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FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat Introduction to Structural Analysis

Copyright © 2011 J. Rungamornrat

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CHAPTER 1

INTRODUCTION TO STRUCTURAL ANALYSIS

This first chapter provides a brief introduction of basic components essential for structural analysis.First, the concept of structural modeling or structural idealization is introduced. This process

involves the construction of a mathematical model or idealized structure to represent a real

structure under consideration. The structural analysis is in fact a subsequent process that is

employed to solve a set of mathematical equations governing the resulting mathematical model to

obtain a mathematical solution. Such solution is subsequently employed to characterize or

approximate responses of the real structure to a certain level of accuracy. Conservation of linear and

angular momentum of a body in equilibrium is also reviewed and a well known set of equilibrium

equations that is fundamentally important to structural analysis is also established. Finally, certain

classifications of idealized structures are addressed.

1.1 Structural Idealization

A real structure is an assemblage of components and parts that are integrated purposely to serve

certain functions while withstanding all external actions or excitations (e.g. applied loads,

environmental conditions such as temperature change and moisture penetration, and movement of

its certain parts such as foundation, etc.) exerted by surrounding environments. Examples of real

structures mostly encountered in civil engineering application include buildings, bridges, airports,

factories, dams, etc as shown in Figure 1.1. The key characteristic of the real structure is that its

responses under actions exerted by environments are often very complex and inaccessible to human

in the sense that the real behavior cannot be known exactly. Laws of physics governing such

physical or real phenomena are not truly known; most of available theories and conjectures are based primarily on various assumptions and, as a consequence, their validity is still disputable and

dependent on experimental evidences.

Figure 1.1: Schematics of some real structures





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Since behavior of the real structure is extremely complex and inaccessible, it necessitates the

development of a simplified or approximate structure termed as an idealized structure. To be more

precise, an idealized structure is a mathematical model or a mathematical object that can be used to

approximate behavior or responses of the real structure to certain degree of accuracy. The main

characteristic of the idealized structure is that its responses are accessible, solvable, and can be

completely determined using available laws of physics and mathematics. The process for obtainingthe idealized structure is called structural idealization or structural modeling. This process

generally involves imposing various assumptions and simplifying the complexity embedded in the

real structure. The idealized structure of a given real structure is in general not unique and many

different idealized structures can be established via use of different assumptions and simplifications.

The level of idealization considered in the process of modeling depends primarily on the required

degree of accuracy of (approximate) responses of the idealized structure in comparison with those

of the real structure. The idealization error is an indicator that is employed to measure the

discrepancy between a particular response of the real structure and the idealized solution obtained

by solving the corresponding idealized structure. The acceptable idealization error is an important

factor influences the level of idealization and a choice of the idealized structure. While a more

complex idealized structure can characterize the real structure to higher accuracy, it at the sametime consumes more computational time and effort in the analysis. The schematic indicating the

process of structural idealization is shown in Figure 1.2.

For brevity and convenience, the term “structure” throughout this text signifies the

“idealized structure” unless stated otherwise. Some useful guidelines for constructing the idealized

structure well-suited for structural analysis procedure are discussed as follows.

Figure 1.2: Diagram indicating the process of structural idealization

1.1.1 Geometry of structure

It is known that geometry of the real structure is very complex and, in fact, occupies space.

However, for certain classes of real structures, several assumptions can be posed to obtain anidealized structure possessing a simplified geometry. A structural component with its length much

Assumptions + Simplification

Governing Physics

Idealization error Idealized solution

Structural analysisRes onse inter retation

Real structure

Complex

&Inaccessible

Idealized structure

Simplified

&Solvable





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larger than dimensions of its cross section can be modeled as a one-dimensional or line member,

e.g. truss, beam, frame and arch shown in Figure 1.3. A structural component with its thickness

much smaller than the other two dimensions can properly be modeled as a two-dimensional or

surface member, e.g. plate and shell structures. For the case where all three dimensions of the

structure are comparable, it may be obligatory to be modeled as a three-dimensional member, e.g.

dam and a local region surrounding the connections or joints.

Figure 1.3: Schematics of idealized structures consisting of one-dimensional members

1.1.2 Displacement and deformation

Hereby, the term deformation is defined as the distortion of the structure while the term

displacement is defined as the movement of points within the structure. These two quantities have a

fundamental difference, i.e., the former is a relative quantity that measures the change in shape or

distortion of any parts of the structure due to any action while the latter is a total quantity that

measures the change in position of individual points resulting from any action. It is worth noting

that the structure undergoing the displacement may possess no deformation; for instance, there is no

change in shape or distortion of the structure if it is subjected to rigid translation or rigid rotation.

This special type of displacement is known as the rigid body displacement . On the contrary, the

deformation of any structure must follow by the displacement; i.e. it is impossible to introduce non-

zero deformation to the structure with the displacement vanishing everywhere.

For typical structures in civil engineering applications, the displacement and deformation

due to external actions are in general infinitesimal in comparison with a characteristic dimension of

the structure. The kinematics of the structure, i.e. a relationship between the displacement and the

deformation, can therefore be simplified or approximated by linear relationship; for instance, the

linear relationship between the elongation and the displacement of the axial member, the linear

relationship between the curvature and the deflection of a beam, the linear relationship between the

rate of twist and the angle of twist of a torsion member, etc. In addition, the small discrepancy

between the undeformed and deformed configurations allows the (known) geometry of the

undeformed configuration to be employed throughout instead of using the (unknown) geometry of the deformed configuration.

FrameTruss

Beam Arch





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1.1.3 Material behavior

The behavior of a constituting material in real structures is extremely complex (i.e. it is generally

nonlinear, nonhomogeneous, anisotropic and time and history dependent) and, as a consequence,

construction of a suitable constitutive model is both theoretically and computationally challenging.

In constitutive modeling, the behavior of materials is generally modeled or approximated via therelationship between the internal force measure (e.g. axial force, torque, bending moment, shear

force, and stress) and the deformation (e.g. elongation, rate of twist, curvature, and strain).

Most of materials encountering in civil engineering applications (e.g. steel and concrete) are

often modeled as an idealized , simple material behavior called an isotropic and linearly elastic

material. The key characteristics of this class of materials are that the material properties are

directional independent, its behavior is independent of both time and history, and stress and strain

are related through a linear function. Only two material parameters are required to completely

describe the material behavior; one is the so-called Young’s modulus denoted by E and the other is

the Poisson’s ratio denoted by . Other material parameters can always be expressed in terms of

these two parameters; for instance, the shear modulus, denoted by G, is given by

)1(2

EG

(1.1)

The Young’s modulus E can readily be obtained from a standard uniaxial tensile test while G is the

elastic shear modulus obtained by conducting a direct shear test or a torsion test. The Poisson’s ratio

can then be computed by the relation (1.1). Both E and G can be interpreted graphically as a slope

of the uniaxial stress-strain curve (- curve) and a slope of the shear stress-strain curve (- curve),

respectively, as indicated in Figure 1.4. The Poisson’s ratio is a parameter that measures the

degree of contraction or expansion of the material in the direction normal to the direction of the

normal stress.

Figure 1.4: Uniaxial and shear stress-strain diagrams

1.1.4 Excitations

All actions or excitations exerted by surrounding environments are generally modeled by vector

quantities such as forces and moments. The excitations can be divided into two different classes

depending on the nature of their application; one called the contact force and the other called the

remote force. The contact force results from the idealization of actions introduced by a direct

E

1

Uniaxial stress-strain curve

G

1

Shear stress-strain curve





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contact between the structure and surrounding environments such as loads from occupants and wind

while the remote force results from the idealization of actions introduced by remote environments

such as gravitational force.

The contact or remote force that acts on a small area of the structure can be modeled by a

concentrated force or a concentrated moment while the contact or remote force that acts over a large

area can properly be modeled by a distributed force or a distributed moment. Figure 4 shows anexample of an idealized structure subjected to two concentrated forces, a distributed force and a

concentrated moment.

Figure 1.5: Schematic of a two-dimensional, idealized structure subjected to idealized loads

1.1.5 Movement constraints

Interaction between the structure and surrounding environments to maintain its stability while

resisting external excitations (e.g. interaction between the structure and the foundation) can

mathematically be modeled in terms of idealized supports. The key function of the idealized

support is to prevent or constrain the movement of the structure in certain directions by means of

reactive forces called support reactions. The support reactions are introduced in the direction where

the movement is constrained and they are unknown a priori; such unknown reactions can generally

be computed by enforcing static equilibrium conditions and other necessary kinematical conditions.

Several types of idealized supports mostly found in two-dimensional idealized structures are

summarized as follows.

1.1.5.1 Roller support

A roller support is a support that can prevent movement of a point only in one direction while

provide no rotational constraint. The corresponding unknown support reaction then possesses only

one component of force in the constraint direction. Typical symbols used to represent the roller

support and support reaction are shown schematically in Figure 1.6.

Figure 1.6: Schematic of a roller support and the corresponding support reaction





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1.1.5.2 Pinned or hinged support

A pinned or hinged support is a support that can prevent movement of a point in both directions

while provide no rotational constraint. The corresponding unknown support reaction then possesses

two components of force in each direction of the constraint. Typical symbols used to represent the

pinned or hinged support and the support reactions are shown schematically in Figure 1.7.

Figure 1.7: Schematic of a pinned or hinged support and the corresponding support reactions.

1.1.5.3 Fixed support

A fixed support is a support that can prevent movement of a point in both directions and provide a

full rotational constraint. The corresponding unknown support reaction then possesses two

components of force in each direction of the translational constraint and one component of moment

in the direction of rotational constraint. Typical symbols used to represent the fixed support and the

support reactions are shown schematically in Figure 1.8.

Figure 1.8: Schematic of a fixed support and the corresponding support reactions

1.1.5.4 Guide support

A guide support is a support that can prevent movement of a point in one direction and provide a

full rotational constraint. The corresponding unknown support reaction then possesses one

component of force in the direction of the translational constraint and one component of moment in

the direction of rotational constraint. Typical symbols used to represent the guide support and the

support reactions are shown schematically in Figure 1.9.

Figure 1.9: Schematic of a guide support and the corresponding support reactions





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1.1.5.5 Flexible support

A flexible support is a support that can partially prevent translation and/or rotational constraints.

The corresponding unknown support reaction is related to the unknown displacement and/or

rotation of the support. Typical symbols used to represent the flexible support and the support

reactions are shown schematically in Figure 1.10.

Figure 1.10: Schematic of a flexible support and the corresponding support reactions

1.1.6 Connections

Behavior of a local region where the structural components are connected is very complicated and

this complexity depends primarily on the type and details of the connection used. To extensively

investigate the behavior of the connection, a three dimensional model is necessarily used to gain

accurate results. For a standard, linear structural analysis, the connection is only modeled as a point

called node or joint and the behavior of the node or joint depends mainly on the degree of force and

moment transfer across the connection.

1.1.6.1 Rigid joint

A rigid joint is a connection that allows the complete transfer of force and moment across the joint.

Both the displacement and rotation are continuous at the rigid joint. This idealized connection is

usually found in the beam or frame structures as shown schematically in Figure 1.11.

Figure 1.11: Schematic of a real connection and the idealized rigid joint

1.1.6.2 Hinge joint

A hinge joint is a connection that allows the complete transfer of force across the joint but does not

allow the transfer of the bending moment. Thus, the displacement is continuous at the hinge joint

while the rotation is not since each end of the member connecting at the hinge joint can rotate freely

from each other. This idealized connection is usually found in the truss structures as shownschematically in Figure 1.12.





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Figure 1.12: Schematic of a real connection and the idealized hinge joint.

1.1.6.3 Partially rigid joint

A partially rigid joint is a connection that allows the complete transfer of force and a partial transfer

of moment across the joint. For this particular case, both the displacement is continuous at the joint

while rotation is not. The behavior of the flexible joint is more complex than the rigid joint and the

hinge joint but it can better represent the real behavior of the connection in the real structure. The

schematic of the partially rigid joint is shown in Figure 1.13.

Figure 1.13: Schematic of an idealized partially rigid joint

1.1.7 Idealized structures

In this text, it is focused attention on a particular class of idealized structures that consist of one-

dimensional and straight components, is contained in a plane, and is subjected only to in-plane

loadings; these structures are sometimes called “two-dimensional” or “plane” structures. Three

specific types of structures in this class that are main focus of this text include truss, beam and

frame.

Figure 1.14: Schematic of idealized trusses





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1.1.7.1 Truss

Truss is an idealized structure consisting of one-dimensional, straight structural components that are

connected by hinge joints. Applied loads are assumed to act only at the joints and all members

possess only one component of internal forces, i.e. the axial force. Examples of truss structures are

shown in the Figure 1.14.

1.1.7.2 Beam

Beam is an idealized structure consisting of one-dimensional, straight members that are connected

in a series either by hinge joints or rigid joints; thus, the geometry of the entire beam must be one-

dimensional. Loads acting on the beam must be transverse loadings (loads including forces normal

to the axis of the beam and moments directing normal to the plane containing the beam) and they

can act at any location within the beam. The internal forces at a particular cross section consist of

only two components, i.e., the shear force and the bending moment. Examples of beams are shown

in Figure 1.15.

Figure 1.15: Schematic of idealized beams

1.1.7.3 Frame

Frame is an idealized structure consisting of one-dimensional, straight members that are connectedeither by hinge joints or rigid joints. Loads acting on the frame can be either transverse loadings or

longitudinal loadings (loads acting in the direction parallel to the axis of the members) and they can

act at any location within the structure. The internal forces at a particular cross section consist of

three components: the axial force, the shear force and the bending moment. It can be remarked that

when the internal axial force identically vanishes for all members and the geometry of the structure

is one dimensional, the frame simply reduces to the beam. Examples of frame structures are shown

in Figure 1.16.

Figure 1.16: Schematic of idealized frames





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1.2 Continuous Structure versus Discrete Structure Models

A continuous structure is defined as an idealized structure where its responses at all points are

unknown a priori and must be determined as a function of position (i.e. be determined at all points

of the structure) in order to completely describe behavior of the entire structure. The primary

unknowns of the continuous structure are in terms of response functions and, as a result, the number of unknowns counted at all points of the structure is infinite. Analysis of such continuous structure

is quite complex and generally involves solving a set of governing differential equations. In the

other hand, a discrete structure is a simplified idealized structure where the responses of the entire

structure can completely be described by a finite set of quantities. This type of structures typically

arises from a continuous structure furnishing with additional assumptions or constraints on the

behavior of the structures to reduce the infinite number of unknowns to a finite number. A typical

example of discrete structures is the one that consists of a collection of a finite number of structural

components called members or elements and a finite number of points connecting those structural

components to make the structure as a whole called nodes or nodal points. All unknowns are forced

to be located only at the nodes by assuming that behavior of each member can be completely

determined in terms of the nodal quantities – quantities associated with the nodes. An example of a

discrete structure consisting of three members and four nodes is shown in Figure 1.17.

Figure 1.17: An example of a discrete structure comprising three members and four nodes

1.3 Configurations of Structure

There are two configurations involve in the analysis of a deformable structure. An undeformed

configuration is used to refer to the geometry of a structure at the reference state that is free of any

disturbances and excitations. A deformed configuration is used to refer to a subsequent

configuration of the structure after experiencing any disturbances or excitations. Figure 1.18 shows both the undeformed configuration and the deformed configuration of a rigid frame.

Figure 1.18: Undeformed and deform configurations of a rigid frame under applied loads

Node 4

Node 3 Node 2

Node 1

Member 1

Member 2

Member 3

vu

X

Y

Deformed configuration

Undeformed configuration





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1.4 Reference Coordinate Systems

In structural analysis, a reference coordinate system is an indispensable tool that is commonly used

to conveniently represent quantities of interest such as displacements and rotations, applied loads,

support reactions, etc. Following subsections provide a clear notion of global and local coordinate

systems and a law of coordinate transformation that is essential for further development.

1.4.1 Global and local coordinate systems

There are two types of reference coordinate systems used throughout the development presented

further in this book. A global coordinate system is a single coordinate system that is used to

reference geometry or involved quantities for the entire structure. A choice of the global coordinate

system is not unique; in particular, an orientation of the reference axes and a location of its origin

can be chosen arbitrarily. The global reference axes are labeled by X, Y and Z with their directions

strictly following the right-handed rule. For a two-dimensional structure, the commonly used,

global coordinate system is one with the Z-axis directing normal to the plane of the structure. A

local coordinate system is a coordinate system that is used to reference geometry or involved

quantities of an individual member. The local reference axes are labeled by x, y and z. This

coordinate system is defined locally for each member and, generally, based on the geometry and

orientation of the member itself. For plane structures, it is typical to orient the local coordinate

system for each member in the way that its origin locates at one of its end, the x-axis directs along

the axis of the member, the z-axis directs normal to the plane of the structure, and the y-axis follows

the right-handed rule. An example of the global and local coordinate systems of a plane structure

consisting of three members is shown in Figure 1.19.

Figure 1.19: Global and local coordinate systems of a plane structure

1.4.2 Coordinate transformation

In this section, we briefly present a basic law of coordinate transformation for both scalar quantities

and vector quantities. To clearly demonstrate the law, let introduce two reference coordinate

systems that possess the same origin: one, denoted by {x1, y1, z1}, with the unit base vectors {i1, j1,

k1} and the other, denoted by {x2, y2, z2}, with the unit base vectors {i2, j2, k2} as indicated in

Figure 1.20. Now, let define a matrix R such that

X

Y

x

y

y

x

y

x





12

332313

322212

312111

212121

212121

212121

coscoscos

coscoscos

coscoscos

kkk jki

jk j j ji

iki jii

R (1.2)

where {11, 21, 31} are angles between the unit vector i2 and the unit vectors {i1, j1, k1},respectively; {12, 22, 32} are angles between the unit vector j2 and the unit vectors {i1, j1, k1},

respectively; and {13, 23, 33} are angles between the unit vector k2 and the unit vectors {i1, j1,

k1}, respectively.

Figure 1.20: Schematic of two reference coordinate systems with the same origin

1.4.2.1 Coordinate transformation for scalar quantities

Let be a scalar quantity whose values measured in the coordinate system {x1, y1, z1} and to the

coordinate system {x2, y2, z2} are denoted by 1 and 2, respectively. Since a scalar quantity

possesses only a magnitude, its values are invariant of the change of reference coordinate systems

and this implies that

21 (1.3)

1.4.2.2 Coordinate transformation for vector quantities

Let v be a vector whose representations with respect to the coordinate system {x1, y

1, z

1} and the

coordinate system {x2, y2, z2} are given by

2

2

z2

2

y2

2

x1

1

z1

1

y1

1

x vvvvvv k jik jiv (1.4)

where { 1

z

1

y

1

x v,v,v } and { 2

z

2

y

2

x v,v,v } are components of a vector v with respect to the coordinate

systems {x1, y1, z1} and {x2, y2, z2}, respectively. To determine the component 2

xv in terms of the

components { 1

z

1

y

1

x v,v,v }, we take an inner product between a vector v given by (1.4) and a unit

vector i2 to obtain

)(v)(v)(vv 211z21

1y21

1x

2x iki jii (1.5)

y1

y2

x1

x2

z1 z2

i1

i2 j1

j2

k1

k2





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Similarly, by taking an inner product between a vector v given by (1.4) and a unit vector j2 and k2,

it leads to

)(v)(v)(vv 21

1

z21

1

y21

1

x

2

y jk j j ji (1.6)

)(v)(v)(vv 21

1

z21

1

y21

1

x

2

x kkk jki (1.7)

With use of the definition of the transformation matrix R given by (1.2), equations (1.5)-(1.7) can

be expressed in a more concise form as

1

z

1

y

1

x

1

z

1

y

1

x

212121

212121

212121

2

z

2

y

2

x

v

v

v

v

v

v

v

v

v

R

kkk jki

jk j j ji

iki jii

(1.8)

The expression of the components { 1z

1y

1x v,v,v } in terms of the components { 2

z2y

2x v,v,v } can readily

be obtained in a similar fashion by taking a vector inner product of the vector v given by (1.4) and

the unit base vectors {i1, j1, k1}. The final results are given by

2

z

2

y

2

x

T

2

z

2

y

2

x

121212

121212

121212

1

z

1

y

1

x

v

v

v

v

v

v

v

v

v

R

kkk jki

jk j j ji

iki jii

(1.9)

where RT is a transpose of the matrix R. Note that the matrix R is commonly termed a

transformation matrix.

1.4.2.3 Special case

Let consider a special case where the reference coordinate system {x2, y2, z2} is simply obtained by

rotating the z1-axis of the reference coordinate system {x1, y1, z1} by an angle . The transformation

matrix R possesses a special form given by

100

0cossin

0sincos

R (1.10)

The coordinate transformation formula (1.8) and (1.9) therefore reduce to

1

z

1

y

1

x

2

z

2

y

2

x

v

v

v

100

0cossin

0sincos

v

v

v

(1.11)

2

z

2

y

2

x

1

z

1

y

1

x

vv

v

1000cossin

0sincos

vv

v

(1.12)





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This clearly indicates that the component along the axis of rotation is unchanged and is independent

of the other two components. The laws of transformation (1.11) and (1.12) can also be applied to

the case of two vectors v and w where v is contained in the x1-y1 plane (and the x2-y2 plane) and w

is perpendicular to the x1-y1 plane (and the x2-y2 plane). More precisely, components of both vectors

v and w in the {x1, y1, z1} coordinate system and in the {x2, y2, z2} coordinate system are related by

2 1

x x

2 1

y y

2 1

z z

v cos sin 0 v

v sin cos 0 v

w 0 0 1 w

(1.13)

1 2

x x

1 2

y y

1 2

z z

v cos sin 0 v

v sin cos 0 v

w 0 0 1 w

(1.14)

1.5 Basic Quantities of Interest

This section devotes to describe two different classes of basic quantities that are involved in

structural analysis, one is termed kinematical quantities and the other is termed static quantities.

1.5.1 Kinematical quantities

Kinematical quantities describe geometry of both the undeformed and deformed configurations of

the structure. Within the context of static structural analysis, kinematical quantities can be

categorized into two different sets: one associated with quantities used to measure the movement or

change in position of the structure and the other is associated with quantities used to measure thechange in shape or distortion of the structure.

Displacement at any point within the structure is a quantity representing the change in

position of that point in the deformed configuration measured relative to the undeformed

configuration. Rotation at any point within the structure is a quantity representing the change in

orientation of that point in the deformed configuration measured relative to the undeformed

configuration. For a plane structure shown in Figure 1.18, the displacement at any point is fully

described by a two-component vector (u, v) where u is a component of the displacement in X-

direction and v is a component of the displacement in Y-direction while the rotation at any point is

fully described by an angle measured from a local tangent line in the undeformed configuration to

a local tangent line at the same point in the deformed configuration. It is important to emphasize

that the rotation is not an independent quantity but its value at any point can be computed when thedisplacement at that point and all its neighboring points is known.

A degree of freedom, denoted by DOF , is defined as a component of the displacement or the

rotation at any node (of the discrete structure) essential for describing the displacement of the entire

structure. There are two types of the degree of freedom, one termed as a prescribed degree of

freedom and the other termed as a free or unknown degree of freedom. The former is the degree of

freedom that is known a priori, for instance, the degree of freedom at nodes located at supports

where components of the displacement or rotation are known while the latter is the degree of

freedom that is unknown a priori. The number of degrees of freedom at each node depends

primarily on the type of nodes and structures and also the internal releases and constraints present

within the structure. In general, it is equal to the number of independent degrees of freedom at that

node essential for describing the displacement of the entire structure. For beams, plane trusses,





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space truss, plane frames, and space frames containing no internal release and constraint, the

number of degrees of freedom per node are 2 (a vertical displacement and a rotation), 2 (two

components of the displacement), 3 (three components of the displacement), 3 (two components of

the displacement and a rotation) and 6 (three components of the displacement and three components

of the rotation), respectively. Figure 1.21 shows examples of both prescribed degrees of freedom

and free degrees of freedom of beam, plane truss and plane frames. The number of degrees of freedom of a structure is defined as the number of all independent degrees of freedom sufficient for

describing the displacement of the entire structure or, equivalently, it is equal to the sum of numbers

of degrees of freedom at all nodes. For instance, a beam shown in Figure 1.21(a) has 6 DOFs {v1,

1, v2, 2, v3, 3} consisting of 3 prescribed DOFs {v1, 1, v3} and 3 free DOFs {v2, 2, 3}; a plane

truss shown in Figure 1.21(b) has 6 DOFs {u1, v1, u2, v2, u3, v3} consisting of 3 prescribed DOFs

{u1, v1, v2} and 3 free DOFs {u2, u3, v3}; and a plane frame shown in Figure 1.21(c) has 9 DOFs

{u1, v1, 1, u2, v2, 2, u3, v3, 3} consisting of 3 prescribed DOFs { u1, v1, v3} and 6 free DOFs {1,

u2, v2, 2, u3, 3}. It is evident that the number of degrees of freedom of a given structure is not

unique but depending primarily on how the structure is discretized. As the number of nodes in the

discrete structure increases, the number of the degrees of freedom of the structure increases.

Figure 1.21: (a) Degrees of freedom of a beam, (b) degrees of freedom of a plane truss, and (c)degrees of freedom of a plane frame

X

v1=0u1=0

u3

v3

u2

v2=0

Y

Node 1 Node 2

Node 3

(b)

(c)

(a)

v1 = 0

1 = 0

v2

2

v3 = 0

3

Y

X

Node 1 Node 2 Node 3

v1=0u1=0

1

3 u3 v3=0u2

v2

2

Y

X

Node 1

Node 2 Node 3





16

Deformation is a quantity used to measure the change in shape or the distortion of a

structure (i.e. elongation, rate of twist, curvature, strain, etc.) due to disturbances and excitations.

The deformation is a relative quantity and a primary source that produces the internal forces or

stresses within the structure. For continuous structures, the deformation is said to be completely

described if and only if the deformation is known at all points or is given as a function of position

while, for discrete structures, the deformation of the entire structure is said to be completelydescribed if and only if the deformation of all members constituting the structure are known. The

deformation for each member of a discrete structure can be described by a finite number of

quantities called the member deformation (this, however, must be furnished by certain assumptions

on kinematics of the member to ascertain that the deformation at every point within the member can

be determined in terms of the member deformation). The quantities selected to be the member

deformation depend primarily on the type and behavior of such member. For instance, the

elongation, e, or a measure of the change in length of a member is commonly chosen as the member

deformation of a truss member as shown in Figure 1.22(a); the relative end rotations {s, e} where

s and e denotes the rotations at both ends of the member measured relative to a chord connecting

both end points as shown in Figure 1.22(b) are commonly chosen as the member deformation of a

beam member; and the elongation and two relative end rotations {e, s, e} as shown in Figure

1.22(c) are commonly chosen as the member deformation of a frame member. It is remarked that

the deformation of the entire discrete structure can fully be described by a finite set containing all

member deformation.

Figure 1.22: Member deformation for different types of members: (a) truss member, (b) beam

member, and (c) frame member

A Rigid body motion is a particular type of displacement that produces no deformation at

any point within the structure. The rigid body motion can be decomposed into two parts: a rigid

translation and a rigid rotation. The rigid translation produces the same displacement at all points

while the rigid rotation produces the displacement that is a linear function of position. Figure 1.23

shows a plane structure undergoing a series of rigid body motions starting from a rigid translation in

the X-direction, then a rigid translation in the Y-direction, and finally a rigid rotation about a pointA´.

L

L´= L + e

y

xL

L´= Ly

x

e

L

L´= L+ e y

x

e

(a) (b)

(c)

s

s





17

Within the context of static structural analysis, the structure under consideration must

sufficiently be constrained to prevent both the rigid body motion of the entire structure and the rigid

body motion of any part of the structure. The former is prevented by providing a sufficient number

of supports and proper directions against movement and the latter is prevented by the proper

arrangement of members and their connections. A structure shown in Figure 1.24(a) is a structure in

Figure 1.23 after prevented all possible rigid body motions by introducing a pinned support at a point A and a roller support at a point B. A structure shown in Figure 1.24(b) indicates that

although many supports are provided but in improper manner, the structure can still experience the

rigid body motion; for this particular structure, the rigid translation can still occur in the X-

direction.

Figure 1.23: An unconstrained plane structure undergoing a series of rigid body motions

Figure 1.24: (a) A structure with sufficient constraints preventing all possible rigid body motions

and (b) a structure with improper constraints

1.5.2 Static quantities

Quantities such as external actions and reactions in terms of forces and moments exerted to the

structure by surrounding environments and the intensity of forces (e.g. stresses and pressure) and

theirs resultants (e.g. axial force, bending moment, shear force, and torque, etc.) induced internally

at any point within the structure are termed as static quantities. Applied load is one of static

quantities referring to the prescribed force or moment acting to the structure. Support reaction is a

term referring to an unknown force or moment exerted to the structure by idealized supports(representatives of surrounding environments) in order to prevent its movement or to maintain its

Y

X

(a) (b)

Y

X

Y

XA

A´ B





18

stability. Support reactions are generally unknown a priori. There are two types of applied loads;

one called a nodal load is an applied load acting to the node of the structure and the other called a

member loads is an applied load acting to the member. An example of applied loads (both nodal

loads and member loads) and support reactions of a plane frame is depicted in Figure 1.25.

Stress is a static quantity used to describe the intensity of force (force per unit area) at any

plane passing through a point. Internal force is a term used to represent the force or momentresultant of stress components on a particular surface such as a cross section of a member. Note

again that a major source that produces the stress and the internal force within the structure is the

deformation. The distribution of both stress and internal force within the member depends primarily

on characteristics or types of that member. For standard one-dimensional members in a plane

structure such as an axial member , a flexural member , and a frame member , the internal force is

typically defined in terms of the force and moment resultants of all stress components over the cross

section of the member – a plane normal to the axis of the member.

Figure 1.25: Schematic of a plane frame subjected to external applied loads

An axial member is a member in which only one component of the internal force, termed as

an axial force and denoted by f – a force resultant normal to the cross section, is present. The axial

force f is considered positive if it results from a tensile stress present at the cross section; otherwise,

it is considered negative. Figure 1.26 shows an axial member subjected to two forces { f x1, f x2} at its

ends where f x1 and f x2 are considered positive if their directions are along the positive local x-axis.

The axial force f at any cross section of the member can readily be related to the two end forces { f x1,

f x2} by enforcing static equilibrium of both parts of the member resulting from an imaginary cut;

this gives rise to f = – f x1 = f x2. Such obtained relation implies that { f , f x1, f x2} are not all independent

but only one of these three quantities can equivalently be chosen to fully represent the internal force

of the axial member.

Figure 1.26: An axial member subjected to two end forces

A flexural member is a member in which only two components of the internal force, termed

as a shear force denoted by V – a resultant force of the shear stress component and a bendingmoment denoted by M – a resultant moment of the normal stress component, are present. The shear

y

x f x1 f x2

y

x f x1 f x2 f f

Node 1

Node 2

Node 3

Node 4





19

force V and the bending moment M are considered positive if their directions are as shown in Figure

1.27; otherwise, they are considered negative. Figure 1.27 illustrates a flexural member subjected to

forces and moments { f y1, m1, f y2, m2} at its ends where f y1 and f y2 are considered positive if their

directions are along the positive local y-axis and m1 and m2 are considered positive if their

directions are along the positive local z-axis. The shear force V and the bending moment M at any

cross section of the member can readily be related to the end forces and moments { f y1, m1, f y2, m2} by enforcing static equilibrium of both parts of the member resulting from an imaginary cut. It can

be verified that only two quantities from a set { f y1, m1, f y2, m2} are independent and the rest can be

obtained from equilibrium of the entire member. This implies in addition that two independent

quantities from { f y1, m1, f y2, m2} can be chosen to fully represent the internal force of the flexural

member; for instance, {m1, m2} is a common choice for the internal force of the flexural member.

Figure 1.27: A flexural member subjected to end forces and end moments.

A frame member is a member in which three components of the internal force (i.e. an axial

force f , a shear force V , and a bending moment M ) are present. The axial force f , the shear force V

and the bending moment M are considered positive if their directions are as indicated in Figure

1.28; otherwise, they are considered negative. Figure 1.28 shows a frame member subjected to a set

of forces and moments { f x1, f y1, m1, f x2, f y2, m2} at its ends where f x1 and f x2 are considered positive if

their directions are along the positive local x-axis , f y1 and f y2 are considered positive if their

directions are along the positive local y-axis and m1 and m2 are considered positive if their

directions are along the positive local z-axis. The axial force can readily be related to the end forces

{ f x1, f x2} by a relation f = – f x1 = f x2 and the internal forces {V , M } at any cross section of the

member can be related to the end forces and end moments { f y1, m1, f y2, m2} by enforcing static

equilibrium to both parts of the member resulting from a cut. It can also be verified that only three

quantities from a set { f x1, f y1, m1, f x2, f y2, m2} are independent and the rest can be obtained from

equilibrium of the entire member. This implies that two independent quantities from { f y1, m1, f y2,

m2} along with one quantity from { f , f x1, f x2} can be chosen to fully represent the internal forces of

the frame member; for instance, { f , m1, m2} is a common choice for the internal force of the frame

member.

Figure 1.28: A frame member subjected to a set of end forces and end moments.

1.6 Basic Components for Structural Mechanics

There are four key quantities involved in the procedure of structural analysis: 1) displacements and

rotations, 2) deformation, 3) internal forces, and 4) applied loads and support reactions. The firsttwo quantities are kinematical quantities describing the change of position and change of shape or

x

y

x

f y1

y

m1

f y2

m2

f y1

m1

f y2

m2 M

V M

V f x1

f x2 f x1

f x2 f f

y

x

f y1

y

xm1

f y2

m2

f y1 f y2

M V M

V m1 m2





20

distortion of the structure under external actions while the last two quantities are static quantities

describing the external actions and the intensity of force introduced within the structure. It is

evident that the displacement and rotation at any constraint points (supports) and the applied loads

are known a priori while the rest are unknown a priori. As a means to solve such unknowns, three

fundamental laws are invoked to establish a set of sufficient governing equations.

1.6.1 Static equilibrium

Static equilibrium is a fundamental principle essential for linear structural analysis. The principle is

based upon a postulate: “the structure is in equilibrium if and only if both the linear momentum and

the angular momentum conserve”. This postulate is conveniently enforced in terms of mathematical

equations called equilibrium equations – equations that relate the static quantities such as applied

loads, support reactions, and the internal force. Note that equilibrium equations can be established

in several forms; for instance, equilibrium of the entire structure gives rise to a relation between

support reactions and applied loads; equilibrium of a part of the structure resulting from sectioning

leads to a relation between applied loads, support reactions appearing in that part, and the internal

force at locations arising from sectioning; and equilibrium of an infinitesimal element of the

structure resulting from the sectioning results in a differential relation between applied loads and the

internal force.

1.6.2 Kinematics

Kinematics is a basic ingredient essential for the analysis of deformable structures. The principle is

based primarily upon the geometric consideration of both the undeformed configuration and the

deformed configuration of the structure. The resulting equations obtained relate the kinematical

quantities such as the displacement and rotation and the deformation such as elongation, rate of

twist, curvature, and strain.

1.6.3 Constitutive law

A constitutive law is a mathematical expression used to characterize the behavior of a material. It

relates the deformation (a kinematical quantity that measures the change in shape or distortion of

the material) and the internal force (a static quantity that measures the intensity of forces and their

resultants). To be able to represent behavior of real materials, all parameters involved in the

constitutive modeling or in the material model must be carried out by conducting proper

experiments.

1.6.4 Relation between static and kinematical quantities

Figure 1.29 indicates relations between the four key quantities (i.e. displacement and rotation,

deformation, internal force, and applied loads and support reactions) by means of the three basic

ingredients (i.e. static equilibrium, kinematics, and constitutive law). This diagram offers an overall

picture of the ingredients necessitating the development of a complete set of governing equations

sufficient for determining all involved unknowns. It is worth noting that while there are only three

basic principles to be enforced, numerous analysis techniques arise in accordance with the fashion

they apply and with quantities chosen as primary unknowns. Methods of analysis can be

categorized, by the type of primary unknowns, into two central classes: the force method and the

displacement method . The former is a method that employs static quantities such as support

reactions and internal forces as primary unknowns while the latter is a method that employs the

displacement and rotation as primary unknowns.





21

Figure 1.29: Diagram indicating relations between static quantities and kinematical quantities

1.7 Static Equilibrium

Equilibrium equations are of fundamental importance and necessary as a basic tool for structural

analysis. Equilibrium equations relate three basic static quantities, i.e. applied loads, support

reactions, and the internal force, by means of the conservation of the linear momentum and the

angular momentum of the structure that is in equilibrium.

The necessary and sufficient condition for the structure to be in equilibrium is that theresultant of all forces and moments acting on the entire structure and any part of the structure

vanishes. For three-dimensional structures, this condition generates six independent equilibrium

equations for each part of the structure considered: three equations associated with the vanishing of

force resultants in each coordinate direction and the other three equations corresponding to the

vanishing of moment resultants in each coordinate direction. These six equilibrium equations can be

expressed in a mathematical form as

0ΣF ; 0ΣF ; 0ΣF ZYX (1.15)

0ΣM ; 0ΣM ; 0ΣM AZAYAX (1.16)

where {O; X, Y, Z} denotes the reference Cartesian coordinate system with origin at a point O and

A denotes a reference point used for computing the moment resultants.

Applied Loads & Support Reactions

(Known and unknown)

F O R C E M E T H O D

D I S P L A C

E M E N T M E T H O D

Internal Forces(Unknown)

Deformation

(Unknown)

Displacement & Rotation(Known and unknown)

Static Equilibrium

Constitutive Law

Kinematics





22

For two-dimensional or plane structures (which are the main focus of this text), there are

only three independent equilibrium equations: two equations associated with the vanishing of force

resultants in two directions defining the plane of the structure and one associated with the vanishing

of moment resultants in the direction normal to the plane of the structure. The other three

equilibrium equations are satisfied automatically. If the X-Y plane is the plane of the structure, such

three equilibrium equations can be expressed as

0ΣM ; 0ΣF ; 0ΣF AZYX (1.17)

It is important to emphasize that the reference point A can be chosen arbitrarily and it can be either

within or outside the structure. According to this aspect, it seems that moment equilibrium equations

can be generated as many as we need by changing only the reference point A. But the fact is these

generated equilibrium equations are not independent of (1.15) and (1.16) and they can in fact be

expressed in terms of a linear combination of (1.15) and (1.16). As a result, this set of additional

moment equilibrium equations cannot be considered as a new set of equations and the number of

independent equilibrium equations is still six and three for three-dimensional and two-dimensional

cases, respectively. It can be noted, however, that selection of a suitable reference point A can

significantly be useful in several situation; for instance, it can offer an alternative form of

equilibrium equations that is well-suited for mathematical operations or simplify the solution

procedures.

To clearly demonstrate the above argument, let consider a plane fame under external loads

as shown in Figure 1.30. For this particular structure, there are three unknown support reactions

{R A, R BX, R BY}, as indicated in the figure, and three independent equilibrium equations (1.17) that

provide a sufficient set of equations to solve for all unknown reactions. It is evident that if a point A

is used as the reference point, all three equations FX = 0, FY = 0 and MAZ = 0 must be solved

simultaneously in order to obtain {R A, R BX, R BY}. To avoid solving such a system of linear

equations, a better choice of the reference point may be used. For instance, by using point B as thereference point, the moment equilibrium equation MBZ = 0 contains only one unknown R A and it

can then be solved. Next, by taking moment about a point C, the reaction R BX can be obtained from

MCZ = 0. Finally the reaction R BY can be obtained from equilibrium of forces in Y-direction, i.e.

FY = 0. It can be noted, for this particular example, that the three equilibrium equations MBZ = 0,

MCZ = 0 and FY = 0 are all independent and are alternative equilibrium equations to be used

instead of (1.17). Note in addition that an alternative set of equilibrium equations is not unique and

such a choice is a matter of taste and preference; for instance, {MBZ = 0,FX = 0,FY = 0},

{MBZ = 0,FY = 0,MDZ = 0}, {MBZ = 0,MCZ = 0,MDZ = 0} are also valid sets.

Figure 1.30: Schematic of a plane frame indicating both applied loads and support reactions

A

B

R A

R BY

R BX

X

Y

CD





23

The number of independent equilibrium equations can further be reduced for certain types of

structures. This is due primarily to that some equilibrium equations are satisfied automatically as a

result of the nature of applied loads. Here, we summarize certain special systems of applied loads

that often encounter in the analysis of plane structures.

1.7.1 A system of forces with the same line of action

Consider a body subjected to a special set of forces that have the same line of action as shown

schematically in Figure 1.31. For this particular case, there is only one independent equilibrium

equation, i.e. equilibrium of forces in the direction parallel to the line of action. The other two

equilibrium equations are satisfied automatically since there is no component of forces normal to

the line of action and the moment about any point located on the line of action identically vanishes.

Truss members and axial members are examples of structures that are subjected to this type of

loadings.

Figure 1.31: Schematic of a body subjected to a system of forces with the same line of action

1.7.2 A system of concurrent forces

Consider the body subjected to a system of forces that pass through the same point as shown inFigure1.32. For this particular case, there are only two independent equilibrium equations

(equilibrium of forces in two directions defining the plane containing the body, i.e. FX = 0 and FY

= 0). The moment equilibrium equation is satisfied automatically when the two force equilibrium

equations are satisfied; this can readily be verified by simply taking the concurrent point as the

reference point for computing the moment resultant. An example of structures or theirs part that are

subjected to this type of loading is the joint of the truss when it is considered separately from the

structure.

Figure 1.32: Schematic of a body subjected to a system of concurrent forces

1.7.3 A system of transverse loads

Consider the body subjected to a system of transverse loads (loads consisting of forces where their lines of action are parallel and moments that direct perpendicular to the plane containing the body)

Line of actionF

1 F

2 F

3

F1

F2

F3

F4

X

Y





24

as shown schematically in Figure 1.33. For this particular case, there are only two independent

equilibrium equations (equilibrium of forces in the direction parallel to any line of actions and

equilibrium of moment in the direction normal to the plane containing the body, i.e. FY = 0 and

MAZ = 0). It is evident that equilibrium of forces in the direction perpendicular to the line of action

is satisfied automatically since there is no component of forces in that direction. Examples of

structures that are subjected to this type of loading are beams.

Figure 1.33: Schematic of a body subjected to a system of transverse loads

An initial step that is important and significantly useful for establishing the correct

equilibrium equations for the entire structure or any part of the structure (resulting from the

sectioning) is to sketch the free body diagram (FBD). The free body diagram simply means the

diagram showing the configuration of the structure or part of the structure under consideration and

all forces and moments acting on it. If the supports are involved, they must be removed and

replaced by corresponding support reactions, likewise, if the part of the structure resulting from the

sectioning is considered, all the internal forces appearing along the cut must be included in the

FBD. Figure 1.34(b) shows the FBD of the entire structure shown in Figure 1.34(a) and Figure

1.34(c) shows the FBD of two parts of the same structure resulting from the sectioning at a point B.

In particular, the fixed support at A and the roller support at C are removed and then replaced by thesupport reactions {R AX, R AY, R AM, R CY}. For the FBD shown in Figure 1.34(c), the internal forces

{FB, VB, MB} are included at the point B of both the FBDs.

1.8 Classification of Structures

Idealized structures can be categorized into various classes depending primarily on criteria used for

classification; for instance, they can be categorized based on their geometry into one-dimensional,

two-dimensional, and three-dimensional structures or they can be categorized based on the

dominant behavior of constituting members into truss, beam, arch, and frame structures, etc. In this

section, we present the classification of structures based upon the following three well-known

criteria: static stability, static indeterminacy, and kinematical indeterminacy. Knowledge of thestructural type is useful and helpful in the selection of appropriate structural analysis techniques.

1.8.1 Classification by static stability criteria

Static stability refers to the ability of the structure to maintain its function (no collapse occurs at the

entire structure and at any of its parts) while resisting external actions. Using this criteria, idealized

structures can be divided into several classes as follows.

1.8.1.1 Statically stable structures

A statically stable structure is a structure that can resist any actions (or applied loads) without lossof stability. Loss of stability means the mechanism or the rigid body displacement (rigid translation

F2

F1

F4

F3

M1

M2

X

Y





25

and rigid rotation) develops on the entire structure or any of its parts. To maintain static stability,

the structure must be properly constrained by a sufficient number of supports to prevent all possible

rigid body displacements. In addition, members constituting the structure must be arranged properly

to prevent the development of mechanics within any part of the structure or, in the other word, to

provide sufficient internal constraints. All “desirable” idealized structures considered in the static

structural analysis must fall into this category. Examples of statically stable structures are shown inFigures 1.3, 1.5 and 1.14-1.16.

Figure 1.34: (a) A plane frame subjected to external loads, (b) FBD of the entire structure, and (c)

FBD of two parts of the structure resulting from sectioning at B.

1.8.1.2 Statically unstable structures

A statically unstable structure is a structure that the mechanism or the rigid body displacement

develops on the entire structure or any of its parts when subjected to applied loads. Loss of stability

in this type of structures may be due to i) an insufficient number of supports as shown in Figure

1.35(a), ii) inappropriate directions of constraints as shown in Figure 1.35(b), iii) inappropriate

P

M

MB

FB

VB

MB

FB

VB

(c)

B

A

C

R AY

R AX R

AM

R CY

(a) (b)

X

Y

R AY

R AX

R

AM

R CY

P

M

P

M





26

arrangement of member as shown in Figure 1.35(c), and iv) too many internal releases such as

hinges as shown in Figure 1.35(d). This class of structures can be divided into three sub-classes

based on how the rigid body displacement develops.

1.8.1.2.1 Externally, statically unstable structures

An externally, statically unstable structure is a statically unstable structure that the mechanism or

the rigid body displacement develops only on the entire structure when subjected to applied loads.

Loss of stability of this type structure is due to an insufficient number of supports provided or an

insufficient number of constraint directions. Examples of externally, statically unstable structures

are shown in Figure 1.35(a) and 1.35(b).

1.8.1.2.2 Internally, statically unstable structures

An internally, statically unstable structure is a statically unstable structure that the mechanism or

the rigid body displacement develops only on a certain part of the structure when subjected to

applied loads. Loss of stability of this type of structure is due to inappropriate arrangement of

member and too many internal releases. Examples of internally, statically unstable structures are

shown in Figure 1.35(c) and 1.35(d).

Figure 1.35: Schematics of statically unstable structures

1.8.1.2.3 Mixed, statically unstable structures

A mixed, statically unstable structure is a statically unstable structure that the mechanism or the

rigid body displacement can develop on both the entire structure and any part of the structure when

subjected to applied loads. Examples of mixed, statically unstable structures are shown in Figure1.36.

(a)

(b)

(c)

(d)





27

Figure 1.36: Schematics of mixed, statically unstable structures

Figure 1.37 clearly demonstrates the classification of idealized structures based upon the static

stability criteria.

Figure 1.37: Diagram indicating classification of structures by static stability criteria

Development of rigid

body displacement?

Statically stable

structures

Statically unstable

structures

Rigid body displacementof entire structure?

Mixed statically

unstable structures

Yes

No

Yes

No

Yes No

Internally statically

unstable structures

Externally statically

unstable structures

Idealized structures

Rigid body displacement

of part of structure?





28

1.8.2 Classification by static indeterminacy criteria

Static indeterminacy refers to an ability or inability to determine static quantities (support reactions

and internal force) at any point within a structure by means of static equilibrium. Using this criteria,

statically stable idealized structures can be divided into several classes as follow.

1.8.2.1 Externally statically determinate structures

An externally, statically determinate structure is a structure that all support reactions can be

determined from static equilibrium. The internal force at any point within the structure can or

cannot be obtained from static equilibrium. Examples of externally, statically determinate structures

are shown in Figure 1.38.

Figure 1.38: Schematics of externally, statically determinate structures.

1.8.2.2 Externally statically indeterminate structures

An externally, statically indeterminate structure is a structure that there exists at least one

component of all support reactions that cannot be determined from static equilibrium. Note that

there is no externally, statically indeterminate structure that the internal force at all points can be

determined from static equilibrium. Examples of externally, statically indeterminate structures are

shown in Figure 1.39.

1.8.2.3 Statically determinate structures

A statically determinate structure is a structure that all support reactions and the internal force at all points within the structure can be determined from static equilibrium. It is evident that a statically

determinate structure must also be an externally, statically determinate structure. Examples of

statically determinate structures are shown in Figures 1.38(a) and 1.38(b).

1.8.2.4 Statically indeterminate structures

A statically indeterminate structure is a structure that there exists at least one component of support

reactions or the internal force at certain points within the structure that cannot be determined from

static equilibrium. This definition implies that all statically stable structures must be either statically

determinate or statically indeterminate. It can be noted that an externally, statically indeterminate

structure must be a statically indeterminate structure. Examples of statically indeterminatestructures are shown in Figures 1.38(c) and 1.39.

(a) (b) (c)





29

Figure 1.39: Schematics of statically indeterminate structures.

1.8.2.5 Internally statically indeterminate structures

An internally statically indeterminate structure is a structure that is externally, staticallydeterminate and, at the same time, statically indeterminate. This implies that all support reactions

of an internally, statically indeterminate can be determined from static equilibrium while there

exists the internal force at certain points within the structure that cannot be determined from static

equilibrium. Examples of internally, statically indeterminate structures are shown in Figures 1.38(c)

and 1.39(a). Figure 1.40 clearly demonstrates the classification of structures based on the static

indeterminacy criteria.

1.8.3 Classification by kinematical indeterminacy criteria

Kinematical indeterminacy referring to the ability or inability to determine kinematical quantities

associated with a structure by means of kinematics or geometric consideration is utilized as a

criterion for classification. A (discrete) structure can therefore be categorized as follows.

1.8.3.1 Kinematically determinate structures

A kinematically determinate structure is a structure in which all degrees of freedom are prescribed

degrees of freedom. With use of additional assumptions on kinematics of a member, the

displacement and deformation at any point within a kinematically determinate structure are known.

An example of this type of structures is given in Figure 1.41(a); all nine degrees of freedom are

prescribed degrees of freedom.

1.8.3.2 Kinematically indeterminate structures

A kinematically indeterminate structure is a structure in which there exists at least one free degree

of freedom. As a result, the displacement and deformation of a kinematically indeterminate

structure are not completely known. The example of this type of structures is given in Figure

1.41(b); for this particular discrete structure, there exist three free degrees of freedom, i.e. {u2, v2,

2}.

Next, we define a term called degree of kinematical indeterminacy of a structure as a total

number of free or unknown degrees of freedom present within that structure. Consistent with this

definition, the degree of kinematical indeterminacy of a kinematically determinate structure is equal

to zero while the degree of kinematical indeterminacy of a kinematically indeterminate structure isalways greater than zero.

(a) (b) (c)





30

Figure 1.40: Diagram indicating classification of structures by static indeterminacy criteria

Determination of reactions from static

equilibrium?

No No

Yes

Yes No

Statically determinate

structure

Externally, statically

determinate

structure

Determination of internal force from

static equilibrium?

Determination of internal force from

static equilibrium?

Externally, statically

indeterminate

structure

Internally statically

indeterminate structure

Statically


Statically


Statically stable structures





31

Figure 1.41: (a) Kinematically determinate structure and (b) kinematically indeterminate structure

1.9 Degree of Static Indeterminacy

The degree of static indeterminacy of a structure, denoted by DI, is defined as a number of

independent static quantities (i.e. support reactions and the internal force) that must be prescribed in

addition to available static equilibrium equations in order to completely describe a static state of the

entire structure (a state where all support reactions and internal forces at any locations within the

structure are known) or, equivalently, to render the structure statically determinate.

From this definition, the degree of static indeterminacy is equal to the number of

independent static unknowns subtracted by the number of independent static equilibrium equations.

Thus, the degree of static indeterminacy of a statically determinate structure is equal to zero whilethe degree of static indeterminacy of a statically indeterminate structure is always greater than zero.

The degree of static indeterminacy is also known as the degree of static redundancy and the

corresponding extra, static unknowns exceeding the number of static equilibrium equations are

termed as the redundants.

1.9.1 General formula for computing DI

The degree of static indeterminacy of a statically stable structure can be computed from the general

formula:

c jma nnnr DI (1.18)

where r a is the number of all components of the support reactions, nm is the number of components

of the internal member force, n j is the number of independent equilibrium equations at all nodes or

joints, and nc is the number of static conditions associated with all internal releases present within

the structure. It is evident that the term r a + nm represents the number of all static unknowns while

the term n j + nc represents the number of all available equilibrium equations (including static

conditions at the internal releases).

1.9.1.1 Number of support reactions

The number of all components of support reaction at a given structure can be obtained using thefollowing steps: 1) identify all supports within the structures, 2) identify the type and a number of

(a) (b)

u1=0

1=0v1=0

u3=0

3=0v3=0

u2=0

2=0v2=0 Node 3

Node 2

Node 1

u2 2 v2

u1=0

1=0v1=0

u3=0

3=0v3=0

Node 3

Node 2

Node 1





32

components of the support reaction at each support (see section 1.1.5), and 3) sum the number of

components of support reactions over all supports.

It is emphasized here that for a beam structure, the component of the support reaction in the

direction of the beam axis must not be counted in the calculation of r a since the beam is subjected

only to transverse loads and there is no internal axial force at any cross section. For instance, the

number of support reactions of the structure shown in Figure 1.42(a), Figure 1.42(b) and Figure1.42(c) is 3, 4, and 8, respectively.

Figure 1.42: Schematics indicating all components of support reaction

1.9.1.2 Number of internal member forces

As clearly demonstrated in subsection 1.5.2, the number of independent components of the internal

force for an axial member , a flexural member , and a two-dimensional frame member are equal to 1,

2 and 3, respectively. Thus, the number of components of the internal forces for the entire structure(nm) can simply be obtained by summing the number of components of the internal forces for all

individual members. It is worth noting that nm depends primarily on both the number and the type

of constituting members of the structure. For instance, nm for the structure shown in Figure 1.42(a)

is equal to 14(1) = 14 since it consists of 14 axial members; n m for the structure shown in Figure

1.42(b) is equal to 2(2) = 4 since it consists of 2 flexural members (by considering all supports as

joints or nodes); and nm for the structure shown in Figure 1.42(c) is equal to 20(3) = 60 since it

consists of 20 frame members (by considering supports and connections between columns and

beams as joints or nodes).

1.9.1.3 Number of joint equilibrium equationsTo compute n j, it is required to know both the number and the type of joints present in the structure.

The number of independent equilibrium equations at each joint depends primarily on the type of the

joint. Here, we summarize standard joints found in the idealized structures.

1.9.1.3.1 Truss joints

A truss joint is an idealized joint used for modeling connections of a truss structure. The truss joint

behaves as a hinge joint so it cannot resist any moment and allows all members joining the joint to

rotate freely relative to each other. Since the truss member possesses only the internal axial force,

when the truss joint is separated from the structure to sketch the FBD, all forces acting to the joint

are concurrent forces as shown in Figure 1.43; in particular, P1 and P2 are external loads and P1, P2 and P3 are internal axial forces from the truss members. As a consequence, the number of

(a) (c)

(b)





33

independent equilibrium equations per one truss joint is equal to 2 (i.e. FX = 0 and FY = 0; see

also subsection 1.7.2).

Figure 1.43: FBD of the truss joint

1.9.1.3.2 Beam joints

A beam joint is an idealized joint used for modeling connections of a flexural or beam structure.

The beam joint behaves as a rigid joint so it can resist the external applied moment and can also

transfer the moment among ends of the members joining the joint. Since the flexural or beam

member possesses only two components of the internal force, i.e. the shear force and the bending

moment, when the beam joint is separated from the structure to sketch the FBD, all forces and

moments acting to the joint form a set of transverse loads as shown in Figure 1.44; in particular, P

and Mo are external loads and V1, M1, V1 and M2 are internal forces from the beam members. As a

consequence, the number of independent equilibrium equations per one beam joint is equal to 2 (i.e.

FY = 0 and MZ = 0; see also subsection 1.7.3).

Figure 1.44: FBD of the beam joint

1.9.1.3.3 Frame joints

A frame joint is an idealized joint used for modeling connections of a frame structure. The frame

joint behaves as a rigid joint so it can resist the external applied moment and can also transfer the

moment among ends of the members joining the joint. Since the frame member possesses three

components of the internal force, i.e. the axial force, the shear force and the bending moment, when

the frame joint is separated from the structure to sketch the FBD, all forces and moments acting to

the joint form a set of general 2D loads as shown in Figure 1.45; in particular, P1, P2 and Mo are

external loads and F1, V1, M1, F2, V2, M2, F3, V3 and M3 are internal forces from the frame

members. As a result, the number of independent equilibrium equations per one frame joint is equal

to 3 (i.e. FX = 0, FY = 0 and MZ = 0).

1.9.1.3.4 Compound joints

A compound joint is an idealized joint used for modeling connections where more than one types of members are connected. When the compound joint is separated from the structure to sketch the

F3

F2

F1 P

1

P2

X

Y

V2

M1

P

M2

V1

Mo

X

Y





34

FBD, all forces and moments acting to the joint can form a set of general 2D loads as shown in

Figure 1.46. As a result, the number of independent equilibrium equations per one compound joint

is generally equal to 3 (i.e. FX = 0, FY = 0 and MZ = 0).

Figure 1.45: FBD of the frame joint

Figure 1.46: FBD of the compound joint

The number of independent joint equilibrium equations of the structure (n j) can simply be

obtained by summing the number of independent equilibrium equations available at each joint.

1.9.1.4 Internal releases

An internal release is a point within the structure where certain components of the internal force

such as axial force, shear force and bending moment are prescribed. Presence of the internal

releases within the structure provides extra equations in addition to those obtained from static

equilibrium. Here, we summarize various types of internal releases that can be found in the

idealized structure.

1.9.1.4.1 Moment release or hinge

A moment release or hinge is an internal release where the bending moment is prescribed equal to

zero or, in the other word, the bending moment cannot be transferred across this point (see Figure

1.47). At the moment release, the displacement is continuous while the rotation or slope is not. For

this particular type of internal releases, it provides 1 additional equation per one hinge, i.e. M = 0 at

the hinge point.

V2

M1 M

2

V1

F2 F

1

V3

M3

F3

P2 M

o

P1

X

Y

X

Y

Truss member Frame member





35

Figure 1.47: Schematics of moment releases or hinges

1.9.1.4.2 Axial release

An axial release is an internal release where the axial force is prescribed equal to zero or, in the

other word, the axial force cannot be transferred across this point (see Figure 1.48). At the axial

release, the longitudinal component of the displacement is discontinuous while the transverse

component and the rotation are still continuous. For this particular type of internal releases, it

provides 1 additional equation per one release, i.e. F = 0 at the axial release.

Figure 1.48: Schematic of axial release

1.9.1.4.3 Shear release

A shear release is an internal release where the shear force is prescribed equal to zero or, in the

other word, the shear force cannot be transferred across this point (see Figure 1.49). At the shear

release, the transverse component of the displacement is discontinuous while the longitudinal

component of the displacement and the rotation are still continuous. For this particular type of internal releases, it provides 1 additional equation per one release, i.e. V = 0 at the shear release.

Figure 1.49: Schematic of shear release

1.9.1.4.4 Combined release

A combined release is an internal release where two or more components of the internal force are prescribed equal to zero (see Figure 1.50). Behavior of the combined release is the combination of

behavior of the moment release, axial release, or shear release. For this particular type of internal

releases, it provides two or more additional equations per one release depending on the number of

prescribed components of the internal force.

Figure 1.50: Schematics of combined release





36

1.9.1.4.5 Full moment release joint

A joint or node where the bending moment at the end of all members jointing that joint is prescribed

equal to zero is termed as a full moment release joint (see Figure 1.51). This joint has the same

behavior and characteristic as the hinge joint. For truss structures, while all joints are full moment

release joints, they provide no additional equation since presence of such joints has been consideredin the reduction of the number of internal forces per member from three to one (i.e. only axial force

is present). For beam or frame structures, presence of a full moment release joint provides n – 1

additional equations where n is the number of member joining the joint; for instance, a full moment

release joint shown in Figure 1.51 provides 4 – 1 = 3 additional equations.

Figure 1.51: Schematic of full moment release joint

1.9.1.4.6 Partial moment release joint

A joint or node where the bending moment at the end of certain but not all members jointing that

joint is prescribed equal to zero is termed as a partial moment release joint (see Figure 1.52). This

type of releases can be found in beam and frame structures. A partial moment release joint provides

n additional equations if the bending moment at the end of n members are prescribed equal to zero;

for instance, a partial moment release joint shown in Figure 1.52 provides 2 additional equations.

Figure 1.52: Schematic of partial moment release joint

The number of static conditions associated with all internal releases present in the structure (n c) can

simply be obtained by summing the number of additional equations provided by each internal

release.

Example 1.1 Determine the degree of static indeterminacy (DI) of the following structures

r a = 2 + 1 = 3

25 truss members nm = 25(1) = 25

14 truss joints n j = 14(2) = 28

No internal release nc = 0

DI = 3 + 25 – 28 – 0 = 0

Statically determinate structure





37

r a = 2 + 1 + 1 + 1 = 5

3 beam members nm = 3(2) = 6

4 beam joints n j = 4(2) = 8

1 moment release nc = 1

DI = 5 + 6 – 8 – 1 = 2 Statically indeterminate structure

r a = 3(3) + 2 + 1 = 12

28 frame members nm = 28(3) = 84

21 frame joints n j = 21(3) = 63


DI = 12 + 84 – 63 – 0 = 33

Statically indeterminate structure

r a = 4(3) = 12

28 frame members and 12 truss members

nm = 28(3) + 12(1) = 96

6 frame joints and 14 compound joints n j = 6(3) + 14(3) = 60


DI = 12 + 96 – 60 – 0 = 48

Statically indeterminate structure

1.9.2 Check of external static indeterminacy

For a given statically stable structure, let r a be the number of all components of the supportreactions, net be the number of independent equilibrium equations available for the entire structure

and ncr be the number of additional static conditions that can be set up without introducing new

static unknowns. The structure is externally, statically determinate if and only if

a et cr r n n (1.19)

and the structure is externally, statically indeterminate if and only if

a et cr r n n (1.20)

This check of external static indeterminacy is essential when the support reactions of the structureare to be determined.





38

In general, for plane structures, the number of independent equilibrium equations that can be

set up for the entire structure (net) is equal to 3, except for beam structures where the number of

independent equilibrium equations reduces to 2 (the equilibrium of forces in the direction along the

beam axis is automatically satisfied). Additional static conditions are typically the conditions

associated with internal releases present within the structure; for instance, points where components

of internal forces are prescribed such as “moment release or hinge”, “shear release”, and “axialrelease”. It is important to note that not all the static conditions can be incorporated in the counting

of ncr but ones that introduce no additional unknowns other than the support reactions can be

counted. These additional equations can be set up in terms of equilibrium equations of certain parts

of the structure resulting from proper sectioning the structure at the internal releases.

To clearly demonstrate the check of external static indeterminacy, let consider a frame

structure as shown in Figure 1.53. For this structure, we obtain r a = 2(2) = 4, nm = 6(3) = 18, n j =

6(3) = 18, nc = 2(1) = 2 DI = 4 +18 – 18 – 2 = 2; thus the structure is statically determinate. In

addition, net = 3 for frame structure and ncr = 1 since one additional equation (without introducing

additional unknowns other than support reactions) can be set up by sectioning at the hinge A and

then enforcing moment equilibrium about the point A of one part of the structure. The static

condition associated with the hinge B cannot be included in ncr since no new equation can be set upwithout introducing additional unknown internal forces along the cut. It is evident that r a = 4 = net +

ncr the structure is externally, statically determinate and, therefore, all support reactions can be

determined from static equilibrium. Since the structure is also statically indeterminate, from the

definition provided above, this implies that the structure is internally, statically indeterminate.

Figure 1.53: Schematic of externally statically determinate structure

1.9.3 DI of truss structures

Consider a statically stable truss structure that consists of m members and n joint. For this particular

structure, we obtain nm = m(1) = m, n j = n(2) = 2n and nc = 0. Upon using the general formula

(1.18), the degree of static indeterminacy of a truss is given by

2nmr DI a (1.21)

It is important to emphasize that there cannot be an internal release at interior points of all truss

members since each member possesses only one component of internal forces; presence of the

(axial) internal release will render the structure statically unstable.

B

A





39

Example 1.2 Determine the degree of static indeterminacy (DI) of the following statically stable

truss structures

r a = 2 + 2 = 4

m = 35 n = 18

DI = 4 + 35 – 2(18) = 3

Statically indeterminate

net = 3

ncr = 0

r a = 4 > net + ncr = 3

Externally statically indeterminate

r a = 2 + 1 = 3 m = 14

n = 8

DI = 3 + 14 – 2(8) = 1


net = 3

ncr = 0

r a = 3 = net + ncr

Externally statically determinate

r a = 2 + 2 = 4

m = 10

n = 7

DI = 4 + 10 – 2(7) = 0


net = 3

ncr = 1

r a = 4 = net + ncr

Externally statically determinate (or

implies from DI as well)

1.9.4 DI of beam structures

Consider a statically stable beam structure that consists of m members and n joint. For this

particular structure, we obtain nm = m(2) = 2m and n j = n(2) = 2n. Upon using the general formula

(1.18), the degree of static indeterminacy of a beam is given by

ca nn)(m2r DI (1.22)

It is important to emphasize that in the determination of r a, components of the support reactions in

the direction parallel to the beam axis must be ignored since there is no internal axial force in any beam members. In addition, the number of members of a given beam is not unique but it depends





40

primarily on the choice of joints or nodes considered; in general, joints are located at the supports

and free ends. However, the choice of joints and members does not affect the final value of DI.


beam structures

r a = 2(2) + 3(1) = 7

m = 4

n = 5

nc = 2

DI = 7 + 2(4 – 5) – 2 = 3


net = 2

ncr = 2

r a = 7 > net + ncr = 4


r a = 2 + 2(1) = 4

m = 3

n = 4

nc = 2

DI = 4 + 2(3 – 4) – 2 = 0


net = 2

ncr = 2

r a = 4 = net + ncr


r a = 2 + 3(1) = 5

m = 4

n = 5

nc = 0

DI = 5 + 2(4 – 5) – 0 = 3


net = 2

ncr = 0

r a = 5 > net + ncr = 2 Externally statically indeterminate

1.9.5 DI of frame structures

Consider a statically stable frame structure that consists of m members and n joint. For this

particular structure, we obtain nm = m(3) = 3m and n j = n(3) = 3n. Upon using the general formula

(1.18), the degree of static indeterminacy of a frame is given by

ca nn)(m3r DI (1.23)

Note that the free end must be treated as a joint or node.





41


frame structures

r a = 2(3) + 2 = 8

m = 7

n = 8 nc = 3 – 1 = 2

DI = 8 + 3(7 – 8) – 2 = 3


net = 3

ncr = 2

r a = 8 > net + ncr = 5


r a = 2 + 1 = 3 m = 7

n = 6

nc = 0

DI = 3 + 3(7 – 6) – 0 = 6


net = 3

ncr = 0

r a = 3 = net + ncr


r a = 3 + 1 = 4

m = 3

n = 4

nc = 1

DI = 4 + 3(3 – 4) – 1 = 0


net = 3

ncr = 1

r a = 4 = net + ncr = 4


1.10 Investigation of Static Stability of Structures

Static stability of the real structure is essential and must extensively be investigated to ascertain that

the structure can maintain its functions and purposes under external actions and excitations without

excessive movement or collapse of the entire structure and its parts. In the structural modeling or

structural idealization, the stability assurance can be achieved by requiring that all suitable idealized

structures must be statically stable. This requirement is also essential in the sense that the

subsequent process of static structural analysis can be performed.

As previously mentioned, loss of stability of the structure can occur either on the entirestructure or on the certain parts. The primary sources of instability are due to an insufficient number





42

of supports provided, inappropriate directions of constraints provided, inappropriate arrangement of

constituting members that forms insufficient internal constraints, or presence of too many of

internal releases. Here, we summarize three basic lemmas that can be used to investigate the static

stability of a given idealized structure.

1.10.1 Lemma 1

From section 1.9, it can be deduced that “if the structure is statically stable, it must be either

statically determinate (DI = 0) or statically indeterminate (DI > 0)”; thus DI of the structure is

nonnegative if the structure is statically stable. This statement is mathematically equivalent to “if DI

< 0, then the structure is statically unstable”. This lemma is simple and can be used to deduce the

instability of the structure by the knowledge of negative DI. It is important to emphasize that, for

the structure possessing DI ≥ 0, the lemma fails to provide information on stability of the structure.

Example 1.5 Use lemma 1 to check instability of the following structures.

Truss structure

r a = 1 + 1 = 2

m = 9

n = 6

DI = 2 + 9 – 2(6) = –1 < 0

Lemma 1 Statically unstable

Frame structure

r a = 1 + 1 +1 = 3

m = 3

n = 4

nc = 0

DI = 3 + 3(3 – 4) – 0 = 0

Lemma 1 No conclusion on stability

Beam structure

r a = 2 + 1 +1 = 4

m = 2

n = 3

nc = 2

DI = 4 + 2(2 – 3) – 2 = 0






43

Frame structure

r a = 3 + 3 +1 = 7

m = 5

n = 6 nc = 1

DI = 7 + 3(5 – 6) – 1 = 3


1.6.2 Lemma 2

The second lemma derives from the definition of static instability: “a structure is statically unstable

if and only if there exists at least one pattern of a rigid body displacement developed within the

structure under a particular action”. This lemma can be used to conclude the instability of thestructure by identifying one mechanics or rigid body displacement.

1.6.3 Lemma 3

The third lemma derives from the definition of static stability of the structure: “a structure is

statically stable if and only if there is no development of a rigid body displacement in any parts of

the structure under any actions”. This lemma can be used to conclude the stability of the structure

by investigating all possible mechanics or rigid body displacement.

Example 1.6 Investigate the static stability of truss structures shown in the figures below. The

structure I is obtained by adding the truss members a and b to the structure I and the structure III isobtained by adding the truss member c to the structure I.

Solution

Structure I: the degree of static indeterminacy (DI) is computed as follow: r a = 2 + 1 +1 = 4, nm =11(1) = 11, n j = 8(2) = 16, nc = 0 DI = 4 + 11 – 16 – 0 = –1 < 0. Thus, from lemma 1, it can be

Structure I Structure II

Structure III

a b

c





44

concluded that the structure I is statically unstable. Note that one can also use lemma 2 to conclude

this instability by sketching the mechanism as shown in the figure below.

Structure II: the degree of static indeterminacy (DI) is computed as follow: r a = 2 + 1 +1 = 4, nm =

13(1) = 13, n j = 8(2) = 16, nc = 0 DI = 4 + 13 – 16 – 0 = 1 > 0. Thus, static stability of the

structure cannot be concluded from the lemma 1. However, by investigating all parts of this

structure, there exists a pattern of rigid body displacement as shown in the figure below. Therefore,

the lemma 2 deduces that the structure is statically unstable.

Structure III: the degree of static indeterminacy (DI) is computed as follow: r a = 2 + 1 +1 = 4, nm =

12(1) = 12, n j = 8(2) = 16, nc = 0 DI = 4 + 12 – 16 – 0 = 0. Thus, static stability of the structure

cannot be concluded from lemma 1. However, by investigating all parts of this structure, there is no

development of rigid body displacement within any parts of the structure. Therefore, lemma 3

deduces that the structure is statically stable and, in addition, the structure is statically determinate

since DI = 0.

Example 1.7 Investigate the static stability of frame structures shown in the figure below. The

structure II and structure III are obtained by adding the horizontal member to the structure I with the

location of the hinge a being above or below the added member.

Structure I

a

Structure II Structure III

a a





45

Solution

Structure I: the degree of static indeterminacy (DI) is computed as follow: r a = 3 + 3 = 6, nm = 4(3)

= 12, n j = 5(3) = 15, nc = 4 DI = 6 + 12 – 15 – 4 = –1 < 0. Thus, from lemma 1, it can be

concluded that the structure I is statically unstable. Note that one can also use lemma 2 to conclude

this instability by sketching the mechanism as shown in the figure below.

Structure II: the degree of static indeterminacy (DI) is computed as follow: r a = 3 + 3 = 6, nm = 7(3)

= 21, n j = 7(3) = 21, nc = 4 DI = 6 + 21 – 21 – 4 = 2 > 0. Thus, static stability of the structure

cannot be concluded from the lemma 1. However, by investigating all parts of this structure, there

exists a pattern of rigid body displacement as shown in the figure below. Therefore, the lemma 2

deduces that the structure is statically unstable.

Structure III: the degree of static indeterminacy (DI) is computed as follow: r a = 3 + 3 = 6, nm = 7(3)

= 21, n j = 7(3) = 21, nc = 4 DI = 6 + 21 – 21 – 4 = 2 > 0. Thus, static stability of the structure

cannot be concluded from lemma 1. However, by investigating all parts of this structure, there is no

development of rigid body displacement within any parts of the structure. Therefore, lemma 3

deduces that the structure is statically stable and, in addition, the structure is statically indeterminate

since DI > 0.

1. For each statically stable truss shown below, determine its degree of static indeterminacy and

also use static indeterminacy criteria to identify whether it belongs to following classes (class 1:externally, statically determinate structures; class 2: externally, statically indeterminate

structures; class 3: internally, statically indeterminate structures; class 4: statically determinate

structures; class 5: statically indeterminate structures). Note that each structure may belong to

several classes.

Exercises

Structure I Structure II

a a





46

2. For each statically stable beam shown below, determine its degree of static indeterminacy and

also use static indeterminacy criteria to identify whether it belongs to following classes (class 1:

externally, statically determinate structures; class 2: externally, statically indeterminate


structures; class 5: statically indeterminate structures). Note that each structure may belong to

several classes.

3. For each statically stable frame shown below, determine its degree of static indeterminacy and

also use static indeterminacy criteria to identify whether it belongs to following classes (class 1:

externally, statically determinate structures; class 2: externally, statically indeterminate


structures; class 5: statically indeterminate structures). Note that each structure may belong toseveral classes.





47




48

4. Investigate the static stability of the structures shown below using Lemma 1, Lemma 2 and

Lemma 3. If the structure is statically unstable, show the sketch of possible rigid body

displacements.

chapter 1 idealizacion

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