Faculty of Engineering

Mechanical Engineering Department

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 1
 Introduction
 This chapter introduces some of the basic concepts on which the direct
stiffness method is derived.
 The linear spring is introduced first because it provides a simple instructive
tool to illustrate the basic concepts.
 We begin with the derivation of the stiffness matrix for a linear-elastic spring
element.
 We next illustrate how to assemble the total stiffness matrix for a structure.
 After establishing the total structure stiffness matrix, we illustrate how to
impose boundary conditions.
 A complete solution including the nodal displacements and reactions is thus
obtained.

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 2
 Direct Stiffness Method
To Derive Spring Element Equations
Step 1: Select the Element Type
Consider the linear spring element (which can be an element in a system of
springs) subjected to resulting nodal tensile forces T

The local 𝑥 axis is directed from node 1 to node 2. The original distance
between nodes before deformation is denoted by L. The material property
(spring constant) of the element is k.
ME 646, Dr. Hafeth Bu Jldain,
A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 3
 Direct Stiffness Method
To Derive Spring Element Equations
Step 2: Select a Displacement Function
We must choose in advance the mathematical function to represent the
deformed shape of the spring element under loading. The most common
functions used are polynomials. 𝒖=𝒂 +𝒂 𝒙 𝟏 𝟐

In general, the total number of coefficients a is equal to the total number of

degrees of freedom associated with the element. Here the total number of
degrees of freedom is two. In matrix form, 𝒂𝟏
𝒖= 𝟏 𝒙 𝒂
𝟐

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 4
 Direct Stiffness Method
To Derive Spring Element Equations
Step 2: Select a Displacement Function
We now want to express 𝒖 as a function of the nodal displacements 𝒅𝟏𝒙
and 𝒅𝟐𝒙 . We achieve this by evaluating 𝑢 at each node and solving for 𝑎1
and 𝑎2 as follows:

𝑥 𝑥
𝑢 = 1− 𝑑 + 𝑑
𝐿 1𝑥 𝐿 2𝑥

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 5
 Direct Stiffness Method
To Derive Spring Element Equations
Step 2: Select a Displacement Function
In matrix form,

Here N1 and N2 are called the shape functions or often called interpolation
functions. In this case, N1 and N2 are linear functions that have the
properties that N1 = 1 at node 1 and N1 = 0 at node 2, whereas N2 = 1 at
node 2 and N2 = 0 at node 1.
ME 646, Dr. Hafeth Bu Jldain,
A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 6
 Direct Stiffness Method
To Derive Spring Element Equations
Step 3: Define the Strain/ Displacement and Stress/Strain Relationships
The deformation of the spring is then represented by:

The stress/strain relationship can be expressed in terms of the force/deformation

relationship instead as:

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 7
 Direct Stiffness Method
To Derive Spring Element Equations
Step 4: Derive the Element Stiffness Matrix and Equations
We now derive the spring element stiffness matrix. By the sign convention for
nodal forces and equilibrium, we have:

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 8
 Direct Stiffness Method
To Derive Spring Element Equations
Step 4: Derive the Element Stiffness Matrix and Equations
In matrix form,

We obtain,

Here k is called the local stiffness matrix for the element.

ME 646, Dr. Hafeth Bu Jldain,
A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 9
 Direct Stiffness Method
To Derive Spring Element Equations
Step 5: Assemble the Element Equations to Obtain the Global
Equations and Introduce Boundary Conditions
The global stiffness matrix and global force matrix are assembled using nodal
force equilibrium equations. This step applies for structures composed of more
than one element such that

where k and f are now element stiffness and force matrices expressed in a
global reference frame.

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 10
 Direct Stiffness Method
To Derive Spring Element Equations
Step 6: Solve for the Nodal Displacements
The displacements are then determined by imposing boundary conditions, such
as support conditions, and solving a system of equations, F = Kd,
simultaneously.

Step 7: Solve for the Element Forces

Finally, the element forces are determined by back-substitution, applied to each
element.

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 11
 Direct Stiffness Method
Example of a Spring Assemblage
Structures such as trusses, building frames, and bridges include basic structural
components connected to form the overall structures.

To analyze these structures, we must determine the total structure stiffness

matrix for the interconnected system of elements.

Consider the system of two springs below:

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 12
 Direct Stiffness Method
Example of a Spring Assemblage
Here we fix node 1 and apply axial forces for F3x at node 3 and F2x at node 2.

The stiffnesses of spring elements 1 and 2 are k1 and k2, respectively.

The nodes of the assemblage have been numbered 1, 3, and 2 for further
generalization because sequential numbering between elements generally does
not occur in large problem

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 13
 Direct Stiffness Method
Example of a Spring Assemblage
 For element 1, using the Eq. developed before, we have:

 And for element 2, we have:

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 14
 Direct Stiffness Method
Example of a Spring Assemblage
 Furthermore, elements 1 and 2 must remain connected at common node 3
throughout the displacement. This is called the continuity or compatibility
requirement. The compatibility requirement yields:

 Based on the fact that external forces must equal internal forces at each
node, we can write nodal equilibrium equations at nodes 3, 2, and 1 as:

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 15
 Direct Stiffness Method
Example of a Spring Assemblage
 Now we can write the displacements in matrix form as:

is called the global nodal displacement matrix.

 And we can write the forces in matrix form as:

is called the global nodal force matrix.

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 16
 Direct Stiffness Method
Example of a Spring Assemblage
Aassembling the Total Stiffness Matrix by Superposition
The element stiffness matrices are given as:

 K is then constructed simply by directly adding terms as:

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 17
 Direct Stiffness Method
Example of a Spring Assemblage
Boundary Conditions:
 Here we have d1x = 0 because node 1 is fixed. F1x is the unknown reaction
and F2x and F3x are known applied loads. Therefore, we can write:

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 18
 Direct Stiffness Method
Example of a Spring Assemblage Solving
 For all homogeneous boundary conditions, we can delete the rows and
columns corresponding to the zero-displacement from the original set of
equations and then solve for the unknown displacements. This
procedure is useful for hand calculations. (computer can be used for
solving the system of simultaneous equations.)

 F1x is not necessarily zero and can be determined once d2x and d3x are
solved for.
ME 646, Dr. Hafeth Bu Jldain,
A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 19
 Example 2.1:
For the spring assemblage with arbitrarily numbered nodes shown in Figure
below, using direct stiffness method obtain (a) the global stiffness matrix, (b)
the displacements of nodes 3 and 4, (c) the reaction forces at nodes 1 and 2,
and (d) the forces in each spring. A force of 5000 lb is applied at node 4 in the
x direction. The spring constants are given in the figure. Nodes 1 and 2 are
fixed.

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 20
 Example 2.1:
(a). We begin by expressing each element stiffness matrix as follows:

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 21
 Example 2.1:
Using the concept of superposition (the direct stiffness method), we obtain the
global stiffness matrix as:

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 22
 Example 2.1:
(b). The global stiffness matrix relates global forces to global displacements as
follows:

Applying the boundary conditions d1x=0 and d2x=0, substituting applied nodal
forces, and deleting the first two rows of {F} and {d} and the first two rows and
columns of [K] corresponding to the zero-displacement boundary conditions,
we obtain:

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 23
 Example 2.1:
(c). To obtain the global nodal forces which include the reactions at nodes 1
and 2, we back-substitute and the boundary conditions d1x=0 and d2x=0. This
substitution yields:

Multiplying matrices and simplifying, we obtain the forces at each node,

From these results, we observe that the sum of the reactions F1x and F2x is
equal in magnitude but opposite in direction to the applied force F4x. This
result verifies equilibrium of the whole spring assemblage.
ME 646, Dr. Hafeth Bu Jldain,
A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 24
 Example 2.1:
Next, we use local element Eqs. to obtain the forces in each element.

By simplifying, we obtain,
A free-body diagram of spring element 1 is shown in Figure (a) below. The
spring is subjected to tensile forces. Also, ^f1x is equal to the reaction force
F1x. A free-body diagram of node 1, Figure (b) shows this result.

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 25
 Example 2.1:

By simplifying, we obtain,

A free-body diagram of spring element 2 is shown in Figure below. The spring

is subjected to tensile forces.

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 26
 Example 2.1:

By simplifying, we obtain,

A free-body diagram of spring element 3 is shown in Figure (a) below. The

spring is subjected to compressive forces. Also, ^f2x is equal to the reaction
force F2x. A free-body diagram of node 2, Figure (b) shows this result.

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 27
 Example 2.2:
For the spring assemblage shown in Figure below, using direct stiffness
method obtain (a) the global stiffness matrix, (b) the displacements of nodes
2–4, (c) the global nodal forces, and (d) the local element forces. Node 1 is
fixed while node 5 is given a fixed, known displacement d = 20 mm. The
spring constants are all equal to k = 200 kN/m

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 28
 Example 2.2:
(a). We express each element stiffness matrix as follows:

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 29
 Example 2.2:
Again, using superposition, we obtain the global stiffness matrix as:

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 30
 Example 2.2:
(b). The global stiffness matrix relates the global forces to the global
displacements as follows:

Applying the boundary conditions d1x=0 and d5x=20 mm (= 0.02 m), deleting
the first and fifth equations corresponding to these boundary conditions, and
substituting known global forces F2x=0, F3x=0, and F4x=0, we obtain:

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 31
 Example 2.2:
We now rewrite Eqs. transposing the product of the appropriate stiffness
coefficient (-200) multiplied by the known displacement (0.02 m) to the left
side.

Solving the above Eqs., we obtain:

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 32
 Example 2.2:
(c). The global nodal forces are obtained by back-substituting the boundary
condition displacements. This substitution yields,

The results yield the reaction F1x opposite that of the nodal force F5x required
to displace node 5 by d = 20 mm. This result verifies equilibrium of the whole
spring assemblage.
ME 646, Dr. Hafeth Bu Jldain,
A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 33
 Example 2.2:
(d). Next, we use local element Eqs. to obtain the forces in each element.

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 34
 Example 2.2:

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 35
 Example 2.3:
Using the direct stiffness method, formulate the global stiffness matrix and
equations for solution of the unknown global displacements and forces. The
spring constants for the elements are k1; k2, and k3; P is an applied force at
node 2.

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 36
 Example 2.3:
Using the direct stiffness method, we formulate the global stiffness matrix.

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 37
 PROBLEMS
For the spring assemblages shown in Figures P2–12 to P2–13, determine the
nodal displacements, the forces in each element, and the reactions. Use the
direct stiffness method for all problems.

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 38
 PROBLEMS
For the spring assemblages shown in Figures P2–15 to P2–16, determine the
nodal displacements, the forces in each element, and the reactions. Use the
direct stiffness method for all problems.

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 39
 Assignment #2a
For the spring assemblages shown in Figures P2–8 to P2–11, determine the
nodal displacements, the forces in each element, and the reactions. Use the
direct stiffness method for all problems.

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 40
 Variational Method
(Potential Energy Approach)
One of the alternative methods often used to derive the element equations
and the stiffness matrix for an element is based on the principle of minimum
potential energy. The principle of minimum potential energy is applicable
only for elastic materials .

Total potential energy (p) is defined as the sum of the internal strain energy
(U) and the potential energy of the external forces (); that is,

𝝅𝒑 = 𝑼 + 𝜴
ME 646, Dr. Hafeth Bu Jldain,
A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 41
 Potential Energy Approach
To Derive Spring Element Equations
Internal Strain Energy (U):
The differential internal work (or strain energy) dU in the spring for a small
change in length of the spring is the internal force multiplied by the change
in displacement through which the force moves, given by: 𝒅𝑼 = 𝑭𝒅𝒙

Now we express F as, 𝑭 = 𝒌𝒙

The differential strain energy becomes, 𝒅𝑼 = 𝒌𝒙𝒅𝒙
𝒙
The total strain energy is then given by, 𝒅𝑼 = 𝒌𝒙𝒅𝒙
𝟎
Upon integration, we obtain, 𝟏 𝟐
𝑼 = 𝒌𝒙
𝟐
ME 646, Dr. Hafeth Bu Jldain,
A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 42
 Potential Energy Approach
To Derive Spring Element Equations
The potential energy of the external forces ():
The potential energy of the external force is lost when the work is done by
the external force and it is given by,
𝜴 = −𝑭𝒙

𝟏 𝟐
Now evaluate The Total Potential Energy (p) as: 𝝅𝒑 = 𝒌𝒙 − 𝑭𝒙
𝟐

Minimization of (p) through standard mathematics by taking the variation

of (p) with respect to x as, 𝜕𝜋𝑝
=0
ME 646, Dr. Hafeth Bu Jldain,
𝜕𝑥
A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 43
 Potential Energy Approach
To Derive Spring Element Equations
We now derive the spring element equations and stiffness matrix using the
principle of minimum potential energy. Consider the linear spring subjected
to nodal forces shown in the Figure.

1 2
𝜋𝑝 = 𝑘 𝑑2𝑥 − 𝑑1𝑥 − 𝑓1𝑥 𝑑1𝑥 − 𝑓2𝑥 𝑑2𝑥
2

1 2 2
𝜋𝑝 = 𝑘 𝑑2𝑥 − 2𝑑2𝑥 𝑑1𝑥 + 𝑑1𝑥 − 𝑓1𝑥 𝑑1𝑥 − 𝑓2𝑥 𝑑2𝑥
2

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 44
 Potential Energy Approach
To Derive Spring Element Equations
The minimization of (p) with respect to each nodal displacement requires
taking partial derivatives of (p) with respect to each nodal displacement such
that: 𝜕𝜋 1
𝑝
= 𝑘 −2𝑑2𝑥 + 2𝑑1𝑥 − 𝑓1𝑥 = 0
𝜕𝑑1𝑥 2
𝜕𝜋𝑝 1
= 𝑘 2𝑑2𝑥 − 2𝑑1𝑥 − 𝑓2𝑥 = 0
𝜕𝑑2𝑥 2

Simplifying, 𝑘 −𝑑2𝑥 + 𝑑1𝑥 = 𝑓1𝑥

𝑘 𝑑2𝑥 − 𝑑1𝑥 = 𝑓2𝑥

ME 646, Dr. Hafeth Bu Jldain,
A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 45
 Potential Energy Approach
To Derive Spring Element Equations
Rewriting them as, 𝑘 𝑑1𝑥 − 𝑑2𝑥 = 𝑓1𝑥

𝑘 −𝑑1𝑥 + 𝑑2𝑥 = 𝑓2𝑥

In matrix form,
𝑘 −𝑘 𝑑1𝑥 𝑓1𝑥
=
−𝑘 𝑘 𝑑2𝑥 𝑓2𝑥

Then, the stiffness matrix for the spring element is,

𝑘 −𝑘
[𝑘 ] =
−𝑘 𝑘

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 46
 Example 2.5:
Obtain the total potential energy of the spring assemblage shown in the Figure
below and find its minimum value. The procedure of assembling element
equations shall be obtained from the minimization of the total potential
energy.

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 47
 Example 2.5:
The total potential energy is given by:
3
(𝑒) 1 2 1 1
𝜋𝑝 = 𝑘1 𝑑3𝑥 − 𝑑1𝑥 − 𝑓1𝑥 𝑑1𝑥 − 𝑓3𝑥 𝑑3𝑥
2
𝑒=1
1 2 2 2
+ 𝑘2 𝑑4𝑥 − 𝑑3𝑥 − 𝑓3𝑥 𝑑3𝑥 − 𝑓4𝑥 𝑑4𝑥
2
1 2 3 3
+ 𝑘3 𝑑2𝑥 − 𝑑4𝑥 − 𝑓4𝑥 𝑑4𝑥 − 𝑓2𝑥 𝑑2𝑥
2

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 48
 Example 2.5:
Upon minimizing (p) with respect to each nodal displacement, we obtain:
𝜕𝜋𝑝 (1)
= −𝑘1 𝑑3𝑥 + 𝑘1 𝑑1𝑥 − 𝑓1𝑥 = 0
𝜕𝑑1𝑥
𝜕𝜋𝑝 (3)
= 𝑘3 𝑑2𝑥 − 𝑘3 𝑑4𝑥 − 𝑓2𝑥 = 0
𝜕𝑑2𝑥
𝜕𝜋𝑝 1 (2)
= 𝑘1 𝑑3𝑥 − 𝑘1 𝑑1𝑥 − 𝑘2 𝑑4𝑥 + 𝑘2 𝑑3𝑥 − 𝑓3𝑥 − 𝑓3𝑥 = 0
𝜕𝑑3𝑥
𝜕𝜋𝑝 2 (3)
= 𝑘2 𝑑4𝑥 − 𝑘2 𝑑3𝑥 − 𝑘3 𝑑2𝑥 + 𝑘3 𝑑4𝑥 − 𝑓4𝑥 − 𝑓4𝑥 = 0
𝜕𝑑4𝑥

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 49
 Example 2.5:
By rearranging the previous equations, we obtain:
(1)
+𝑘1 𝑑1𝑥 −𝑘1 𝑑3𝑥 = 𝑓1𝑥

(3)
+𝑘3 𝑑2𝑥 − 𝑘3 𝑑4𝑥 = 𝑓2𝑥

1 (2)
−𝑘1 𝑑1𝑥 + 𝑘1 + 𝑘2 𝑑3𝑥 − 𝑘2 𝑑4𝑥 = 𝑓3𝑥 + 𝑓3𝑥

2 (3)
−𝑘3 𝑑2𝑥 − 𝑘2 𝑑3𝑥 + 𝑘2 + 𝑘3 𝑑4𝑥 = 𝑓4𝑥 + 𝑓4𝑥

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 50
 Example 2.5:
In matrix form become: (1)
𝑓1𝑥
𝑘1 0 −𝑘1 0 𝑑1𝑥 𝐹1𝑥
(3)
0 𝑘3 0 −𝑘3 𝑑2𝑥 𝑓 2𝑥 𝐹2𝑥
= =
−𝑘1 0 𝑘1 + 𝑘2 −𝑘2 𝑑3𝑥 (1)
𝑓3𝑥 + 𝑓3𝑥
(2) 𝐹3𝑥
0 −𝑘3 −𝑘2 𝑘2 + 𝑘3 𝑑4𝑥 (2) (3) 𝐹4𝑥
𝑓4𝑥 + 𝑓4𝑥
By substituting numerical values we obtain,
1000 0 −1000 0 𝑑1𝑥 𝐹1𝑥
0 3000 0 −3000 𝑑2𝑥 𝐹2𝑥
=
−1000 0 3000 −2000 𝑑3𝑥 𝐹3𝑥
0 −3000 −2000 5000 𝑑4𝑥 𝐹4𝑥
When we apply the boundary conditions and substitute F3x = 0 and F4x = 5000 lb,
the solution is identical to that of Example 2.1.
ME 646, Dr. Hafeth Bu Jldain,
A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 51
 PROBLEMS
Using P.E. approach to solve next Problems.

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 52
 Assignment #2b
Using P.E. approach to solve next Problems.

ME 646, Dr. Hafeth Bu Jldain,

A First Course in the Finite Element Method, Fourth Edition, Daryl L. Logan 53
Thank you
54