CHAPTER ONE
INTRODUCTION
1.1
Background
Structural analysis of arbitrary shaped structures has been of interest to structural designers for
several decades. Dynamic behavior of these structures is strongly dependent on boundary
conditions, geometrical shapes, material properties, different theories, and various complicating
effects. Closed-form solutions are possible only for a limited set of simple boundary conditions
and geometries. For analysis of arbitrary shaped structures, several numerical methods, such as
finite element method, finite difference method, boundary element method, and so on, are
usually applied. In this report, interest has been put on the methods of calculation to find out the
deflection, slope, shear force and moment in structural members by taking advantage of the
Transfer Matrix Method.
The primary use of transfer matrices is in deriving stiffness matrices for beam elements.
However, for line like structural systems such as structures made of beam elements, responses
can be calculated directly from the transfer matrices. The displacement method is based on the
elastic theory, where it can be assumed that most structures behave like complex elastic springs.
The complex spring is subdivided into a number of simple springs, which can be readily
analyzed, and then by considering equilibrium and compatibility at the boundaries, or nodes, of
these simple elastic springs, the entire structure can be represented by a large number of
simultaneous equations. Solutions of these simultaneous equations results in the displacement at
these nodes. (John Case, 1999) In contrast to the displacement method, for which the equations
represent the conditions of equilibrium, the transfer matrix method solution for a whole structure
satisfies the equilibrium conditions, kinematic admissibility and the material law simultaneously.
(Dilkey, 1994)
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A generally-shaped machine member requires, in general, complicated procedures to produce the
state of stress at each point owing to the static forces, thermal and dynamic loads that the
member sustains in service. This is a very complicated problem, in general and the designer
seeks always ways to achieve practical solutions to engineering accuracy. Fortunately, most
machine components have geometry which allows such solutions to be found by way of
modeling these components appropriately with simpler ones.
One such simplification is possible when the dimensions of the component along one direction
are much greater than the ones perpendicular to it. We speak then of linear members which have
the form of a rod when the length is several times greater than its maximum thickness. There is
no specific rule to decide just when a long component can be considered a rod. It also depends
on the form of loading and the accuracy required. In general, a factor length/thickness of more
than 10 can be considered as adequate. Such components are usually shafts, bolts, power screws,
hooks, springs, torsion bars, etc. Some of them can be computed with strength of materials
methods, as prismatic bars in tension/compression, bending, shear, torsion. Stress analysis then is
relatively simple.
In many applications, however, machine components have more complicated geometry and the
designer needs adequate solutions which cannot be obtained with strength of materials methods.
Fortunately, computer methods come to the rescue and engineering solutions can be achieved for
fairly complicated geometries. Examples are stepped shafts, machine frames, supports, steel
structures, etc.
1.2
Objective
The overall objective of this project is to develop a computer program which uses the
transfer matrix method to analyze beams and shafts subjected to various loading.
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CHAPTER TWO
LITERATURE REVIEW
2.1 The transfer matrix method
The transfer matrix method was first introduced in an elementary form by Holzer, in the 1920’s
for torsional vibration of rotating shafts. It makes use of the fact that in a large class of design
problems, some structural member is designed along a line and the behavior at every point of the
system is influenced by the behavior at neighboring points only. Typical examples are beams,
shafts, piping systems, etc. (Dimarogonas, 1989)
The transfer matrix method is an approach to matrix structural analysis that uses a mixed form of
the element force-displacement relationship and transfers the structural behavior parameters, the
joint forces and displacement, from one end of the structures of line element to the other. An
advantage of transfer matrix method is that it produces systems of equation to be solved that are
quite small in comparison with those produced by the stiffness method. A disadvantage is the
extensive sequence of operations that are required on a small matrix.
The transfer-matrix method is used when the total system can be broken into a sequence of
subsystems that interact only with adjacent subsystems. This method is suitable for line
structures such as arches and cables. To implement the transfer matrix method, we need a
relationship that gives the state of forces and displacements at one end of the element in terms of
force and displacement at the other end.
If we are to assume a beam, say fixed at both ends, which consist of n-1 prismatic beam elements
of different cross-section as in Figure 1.1
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Pi
Pi +1
Element (n-1)
Node (n – 2)
1
2
n
n-1
2
n-2
1
n-1
Figure 2.1: A Stepped Shaft with point loadings
Thus the beam has n-1elements with constant moments of inertia and n nodes, points (or planes)
which define the beginning or the end of a uniform beam element.
To fully describe the situation at each node on the vertical plane, we need to know four
quantities: The deflection y, the slope θ, the moment M and the shear force V. The element i of a
beam element between nodes i and i+1 is shown in Figure 2.2 with the sign conventions usual in
statics. That is, (John Case, 1999)
i.
Downward vertical forces are positive.
ii.
Sagging bending moments are positive.
iii.
Clockwise shearing forces are positive.
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V2
V1
θ2
M2
M1
θ1
y2
y1
L1
Figure 2.2: Free body diagram of the stepped shaft
These four quantities can be arranged in a vector form z = {yθMV } which, because it describes
the state of affairs of node j, is called ‘state vector’ and to designate the node j the state vector is
written with a subscript ‘j’. (Dimarogonas, 1989)
Suppose that at the node 1 the state vector is
{z1} = {y1θ1M1V1}
(2.1)
but yet unknown. If no force is acting between nodes 1 and 2, the deflection, slope, moment and
shear at node, from simple beam theory (John Case, 1999), will be:
Equilibrium equation of statics
=0
ΣM = 0
V1 = -V2
(2.2)
M = M + VL
(2.3)
Deflection equation (John Case, 1999)
M = V x+ M
(2.4)
= M
(2.5)
EI
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Hence
EI
= M1x +
+
EIy =
+A
(2.6)
+ Ax + B
where A and B are constants of integration.
(2.7)
Applying boundary equations for the element,
At x=0, y = y1 and θ = θ1
A = EIθ1 and
B = EIy1
Hence the slope and deflection at x = l1 termed θ2 and y2 will be given by
l
2
EIθ =
l + 121 + EIθ
(2.8)
EIy2 =
+
(2.9)
+ EIθ l + EIy1
Which for computational convenience maybe written as
y2 =
+ l1θ1 +
+
(2.10)
θ2 =
θ1 +
+
(2.11)
M2 =
V2 =
1
=
M1 + l1V1
(2.12)
V1
(2.13)
1
(2.14)
Equations (2.12) and (2.13) express the equilibrium of the forces and moments on the beam and
equations (2.10) and (2.11) give the deflections and slope due to these moments and forces.
Equation (2.14) is an identity 1 = 1 and is added for computational convenience.
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The above set of equations can be written in the following matrix form
{z2} = [L1]{z1}
⎡1
⎢
⎢
Where [L1] =⎢0
⎢0
⎢0
⎣0
2
2
1
0
0
0
3
6
2
2
1
0
0
1
0
(2.15)
0⎤
⎥
0⎥
⎥
0⎥
0⎥
1⎦
(2.16)
The subscript ‘1’ of the matrix indicates that the quantity l, E and I are properties of the first
element.
It is evident from equation 2.15, the state vector at node 2 is the state vector at node 1 multiplied
by a 5 x 5 matrix L, which depends on the element properties only and it is well known. This
matrix transferred the state from node 1 to node 2 and therefore it is called ‘transfer matrix’.
As would be shown later, for every element of the beam there exists one known transfer matrix
L. The procedure is repeated for subsequent elements and the state vectors determined.
For example,
{z2} = [L1] {z1}
{z3} = [L2] {z2} = [L2] [L1] {z1}
(2.17)
{z4} = [L3] {z3} = [L3] [L2] [L1] {z1}
At the nodes, the state vector as we approach the node from left and right is the same. However
if at the node we have a static force F, this is not true as seen in Figure 2.3
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VL
P
1
ML
MR
2
VR
Figure2.3: Node Equilibrium
For a small length about the node, the deflection, slope and moment remain the same, but in
order to maintain equilibrium, we must have VR = VL + P, where with superscript ‘L’ designate
the situation at the left of the node and ‘R’ refers to the situation at the right of the node.
(Dimarogonas, 1989) This can be written as
=
θ
= θ
MR = ML
(2.18)
VR = VL + P
1=1
And in matrix form as
{
}=[ ]{ }
(2.19)
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Where
1
⎡0
⎢
[ ] = ⎢0
⎢0
⎣0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0⎤
⎥
0⎥
⎥
1⎦
(2.20)
L matrices refer to a beam element and they are known as ‘Field Matrices’. P matrices refer to a
nodal point and they are called ‘nodal’ or ‘point matrices’.
Finally, the sequence of equation (2.19) can be completed as follows
{
} = [P2] { } = [P2] [L1] [P1] { }
(2.21)
{
} = [P3] [L2] {
(2.22)
} = [P3] [L2] [P2] [L1] { }
Or simply
{
} = [A] { }
(2.23)
Where A is a 5 x 5 square matrix of the product of all 5x5 matrices of the element and point
matrices. This can be written in matrix notation as
⎧
⎪
⎫
⎪
⎨ ⎬
⎪ ⎪
⎩1⎭
⎡
⎢
= ⎢
⎢
⎣
⎤⎧
⎥⎪
⎥
⎥⎨
⎪
⎦⎩ 1
⎫
⎪
(2.23a)
⎬
⎪
⎭
This gives rise to four equations with eight unknowns. That is
,
,
,
,
,
,
.
However, applying boundary conditions for a fixed stepped shaft at both ends,
=
=0
=
=0
Therefore leaving us with four unknowns (
,
,
) which can be solved by the four
equations above.
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After computation of these unknowns, we can obtain the state vectors at the nodes, and thus the
deflection of the beam from equations (2.21) and (2.22), applied successively from left to right.
For the stepped shaft in consideration,
⎧
⎪
⎨
⎪
⎩
{
⎡1
⎢
⎫
⎪
⎢
= ⎢0
⎬
⎢0
⎪
⎢0
1⎭
⎣0
1
1
0
0
0
2
1
2
3
1
0⎤
6
⎥
0⎥
⎥
0⎥
0⎥
1⎦
2
1
1
2
1
0
0
1
1
0
⎧
⎪
⎫
⎪
⎨ ⎬
⎪ ⎪
⎩1⎭
} = [P2] { } = [P2] [L1] [P1] { }
1
⎡0
⎧ ⎫
⎪ ⎪
⎢
{ }=
= ⎢0
⎨ ⎬
⎢0
⎪ ⎪
⎩1⎭
⎣0
0
1
0
0
0
0
0
1
0
0
0 0
0 0⎤
⎥
0 0⎥
1 P⎥
0 1⎦
⎡1
⎢
⎢0
⎢
⎢0
⎢0
⎣0
1
1
0
0
0
2
1
2
3
1
6
2
1
1
2
1
0
0
1
1
0
0⎤
⎥
0⎥
⎥
0⎥
0⎥
1⎦
⎧
⎪
⎫
⎪
⎨ ⎬
⎪ ⎪
⎩1⎭
Similarly, considering the second element of the stepped shaft, it is evident that the field matrix
[L2] is analogous to that of the first element due to the similarity in the loading. Hence,
{ } = [L2] {
{
}
(2.24)
} = [P3] { } = [P3] [L2] {
}
(2.25)
In matrix notation, equation (2.24) is written as
⎧
⎪
{ }=
⎨
⎪
⎩
⎡1
⎢
⎫
⎪
⎢
= ⎢0
⎬
⎢0
⎪
⎢0
1 ⎭
⎣0
2
1
0
0
0
2
2
2
3
2
6
2
2
2
2
1
0
0
2
1
0
0⎤
⎥
0⎥
⎥
0⎥
0⎥
1⎦
⎧
⎪
⎫
⎪
⎨ ⎬
⎪ ⎪
⎩1⎭
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1
⎡0
⎢
{ } = ⎢0
⎢0
⎣0
0
1
0
0
0
0
0
1
0
0
0 ⎡1
0 ⎤⎢
⎥ 0
0 ⎥⎢
⎥ ⎢0
1 ⎦ ⎢0
⎣0
0
0
0
1
0
1
0
0
0
1
0
0
1
0
0⎤
⎥
0⎥
0⎥
0⎥
1⎦
⎧
⎪
⎫
⎪
⎨ ⎬
⎪ ⎪
⎩1⎭
Considering element 3 of the stepped shaft, the nodal vector can be written as,
{ } = [L3] {
{
}
(2.26)
} = [P4] { } = [P4][L3][P3][L2][P2][L1]{z1}
⎡1
⎢
⎢
Where [L3] = ⎢0
⎢0
⎢0
⎣0
2
3
3
3
3
2
6
2
3
3
1
0
0
0
2
1
0
0
3
1
0
(2.27)
0⎤
⎥
0⎥
⎥
0⎥
0⎥
1⎦
And so forth for the other elements.
Thus for our seventh element, the state vector of the eighth node, shall in correspondence to the
above foregoing be given by
{ } = [L7] {
{
}
(2.28)
} = [L8] { } = [L7][P7][L6][P6][L5][P5][L4][P4][L3][P3][L2][P2][L1]{z1}
(2.29)
In general, the transfer matrix is written as
⎡1
⎢
⎢
[Li] = ⎢0
⎢0
⎢0
⎣0
2
2
1
0
0
0
3
6
2
2
1
0
0
1
0
0⎤
⎥
0⎥
⎥
0⎥
0⎥
1⎦
Equation (2.29) can also be written as
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{
} = [A] {z1}
Where [A] = [L7][P7][L6][P6][L5][P5][L4][P4][L3][P3][L2][P2][L1]
Using equation (2.23a)
⎧
⎪
⎫
⎪
⎡
⎢
= ⎢
⎨ ⎬
⎢
⎪ ⎪
⎣
⎩1⎭
⎤
⎥
⎥
⎥
⎦
⎧
⎪
⎫
⎪
⎨ ⎬
⎪ ⎪
⎩1⎭
Applying boundary conditions for the fixed end stepped shaft,
Deflection y = y = 0
Slope θ1 = θ = 0
Thus, in matrix notation
0
⎧ ⎫
⎪0⎪
⎨ ⎬
⎪ ⎪
⎩1⎭
=
⎡
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎦
0
⎧ ⎫
⎪0⎪
⎨ ⎬
⎪ ⎪
⎩1⎭
Or in algebraic form
0 = a13M1 + a14V1 + a15
0 = a23M1 + a24V1 + a25
M8 = a33M1 + a34V1 + a35
(2.30)
V8 = a43M1 + a44V1 +a45
1 = a53M1 + a54V1 +a55
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The last equation above is used to detect any error that would have occurred. Solving the first
two equations in equation (2.30), the following solutions are obtained
V1 =
M1=
From these values of V1 and M1, values of M8 and V8 can be determined from equation 2.30.
Having determined the state vector at node 1, the state vector at node 2 and node 3 are calculated
as shown below:
{
} = [P2] { } = [P2] [L1] [P1] { }
{
} = [P3] { } = [P3] [L2] {
}
This is repeated so as to determine all the other state vectors at all the nodes. (Dimarogonas,
1989)
A computer program, using C++ has been developed to aid in the above analysis, for a stepped
shaft loaded as shown in Figure 2.1 but for a general number of nodal elements.
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CHAPTER THREE
3.1 ELEMENT LOADINGS
In general, for an element, four distinct types of loadings can be identifiable, namely

Point Loading

Uniformly Distributed Loading (UDL)

Linearly Distributed Loading (LDL)

Moment Loading
Often, these loadings act alone, or in pairs, or in triplets or all at the same time on an element,
giving 15 possible combinations as explained below.
3.1.1 LOADINGS ON A TYPICAL ELEMENT

Point Loading Only

Uniformly Distributed Loading Only (UDL)

Linearly Distributed Loading Only (LDL)

Moment Loading Only

UDL plus Point Loading Only

Point Loading plus Moment Loading Only

UDL plus LDL Only

UDL plus Moment Loading Only

Point Loading plus LDL Only

Moment Loading plus LDL Only

Point Loading plus UDL plus Moment Loading Only

UDL plus LDL plus Point Loading Only

UDL plus LDL plus Moment Loading Only

Point Loading plus Moment Loading plus LDL Loading Only

Point Loading plus UDL plus LDL plus Moment Loading Only
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If the loadings occur in pairs, triplets, or all at the same time, then the principles of linear
superposition may be introduced to determine the total effect of the Loadings.
3.2 TRANSFER MATRICES FOR THE VARIOUS LOAD COMBINATIONS ON A
TYPICAL BEAM ELEMENT
3.2.1 POINT LOADING ONLY
Pi+1
y
Pi
Mi+1
Mi
vi+1
vi
yi+1
yi
x
li
x
Figure 3.1: The free body diagram for a point loaded element
The point load Pi is to the left of element while the point load Pi+1 is to the right of the element, li
is the length of the beam element. The P loads act downwards, hence they are positive following
the sign convention.
Analysis of the beam can be described by
{
{
} = [ ]{
} = [ ]{
},
(3.1)
},
(3.2)
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{
}=[
][ ][ ]{
},
(3.3)
Where, [ ] is the field matrix, and is obtained as follows:
Consider the above free body diagram of a point loaded beam element. By applying equilibrium
equations of statics (John Case, 1999)
=−
(3.4)
=
+
(3.5)
=
=
=
+
+
=
(3.6)
+
+
(3.7)
+
+
(3.8)
Where A and B are constants of integration. Applying finite element boundary conditions, such
that at
= 0,
=
,
=
Then we find that
=
,
=
,
=
+
=
+
+
=
+
+
+
(3.9)
2
Giving the transfer matrix
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⎡1
⎢
⎢
[ ] = ⎢0
⎢0
⎢0
⎣0
1
0
0
0
2
6
2
1
0
0
2
0⎤
⎥
⎥
0
⎥
0⎥
0⎥
1⎦
1
0
Hence
[
][ ][ ]
Gives
1
⎡0
⎢
⎢0
⎢0
⎣0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
⎡1
0
⎢
0 ⎤ ⎢
⎥
0 ⎥ ∗ ⎢0
−
⎥ ⎢0
1 ⎦ ⎢0
⎣0
1
0
0
0
2
6
2
1
0
0
2
0⎤ 1
⎥ ⎡
⎥ 0
0 ∗ ⎢0
⎥ ⎢
0⎥ ⎢0
0⎥ ⎣0
1⎦
1
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
−
1
⎤
⎥
⎥
⎥
⎦
Giving the general transfer matrix to be
⎡1
⎢
⎢
0
⎢
⎢0
⎢0
⎣0
1
0
0
0
2
6
2
1
0
0
2
1
0
−
−
−
6
2
−
−
1
⎤
⎥
⎥
⎥
⎥
⎥
⎦
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3.2.2 FOR UNIFORMLY DISTRIBUTED LOADING ONLY
y
wi
Mi+1
θi+1
mi+1
Mi
θi
vi+1
vi
yi+1
yi
x
x
li
Figure 3.2: The free body diagram for a UDL loaded beam element
Equilibrium equation of statics (John Case, 1999)
=−
+
(3.10)
Deflection equation (Dimarogonas, 1989)
=
At
+
−
=
=
= ,
=
=
=
+
(3.11)
+
−
(3.12)
−
+
(3.13)
−
+
(3.14)
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=
+
−
+
+
(3.15)
Where A and B are constants of integration, and li is the length of the beam element. Applying
finite element boundary conditions that at
= 0,
=
,
=
Then
=
,
=
,
Giving the field matrix
⎡1
⎢
⎢
⎢0
[ ]=⎢
⎢0
⎢
⎢0
⎣0
1
2
6
2
2
0
1
0
0
0
0
1
0
⎤
⎥
⎥
−
6 ⎥
⎥
⎥
−
2 ⎥
⎥
−
1 ⎦
−
24
This is the total transfer matrix for a beam element loaded with a UDL only
19
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3.2.3 FOR LINEARLY DISTRIBUTED LOADIG (LDL) ONLY
y
wi
Mi+1
n
θi+1
mi+1
Mi
θi
vi+1
Vi
yi+1
yi
x
li
x
Figure 3.3: The free body diagram for an LDL loaded beam element
Equilibrium equation of statics gives
=−
+
(3.16)
From trigonometry,
= , where
The LDL at
is the LDL at
can be modeled as a point load of magnitude
and this acts at from the
section (a property of the shape of the loading).
Therefore, we can write
=
Deflection equation yields
20
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=
+
−
In essence,
(3.17)
is the slope of the linearly distributed load.
=
=
+
= ,
At
−
=
=
+
=
+
+
+
(3.18)
−
+
−
(3.19)
+
+
(3.20)
Where A and B are constants of integration.
Applying finite element boundary conditions that at
= 0,
=
,
=
Hence
=
,
=
,
And
=
+
+
=
+
+
−
+
(3.21)
−
(3.22)
Giving the transfer matrix to be
21
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⎡1
⎢
⎢
⎢0
[ ]=⎢
⎢0
⎢
⎢
⎢0
⎣0
1
2
6
2
2
0
1
0
0
1
0
0
0
−
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
120
−
24
−
−
6
2
1
3.2.4 FOR MOMENT LOADING ONLY
y
Ni+1
Ni
Mi+1
θi+1
mi+1
Mi
θi
vi+1
vi
yi+1
yi
x
li
x
Figure 3.4: The free body diagram for a moment loaded beam element
Where
and
are clockwise moments applied at the nodes and + 1 respectively, and in
matrix notation
22
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1
⎡0
⎢
[ ] = ⎢0
⎢0
⎣0
1
⎡0
⎢
] = ⎢0
⎢0
⎣0
[
⎡1
⎢
⎢
[ ] = ⎢0
⎢0
⎢0
⎣0
0
1
0
0
0
0
1
0
0
0
1
0
0
0
0
0
1
0
0
0
0
1
0
0
0
0
0
1
0
0
0⎤
⎥
⎥
0⎥
1⎦
0
0
0
1
0
0
0 ⎤
⎥
⎥
0 ⎥
1 ⎦
2
6
2
1
0
0
2
0⎤
⎥
⎥
0
⎥
0⎥
0⎥
1⎦
1
0
This can therefore be written as
{
}=[
][ ]{
},
(3.23)
Where
{
} = [ ]{
},
(3.24)
Hence
{
}=[
][ ][ ]{
},
(3.25)
And the total transfer matrix becomes
[
][ ][ ], which is found to be
⎡1
⎢
⎢
0
⎢
⎢0
⎢0
⎣0
1
0
0
0
2
6
2
1
0
0
2
−
2
−
−(
1
0
+
0
1
⎤
⎥
⎥
⎥
)⎥
⎥
⎦
23
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3.2.5 FOR UDL PLUS POINT LOADING ONLY
Pi+1
Pi
y
wi
Mi+1
θi+1
mi+1
Mi
θi
vi+1
vi
yi+1
yi
x
li
x
Figure 3.5 the free body diagram for a UDL loaded beam element plus Point load.
By principles of linear superposition, the state vector equation can be written as,
{
} = [ ]{
{
}=[
},
][ ]{
(3.26)
},
(3.27)
But from equation 3.2
{
} = [ ]{
},
Hence
{
}=[
][ ][ ]{
},
(3.28)
Where [ ] is the field matrix for a UDL beam element
The total transfer matrix is thus [
][ ][ ] and is found to be
24
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1
⎡0
⎢
⎢0
⎢0
⎣0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
⎡1
0
⎢
0 ⎤ ⎢0
⎥
0 ⎥∗⎢
⎥ ⎢0
1 ⎦ ⎢0
⎣0
−
1
−
0
1
0
0
0
0
⎡1
⎢
⎢
⎢0
= ⎢
⎢0
⎢
⎢0
⎣0
⎤
⎥ ⎡1
⎥ ⎢0
⎥ ∗ ⎢0
⎥ ⎢0
⎥ ⎣0
⎦
1
−
1
0
−
1
2
6
2
2
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0⎤
⎥
0⎥
⎥
1⎦
) ⎤
⎥
⎥
−(
+
) ⎥
6
2
⎥
−(
+
) ⎥
⎥
2
−(
+ −
)⎥
⎦
1
−(
1
0
24
+
6
3.2.6 FOR POINT LOAD PLUS MOMENT LOADING ONLY
y
Pi+1
Pi
Ni+1
Ni
Mi+1
θi+1
mi+1
Mi
θi
vi+1
vi
yi+1
yi
x
li
x
Figure 3.6: The free body diagram for a point load plus a moment loaded beam element
Analysis of the element gives,
25
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{
} = [ ]{
{
} = [ ][ ]{
{
}=[
Hence {
Where
},
][
}=[
(3.29)
},
(3.30)
]{
][
},
(3.31)
][ ][ ][ ]{
},
[
1
⎡0
⎢
] = ⎢0
⎢0
⎣0
0
1
0
0
0
0
0
1
0
0
0
0
0
0
0 −
1
0
0
1
⎤
⎥
⎥
⎥
⎦
[
1
⎡0
⎢
] = ⎢0
⎢0
⎣0
0
1
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1 −
0
1
⎤
⎥
⎥
⎥
⎦
2
6
2
1
0
0
2
⎡1
⎢
⎢
[ ] = ⎢0
⎢0
⎢0
⎣0
1
0
0
0
(3.32)
1
0
0⎤
⎥
⎥
0
⎥
0⎥
0⎥
1⎦
1
⎡0
⎢
[ ] = ⎢0
⎢0
⎣0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
−
0
1
⎤
⎥
⎥
⎥
⎦
1
⎡0
⎢
[ ] = ⎢0
⎢0
⎣0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
−
1
⎤
⎥
⎥
⎥
⎦
This gives the transfer matrix for this element as [
][
][ ][ ][ ] which when multiplied
gives,
26
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⎡1
⎢
⎢
⎢0
⎢
⎢0
⎢0
⎣0
1
0
0
0
2
6
2
1
0
0
2
) ⎤
⎥
⎥
−(
+
) ⎥
2
⎥
−( +
+
)⎥
−( −
) ⎥
⎦
1
−(
1
0
2
+
6
3.2.7 FOR LDL PLUS UDL ONLY
y
wPii+1
n
ui
Mi+1
θi+1
mi+1
Mi
θi
vi+1
vi
yi+1
yi
x
li
x
Figure 3.7: The free body diagram for a UDL beam element plus LDL load only
Equilibrium equation of statics gives (John Case, 1999)
=
+
−
(3.33)
27
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The moment equation gives
=
+
−
=
=
= ,
At
−
+
=
=
−
−
+
+
=
(3.34)
−
−
+
−
−
−
(3.35)
−
(3.36)
+
(3.37)
+
+
(3.38)
Where A and B are constants of integration. Applying finite element boundary conditions that at
= 0,
=
,
=
Hence
=
,
=
,
And
=
+
+
−
−
=
+
+
+
−
(3.39)
−
(3.40)
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Which combined gives the transfer matrix to be
⎡1
⎢
⎢
⎢0
[ ]=⎢
⎢0
⎢
⎢
⎢0
⎣0
1
2
6
2
2
0
1
0
0
1
0
0
0
)⎤
⎥
⎥
−(
+
) ⎥
24
6
⎥
−(
+
) ⎥
6
2
⎥
⎥
−(
+
) ⎥
2
⎦
1
−(
+
120
24
3.2.8 FOR UDL PLUS MOMENT LOADING ONLY
y
wi
Ni+1
Mi+1
Ni
θi+1
mi+1
Mi
θi
Vi+1
Vi
yi+1
yi
x
li
x
Figure 3.8: The free body diagram for a UDL plus a moment loaded beam element
Note that [ ] is the field matrix of a UDL loaded beam element. Using superposition principles,
the state vector can be written as
{
{
} = [ ]{
} = [ ]{
},
(3.41)
},
(3.42)
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{
}=[
Hence {
]{
}=[
},
(3.43)
][ ][ ]{
},
(3.44)
This gives the transfer matrix for the element as
[
][ ][ ]. Upon multiplying, it gives the transfer matrix as
⎡1
⎢
⎢
⎢0
⎢
⎢0
⎢
⎢0
⎣0
1
2
6
2
2
0
1
0
0
0
0
⎤
⎥
⎥
−(
+
) ⎥
6
⎥
−(
+ +
)⎥
⎥
2
⎥
−(
)
⎦
1
−(
1
0
+
24
)
2
3.2.9 FOR POINT LOADING PLUS LDL ONLY
Pi+1
y
wi
Pi
Mi+1
n
θi+1
mi+1
Mi
θi
Vi+1
Vi
yi+1
yi
x
li
x
Figure 3.9: The free body diagram for an LDL plus point loading beam element
30
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Note that [ ] is the field matrix of an LDL loaded beam element. Using superposition
principles, we can write
{
} = [ ]{
},
{
} = [ ]{
},
{
}=[
]{
Hence {
}=[
(3.45)
(3.46)
},
(3.47)
][ ][ ]{
},
(3.48)
This gives the transfer matrix for the element as[
][ ][ ].
Upon multiplying it gives the transfer matrix as
⎡1
⎢
⎢
⎢0
⎢
⎢0
⎢
⎢
⎢0
⎣0
1
2
6
2
2
0
1
0
0
1
0
0
0
) ⎤
⎥
⎥
−(
+
) ⎥
24
2
⎥
−(
+
) ⎥
6
⎥
⎥
−( +
−
)⎥
2
⎦
1
−(
120
+
6
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3.2.10 FOR MOMENT LOADING PLUS LDL LOADING ONLY
y
wi
Ni
Ni+1 Mi+1
n
θi+1
mi+1
Mi
θi
Vi+1
Vi
yi+1
yi
x
li
x
Figure 3.10: The free body diagram for an LDL load plus moment loading beam element
Note that [ ] is the field matrix of an LDL loaded beam element. Using superposition
principles, we can write
{
} = [ ]{
},
{
} = [ ]{
},
{
}=[
]{
Hence {
}=[
(3.49)
(3.50)
},
][ ][ ]{
(3.51)
}
(3.52)
This gives the transfer matrix for the element as
[
][ ][ ].
32
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Upon multiplying it gives the transfer matrix as
⎡1
⎢
⎢
⎢0
⎢
⎢
⎢0
⎢
⎢
⎢0
⎣0
1
2
6
2
2
0
1
0
0
1
0
0
0
−
+
120
−
+
24
−
+
6
−
2
2
1
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
3.2.11 FOR POINT LOADING PLUS UDL PLUS MOMENT ONLY
Pi+1
y
Pi
wi
Ni+1
Mi+1
Ni
θi+1
mi+1
Mi
θi
Vi+1
Vi
yi+1
yi
x
li
x
Figure 3.11: The free body diagram for a UDL plus point loaded plus a moment loaded beam element
Note that [ ] is the field matrix of an UDL loaded beam element. Using superposition
principles, we can write
{
} = [ ]{
{
} = [ ][ ]{
{
}=[
},
][
(3.53)
},
(3.54)
]{
},
(3.55)
33
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Hence {
}=[
][
][ ][ ][ ]{
},
(3.56)
This gives the transfer matrix for the element as
[
][
][ ][ ][ ].
Upon multiplying gives the transfer matrix as
⎡1
⎢
⎢
⎢0
⎢
⎢
⎢0
⎢
⎢0
⎣0
1
2
6
2
2
0
1
0
0
0
0
−
24
−
−
1
0
+
+
6
+
2
−(
+
2
+
+
+
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
6
2
+
−
)
1
3.2.12 FOR UDL PLUS LDL PLUS POINT LOADING ONLY
Pi+1
wPii+1
Pi
y
n
ui
Mi+1
θi+1
mi+1
Mi
θi
vi+1
vi
yi+1
yi
x
li
x
Figure 3.12: The free body diagram for a UDL plus LDL plus point loaded beam element only
34
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Note that [ ] is the field matrix of an UDL plus LDL loaded beam element. Using superposition
principles, we can write
{
} = [ ]{
},
{
} = [ ]{
},
{
}=[
]{
Hence {
}=[
(3.57)
(3.58)
},
(3.59)
][ ][ ]{
},
(3.60)
This gives the transfer matrix for the element as[
][ ][ ].
Upon multiplying gives the transfer matrix as
⎡1
⎢
⎢
⎢0
⎢
⎢
⎢0
⎢
⎢
⎢0
⎣0
1
2
6
2
2
0
1
0
0
1
0
0
0
−
120
−
24
−
−
6
+
+
+
24
2
+
6
+
2
+
1
+
6
2
+
−
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
35
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3.2.13 FOR UDL PLUS LDL PLUS MOMENT LOADING ONLY
y
wPii+1
n
ui
Ni+1
Ni
Mi+1
Mi
Vi+1
Vi
yi+1
yi
x
x
li
Figure 3.13: The free body diagram for a UDL plus LDL plus moment loaded beam element only
Note that [ ] is the field matrix of an UDL plus LDL loaded beam element. Using superposition
principles, we can write
{
} = [ ]{
},
{
} = [ ]{
},
{
}=[
]{
Hence {
}=[
(3.61)
(3.62)
},
][ ][ ]{
(3.63)
},
(3.64)
This gives the transfer matrix for the element as[
][ ][ ].
36
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Upon multiplying gives the transfer matrix as
⎡1
⎢
⎢
⎢0
⎢
⎢
⎢0
⎢
⎢
⎢0
⎣0
1
2
6
2
2
0
1
0
0
1
0
0
0
−
−
−
+
120
+
24
6
+
−
+
24
+
6
+
2
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
2
+
+
2
1
3.2.14 FOR POINT LOADING PLUS MOMENT LOADING PLUS LDL LOADING ONLY
Pi+1
y
Pi
wi
Ni+1
n
Ni
Mi+1
Mi
Vi+1
Vi
yi+1
yi
x
li
x
Figure 3.14: The free body diagram for an LDL loaded plus point loading plus moment loading beam element
Note that [ ] is the field matrix of an LDL loaded beam element. Using superposition
principles, we can write
{
} = [ ]{
{
} = [ ][ ]{
},
(3.65)
},
(3.66)
37
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{
}=[
Hence {
][
}=[
]{
},
][
(3.67)
][ ][ ][ ]{
},
(3.68)
This gives the transfer matrix for the element as[
][
][ ][ ][ ].
Upon multiplying the transfer matrix is given as
⎡1
⎢
⎢
⎢0
⎢
⎢
⎢0
⎢
⎢
⎢0
⎣0
1
2
6
2
2
−
−
0
1
−
0
0
1
0
0
0
+
120
+
24
−
2
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
2
+
2
+
6
+
6
+
+
+
−
1
3.2.15 FOR POINT LOAD PLUS UDL PLUS LDL PLUS MOMENT LOADING ONLY
This is the upper limit case of all the possible loadings acting at the same time on a finite
beam element for the considered possibilities.
Pi+1
y
Pi
wPii+1
n
ui
Ni+1
Ni
Mi+1
vi+1
Mi vi
yi+1
yi
x
li
x
Figure 3.15: The free body diagram for an LDL loaded plus point loading plus UDL plus moment loading beam
38element
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Note that [ ] is the field matrix of an UDL plus LDL loaded beam element. Using superposition
principles, we can write
{
} = [ ]{
{
} = [ ][ ]{
{
}=[
Hence {
},
(3.69)
},
][
(3.70)
]{
}=[
},
][
(3.71)
][ ][ ][ ]{
},
(3.72)
This gives the transfer matrix for the element as[
][
][ ][ ][ ]. (Dimarogonas, 1989)
Upon multiplying the transfer matrix is given as
⎡1
⎢
⎢
⎢0
⎢
⎢
⎢0
⎢
⎢
⎢0
⎣0
1
2
6
2
2
0
1
0
0
1
0
0
0
−
−
−
+
120
+
24
6
−
+
2
+
24
+
6
2
+
2
+
+
+
+
1
+
6
2
+
−
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
39
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CHAPTER FOUR
4.1 ALGORITHM DESIGN FOR SHAFTS AND BEAMS
The algorithm of the transfer matrix method is defined in two principal steps:-

The unknown initial parameter is defined by the matrix of initial parameters which are
transferred using the matrix multiplication of transfer and nodal matrices into the end
point of the applied simulation. The boundary condition in the end point, implemented in
the matrix of boundary conditions, define the set of algebraic equations for determination
of the unknown initial parameters.

The calculated initial parameters are put into the state vector in initial point of simulation
and repeated multiplications with transfer and nodal matrices determine the set of state
vectors in the other nodes on the beam.
Given:

The stepped shaft is loaded as shown in figure 1.1 and is fixed at both ends.
The fixed shaft is statically indeterminate hence cannot be solved by simple equations of
statics. Couples are supplied at the ends by the resistance of the supports to deformation. The
couples are termed fixed-end moments, and the main problem of the built in beam is the
determination of these couples. (John Case, 1999)

The shaft has n number of elements to be specified by the end user.
An element is a loaded section carrying either a constant uniformly distributed load (UDL), a
constant linearly distributed load (LDL), or is demarcated by point loading or moment
loading or carries a combination of these.
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Required:

Read in the number of elements, n.

Calculate the moment of inertia of each element
Define the geometry of the cross sectional shape of each element which are rectangular and
circular for this program.

Read in the point loads acting at each element

Read in the modulus of elasticity

Read in the lengths of each element.

Generate both field and nodal matrices for each element

Multiply all the generated matrices to determine the universal matrix [A]

Calculate the unknowns M1, V1,
and
These are the moments and shear forces at the ends.

Determine the state vector at each node
Uses the state vector calculated for the first node and multiplies with the transfer matrix of
the first element to determine the state vector at the second node. Use the state vector at the
second node multiplied by the transfer matrix of the second element to determine the state
vector at the third node. The process is repeated till the last node.
4.2 PSEUDOCODE
1. Begin
2.
Declare all the variable identifiers namely number of elements, n, length of each
element li, moment of Inertia of each element, Ii, Point loads of each element pi , UDL
of each element, LDL of each element wi and moment of each element Mi, Modulus
of elasticity Ei of each element and index i.
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3. Initialize the variables,

n=0

l1, l2, l3….ln = 0

I1, I2, I3…..In = 1

p1, p2, p3…pn = 0

Ei = 1

i=0
4. Read in the number of elements, n
5. Read in the modulus of Elasticity, E.
6. Repeat
Choose the cross sectional shape of each element
a. Rectangular
b. Circular
If (a), Read in breadth (b) and height (d)
3
Calculate = 12
Else if (b), read in diameter, D
4
Calculate = 4
Write I
Index = index + 1
Until index < n+1
7. Repeat

Input the point load acting at node i

Read in the point load acting at node i

Index = index + 1

Until index < n+1;
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8. Repeat

Input the length of the element i

Read in the length of the element i

Index = index + 1

Until index < n+1;
9. Read in all the field and nodal matrices
10. Calculate the matrix [A]
11. Determine the unknowns M1, V1,
and
12. Determine all the required state vectors
13. End
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4.3 FLOWCHART
Begin
Declare the variables:
Number of elements, lengths, Young’s Modulus,
moment of Inertia, point loads, UDL, LDL, moment
load and index
Initialize the variables
l1, l2, l3….ln = 0;
Ei = 1;
i = 1;
I1 , I2, I3…..In = 1;
p1 ,p2, p3…pn = 0;
ui, wi and Mi = 0;
n = 0.
Input the number
of elements, n
n>=1
Yes
No
A
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A
Rectangle
Shape of
Element
Input breadth (b) and height (d)
Circular
Input diameter (D)
Read in breadth (b) and height
(d)
Read in diameter (D)
I=
I=
Display I
Index = index +1
Index>n+1
No
Yes
B
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B
Input the length
of the element
Read in the length
of the element
Index = index +1
Index>n+1
No
Yes
Input the load
of element
Read in the
load
Index = index +1
D
C
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C
D
Index>n+1
No
Yes
Input the Young’s Modulus
of the element
Read in the
Young’s modulus
Index = index +1
Index>n+1
No
Yes
Input all the field
and point matrices
for all the elements
E
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E
Calculate Matrix [A]
Calculate M1, V1, Mn and Vn
Calculate all state vectors at
each node
Display the
solution
Stop
Figure 4.1: Flowchart for calculating the slope, displacement, moment and reaction of a beam
section using the transfer matrix method
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4.4 COMPUTER ANALYSIS USING C++
4.4.1 About C++
The C++ language was conceived and developed in the early 1980’s by Dr. Bjarne Stroustrup at
AT & T’s Bell Labs in New Jersey. It is an object-oriented programming language that is quickly
replacing C as the predominant programming language in terms of flexibility, speed, portability
and power. (McGregor, 1999)
The four major steps in creating the C++ program are
1. Deciding what the program is supposed to do
2. Figuring out how you want the program to do it.
3. Writing the source code for the program.
4. Debugging and testing the program
4.4.2 The design of the C++ program for beams and shafts
The following is the design of the computer program that analyses the loading of beams and
shafts using the transfer matrix method.
The first entry is inputting the number of elements in the beam or shaft which is done in the main
menu function. See appendix (code for beams and shafts): lines 46 and 47.
The second step is choosing the type of loading for each element starting with the last element.
See appendix (code for beams and shafts): lines 322-398.
For every loading selected above the user is prompted to choose the geometry of the element to
calculate the second moment of area I for two scenarios only. That is, circular or rectangular.
The second moment of area was coded in a ‘function’ which performs its task using its own
internal data, along with any data passed into the function.
See appendix (code for beams and shafts) lines 101-119.
long double momentOfInertia()
{
char choice;
do
{
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cout<<" \n\n\n\tCHOOSE THE CROSS-SECTIONAL SHAPE OF THE ELEMENT ";
cout<<" \n\n\tCHOOSE CHARACTER <C or R> AND PRESS ENTER";
cout<<" \n\n\t<C>: FOR CIRCULAR CROSS-SECTIONAL SHAPE " ;
cout<<" \n\n\t<R>: FOR RECTANGULAR CROSS-SECTIONAL SHAPE: ";
cin>>choice;
switch(choice)
{
case 'C':
case 'c':
return
circularMomentOfInertia();
break;
case 'R':
case 'r':
return
rectangularMomentOfInertia();
break;
default:
cerr<<" \n\n\aINVALID INPUT..........CHOOSE AGAIN\a\a";
break;
}
}
while((choice!='C')&&(choice!='c')&&(choice!='R')&&(choice!='r'));
return 0;
}
During coding of the above function, the ‘switch statement’ was used since it allows a program
to determine the best course of execution based on an expression value. (McGregor, 1999)
After inputting the loads on all the elements the transfer matrices are formed. Then these transfer
matrices are multiplied sequentially from the last element to the first element. To achieve this
matrix multiplication, a for loop is used since it executes the multiplication statement a specified
number of times i.e. n times. (McGregor, 1999) See appendix (code for beams and shafts) lines
59-83.
long double totalTransferMatrix(int n)
/* this is a function that loads the beam/shaft with the various loadings
and returns a universal product matrix to determine state of node 1*/
{
long double p[5][5];/* a temporary storage product matrix */
int index, i,j,k;
index=n;
ofstream fout("loading.cpp",ios::left);
//open a file to hold the loadings for all the elements
do
{
cout<<"\n\nCHOOSE THE LOADING ON ELEMENT :"<<index;
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mainMenu();
cout<<"\n\n________________________________________________________________________________
______________\n";
if(n==1)
{
/*assign the loading matrix to product matrix*/
for(i=0; i<5; i++)
{
for(j=0; j<5; j++)
{
productMatrix[i][j]=loading[i][j];
}
}
}
else
{
/**
Assign productMatrix=productMatrix * returnedMatrix */
for(i=0; i<5; i++)
for(j=0; j<5; j++)
{
p[i][j] = 0;
for(k=0; k<5; k++)
// same as z[i][j] = z[i][j] + x[i][k] * y[k][j];
p[i][j] += productMatrix[i][k] * loading[k][j];
}
/*assign the product to productMatrix*/
for(i=0; i<5; i++)
{
for(j=0; j<5; j++)
{
productMatrix[i][j]=p[i][j];
}
}
}
index=index-1;
}while(index>0);
fout.close();//close the file holding the loadings
for (i=0; i<5; i++)
{
cout<<"\n\n";
for(j=0; j<5; j++)
// display the result...
cout<<"\t"<<productMatrix[i][j];
}
return productMatrix[i][j];
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Figure 4.2 Screen output for choosing element loadings
The user also inputs the modulus of elasticity and the length of the element. From the properties
of the element (length of element, modulus of elasticity and cross sectional shape), the transfer
matrix of the element is displayed.
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Figure 4.3 Screen output showing the transfer matrix of the element
If there were two elements as implied by figure 4.2, the same procedure is repeated for the first
element, and the transfer matrix determined.
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Figure 4.4 Screen output showing the reactions and moments
The transfer matrices of the last element and the first element are multiplied to obtain the
product of the transfer matrices as shown in figure 4.4.
The results display the reactions and moments at the ends of the beam.
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CHAPTER FIVE
VALIDATION
Example 5.1
Consider a beam fixed at both ends carrying a uniformly distributed load of 1 kN/m for a length
of 3 meters and another uniformly distributed load of 2 kN/m for a length of 2 meters. The
loading of the beam is as shown in figure 4.1. (E = 200 GN/m2, circular cross section shape of
25mm diameter)
2kN/m
1kN/m
3m
2m
1ST element
2ND element
Figure 5.1: A beam with uniformly distributed load
Consider the first element,
y
w1
M2
M1
x
l1
x
Figure 5.2: Free body diagram of the first element
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Applying static equations for the element (Singh, 1999)
+
=
=
(4.1)
−
(4.2)
M = V x+ M −
(4.3)
= -Mx = M − V x +
EI
(4.4)
Hence
EI
= M1x -
+
+A
(4.5)
-
+
+ Ax + B
(4.6)
EIy =
Applying boundary equations for the element,
At x=0, y = y1 and θ = θ1
A = EIθ1 and
B = EIy1
Substituting back to equations (4.5) and (4.6)
y2 = y1 + l1θ1 +
+
−
θ2 =
+
−
θ1 +
(4.7)
(4.8)
2
M2 =
M1 + l1V1 − 21 1
(4.9)
In matrix notation,
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⎧
⎪
⎨
⎪
⎩
⎡1
⎢
⎫
⎢0
⎪
= ⎢
⎢0
⎬
⎪
⎢
1⎭
⎢0
⎣0
2
1
1
3
1
2
6
2
1
1
1
2
0
0
0
1
0
0
1
1
0
4
− 241 1 ⎤
3⎥
1 1⎥
−
6
⎧
⎪
⎫
⎪
⎥
⎬
− 2 ⎥ ⎨
⎪
⎥ ⎪
⎩1⎭
−
⎥
1 ⎦
2
1 1
Similarly, the field transfer matrix for the second element is
⎡1
⎢
⎢0
[L2] = ⎢
⎢0
⎢
⎢0
⎣0
2
2
2
3
2
2
6
2
2
2
1
4
− 242 2 ⎤
−
2
0
0
0
1
0
0
3⎥
2 2⎥
6
⎥
− 2 ⎥
⎥
−
⎥
1 ⎦
2
2 2
2
1
0
{ }=[ ]{ }
{z3} = [L2] {z2} = [L2] [L1] {z1}
Where { }, { }
⎧
⎪
⎨
⎪
⎩
⎡1
⎢
⎫ ⎢
⎪ 0
=⎢
⎬ ⎢0
⎪
1⎭ ⎢
⎢0
⎣0
=
=
2
1
0
0
0
{z3} are state vectors at nodes 1, 2 and 3 respectively.
2
2
2
3
2
6
2
2
2
2
1
0
0
2
1
0
4
− 242 2 ⎤ ⎡1
3⎥
2 2⎥
⎢
− 6 ⎢0
⎥⎢
2
2 2⎥⎢
− 2
0
⎥⎢
−
⎥ ⎢0
1 ⎦ ⎣0
1
2
1
2
6
2
1
1
1
0
0
0
3
1
2
1
0
0
1
1
0
4
− 241 1 ⎤
−
3⎥
1 1⎥
6
⎧
⎪
⎫
⎪
⎥
⎬
− 2 ⎥⎨
⎪
⎥⎪
⎩1⎭
−
⎥
1 ⎦
2
1 1
(4.10)
(Ryder, 1969)
∗
.
= 1.1917475e-8 m4
Substituting the values of E and I into equation (4.10)
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⎧
⎪
1
⎫
⎪
⎡0
⎢
= ⎢0
⎨ ⎬
⎢0
⎪ ⎪
⎣0
⎩1⎭
1
3
1
0
0
0
⎡0
⎢
∗ ⎢0
⎢0
⎣0
2
1
0
0
0
5.2159 − 4
5.2159 − 4
1
0
0
1.117342 − 3
7.82278 − 4
1
0
0
3.47679 − 4
5.2159 − 4
2
1
0
1.117342 − 3
1.117342 − 3
3
1
0
−3.47679 − 4
− 6.95358 − 4⎤
⎥
−4
⎥
−4
⎥
⎦
1
−8.80063 − 4
− 1.117342 − 3⎤
⎥
−4.5
⎥
−3
⎥
⎦
1
⎧
⎪
⎫
⎪
⎨ ⎬
⎪ ⎪
⎩1⎭
the product of the transfer matrices is given by
1
⎡0
⎢
= ⎢0
⎨ ⎬ ⎢0
⎪ ⎪
⎩ 1 ⎭ ⎣0
⎧
⎪
⎫
⎪
5 3.25949 − 3
1 1.3038 − 3
0
1
0
0
0
0
5.43249 − 3 −6.96445 − 3
⎧
3.25949 − 3 − 5.78017 − 3⎤ ⎪
⎥
5
−14.5
⎥
⎥⎨
1
−7
⎪
⎦⎩ 1
0
1
⎫
⎪
⎬
⎪
⎭
Applying the boundary conditions for both ends fixed
=
0
⎧ ⎫
⎪0⎪
=0
1
⎡0
⎢
= ⎢0
⎨ ⎬ ⎢0
⎪ ⎪
⎩ 1 ⎭ ⎣0
=
5 3.25949 − 3
1 1.3038 − 3
0
1
0
0
0
0
=0
0
5.43249 − 3 −6.96445 − 3
⎧
⎤
3.25949 − 3 − 5.78017 − 3 ⎪ 0
⎥
5
−14.5
⎥
⎥⎨
1
−7
⎪
⎦⎩ 1
0
1
⎫
⎪
⎬
⎪
⎭
In algebraic form,
0 = 3.25949 − 3
0 = 1.3038 − 3
+ 5.43249 − 3
+ 3.25949 − 3
− 6.96445 − 3
− 5.78017 − 3
Solving simultaneously,
V1 = 2.756 kN and M1 = -2.45673 kNm
V3 = -4.244 kN and M3 = -3.17674 kNm
The results are similar to those shown in the screen output of figure 4.4.
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Example 5.2
A fixed beam is loaded as shown in figure 5.3. The end reactions and fixing moments are
determined. E = 200Gpa, I = 3600cm4 (Singh, 1999)
20 kNm
5kN/m
B
A
2m
1.5m
4m
Figure 5.3: A fixed beam loaded with uniformly distributed load and moment loading
Using Macaulay’s method (Singh, 1999), the free body diagram of the fixed beam is as shown
below.
X
20 kNm
5kN/m
B
A
MB
MA
2m
RA
1.5m
4m
RB
x
x
Figure 5.4: Free body diagram of the fixed beam
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Consider a distance x-x at a distance x from A, then
=
EI
EI
+
=−
( − 2) −
= −
−
− ( − 3.5)
( − 2) +
=
− 20( − 2) +
=
− 10( − 2) +
(4.11)
+ ( − 3.5)
(4.12)
+ ( − 3.5) +
+
( − 3.5) +
(4.13)
+
(4.14)
Apply boundary conditions
At x = 0, y = 0 and
=0
Therefore
=0
At
= ,
=0
=0
Substituting into equations (4.13) and (4.14)
0=
0=
( . )
( . )
− 20(7.5 − 2) +
(7.5) + (7.5 − 3.5)
− 10(7.5 − 2) +
.
+
(7.5 − 3.5)
(4.15)
(4.16)
Solving the two simultaneous equations,
RA = 1.04596 kN and MA = 11.466 kNm
From equilibrium equation of statics,
RA + RB = 20kN
RB = 20 - 1.04596 = 18.957 kN
Taking moments about B
MB + 20 + 7.5RA = MA + 40
MB = 11.466 + 40 - (7.5*1.04596) – 20
MB = 23.644 kNm
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Using the computer program to analyze the loading of the beam, the following output is obtained
Figure 5.5 Screen output for validation of beam in example 5.2
Table 5.1 Comparison of Macaulay’s results and transfer matrix output for figure 5.3
Macaulay’s results
Transfer matrix output
Reaction at A
1042.96 N
1042.96 N
Reaction at B
18957 N
18957 N
Moment at A
11466 Nm
11466.7 Nm
Moment at B
23644.4 Nm
23644.4 Nm
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Example 5.3
A fixed beam is loaded as shown in figure 5.5. The end reactions and fixing moments are
determined. E = 200Gpa, I = 3600cm4
2kN
20kN/m
2kN
10 kNm
B
A
2m
1m
1.5m
3m
RB
RA
Figure 5.5: A fixed beam loaded with linearly distributed load, a moment and point loadings
Using Macaulay’s method (Singh, 1999), the free body diagram of the fixed beam is as shown
below.
2kN
X
2kN
20kN/m
10 kNm
MA
MB
B
A
2m
1m
1.5m
3m
RB
x
RA
x
Figure 5.6: Free body diagram of the fixed beam
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Consider a distance x-x at a distance x from A, then
=
EI
EI
+
( − 2) −
=−
=−
−
−
( − 4.5) − ( − 2) − ( − 3)
( − 2) +
=
−
( − 2) +
+
=
−
( − 2) +
+
+
( − 4.5) + ( − 2) + ( − 3)
(4.17)
(4.18)
( − 4.5) + ( − 2) + ( − 3) +
( − 4.5) + ( − 2) + ( − 3) +
(4.19)
+
(4.20)
Apply boundary conditions
At x = 0, y = 0 and
=0
Therefore
=0
At
= ,
=0
=0
Substituting into equations (4.19) and (4.20)
( . )
0=
0=
0=
( − 2) +
−
0=
−
6
( − 4.5) + ( − 2) + ( − 3)
+
− 10(7.5 − 2) +
−
( . )
2
−
( − 2) +
(7.5 − 2) +
(7.5) +
2
+
.
360
+
(7.5 − 4.5) + (7.5 − 2) + (7.5 − 3)
( − 4.5) +
6
( − 2) +
6
(4.21)
( − 3)
(7.5 − 4.5) + (7.5 − 2) + (7.5 − 3)
(4.22)
Solving the two simultaneous equations,
RA = 3.397 kN and MA =10.337 kNm
From equilibrium equation of statics,
RA + RB = 34 kN
RB = 34 – 3.397 = 30.603 kN
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Taking moments about B
MB + 10 + 7.5RA = MA + (5.5 * 2) + (4.5*2) + 30
MB = 10.337 + 11 + 9 + 30 - (25.775 + 10)
MB = 24.856 kNm
Using the computer program to analyze the loading of the beam, the following output is obtained
Figure 5.6 Screen output for validation of beam in example 5.3
Table 5.2 Comparison of Macaulay’s results and transfer matrix output for figure 5.5
Macaulay’s results
Transfer matrix output
Reaction at A
3397 N
3396.74 N
Reaction at B
30603 N
30603.3 N
Moment at A
10337 Nm
10337.8 Nm
Moment at B
24856 Nm
24862.2 Nm
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CHAPTER SIX
6.1 CONCLUSION
Although the transfer matrix method is a numerical approach, the results obtained compare
closely to the analytical approach adopted by Macaulay’s as shown in tables 5.1 and 5.2.
The transfer matrix method applied to the analysis of beams and shafts involves a lot of iterative
multiplication of matrices and thus can be programmed to make it faster and easier to give
results.
The transfer-matrix method is used when the total system can be broken into a sequence of
subsystems that interact only with adjacent subsystems. To implement the Transfer matrix
method, a relationship was set that gives the state vector (displacements, slope, moments and
shear forces) at one end of the element in terms of state vector at the other end. Therefore, the
Transfer matrix method is an approach to matrix structural analysis that uses a mixed form of the
element force-displacement relationship and transfers the structural behavior parameters, the
joint forces, displacement, slope and moments from one line like structural systems to the other.
6.2 RECOMMENDATION
 Future Students to develop this program further by analyzing beams of different cross
sections other than rectangular and circular. These include, I sections, T sections and
even channel sections
 Future students to develop the program further so that the program calculates and
displays the state vectors at each node as well as shear stresses, normal stresses and
strains.
 As much as the programming language of C++ is flexible to C language, it is very
cumbersome just forming the matrices. A simple solution is to choose Matlab as the
programming language because the matrices are already incorporated in the system. Java
would also be a suitable program because it provides routine to graphically display the
problem.
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BIBLIOGRAPHY
1. Dilkey, W. (1994). Formulas for stress, strain and structural matrices. New York: Wiley
publishers.
2. Dimarogonas, A. (1989). computer aided machine design. St. Louis: Prentice Hall.
3. John Case, L. C. (1999). Strength of materials and structures. London: Arnold.
4. McGregor, R. (1999). Using C++. Indianapolis, USA: Prentice Hall of India.
5. Ryder, G. (1969). Strength of Materials. Shrivenham: Macmillan.
6. Singh, D. (1999). Strength of materials. Delhi: Khanna publishers.
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