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From Structural Analysis to Finite Element Method

Dhiman Basu IIT Gandhinagar

-------------------------------------------------------------------------------------------------------------------- Acknowledgement

Following text books were consulted while preparing this lecture notes: Zienkiewicz, O.C. and Taylor, R.L. (2000). “The Finite Element Method”, Vol. 1: The Basis,

Fifth edition, Butterworth-Heinemann. Yang, T.Y. (1986). “Finite Element Structural Analysis”, Prentice-Hall Inc. Jain, A.K. (2009). “Advanced Structural Analysis”, Nem Chand & Bros. --------------------------------------------------------------------------------------------------------------------- 1.0 Introduction

Analysis of a civil engineering structure, for example, a rigid jointed frame is often performed using center-line element model, wherein cross-sectional properties are lumped onto the center-line of the element. Analysis procedure is usually displacement based, and direct stiffness method is generally adopted. Note that center-line model is restricted to a very simple configuration and cannot be applied in most cases including varying cross-sections, with opening/discontinuity and with two or three dimensional effects. In such cases, finite element method (FEM) is widely used to capture the reasonable behavior. If FEM is applied to those simple cases wherein center-line model provides reasonable solution, results will be identical. Therefore, some similarity exists in principle between the center-line model (or conventional structural analysis) and FEM. Objective of this lecture is to explore that similarity and illustrate the transition path from conventional structural analysis to FEM.

In what follows next, first equilibrium of a beam element is considered and effect of orthogonal transformation is illustrated. Second, conventional structural analysis using direct stiffness method is discussed followed by a numerical illustration. Third, conventional structural analysis is revisited so as to describe the passage to FEM. Fourth, concept of FEM is briefly discussed to address the similitude with conventional structural analysis followed by a numerical example. Finally, a problem statement is made that describe the generality of FEM approach.

2.0 Element Stiffness Matrix

Consider a two-noded beam element. Neglecting the axial deformation for now, joint displacements (in generalized sense, including rotations as well) and joint forces (including moments also in a generalized sense) are shown in their positive sense in Figure 1.

Using the principle of elementary structural mechanics, equilibrium of the beam element can be expressed as

2  

{ } { }

2 2

1 1

1 1

2 22 2

2 2

12 6 12 6

6 64 2

12 6 12 6

6 62 4

e e eq K a

L L L LY vM EI L LY vL

L L L LM

L L

θ

θ

⎡ ⎤= ⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪−⎢ ⎥⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪− − −⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

(1)

Axial deformation, if considered (Figure 2), under the assumption of small deformation remains uncoupled with respect to others and the equilibrium of the element takes the form as follows:

{ } { }

3 2 3 21

1

2 21

2

2

23 2 3 2

2 2

0 0 0 0

12 6 12 60 0

6 4 6 20 0

0 0 0 0

12 6 12 60 0

6 2 6 40 0

e e eq K a

EA EAL L

EI EI EI EIX L L L LY EI EI EI EIM L L L LX EA EA

L LYEI EI EI EIM

L L L LEI EI EI EIL L L L

⎡ ⎤= ⎢ ⎥⎣ ⎦⎡⎢ −⎢⎢⎢ −⎢⎧ ⎫⎪ ⎪⎪ ⎪ ⎢⎪ ⎪ ⎢⎪ ⎪⎪ ⎪ ⎢⎪ ⎪ −⎪ ⎪ ⎢⎪ ⎪⎪ ⎪ ⎢=⎨ ⎬ ⎢⎪ ⎪ ⎢⎪ ⎪ −⎪ ⎪ ⎢⎪ ⎪⎪ ⎪ ⎢⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪ − − −⎩ ⎭

−⎣

1

1

1

2

2

2

uv

uv

θ

θ

⎤⎥⎥⎥⎥⎥ ⎧ ⎫⎪ ⎪⎪ ⎪⎥⎪ ⎪⎥⎪ ⎪⎪ ⎪⎥⎪ ⎪⎪ ⎪⎥⎪ ⎪⎪ ⎪⎥⎨ ⎬⎥⎪ ⎪⎥⎪ ⎪⎪ ⎪⎥⎪ ⎪⎪ ⎪⎥⎪ ⎪⎢ ⎥⎪ ⎪⎪ ⎪⎢ ⎥⎪ ⎪⎩ ⎭⎢ ⎥

⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎦

(2)

Matrix eK⎡ ⎤⎢ ⎥⎣ ⎦ is known as the stiffness matrix. Any element, say thij , of the stiffness matrix indicates the

force developed along the direction of thi degree of freedom due to a unit displacement along the direction of thj degree of freedom while all others degrees of freedom are held restrained.

Figure 1: Straight beam element without axial degrees of freedom

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Figure 2: Straight beam element with axial degrees of freedom

3.0 Orthogonal Transformation

In previous section, degrees of freedom are chosen along and normal to the axis of the element. However, constituting elements in a structure may have their axes oriented along any arbitrary direction (Figure 3). In such cases, two different coordinate systems are usually considered namely, local and global. In local coordinate system, one of the axes is chosen along the axis of the element and hence, every element has its own local system. On the other hand, global coordinate system is unique to the overall structure and analysis of deformation is carried out in this coordinate system. Note that, local coordinate system of any element is related to the global coordinate system of the structure through a rotation only. Structural analysis often requires transformation of vector from one coordinate system to another.

Figure 3: Frame with arbitrary orientation of element axis

Consider a local coordinate system which is rotated through an angle θ with respect to the global coordinate system as shown in Figure 4. The coordinate of the point P is ( ),x y and ( )' ',x y in local and

global systems, respectively. By simple geometry, it can be shown that

{ } [ ]{ }'

' '

'

cos sin 0sin cos 00 0 1

x xy y

θ θθ θ δ λ δ

θ θ

⎧ ⎫⎧ ⎫ ⎡ ⎤⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎢ ⎥⎪ ⎪⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥= − ⇒ =⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎪ ⎪⎩ ⎭ ⎣ ⎦⎪ ⎪⎩ ⎭

(3)

Since [ ] [ ]1 Tλ λ−= , this is also noted as orthogonal transformation.

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Figure 4: Orthogonal transformation

4.0 Element Equilibrium in Global Coordinate System

Consider an element with axis oriented at an angle φ with respect to the global coordinate system along the counter clockwise direction as shown in Figure 5; over bar denotes the local coordinate system. Displacements and forces acting at the joints in both coordinate systems are related as follows:

{ }

1 11 1

1 11 1

1 11 1

2 22 2

2 22 2

2 22 2

cos sin 0 0 0 0sin cos 0 0 0 00 0 1 0 0 00 0 0 cos sin 00 0 0 sin cos 00 0 0 0 0 1

eL e

u Xu Xv Yv Y

a TMMu Xu Xv Yv Y

MM

φ φφ φ

θθφ φφ φ

θθ

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ = ⎢⎣⎢ ⎥ ⎢ ⎥⎢ ⎥= ⇒⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

{ }{ } { }

eG

eL e eG

a

q T q

⎥⎦⎡ ⎤= ⎢ ⎥⎣ ⎦

(4)

Employing Eq (4) into Eq (2), equilibrium relation in the global coordinate system can be derived as follows:

{ } { } { } { } { } { }{ } { } and

TeL eL eL e eG eL e eG eG e eL e eG

TeG eG eG eG e eL e

q K a T q K T a q T K T a

q K a K T K T

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= ⇒ = ⇒ =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

(5)

Note that the size of the matrices remains same in this transformation.

5.0 Direct Stiffness Matrix Method of Analysis

Step-1: Element Equilibrium in Local Coordinate

For any typical element (say ith), develop the equilibrium equation in local coordinate:

{ } { }6 1 6 16 6

ii ieL eL eLq K a× ××

⎡ ⎤= ⎢ ⎥⎣ ⎦ (6)

Consider { }6 1

ieLq×

as the negative of the fixed end forces due to span loading.

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Figure 5: Local to global coordinate transformation

Step-2: Element Equilibrium in Global Coordinate

Transform the equilibrium equation from local to global coordinate.

{ } { }

{ } { } { } { }6 1 6 16 6

6 1 6 1 6 1 6 16 6 6 6 6 6 6 6 6 6 6 6

, ,

ii ieG eG eG

i T i T Ti i i ieG ei eL ei eG ei eL eG ei eL

q K a

K T K T q T q a T a

× ××

× × × ×× × × × × ×

⎡ ⎤= ⎢ ⎥⎣ ⎦

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

(7)

Step-3: Element Equilibrium in Expanded Global Coordinate

Note that the degrees of freedom considered in Eq (7) are oriented along the global coordinate system and hence, a connectivity matrix must be formed that relates each of these to the global degrees of freedom. With the help of connectivity matrix, Eq (7) can be expanded to the global size of the problem, which is 3N and N is the number of nodes including that are restraint.

{ } { }3 1 3 13 3

ii iExp eG Exp eG Exp eG

N NN Nq K a

× ××⎡ ⎤= ⎢ ⎥⎣ ⎦ (8)

Step-4: Assemble Element Equilibrium in Expanded Global Coordinate

Assemble Eq (8) for elements. Let the number of elements be M. Then

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{ } { }( )3 1 3 13 31 1

M M ii iExp eG Exp eG Exp eG

N NN Ni i

q K a× ××

= =

⎡ ⎤= ⎢ ⎥⎣ ⎦∑ ∑ (9)

After assembling Eq (9) and taking into account the externally applied joint forces { }*

3 1Nq

×in global

coordinate, it may be written as

{ } { } { }*

3 1 3 1 3 13 3

G G G

N N NN Nq q K a

× × ××⎡ ⎤+ = ⎢ ⎥⎣ ⎦ (10)

Here, 3 3

G

N NK

×⎡ ⎤⎢ ⎥⎣ ⎦ , { }

3 1

G

Nq

× { }

3 1

G

Na

×represent the stiffness matrix, force vector associated with span

loading and deformation vector, respectively, in global coordinate but without accounting for the effect of restraint.

Step-5: Effect of Restraints

Identify the restraint degrees of freedom and remove the associated rows and columns from the stiffness matrix. Also remove the associated elements from the force and deformation vectors. Let the problem be reduced to a size S and Eq (10) is more formally written as

{ } [ ] { }1 1S S S Sq K a

× × ×= (11)

Step-6: Solution for Displacement

Eq (11) can be solved for the displacement vector using any standard procedure.

{ } [ ] { }1

1 1S S S Sa K q−

× × ×= (12)

Next, inserting zeros into the computed displacement vector for the restraint degrees of freedom, the

displacement vector { }3 1

G

Na

×is formed.

Step-7: Solution for Element Response

Using the same connectivity matrix as used in Step-3, nodal displacement vector of any element in global

coordinate, i.e., { }6 1

ieGa×

is extracted. Associated vector in local coordinate can be calculated as

{ } { }6 1 6 16 6

i ieL ei eGa T a× ××

⎡ ⎤= ⎢ ⎥⎣ ⎦ (13)

Member end force for the element is then calculated as

{ } { } { }6 1 6 1 6 16 6

ii i ieL eL eL eLF K a q× × ××

⎡ ⎤= −⎢ ⎥⎣ ⎦ (14)

Note that member end force vector includes the effect of span loading through the fixed end forces. In order to calculate the member forces at any point within the span, superimpose the span loading on the end forces calculated through Eq (14).

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Step-8: Calculation of Reaction Forces

Instead of removing the rows and column associated with the restraint degrees of freedom, entire degrees of freedom and equilibrium equations in global coordinate may be rearranged as

0r rr rs r

s sr ss s

q K K aq K K a

⎧ ⎫ ⎡ ⎤⎧ ⎫=⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎣ ⎦⎩ ⎭ (15)

Here the subscript r and s denote the restraint and unrestraint degrees of freedom. Note, ssK in Eq (15) is

same as [ ]S SK

× is Eq (11). Reactions can now be calculated as

r rs sq K a= (16)

where sa is computed in Eq (12). If displacements along the restraint degrees of freedom are specified non-zero,

[ ] { }1s ss s sr r

r rr r rs s

a K q K aq K a K a

−= −

= + (17)

5.1 Brief Numerical Example for Direct Stiffness Method

Consider the frame as shown in Figure 6, EA=8000 kN/m2 and EI= 20000 kNm2.

Figure 6: Numerical example

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Element stiffness matrix in global coordinate:

1 11 1

6 6 6 6 6 6 6 6

TeG e eL eK T K T× × × ×

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

1 2 3 4 5 6 Global 1 2 3 4 5 6 Local 1715.2 1 1 -153.6 1804.8 Sym 2 2 -2880 3840 16000 3 3 -1715.2 153.6 2880 1715.2 4 4 153.6 -1804.8 -3840 -153.6 1804.8 5 5 -2880 3840 8000 2880 -3840 16000 6 6

2 22 2

6 6 6 6 6 6 6 6

TeG e eL eK T K T× × × ×

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

4 5 6 7 8 9 Global 1 2 3 4 5 6 Local 2000 1 4 0 8437.5 Sym 2 5 0 16875 45000 3 6 -2000 0 0 20000 4 7 0 -8437.5 -16875 0 8437.5 5 8 0 16875 22500 0 -16875 45000 6 9

Unrestrained global stiffness matrix 3 3

G

N NK

×⎡ ⎤⎢ ⎥⎣ ⎦

1 2 3 4 5 6 7 8 9 Global 1715.2 1 -153.6 1804.8 Sym 2 -2880 3840 16000 3 -1715.2 153.6 2880 1715.2

+2000 4

153.6 -1804.8 -3840 -153.6 +0

1804.8 +8437.5

5

-2880 3840 8000 2880 +0

-3840 +16875

16000 +45000

6

-2000 0 0 20000 7 ZERO 0 -8437.5 -16875 0 8437.5 8 0 16875 22500 0 -16875 45000 9

Removing the rows and column associated with the restraint degrees of freedom,

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1 2 3 4 5 6 7 8 9 Global 1715.2 1 -153.6 1804.8 Sym 2 -2880 3840 16000 3 -1715.2 153.6 2880 1715.2

+2000 4

153.6 -1804.8 -3840 -153.6 +0

1804.8 +8437.5

5

-2880 3840 8000 2880 +0

-3840 +16875

16000 +45000

6

-2000 0 0 20000 7 ZERO 0 -8437.5 -16875 0 8437.5 8 0 16875 22500 0 -16875 45000 9

sK =

16000 2880 1715.2

+2000

-3840 -153.6 +0

1804.8 +8437.5

8000 2880 +0

-3840 +16875

16000 +45000

Similarly, fixed end forces can be assembled and accounting for the joint load, global force vector in unrestraint and restraint conditions are:

6.0 -8.0 -12.5 -12.5 6.0 6.0 -23.0 and -23.0 12.5 12.5 0 0 0

Unrestrained degrees of freedom may be solved as sa = {-0.00356, 0.00275, -0.0058, 0.00178}T.

Member end forces in global coordinates are obtained as { } { } { }6 1 6 1 6 16 6 6 6

i Ti i ieG eG eG ei eLF K a T q× × ×× ×

⎡ ⎤ ⎡ ⎤= −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ : 

For member 1: {-6.49, 12.07, 0, -5.51, 3.93, 17.74}T and for member 2: {5.50, -3.93, -17.72, -5.51, 18.93, -58.0}T. These forces are shown in Figure 6d.

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6.0 Revisiting Stiffness Matrix for Structural Analysis and FEM

In structural analysis, as defined earlier, one column of a stiffness matrix is directly computed by applying unit displacement along the direction of a particular degree of freedom while restraining the others and then calculating the forces developed along the direction of all degrees of freedom. In this section, we will revisit the same concept from a different perspective.

6.1 Shape Functions

Considering only flexural deformation, equilibrium equation of a beam element of constant flexural rigidity in the unloaded region is given by

4

4 0vx

∂=

∂ (18)

The solution of Eq (18) may be expressed through a cubic polynomial as

( ) 2 31 2 3 4v x x x xα α α α= + + + (19)

where iα ’s are arbitrary constants and can be evaluated through the boundary conditions. These boundary conditions are:

1 1

2 2

and at 0

and at

vv v xxvv v x Lx

θ

θ

∂= = =

∂∂

= = =∂

(20)

Utilizing Eq (20) into Eq (19) and arranging the resulting equations in matrix form, it may be shown that

1 1

1 22 3

2 32

2 4

1 0 0 00 1 0 010 1 2 3

v

v L L LL L

αθ α

αθ α

⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎣ ⎦⎩ ⎭ ⎩ ⎭

(21)

Inverting Eq (21),

{ } [ ]{ }

31 1

32 1

3 2 23 2

4 2

0 0 00 0 013 2 32 2

vLL

H avL L L L L

L L

αα θ

ααα θ

⎡ ⎤⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥= ⇒ =⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪− − −⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎣ ⎦

(22)

Substituting iα ’s from Eq (22) into Eq (19) and thereafter rearranging the resulting equation, it may be shown that

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( ) ( ) ( ) ( ) ( )

( )

( )

( )

( )

1 1 1 2 2 3 2 4

2 3

1

2

2

2 3

3

2

4

1 3 2

1 2

3 2

v x v f x f x v f x f x

x xf xL L

x xf x xL L

x xf xL L

x xf x xL L

θ θ= + + +

⎛ ⎞ ⎛ ⎞⎟ ⎟⎜ ⎜= − +⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠

⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥⎟ ⎟⎜ ⎜= − +⎟ ⎟⎜ ⎜⎢ ⎥⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦⎛ ⎞ ⎛ ⎞⎟ ⎟⎜ ⎜= −⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠

⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥⎟ ⎟⎜ ⎜= − +⎟ ⎟⎜ ⎜⎢ ⎥⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

(23)

The functions if ’s are essentially the shape functions and derived from the equilibrium of the unloaded beam element of constant flexural rigidity. Variation of these shape functions over the length is shown in Figure 7.

Figure 7: Shape functions for a beam with constant flexural rigidity

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Now consider, for example, 1 1 2 21.0, 0, 0, 0,v vθ θ= = = = which corresponds to ( ) ( )1v x f x= . Recall

that, this is also the case of generating the first column of stiffness matrix. Therefore, ( )1f x represents the

displacement profile when unit displacement is applied along the first degree from freedom while others are held restraint. Similar interpretation holds for other shape functions also. Hence, displacement at any point over the span can be readily calculated if the nodal displacements and all the associated shape functions are known.

6.2 Castigliano’s Theorem

Consider an elastic system subjected to a set of conservative forces , 1, .iP i n= Let iΔ denotes displacement along the direction of iP and at its point of application. Let us consider two different cases.

In case-1, only thi force iP is applied and { }iΔ denotes the resulting displacement at all the n points but

along the direction of force as described in iΔ . In case-2, all the forces except thi are applied and { }ˆiΔ

denotes the resulting displacement at all the n points but along the direction of force as described in iΔ . Now, if case-1 loading is applied first followed by case-2, the work done on the body or strain energy is given by

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1 1ˆ ˆ2 2

n

i i i i j jjj i

U P P P=≠

= Δ + Δ + Δ∑ (24)

If an infinitesimal virtual deformation profile iaδ (compatible with the restraint) is applied in between case-1 and case-2, then the strain energy will be given by

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1 1ˆ ˆ2 2

n

i i i i i i j jjj i

U P P a P Pδ=≠

= Δ + + Δ + Δ∑ (25)

Therefore, change in strain energy is

i i ii

UU P a Pa

ΔΔ δδ

= ⇒ = (26)

Considering 0iLt aδ → ,

ii

UPa

∂=

∂ (27)

Eq (27) is known as the Castigliano’s first theorem.

6.3 Application of Castigliano’s Theorem

Assuming only flexural deformation, strain energy of the beam element shown in Figure 1 can be expressed as

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22

202

LEI vU dxx

⎛ ⎞∂ ⎟⎜ ⎟= ⎜ ⎟⎜ ⎟⎜∂⎝ ⎠∫ (28)

Noting

( ) ( ) ( ) ( ) ( )'' '' '' '' ''1 1 1 2 2 3 2 4v x v f x f x v f x f xθ θ= + + + (29)

and employing Castigliano’s theorem we may write, for example,

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

2 2'' '' '' '' ''

1 1 1 1 2 2 3 2 4 12 21 10 0

'' '' '' '' '' '' '' ''1 1 1 1 1 2 2 1 3 2 1 4

0 0 0

L L

L L L

U v vY EI dx EI v f x f x v f x f x f x dxv x v x

v EI f x f x dx EI f x f x dx v EI f x f x dx EI f x f x

θ θ

θ θ

⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂⎟ ⎟⎜ ⎜ ⎡ ⎤⎟ ⎟= = = + + +⎜ ⎜⎟ ⎟ ⎢ ⎥⎣ ⎦⎜ ⎜⎟ ⎟⎜ ⎜∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡= + + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

∫ ∫

∫ ∫ ∫0

11 1 12 1 13 2 14 2

L

dx

K v K K v Kθ θ

⎤⎢ ⎥⎣ ⎦

= + + +

∫ (30)

Similarly, other three nodal force can also be derived leading to the thij element of stiffness matrix as

( ) ( )'' ''

0

L

ij i jK EI f x f x dx⎡ ⎤= ⎢ ⎥⎣ ⎦∫ (31)

Using the shape functions developed in Eq (23) into Eq (31) and carrying out the necessary integration over the length, one may derive the stiffness matrix which is same as that shown in Eq (1). For example,

'' '' 22 3 2 3

11 2 3 30 0

6 12 121 3 2 1 3 2L Lx x x x x EIK EI dx EI dx

L L L L L L L

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜ ⎢ ⎥= − + − + = − + =⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎢ ⎥ ⎢ ⎥⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦∫ ∫ (32)

6.4 Application of Rayleigh Ritz Method

Expression of strain energy involves derivative/integration of displacement profile and hence that of shape functions, which becomes complicated for more complex finite elements. To overcome such problem, Rayleigh Ritz method is illustrated below with respect to the same beam element. Assuming the same cubic polynomial displacement profile as considered before and substituting its second derivative into the expression of strain energy, we may write

( ) ( )

( ) ( )

2 3 ''1 2 3 4 3 4

2 2 2 2 33 4 3 3 4 4

0

2 6

2 6 2 6 62

L

v x x x x v x x

EIU x dx EI L L L

α α α α α α

α α α α α α

= + + + ⇒ = +

= + = + +∫ (33)

Eq (33) is quadratic of the coefficient iα ’s and can be expressed as

14  

{ } { } { }

1

21 2 3 4 2

32 3

4

0 0 0 00 0 0 01 10 0 4 62 20 0 6 12

TU kEIL EILEIL EIL

αα

α α α α α ααα

⎧ ⎫⎡ ⎤⎪ ⎪⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎪ ⎪⎪ ⎪ ⎡ ⎤⎢ ⎥= =⎨ ⎬ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎪ ⎪⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎪ ⎪⎪ ⎪⎣ ⎦⎩ ⎭

(34)

Here, elements of k⎡ ⎤⎢ ⎥⎣ ⎦ are obtained through

2

iji j

Ukα α∂

=∂ ∂

(35)

Substituting { }α from Eq (22) into Eq (34), strain energy can be expressed as

{ } [ ] [ ]( ){ }12

T TU a H k H a⎡ ⎤= ⎢ ⎥⎣ ⎦ (36)

Another way to formulate the strain energy is to use the nodal force and displacement vectors as follows:

{ } { } [ ]{ }

1

11 1 2 2

2

2

1 12 2

T

YM

U v v a K aYM

θ θ

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪= =⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

(37)

Here [ ]K is the element stiffness matrix, which can be expressed by equating the strain energy from Eq

(36) and Eq (37) as

[ ] [ ] [ ]TK H k H⎡ ⎤= ⎢ ⎥⎣ ⎦ (38)

Substituting [ ]H from Eq (22) and k⎡ ⎤⎢ ⎥⎣ ⎦ from Eq (34) into Eq (38), element stiffness matrix can be

obtained as

[ ]

3 3

3 3

23 32 2 2 2

2 3

2 2

2 2

0 0 0 00 0 0 0 0 00 0 0 00 0 0 0 0 01 10 0 4 63 2 3 3 2 30 0 6 122 2 2 2

12 6 12 6

6 64 2

12 6 12 6

6 62 4

TL L

L LK

EIL EILL LL L L L L L L LEIL EILL L L L

L L L L

EI L LL

L L L L

L L

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎢ ⎥− − − − − −⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥− −⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

⎡⎢ −⎢⎢⎢

−⎢=

− − −

−⎣

⎤⎥⎥⎥⎥⎥

⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎦

(39)

15  

The stiffness matrix is same as that obtained using elementary structural mechanics or by using Castigliano’s theorem.

While deriving the stiffness matrix using either Castigliano’s theorem or Rayleigh Ritz method, axial deformation is not considered. Under the assumption of small deformation problem, axial degrees of freedom are uncoupled with flexure and shear, and can be included in the stiffness matrix with an additional pair of shape functions.

7.0 Finite Element Method: A Preliminary Revisit

7.1 Displacement Functions

Consider for simplicity a plane stress (or two dimensional in a loose sense) problem as shown in Figure 8. The structure is discretized into several area elements and a typical element is denoted as ‘e’. Nodal points of the element are indicated by i, j, m etc. A typical node has two displacements, one along each

orthogonal direction, and for the thi node, it is denoted as { }T

i xi yia u u= . Here over bar denotes a

vector quantity to make distinction from a scalar. Similarly, the displacement at any point within the

element may be denoted as ( ) ( ){ }, ,T

xi yiu u x y u x y= .

Displacement u may be approximated as

ˆ ......

e

ie e

k k i j jk

au u N a N N a Na

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎡ ⎤≈ = = =⎨ ⎬⎢ ⎥⎣ ⎦⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

∑ (40)

The functions iN , jN should be chosen as to give appropriate displacements at the respective nodes. In

other words

( ) 1,

0i j j ij

i jN x y

i jδ

⎧ =⎪⎪= =⎨⎪ ≠⎪⎩ (41)

Understanding that nodal displacement at a particular node is a vector quantity involving displacement along any set orthogonal directions, their spatial derivative etc, we may now drop the over bar from the representation of vector quantities and Eq (40) is restated as

ˆ eu u Na≈ = (42)

Further, unless otherwise specifically stated, we do not restrict the discussion within the domain of plane stress problem.

16  

Figure 8: A plane stress problem

7.2 Strain-Displacement Relation

In most cases, using the principle of mechanics, it is possible to express the strain tensor at any point within a finite element in terms of the spatial derivative of displacements at that point. In a plane stress problem, for example, strain vector (constituted from the independent elements of the strain tensor) can be expressed as

{ }

0

0

x

xxxy

yyy

xyyx

ux x

uuuy y

uuy xy x

εε ε

ε

⎧ ⎫ ⎡ ⎤⎪ ⎪∂ ∂⎪ ⎪ ⎢ ⎥⎪ ⎪⎪ ⎪ ⎢ ⎥∂⎪ ⎪ ∂⎧ ⎫ ⎢ ⎥⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎢ ⎥⎧ ⎫∂⎪ ⎪ ⎪ ⎪⎪ ⎪ ∂⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥= = =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪∂ ∂⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪⎩ ⎭ ∂ ∂ ∂⎪ ⎪∂ ⎢ ⎥⎪ ⎪+⎪ ⎪ ⎢ ⎥⎪ ⎪ ∂ ∂∂ ∂ ⎣ ⎦⎪ ⎪⎩ ⎭

(43)

Here, xu and yu denote the displacement at the point along two orthogonal directions. In general, Eq (43)

may be written as

{ } { } [ ]{ }ˆ S uε ε≈ = (44)

and upon substituting Eq (42)

17  

{ } { } [ ]{ } [ ][ ]{ } [ ]{ }[ ] [ ][ ]

ˆ e eS u S N a B a

B S N

ε ε≈ = = =

= (45)

7.3 Constitutive Relation

Assuming linear elastic behavior, general stress strain relation can be expressed as

{ } [ ]{ } { }0 0Dσ ε ε σ= − + (46)

Here, { }σ is the strain vector (constituted from the independent elements of the stress tensor) and, { }0σ

and { }0ε stand for the residual stress and initial strain, respectively. For a plane stress problem,

{ } [ ]( )

2

1 0 and 1 0

10 0 1 2

xx

yy

xy

EDσ ν

σ σ νν

τ ν

⎧ ⎫ ⎡ ⎤⎪ ⎪⎪ ⎪ ⎢ ⎥⎪ ⎪⎪ ⎪ ⎢ ⎥= =⎨ ⎬ ⎢ ⎥⎪ ⎪ −⎪ ⎪ ⎢ ⎥−⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎣ ⎦⎩ ⎭

(47)

where, E and ν are the Young’s Modulus and Poisson’s Ratio, respectively.

7.4 External Loading

Three different types of loadings are considered. First, distributed body forces of intensity b ; this force is considered acting on per unit volume of the material. Second, distributed traction of intensity t ; this force is considered acting on per unit surface area. Third, external concentrated forces acting on the nodal points.

7.5 Derivation of Element Equilibrium

The simplest procedure of establishing element equilibrium is to apply an admissible virtual displacement and equate the external and internal work done. Let { }eaδ be the virtual displacement vector at the

nodes. At any point within the element, the associated displacement and strain vectors may be expressed as

{ } [ ]{ } { } [ ]{ } and e eu N a B aδ δ δε δ= = (48)

Work done per unit volume by the internal stress and body force may be expressed as

{ } { } { } { }T Tu bδε σ δ− . Further, work done by the surface loading per unit surface area is { } { }Tu tδ− . Noting that total work done over the entire element is zero, we write

{ } { } { } { } { } { } 0e e

T T Te e

V A

u b dV u t dAδε σ δ δ⎡ ⎤ ⎡ ⎤− − =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦∫ ∫ (49)

Substituting Eq (48), Eq (49) can be simplified to

18  

{ } [ ] { } { } [ ] { } { } [ ] { }

{ } [ ] { } [ ] { } [ ] { }

0

e e

e e

T T TT T Te e ee e

V A

T T T Tee e

V A

a B a N b dV a N t dA

a B N b dV N t dA

δ σ δ δ

δ σ

⎡ ⎤ ⎡ ⎤= − −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

⎛ ⎞⎟⎜ ⎡ ⎤ ⎡ ⎤ ⎟⎜= − − ⎟⎜ ⎢ ⎥ ⎢ ⎥ ⎟⎜ ⎣ ⎦ ⎣ ⎦ ⎟⎟⎜⎝ ⎠

∫ ∫

∫ ∫ (50)

Therefore, element equilibrium can be expressed as

[ ] { } [ ] { } [ ] { } 0e e

T T Te e

V A

B N b dV N t dAσ⎡ ⎤ ⎡ ⎤− − =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦∫ ∫ (51)

Further substituting the constitutive relation Eq (46) and subsequently Eq (45), equilibrium equation can be rewritten as

[ ] [ ]{ } [ ] [ ]{ } [ ] { } [ ] { } [ ] { }

[ ] [ ][ ] { } [ ] [ ]{ } [ ] { } [ ] { } [ ] { }

0 0

0 0

0

=

e e e e e

e e e e e

T T T T Te e e e e

V V V V A

T T T T Tee e e e e

V V V V A

B D dV B D dV B dV N b dV N t dA

B D B dV a B D dV B dV N b dV N t dA

ε ε σ

ε σ

⎡ ⎤= − + − − ⎢ ⎥⎣ ⎦

⎛ ⎞⎟⎜ ⎡ ⎤⎟⎜ − + − −⎟⎜ ⎢ ⎥⎟⎜ ⎣ ⎦⎟⎟⎜⎝ ⎠

∫ ∫ ∫ ∫ ∫

∫ ∫ ∫ ∫ ∫ (52)

The element equilibrium equation can be expressed in more formal way as follows:

{ } { }[ ] [ ][ ]

{ } [ ] [ ]{ } [ ] { } [ ] { } [ ] { }0 0

e

e e e e

e e e

Tee

V

T T T Tee e e e

V V V A

q K a

K B D B dV

q B D dV B dV N b dV N t dAε σ

⎡ ⎤= ⎢ ⎥⎣ ⎦⎡ ⎤ =⎢ ⎥⎣ ⎦

⎡ ⎤= − + + ⎢ ⎥⎣ ⎦

∫

∫ ∫ ∫ ∫

(53)

7.6 Overall Analysis

Eq (53) is of the same form as discussed in conventional structural analysis (Eq (1) or Eq (2)) . Therefore, all the steps described in direct stiffness method of conventional structural analysis including local to global transformation, expansion in global coordinate, assembly (taking into account the externally applied concentrated nodal force vector { }*q ), and applying physical restraints to the global degrees of

freedom will be followed after Eq (53) to get the complete solution for unrestraint nodal displacements. Once the nodal displacement vector { }ea associated with an element is extracted, stress vector at any

point within the element may be calculated as

{ } [ ][ ]{ } [ ]{ } { }0 0eD B a Dσ ε σ= − + (54)

Therefore, FEM follows essentially the same principle as that of the conventional structural analysis but the procedure of formulating governing equilibrium equations, for example stiffness matrix etc., is different. Use of approximate shape functions while formulating the equilibrium equation allows FEM to

19  

be applicable to all possible problems, whereas conventional structural analysis uses exact shape function and its applicability is restricted to such simple cases for which exact shape functions exist.

7.7 Finite Element Method without Assembling Element Equilibrium

Unlike direct stiffness method, equilibrium of the entire structure without assembling the element equilibrium equations is often considered in conventional method. Such is a case, usually applies to a relatively simple configuration, wherein stiffness matrix of the entire structure based on unrestraint degrees of freedom is derived in global coordinate. FEM can also be derived from that perspective without assembling the element equilibrium equations.

In order to illustrate that, a set of admissible discretization as in the previous case is considered. All the concentrated loads are assumed to be applied through the nodal points only. Let { }a denotes a vector

listing the displacement of all the nodal points and N⎡ ⎤⎢ ⎥⎣ ⎦ denotes the associated shape functions such that

displacement at any point is given by

{ } { }u N a⎡ ⎤= ⎢ ⎥⎣ ⎦ (55)

These shape functions are different than what were assumed previously. Consider, for example, j as a

nodal point common to a set of elements { }e . Shape function associated with the nodal displacement at

j is jN . If the point at which displacement to be approximated lies outside all the elements included in

{ }e , then 0jN = . Otherwise, ej jN N= where the point belongs to the element e and e

jN is the shape

function of the thj node of element e , as defined in previous approach. As before,

{ } [ ]{ } [ ] { } { }S u S N a B aε ⎡ ⎤ ⎡ ⎤= = =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ (56)

We now drop the over bar with an understanding that the quantities like shape functions etc. are defined over the whole region.

Applying any admissible virtual displacement { }aδ and equating the external and internal work done, we may write

{ } { } { } { } { } { } { } { }*T T T T

V A V

a q u b dV u t dA dVδ δ δ δε σ+ + =∫ ∫ ∫ (57)

Now taking the variation of Eq (55) and Eq (56) for the virtual quantities, and substituting into Eq (57), we write

{ } { } [ ] { } [ ] { } { } [ ] { }*T T T T T

V A V

a q N b dV N t dA a B dVδ δ σ⎛ ⎞⎟⎜ ⎟⎜ + + =⎟⎜ ⎟⎜ ⎟⎝ ⎠

∫ ∫ ∫ (58)

Cancelling out the virtual deformation and, substituting the constitutive relation and Eq (56),

20  

{ } [ ] { } [ ] { } [ ] [ ][ ] { } [ ] [ ]{ } [ ] { }

[ ] [ ][ ] { } { } [ ] { } [ ] { } [ ] [ ]{ } [ ] { }

[ ]{ } { } { }

*0 0

*0 0

*

,

,

T T T T T

V A V V V

T T T T T

V V A V V

q N b dV N t dA B D B dV a B D dV B dV

or B D B dV a q N b dV N t dA B D dV B dV

or K a q q

ε σ

ε σ

⎛ ⎞⎟⎜ ⎟⎜+ + = − +⎟⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞⎟⎜ ⎟⎜ = + + + −⎟⎜ ⎟⎜ ⎟⎝ ⎠

= +

∫ ∫ ∫ ∫ ∫

∫ ∫ ∫ ∫ ∫ (59)

Clearly, Eq (59) represents the similar form that obtained after assembling Eq (53) for all the elements, wherein i) { }q is same as the assembled equivalent nodal force vector, ii) assembly of integration over

elements is same as the integration over the whole region and iii) [ ]K is same as the assembled element

stiffness matrices.

7.8 Finite Element Formulation from the Minimization of Total Potential Energy

Virtual work principle is used in formulating the FEM in previous two cases. In this section, admissible virtual displacement is considered as the variation of the real displacement. Therefore, { }aδ and hence,

{ }uδ and { }δε are the variation of real quantities. Denoting W and U as the potential energy of the

external load and strain energy, respectively, Eq (57) can be rewritten as

{ } { } { } { } { } { } { } { }

( ) ( )

*

, 0

T T T T

V A V

a q u b dV u t dA dV

or W U U W

δ δ ε σ

δ δ δ δ Π

⎛ ⎞⎟⎜ ⎟⎜ + + =⎟⎜ ⎟⎜ ⎟⎝ ⎠

− = ⇒ + = =

∫ ∫ ∫ (60)

Here Π is the total potential energy, which is stationary per Eq (60). Therefore, finite element formulation can also be derived by setting

1 2

. . 0T

a a aΠ Π Π⎧ ⎫⎪ ⎪∂ ∂ ∂⎪ ⎪= =⎨ ⎬⎪ ⎪∂ ∂ ∂⎪ ⎪⎩ ⎭

(61)

This is the well known Rayleigh Ritz method and illustrated earlier in context with conventional structural analysis of a beam element.

7.9 Example of FEM Formulation on a Beam Element

Assuming only flexural deformation governs, stress-strain relation can be considered in as generalized

sense as moment-curvature relation. Hence, generalized strain is 2

2

d vdx

ε κ≡ =− and generalized stress is

2

2

d vM EIdx

σ ≡ =− . Clearly, Dσ ε= with D EI= . The beam is discretized and consider an element e

with nodes i and j . Displacement at any point within this element can be approximated as

{ } [ ]{ }eu N a= . Since the strain involves second derivative of displacement, it is necessary that both v

21  

and dvdx

be continuous between the element, which can be enforced by incorporating them into the nodal

displacement vector. Therefore, { } { }T

Tei i i

dva v vdx

θ⎧ ⎫⎪ ⎪⎪ ⎪= =⎨ ⎬⎪ ⎪⎪ ⎪⎩ ⎭

.

Since, the element has two nodes each with two variables, assuming a cubic polynomial displacement profile, as shown in Eq (19), the shape functions can be derived as shown in Eq (23). Therefore,

( ) ( ) ( ) ( )1 2 3 4, , ,i jN f x f x N f x f x⎡ ⎤ ⎡ ⎤= =⎣ ⎦ ⎣ ⎦

Consequently,

( ) ( ) ( ) ( )'' '' '' ''1 2 3 4, , ,i jB f x f x B f x f x⎡ ⎤ ⎡ ⎤= − − = − −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Noting that [ ] ( ) ( ) ( ) ( )'' '' '' ''1 2 3 4i jB B B f x f x f x f x⎡ ⎤⎡ ⎤= = − − − −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ , the element stiffness matrix is given

by

[ ] [ ][ ] [ ] ( )[ ]e

T Tee

V L

K B D B dV B EI B dx⎡ ⎤ = =⎢ ⎥⎣ ⎦ ∫ ∫

Any element of this is given by

[ ] ( )[ ] ( ) ( ) ( ) ( ) ( )'' '' '' ''Tei j i j

L L L

K B EI B dx f x EI f x dx EI f x f x dx⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = − − =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦∫ ∫ ∫

This is same as what has been derived with conventional structural analysis.

8.0 General Problem Statement

Consider a domain Ω enclosed by a boundary Γ . A set of coupled differential equations of the form

( ) ( ) ( ){ } { }1 2, ,... 0T TA u A u A u= = within the domain Ω defines the problem statement together with

certain boundary conditions of the form ( ) ( ) ( ){ } { }1 2, ,... 0T TB u B u B u= = defined on the boundary Γ .

The objective is to find an unknown solution u that satisfies the differential equations and the boundary conditions. FEM attempts to find the solution of this problem using a variety of techniques.