C I V I L E N G I N E E R I N G Structural Analysis Worsak Kanok Nukulchaii

CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai

STRUCTURAL ANALYSIS
Worsak Kanok-Nukulchai
Asian Institute of Technology, Thailand

Keywords: Structural System, Structural Analysis, Discrete Modeling, Matrix Analysis
of Structures, Linear Elastic Analysis.

Contents

1. Structural system
2. Structural modeling.
3. Linearity of the structural system.

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4. Definition of kinematics

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5. Definitions of statics
6. Balance of linear momentum
7. Material constitution

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8. Reduction of 3D constitutive equations for 2D plane problems.

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9. Deduction of Euler-Bernoulli Beams from Solid.
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10. Methods of structural analysis
11. Discrete modeling of structures
12. Matrix force method AP
13. Matrix displacement method
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14. Trends and perspectives
Glossary
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Bibliography
Biographical Sketch
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To cite this chapter
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Summary
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This chapter presents an overview of the modern method of structural analysis based on
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discrete modeling methods. Discrete structural modeling is suited for digital
computation and has lead to the generalization of the formulation procedure. The two
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principal methods, namely the matrix force method and the matrix displacement method,
are convenient for analysis of frames made up mainly of one-dimensional members. Of
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the two methods, the matrix displacement method is more popular, due to its natural
extension to the more generalized finite element method. Using displacements as the
primary variables, the stiffness matrix of a discrete structural model can be formed
globally as in the case of the matrix displacement method, or locally by considering the
stiffness contributions of individual elements. This latter procedure is generally known
as the direct stiffness method. The direct stiffness procedure allows the assembly of
stiffness contributions from a finite number of elements that are used to model any
complex structure. This is also the procedure used in a more generalized method known
as the Finite Element Method

©Encyclopedia of Life Support Systems (EOLSS)


1. Structural System

Structural analysis is a process to analyze a structural system in order to predict the
responses of the real structure under the excitation of expected loading and external
environment during the service life of the structure. The purpose of a structural analysis
is to ensure the adequacy of the design from the view point of safety and serviceability
of the structure. The process of structural analysis in relation to other processes is
depicted in Figure 1.

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Figure 1. Role of structural analysis in the design process of a structure.
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A structural system normally consists of three essential components as illustrated in
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Figure 2: (a) the structural model; (b) the prescribed excitations; and (c) the structural
responses as the result of the analysis process. In all cases, a structure must be idealized
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by a mathematical model so that its behaviors can be determined by solving a set of
mathematical equations.

Figure 2. Definition of a structural system



A structural system can be one-dimensional, two-dimensional or three-dimensional
depending on the space dimension of the loadings and the types of structural responses
that are of interest to the designer. Although any real-world structure is strictly three-
dimensional, for the purpose of simplification and focus, one can recognize a specific
pattern of loading under which the key structural responses will remain in just one or
two-dimensional space. Some examples of 2D and 3D structural systems are shown in
Figure 3.

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Figure 3. Some examples of one, two and three-dimensional structural models.

2. Structural Modeling

A structural (mathematical) model can be defined as an assembly of structural members
(elements) interconnected at the boundaries (surfaces, lines, joints). Thus, a structural
model consists of three basic components namely, (a) structural members, (b) joints
(nodes, connecting edges or surfaces) and (c) boundary conditions.



(a) Structural members: Structural members can be one-dimensional (1D) members
(beams, bars, cables etc.), 2D members (planes, membranes, plates, shells etc.) or in the
most general case 3D solids.

(b) Joints: For one-dimensional members, a joint can be rigid joint, deformable joint or
pinned joint, as shown in Figure 4. In rigid joints, both static and kinematics variables
are continuous across the joint. For pinned joints, continuity will be lost on rotation as
well as bending moment. In between, the deformable joint, represented by a rotational
spring, will carry over only a part the rotation from one member to its neighbor offset
by the joint deformation under the effect of the bending moment.

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Figure 4. Typical joints between two 1D members.

(c) Boundary conditions: To serve its purposeful functions, structures are normally
prevented from moving freely in space at certain points called supports. As shown in
Figure 5, supports can be fully or partially restrained. In addition, fully restrained
components of the support may be subjected to prescribed displacements such as
ground settlements.



Figure 5. Boundary conditions of supports for 1D members.

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3. Linearity of the Structural System

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Assumptions are usually observed in order for the structural system to be treated as
linear:

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(a) The displacement of the structure is so insignificant that under the applied loads, the

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deformed configuration can be approximated by the un-deformed configuration in
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satisfying the equilibrium equations.
(b) The structural deformation is so small that the relationship between strain and
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displacement remains linear.
(c) For small deformation, the stress-strain relationship of all structural members falls
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in the range of Hooke’s law, i.e., it is linear elastic, isotropic and homogeneous.
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As a result of (a), (b) and (c), the overall structural system becomes a linear problem;
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consequently the principle of superposition holds.

4. Definition of Kinematics
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a) Motions
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As shown in Figure 6, the motion of any particle in a body is a time parameter family of
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its configurations, given mathematically by
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^
x = x ( P, t ) (1)
~ ~

ˆ
where x is a time function of a particle, P, in the body. During a motion, if relative
positions of all particles remain the same as the original configuration, the motion is
called a “rigid-body motion”.

b) Displacement

Displacement of a particle is defined as a vector from its reference position to the new
position due to the motion. If P0 is taken as the reference position of P at t = t0, then the
displacement of P at time t = t1 is



^ ^
u ( P, t1 ) = x( P, t1 ) − x( P, t0 ) (2)
~ ~ ~

c) Deformation

The quantitative measurement of deformation of a body can be presented in many
forms:

(1) Displacement gradient matrix ∇ u
~

⎡ u1,1 u1,2 u1,3 ⎤
⎢ ⎥

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∇u = ⎡ui , j ⎤ = ⎢u2,1 u2,2
⎣ ⎦ u2,3 ⎥ (3a)
⎢ ⎥

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⎢ u3,1 u3,2
⎣ u3,3 ⎥
⎦

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Note that this matrix is not symmetric. O

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(2) Deformation gradient matrix F
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F = I + ∇u or Fij = δ ij + ui , j AP (3b)
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where I is the identity matrix and δ ij is Kronecker delta. Like the displacement
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gradient matrix, the deformation gradient matrix is not symmetric.
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Figure 6. Motion of a body and position vectors of a particle.

(3) Infinitesimal strain tensor (for small strains)

1
ε ij = (ui , j + u j ,i ) (4)
2



⎡ 1 1 ⎤
⎢ u1,1 (u + u2,1 ) (u1,3 + u3,1 ) ⎥
⎡ ε11 ε12 ε13 ⎤ ⎢ 2 1,2 2
⎥
⎢ ⎥
ε 23 ⎥ = ⎢ (u + u3,2 ) ⎥
1
⎢ε 21 ε 22 ⎢
u2,2
2 2,3 ⎥
(5)
⎢ε 31 ε 32
⎣ ε 33 ⎥ ⎢
⎦ ⎥
⎢( SYM ) u3,3 ⎥
⎢
⎣ ⎥
⎦

The strain tensor is symmetric. Its symmetric shear strain components in the opposite
off-diagonal positions can be combined to yield the engineering shear strain γ ij as
depicted in Figure 7 as,

γ ij = ε ij + ε ji = ui, j + u j ,i

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(6)

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Figure 7. Engineering shear strain γ ij
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Since ε is symmetric, there is no need to work with all the 9 components, strain tensor
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~
is often rewritten in a vector form as
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ε = [ε11 ε 22 ε 33 γ 12 γ 23 γ 31 ]T (7)
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~
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(4) Rotational Strain tensor, illustrated in Figure 8, is defined as
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1
ωij = (u j ,i − ui , j )
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(8a)
2
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or in matrix form:

⎡ 1 1 ⎤
⎢ 0 (u − u ) (u3,1 − u1,3 ) ⎥
2 2,1 1,2 2
⎢ ⎥
ω = ⎢ (u1,2 − u2,1 ) (u3,2 − u2,3 ) ⎥
1 1
0 (8b)
~ ⎢2 2 ⎥
⎢1 1 ⎥
⎢ (u1,3 − u3,1 ) (u − u ) 0 ⎥
⎢2
⎣ 2 2,3 3,2 ⎥
⎦



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Figure 8. Illustration of a rotational strain ω 12
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Some notes on the properties of ε and ω
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~ ~
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a) One can easily show that
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ui, j = ε ij − ωij (9)
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and that a displacement gradient is decomposable into 2 parts ε and − ω .
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~ ~
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b) If ui, j = u j ,i then ω = 0 (pure deformation)
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~
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c) If ui , j = −u j ,i then ε = 0 (pure rotation)
~
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5. Definitions of Statics

Stress is defined as internal force per unit deformed area, distributed continuously
within the domain of a continuum that is subjected to external applied forces. When the
deformation is small, stress can be approximated by internal force per un-deformed area
based on its original configuration. This type of stress is represented by the Cauchy
stress tensor in the form of

⎡σ 11 σ 12 σ 13 ⎤
⎡ ⎤ ⎢
σ = ⎣σ ij ⎦ = ⎢σ 21 σ 22 σ 23 ⎥
⎥ (10)
⎢σ 31 σ 32 σ 33 ⎥
⎣ ⎦



As illustrated in Figure 9, the two subscript indices i and j corresponding to xi and x j
refer to the normal vector of the surface and the direction associated with the acting
internal force.

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Figure 9. Definition of Cauchy Stress Tensor.
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6. Balance of Linear Momentum
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Based on the 2nd law of Newton, the rate of linear momentum change equals the total
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force applied in the same direction. Thus, at any deformed state Bt of a deformable
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body, the rate of total linear momentum change on the left-handed side equals to body
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force, b , over the whole body Bt and traction force τ on the boundary ∂Bt respectively.
B
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d
∫ ρ u dv = ∫ ρt b dv + ∫ τ da
dt B t ~
(11)
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~ ~
t B t ∂B t
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where ρt is the unit mass of the body at the Bt state and u is the velocity of a particle in
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the body. In componential form, using the relationship between the Cauchy stress tensor
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and the traction as illustrated in Figure 10, one can present Eq.(11) in tensor notation as
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d
∫ ρ u dv = ∫ ρt b j dv + ∫ σ ij ni da
dt B t j
(12)
t B t ∂B t

where ρt is mass density, Bt is a deformed body at time t, ∂ Bt is its boundary and n is
B

its unit vector. Eq.(11) is the global balance of linear momentum. It applies to a body at
the state of Bt (deformed configuration corresponding to a time station t). Integrals over
B

Bt and ∂ Bt are not explicit as the deformed configuration itself depends on u (t ) . This is
B

~
therefore a nonlinear problem.



Figure 10 Relationship between Cauchy stress tensor and traction.

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For the case of small displacements and infinitesimal strains, Bt ≈ B0 and δ Bt ≈ δ B0 .

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Then, all integrals can be carried over Bo and ∂ B0 , Eq.(11), which governs the body as
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a whole, can be localized to obtain equations of motion which govern any particle
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within the body. This is achieved through the use of the Gauss theorem which

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transforms surface integral to a volume integral, i.e.,
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∫ σ ij ni da = ∫ σ ij ,i dv AP (13)
∂ Bo Bo
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Hence, Eq.(12) is reduced to
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∫ ρo u j dv = ∫ ρo b j dv + ∫ σ ij ,i dv
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(14)
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Bo Bo Bo
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Consequently, the tensor form of local balance of linear momentum, also known as the
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equation of motion is
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ρo u j = ρo b j + σ ij ,i (15a)
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The left-handed side term is inertia force and the two right-handed side terms are
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respectively the body force and the internal force all in xj direction.

7. Material Constitution

So far, the balance of linear momentum is applicable to any solid material. In the
present scope, only linear, elastic, isotropic and homogeneous material is considered. In
most practical structures under service loads, the material deformation is limited to
infinitesimal small strains. Under this situation, its stress-strain relationship can be
assumed to vary linearly, as illustrated in Figure 11. The standard Hooke’s three-
dimensional stress-strain relationship can be expressed as



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Figure 11. Stress-strain relationship in most structural materials can be assumed linear if
strain is limited to a small magnitude.

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⎧σ 11 ⎫ ⎡λ + 2μ λ λ 0 ⎤ ⎧ ε11 ⎫

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0 0
⎪σ ⎪ ⎢ ⎥⎪ ⎪
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⎪ 22 ⎪ ⎢ λ λ + 2μ λ 0 0 0 ⎥ ⎪ε 22 ⎪
⎪σ 33 ⎪ ⎢ λ
⎪ ⎪ λ λ + 2μ 0 0 0 ⎥ ⎪ε 33 ⎪
⎪ ⎪
⎬=⎢
⎨
⎪σ 12 ⎪ ⎢ 0 0 0 2μ 0
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⎥ ⎨ ⎬ (15c)
0 ⎥ ⎪ε12 ⎪
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⎪σ 23 ⎪ ⎢ 0 0 0 0 2μ 0 ⎥ ⎪ε 23 ⎪
⎪ ⎪ ⎢ ⎥⎪ ⎪
⎩σ 31 ⎭ ⎣ 0 2 μ ⎦ ⎩ε 31 ⎪
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⎪ ⎪ ⎪ ⎭
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0 0 0 0
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or
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σ ij = λδ ij ε kk + 2με ij (15d)
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in which the two Lame’s constants

νE
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λ= and μ = .
(1 + ν )(1 − 2ν ) 2(1 + ν )
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8. Reduction of 3D Constitutive Equations for 2D Plane Problems

There are two special cases of 3D solid that can be reduced to 2D plane problems, as
follows:

a) Plane strain problem

When the longitudinal dimension of a prismatic solid is relatively long, strains in the
longitudinal direction can be negligible as shown in Figure 12.



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Figure 12. Plane strain reduction from 3D prismatic solid

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From the 3D constitutive equations in Eq.(15c), by prescribing ε 33 = ε13 = ε 23 = 0 , the
remaining constitutive equation can be expressed as

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⎧σ 11 ⎫ ⎡λ + 2μ λ ⎤ ⎧ ε11 ⎫
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0
⎪ ⎪ ⎢ ⎥ ⎪ε ⎪
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⎨σ 22 ⎬ = ⎢ λ λ + 2μ 0 ⎥ ⎨ 22 ⎬ (16a)
⎪ ⎪ 2μ ⎥ ⎩ε12 ⎪
⎪
⎩σ 12 ⎭ ⎢ 0
⎣ 0 ⎦ ⎭ AP
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or
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σ ij = λδ ij ε kk + 2με ij (i , j = 1, 2) (16b)
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Note that while σ 22 + σ 33 is not zero due to the Poisson’s ratio effect and can be
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obtained as
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σ 33 = λ (ε11 + ε 22 ) (16c)
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(b) Plane stress problem
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When the transverse dimension of a solid is relatively small as illustrated in Figure 13,
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say in x3 direction, the transverse stress components, namely, σ 33 , σ 13 and σ 23 can be
assumed negligible. Consequently the only stress components need to be considered in
plane stress problems are σ 11 , σ 22 , σ 12 all acting in the plane.

Referring to the third equation of Eq.(15), in views of the zero value of σ 33 , one can
λ
obtain ε 33 = − (ε11 + ε 22 ) . Then substituting ε 33 in terms of ε11 and ε 22 in the
λ + 2μ
first and second equation of Eq.(15) leads to:



⎧σ 11 ⎫ ⎡λ + 2 μ λ 0 ⎤ ⎧ ε11 ⎫
⎪ ⎪ ⎢ ⎥⎪ ⎪
⎨σ 22 ⎬ = ⎢ λ λ + 2μ 0 ⎥ ⎨ε 22 ⎬ (17a)
⎪ ⎪ ⎢ ⎥⎪ ⎭
2 μ ⎦ ⎩ε12 ⎪
⎩σ 12 ⎭ ⎣ 0 0

or

σ ij = λδ ij ε kk + 2με ij (i,j = 1,2) (17b)

where

2λμ νE
λ= = (18)

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λ + 2μ 1 − ν 2

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Note that the form of the constitutive equation for the plane stress problems is exactly the
same as that for the plane strain problems, except that λ is used instead of λ .

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Table 1 summarizes the three sets of equations of which the number is exactly the same as
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the number of unknowns in both 2D and 3D cases. To solve for the unknowns, one can
combine the set of equations to obtain a single governing equation in terms of the
displacement unknowns, u j , as follows: AP
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ρo u j = ρo b j + (λ + μ )uk ,kj + μ u j ,kk (19)
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This governing equation is a second-order partial differential equation with respect to
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space and time.
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Figure 13. Non-zero stress components in 2D plane stress problems



9. Deduction of Euler-Bernoulli Beams from Solid

Thin beam is one of the most common 1D components of a structural system, defined
by its bending capability about the beam axis. Thin beam is governed by Euler-
Bernoulli beam theory, based on the following assumptions:

Assumption 1. Transverse dimensions of a beam are small in comparison with the
longitudinal dimension; hence, normal stresses in the two transverse directions are
assumed to be negligible as illustrated in Figure 14.

Assumption 2. Shear distortion in a thin beam is negligible. Therefore, its plane sections
which are normal to the beam axis remain plane and normal to the deformed beam axis
as illustrated in Figure 15.

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Equations 2D 3D
1 Local Balance of Linear Momentum 2 3
ρo u j = ρo b j + σ ij ,i ( j = 1, n )

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2 Constitutive Law
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σ ij = λ δ ij ε kk + 2με ij (i, j = 1, n)

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3 Strain-Displacement Relations 3 6
1
ε ij = (ui , j + u j ,i )(i, j = 1, n) AP
2
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Total number of equations 8 15
Unknowns
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1 Displacements ui (i = 1, n) 2 3

Strains ε ij (i, j = 1, n)
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2 3 6
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3 Stresses σ ij (i, j = 1, n) 3 6
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Total number of unknowns 8 15
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Table 1. Summary Of Equations And Unknowns For 2D/3D Solids
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Figure 14. All transverse stress components in thin beam are negligible.



Figure 15. Plane section remains plane in thin beam

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(a) Effect of Assumption 1

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From the 3D constitutive equations

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⎧σ 11 ⎫ ⎡λ + 2μ λ λ 0
O 0 0 ⎤ ⎧ ε11 ⎫

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⎪σ ⎪ ⎢ ⎪ ⎪
⎪ 22 ⎪ ⎢ λ λ + 2μ λ 0 ⎥ ⎪ε 22 ⎪
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0 0 ⎥
⎪σ 33 ⎪ ⎢ λ
⎪ ⎪ λ λ + 2μ 0 0 0 ⎥ ⎪ε 33 ⎪
⎪ ⎪
⎨ ⎬=⎢ ⎥⎨ ⎬
AP (20)
⎪σ 12 ⎪ ⎢ 0 0 0 2μ 0 0 ⎥ ⎪ε12 ⎪
⎪σ 23 ⎪ ⎢ 0 2μ 0 ⎥ ⎪ε 23 ⎪
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0 0 0
⎪ ⎪ ⎢ ⎥⎪ ⎪
⎪σ 31 ⎪ ⎣ 0
⎩ ⎭ 0 0 0 0 2 μ ⎦ ⎪ε 31 ⎪
⎩ ⎭
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C

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Eliminating (σ 22 + σ 33 ) between the first equation of Eq.(20), i.e.,
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σ 11 = (λ + 2μ )ε11 + λ (ε 22 + ε 33 )
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(21)
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and the summation of the second and the third equations, i.e.,
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σ 22 + σ 33 = 2λε11 + 2(λ + μ )(ε 22 + ε 33 ) = 0 (22)
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leads to

(3λ + 2μ )
σ 11 = με11 (23)
(λ + μ )

νE E
Substituting λ = and μ = in Eq.(23) yields
(1 − 2ν )(1 + ν ) 2(1 + ν )

σ 11 = Eε11 (24)



−λ
Note that (ε 22 + ε 33 ) = ε = −2νε11 due to the Poisson’s ratio effect.
λ + μ 11

(b) Stress resultants

N11 = ∫ σ 11dA (Normal stress resultant) (25)
A

N12 = ∫ σ 12 dA (Transverse shear resultant) (26)
A

M = ∫ σ 11 x2 dA (Bending moment) (27)

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(c) Effect of Assumption 2.

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AP
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Figure 16. Displacement field can be represented by the displacements and rotation of
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the corresponding plane section based on thin beam. assumptions.
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From Figure 16, the longitudinal displacement field at any point in the beam body can
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be expressed in terms of the longitudinal displacement at the corresponding point on the
beam axis and the effect of the rotation of the plane section, i.e.,
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u1 ( x1 , x2 ) = U1 ( x1 ) − x2θ (28)
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In views of the assumption that plane section remains plane and normal to beam axis,
one can show that the rotation of the plane section equal the slope of the beam axis, i.e.,
dU 2 dU 2
θ= . Substituting θ with leads to
dx1 dx1

dU 2
u1 ( x1 , x2 ) = U1 ( x1 ) − x2 (x ) (29)
dx1 1



If the normal strains in the transverse direction are negligible, the transverse
displacement field in the beam body is equal to the corresponding transverse
displacement of at the beam axis, i.e.,

u2 ( x1 , x2 ) = U 2 ( x1 ) (30)

(d) Strains resultants

Measurements of deformation in a thin beam are made on a small segment of the beam in
the form of strain resultants. The main strain resultant is the normal strain of a small
segment lim ΔX1 , i.e.,
ΔX1 →0

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du1
ε11 ( x1 x2 ) ≡ (31)

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dx1

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Substituting Eq.(29) in Eq.(31) leads to O

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d 2U 2
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dU1
ε11 ( x1 x2 ) ≡ ( x1 ) − x2 2
( x1 ) (32)
dx1 dx1

The second thin beam assumption of zero shear can be confirmed from
AP
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∂ u1 ∂ u2 ⎧ dU 2 ⎫ dU 2
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∂ ⎪ ⎪
γ 12 ( x1 x2 ) = + = ⎨U1 − x2 ⎬+ =0 (33)
∂ x2 ∂ x1 ∂x2 ⎪
⎩ dx1 ⎪ dx1
⎭
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(e) Constitutive equations
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At the material level, the relationship between the only pair of the stress and strain tensors is
given in Eq.(24), i.e., σ 11 = Eε11 . Applying ∫ ( )dA on both sides of Eq.(24) in views of
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A
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Eq.(32) leads to
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⎛ ⎞ dU ( x ) d 2U 2 dU ( x )
N11 = ∫ Eε11 ( x1 , x2 )dA = E ⎜ ∫ dA ⎟ 1 1 + ∫ x2 dA. = EA 1 1 (34)
⎜ ⎟ 2
A ⎝ A ⎠ dx1 A dx1 dx1

h/2
Note that ∫ x2 dA ≡ 0 i.e. ∫ bdx2 = 0 when the origin of x2 axis is located at the
A −h / 2
centroid of the plane section. Applying
2 x d A

on both sides of Eq.(24) in views of Eq.(32) leads to



dU1 d 2U 2 d 2U 2
M = ∫ σ 11 x2 dA = E ( ∫ x2 dA) − E ( ∫ x2 dA)
2
2
= EI 2
(35)
A A
dx1 A dx1 dx1

where the sectional moment of inertia, I3 = ∫ x2 dA .
2

(f) Beam equation of motion

The equations of motion, Eq.(15), are applicable to any particle within a continuum.
These equations will be used to derive a set of governing equations for thin beam
members.

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(i) Based on Eq.(15), by applying the cross-sectional integration over both sides of the
equation of motion in x1 direction, the longitudinal equation of motion can be derived

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as follows:

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∫ ρ0u1dA = ∫ ρ0b1dA + ∫ (σ 11,1 + σ 21,2 + σ 31,3 )dA
O (36)

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A A A
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In view of Eq.(29) and the definition of stress resultants in Equations 25 and 26, Eq.(36) can
be further expanded as follows: AP
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dU 2 ∂σ ∂σ
∫ ρ0 [U1 − x2 ]dA = F1 + ∫ 11 dA + ∫ 21 dA
∂ x1 σ x2
(37)
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dx1
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A A A
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dN11 dN 21
ρ0 AU1 = F1 + + (38)
dx1 dx2
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As N 21 is not a function of x2 , the last term on the right-handed side can be omitted,
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thus,
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dN11
ρ0 AU1 = F1 + (39)
dX1
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(ii) Based on Eq.(15), by applying ∫ () x2 dA over both sides of the equation, the bending
A
equation of motion can be derived as follows:

∫ ( ρ0u1 ) x2 dA = ∫ ( ρ0b1 ) x2 dA + ∫ (σ 11,1 + σ 21,2 ) x2 dA (40)
A A A

Substituting Eq.(29) and considering zero value of the first term on the right-handed
side yield



⎡ dU 2 ⎤ ∂σ 11 ∂σ
∫ ρ0 ⎢U1 − x2
⎣
⎥ x2 dA = 0 + ∫
dx1 ⎦ ∂ x1
x2 dA + ∫ 21 x2 dx2 b
∂ x2
(41)
A A A

Omitting the zero terms and integrating by parts the last term of Eq.(41) lead to

dU 2 d d
− ρ0 ∫ x2 dA
2
= ∫ σ11 x2 dA + ∫ dx2 (σ 21 x2 )dx2b − ∫ σ 21dA (42)
A
∂ x1 dx1 A A A

Using the definitions in Eqs. (26) and (27) and in view of the vanishing second term on the
right-handed side, the bending equation of motion can be obtained as

S
dU 2 dM
− ρ0 I 3 = − N12

LS
(43)
dx1 dx1

R
∫ ()dA
(iii) Based on Eq.(15), by applying
O in x2 direction over both sides of the

TE
A
-E
equation, the transverse equation of motion can be derived as follows:

∂σ 12 ∂σ 22 ∂σ 12 AP
∫ ρo u2 dA = ∫ ρ0b2 dA + ∫ ( ∂ x1 +
∂ x2
+
∂ x3
) dA (44)
O

A A A
H
C

Removing the zero terms and defining the vertical distributed unit load q = ∫ ρ0 b2 dA
C
ES

A
lead to
E

d
∫ ρ0U 2 dA = q + dx1 ∫ σ12 dA
N

(45)
PL

A
U

Finally, upon substituting the definition of the sectional shear term from Eq.(26), the
M

transverse equation of motion for thin beam can be obtained as
SA

dN12
ρ0 AU 2 = q + (46)
dx1

Table 2 summarizes the three sets of equations that govern the behaviors of thin beam
problems. The number of equations is seven which is exactly the same as the number of
unknowns. To solve for the unknowns, one can combine the three set of equations to obtain
two sets of governing equations in terms of the displacement unknowns as follows:

d 2U1
EA 2
+ F1 = ρ 0 AU1 (47)
dx1



d 4U 2 dθ
EI 3 4
− q = ρ0 I 3 − ρ0 AU 2 (48)
dx1 dx1

For static problems, the time-dependent terms can be omitted, and the equations of
motions reduce to the equilibrium equations of thin beams as follows:

d 2U1
EA + F1 = 0 (49)
dx1

d 4U 2
EI 3 4
−q =0 (50)

S
dx1

LS
10. Methods of Structural Analysis

R
To design safe structures, structural engineers must fully understand the structural
O
behaviors of these structures. In the long past, structural engineers gained the

TE
knowledge into the structural behaviors by carrying out experimentations using a
-E
physical model of the real structure in the laboratory. Based on the test results, the
behaviors of the prototype structure can be understood and generalized. However, the
AP
physical modeling has its limitations as it is expensive and time-consuming. Thus,
mathematical modeling has been a viable alternative.
O

H
Equations Number
C

1 Local Balance of Linear Momentum 3
C

dN11
ES

+ F1 = ρ0 AU1
dx1
E
N

PL

dM
− N12 = ρ0 I 3θ
U

dx1
M

dN12
+ q = ρ0 AU 2
SA

dx1

2 Constitutive Law 2
N11 = EAε11

M = − EI 3φ
3 Strain-Displacement Relations 2



dU1
ε11 =
dx1
d 2U 2
φ = 2
dx1
Total number of equations 7
Unknowns (Figure 17) Number
1 Displacements U ,U 2
1 2
2 Strain resultants ε11 , φ 2
3 Stress resultants M , N11 , N12 3
Total number of unknowns 7

S
Table 2. Summary of Equations and Unknowns for Thin Beams

LS

R
O

TE
-E

AP
O

H
C

Figure 17. Definition of unknown variables in a thin beam segment
C
ES

As shown in the diagram of Figure 18, mathematical modeling is an idealization of a
specific structural behavior that is expressed by mathematical equations. For simple
E

structures, mathematical model is a convenient tool for predicting a specific behavior of
N

the structures. For more complex structural system, a structural model may consist of an
PL

assembly of structural members. Analytical solution of its mathematical model requires
U

solving a large-scale boundary value problem. Thus, the solution of large-scale
structural system based on mathematical modeling is rather impractical, if not
M

impossible at all.
SA

After the advent of computer technology, discrete modeling has become an industrial
standard in structural analysis. Like mathematical modeling, the formulation of discrete
structural model is based on a mathematical foundation, to enable its large-scale
solution with the help of computational tools. Basically, each structural member in the
structural model must be discretized so that its key properties can be represented at
selected discrete variables associated with the member. The assembly of member
contributions leads to a sizable set of algebraic equations associated with these
variables, which can be systematically solved using computers.



S
LS
Figure 18. Three modeling techniques and their relationships.

R
11. Discrete Modeling of Structures O

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(a) Definition of discrete structures
-E

In discrete structural model, finite number of control points (joints) will be designated
AP
and corresponding discrete variables are assigned to these joints, for which the solution
will be obtained. Interior (member) variables are treated as dependent variables, which
O

would be determined once the joint variables are known. The difference in the classical
H
continuous model and the discrete model is demonstrated in Figure 19.
C

C
ES

E
N

PL
U

M
SA

Figure 19. Continuous model versus discrete model.

Therefore a discrete structure is a network of joints or nodes interlinked by 1D, 2D or
3D deformable members or any of their combinations. Joints are usually established for
a member at points of geometric discontinuities and applied loads.

(b) Definition of discrete quantities

In the continuous system, one deals with a particle within the domain of the continuum.
Interesting quantities in a continuum can be categorized as statics and kinematics.



Statics consist of external force and internal force in the form of stress tensors, while
kinematics consist of displacements and deformation in the form of strain tensors.

In the discrete system, an element is associated with a set of selected nodes, only
through which the element interacts with other elements. The external work done
associated with each of these nodes is the product of ‘a vector of force, R ’ and its
corresponding ‘vector of displacement, r ’. The internal work done of an element is
represented by the product of ‘a vector of action, S ’ and the corresponding ‘vector of
deformation, v ’.

(c) Discrete representation of internal work done:

S
If a structural member is free of interior member loads, the distribution of stresses will
follow a specific pattern. Consequently, a set of discrete actions ( S ) and deformations

LS
( v ) can be defined to represent the total internal work done, i.e.,

R
∫σ
T
ε dV = S T v
O (51)

TE
B0
-E

Generally, for a 2D beam member, it can be shown that three quantities are needed for
AP
the pair of action and deformation for their product to represent the overall internal
work done as
O

H
⎧θ 1 ⎫ ⎧ v1 ⎫
C

⎪ ⎪
⎪ ⎪ ⎪ ⎪
⎪ 2⎪
∫ σ T ε dV = M 1M 2 N ⎨θ 2 ⎬ = S1S 2 S 3
C

⎨v ⎬
ES

(52)
B0 ⎪Δ⎪ ⎪ 3⎪
⎪ ⎪
⎩ ⎭ ⎪v ⎪
⎩ ⎭
E
N

PL

where the three action terms M 1M 2 N are the bending moments at the two ends of
U

the beam and its axial force respectively, while the corresponding three deformation
M

terms θ 1 θ 2 Δ are the two chord rotations and the axial elongation respectively, as
shown in Figure 20.
SA



S
Figure 20 Definition of actions and deformations for 2D bending members

LS
It could be shown that the relationship between and , i.e., , can be obtained as

R
⎡ L 1 L 1 O ⎤
⎧v 1⎫ ⎢ 3EI + GA′L − +
6 EI GA′L
0 ⎥ 1
⎧ ⎫

TE
⎪ ⎪ ⎢ L ⎥ ⎪S ⎪
-E
⎪ 2⎪ ⎢ ⎪ ⎪
0 ⎥ ⎨S 2 ⎬
1 L 1
⎨v ⎬ = ⎢ − + +
⎥
(53)
⎪ 3 ⎪ ⎢ 6 EI GA′L 3EI GA′L
⎥⎪ 3⎪ AP
⎪v ⎪
⎩ ⎭ ⎢ L ⎥ ⎪S ⎪
⎩ ⎭
0 0
⎢
⎣ ⎥
AE ⎦
O

H
C

The corresponding beam stiffness k can be obtained as k = f −1
C
ES

12. Matrix Force Method
E

Matrix Force Method employs the virtual force principle to form a system of flexibility
N

PL

equations. In Figure 21, a diagram shows the relationship between statics R and S , and
U

that between the kinematics v and r , as well as the constitutive equation linking S and
v . Note that if one defines S = BR , one can prove by using virtual principle that
M

r = BT v .
SA

If the size of S is exactly the same as the number of equilibrium equations, this is the
case of statically determinate structures. The solution of S can be obtained explicitly
from the equilibrium equations alone, i.e., S = BR . Subsequently, other part of the
solution can be determined one after another.

If the size of S is greater than the number of equations, this is the case of statically
indeterminate structures. The solution of S cannot be determined directly from the
equilibrium equations alone. The three sets of equations, namely the equilibrium, the
material constitution and the compatibility, must be combined to obtain a system of
flexibility equations, r = F R to obtain r . Subsequently other parts of the solution, v



and S can be determined. The formulation procedure for the matrix force method
applicable to statically indeterminate structures will be presented next.

S
LS

R
O
Figure 21. Notation of quantities and their relationships in matrix force method

TE
-E
For the purpose of demonstration, a simple indeterminate structure with 3 degrees of
statically indeterminacy under joint loads R , temperature change ΔT and prescribed
ˆ ˆ
AP
displacements x and q will be used in the formulation of the flexibility equations based
O

on the matrix force method.
H
C

The following step by step formulation procedure is used:
C
ES

(a) Define an associated determinate base structure.
E

A statically determinate base structure will be selected after releasing kinematics
N

corresponding to the selected redundants X as shown on the right-handed side of
PL

Figure 22.
U

(b) Decomposition of the structure into 3 base structures.
M

In Figure 23, the indeterminate structure is decomposed into 3 cases of the base
SA

structure under (a) applied joint loads, (b) the redundant forces and (c) thermal changes.
The sum of the kinematics corresponding to the redundants must be compatible to their
real values of the total structure before their release. Thus, there are three compatibility
equations to be solved for the three unknown redundants X . The total value of each
quantity is the sum of its contributions from the three cases, as listed in Table 3.



S
Figure 22. A prototype indeterminate structure with 3 degrees of static indeterminacy

LS
and its base structure after 3 redundants are removed.

R
O

TE
-E

AP
O

H
C

Figure 23. The superposition of three cases of base structure to satisfy the real
C
ES

kinematics corresponding to the redundants.
E
N

PL

Statics Kinematics
U

Actions S = S R + S X ( ST = 0) Deformations v = vR + v X + vT
r = rR + rX + rT
M

Forces R Displacements
Redundants X Kinematics of X x = xR + x X + xT
ˆ
SA

Reactions Q = QR + QX (QT = 0) Settlements ˆ
q

Table 3. List of statics and kinematics

(c) Determination of the transformation matrices ( BR , BX , DR and DX ) and
thermal strain ( VT )

As the base structure is statically determinate, actions and reactions can be obtained
easily using the equilibrium equation alone. As a result, the transformation matrices



BR , BX , DR and DX can be obtained, as well as the deformation VT due to the free
expansion effect of the temperature change, as illustrated in Figure 24.

S
LS
Figure 24. Analysis of 3 systems of statically determinate structures

R
(d) Determining the unknown redundants X using virtual force method:
O

TE
Applying virtual forces δ X corresponding to the redundants on the base structure as the
-E
virtual force system, Figure 25, one can obtain the equality of external and internal
virtual works as: AP
O

δ WE = δ WI
H
δ X T x + δ QT q = δ S T v
C

ˆ X ˆ X

( ) ( ) ( vR + vX + vT )
C

δ X T x + δ X T DT q = δ X T B T
ES

ˆ X ˆ X

ˆ X ˆ (
x + DT q = BT f S R + f S X + vT ) (54)
E

X
N

x + DT q = ⎡ BT f BR ⎤ R + ⎡ BT f BX ⎤ X + BT vT
PL

ˆ X ˆ ⎣ X ⎦ ⎣ X ⎦ X
U

x + DT q = FXR R + FXX X + BT vT
ˆ X ˆ X
M

x + xq = xR + x X + xT
ˆ ˆ
SA



Figure 25. Application of virtual force to obtain kinematics corresponding to the
redundants.

The last equation assures that the prescribed kinematics x + xq at the supports are
ˆ ˆ
satisfied by the summation of the three cases. If there is no support settlement, both
ˆ ˆ
x and xq are zero. Physically the size of X must be adjusted to ensure this condition;
thus, this equation is solved for the value of X , i.e.,

⎣ ⎦ {
⎡ FXX ⎤ { X } = x + DT q − FXR R − BT vT → X
ˆ X ˆ X } (55)

Note the two subscripts of the flexibility matrix FXR = BT f BR . The first refers to the

S
X
resulting displacement vector x while the second refers to the applying force vector R .

LS
Similarly, the two subscripts in the flexibility matrix FXX = BT f BX refer to the
X

R
displacement x and the force vector X respectively. O

TE
(e) Obtain other parts of the solution.
-E

Similar to Eq.(54), one obtains by virtual force principle the following:
AP
O

r + DR q = FRR R + FRX X + BR vT
T
ˆ T
(56)
H
C

where FRR = BR f BR and FRX = BR f BX .
T T
C
ES

Once the unknown redundants X are solved, the remaining quantities can be obtained
E

as shown in Table 4. Finally, the step by step procedure of the Matrix Force Method is
N

summarized in Table 5.
PL
U

Statics Kinematics
M

Actions S = BR R + BX X Deformations v = f BR R + f BX X + vT
r = FRR R + FRX X + BR vT − DR q
SA

T T
R ˆ
Forces Displacements

Redundants X
Kinematics x = FXR R + FXX X + BT vT − DT q
ˆ X X ˆ
of X

Reactions Q = DR R + DX X Settlements ˆ
q

Table 4. Final results of statics and kinematics after X is solved.

1 ˆ
Define a base structure and identify X and x
2 ˆ
Define f , R, r , S , v and Q (if there is settlement q )



3 Form BR , BX , DR , DX and vT (if temperature changes are given)
4 Form FRR , FRX , FXR and FXX
5
Solve for X from ⎣ ⎦ {
⎡ FXX ⎤ { X } = x + DT q − FXR R − BT vT
ˆ X ˆ X }
6 Obtain the rest from Table 4.

Table 5. Step by step Procedure for Matrix Force Method

13. Matrix Displacement Method

Matrix Displacement Method employs the virtual displacement principle to form a
system of stiffness equations. Figure 26 shows a diagram of the relationship between

S
kinematics v and r and the statics R and S , as well as the constitutive equation
linking S and v . Note that if one defines v = a r , one can prove by using virtual

LS
principle that R = aT S .

R
O
In a structure in which the deformation v can be obtained directly from known

TE
displacement vector r , this structure is referred as a kinematically determinate structure.
-E
In usual cases, displacements in a structure are unknowns and therefore most practical
structures are kinematically indeterminate. AP
O

In Matrix Displacement Method, it is necessary to identify the degree of kinematic
indeterminacy ( o KI ), also known as the number of degrees of freedom (dof). In frame
H
C

structures, this refers to the number of independent discrete displacements, i.e., all
displacements that vary varied independently. In a 2D frame structure, each rigid joint
C
ES

has 3 degrees of freedom, while in 3D frame structure, each rigid joint has 6 degrees of
freedom. If there are internal constraints, some joint degrees of freedom could become
E

inter-dependent. Thus, the size of independent degrees of freedom will reduce. The need
N

to eliminate dependency among joint degrees of freedom is to avoid solving a singular
PL

equation system.
U

Figure 26 illustrates the deformation modes associated with each degree of freedom
M

independent of the others. When constraints are prescribed in the system, the number of
degrees of freedom will be reduced. Figure 27 illustrates the same structure except that
SA

three constraints are introduced, i.e., the three members are not allowed to elongate or
shorten after assuming that their axial deformation is negligible. Under these constraints,
deformation modes in (a) (b) (d) and (e) in Figure 26 are not feasible, while one new
deformation mode arises as shown in Figure 27. Theoretically, the degree of kinematic
indeterminacy can be calculated from the number of joint degrees of freedom less the
number of internal constraints. In contrary to the force method which derives kinematics
from statics, the displacement method derives statics from kinematics. The relationships
among the displacement r , the deformation v , the action S and the nodal force R are
depicted in Figure 28.



For the purpose of demonstration, a simple kinematically indeterminate structure under
joint loads R , temperature change ΔT and prescribed displacements x and q as shown
ˆ ˆ
in Figure 29 will be used in the formulation of the stiffness equations based on the
matrix displacement method. The following step by step formulation procedure is used:
ˆ
(a) Define R, r , S , v, X , x, Q and q (Figure 29).

(b) Decompose the total structure to separate the effects of r , x and ΔT
ˆ

As shown in Figure 30, the independent effects of r , x and ΔT are considered
ˆ
separately and their contributions into each quantity of statics and kinematics will be
superposed as listed in Table 6.

S
(c) Derive the matrices ar , ax , cr , cx , k , vT , qT and ST

LS

R
O

TE
-E

AP
O

H
C

C
ES

E
N

PL
U

M

Figure 26. Degree of Kinematic Indeterminacy equals joint degrees of freedom if there
is no constraint in the system.
SA



S
Figure 27. Degree of Kinematic Indeterminacy equals joint degrees of freedom less

LS
number of constraints.
Based on the geometric change, one can obtain the deformed elastic curve as a result of

R
O
a unit displacement of r by which ar and cr can be derived. Similarly for the effect of

TE
ˆ
settlement x , one can obtain a x and cx . For the effect of temperature change, r and
-E
x have to be clamped; thus ST can be obtained while deformation vT is normally zero.
However, if constraint exists by means of allowing for qT , vT can be non-zero. Figure
AP
30 presents the symbolic results of all these quantities. The member stiffness matrix
O

k relating v to S can be obtained from f −1 .
H
C

C
ES

E
N

PL
U

M
SA

Figure 28. Notation of quantities and their relationships in matrix displacement method

(d) Determination of r by virtual displacement method.

Using the virtual displacement system on the right-handed side of Figure 31, applying
virtual displacement δ r corresponding to the external force R , one can obtain the
equality of external and internal virtual works as:



δ WE = δ WI
δ r T R + δ qr Q = δ vr S
T T

( ) (
δ r T R + δ r T cr Q = δ r T ar
T T
) ( Sr + S x + ST )
R + cr Q = ar ( k vr + k vx + ST )
T T
(57)

R + cr Q = ⎡ ar k ar ⎤ r + ⎡ ar k ax ⎤ x + ar ST
T T T
ˆ T
⎣ ⎦ ⎣ ⎦
R + cr Q = K rr r + K xx x + ar ST
T
ˆ T

R + Rq = Rr + Rx + RT

S
LS

R
O

TE
-E

AP
O

Figure 29 Quantities needed in the formulation of stiffness equations by Matrix
Displacement Method.
H
C

C
ES

E
N

PL
U

M
SA

Figure 30. The superposition of effects due to three cases of kinematic excitation



Figure 31. Application of virtual displacement δ r to obtain external force R .

The last equation above equates external forces to internal forces as individual effects of

S
the three cases. Note that Kij = ai k a j is a stiffness matrix. The first subscript in the

LS
stiffness matrices Kij refer to the internal force result while the second subscript refers
ˆ
to the displacement cause. Thus, the term K rx x will result in Rx which corresponds to

R
O
the contribution to internal force vector R as a result of applying the settlement vector

TE
ˆ
x.
-E
Once the displacement vector r is obtained from the system of stiffness equations, i.e.,

{ }
⎡ K rr ⎤ {r} = R + cr Q − K xx x − ar ST
⎣ ⎦
T
ˆ T
AP (58)
O

H
the remaining quantities can be obtained as shown in Table 6. Finally, the step by step
C

procedure of the Matrix Displacement Method is summarized in Table 7.
C
ES

14. Trends and Perspectives
E

Discrete approach of structural modeling is suited for digital computation and has lead
N

PL

to the generalization of the formulation procedure. The two principal methods, namely
U

the matrix force method and the matrix displacement method, are convenient for
analysis of frames made up mainly of one-dimensional members. Of the two methods,
M

the matrix displacement method is more popular, due to its natural extension to the
more generalized finite element method.
SA

Kinematics Statics

v = ar r + ax x + vT S = k ar r + k ax x + ST
Deformation Action

Displacement r Force
R = K rr r + K xx x + ar ST − cr Q
ˆ T T

Settlement ˆ
x Reaction
X = K xr r + K xx x + aT ST − cT Q
x x

Dependent dof q = qr + qx + qT Dependent force Q



Table 6. Final results of statics and kinematics after r is solved.

1 o
Identify KI or degrees of freedom of the problem and select appropriate vector of
r
2 ˆ
Define k R, r , S , v , X , x , Q and q
3 Form ar , ax , cr , cx and vT , ST (if temperature changes are given)
4 Form K rr , K rx , K xr and K xx
5
Solve for r from ⎣ ⎦ {
⎡ K rr ⎤ {r} = R + cr Q − K xx x − ar ST
T
ˆ T }
6 Obtain the rest from Table 7.

S
Table 8 Step by step Procedure for Matrix Displacement Method

LS
Using displacements as the primary variables, the stiffness matrix of a discrete

R
structural model can be formed globally as in the case of the matrix displacement
O
method, or locally by considering the stiffness contributions of individual elements.

TE
This latter procedure is generally known as the direct stiffness method. The direct
-E
stiffness procedure allows the assembly of stiffness contributions from a finite number
of elements that are used to model any complex structure. The basic procedure for
AP
developing the stiffness of an element is to assume appropriate shape functions that
represent the displacement field over the element. Based on these assumed displacement
O

fields, the element stiffness can be formulated in terms of nodal degrees of freedom.
H
C

The direct stiffness process was in fact the main reason how the powerful general-
C
ES

purposed finite element programs can be developed for any discrete structural model.
Today, many commercial software packages are widely available for linear elastic
analysis based on the stiffness method. These commercial software packages may help
E

reduce the work load of structural engineers, but will not decrease their accountability
N

PL

on the reliability the analysis results. Thus, it is important that structural engineers
U

understand the fundamentals of the analysis, whether the analysis is performed by the
engineers themselves or with the help of a computer program.
M

Glossary
SA

Base structure: The choice of a statically determinate structure reduced from the
original statically indeterminate structure after removing the
redundant kinematics.
Boundary The imposition of either force or displacement along a boundary
condition: of the structure
Cauchy stress: A definition of stress associated with the deformed configuration.
Compatibility: The term used to describe the conformity between the strains and
the displacement field of a deformed structure.
Constitutive The equations used to describe the relationship between stresses
equations: and strains of a material.
Deformation The gradient of deformation in the neighborhood of a particle.
gradient:



Deformation: A change in the relative positions of particles of a body.
Degrees of Displacement variables associated with nodes of a discrete
freedom: structure.
Direct stiffness A method to directly assemble the contributions of individual
method: members or elements of a structure to form the stiffness matrix of
a structure.
Discrete model of A representation of the structure to be analyzed by a computer
structures: method by which the primary variables are determined
numerically.
Discrete A method to model structures by a finite number of elements
modeling: (members) that are associated with a set of discrete variables to be
determined numerically.
Displacement: The disposition measurement of a particle in a body or a structure.

S
Elastic: Property of a material that rebounds to its original position from

LS
deformation upon the release of the applied force.
Engineering shear A measurement of shear distortion of a body commonly used in
strain: engineering.

R
Equilibrium:
all directions.
O
A state in which the internal forces balance all external forces in

TE
-E
Euler-Bernoulli A beam theory by which the plane section of a beam remains
beams: plane and normal to the neutral axis.
Excitation: Any disturbance to the structure including external forces,
AP
prescribed displacement, accelerations, etc.
O

Finite element A method to model structures by a finite number of typical
method: elements associated with a set of discrete variables to be
H
C

determined numerically.
Infinitesimal A linearized strain after neglecting the nonlinear high-order terms
C
ES

strain: due to the smallness of the strain.
Internal work Work done by the product of stress and strain.
E

done:
N

Isotropic: A description of material property that is invariant to all
PL

directions.
U

Joints: A point where two or more one-dimensional members meet, about
which a set of degrees of freedom are specified..
M

Kinematics: Quantities associated with geometry, the position changes or the
SA

deformation of geometry. This term is used in opposition to the
term “statics”.
Linear elastic A structural analysis that assumes linear stress-strain relationship
analysis: for all elastic structural members.
Mathematical A mathematical representation of the structure.
model:
Matrix analysis of A term used to describe structural analysis of discrete structural
structures: system using matrix operations.
Matrix An analysis of discrete structural model of which displacements
displacement are primary variables.
method:
Matrix force An analysis of discrete structural model of which force parameters


C I V I L E N G I N E E R I N G Structural Analysis Worsak Kanok Nukulchaii

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