
v f /vc u f /uc θ f /θc v f /vc u f /uc (du f /ds)/(duc/ds)
4 1.0133 1.0000 1.0000 1.5040 0.6906 0.5284
10 1.0053 1.0000 1.0000 1.5367 0.7406 0.5682
20 1.0013 1.0000 1.0000 1.5451 0.7582 0.5791
50 1.0002 1.0000 1.0000 1.5497 0.7690 0.5851
100 1.0000 1.0000 1.0000 1.5511 0.7726 0.5870
200 1.0000 1.0000 1.0000 1.5519 0.7745 0.5879
500 1.0000 1.0000 1.0000 1.5523 0.7756 0.5885
1000 1.0000 1.0000 1.0000 1.5524 0.7759 0.5887
Table 4.4: Comparison of the two different finite element solution models loaded with
tip radial force, of a quarter circular cantilever beam with Castigliano’s energy solu-
tions for Euler-Bernoulli Beam.
accurate as for the Timoshenko beam, for Euler-bernoulli beam subject to bend-
ing moments only. The beam cannot represent the zero radial shear strains as-
sumption by our use of low order interpolation functions, hence we used reduced
integration, but still the results are not very accurate. Considerable improvement
in radial approximation is noted, where we used cubic hermitian polynomials to
58
approximate the displacements u, while the cross sectional rotation approxima-
tion is not better than for the Timoshenko beam.
4.11 Points to Note: Timoshenko curved beam strains
definition
We have explained how the equations governing the deformation of a curved
Euler-Bernoulli beam are derived, based on the derivation given by (Oden, [5]).
To consider the shearing effect, hence the Timoshenko beam term, we have used
the stress tensor σsy defined from the same source to formulate the differential
equation governing the shear deformation. These strains are the conventional
strains definitions used in the deformation of a curved shearable beam, for exam-
ple (Prathap and Ramesh, [26] ), and (Zienkiewicz et al. [30]) use this definition
of strain.
Based on minimum potential energy of a deformed beam, we have calculated
the displacement of a loaded beam under boundary constrains. While the energy
strains of the curved Euler-Bernoulli beam are correct, the representation of the
Timoshenko curved beam energy strains, based on the definition we have used
here are not according to studies by (Day, Potts, [31]) entirely correct. The strain
definition used here is
ε =dvds− u
R(4.74a)
χ = −d2uds2 −
dvRds
(4.74b)
γ = −θ +duds
(4.74c)
We have used rotation due to bending θ as a degree of freedom, where the cur-
vature changes due to bending is defined therefore as
χ = −dθds− dv
Rds(4.75)
Day and Potts, [31], while considering the rigid-body displacements, have showed
that θ = du/ds is not equal to the physical rotation of the cross section for a
59
shearable beam and hence is not a suitable parameter to be used as a nodal de-
gree of freedom, whose continuity is to be imposed.
We can explain this in other way as: we know that for a straight beam, the ro-
tation is a function depending only on the vertical displacement u, but when
the beam is initially curved, the tangential displacement will also now affect
the cross sectional rotations, therefore the total rotation θ at a point becomes
du/ds+ v/R.
Therefore the appropriate choice of nodal degree of freedom, instead of θ , would
be θ + v/R, hence we can define a single variable for example θ = θ + u/R so
that now the strains for the curved timoshenko beam will instead be defined as
ε =dvds− u
R(4.76a)
χ =dθds
(4.76b)
γ = −θ +duds
+vR
(4.76c)
This definition is used in many problems relating to shearable curved beams, and
is not a new concept.
If we now compare this definition of strains with the conventional strains we
have previously used for the example two, we get the following results in table
(4.5) .
R/b 1 Element - Days and Pott 1 Element - Present Element
v f /vc u f /uc θ f /θc v f /vc u f /uc θ f /θc
4 1.2386 0.6110 1.1985 1.2386 0.6110 0.7117
10 1.1914 0.5945 1.1985 1.1914 0.5945 0.7306
20 1.1775 0.5885 1.1985 1.1775 0.5885 0.7361
50 1.1696 0.5847 1.1985 1.1696 0.5847 0.7393
100 1.1670 0.5835 1.1985 1.1670 0.5835 0.7403
200 1.1657 0.5828 1.1985 1.1657 0.5828 0.7408
500 1.1649 0.5825 1.1985 1.1649 0.5825 0.7411
1000 1.1647 0.5823 1.1985 1.1647 0.5823 0.7412
Table 4.5: Comparison of the two different strain definition for fem solution models
loaded with tip radial force, of a quarter circular cantilever beam with Castigliano’s
energy solutions for Timoshenko Beam.
60
We can note that there is a difference in values approximated for the displace-
ments when using the two definitions of strains especially in cross sectional rota-
tions. We still do not achieve convergence to the exact value of unity, even with
this definition of strain, as we still use low-order linear basis to interpolate the
displacements.
When we use the conventional definition of strains for a shearable curved beam,
then we use an inappropriate parameter θ = θ − u/R as a nodal degree of free-
dom. According to (Day, Potts, [31]), this parameter when chosen as the rotation
degree of freedom is dependent on the geometry and orientation of the elements,
while the rotation, θ , is uniquely defined at all points.
Therefore for normalized displacements approximation for strains definition in
(4.76), with the conventional strain definition (4.74) for the quarter-circular can-
tilever beam in figure (4.21), we get the following error in figure (4.23)in our
approximation
0.5 1 1.5 2 2.5 3 3.50
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
Tangential, Radial displacements and Cross Sectional rotation errors
Max
imum
Abs
olut
e E
rror
Strain displacement errors for a Timoshenko curved beam
TangentialDisplacementError x 108
RadialDisplacementError x 108
cross sectionalrotation error
Figure 4.23: Maximum Absolute Error for Tangential, Radial displacements and Cross
sectional Rotation when the conventional strains of a Timoshenko curved beam are
used
The rotation due to bending, as a degree of freedom for a timoshenko beam
results into incorrect results, and this is magnified in the cross sectional rotations,
as the tangential and radial displacements are approximated with a very small
and negligible error, while the approximations for cross sectional rotations have
a large error.
61
4.12 Summary and Conclusion
We have investigated the definition and consequently the difference between the
deformation of a beam subjected to external loading for an Euler-Bernoulli beam,
and a beam under Timoshenko beam theory in chapter 2. Introduction to finite
elements is covered in chapter three, while the curved beam is introduced in
chapter4.
In order to investigate how a ring deformers when subjected to external loads and
moments, and from the definitions given herein, cited from references quoted,
we have noted that the deformation of a curved beam, unlike the straight beam
is subject to nonlinear variation of strain distribution even though the cross sec-
tion of the beam remains plane after deformation. Further more, the equations
governing the deformation of the curved beam are coupled. This makes the dis-
placement of a curved beam to be a bit more difficult to tackle than the straight
beam, but there is an advantage to that, as curved beams are efficient in load
transfer due to inclusion of bending, shearing and membrane action in the trans-
fer.
We have implemented the equations of the curved beam to study their displace-
ment when subjected to external loads in finite element method, and have as-
sumed linear basis functions, for Timoshenko beam, while for Euler-Bernoulli
beam, have assumed linear basis functions for tangential displacements, and cu-
bic hermitian for radial displacement. To remedy shear and membrane locking,
we have used 1− point gaussian integration for the shear strain and membrane
strain.
The numerical results we have gotten, we have compared with already existing
different beam elements that gives very accurate approximation, and for the sec-
ond example that is used is compared with the analytical solution,
For the choice of the basis functions used, the results are not an accurate approx-
imation of the exact solution.
From our results,
(a) Although there is notable improvement in the convergence when we used re-
duced integration, (4.18) the finite element solution does not converge to the
analytical solution for our test examples with only use of one element, instead it
converges to a value lower that the expected.
We choose to use one element because the elements we are using for compari-
son, already existing in literature have been proven to be very accurate with just
the use of one element, hence we use this as a comparison bench for the curved
62
beam element we have used here.
(b) The use of reduce integration comes along with its disadvantages. When used,
reduced integration admits deformation modes which are able to occur without
any changes in strain energies (hence the name zero - energy modes).This results
because the nodal displacements when using reduced integration are now disas-
sociated from the gauss points, therefore there is no stiffness associated with this
mechanism and these can lead to considerably inaccurate solutions.
(c) Errors especially in rotation of the cross section arises from the use of the in-
correct definitions of a Timoshenko beam strains, and we have with reference to
[31] given the proper definition of strains.
(d) Therefore though they are the most elementary and simple finite elements to
implement, the finite elements implemented here for a curved beam under both
Timoshenko and Euler-Bernoulli beam theory are not the most appropriate ele-
ments to use when implementing in practical problems, as they do not give very
accurate results, and are prone to severe shear and membrane locking.
63
Appendix
.1 Matlab Programmes
.1.1 Common programmes and Parameters for both beam theories
%-------------------------------------------------------------------
clear;
clear figure
clc
nel = 30; %Number of elements
nnel = 2; %Number of nodes per element
ndof = 3; %Number of degrees of freedom per node
nnode = (nnel-1)*nel+1; %Total number of nodes
edof = nnel*ndof; % Total number of degrees of freedom per element
sdof = nnode*ndof; % Total number of degrees of freedom in all elements
%-------------parameters--------------------------------------------
E = 27.6*10^6; %Young’s Modulus of Elasticity (Stainless steel)
G = 3.8*10^6; % Shear Modulus of Elasticity (Stainless steel)
b = 1; % Breadth of the beam
A = pi*(b/2)^2 %Cross sectional Area (Circular cross section)
I = pi*(b/2)^4/4;%Area moment of inertia of cross section
tleng = 40; %Beam span length
h = tleng/nel; %Mesh Size
R = 15; %Radius
theta = pi; %Subtended Angle of the beam
cf = 5/6; %Shear correction factor
Q = 500; %External load
u_m=Q; v_m=Q; w_m=Q; %External Tangential, Radial and Moment respectively.
%----ESSENTIAL--BOUNDARY--CONDITIONS--------------------------------------
%Example boundary for a clamped beam on both ends, which means that
%the first three and the last three dofs values is zero.
bcdof = [1,2,3,nel*3+1,nel*3+2,nel*3+3];
bcval = [0, 0, 0,0,0,0];
64
%-------------LOCATION MATRIX---------------------------------------------
function[locmatrix] = eldof(iel,nnel,ndof,nel)
%compute dofs associated with each element
%locmatrix = matrix system of dof vector associated with element iel
%iel = element number whose global dofs are to be determined
edof = nnel* ndof;
count = (iel -1)* (nnel -1)*ndof;
for i = 1:edof
locmatrix(i) = count + i; %gives location of an element at global level
end
%----------APPLY ESSENTIAL BCS TO THE MATRIX EQN---Kd = F --------------
function[kk,ff] = febc(kk,ff,bcdof,bcval,nel)
n = length(bcdof);
sdof = size(kk);
for i = 1:n
c = bcdof(i); % gives the dofs where bcs are to be applied.
for j = 1:sdof
kk(c,j) = 0; %NOTE
end
kk(c,c) = 1;
ff(c) = bcval(i);
end
NOTE: Removing the equations corresponding to zero boundary
conditions is very convenient for small matrices,because it reduces
down on the number of equations to solve. But when we were
implementing in Matlab the equations have to be rearranged. With
rearrangement it was taking a bit more time than solving the equations.
To apply boundary conditions without rearranging the equations we set
to zero rows and columns corresponding to prescribed zero displacements
as well as the in the force vector too. We then place ones on the
diagonal to maintain non-singularity.
%----------GAUSS INTEGRATION --POINTS---(Adopted from Kwon,Bang [6],pp 178;)-------
function [point1,weight1]= gaussint(ngl,h)
% ngl - number of integration points
65
% point1 - vector containing integration points
% weight1 - vector containing weighting coefficients
%-------------------------------------------------------------------
point1=zeros(ngl,1);
weight1=zeros(ngl,1);
%find corresponding integration points and weights
if ngl==1 % 1-point quadrature rule
point1(1)=0.0;
weight1(1)=2.0;
elseif ngl==2 % 2-point quadrature rule
point1(1)=-0.577350269189626;
point1(2)=-point1(1);
weight1(1)=1.0;
weight1(2)=weight1(1);
elseif ngl==3 % 3-point quadrature rule
point1(1)=-0.774596669241483;
point1(2)=0.0;
point1(3)=-point1(1);
weight1(1)=0.555555555555556;
weight1(2)=0.888888888888889;
weight1(3)=weight1(1);
elseif ngl==4 % 4-point quadrature rule
point1(1)=-0.861136311594053;
point1(2)=-0.339981043584856;
point1(3)=-point1(2);
point1(4)=-point1(1);
weight1(1)=0.347854845137454;
weight1(2)=0.652145154862546;
weight1(3)=weight1(2);
weight1(4)=weight1(1);
end
.1.2 Curved Beam - Euler-Bernoulli beam theory
%---------INITIALIZE EMPTY STIFFNESS, LOAD and LOCATION MATRICES----------
ff =zeros(sdof,1);
kk = zeros(sdof,sdof);
66
locmatrix = zeros(nnel*ndof,1); %set up location matrix, to renumber the
% dofs when doing assembly into global
% system.
for iel = 1:nel %Loop over the elements
locmatrix = eldof(iel,nnel,ndof);
%Element stiffness matrix-------------------------------
kc = eulerelemetmatrix(E,I,G,A,cf,R,h,theta,ndof);
%Element load vector-----------------------------------
fs = eulerelementvector(u_m,v_m,h,w_m,R,theta,ndof);
%Global load vector------------------------------------
ff = eulerglobalvector(ff,fs,locmatrix,nel);
%Global stiffness matrix------------------------------
kk = eulerglobalmatrix(kk,kc,locmatrix);
end
[kk,ff] = febc(kk,ff,bcdof,bcval); %Implement Essential Boundary conditions
sol = kk\ff; % Solve for displacement
%-------------DETERMINATION OF ELEMENT STIFFNESS MATRIX ----------------
function[kc] = eulerelemetmatrix(E,I,G,A,cf,R,h,theta,ndof)
%-------------Rotation matrix-------------------------------------------
rot = [cos(theta) sin(theta) 0 0 0 0;...
-sin(theta) cos(theta) 0 0 0 0;...
0 0 1 0 0 0;...
0 0 0 cos(theta) sin(theta) 0;...
0 0 0 -sin(theta) cos(theta) 0;...
0 0 0 0 0 1];
[axial ] = euleraxial(R,h,ndof); %Axial Strain Stiffness Matrix
%Contribution
[bend,Bc ] = eulerbend(R,h,ndof); %Bending Strain Stiffness Matrix
%Contribution
ktt = (E*A/2)*axial + (E*I/2)*bend;
kc = rot’*ktt*rot;
%--------------AXIAL- STRAIN--MATRIX-----------------------------
function[axial ] = euleraxial(R,h,ndof)
ngl = 1; % 1- point integration
[point1,weight1] = gaussint(ngl,h); % extract integration points as weights
axial = zeros(ndof*2,ndof*2); for int=1:ngl
x = point1(int);
67
wt = weight1(int);
%---------Linear---Interpolation functions----------------------
s_1 = 1/2*(1 - x); s_2 = 1/2*(1 + x);
st_1 = -1/2; st_2 = 1/2; % 1st derivative of the shape functions
%---------------cubic--Hermitian Interpolation functions---------
c_1 = 1/4*(2- 3*x + x^3); c_2 = 1/4*(1 - x - x^2 + x^3);
c_3=1/4*(2+3*x - x^3); c_4 = 1/4*(-1-x + x^2 + x^3);
------First derivative----------------------------------------
c_u = [(-3/4+3/4*x^2)*2/h ,-1/4-1/2*x+3/4*x^2, (3/4-3/4*x^2)*2/h,
-1/4+1/2*x+3/4*x^2];
------Second derivative--------------------------------------
c_uu = [(3/2*x)*4/h^2, (-1/2+3/2*x)*(2/h), (-3/2*x )*4/h^2,
1/2+3/2*x*(2/h)];
c_u1 = 6*x/h^2; c_u2 = (-1 + 3*x)/h; c_u3 = -6*x/h^2; c_u4 = (1 +
3*x)/h;
%-------Axial strain Matrix------------------------------------------
a_ns = [ 2*st_1/h, -c_1/R, -c_2/R, 2*st_2/h, -c_3/R, -c_4/R ];
A = a_ns’* a_ns;
for i = 1: ndof*2
for j = 1:ndof*2
axial(i,j) = axial(i,j) + A(i,j)*wt*h/2;
end
end
end
%--------------BENDING- STRAIN--MATRIX---------------------------
function[bend,Bc ] = eulerbend(R,h,ndof)
ngl = 4 ; % 4 - point integration
[point1,weight1]=gaussint(ngl,h); bend = zeros(ndof*2,ndof*2);
Bc = zeros(ndof*2,ndof*2);
for int=1:ngl
x=point1(int);
wt=weight1(int);
%---------Linear---Interpolation functions----------------------
s_1 = 1/2*(1 - x);
s_2 = 1/2*(1 + x);
---1st derivative of the shape functions----------------
68
st_1 = -1/2;
st_2 = 1/2;
%---------------cubic--Hermitian Interpolation functions---------
c_1 = 1/4*(2- 3*x + x^3); c_2 = 1/4*(1 - x - x^2 + x^3);
c_3=1/4*(2+3*x - x^3); c_4 = 1/4*(-1-x + x^2 + x^3);
------First derivative----------------------------------------
c_u = [(-3/4+3/4*x^2)*2/h ,-1/4-1/2*x+3/4*x^2, (3/4-3/4*x^2)*2/h,
-1/4+1/2*x+3/4*x^2];
------Second derivative--------------------------------------
c_uu = [(3/2*x)*4/h^2, (-1/2+3/2*x)*(2/h), (-3/2*x )*4/h^2,
1/2+3/2*x*(2/h)];
c_u1 = 6*x/h^2; c_u2 = (-1 + 3*x)/h; c_u3 = -6*x/h^2; c_u4 = (1 +
3*x)/h;;
%-------Bending strain matrix ----------------------------------
b_ns = [-st_1*2/(h*R), -c_u1, -c_u2, -st_2*2/(h*R), -c_u3, -c_u4];
B = b_ns’* b_ns;
for i = 1: ndof*2
for j = 1:ndof*2
bend(i,j) = bend(i,j) + B(i,j)*h/2*wt;
Bc(i,j) = Bc(i,j) + P_e(i,j)*h/2*wt; % Stiffness Matrix
%contribution for Robin type boundary condition,
end
end
end
%--------------GLOBAL ELEMENT MATRIX--------------------------------
function[kk] = eulerglobalmatrix(kk,kc,locmatrix)
N = length(locmatrix); % # of dofs per element
for i = 1:N % gives element row position
ii = locmatrix(i) ;%global row
for j = 1:N % element column position
jj = locmatrix(j); % global column
kk(ii,jj) = kk(ii,jj)+kc(i,j); %assemble global matrix
end
end
%-------ELEMENT---LOAD----VECTOR-------------------------------------
69
function[fs] = eulerelementvector(u_m,v_m,h,w_m,R,theta,proc,ndof)
rot = [cos(theta) sin(theta) 0 0 0 0;...
-sin(theta) cos(theta) 0 0 0 0;...
0 0 1 0 0 0;...
0 0 0 cos(theta) sin(theta) 0;...
0 0 0 -sin(theta) cos(theta) 0;...
0 0 0 0 0 1];
Ax = eulergaussvtwo(R,h,ndof);
Bn = eulergaussvfour(R,h,ndof);
ftt = [u_m*Ax + v_m*Bn];
fs = rot*ftt;
%---------------TANGENTIAL DISPLACEMENT FUNCTION CONTRIBUTION----------
function[Ax] = eulergaussvtwo(R,h,ndof)
ngl = 2; %two - point integration
[point1,weight1]= gaussint(ngl,h);
Ax = zeros(ndof*2,1);
for int = 1:ngl
x = point1(int);
wt = weight1(int);
%---------Liner shape functions--------------------------------------------
s_1 = 1/2*(1 - x);
s_2 = 1/2*(1 + x);
st_1 = -1/2;
st_2 = 1/2;
%--------Load vector calculation------------------------------------------
a = [s_1, 0, 0 , s_2, 0, 0]’;
for i = 1: ndof*2
Ax(i) = Ax(i) + a(i)*wt*h/2;
end
end
%---------------RADIAL DISPLACEMENT FUNCTION CONTRIBUTION-----------
function[Bn] = eulergaussvfour(R,h,ndof)
ngl = 4;
[point1,weight1]= gaussint(ngl,h);
Bn = zeros(ndof*2,1);
for int = 1:ngl
x = point1(int);
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wt = weight1(int);
%-----------cubic hermitian interpolation functions------------------
c_1 = 1/4*(2-3*x+x^3);
c_2 = 1/4*(1-x-x^2+x^3);
c_3 = 1/4*(2 +3*x-x^3);
c_4 = 1/4*(-1-x + x^2 + x^3);
%------------Load vector calculation-----------------------------------
b = [0, c_1, c_2, 0, c_3, c_4 ]’;
for i = 1: ndof*2
Bn(i) = Bn(i) + b(i)*wt*h/2;
end
end
%--------------GLOBAL ELEMENT MATRIX----------------------------------
function[ff] = eulerglobalvector(ff,fs,locmatrix,nel)
N = length(locmatrix);
for i = 1:N % gives local vector row position
j = locmatrix(i); % global position
ff(j)= ff(j)+ fs(i); %assembled matrix
end
.1.3 Curved Beam - Timoshenko Beam Theory
%---------INITIALIZE EMPTY STIFFNESS, LOAD and LOCATION MATRICES---------------
ff = zeros(sdof,1);
kk = zeros(sdof,sdof);
locmatrix = zeros(nnel*ndof,1); %set up location matrix, to renumber the
% dofs when doing assembly into global
% system.
for iel = 1:nel %Loop over the elements
locmatrix = eldof(iel,nnel,ndof);
%Element stiffness matrix----------------------------
Kel = shearelementmatrix(E,I,A,G,cf,R,h,theta,ndof,tleng);
%Element load vector-----------------------------------
Fel = shearelementvector(u_m,v_m,w_m,h,R,theta,ndof,tleng,cf,E,A,I,G);
%Global load matrix----------------------------------
F = shearglobalvector(F,Fel,locmatrix,nel);
%Global stiffness matrix------------------------------
71
KK = shearglobalmatrix(KK,Kel,locmatrix);
end
[kk,ff] = febc(kk,ff,bcdof,bcval); %Implement Essential Boundary conditions
sol = kk\ff; % Solve for displacement
%-------------DETERMINATION OF ELEMENT STIFFNESS MATRIX ----------------
function[Kel]=shearelementmatrix(E,I,A,G,cf,R,h,theta,ndof,tleng)
%-------------Rotation matrix-------------------------------------------
rot = [cos(theta) sin(theta) 0 0 0 0;...
-sin(theta) cos(theta) 0 0 0 0;...
0 0 1 0 0 0;...
0 0 0 cos(theta) sin(theta) 0;...
0 0 0 -sin(theta) cos(theta) 0;...
0 0 0 0 0 1];
%----Energy strains --stiffness matrix contribution ---------------------
[axial, shear] = axialnshear(R,h,ndof); %Axial Strain Stiffness Matrix
%Contribution
[bend,Bc] = bendenergy(R,h,ndof); %Bending Strain Stiffness Matrix
%Contribution
kel = (E*A/2)*axial + (E*I/2)*bend + (cf*G*A/2)*shear;
kc = rot’*ktt*rot;
Kel = rot’*kel*rot;
%--------------AXIAL and SHEAR STRAIN MATRICES----------------------
function[axial, shear] = axialnshear(R,h,ndof)
ngl= 1; % 1- point integration
[point1,weight1]=gaussint(ngl,h);
axial = zeros(ndof*2,ndof*2);
shear = zeros(ndof*2,ndof*2);
for int=1:ngl
x=point1(int);
wt=weight1(int);
%---------Linear---Interpolation functions----------------------
s_1 = 1/2*(1 - x);
s_2 = 1/2*(1 + x);
-------1st derivative of the shape functions-------------------
st_1 = -1/2;
st_2 = 1/2;
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%-------Strain Matrix------------------------------------------
a_x = [2/h*st_1, -s_1/R, 0 , 2/h*st_2, -s_2/R, 0];
s_h = [0, 2/h*st_1, -s_1, 0, 2/h*st_2, -s_2 ];
A_e = a_x’* a_x;
S_e = s_h’* s_h; for i = 1: ndof*2
for i = 1: ndof*2
for j = 1: ndof*2
axial(i,j) = axial(i,j) + A_e(i,j)*h/2*wt;
shear(i,j) = shear(i,j) + S_e(i,j)*h/2*wt;
end
end
end
%--------------BENDING STRAIN MATRIX---------------------------
function[bend,Bc ] = bendenergy(R,h,ndof)
ngl = 2 ; % 2 - point integration
[point1,weight1]=gaussint(ngl,h);
bend = zeros(ndof*2,ndof*2);
Bc = zeros(ndof*2,ndof*2);
for int=1:ngl
x=point1(int);
wt=weight1(int);
%---------Linear---Interpolation functions----------------------
s_1 = 1/2*(1 - x);
s_2 = 1/2*(1 + x);
% -------1st derivative of the shape functions-----------------
st_1 = -1/2;
st_2 = 1/2;
%----strain matrix ----------------------------------------
a_x = [2/h*st_1, -s_1/R, 0 , 2/h*st_2, -s_2/R, 0];
s_h = [0, 2/h*st_1, -s_1, 0, 2/h*st_2, -s_2 ];
p = [N_1,0,0,N_2,0,0]; % Boundary contribution to the stiffness matrix
A_e = a_x’* a_x;
S_e = s_h’* s_h;
P_e = 1/R*p’*p;
for i = 1: ndof*2
for j = 1:ndof*2
bend(i,j) = bend(i,j) + B_e(i,j)*h/2*wt;
73
Bc(i,j) = Bc(i,j) + P_e(i,j)*h/2*wt; % Stiffness Matrix
%contribution for Robin type natural boundary condition,
%example when the moment is zero,
then du/dx - v = 0;
end
end
end
%--------------GLOBAL ELEMENT MATRIX--------------------------------
function[kk] = eulerglobalmatrix(kk,kc,locmatrix)
N = length(locmatrix);
for i = 1:N % gives element row position
ii = locmatrix(i) ;%global row
for j = 1:N % element column position
jj = locmatrix(j); % global column
kk(ii,jj) = kk(ii,jj)+kc(i,j); %assemble global matrix
end
end
%-------ELEMENT---LOAD----VECTOR-------------------------------------
function[Fel] = shearelementvector(u_m,v_m,w_m,h,R,theta,meth,ndof,tleng,cf,E,A,I,G)
rot = [cos(theta) sin(theta) 0 0 0 0;...
-sin(theta) cos(theta) 0 0 0 0;...
0 0 1 0 0 0;...
0 0 0 cos(theta) sin(theta) 0;...
0 0 0 -sin(theta) cos(theta) 0;...
0 0 0 0 0 1];
[Ax,Bn,Mn] = sheargaussvector(R,h,ndof);
fst = [u_m*Ax + v_m* Bn + w_m*Mn];
Fel = rot*fst;
%-------Gauss--Numerical integration ---------------------------------
function[Ax,Bn,Mn] = sheargaussvector(R,h,ndof)
ngl = 2;
[point1,weight1]= gaussint(ngl,h); % extract integration points as weights
Ax = zeros(ndof*2,1);
Bn = zeros(ndof*2,1);
Mn = zeros(ndof*2,1);
for int=1:ngl
x=point1(int); wt=weight1(int);
74
%-------Linear Basis Functions-----------------------------------------
s_1 = 1/2*(1 - x);
s_2 = 1/2*(1 + x);
st_1 = -1/2;
st_2 = 1/2;
%------------Load vector calculation-----------------------------------
ax = [s_1, 0, 0 , s_2, 0, 0]’;
bn = [0, s_1 , 0, 0, s_2, 0]’;
sh = [0 , 0 , s_1, 0, 0, s_2]’;
for i = 1: ndof*2
Ax(i) = Ax(i) + ax(i)*wt*h/2;
Bn(i) = Bn(i) + bn(i)*wt*h/2;
Mn(i) = Mn(i) + sh(i)*wt*h/2;
end
end
%--------------GLOBAL FORCE VECTOR------------------------------------
function[ff] = shearglobalvector(F,Fel,locmatrix,nel)
N = length(locmatrix); % # of dofs per element
for i = 1:N % gives local vector row position
j = locmatrix(i); % global position
F(j)= F(j)+ Fel(i); %assembled matrix
end
75
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Declaration of authorship
I hereby declare
• that I have written this thesis without any help from others and without the use
of documents and aids other than those stated in the bibliography,
• that I have mentioned all used sources and that I have cited them correctly.
Signature: Date:
79