Theory of Structures-1 B. E. 3233, 1st Term 3rd Class ... · Text Book Elementary theory of...
Transcript of Theory of Structures-1 B. E. 3233, 1st Term 3rd Class ... · Text Book Elementary theory of...
Theory of Structures-1
B. E. 3233, 1st Term
3rd Class (2017-2018)
Theory of Structures-1 CH-1: Introduction
1Q-L ( 2017-2018) Ch.1-
Theory of Structures
Lecturers: Theory of Structures Committee.
Supervised by: Dr. Qais Abdul Majeed Hassan
Text Book Elementary theory of structures, by Yuan-Yu Hsieh
Chapter One – Introduction
1. Introduction:
1.1. Engineering Structures:
Engineering structures include a wide variety of systems that can support loads
such as buildings, bridges, dams, aircrafts, etc., which are built to perform their
primary functions (for example, habitation, transportation, storage, etc.).
Designing a structure involves many considerations, the major two objectives are:
1) The structure must meet the performance requirements.
2) The structure must carry loads safely.
1.2. Definition of the Theory of Structures:
The Theory of Structures includes the design and analysis of engineering
structures, where the concentration will be focused on the fundamentals rather than
the details of design.
The complete design of a structure follows the following stages:
1) Developing a general layout.
2) Investigating the loads.
3) Stress analysis.
4) Selection of elements.
5) Drawing and detailing.
These five stages are interrelated and may be subdivided and modified. In most
cases they must be carried out simultaneously.
The subject that matters in our study is “Stress Analysis” and its relation with
loadings.
1.3. Classification of the Theory of Structures:
Structural theories may be classified from various points of views such as:
1) Static versus Dynamics:
Ordinary structures are designed under static loads. Dead load and
snow load are static loads that cause no dynamic effects on structures.
Some live loads, such as vehicles moving on bridges are assumed as
concentrated static load systems. They do cause impact on structures
but the dynamic effects are treated as a function of the moving loads to
simplify the design.
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The specialized branch that deals with the dynamic effects on structures
is structural dynamics.
2) Plane versus Space:
No structure is really planer, but some structural elements such as
beams, trussed bridges and rigid frame buildings are usually analyzed
as plane problems.
On the other hand, some structures, such as towers and domes, the
stresses are interrelated between members that lie in different planes;
these structures are considered as space frameworks under non-
coplanar force system.
3) Linear versus Nonlinear:
In linear structures the relationship between the applied loads and the
resulting deformations are assumed to be linear, this assumption is
based on the following:
a) The material of the structure is elastic and obeys Hook’s law.
b) The geometry changes are small and can be neglected when
calculating stresses.
In nonlinear structures the relationship between the applied loads and
the resulting deformations are nonlinear, this relationship exists under
one of the following conditions:
a) The material of the structure is inelastic.
b) The material is within the elastic range, but geometry changes are
significantly large during the application of loads.
Nonlinear behavior of structures is studied within the plastic analysis
and buckling of structures.
4) Statically Determinate versus Statically Indeterminate:
In statically determinate structures the structural analysis can be
performed by statics alone otherwise the structure is called
indeterminate. The analysis of the latter is performed using static
equations together with the equations furnished by the geometry of the
elastic curve of the structure in linear analysis.
5) Force versus Displacement:
Structural analysis can be divided into two methods: force method and
displacement method. In the former, the forces are treated as the basic
unknowns and the displacement are expressed in terms of forces;
whereas in the displacement method the displacement is the
fundamental unknowns and the forces are expressed in terms of
displacements.
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In matrix analysis of linear structures, the force method is referred to as
flexibility method and the displacement method is called the stiffness
method.
1.4. Scope of this Course:
The three major types of basic structures, as shown in Fig. (1.1), that will be
discussed throughout this course are as follows:
1) Beams:
Which are straight members subjected only to transverse loads. A beam
is completely analyzed when the values of bending moment and shear
are determined.
2) Trusses:
A truss is composed of members connected by frictionless hinges or
pins where the loads are concentrated at the joints. Each truss member
is considered as a two-force member subjected to axial forces only.
3) Rigid Frames:
Members in rigid frames are connected together by rigid joints capable
of resisting moment, shear and axial forces.
(a) Beam
(c) Frame Structure
(b) Trussed Bridge
Fig. (1.1) - Various structural forms
To cover all aspect of the theory of structures, for under graduate students, two or
more courses are needed, as it is carried out in most civil engineering departments
around the world, but since our study was set on one course only, we will try to
cover the most important subjects to cover the analysis of both statically
determinate and indeterminate structures.
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Chapter Two – Stability and Determinacy of Structures
2. Review:
2.1. Equations of Equilibrium for a Coplanar Force System:
A structure is said to be in equilibrium if under the action of external forces it
remains at rest relative to earth.
Since this course is confined to planar structures and all the forces systems are
coplanar, then the balanced coplanar force system must satisfy the following three
equations:
0xF , 0yF and 0aM …… (2-1)
Where:
xF is the summation of the x component of each force in the system.
yF is the summation of the y component of each force in the system.
The x and y subscripts indicate two perpendicular directions in the Cartesian
coordinate system.
aM is the summation of moment about any point a in the plane due to each
force in the system.
There are two simple special cases of equilibrium:
2.1.1. The Two-force member:
If a body is subjected to two external forces applied at points a and b and the body
is in equilibrium then the forces should be directed along the line ab and must be
equal in magnitude and opposite in direction, as shown in Fig. (2.1).
Fig. (2.1) Fig. (2.2)
2.1.2. The three-force member:
If a body is subjected to three external forces applied at points a, b and c and the
body is in equilibrium then the forces must be concurrent at a common point, O, as
shown in Fig. (2.2).
2.2. Support Reaction:
Structures are either partially or completely restrained so that they cannot move
freely in space. These restraints are provided by supports. The first step in
structural analysis is to take the structure without the supports and calculate the
forces, known as reactions, exerted on the structure by the supports. The reactions
Fc
Fa
Fb
Fb
a
b c O
o
Fa
Fb
a b
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are considered part of the external forces and are to balance the other external
loads in a state of equilibrium. There are mainly three types of supports, roller,
hinge and fixed supports.
In an idealized state, the resultant of all forces may be represented by a single force
specified by three elements:
1) The point of application.
2) The direction.
3) The magnitude.
In analysis, the direction means the slope of the action line, while the magnitude of
the force may be positive or negative; therefore, in mentioning the reaction force,
both the numerical magnitude and the sense of the action line must be
indicated.
2.2.1. Hinge or Pin Support:
A hinge support is represented by the symbols shown in Fig. (2.3-a), it can resist a
force in any direction but cannot resist the moment of the force about the
connecting point, as illustrated in Fig. (2.3-b). (Two unknowns and one degree of
freedom).
Fig. (2.3)
2.2.2. Roller Support:
A roller support is represented by the symbols shown in Fig. (2.4-a); the reaction
acts perpendicular to the surface through the center of the connecting pin; it cannot
resist moment and lateral force along the surface of the support as illustrated in
Fig. (2.4-b). (One unknown and two degrees of freedom).
Fig. (2.4)
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2.2.3. Fixed Support:
A fixed support is represented by the symbols shown in Fig. (2.4-a); it is capable of
resisting force in any direction and moment of force about the connecting end, thus
preventing the end of the member from both translation and rotation, as illustrated
in Fig. (2.5-b). (Three unknowns and zero degree of freedom).
Fig. (2.5)
2.3. Equations of Conditions or Construction:
Simple structures such as beams, trusses and rigid frames may be considered as
one rigid body sustained in space by a number of supports.
A compound form of a structure, mounted on a number of supports, may be built if
more than one simple structure is connected together by hinges, links or rollers.
For both the simple and compound structures, the external force system, external
loads plus support reactions, must satisfy the equations of equilibrium, if the
structure is at rest, 0xF , 0yF and 0aM .
In the compound type, the connecting devices enforce more restrictions on the
force system acting on the structure, thus providing additional equations of static
to supplement the equations of equilibrium; these equations are called equations
of conditions or construction, c, (c=1 for a hinge, c=2 for a roller and c=0 for a
beam without internal connection).
2.3.1. Internal Hinge:
0xFR
0 yFR
0FM
c=1
2.3.2. Internal Roller:
0 yFR
0xFR
0FM
c=2
F
(b)
Ry
Rx
Mo
o
(a)
F
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2.4. Stability and Determinacy of Structures with Respect to Supports:
Stability of structures is affected by the number and arrangement of the supports.
The structure is said to be stable or unstable if any of the following cases
occur:
1) Two elements of reaction supplied by supports are not sufficient to ensure the
stability of a rigid body, because these two reactions are either collinear, Fig.
(2.6-a), concurrent, Fig. (2.6-b) or parallel, Fig. (2.6-c), then the structure is
considered unstable, not because of the insufficient number of support
element, but because of statical instability.
Fig. (2.6)
2) At least three elements of reaction are necessary to restrain a body in stable
equilibrium. The cases shown in Fig. (2.7- a, b and c) illustrate rigid bodies
subjected to restraints by three elements of reaction, these restraints can be
solved by the three available equilibrium equations, 0xF , 0yF and
0aM , then the system is said to be statically stable and determinate.
Fig. (2.7)
(a) (b)
(c)
(f)
(e)
(d)
o
(b)
(c)
(a)
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3) If there are more than three elements of reaction, as shown in
Fig. (2.7- d, e and f), and the number of unknown elements is more than the
number of equations of static equilibrium, 0xF , 0yF and 0aM ,
then the system is said to be statically indeterminate, with regard to reaction
of support, if stable.
4) Indeterminate structures are introduced according to their degree of
indeterminacy, m, which can be calculated due to the excess number of
unknown elements, n, and the total number of reaction elements, r, as
described in equation (2-2). For the structure in Fig. (2.9- a), the total number
of reaction elements, r=4, the excess number of unknown elements, n=r-3=4-
3=1, therefore the structure is indeterminate to the 1st degree.
3rnm …… (2-2)
Summary of the main points of stability and determinacy are as follows:
1) If the number of unknown elements of reactions is less than three, the
equations of static equilibrium are not satisfied, then the structure is said to
be unstable.
2) If the number of unknown elements of reactions is equal to three, the
equations of static equilibrium are satisfied provided that there is no external
or internal geometric instability involved, then the structure is said to be
stable and determinate.
5) If the number of unknown elements of reactions is more than three, provided
that there is no external or internal geometric instability involved, then the
structure is said to be stable and indeterminate to the mth
degree. The degree
of indeterminacy is calculated following equation (2-2).
2.5. Cases of External Geometric Instability:
The three elements of reaction are necessary to restrain a body in equilibrium but
not sufficient for making a structure stable, such cases are referred to as external
geometric instability. These cases occur due to the following:
2.5.1. The lines of all reactions are all parallel, Fig. (2.8- a).
2.5.2. The lines of all reactions are concurrent at point o, Fig. (2.8- b).
Fig. (2.8)
(a)
(b)
o
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Sometimes the inadequacy of the arrangement of members causes instability, such
as the case shown in Fig. (2.9), such cases are referred to as internal geometric
instability. When this instability occurs the structure will collapse.
The structure in Fig. (2.9- a) is stable and indeterminate, but if an internal hinge is
replaced at the point of applied load, as in Fig. (2.9- b), the structure will be
unstable.
Fig. (2.9)
2.6. Cases of Internal Geometric Instability
2.6.1. Three reaction element hinges on the same line of action.
r=6, c=2
r , c+3
6>5
Indeterminate, but unstable (Internal
instability)
(Three hinges on the same line of action)
(b)
(a)
r=7, c=2
r , c+3
7>5
Stable and Indeterminate to the 2nd
degree
r=4, c=1
r = c+3
4 = 1+3
4=4
Determinate, but unstable
(Internal instability)
r=7, c=2
r > c+3
7 = 2+3
7>5
Indeterminate to the 2nd
degree,
but unstable (Internal instability)
(b)
(a)
Changing the arrangement of support will transform the structure
from instable to stable as shown in the following:
Stable and Indeterminate to the 2nd
degree
(a)
(b)
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2.6.2. Compatibility of movement in all or some parts of the structure.
2.6.3. Internal geometric instability due to lack of resistance in truss
panels w/o diagonal members.
b=13, r=3, j=8
b+r = 2j
16 = 16
Stable and Determinate
Rearrangement of bars causes internal
geometric instability as shown:
b=13, r=3, j=8
b+r = 2j
16 = 16
But geometrically unstable because
there is no bar to carry the vertical force
in panel (1) where the diagonal were
omitted.
The same can be viewed in panel (2)
b=6, r=4, j=5
b+r = 2j
10 = 10
But geometrically unstable due to lack
of lateral resistance.
1
2
(b)
r=5, c=2
r = c+3
5 = 2+3
5=5
Determinate, but unstable
(Internal instability)
After changing the arrangement of
support: r=5, c=2
r = c+3
5 = 2+3
5=5
Stable and Determinate
(a)
Horizontal Movement
r=7, c=2
r > c+3
7 > 2+3
7>5
Stable and Indeterminate to the 2nd
degree
Vertical Movement
Horizontal
Movement
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2.7. Stability and Determinacy of Beams:
Let (r) be the number of reaction elements and (c) is the number of equations of
condition, then the beam is said to be:
1) Unstable if (r < c+3).
2) Stable and statically determinate, provided that there is no external or
internal geometric instability involved, if (r = c+3).
3) Stable and statically indeterminate, provided that there is no external or
internal geometric instability involved, if (r > c+3), the degree of
indeterminacy would be [m = r – (c+3)].
Examples:
Structure, Beam r c c+3 m Classification
7 2 5 2
Stable and
Indeterminate to
the 2nd
degree
Internal Geometric
Instability Unstable
3 m
4 m
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2.8. Stability and Determinacy of Trusses:
Let (r) be the number of reaction elements, (b) is the number of bars, and (j) is the
number of joints, leading to the conclusion that (b+r) is the total number of
unknowns and (2j) is the total number of equilibrium equations, then the truss is
said to be:
1) Unstable if (b+r < 2j).
2) Stable and statically determinate, provided that there is no external or
internal geometric instability involved, if (b+r = 2j).
3) Stable and statically indeterminate, provided that there is no external or
internal geometric instability involved, if (b+r > 2j), the degree of
indeterminacy would be [m = (b+r) – 2j].
Note: for stability check the same rules used for beams can be applied.
Examples:
Structure, Truss b r b+r j 2j Classification
11 3 14 7 14 Stable and
Determinate
14 3 17 8 16
Stable and
Indeterminate
to the 1st
degree
Parallel Reaction Elements Unstable
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Structure, Truss b r b+r j 2j Classification
2.9. Stability and Determinacy of Frames:
Let (r) be the number of reaction elements, (b) is the number of members, (j) is the
number of joints and (c) is the total number of the equations of conditions, leading
to the conclusion that (3b+r) is the total number of unknowns and (3j+c) is the
total number of equilibrium equations, then the frame is said to be:
1) Unstable if (3b+r < 3j+c).
2) Stable and statically determinate, provided that there is no external or
internal geometric instability involved, if (3b+r = 3j+c).
3) Stable and statically indeterminate, provided that there is no external or
internal geometric instability involved, if (3b+r > 3j+c), the degree of
indeterminacy would be [m = (3b+r) – (3j+c)].
Note-1: for stability check the same rules used for beams can be applied.
Note-2: if there are more than two members connected by an internal hinge, then
the total number of the equations of conditions, c, will be as in equation (2-3):
c= No. of members at hinge - 1 …… (2-3)
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Examples:
Structure, Frame b r 3b+r j c 3j+c Classification
10 9 39 9 0 27
Stable and
Indeterminate
to the 12th
degree
10 9 39 9 6 33
Stable and
Indeterminate
to the 6th
degree
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Structure, Frame b r 3b+r j c 3j+c Classification
Note-3:
There is another easier approach to calculate the degree of indeterminacy of
frames; the frame members should be cut in a way to reduce the structure to
several simple statically determinate parts, the number of restrains removed to
accomplish this result gives the degree of determinacy of the frame, as shown in
equation (2-4).
cdam 3 …… (2-4)
Where:
m = is the degree of indeterminacy of the frame.
a = is the number of cut members.
3 rd
r = is the number of reaction elements within the section.
c = is the number of equations of condition, as in equation (2-3).
Structure, Frame a 3a c r-
part
d d m
Classification
4 12 0
3
3
3
0
0
0
0 12
Stable and
Indeterminate
to the 12th
degree
4 12 3
3
3
3
0
0
0
0 9
Stable and
Indeterminate
to the 9th
degree
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Chapter Three – Influence Lines for Statically Determinate Structures
3. The Concept of Influence Lines:
In designing a structure, loads acting on that structure must be established before
the stress analysis can be made. For a static structure two kinds of loads are
important, dead load and live load. The former is constant throughout the structure
life, while the latter may vary in position on the structure. In designing any specific
part of the structure attention should be paid to the placement of the live load that
will cause maximum live stresses for that part.
It is not necessary that a structure is subjected to a single set of loads all of the
time. For example, the single-lane bridge deck shown in Fig. (3.1) may be
subjected to one set of a loading at one time, Fig. (3.1-a), and the same structure
may be subjected to another set of loading at a different time, Fig. (3.1-b). It
depends on the number, position and weight of vehicles moving on the bridge.
Fig. (3.1) - Loading Condition on a Bridge Deck at Different Times
The variation of load on a structure results in variation in the response of the
structure. Thus, multiple sets of loading require multiple sets of analysis in order to
obtain the critical response parameters.
Influence lines offer a quick and easy way of performing multiple analyses for a
single structure. Response parameters such as shear forces or bending moment at a
point or reaction at a support for several load sets can be easily computed using
influence lines.
An influence line is a diagram which presents the variation of a certain response
parameter, such as a reaction of a support, shear force or bending moment at a
point, due to the variation of the position of a unit concentrated load along the
length of the structural member.
For the beam shown in Fig (3.2), consider a unit downward concentrated load is
moving from point A to point B. Assume that load to be a wheel of unit weight
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moving along the length of the beam. The magnitude of the vertical support
reaction at A, RA, will change depending on the location of this unit load. The
influence line for RA, (Fig. 3.2-b), show the value of RA for different locations of
the moving unit load. From the ordinate of the influence line at C, it is found that
RA = 0.5 when the unit load reaches point C.
(a)
(b)
(c)
Fig. (3.2) - Influence Line of RA for Beam AB
3.1. Construction of Influence Lines using Equilibrium Methods:
The most basic method of obtaining the influence line for a specific response
parameter is to solve the static equilibrium equations for various locations of the
unit load. The general procedure for constructing an influence line is as follows:
1) Define the positive direction of the response parameter under consideration
through a free body diagram of the whole system, sign convention.
For flexure members, beams and frames:
Axial Force, N
Tension, +ve
Shear Force, V
Clockwise, +ve
Bending
Moment, M
Compression on top,
+ve
For axial members, trusses:
Axial Force, N
Tension, +ve
2) Consider a generic location for the unit load, at distance x, then solve for the
equilibrium of the whole system to obtain the response parameter for that
location of the unit load with respect to x. by substituting the values of x along
the structure, the influence line for that parameter is obtained.
(+) M
(+) V
(+) N
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Example 3-1
Draw the influence line for the vertical reaction at A and B of beam AB shown in
Fig. (3.3).
I.L. for RA and RB:
Assume a unit load (UL=1) is moving
along beam AB and is located at distance x
from support A, 0<x<10.
0AM
10
0101
xR
Rx
B
B
0yF
101
110
1
xR
xRRR
A
ABA
By substituting the values of x from 0 to 10
the influence line for either RA or RB can
be obtained.
Fig. (3.3)
Example 3-2
Draw the influence line for the shear and moment at point C located 2 m from
support A of beam AB shown in Fig. (3.4).
1) I.L. for VC:
Assume a unit load (UL=1) is moving
along beam AB. Divide the beam into two
parts, from 0 to 2 and from 2 to 10.
Assume the unit load is located at distance
0<x<2, and solve for Vc. Then assume that
the unit load is located at distance 2<x<10
and solve for Vc.
Fig. (3.4)
I.L. VC
I.L. MC
10 m
10 m
2 m
C
0.8
0.2
1.6
Ex: 3-2
I.L. RA
I.L. RB
10 m
10 m
1
1
0
0
Ex: 3-1
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for 0<x<2:
0yF
01 CA VR
0110
1 CVx
10
xVC
2.010
22@
00@
C
C
Vx
Vx
for 2<x<10:
0yF
01 CB VR
101
xVC
010@
8.02@
C
C
Vx
Vx
2) I.L. for MC:
for 0<x<2:
0CM
0221 AC RxM
xM
xx
xRM
C
AC
8.0
210
2222
6.128.02@
00@
C
C
Mx
Mx
for 2<x<8:
0CM
0821 BC RxM
xM
xxRM
C
BC
2.02
2110
828
028.010@
6.12@
C
C
Mx
Mx
C B
8 m
M
UL=1
M
RB
Vc
Mc
x
M
x
M
A C
VC
Mc
2 m
M
UL=1
M
RA
x
M
A C
VC
Mc
2 m
M
UL=1
M
RA
x
M
C B
8 m
M
UL=1
M
RB
Vc
Mc
x
M
x
M
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3.2. Important Steps to be Followed to Construct Influence Lines :
1) The structure may be divided into pieces, the ends of each piece is either a
free end or an internal hinge, as shown in Fig. (3.5).
Fig. (3.5)
2) To draw the influence lines for the reactions at supports, the moment at the
fixed end or the shear force at an internal hinge the structure must be divided
into two pieces, as shown in Fig. (3.5-a).
3) To draw the influence lines for the moment and shear at any other point on
the structure which must be divided into three pieces according to the
position wanted, as shown in Fig. (3.5-c).
4) The supports are assumed to be fixed points where the IL for them should be
equal to zero.
5) If the IL for two points were known then the line for that parameter can be
drawn, a line can be drawn using two points or a point and a slope.
6) The structure part cannot be broken or bended but it can be tended.
7) The internal hinge is able to move if the adjoining part allowed that
movement.
3.3. Construction of Influence Lines for Beams using a Simple Fast
Procedure:
There is a fast procedure to draw the influence lines for beams, as shown below:
3.3.1. Influence Lines for Reactions at Supports:
Lift the support, upward, one unit and keep the other supports fixed with zero
values, as shown in Fig. (3.6).
Fig. (3.6)
I.L. RB
1
0
I.L. RA
1
A
10
m
B
10
m
0
1 2
(a)
1 4 2 3
(b)
C
1 2 3
(c)
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-6
3.3.2. Influence Lines for Shear at Supports:
The influence line for the shear at supports is drawn depending whether the shear
calculated is on the left or right side of the support, ss follows:
3.3.2.1. Influence Lines for Shear at left side of Support, , VL:
The support will be fixed at zero value and the left side will be pulled down for one
unit, keeping in mind that the portion before and after the support should be
parallel, having the same slope, as shown in Fig. (3.7-a).
Fig. (3.7)
3.3.2.2. Influence Lines for Shear at right side of Support, , VR:
The support will be fixed at zero value and the right side will be lifted upward for
one unit, keeping in mind that the portion before and after the support should be
parallel, having the same slope, as shown in Fig. (3.7-b).
3.3.3. Influence Lines for Shear at Any Point within the Beam:
The influence line for the shear at any point is drawn as follows: Lift the right part
of the cut member, upward, with an amount equal to (length of right cut/length of
beam); then pull down the left part of the cut member, downward, with an amount
equal to (length of left cut/length of beam); keeping the distance between the two
points of cut equal to one unit, the two points of shear at a cut are separated with
one unit, as shown in Fig. (3.8).
Fig. (3.8)
a. If the right side of the shear cut is fixed then the left side will be pulled down
for one unit.
b. If the left side of the shear cut is fixed then the right side will be lifted
upward for one unit.
1 I.L. VB
A B C D
a b L
a/L
b/L
(a) (b)
A B C D
A B C D
I.L. VC-L I.L. VC-R
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-7
3.3.4. Influence Lines for Bending Moment at Any Point within the Beam:
The influence line for the bending moment at any point is drawn as follows: Lift
the point upward, with an amount equal to (a×b/L), where a is the length of the left
portion, b is the length of the right portion and L is the total length of the beam, as
shown in Fig. (3.9).
Fig. (3.9)
a. If the right side of the moment cut is fixed then the left side will be pulled
down with 45o, as shown in Fig. (3.10-a).
b. If the left side of the moment cut is fixed then the right side will be pulled
down with 45o, as shown in Fig. (3.10-b).
c. At any unfixed point, the two points of moment stick together and are lifted
upwards with an amount equal to (a×b/L).
Fig. (3.10)
3.3.5. Influence Lines for the Bending Moment at Supports:
The influence line for the moment at supports is drawn as follows:
The support will be fixed at zero value and the unfixed side of the structure, either
the left or the right, will be pulled down with 45o, as shown in Fig. (3.11).
Fig. (3.11)
I.L. MC
A
B C D
45
o
I.L. MC 45
o
A
B C D
E D
(a)
I.L. MD 45
o
A
B C D
E D
(b)
I.L. MB
A B C D
a b L
ab/L
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-8
Ex: 3-3
For the beam shown, draw the I.L. for following:
1) The reactions at A and C.
2) The shear at C, F, G, I, J, K and L.
3) The bending moment at A, C, F, I, J, K and L
I.L. VC-L
-1 -1 +1
I.L. VC-R
+1
I.L. VF-L
-1 -1
+1
I.L. VF-R
+2/3
+2
I.L. RC
+1
+1
-1
I.L. RA
+1
-1
I.L. VG-L
+2/3
-2/3
+1 +1
I.L. VG-R
A B C D E F G H
3 m
1 m
1 m
1 m
1 m
1 m
1 m
1 m
2 m
2 m
4 m
I J K L Ex: 3-3
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-9
A B C D E F G H
3 m
1 m
1 m
1 m
1 m
1 m
1 m
1 m
2 m
2 m
4 m
I J K L
I.L. VJ
+1 +1
I.L. VK
-1 -1
I.L. VL
-2/3
+1/3 +2/3
+1/2
-1/2
I.L. VI
-1
45o
-3
I.L. MA
+3
45o
-2
I.L. MC
45o
-2
I.L. MF
Ex: 3-3
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-10
A B C D E F G H
3 m
1 m
1 m
1 m
1 m
1 m
1 m
1 m
2 m
2 m
4 m
I J K L
-1
I.L. MI
+1/2
45o
-1
I.L. MJ
45o
-1
I.L. MK
Ex: 3-3
45o
-2/3
I.L. ML
+2/3
-4/3
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-11
A B C D E F G H
3 m
1 m
1 m
1 m
1 m
1 m
1 m
1 m
2 m
2 m
4 m
I J K L Ex: 3-3
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-12
A B C D E F G H
3 m
1 m
1 m
1 m
1 m
1 m
1 m
1 m
2 m
2 m
4 m
I J K L Ex: 3-3
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-13
A B C D E F G H
3 m
1 m
1 m
1 m
1 m
1 m
1 m
1 m
2 m
2 m
4 m
I J K L Ex: 3-3
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-14
3.4. Calculating the Maximum Effect due to Concentrated or Uniformly
Distributed Loads:
To calculate the maximum effect of any parameter due to either a concentrated
load or a uniformly distributed load, follow the following procedures:
3.4.1. Due to a Concentrated Load:
After drawing the influence line for the parameter desired, the value of the
concentrated load must be multiplied by the maximum value of the influence line.
If the positive maximum effect is required then the load would be multiplied by the
maximum positive value, the same thing is for the maximum negative effect.
3.4.2. Due to a Uniformly Distributed Load:
The uniformly distributed loads are either dead or live loads.
3.4.2.1. Due to a Dead Uniformly Distributed Load, wDL:
Multiply the value of load, (wDL) by the algebraic sum for the values of area of
influence line drawing, under that load, for both positive and negative areas.
3.4.2.2. Due to a Live Uniformly Distributed Load, wLL, covering any Length:
Calculate the positive and negative areas of the influence line, under that load,
separately; if the maximum positive effect is desired then multiply the value of the
load by the positive area; the same is followed if the maximum negative effect is
desired.
3.4.2.3. Due to a Live Uniformly Distributed Load, wLL, with Specific length:
There are two situations for this kind of loading:
1) The shape of the influence line, including the maximum value, may be a
rectangle or a right-angle triangle; the uniformly distributed load must be
located on the area at the location of maximum value, as shown in Fig. (3.12).
Fig. (3.12)
2) The shape of the influence line, including the maximum value, may be a
scalene triangle; the uniformly distributed load must be located on the area so
that the length A will produce the same value on the influence line drawing, y,
as shown in Fig. (3.13).
The maximum value of the parameter due to the live load would be equal to
the dashed area multiplied by w.
w
A I.L.
L2
A
w
A2 y I.L.
L1
w
A
A1 y1 y2
A1 =0.5(y1+y2) A
A2 = Ay
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-15
Fig. (3.13)
Ex: 3-4, (1998-1999)
The beam shown in Fig. (3.14) is on three supports but has one internal hinge, the
beam has also an overhang. Draw the influence line (I.L.) for the bending moment
at section (x-x), when a unit load transverse (or moves on) this beam, the self-
weight of the beam is estimated to be (18 kN/m) of length, a uniform live load of
22 (kN/m) of length can occupy any length and region of the beam. Find the
maximum bending moment at section (x-x), (both max. (+ve) and max. (–ve)).
Solution:
31321 AAwAAAwM LLDL
ve
xx
4
33
2
1312
2
122
4
33
2
1212
2
1312
2
118
ve
xxM
mkNMve
xx .54975.42025.1288
91822
8
9121818
2321 AwAAAwM LLDL
ve
xx
Ex: 3-4
75.0y8
y
3
y2y
6
3
4
y3
12
66
l
bay 3
232
21
x
y1
x
75.0y8
y
3
y2y
6
3
4
y3
12
66
l
bay 3
232
21
6 m
10
m
A1
y
3
6 m
10
m
6 m
10
m
4 m
10
m
8 m
10
m
3 m
10
m
6*6/12=3
A3
y2
A
2
I.L. Mx-x
y3
A1
A2
A3
6 m
10
m
y1
6
*
6/
1
2
=
3 Fig. (3.14)
w
A
L
a b
I.L.
(x) (A-x)
y y c
x = A a / L
y = c (1 – A / L)
Dashed Area = Ac (1 - A / 2 L)
The maximum value of the parameter due to the live load would
be equal to the dashed area multiplied by w.
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-16
2122
122
4
33
2
1212
2
1312
2
118M
ve
xx
m.kN75.13726425.1281222125.7182122
122125.1121818M
ve
xx
3.5. Influence Lines for Girders with Stringer and Floor-Beam Systems:
In bridge construction, where a long-span girder is used, the live loads are not
applied directly to the main girder, but they are transmitted from the stringer (slab)
to the girder by a floor beam system as the one shown in Fig. (3.15, a and b).
(a)
(b)
Fig. (3.15)
3.5.1. Influence Lines for the Upper Segment of Girders with Stringer and
Floor-Beam Systems:
Consider the system shown in Fig (3.16) which consists of five panels each of (6 m)
long. To draw the influence lines for the upper segment, the same past rules for
beams are applied as shown in the following example:
Ex: 3-5:
Draw the IL for R2, VH, V3, M3, Vx, Mx on the upper beam.
Ex: 3-5
6 m
6 m
6 m
3 m
3 m
3 m
3 m
12 m
0
M 1
M
2
M
3
M
4
M 5
M A
M B
M C
M
D
M
E
M
Hinge
M H
M
Gap
M x
M x
M
3 m
M
Stringer (Slab)
M Floor Beam
M
Girder
M
Stringer (Slab)
Girder
Floor Beam
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-17
I.L. R2-Top
+1
I.L. VH
-1
I.L. M3-Top
-3
45o
+1
I.L. V3-R-Top
+1/2
I.L. Mx
-3/2 -3/2
+3/2
I.L. V3-L-Top
-1 -1
I.L. Vx
+1/2 +1/2
-1/2 -1/2
Ex: 3-5
6 m
6 m
6 m
3 m
3 m
3 m
3 m
12 m
0
M 1
M
2
M
3
M
4
M 5
M A
M B
M C
M
D
M
E
M
Hinge
M H
M
Gap
M x
M x
M
3 m
M
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-18
3.5.2. Influence Lines for the Lower Segment of Girders with Stringer and
Floor-Beam Systems:
Ex: 3-6-a:
For the same structure of Ex: 3-5, draw the IL for RB and V2-3-Bottom.
To construct the influence lines for those two parameters in the lower beam, the
following steps should be followed:
1) Draw the required influence line for the lower beam only, the girder,
following the previous rules, regardless of anything above it. (Drawn in
dotted lines), (the Premier Drawing, PD).
2) On that plot, transfer the effect of the connection point between the upper and
lower segments of the system (Floor Beam) and find their coordinates by
proportions.
3) Transfer the effect of the upper segments (Slab) based on the supporting points
from the previous step (Floor Beam), connecting the selected points with
regards of the shape of the upper segments (Slab), (the Final Drawing, FD).
Ex: 3-6-b: (Do it as homework-HW-1)
For the same structure of Ex:5, draw the IL for Rc-Bottom and My-y-Bottom.
I.L. RB-Bottom, PD
+1 +9/7
-1/7
I.L. RB-Bottom, FD
-5/28 -1/7
+9/7
I.L. V2-3-Bottom, PD
-1/7
+4/7
-3/7
+2/7
I.L. V2-3-Bottom, FD
-1/7
+4/7
-2/7
+2/7
-5/28
Ex: 3-6-a
6 m
6 m
6 m
3 m
3 m
3 m
3 m
12 m
0
M 1
M
2
M
3
M
4
M 5
M A
M B
M
C
M
D
M
E
M
Hinge
M H
M
Gap
M x
M x
M
3 m
M
3 m
M
y
M
y
M
Stringer (Slab)
M
Floor Beam
M
Girder
M
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-19
Ex: 3-7:
For the structure shown below, draw the IL for RB, VB-L, VB-R and MB, then find
the maximum bending moment at point B due to a concentrated load of 90 kN and
a uniformly distributed live load of 15 kN/m and a uniformly distributed dead load
of 10 kN/m.
H
i
n
g
e
M
1
M
2
M 3
M A
M
B
M C
M
Gap
M
Hinge
M
2 m
2 m
2 m
4 m
2 m
2 m
2 m
2 m
2 m
Hinge
M
I.L. RB
1
M y1= +1/3
m
y2= +5/3
m
I.L. RB
z1= +10/3
m
z1= -2/3
m
x
M
Ex: 3-7
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-20
3.6. Influence Lines for Trusses:
The general procedure to draw the influence lines for trusses is by using the
equilibrium method; but first, the idea to solve the problem must be found, the
axial forces of the truss member required, either by the section method or the joint
method, then the following procedure is followed:
1) If the solution is by the section method:
Select two points before the section and two after it then calculate the force in the
member for each point; based on these results the IL for that member is drawn.
Note:
In the section method if the solution includes summation of moments around a
point that lies over, under, left or right the section, then three point will be
satisfying to draw the IL, one point before the section, one after the section and the
point of section.
2) If the solution is by the joint method:
The joint must be on the line of loads. Assume that the joint is loaded with a UL
then find the force in the member required, and then the joint is assumed to be
unloaded, and the force in the member required is found, after that the IL can be
drawn.
Ex: (3-8), Final (1997-1998)
Draw the influence for members a and b in the truss shown in figure below.
To solve this problem think about an idea for solution.
For member a:
If the joint method is used, joint D is analyzed to find GD, then 0yF for joint
G, the force in member (a) can be determined.
1) UL @ D
10 GDFy
From joint G
kNa
aFy
46.2
094
410
22
a
M
b
M
A
3 m
M
3 m
M
3 m
M 3 m
M 4 @ 3 m = 12 m
M
4 m
M 1
B C D E
J I H G F 1
1
Ex: 3-8
D
GD
1.0
M
G
a
GD
9 4
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-21
2) UL not @ D
00 GDFy
From joint G
kNa
aFy
0
094
40
22
For member b:
To solve member b the section n method is used, section (1-1) then 0AM , the
force in member (b) can be determined.
Since point (A) does not lie above, under, left or right sec (1-1), therefore 4 point
are necessary for the solution, two before the section, A and B and two after the
section, C and E.
1) UL @ A
kNbMA 00
2) UL @ B
kNbMA 75.00
3) UL @ C
kNbMA 00
4) UL @ E
kNbMA 00
Sec (1-1)
M
a
M
b
M
A B
J I
AH
AF
BC
a
M
b
M
C D E
H G F
BC
AF
AH
D
GD G
a
GD
9 4
a
M
b
M
A
4 @ 3 m = 12 m
M
4 m
M
B C D E
J I H G F
I.L. Fb
-0.75
M
I.L. Fa
-2.46
M
Ex: 3-8
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-22
3.7. Influence Lines for Girders with Stringer and Floor-Beam Systems in
Trusses:
For girders in trusses the UL moves on the girder and the loads are transformed to
the truss through the connecting points between the girder and the truss.
Notes:
1) If the girder was regular then the truss will be drawn without the girder,
because the supports of the girder will be the same joints for the truss, but if the
girder was irregular, it should be drawn with the truss to show the points of
connection between them.
2) If the demand was to draw the IL of a parameter within the upper beam of the
girder, the simple fast procedure could be applied, but for the bottom part the
traditional method will be used.
Ex: (3-9)
Draw the influence for members DB of the structure shown in figure below, where
the UL moves across the floor-beam.
Solution:
First of all find an idea for the solution of the truss member, DB.
1) Analyze joint F, 0yF , find AF.
2) Sec (1-1), 0yF , find DB.
Then select the points to apply the UL, one before the section, D, and three after
the section, E, F and G, the load would be applied on the upper joints of the truss.
1) UL @ D:
From joint F:
00 AFFy
From sec (1-1)
00 DBFy
Ex: 3-9
1
A
3 m
M
3 m
M
4 m
M
B C
E D F G
2 m
M
2 m
M
2 m
M
1 m
M 1 m
M
1 m
M
1 m
M
H I
Ex: 3-9
A
3 m
M
3 m
M
4 m
M
B C
E D F G
1.0
RA
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-23
2) UL @ E:
From whole truss:
3
20 AC RM
From joint F:
00 AFFy
From sec (1-1)
6
50 DBFy
3) UL @ G:
From whole truss:
00 AC RM
From joint F:
00 AFFy
From sec (1-1)
00 DBFy
4) UL @ F:
From whole truss:
3
10 AC RM
From joint F:
20 AFFy
From sec (1-1)
6
50 DBFy
3
5
2
6
5
6
5
3
6
5
1
1
y
y
3
5
2
6
5
6
5
3
6
5
2
2
y
y
9
10
3
5,
3232
23 yyyy
Ex: 3-9
A
3 m
M
3 m
M
4 m
M
B C
E D F G
1.0
RA
Ex: 3-9
A
3 m
M
3 m
M
4 m
M
B C
E D F G
1.0
RA
Ex: 3-9
A
3 m
M
3 m
M
4 m
M
B C
E D F G
1.0
RA
1.0
F
AF
EF FG
A AB
DE D
AF
DB
RA Sec (1-1)
Ex: 3-9
A B
C
E D F G
H I
+5/6
M
-5/6
M +5/3
M
-5/3
M
y1
m y3
m y2
m
I.L. DB
-10/9
M
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-24
Ex: (3-10)
Draw the influence for the floor beam reaction at (u3) and the force in members (a)
for the structure shown in figure below, where the UL moves across the floor-beam
from (u0) to (u4).
Solution:
1) For the reaction at (u3)
To draw the IL for the reaction at (u3), the simple fast method may be used. Lift up
the support at (u3) one unit and connect the supporting points with regard to the
shape of the upper beam.
2) For member (a)
First of all find an idea for the solution of the truss member, (a):
From section (1-1), 0yF , find (a).
Then select the points to apply the UL, two before the section, (u0) and (u1), and
two after the section, (u2) and (u4), the load would be applied on the upper joints of
the truss.
Ex: 3-10
1 2 m
M
A
4 @ 4 m = 16 m
M
3 m
M
B
u0 1
1
a
M
u1 u2 u4 u3
Gap
Hinge 2 m
M
E
Ex: 3-10
1 2 m
M
A
4 @ 4 m = 16 m
M
3 m
M
B
u0 1
1
a
M
u1 u2 u4 u3
Gap
Hinge 2 m
M
E
-3/2
M
E
+3/2
M
E
+1
M
E I.L. (u3)
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-25
1) UL @ (u0):
From the whole truss:
0.10 yB AM
00 yy BF
From sec (1-1)
00 aFy
2) UL @ (u1):
From the whole truss:
4
30 yB AM
4
10 yy BF
From sec (1-1)
12
50
4
1
5
30
aaFy
3) UL @ (u2):
From the whole truss:
2
10 yB AM
2
10 yy BF
From sec (1-1)
6
50
2
1
5
30
aaFy
4) UL @ (u4):
From the whole truss:
0.10 yA BM
00 yy AF
From sec (1-1)
00 aFy
Then apply these values with dotted lines as shown in the premier drawing, PD,
below.
Transfer the effect of the upper segments (Slab) based on the supporting points
from (Floor Beam), connecting the selected points with regards of the shape of the
upper segments (Slab), as shown in the Final Drawing, FD, below.
u0 1
1
a
M
u1 u2 u4 u3
1.0
Ex: 3-10
A B
u0 1
1
a
M
u1 u2 u4 u3
1.0
Ex: 3-10
A B
u0 1
1
a
M
u1 u2 u4 u3
1.0
Ex: 3-10
A B
u0 1
1
a
M
u1 u2 u4 u3
1.0
Ex: 3-10
A B
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-26
.
24
25
8
6
5
101
1 yy
8
54
62
2 yy
12
5
8
6
5
43
3 yy
Ex: 3-10
1 2 m
M
A
4 @ 4 m = 16 m
M
3 m
M
B
u0 1
1
a
M
u1 u2 u4 u3
Gap
Hinge 2 m
M
E
-5/12
M
E
+5/6
M
E
y1
M
E -5/12
M
E
+5/6
M
E
y2
m
e
y3
I.L. (a), PD
I.L. (a), FD
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-27
3.8. Moving Loads:
If a maximum effect of a parameter is required due to a movement of a group of
loads, first of all the influence line of that parameter should be constructed, then
the group of loads should be placed on the maximum point of the IL, each at a
time, and the parameter should be calculated due to this position. The maximum
result is considered the maximum effect required.
Ex: (3-11)
Find the maximum upward and downward reaction of support (A) due to the
moving loads shown, if they move from left to right.
Solution:
Ex: (3-11)
A B
Gap
M
Hinge
M
Hinge
M 4 m
M
2 m
M
4 m
M
2 m
M
2 m
M
2 m
M
2 m
M
2 m
M
20
0
kN
M 30
00
kN
M 10
kN
M
2 m
M
1
I.L. RA-V-AB
-1
1
1
I.L. RA-V
-1
-2
1
-2
0.5
1.5
-1
1
-1
2 -2
2
0.5
1.5
-1
1
-1
I.L. RA-V
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-28
A B
Gap
M
Hinge
M
Hinge
M 4 m
M
2 m
M
4 m
M
2 m
M
2 m
M
2 m
M
2 m
M
2 m
M
20
0
kN
M 30
00
kN
M 10
kN
M
2 m
M
1 0.5
1.5
-1
I.L. RA-V
2
-1
-2
1
20
0
kN
M 30
00
kN
M 10
kN
M
kN851205.130210RMLmaxA
20
0
kN
M 30
00
kN
M 10
kN
M
20
0
kN
M 30
00
kN
M 10
kN
M
kN905.120230010RMLmaxA
20
0
kN
M 30
00
kN
M 10
kN
M
kN30220030110RMLmaxA
kN50020130210RMLmaxA
20
0
kN
M 30
00
kN
M 10
kN
M
kN90120230110RMLmaxA
20
0
kN
M 30
00
kN
M 10
kN
M
kN70220130010RMLmaxA
kN90RMLmaxA
kN90RMLmaxA
Ex: (3-11)
Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures
Q-L (2017-2018) Ch.3-29
Ex: (3-12) (Do it as homework, HW-2)
If the self-weight of the girder in the previous example is estimated to be (20
kN/m) of length, and a uniform live load of 10 (kN/m) of length, that can occupy
any length and region of the girder. Find the maximum vertical reaction at support
(A) due to the moving loads shown as well as the dead and live loads applied,
(upward and downward).
A B
Gap
M
Hinge
M
Hinge
M 4 m
M
2 m
M
4 m
M
2 m
M
2 m
M
2 m
M
2 m
M
2 m
M
20
0
kN
M 30
00
kN
M 10
kN
M
2 m
M
1
I.L. RA-V
2
-1
-2
1 0.5
1.5
-1
A1 A2
A3
HW-2
wD = 20 kN/m
wL = 10 kN/m
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-1
Chapter Four – Elastic Deformation of Structures
4. Deformation of Structures:
4.1. Introduction
The calculation of elastic deformations for structures, both the linear deformation
of points, (∆) and the rotational deformations of lines (slopes), (θ), from their
original positions, is very important in the analysis, design and construction of
structures.
There are several methods for computing elastic deformations; the most significant
in structural analysis are the following:
1) The method of virtual work (Unit-Load method).
2) Castigliano’s theorem.
3) The conjugate-beam method.
In this chapter, the first method for computing deformations of structures will be
discussed.
4.2. The Method of Virtual Work (Unit-Load Method)
In this method, in order to find an expression for the deformation, displacement or
rotation, at any point of the structure, a unit virtual load or moment is applied at
that point in the direction of deformation, then by applying the principle of virtual
work that states: “If a system in equilibrium under a system of forces undergoes a
deformation, the work done by the external forces (P) equals the work done by the
internal stresses due to those forces, (σP)”.
External Work = Internal Work
Work = Force × Displacement
Work = Moment × Rotation
Deformations
Displacements, ∆
Rotations, θ
Vertical
Displacements
(Deflections, ∆V)
Horizontal
Displacements, (∆H)
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-2
4.2.1. The Unit-Load Method for beams and Frames:
External Work = Internal Work
dLu1
Where:
1= external virtual force of unity
∆= external displacement
u= internal force, dAI
ymdAu
dL= Change in length of internal
strip, dxEI
yMdx
EdxdL
dxEI
yMdA
I
ym
IdxIE
mMdAydx
IE
mM2
2
2
dxIE
mM
If the deformation is a rotation then the rotation will be:
dxIE
mM
Where:
M= is the moment in the structure due to the applied loads.
m= is the moment in the structure due to a unit load or a unit moment applied at
the section where the displacement or rotation is required.
EI= material properties.
dx= infinitesimal length of the structure, as shown in Fig. (4.1- a and b).
Fig. (4.1)
M A
10
m
B
10
m C
10
m
1
m A
10
m
B
10
m
C
10
m (a)
(b)
m y
10
m B
10
m
A
10
m
C
10
m
1
dA
u u
dx
M y
10
m B
10
m
A
10
m
C
10
m
∆C
dL Fig. (3.7)
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-3
Ex (4-1):
Find the maximum deflection at the free end of the beam shown below.
Solution:
dxIE
mMVB
IE8
Lw0
IE8
Lw
4
x
IE2
wdx
IE2
xwdx
IE
x2
wx44
0
L4L
0
3L
0
2
VB
IE8
Lw4
VB
4.2.1.1. The Basic Steps of the Unit-Load Method for Beams and Frames:
The basic steps to be followed for finding the displacement or slope of beams or
frames by the virtual work method (Unit-Load method) are summarized as:
1) Depending on the number of deformations required, additional virtual
structures must be created each with a unit-load or a unit-moment at the
location of the deformation.
2) From the actual structure, compute the bending moment (M) due to the applied
external forces.
3) From each virtual structure, compute the bending moment (m) due to the unit
load or the unit moment applied in the direction of the required deformation or
slope.
4) Compute the integral dxIE
mM
over the entire members of the beam or frame
which will provide the desired deformation.
5) The bending moment shall be taken as positive if sagging and negative if
hogging (in the case of beams).
A
10
m
M
L
10
m
w
10
m
1
1
0
m
1
1
0
m
B
10
m
w
10
m
x
10
m
B
10
m
M
2
wx
2
xwxM
2
m
L
10
m
1
10
m
A
10
m
1
1
0
m
1
1
0
m
B
10
m
1 x
10
m
B
10
m
m
xx1m
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-4
6) The positive dxIE
mM
implies that the desired displacement is in the
direction of the applied unit load and negative quantity will indicate that the
desired displacement is in the opposite direction of the applied unit load.
Ex (4-2):
For the structure shown below, find the deflection at points B and C due to the
applied loading.
Solution:
1) To solve this problem with the U-L method, a new virtual structure must be
created for each deformation needed, therefore, three structure will be needed
as follows:
a) The actual structure with
the original applied loads
to calculate M.
b) A new virtual structure,
w/o the applied loads,
but with a unit virtual
load applied at (B), to
find (m1) for the
calculation of (∆B).
c) Another new virtual
structure, w/o the
applied loads, but with a
unit virtual load applied
at (C), to find (m2) for
the calculation of (∆C).
2) For each structure calculate the reactions due to the loads applied.
3) Each structure should be divided into portions based on the following:
A change in loadings.
A change in section properties.
A change in direction (for frames).
For this example, the structure will be divided into three portions for (∆B) and two
portions for (∆C) based on the changes in loadings and section properties.
2 m
10
m
2 m
10
m
4 m
10
m
A
10
m
B
10
m
C
10
m
D
10
m 2I
10
m
I
10
m
100 kN
10 m
M
2 m
10
m
2 m
10
m
4 m
10
m
A
10
m
B
10
m
C
10
m
D
10
m 2I
10
m
I
10
m
100 kN
10 m
m1
2 m
10
m
2 m
10
m
4 m
10
m
A
10
m
B
10
m
C
10
m
D
10
m 2I
10
m
I
10
m
1 kN
10 m
m2
2 m
10
m
2 m
10
m
4 m
10
m
A
10
m
B
10
m
1
k
N
10
m
C
10
m
D
10
m 2I
10
m
I
10
m
1 kN
10 m
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-5
For (∆B): AB (change in loading), BC (change in loading), CD (change in I).
For (∆C): AC and CD (change in loading and I).
x
A
10
m
0.75 kN
10 m
x75.0m AB1
2I
10
m
I
10
m
0.75 kN
10 m
0.25 kN
10 m
x x x
2 m
10
m
2 m
10
m
4 m
10
m
A
10
m
B
10
m
C
10
m
D
10
m
m1-AB
2I
10
m
m1-DC
1 kN
10 m
m1-BC
x25.05.1m
x1x275.0m
BC1
BC1
A
10
m
0.75 kN
10 m
x
2 m
10
m
1 kN
10 m
x D
10
m
0.25 kN
10 m
x25.0m DC1
m1 – with Unit-Load at B
A
10
m 2 m
10
m
2 m
10
m
4 m
10
m
B
10
m
C
10
m
D
10
m 2I
10
m
I
10
m
100 kN
10 m
50 kN
10 m
50 kN
10 m
MA
B x x
x MBC MDC
A
10
m
50 kN
10 m
x x250MBC
2 m
10
m
B
10
m
x
A
10
m
50 kN
10 m
x50MAB
x D
10
m
50 kN
10 m
x50MDC
M – Under Applied Loading
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-6
4) To facilitate the solution arrange a table to include the information for each
part as follows:
for ∆B:
Member Origin Limit EI
M 1m
EImM 1
AB
A
20 EI1
x50
x75.0
EI
x5.372
BC B
20
EI1
x250 x25.05.1
EI
x5.12x501502
CD D
40
EI2
x50
x25.0 EI
x25.6
EI2
x5.1222
4
0 CDCD
CD1CD2
0 BCBC
BC1BC2
0 ABAB
AB1AB
B dxIE
mMdx
IE
mMdx
IE
mM
4
0
2
0
2
0
B dxIE2
x25.0x50dx
IE
x25.05.1x250dx
IE
x75.0x50
4
0
22
0
22
0
2
B dxEI
x25.6dx
EI
x5.12x50150dx
EI
x5.37
x
A
10
m
0.5 kN
10 m
x5.0m AB2
x5.01m
x25.0m
BC2
BC2
A
10
m
0.5 kN
10 m
x
2 m
10
m
B
10
m
x D
10
m
0.5 kN
10 m
x5.0m DC2
2I
10
m
I
10
m
0.5 kN
10 m
0.5 kN
10 m
x x
2 m
10
m
2 m
10
m
4 m
10
m
A
10
m
B
10
m
C
10
m
D
10
m
m2-AC
2I
10
m
m2-DC
1 kN
10 m
x
m2 – with Unit-Load at C
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-7
4
0
32
0
322
0
3
B3
x25.6
3
x5.12
2
x50
1
x150
3
x5.37
EI
1
4
0
32
0
322
0
3
B3
425.6
3
25.12
2
250
1
2150
3
25.37
EI
1
EI
600333.133667.366100
EI
1333.133333.33100300100
EI
1B
EI
600B
For ∆C:
Member Origin Limit EI
M 2m
Mm2/EI
AB
A
20 EI1
x50
x5.0
EIx25
2
BC B
20
EI1
x250 x5.01
EI
x25x1001002
CD D
40
EI2
x50
x5.0 EI
x5.12
EI2
x2522
4
0 CDCD
CD2CD2
0 BCBC
BC2BC2
0 ABAB
AB2AB
C dxIE
mMdx
IE
mMdx
IE
mM
4
0
2
0
2
0
C dxIE2
x5.0x50dx
IE
x5.01x250dx
IE
x5.0x50
4
0
22
0
22
0
2
C dxEI
x5.12dx
EI
x25x100100dx
EI
x25
4
0
32
0
322
0
3
C3
x5.12
3
x25
2
x100
1
x100
3
x25
EI
1
4
0
32
0
322
0
3
C3
45.12
3
225
2
2100
1
2100
3
225
EI
1
EI
800667.266667.66200200667.66
EI
1C
EI
800C
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-8
Notes:
1) The limits for the integration will depend on the origin selected; either the
beginning or the end of the portion, and the limit will start from zero till the
end of that portion.
2) The beginning of the x distance must be taken either at the beginning of the
portion or at the end, the easiest way will be preferred.
3) If the actual structure is under the effect of one applied load and the
deformation needed to be calculated is at the same location of the applied load,
then there is no need to calculate the moments on the virtual structure, because
it will be equal to the moment due to the applied load divided by the value of
that load, this situation is observed in the previous example with (∆C).
100
MmM
100
Mm
2
22
4
0 CD
CD2
2
0 BC
BC2
2
0 AB
AB2
C dxEI
mMdx
EI
mMdx
EI
mM
4
0 CD
CD
22
0 BC
BC
22
0 AB
AB
2
C dxEI
100Mdx
EI
100Mdx
EI
100M
4
0
22
0
22
0
2
C dxEI2100
x50dx
EI100
x250dx
EI100
x50
4
0
22
0
2
2
0
2
C dx2
xdxxx44dxx
EI100
2500
4
0
32
0
322
0
3
C32
x
3
x
2
x4x4
3
x
EI
25
32
4
3
2
2
2424
3
2
EI
253323
C
32EI
25
6
64
3
888
3
8
EI
25C
EI
800C
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-9
Ex (4-3):
For the structure shown below, find the horizontal displacement at (d), the vertical
displacement at (b) and the rotation of (b) due to the applied loading.
Solution:
To solve this problem with the U-L method, the following steps are followed:
1) Since three deformations are required, then,
three virtual structures are needed in
addition to the actual one with the applied
loads, making a total number of four
structures.
a) The actual structure with the original
applied loads to calculate M.
b) The 1st virtual structure, w/o the applied
loads, but with a unit virtual load applied at
(d), to find (m1) for the calculation of
(∆H-d).
c) The 2nd
virtual structure, w/o the applied
loads, but with a unit virtual load applied at
(b), to find (m2) for the calculation of (∆V-b).
I
1
0
m
a
1
0
m
b
1
0
m
c
1
0
m
d
1
0
m e
1
0
m
2I
1
0
m
3I
1
0
m
2 m
10
m
4 m
10
m
4 m
10
m
1 m
10
m 3 m
10
m
10 kN
10 m
I
1
0
m
a
1
0
m
b
1
0
m
c
1
0
m
d
1
0
m e
1
0
m
2I
1
0
m
3I
1
0
m
2 m
10
m
4 m
10
m
4 m
10
m
1 m
10
m 3 m
10
m
10 kN
10 m
M
I
1
0
m
a
1
0
m
b
1
0
m
c
1
0
m
d
1
0
m e
1
0
m
2I
1
0
m
3I
1
0
m
2 m
10
m
4 m
10
m
4 m
10
m
1 m
10
m 3 m
10
m
1 kN
10 m
m1
I
1
0
m
a
1
0
m
b
1
0
m
c
1
0
m
d
1
0
m e
1
0
m
2I
1
0
m
3I
1
0
m
2 m
10
m
4 m
10
m
4 m
10
m
1 m
10
m 3 m
10
m
1 kN
10 m
m2
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-10
d) The 3rd
virtual structure, w/o the applied
loads, but with a unit virtual moment
applied at (b), to find (m3) for the
calculation of (θb).
2) For each structure find the reactions and the
required moments for each portion.
The reactions are determined using the
ordinary equations of equilibrium,
0Fand,0F,0M yx
3) Each structure should be divided into
portions based on a change in loadings,
section properties and direction.
For this example, the structure will be
divided into 4 portions for the actual
structure and 3 portions for each of the three
virtual structures, (m1, ∆H-d), (m1, ∆V-b) and
(m1, θb) based on the changes in directions,
loadings and section properties.
For the actual structure, (M):
ab (change in direction and I).
bc (change in direction, in loading and I).
bd (change in direction, in loading and I).
be (change in direction and I).
For the 1st virtual structure, (m1):
ab, bd and be (change in direction and I).
For the 2nd
virtual structure, (m2):
ab (change in direction and I), bd (change in
direction, in loading and I) and be (change
in direction and I).
For the 3rd
virtual structure, (m3):
ab (change in direction and I), bd (change in
direction, in loading and I) and be (change
in direction and I).
m3
I
1
0
m
a
1
0
m
b
1
0
m
c
1
0
m
d
1
0
m e
1
0
m
2I
1
0
m
3I
1
0
m
2 m
10
m
4 m
10
m
4 m
10
m
1
kN.
m
10
m
1
kN.
m
10
m
1 m
10
m 3 m
10
m
1 kN.m
10 m
1 kN.m
10 m
I
1
0
m
a
1
0
m
b
1
0
m
c
1
0
m
d
1
0
m e
1
0
m
2I
1
0
m
3I
1
0
m
2 m
10
m
4 m
10
m
4 m
10
m
1 m
10
m 3 m
10
m
10 kN
10 m
10 kN
5 kN
5 kN
I
1
0
m
a
1
0
m
b
1
0
m
c
1
0
m
d
1
0
m e
1
0
m
2I
1
0
m
3I
1
0
m
2 m
10
m
4 m
10
m
4 m
10
m
1 m
10
m 3 m
10
m
1 kN
10 m
1 kN
0.75 kN
0.75 kN
I
1
0
m
a
1
0
m
b
1
0
m
c
1
0
m
d
1
0
m e
1
0
m
2I
1
0
m
3I
1
0
m
2 m
10
m
4 m
10
m
4 m
10
m
1
kN.
m
10
m
1 m
10
m 3 m
10
m
1 kN.m
10 m
1 kN.m
10 m
0 kN
0.125 kN
0.125 kN
I
1
0
m
a
1
0
m
b
1
0
m
c
1
0
m
d
1
0
m e
1
0
m
2I
1
0
m
3I
1
0
m
2 m
10
m
4 m
10
m
4 m
10
m
1 m
10
m 3 m
10
m
1 kN
10 m
0 kN
0.5 kN
0.5 kN
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-11
4) Tabulate the results to facilitate solution to include the information for each
portion as follows:
Member Origin Limit EI
M 1m
2m
3m
ab
a
50 EI1
x2
0
5
x2
10
x
bc
c
10 EI3 x10
x2
0
0
cd
d
20 EI3 0
- -
-
be
e
40 EI2 x5 4
x3
2
x
8
x
Members M m1
ab
x2x5
45x
5
310M
0x5
475.0x
5
31m1
m2 m3
x4.0x5
45.0m2
x1.0x5
4125.0m1
m3
10
m
x
1
0
m
a
1
0
m
3
1
0
m
4
1
0
m
(4/5) x
m
10 m
(3/5) x
m
10 m
V
10
m
N
10
m
0 kN
0.125
kN
m2
10
m
x
1
0
m
a
1
0
m
3
1
0
m
4
1
0
m
(4/5) x
m
10 m
(3/5) x
m
10 m
V
10
m
N
10
m
0 kN
0.5 kN
m1
10
m
x
1
0
m
a
1
0
m
3
1
0
m
4
1
0
m
(4/5) x
m
10 m
(3/5) x
m
10 m
V
10
m
N
10
m
1 kN
0.75 kN
M
10
m
x
1
0
m
a
1
0
m
3
1
0
m
4
1
0
m
(4/5) x
m
10 m
(3/5) x
m
10 m
V
10
m
N
10
m
10 kN
5 kN
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-12
Members M m1
bc
x10M
x21m1
m2 m3
0m2
0m3
Members M m1, m2 and m3
cd
0M
Since M is zero, then no need to
calculate m1, m2 and m3
Members M m1
be
x5M
x75.0m1
m2 m3
x5.0m2
x125.0m3
b
1
0
m
c
1
0
m
d
1
0
m
2 m
10
m
m3
10
m
V
10
m
N
10
m
x
1
0
m
b
1
0
m
c
1
0
m
d
1
0
m
2 m
10
m
m2
10
m
V
10
m
N
10
m
x
1
0
m
1 kN
b
1
0
m
c
1
0
m
d
1
0
m
2 m
10
m
m1
10
m
V
10
m
N
10
m
x
1
0
m
10 kN
b
1
0
m
c
1
0
m
d
1
0
m
2 m
10
m
M
10
m
V
10
m
N
10
m
x
1
0
m
x
1
0
m
x
1
0
m
e
1
0
m
4 m
10
m
1
kN.
m
10
m
1
V
10
m
N
10
m
m3
10
m
0.125 kN
x
1
0
m
e
1
0
m
4 m
10
m
1
kN.
m
10
m
1
V
10
m
N
10
m
m2
10
m
0.5 kN
x
1
0
m
e
1
0
m
4 m
10
m
1
kN.
m
10
m
1
kN.
m
10
m
V
10
m
N
10
m
m1
10
m
0.75 kN
x
1
0
m
e
1
0
m
4 m
10
m
1
kN.
m
10
m
1
kN.
m
10
m
M
10
m
V
10
m
N
10
m
5 kN
c
1
0
m
d
1
0
m
2 m
10
m
M
10
m
V
10
m
N
10
m
x
1
0
m
x
1
0
m
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-13
Now to find the deformations required in the problem:
1)
4
0 be
be1be2
0 cd
cd1cd1
0 cb
cb1cb5
0 ab
ab1ab
dH dxEI
mMdx
EI
mMdx
EI
mMdx
EI
mM
IE
mM
4
0
2
0
cd11
0
5
0
dH dxEI2
x75.0x5dx
EI3
m0dx
EI3
x2x10dx
EI
0x2
4
0
21
0
2
dH dxEI2
x75.30dx
EI3
x10x200
3
4
2
75.3
3
1
3
10
2
1
3
20
EI
1
3
x
2
75.3
3
x
3
10
2
x
3
20
EI
13324
0
31
0
32
dH
EI
444.444011.133.3
EI
1
6
240
9
10
6
20
EI
1dH
EI
44.44dH
2)
4
0 be
be2be2
0 cd
cd2cd1
0 cb
cb2cb5
0 ab
ab2ab
bV dxEI
mMdx
EI
mMdx
EI
mMdx
EI
mM
IE
mM
4
0
2
0
cd21
0
5
0
bV dxEI2
x5.0x5dx
EI3
m0dx
EI3
0x10dx
EI
x4.0x2
4
0
25
0
24
0
5
0
bV dxEI2
x5.2dx
EI
x8.0dx
EI2
x5.0x500dx
EI
x4.0x2
4
0
35
0
3
bV6
x5.2
3
x8.0
EI
1
EI
6067.2633.33
EI
1
6
45.2
3
58.0
EI
133
bV
EI
60bV
3)
4
0 be
be3be2
0 cd
cd3cd1
0 cb
cb3cb5
0 ab
ab3ab
b dxEI
mMdx
EI
mMdx
EI
mMdx
EI
mM
IE
mM
4
0
2
0
cd31
0
5
0
b dxEI2
x125.0x5dx
EI3
m0dx
EI3
0x10dx
EI
x1.0x2
4
0
35
0
34
0
25
0
2
b6
x625.0
3
x2.0
EI
1dx
EI2
x625.0dx
EI
x2.0
667.6333.8
EI
1
6
40
3
25
EI
1
6
4625.0
3
52.0
EI
133
b
EI
67.1b
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-14
Ex (4-4): (1998)
For the structure shown below, use the virtual work
method to find the deformation of point (D) due to
the applied loading; noting that EI constant.
Solution:
To solve this problem with the V-W method (U-L method), the following steps are
followed:
1) Since only one deformation is required, then, only one virtual structure is
needed in addition to the actual one with
the applied loads, making a total number
of two structures.
a) The actual structure with the original
applied loads to calculate M.
b) The virtual structure, w/o the applied
loads, but with a unit virtual load applied
at (D), in the direction of the movement
of the roller (horizontally), to find (m)
for the calculation of (∆D).
2) For each structure find the reactions and
the required moments for each portion.
The reactions are determined using the
ordinary equations of equilibrium,
0Fand,0F,0M yx
3) Each structure should be divided into portions based on a change in loadings
and direction.
For this example, the structure will be divided into 3 portions for both the
actual and virtual structures.
For the actual structure, (M):
AB, BC and BD (change in direction and loading).
For the 1st virtual structure, (m):
AB, BC and BD (change in direction and loading).
4) Tabulate the results to facilitate solution to include the information for each
portion as follows:
4 m
10
m
10 kN
10 m
A
1
0
m
B
1
0
m
C
1
0
m
D
1
0
m
3 m
10
m
3 m
10
m
0 kN
0 kN
10 kN
M 4 m
10
m
10 kN
10 m
A
1
0
m
B
1
0
m
C
1
0
m
D
1
0
m
3 m
10
m
3 m
10
m
m 4 m
10
m 1 kN
10 m
A
1
0
m
B
1
0
m
C
1
0
m
D
1
0
m
3 m
10
m
3 m
10
m
1 kN
2/3 kN
2/3 kN
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-15
Member Origin Limit EI
M
m
AB
A
30 EI
0
-
BC
C
30 EI
x10
0
BD
D
50 EI
x6
x52
Member M m
AB
0M
calculatetoneedNom
BC
x10M
0m
BD
x6x5
310M
x
5
2x
5
3
3
2x
5
41m
5) Now to find the deformations required in the problem:
5
0
BDBD
3
0
BCBC
3
0
ABABD dxmMdxmMdxmMEI
1
IE
mM
5
0
3
0
BC
3
0
ABD dxx5
2x6dx0Mdxm0
EI
1
IE
mM
EI
1005
5
4
EI
1
3
x
5
12
EI
1dxx
5
12
EI
1
IE
mM5
0
3
5
0
35
0
2
D
10 kN
(4/5) x
10 m
(3/5) x
10 m
D
1
0
m
x
1
0
m
V
10
m
N
10
m
M
10
m
C
1
0
m
x
1
0
m
m
V
10
m
N
10
m
C
1
0
m
x
1
0
m
M
10
m V
10
m
N
10
m
10 kN
0 kN
0 kN
A
1
0
m
B
1
0
m
x
1
0
m
m
0
m V
10
m
N
10
m
0 kN
0 kN
A
1
0
m
B
1
0
m
x
1
0
m
M
10
m V
10
m
N
10
m
2/3 kN
(4/5) x
10 m
(3/5) x
10 m
D
1
0
m
x
1
0
m
V
10
m
N
10
m
M
10
m
1 kN
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-16
Note:
If the rotation in an internal hinge is required, the problem will be solved twice,
because the internal hinge has two different rotations, one on each side of the
internal hinge.
So the problem will be solved in two stages:
1) By applying a unit moment on one side of the internal hinge and calculating (θ)
based on that.
2) By applying a unit moment on the other side of the internal hinge and the other
(θ) will be calculated.
Ex (4-5): (1998)
For the structure shown below; use the virtual work method to find the deformation
of point (B) (represented by the vertical displacement and rotation) due to the
applied loading.
Solution:
1) Find (M) from the actual structure with the original applied loads.
2) To find the rotation of the internal hinge at point B, two rotations are included:
(θBC) and (θBA).
For (θBC) a unit moment is applied at the hinge on the side of member bc,
(m1).
C
1
0
m
20 kN
10 m
A
1
0
m
B
1
0
m
2 m
10
m
2 m
10
m
3 m
10
m
3 m
10
m
4 kN/m
10 m
2I
1
0
m
I
1
0
m
D
1
0
m
M
3 m
10
m
A
1
0
m
B
1
0
m
4 kN/m
18 kN.m
10 m
1 kN.m
10 m
12 kN
10 kN
0 kN
10 kN
C
1
0
m
20 kN
10 m
3 m
10
m
10 kN
10 kN 2 m
10
m
2 m
10
m
0 kN
D
1
0
m
B
1
0
m
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-17
For (θAB) a unit moment is applied at the hinge on the side of member AB,
(m2).
3) To find the vertical displacement of point B, (∆V-B), a unit load is applied at
point B, in the vertical direction.
4) Tabulate the results to facilitate solution to include the information for each
portion as follows:
Member Origin Limit EI
M 1m
2m
3m
AB
B
30 EI1
2
x2
0
5x2 0
BD
B
5.20 EI2 x8
5x1 0
0
DC
C
5.20 EI2 x8
5x 0 0
(m3)
C
1
0
m
3 m
10
m
0 kN
0 kN
2 m
10
m
2 m
10
m
0 kN
D
1
0
m
B
1
0
m
3 m
10
m
A
1
0
m
B
1
0
m 0 kN
0 kN
0 kN
1 kN
0 kN.m
10 m
1 kN.m
10 m
C
1
0
m
A
1
0
m
B
1
0
m
1 kN
1 kN
(m2)
C
1
0
m
3 m
10
m
0 kN
0 kN
2 m
10
m
2 m
10
m
0 kN
D
1
0
m
B
1
0
m
C
1
0
m
A
1
0
m
B
1
0
m
1 kN.m
10 m
1 kN.m
10 m
3 m
10
m
A
1
0
m
B
1
0
m 0 kN
0 kN
0 kN
0 kN
1 kN.m
10 m
1 kN.m
10 m
1 kN.m
10 m
1 kN.m
10 m
(m1)
C
1
0
m
3 m
10
m
1/4 kN
1/4 kN 2 m
10
m
2 m
10
m
0 kN
D
1
0
m
B
1
0
m
1 kN.m
10 m
1 kN.m
10 m
C
1
0
m
A
1
0
m
B
1
0
m
1 kN.m
10 m
1 kN.m
10 m 3 m
10
m
A
1
0
m
B
1
0
m 0 kN
1/4 kN
0 kN
1/4 kN
0 kN.m
10 m
1 kN.m
10 m
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-18
1 kN.m
10 m
1 kN.m
10 m
Calculating (m2) for each Portion
0m CD2
C
1
0
m
(4/5) x
10 m
1
kN.m
10 m
x
10
m
1
k
N.
m
10
m
(3/5) x
10 m
1
kN.m
10 m
D
1
0
m
V
m2
N 0 kN
10 m
3
10
m
4
10
m
0m BD2
0 kN
(4/5) x
10 m
1
kN.m
10 m
x
10
m
1
k
N.
m
10
m
(3/5) x
10 m
1
kN.m
10 m
D
1
0
m
B
1
0
m
3
10
m
V m2 N
4
10
m
1m AB2
B
1
0
m
m2
10 m
1 kn.m
10 m
0 kN
N
x
10
m
1
k
N.
m
10
m
V
2
AB
AB
x2M
2
xx4M
x8M
x5
410M
BC
BD
10 kN
(4/5) x
10 m
1
kN.m
10 m
x
10
m
1
k
N.
m
10
m
(3/5) x
10 m
1
kN.m
10 m
D
1
0
m
B
1
0
m
3
10
m
V M N
4
10
m
x8M
x5
410M
CD
CD
B
1
0
m
4 kN/m
M
10 m
1 kN.m
10 m
10 kN
N
x
10
m
1
k
N.
m
10
m
V
C
1
0
m
(4/5) x
10 m
1
kN.m
10 m
x
10
m
1
k
N.
m
10
m
(3/5) x
10 m
1
kN.m
10 m
D
1
0
m
V
M
N 10 kN
10 m
3
10
m
4
10
m
Calculating (M) for each Portion
Calculating (m1) for each Portion 0m AB1
B
1
0
m
m1
10 m
1 kn.m
10 m
1/4 kN
N
x
10
m
1
k
N.
m
10
m
V
x2.0x5
4
4
1m CD1
C
1
0
m
(4/5) x
10 m
1
kN.m
10 m
x
10
m
1
k
N.
m
10
m
(3/5) x
10 m
1
kN.m
10 m
D
1
0
m
V
m1
N 1/4 kN
10 m
3
10
m
4
10
m
x8.0x5
4
4
11m BD1
1/4 kN
(4/5) x
10 m
1
kN.m
10 m
x
10
m
1
k
N.
m
10
m
(3/5) x
10 m
1
kN.m
10 m
D
1
0
m
B
1
0
m
3
10
m
V m1 N
4
10
m
1 kN.m
10 m
1 kN.m
10 m
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-19
5) Now to find the deformations required in the problem:
To find (θBC):
5.2
0 CD
CD1CD5.2
0 BD
BD1BD3
0 AB
AB1AB
BC dxEI
mMdx
EI
mMdx
EI
mM
IE
mM
5.2
0
5.2
0
3
0
2
BC dxEI2
5
xx8
dxEI2
5
x1x8
dxEI1
0x2
5.2
0
25.2
0
5.2
0
2
BC dx5
x4dx
5
x4dxx40
EI
1
EI
5.125.2
EI
2
2
x4
EI
1dxx4
EI
1 2
5.2
0
25.2
0
BC
To find (θBA):
5.2
0 CD
CD2CD5.2
0 BD
BD2BD3
0 AB
AB2AB
BA dxEI
mMdx
EI
mMdx
EI
mM
IE
mM
5.2
0
5.2
0
3
0
2
BA dxEI2
0x8dx
EI2
0x8dx
EI1
1x2
EI
18
3
32
EI
1
3
x2
EI
1dx
EI1
1x2 33
0
33
0
2
BA
To find (∆V-B):
5.2
0 CD
CD3CD5.2
0 BD
BD3BD3
0 AB
AB3AB
BV dxEI
mMdx
EI
mMdx
EI
mM
IE
mM
0dx
EI2
0x8dx
EI2
0x8dx
EI1
0x2 5.2
0
5.2
0
3
0
2
BV
Calculating (m3) for each Portion
0m CD3
C
1
0
m
(4/5) x
10 m
1
kN.m
10 m
x
10
m
1
k
N.
m
10
m
(3/5) x
10 m
1
kN.m
10 m
D
1
0
m
V
m3
N 0 kN
10 m
3
10
m
4
10
m
0m BD3
0 kN
(4/5) x
10 m
1
kN.m
10 m
x
10
m
1
k
N.
m
10
m
(3/5) x
10 m
1
kN.m
10 m
D
1
0
m
B
1
0
m
3
10
m
V m3 N
4
10
m
0m AB3
B
1
0
m
m3
10 m
1 kn.m
10 m
0 kN
N
x
10
m
1
k
N.
m
10
m
V
1 kN
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-20
Ex (4-6):
For the structure shown below; use the virtual work method to find the rotation of
support (A) due to the applied loading.
Solution:
To solve this problem with the V-W method (U-L method), the following steps are
followed:
1) Since only one deformation is
required, the rotation of support
(A), then, only one virtual
structure is needed in addition to
the actual one with the applied
loads, making a total number of
two structures, the virtual
structure with a unit moment
applied at support (A), then find
the reaction for both structures.
2) Find (M) from the actual
structure with the original
applied loads.
3) Find (m) from the virtual
structure with a unit moment
applied at support (A).
4) Tabulate the results to facilitate
solution to include the information for each portion as follows:
Member Origin Limit EI
M
m
AB
A
230
EI3 27
x104
27
x1
BC B
40 EI2
2x2x
7
1488
7
x
CD D
30 EI1 - 0
CE E
20
EI2
- 0
3I
1
0
m
I
1
0
m
C
1
0
m
A
1
0
m
B
1
0
m
E
1
0
m
D
1
0
m
3 m
10
m
2 m
10
m
3 m
10
m
4 m
10
m
4 kN/m
10 m 20 kN
10 m
2I
1
0
m
m
x
1/7 kN
x
3I
1
0
m
x
I
1
0
m
C
1
0
m A
1
0
m
B
1
0
m
E
1
0
m
D
1
0
m
3 m
10
m
2 m
10
m
3 m
10
m
4 m
10
m
2I
1
0
m
1 kN.m
10 m
1 kN.m
10 m
0 kN
1/7 kN
x x
x x
x
M
x x
20 kN
x
36/7 kN
3I
1
0
m
I
1
0
m
C
1
0
m A
1
0
m
B
1
0
m
E
1
0
m
D
1
0
m
3 m
10
m
2 m
10
m
3 m
10
m
4 m
10
m
4 kN/m
10 m 20 kN
10 m
2I
1
0
m
204/7
kN
x x
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-21
5) To find the rotation of support (A), (θA), apply dxIE
mM
for all portions.
2
0 CE
CECE3
0 CD
CDCD4
0 BC
BCBC23
0 AB
ABAB
A dxEI
mMdx
EI
mMdx
EI
mMdx
EI
mMdx
IE
mM
4
0
2
23
0
A dxEI2
7
xx2
7
x1488
dxEI3
27
x1
27
x104
23
0
32
4
0
2
23
0
A x7
1x
49
74dxx
7
4dxx
294
104dxx
221
104
EI
1
4
0
43223
0
323
0
2
A4
x
7
1
3
x
49
74
2
x
7
4
3
x
294
104
2
x
221
104
EI
1
4
4
7
1
3
4
49
74
2
4
7
4
3
23
294
104
2
23
221
104
EI
143232
A
EI
07.41143.9218.32571.4005.9517.31
EI
1
7
64
147
4736
7
32
249
624
27
312
EI
1A
4.2.2. The Unit-Load Method for Trusses:
Since trusses are axial members, therefore the deformations in such members are
determined as follows:
dLuor
Where:
u = Represents the force in truss members due to a unit load applied at the joint
where the deformation is required.
dL= Represents the change in length of the truss member due to external effect
such as (Applied Loading, Temperature Changes and Fabrication Error)
dL may be calculated as follows:
1) Due to the Applied Loading:
nCompressiovedL
TensionvedL
EA
LSdL
S = Represents the force in truss members due to the applied loads.
EA
LuSor
2) Due to Temperature Changes:
eTemperaturindecreasevedL
eTemperaturinincreasevedLtLdL
α = Represents the coefficient of thermal expansion of the material.
0 0
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-22
tLuor
3) Due to Fabrication Error:
shorttooismembervedL
longtooismembervedLedL
e = Represents the fabrication error in length.
euor
Generally, the deformation in the truss will be the sum of all deformations:
eutLu
EA
LuSor
4.2.2.1. The Basic Steps of the Unit-Load Method for Trusses:
The basic steps to be followed for finding the displacement or slope of trusses by
the virtual work method (Unit-Load method) are summarized as:
1) Compute the axial force in various members (S) due to the applied external
forces.
2) Compute the axial force in various members (u) due to a unit load applied in
the direction of the required displacement of the point.
3) Compute the product EA
LuS
for all members.
4) The summation
EA
LuS
will provide the desired displacement.
5) The axial force shall be taken as positive if tensile and negative if
compressive.
6) The positive implies that the desired displacement is in the direction of the
applied unit load and negative quantity will indicate that the desired
displacement is in the opposite direction of the applied unit load.
General Notes Regarding Deformations in Trusses:
1) If the requirement was to find the vertical, horizontal or inclined displacement,
a unit load is applied at the location and in the direction required.
2) If the absolute displacement is required, the problem will be solved twice, first
to find (∆H), then to find (∆V), then the absolute displacement will be:
22VH
3) If the relative displacement between two points is required, two unit loads in
opposite direction must be applied on the line of action connecting the two
points where the relative displacement is required, in that case all reactions will
be zero because the applied forces are equal and opposite, then after
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-23
determining S and u, the displacement is calculated by:
EA
LuS
, as
shown in the examples below:
4) If the rotation of a truss member is required, a couple must be applied
perpendicularly on both sides of that member, then the rotation will be
lengthmember
ntdisplaceme
, as shown in the following example:
For the previous example to calculate (θAE):
A couple is applied on both sides of AE.
(The Rotation of Member (AE) is Required)
A B C
F
E D
L
L L
(S) (u)
A B C
F
E
D
L
L L 1 kN
1 kN
(+ve → Convergence)
A B C
F
E D
L
L L
(u)
1 kN
1 kN A B C
F
E
D
L
L L
(u)
1 kN
1 kN A B C
F
E D
L
L L
(S)
(The Relative Displacement between (B) and (E) is Required)
(-ve → Divergence)
A B C
F
E D
L
L L
(S)
A B C
F
E
D
L
L L
(u)
1 kN
1 kN
(The Relative Displacement between (A) and (D) is Required)
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-24
To determine the reactions of the supports the couple is converted into a
moment equating (1×member length), after that the moment is turned into a
couple again.
Find S from applied loads, u from couple, then the displacement (∆AE) is
obtained,
EA
LuSAE
.
Finally, (θAE) is found, AE
AEAE
L
Ex (4-7):
For the truss shown in figure use the unit
load method to do the following:
1) Find the relative displacement
between (B) and (c) along the line
joining them, noting that (L/A=2), is
constant.
2) Find the rotation of member (aB).
Solution:
To solve this problem with the U-L
method, the following steps are
followed:
1) Compute the axial forces in various
members, (S), due to the applied
external forces.
2) Since the relative displacement
between (B) and (c) is required,
apply two unit opposite loads along
the line joining them, then compute
the axial forces in various members,
(u1), due to the unit load applied.
3) Since the rotation of truss member,
(aB), is required, couple is applied
on both sides of (aB), then compute
the axial forces in various members,
(u2), due to the unit load applied
4) Compute the product EA
LuS
for all
members.
5) The summation
EA
LuS
will
provide the required deformations.
90 kN
8 m
L
B
a b
C
12 m 12 m 12 m
c d 8 m
L
(S)
90 kN
8 m
L
B
a
C
12 m 12 m 12 m
b c d 8 m
L
45 kN
1 kN
(u1)
8 m
L
B
a
C
12 m 12 m 12 m
b c d 8 m
L
1 kN
0 kN 0 kN
4 3
1 kN
(u2)
8 m
L
B
C
12 m 12 m 12 m
a
b c d 8 m
L
1 kN
5/6
kN
5/6 kN
4 3
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-25
6) Tabulate the results.
Member S u1 u2 Su1 Su2
ab -67.5 0 1.25 0 -84.4
bc -67.5 -3/5 1.25 40.5 -84.4
cd -67.5 0 1.25 0 -84.4
aB 112.5 0 -0.75 0 -84.4
bB -135 -2/5 5/6 54 -112.5
bC 0 3.6/5 0 0 0
cC 0 -4/5 0 0 0
dC 81.125 0 -1.5 0 -121.7
BC 81.125 -3.6/5 -1.5 -58.41 -121.7
∑ 36.09 -693.5
7) Determine the deformations:
1)
E
18.7209.36
E
209.36
AE
LuS
AE
L1Bc
2) rad
E
35.695.693
E
2
20
1uS
AE
L
L
1
EA
LuS
L
12
aB
2
aB
aB
Since the couple applied was clockwise, and (θaB) obtained was (–ve), then
(θaB) will be counter clockwise, CCWrad
E
35.69aB
.
Ex (4-8):
Resolve the second part of the previous
example if the external effect was:
1) A drop in temperature of about (50oC) in
the lower bars.
2) Due to the following fabrication errors:
Bar (BC) was 3 mm too long.
Bar (Ba) was 2 mm too short.
Solution:
1) Due to a drop in temperature, (50oC):
tLdL
Member L u2 Lu2
ab 12 1.25 15
bc 12 1.25 15
cd 12 1.25 15
∑ 45
8 m
L
B
a b
C
12 m 12 m 12 m
c d 8 m
L
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-26
rad5.11245
20
50uL
L
ttuL
L
12
aB
2
aB
aB
Since the couple applied was clockwise, and (θaB) obtained was (–ve), then
(θaB) will be counter clockwise, CCWrad5.112aB
.
2) Due to fabrication error:
shorttooismembervedL
longtooismembervedLedL
Take only the members with the fabrication errors, (BC) and (Ba).
Member e u2 eu2
BC +3 -1.5 -4.5
Ba -2 -0.75 1.5
∑ -3.0
rad20
3
L
ue
aB
2
aB
Since the couple applied was clockwise, and (θaB) obtained was (–ve), then
(θaB) will be counter clockwise, CCWrad
20
3aB
.
If the total effect on (θaB) was required, it will be the summation of all types of
external effect: (applied load drop in temperature + fabrication error)
rad20
35.112
E
693eutLu
EA
LuSaB
Ex (4-9):
Compute the horizontal and vertical movement
of point (C) if bar (AC) was (1 cm) too long.
Solution:
Since the fabrication error in member AC,
then the length of AC must be found:
The triangle ACD:
53sin
CD
90sin
AD
37sin
AC
m4.2AC1
4
5
3
AC
The triangle ACa:
m44.15
4.23Aa
AC
Aa
5
353cos
3 m
L
B
A
C
4 m
D
4 3 4
3
A
C
4 m
D
4 3 4
3
a
53o 37
o
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-27
Since the both the horizontal and vertical
movements of point C are required then two
virtual structures are needed.
Apply a vertical unit load at point C and then
compute the axial forces in various members,
(u1), due to that unit load.
Apply a horizontal unit load at point C and
then compute the axial forces in various
members, (u2), due to that unit load.
Due to fabrication error:
shorttooismembervedL
longtooismembervedLedL
ueudL
cm8.08.01CV
cm6.06.01CH
Ex (4-10):
If the length of each bar of the truss shown in figure
is (1000 mm) and EA is equal to k for tension
members and 2k for compression members,
calculate the vertical displacement at point C.
barsncompressiokNfor80000k2
barstensionforkN40000kEA
Solution:
Since all bars have the same length then the angles
between each two bars is equal to 60o.
Let point E be the center of AC
m866.02
3AE
1
AE
AB
AE
2
360sin
Since there is only one load on the structure and is
applied at the same location and direction of the
deformation required, point C, then the forces in the
bars of the virtual structure, u, will be equal to the forces of the bars in the actual
structure divided by the value of the load.
80
S
P
Su
3 m
L
B
A
C
4 m
D
4 3 4
3
1 kN
(u1)
0.36 kN 0.64 kN
3 m
L
B
A
C
4 m
D
4 3 4
3
1 kN
(u2)
0.48 kN 0.48 kN
1 kN
P=80 kN
L
B
A C
D
D
P=80 kN
L
B
A C
D
D
60o 60
o 60
o
60o 60
o
60o
0.866 m
L
0.866 m
L
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-28
To find the forces in the structural members due
to the applied loads, S, at fist find all reactions
due to the applied loads:
1) Reactions:
0M A
kN160D0866.0D866.0280 yy
0Fy
kNo8A080160A yy
0A0F xx
2) Bar Forces:
Start with Joint C:
0Fx
CDCB
030cosCD30cosCB
0Fy
kN80CD
kN80CB
30sin2
80CB8030sinCB2
08030sinCB30sinCB
CDCB
08030sinCD30sinCB
Joint A:
The same procedure leads to:
kN80AD
kN80AB
Joint B:
0Fy
kN80BD
0BD5.0805.080
0BD30sinBA30sinBC
3) 80
Su
EA
LuSCV
P=80 kN
L
B
A C
D
D
0.866 m
L
0.866 m
L 160 kN
L
80 kN
L
0 kN
L
Forces in Truss Elements, S, due to
External Applied Loads
C
80 kN
L
CB = 80 kN
60o
CD= -80 kN
L
60o
BA=80 kN
L BD = 80 kN
B
BC=80 kN
L 60
o
30o 30
o
60o
BA=-80 kN
L
BD = 80 kN
BC=-80 kN
L
60o
D 30
o 30
o
A
AD = 80 kN
L
AB = 80 kN
60o
80 kN
L
P=80 kN
L
B
A C
D
D
0.866 m
L
0.866 m
L 160 kN
L
80 kN
L
0 kN
L
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-29
m007.0AE80
LS
EA
L80
SS 2
CV
Member AE L S u=S/80 SuL/AE
AB 40000 1 80 1 0.002
BC 40000 1 80 1 0.002
CD 80000 1 -80 -1 0.001
DA 80000 1 -80 -1 0.001
BD 80000 1 -80 -1 0.001
∑ 0.007
HW (4-1):
If the area of each bar of the truss shown in
figure is (2700 mm2) and the movement of
roller (D) was limited to (50 mm), calculate
the value of (w).
Notes for Solution:
To find the deformation of the roller at D,
a unit load is applied at that point and in
the direction of the roller’s movement.
Find the reactions, S (in terms of w) and
u, then w will be found by applying
m05.0
EA
LuSD
.
Ex (4-11):
For the truss shown in the figure below, if (E=2×104 kN/cm
2) and the area of each
vertical member is (4 cm2), each horizontal member is (6 cm
2) and each diagonal
member is (10 cm2), find the following:
1) The vertical displacement at (B).
2) The vertical displacement at (D).
3) The relative displacement between (B & H).
4) The rotation of member (GH).
5) The vertical displacement at (B) due to a rise in temperature of (40oC),
α=12×10-6
/oC.
10 m
L
B A C
D
D E
D
F
D
2w
L
w
L
10 m
L
10 m
L
10 m
L
100 kN
L
B
A
C
D
D
E F G H
4 m
L
3 m
L
3 m
L
3 m
L
3 m
L
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-30
Solution:
1) The vertical displacement at (B):
Since the actual structure has only one load at the same location and direction
of the displacement required, then the values of the forces in the truss members
of the virtual structure will be equal to the values of the forces in the truss
members of the actual structure divided by the load,
100
Su1
.
A
L100S
E
1
EA
LuS 2
1
BV
A100
LS
E
12
BV
Member L
(cm)
Area
(cm2)
Horizontal 300 6
Vertical 400 4
Diagonal 500 10
Member A, cm2
L, cm S, kN u1=S/100 S u1
L/A
AB 6 300 0 0 0.000
BC 6
300 0 0 0.000
CD 6
300 0 0 0.000
EF 6
300 -37.5 -0.375 703.125
FG 6
300 -37.5 -0.375 703.125
GH 6
300 0 0 0.000
EA 4 400 -50 -0.5 2500.000
FB 4
400 0 0 0.000
GC 4
400 -50 -0.5 2500.000
HD 4
400 0 0 0.000
EB 10 500 62.5 0.625 1953.125
BG 10
500 62.5 0.625 1953.125
GD 10
500 0 0 0.000
∑ 10312.500
cm515625.05.1031220000
1
A
LuS
E
1
EA
LuS 11
BV
100 kN
L
B A C
D
D
E F G H
4 m
L
3 m
L
3 m
L
3 m
L
3 m
L
50 kN
L
50 kN
L
0 kN
L
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-31
2) The vertical displacement at (D):
To find this deformation a new virtual structure is needed with a unit vertical
load at D, then find the reaction and u2.
Member A, cm2
L, cm S, kN u2 S u2
L/A
AB 6 300 0 0 0.000
BC 6
300 0 -0.75 0.000
CD 6
300 0 -0.75 0.000
EF 6
300 -37.5 0.375 -703.125
FG 6
300 -37.5 0.375 -703.125
GH 6
300 0 0 0.000
EA 4 400 -50 0.5 -2500.000
FB 4
400 0 0 0.000
GC 4
400 -50 -1.5 7500.000
HD 4
400 0 0 0.000
EB 10 500 62.5 -0.625 -1953.125
BG 10
500 62.5 0.625 1953.125
GD 10
500 0 1.25 0.000
∑ 3593.75
A
LuS
E
1
EA
LuS 22
DV
cm179688.075.359320000
1DV
100 kN
L
(S)
L
B A C
D
D
E F G H
4 m
L
3 m
L
3 m
L
3 m
L
3 m
L
50 kN
L
50 kN
L
0 kN
L
(S)
L
0 kN
L
1 kN
L
B A C
D
D
E F G H
4 m
L
3 m
L
3 m
L
3 m
L
3 m
L
1.5 kN
L
0.5 kN
L
0 kN
L
(u2)
L
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-32
3) The relative displacement between (B & H):
To find this deformation a new virtual structure is needed with two opposite
unit loads applied along the line between (B & H), then find the reaction and
u3.
Member A, cm2
L, cm S, kN u3 S u3
L/A
AB 6 300 0 0 0
BC 6
300 0 0.42 0
CD 6
300 0 0.42 0
EF 6
300 -37.5 0 0
FG 6
300 -37.5 0 0
GH 6
300 0 0.83 0
EA 4 400 -50 0 0
FB 4
400 0 0 0
GC 4
400 -50 0 0
HD 4
400 0 0.55 0
EB 10 500 62.5 0 0
BG 10
500 62.5 0.69 2156.25
GD 10
500 0 -0.69 0
∑ 2156.25
A
LuS
E
1
EA
LuS 33
H&BlativeRe
cm107813.025.215620000
1H&BlativeRe
4) The rotation of member (GH):
To find this deformation a new virtual structure is needed with unit couple
applied on both ends of member (GH), then find the reaction and u4.
4 m
L
100 kN
L
(S)
L
B A C
D
D
E F G H
4 m
L
3 m
L
3 m
L
3 m
L
3 m
L
50 kN
L
50 kN
L
0 kN
L
(S)
L
0 kN
L
1 kN
L
B A C
D
D
E F G H
3 m
L
3 m
L
3 m
L
3 m
L
0 kN
L
0 kN
L
0 kN
L
(u3)
L
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-33
Member A, cm2
L, cm S, kN u4 S u4
L/A
AB 6 300 0 0 0
BC 6
300 0 -0.75 0
CD 6
300 0 -0.75 0
EF 6
300 -37.5 0.375 -703.125
FG 6
300 -37.5 0.375 -703.125
GH 6
300 0 0 0
EA 4 400 -50 0.5 -2500
FB 4
400 0 0 0
GC 4
400 -50 -0.5 2500
HD 4
400 0 -1 0
EB 10 500 62.5 -0.625 -1953.125
BG 10
500 62.5 0.625 1953.125
GD 10
500 0 1.25 0
∑ -1406.250
A
LuS
EL
1
EA
LuS
L
1 4
GH
4
GH
GH
rad000234375.025.1406
20000300
1GH
5) The vertical displacement at (B) due to a rise in temperature of (40oC),
α=12×10-6
/oC:
To find this deformation use the same data of u1for all member subjected to
temperature.
Lut 1BV
100 kN
L
(S)
L
B A C
D
D
E F G H
4 m
L
3 m
L
3 m
L
3 m
L
3 m
L
50 kN
L
50 kN
L
0 kN
L
(S)
L
0 kN
L
4 m
L
1 kN
L
B A C
D
D
E F G H
3 m
L
3 m
L
3 m
L
3 m
L
0 kN
L
0 kN
L
0 kN
L
(u4)
L
1 kN
L
Theory of Structures-1 CH-4: Deformation of Structures
Q-L (2017-2018) Ch.4-34
Member L, cm S, kN u1=S/100 u1 L
AB 300 0 0 0
BC 300 0 0 0
CD 300 0 0 0
EF 300 -37.5 -0.375 -112.5
FG 300 -37.5 -0.375 -112.5
GH 300 0 0 0
EA 400 -50 -0.5 -200
FB 400 0 0 0
GC 400 -50 -0.5 -200
HD 400 0 0 0
EB 500 62.5 0.625 312.5
BG 500 62.5 0.625 312.5
GD 500 0 0 0
∑ 0.0
Lut 1BV
00401012Lut
6
1BV
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-1
Chapter Five
Analysis of Statically Indeterminate Structures by the
Method of Consistent Deformations
5. The Method of Consistent Deformations:
5.1. Introduction
Most structures in the real world are statically indeterminate structures, which can
be identified when:
No. of unknown Reactions or Internal forces > No. of equilibrium equations
For such structures, Statics (equilibrium) alone is not sufficient to conduct
structural analysis. Compatibility and material information are essential.
Compatibility means that the structure must fit together – no gaps can exist – and
the deflected shape must be consistent with the constraints imposed by the
supports.
There are two approaches of analysis for statically indeterminate structures
depending on how equations of equilibrium, load displacement and compatibility
conditions are satisfied:
1) The force-based method of analysis.
2) The displacement-based method of analysis.
The method of consistent deformations is one of the earliest methods available for
the analysis of statically indeterminate structures, using the first approach of
analysis, the force-based method.
This method involves analyzing the indeterminate structure as a stable determinate
one, as shown in these steps:
1) Check the determinacy of the original structure and if it is found as
indeterminate, determine the degree of indeterminacy.
2) Remove enough restraints (the same number as the degree of determinacy)
from the indeterminate structure to make it stable and statically determinate.
The structure after removing the restraints must be stable and determinate.
The removed restraints are called “redundant restraints”.
The sense of the redundant can be arbitrarily assumed.
After removing the restraint corresponding to the redundant, the structure
obtained is called “primary structure”.
3) Draw a diagram of the primary structure, and number it as (0), then calculate
(M) or (S), according to the type of structure, whether flexure or axial, with
external loads only.
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-2
4) Draw a diagram of a virtual structure, for each redundant, and number it
according to the number of the redundant, by applying a unit load or a unit
moment, depending on the type of the redundant, then calculate (m) or (u),
according to the type of structure, whether flexure or axial.
The number of the virtual structures is equal to the number of redundant, for
example, if there is only one redundant then only one virtual structure.
5) Write the compatibility equations in terms of the deformations of the primary
and virtual structures in (3) and (4).
6) By solving the compatibility equations, the redundant can be obtained.
7) Determine the remaining support reactions on the structure by satisfying
equilibrium requirements.
5.2. Definitions:
1) Redundant:
Refer to the number of unknowns that make the indeterminate structure stable and
determinate after removing them.
Note:
The redundant removed must keep the structure stable otherwise the method
cannot be applied.
2) Compatibility Equations:
The equations created for the problem by using the method of consistent
deformations:
Si....ijXijXijX0i 321
Where:
0i : represents the deformation of point (i), at the location where the redundant
was removed, in the primary structure, due to the applied loads.
ij : represents the deformation of point (i) due to a unit load applied at point (j).
321 X,X,X : represent the redundant restraints removed from the indeterminate
structure to make it stable and determinate.
Si : represents the equivalent deformation for the structure at point (i), the
location where the redundant was removed.
5.3. Analysis of Indeterminate Beams and Frames with the Method of
Consistent Deformations:
Applying this method to solve indeterminate beams and frames is shown in the
following example:
1) The beam shown in figure is stable and indeterminate to the 2nd
degree;
therefore, two redundant restrains must be removed in order to make it stable
and determinate. This structure is called the original structure.
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-3
2) The two redundant to be removed in order to make the beam stable and
determinate are the rollers at (C) and (D), referred to as (X1) and (X2)
respectively.
The structure after the removal of the redundants is called the primary
structure.
Determine the deformations at (C) and (D) of the primary structure due to the
applied loads, (∆10) and (∆20), respectively.
3) Remove the applied loads and apply
a unit load at (C) and determine the
deformations at (C) and (D) of the
first virtual structure, due to the
unit load, (δ11) and (δ21),
respectively.
4) Remove the applied loads and apply
a unit load at (D) and determine the
deformations at (C) and (D) of the
second virtual structure, due to
the unit load, (δ12) and (δ22),
respectively.
P1 P2
w
A B D C
1 m 1 m 1 m 1 m 2 m
∆10 ∆20
(The Primary Structure,
M)
w
P1 P2
X2 X1
A B D C
1 m 1 m 1 m 1 m 2 m
(The Original Structure)
A B
D C
δ11 δ21
(1st Virtual Structure, m1)
1 kN
A B
D C
δ12 δ22
(2nd
Virtual Structure, m2)
1 kN
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-4
5) The deformation at (C) and (D) in the original structure, at the supports, are
equal to zero (unless otherwise stated), therefore, [the sum of the deformations
at each support for the primary, first and second virtual structures will be equal
to zero (or otherwise stated)]. The equations representing this statement are
shown as follows:
022X21X20
012X11X10
21
21
Where:
dxEI
mM20dx
EI
mM10
21
dxEI
mm22dx
EI
mm11
2211
dxEI
mm2112
21
6) By solving these equations the values of (X1) and (X2) will be found.
7) By satisfying equilibrium requirements, the remaining support reactions on the
structure can be determined.
Note:
The support with a given deformation should be selected as the redundant, for
example, if in the beam shown in figure below with a fixed support at (A), if it was
stated in the question that support (A) suffers of a rotational slip of some amount
then the moment at support (A), (MA) will be the redundant.
Supports in Original Structure
(Before Removing the Redundant)
Supports in Primary Structure
(After Removing the Redundant)
X1
P1 P2 w
B C A
1 m 2 m 1 m 2 m
P1 P2 w
B C
A 1 m 2 m 1 m 2 m
X1
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-5
Supports in Original Structure
(Before Removing the Redundant)
Supports in Primary Structure
(After Removing the Redundant)
5.3.1. Examples on Analyzing Statically Indeterminate Frames with the
Method of Consistent Deformations:
Ex: (5-1):
For the structure shown in figure, use the
consistent deformations method to do the
following:
1) Find the reaction of the roller at B.
2) Draw the diagrams for axial force, shear force
and bending moment for all members.
X1
X1
X1
X1
X1
B
L
L/2 (EI Constant)
A
D
C
P
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-6
Solution:
1) To find the reaction of the roller at B:
1) Check the determinacy of the original structure.
The original structure is a frame structure: cj3rb3
1213
12cj30c,4j
13rb34r,3bThe original structure is indeterminate to
the 1st degree.
2) Choose the redundant that by removing it the structure will remain stable and
determinate.
Let X1 be (RB).
3) Draw a diagram of the primary structure, and number it as (0), then calculate
all reactions and (M) due to external loads only, for each portion of the
structure.
4) Draw a diagram of the virtual structure, by applying a unit load in the direction
of the redundant, and number it as (1), then calculate all reactions and (m1) due
to the unit load applied, for each portion of the structure.
B
L
L/2 (EI Constant)
A
D
C
P
X1
B
L
L/2 (EI Constant)
A
D
C
P
(The Original Structure) (The Primary Structure)
(The Primary Structure, 0, M)
L/2
B
L
(EI Constant)
A
D
C
P
MA=PL Ay=P x
x x
M
10
m
M
10
m
M
10
m
(The Virtual Structure, 1, m1)
L/2
L
(EI Constant)
B
A
D
C MA=0
Ay=1
1 kN
x
x
x
m1
m1
10
m
m1
10
m
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-7
Member Origin Limit M 1m
AC A
L0
PLPx
x
CD D
2L0
0 L
DB
B
L0
0
x
5) Write the compatibility equations in terms of the deformations of the primary
and virtual structures, then solve them:
1S11X10 1
dxEI
mM10
1
dxEI
mm11
11
L
0
2
L
0
1dxPLxPx
EI
1dx
EI
xPLPxdx
EI
mM10
EI6
PL
2
PL
3
PL
EI
110
333
L
0
2L
0
L
0
11dxxxdxLLdxxx
EI
1dx
EI
mm11
3
L
2
LL
3
L
EI
1
3
xxL
3
x
EI
111
32
3L
0
32L
0
2
L
0
3
EI6
L7
3
L
2
L
3
L
EI
111
3333
B3
3
1
3
1
3
R7
P
L7
EI6
EI6
PLX0
EI6
L7X
EI6
PL
2) To draw the diagrams for axial force, shear force and bending moment for all
members, all reaction must be calculated using equilibrium equations, based on
(RB=P/7↓):
For the whole structure:
PLM0PLM0M AAA
0A0F xx
7
P8A0
7
PPA0F yyy
(The Original Structure)
B
L
L/2
A
D
C
P
P/7
Ay=8P/7
MA=PL
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-8
Ex: (5-2):
Use the method of consistent deformations
to analyze the frame shown in figure below
to find the reactions due to the combined
effect of external loading and a vertical
settlement of support (A) of (18 mm ↓),
noting that (E=2×105
kN/m2) and
( I=3×10-4
m4).
Solution:
1) Check the determinacy of the original structure.
The original structure is a frame structure: cj3rb3
1215
12cj30c,4j
15rb36r,3bThe original structure is indeterminate to
the 3rd
degree.
C A
Ay=8P/7 8P/7
MA=PL PL/7
C
D
8P/7
PL/7
PL/7
8P/7
B
L
D P
P/7
8P/7
PL/7
+
8P/7
8P/7
Axial Force
Diagram
10 m
-
+
-8P/7 -8P/7
P/7 P/7
Shear Force
Diagram
10 m
PL/7
-
-PL
-
+
+
PL/7
PL/7
-PL/7
Bending Moment
Diagram
10 m
B
6 m
5 m
A D
C
18 mm
I I
2I
10 kN/m
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-9
2) Choose three redundant that by removing them the structure will be
determinate and stable.
Since support (A) is suffering from vertical settlement then remove that
support and its three components will be the redundant needed; let (X1) be
(Ax), (X2) be (Ay) and (X3) be (MA).
3) Draw a diagram of the primary structure, and number it as (0), then calculate
all reactions and (M) due to external loads only, for each portion of the
structure.
4) Draw a diagram for each of the virtual structures, by applying a unit load, or a
unit moment, in the direction of each redundant removed, and number them as
(1, 2 and 3), then calculate all reactions and (m1, m2, and m3) due to the unit
load, or unit moment, applied, for each portion of the structure.
Member Origin EI Limit M
1m
2m 3m
AB
A
EI1
50
0
x
0 1
BC B
EI2
60
2
x102
5
x
1
CD D
EI1
50
180
x
6
1
x
B
D
C
I I
2I
m3
1 kN.m MD=1 kN.m
A
X3
B
D
C
I I
2I
m2
1 kN
MD=6 kN.m
Dy=1 kN
A
X3
B
D
C
I I
2I
m1
1 kN
Dx=1 kN
A
X3
B
D
C
I I
2I
10 kN/m
M
MD=180 kN.m
Dy=60 kN
A
X3
x x x
x
x x x x x x x
X1
B
A
X3
D
C
I I
2I
10 kN/m
X2 X3
(The Original Structure)
B
D
C
I I
2I
10 kN/m
(The Primary Structure)
A
X3
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-10
5) Write the compatibility equations in terms of the deformations of the primary
and virtual structures, then solve them to find X1, X2 and X3:
1S13X12X11X10 321
2S23X22X21X20 321
3S33X32X31X30 321
dxEI
mM30,dx
EI
mM20,dx
EI
mM10
321
dxEI
mm33,dx
EI
mm22,dx
EI
mm11
332211
dxEI
mm3223,dx
EI
mm3113,dx
EI
mm2112
323121
5
0
16
0
15
0
1dx
EI1
mMdx
EI2
mMdx
EI1
mM10
5
0
6
0
25
0
dxEI1
x180dx
EI2
52x10dx
EI1
x010
5
0
6
0
2
5
0
6
0
2
dxx180dxx2
25
EI
1dx
EI1
x180dx
EI2
x2510
EI
31502250900
EI
1
2
5180
6
625
EI
1
2
x180
6
x25
EI
110
235
0
26
0
3
5
0
26
0
25
0
2dx
EI1
mMdx
EI2
mMdx
EI1
mM20
5
0
6
0
25
0
dxEI1
6180dx
EI2
x2x10dx
EI1
0020
5
0
6
0
45
0
6
0
3x1080
4
x5.2
EI
1dx1080dxx5.2
EI
120
EI
62105400810
EI
151080
4
65.2
EI
120
4
5
0
36
0
35
0
3dx
EI1
mMdx
EI2
mMdx
EI1
mM30
5
0
6
0
25
0
dxEI1
1180dx
EI2
12x10dx
EI1
1030
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-11
5180
6
65
EI
1x180
6
x5
EI
1dx
EI1
180dx
EI2
x530
35
0
6
0
35
0
6
0
2
EI
1080900180
EI
130
5
0
6
0
5
0
11dx
EI1
xxdx
EI2
55dx
EI1
xxdx
EI
mm11
6
0
5
0
2
5
0
2
6
0
5
0
2dx
2
25dxx2
EI
1dxxdx
2
25dxx
EI
111
EI
33.1587533.83
EI
1
2
625
3
52
EI
1
2
x25
3
x2
EI
111
36
0
5
0
3
5
0
6
0
5
0
22dx
EI1
66dx
EI2
xxdx
EI1
00dx
EI
mm22
5
0
6
0
35
0
6
0
222
x366
x
EI
1dx36dx
2
x
EI
1dx
EI
mm22
EI
21618036
EI
1536
6
6
EI
122
3
5
0
6
0
5
0
33dx
EI1
11dx
EI2
11dx
EI1
11dx
EI
mm33
EI
135
2
65
EI
1x
2
xx
EI
1dx1dx
2
1dx1
EI
133
5
0
6
0
5
0
5
0
6
0
5
0
5
0
6
0
5
0
21dx
EI1
6xdx
EI2
x5dx
EI1
0xdx
EI
mm2112
5
0
26
0
25
0
6
02
x6
4
x5
EI
1dxx6dx
2
x5
EI
12112
EI
1207545
EI
1
2
56
4
65
EI
12112
22
5
0
6
0
5
0
31dx
EI1
1xdx
EI2
15dx
EI1
1xdx
EI
mm3113
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-12
5
0
26
0
5
0
25
0
6
0
5
02
x
2
x5
2
x
EI
1dxxdx
2
5dxx
EI
13113
EI
405.12155.12
EI
13113
5
0
6
0
5
0
32dx
EI1
16dx
EI2
1xdx
EI1
10dx
EI
mm3223
56
4
6
EI
1x6
4
x
EI
1dx6dx
2
x
EI
13223
25
0
6
0
25
0
6
0
EI
39309
EI
13223
0EI
13X
EI
39X
EI
40X
EI
1080
018.0EI
39X
EI
216X
EI
120X
EI
6210
0EI
40X
EI
120X
EI
33.158X
EI
3150
321
321
321
22.8X
30X
91.4X
1080X13X39X40
92.6208X39X216X120
3150X40X120X33.158
3
2
1
321
321
321
Note:
Sometimes the problem is provided with more than one displacement of a
support, to solve this kind of problem, the equivalent deformation of that
support, must be added to the displacement calculated.
atULtodueRSpportSi Support Redundant
If the ( atULtodueRSupport Redundant) is in the same direction with the
deformation, (∆support or θsupport), substitute the value of reaction with (+ve)
sign if not then the sign should be (–ve).
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-13
Ex: (5-3):
Use the method of consistent deformations to
analyze the frame shown in figure below if
the of external effects on support (C)
include, (1 cm) horizontal settlement, (2 cm)
vertical settlement and (0.002 rad), noting
that (EI=6×104
kN/m2), and support (A)
moved perpendicular to the roller surface a
displacement of (1 cm ), (→+, ↑+).
Solution:
1) Check the determinacy of the original structure.
The original structure is a frame structure: 1213cj3rb3 The
original structure is indeterminate to the 1st degree.
2) Choose the redundant that by removing it the structure will be determinate and
stable, let support (A) be the redundant needed; therefore, (X1) will be (RA).
3) Draw a diagram of the primary structure, and number it as (0), then calculate
all reactions and (M) due to external loads only, for each portion of the
structure.
4) Draw a diagram of the virtual structure, by applying a unit load in the direction
of the redundant, Roller A, and number it as (1), then calculate all reactions
and (m1), due to that unit load, for each portion of the structure.
Member Origin Limit M
1m
AB
A
40
0
x21
BC B
30
x10
24x
B
4 m
A
C
I
I
10 kN
1
1
0.002 rad
1 cm 2 cm
3 m
X1=RA
(The Original Structure)
B
4 m
A
C
I
I
10 kN
3 m
MC
Cy
Cy
(The Primary Structure)
1
1
B
4 m
A
C
I
I
10 kN
1
1
0.002 rad
1 cm 2 cm
3 m
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-14
5) Write the compatibility equations in terms of the deformations of the primary
and virtual structures, then solve them:
Cyx1 M002.0C100
2C
100
1
100
11S11X10
Cyx M002.0C100
2C
100
1
100
11S
2
1002.0
2
1
100
2
2
101.001.01S
009826.0001414.001414.00071.001.01S
3
0
2
3
0
1dxx40x10
EI2
1dx
2
4xx10
EI
1dx
EI
mM10
18090
EI2
1
2
340
3
310
EI2
1
2
x40
3
x10
EI2
110
233
0
23
EI
92.190
EI2
27010
3
0
2
4
0
2
3
0 1
24
0
22
1 dx16x8xdxxEI2
1dx
2
4xdx
2
x
EI
1dx
EI
m11
316
2
38
3
3
3
4
EI2
1x16
2
x8
3
x
3
x
EI2
111
2333
0
234
0
3
EI
165.574836933.21
EI2
111
009826.0EI
165.57X
EI
92.1901
kN7.136531.13X
56.589X165.5792.190
EI009826.0X165.5792.190
1
1
1
1
1 1 kN
B
4 m
A
C
I
I
3 m
m1
(The Virtual Structure, m1)
21
1 kN 21
1 kN
21Cx
1 kN
21C y
1 kN
21MC
1 kN
x
x
B
4 m
A
C
I
I
10 kN
3 m
MC=30 kN.m Cy=10 kN
Cx=0 kN
M
(The Primary Structure, M)
x
x
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-15
6) All reactions can be calculated
using equilibrium equations, based
on (RA=13.7 kN )
kN7.9Cx
0C2
1R0F xAx
kN7.19C
0C102
1R0F
y
yAy
0MA
03C4CM yxC
m.kN3.2047.937.19MC
5.4. Analysis of Indeterminate Trusses with the Method of Consistent
Deformations:
The degree of indeterminacy of trusses can be calculated as follows:
1) The total indeterminacy degree:
j2rbm .Tot
2) The external indeterminacy degree:
3crm .Ext
Shows how many reactions should be considered as redundant; (c) is the point
in the truss that when separated the truss will be divided into two trusses.
3) The internal indeterminacy degree:
3crj2rbmmm .EXT.Tot.Int
Shows how many internal members, (Truss bars), should be considered as
redundant.
The following simple example shows how this degree of indeterminacy for trusses
is calculated:
Ex: (5-4):
Use the method of consistent deformations to analyze the truss shown in figure.
A
B
C
D
10 kN
1
1
B
4 m
A
C
I
I
10 kN
3 m
MC=20.3 kN.m
Cy=19.7 kN
Cx=9.7 kN
RA RA=13.7 kN
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-16
Solution:
1) Check the degree of indeterminacy of the truss:
28104246j2rbm .Tot
The Truss is indeterminate to the 2nd
degree, total
indeterminacy.
13043crm .Ext
There is one external reaction as redundant.
112mmm .Ext.Tot.Int
There is one internal member as redundant.
2) Choose the horizontal reaction of support (A) as the
first redundant, and member (AB) as the second
redundant.
3) Draw a diagram of the primary structure, and
number it as (0), then calculate all reactions and (S)
due to external loads only, for each bar of the truss.
4) Draw a diagram of the virtual structures needed, by applying a unit load in the
direction of the redundant, and number them (1, 2, ….), then calculate all
reactions and (u1, u2, ….), due to that unit load, for each bar of the truss.
5) Write the compatibility equations in terms of the deformations of the primary
and virtual structures, then solve them:
1S12X11X10 21
2S22X21X20 21
EA
LuS20
EA
LuS10
21
EA
Luu2112
EA
Lu22
EA
Lu11
212
2
2
1
10 kN
A
B
C
D
(S) (u2)
A
B
C
D 1 kN
(u1)
A
B
C
D 1 kN
A
B
C
D X1
X2
10 kN
A
B
C
D
10 kN
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-17
Ex: (5-5):
Use the method of consistent deformations to analyze the truss shown in figure
below, due to a rise in temperature of (40oC) for bars (AB) and (BE) in addition to
the external applied loads, and support (E) settles vertically (1 cm); noting that
(EA=1×105 kN) and (α=12×10
-6/oC).
Solution:
1) Check the degree of indeterminacy of the truss:
210125248j2rb.detIn.T
The Truss is indeterminate to the 2nd
degree, total
indeterminacy.
13043cr.detIn.Ext
There is one external reaction as a redundant.
112.detIn.Ext.detIn.T.detIn.Int
There is one internal member as a redundant.
2) Since support (E) settles (1 cm), then choose the
vertical reaction (Ey) as the first redundant, and
member (CB) as the second redundant.
3) Draw a diagram of the primary structure, and number
it as (0), then calculate all reactions and (S) due to
external loads only, for each bar of the truss.
4) Draw a diagram of the virtual structures needed, by applying a unit load in the
direction of the redundant, and give number them (1, 2, ….), then calculate all
reactions and (u1, u2, ….), due to that unit load, for each bar of the truss.
(S)
D E
A
C
80 kN
B
4 m
3 m 3 m
60 kN
60 kN
80 kN
(u1)
D
3 m
E
A
C
1 kN
B
3 m
1 kN
1.5 kN
1.5 kN
(u2)
A
D
3 m
E C
B
3 m
1
0 kN
0 kN
0 kN 1
D E
A
C
80 kN
B
4 m
3 m 3 m
D E
A
C
80 kN
B
4 m
3 m 3 m
D E
A
C
80 kN
B
4 m
3 m 3 m
X1
X2
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-18
5) Write the compatibility equations in terms of the deformations of the primary
and virtual structures, then solve them:
1S12X11X10 21
2S22X21X20 21
LutEA
LuS20Lut
EA
LuS10 2
2
1
1
EA
Luu2112
EA
Lu22
EA
Lu11
212
2
2
1
Member L,
m S u1 u2 Su1L Su2L (u1)
2 L (u2)
2 L u1 u2 L
AB 3 0 -0.75 -0.6 0
0
1.6875
1.08
1.35
CD 3 0 0 -0.6 0
0 0 1.08 0
DE 3 -60 -0.75 0 135
0 1.6875 0 0
AC 4 0 0 -0.8 0
0
0
2.56
0
BD 4 80 1 -0.8 320
-256
4
2.56
-3.2
AD 5 -100 -1.25 1 625
-500
7.8125
5
-6.25
BC 5 0 0 1 0
0
0
5
0
BE 5 0 -1.25 0 0
0
7.8125
0
0
∑ 1080
-756
23
17.28
-8.1
m00672.000408.00108.010
25.1575.034010121080101
110
LutLuSEA
110
6
5
11
m008424.0000864.000756.020
056.03401012756101
120
LutLuSEA
120
6
5
22
m00023.0
101
23
EA
Lu11
5
2
1
m000081.0101
1.8
EA
Luu2112
5
21
m0001728.0
101
28.17
EA
Lu22
5
2
2
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-19
1......328X1.8X23
1000328.0X000081.0X00023.0
01.0X000081.0X00023.000672.0
21
5
21
21
2......4.842X28.17X1.8
10008424.0X0001728.0X000081.0
0X0001728.0X000081.0008424.0
21
5
21
21
kN4.50X
kN5.3X
2
1
6) After finding X1 and X2, to find the real values for the bar forces, substitute
theses values into the following equation:
k
1j
jj XuSF
Where:
F the real force in truss bar.
S the force in truss bar after removing the redundant, due to external loads.
ju the force in truss bar due to a unit load replacing the redundant.
jX the value of the redundant obtained.
TenskN865.324.506.05.375.00FAB
TenskN24.304.506.05.300FCD
CompkN375.574.5005.375.060FDE
TenskN32.404.508.05.300FAC
TenskN82.1164.508.05.3180FBD
CompkN025.1464.5015.325.1100FAD
TenskN375.44.5005.325.10FBE
CompkN4.504.5015.300FBC
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-20
Ex: (5-6):
Use the method of consistent deformations to analyze the truss shown in figure
below, due to a rise in temperature of (60oF) for bars (aB), (BC) and (Cd); noting
that (E= 30×102 kip/in
2), (A=10 in
2) and (α=1/150000/
oF).
Solution:
1) Check the degree of indeterminacy of the
truss:
62310j2rb.detIn.T
11213.detIn.T
The Truss is indeterminate to the 1st
degree, total indeterminacy.
03033cr.detIn.Ext
There is no external reaction as a redundant.
101.detIn.Ext.detIn.T.detIn.Int
There is one internal member as a redundant.
2) Choose member (bC) as the redundant,
(X1).
3) Draw a diagram of the primary structure,
and number it as (0), then calculate all
reactions and (S) due to external loads only,
for each bar of the truss.
4) Draw a diagram of the virtual structure, by applying a unit load in the direction
of the redundant, and give it number (1), then calculate all reactions and (u1),
due to that unit load, for each bar of the truss.
0 kN
(S)
20 in
a d
C B
15 in 15 in 15 in
b c
0 kN 0 kN (u1)
20 in
a d
C B
15 in 15 in 15 in
b c
1
1 0 kN
0 kN 0 kN
20 in
a d
C B
15 in 15 in 15 in
b c
20 in
a d
C B
15 in 15 in 15 in
b c
20 in
a d
C B
15 in 15 in 15 in
b c
X1
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-21
5) Write the compatibility equations in terms of the deformations of the primary
and virtual structures, then solve them:
1S11X10 1
LutEA
LuS10 1
1
EA
Lu11
2
1
Member L, in S u1 S u1 L (u1)2
L
ab 15 0 0 0 0
bc 15 0 -0.6 0 5.4
cd 15 0 0 0 0
BC 15 0
-0.6 0 5.4
Bb 20 0
-0.8 0 12.8
Cc 20 0
-0.8 0 12.8
aB 25 0
0 0 0
Bc 25 0
1 0 25
Cd 25 0
0 0 0
bC 25 0
1 0 25
∑ 0 86.4
Members subjected to temperature are: (aB, BC and Cd)
LutLuSEA
110 11
in0036.010
250156.025060150000
1Lut10 1
in00288.0101030
4.86
EA
Lu11
2
2
1
Tenskip25.1X0X00288.00036.0 11
6) After finding X1, to find the real values for the bar forces, substitute this value
into the following equation:
1111 XuXuSF
kN025.10Fab
CompkN75.025.16.0Fbc
kN025.10Fcd
CompkN75.025.16.0FBC
20 in
a d
C B
15 in 15 in 15 in
b c
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-22
CompkN125.18.0FBb
CompkN125.18.0FCc
kN025.10FaB
TenskN25.125.11FBc
kN025.10FCd
TenskN25.125.11FbC
Ex: (5-7):
Analyze the truss shown by the method of consistent deformations; Assume that
(E= 2×104 kN/cm
2) and (A=25 cm
2) for all members.
Solution:
1) Check the degree of indeterminacy of the
truss:
4236j2rb.detIn.T
189.detIn.T
The Truss is indeterminate to the
1st degree, total indeterminacy.
03033cr.detIn.Ext
There is no external reaction as a redundant.
101.detIn.Ext.detIn.T.detIn.Int
There is one internal member as a redundant.
2) Choose member (bc) as the redundant, (X1).
3) Draw a diagram of the primary structure, and
number it as (0), then calculate all reactions
and (S) due to external loads only, for each
bar of the truss.
4) Draw a diagram of the virtual structure, by
applying a unit load in the direction of the
redundant, and give it number (1), then
a
d
b
c 4 m
3 m
12 kN
a
d
b
c 4 m
3 m
12 kN
a
d
b
c 4 m
3 m
12 kN
X1
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-23
calculate all reactions and (u1), due to that unit load, for each bar of the truss.
5) Write the compatibility equations in terms of the deformations of the primary
and virtual structures, then solve them:
1S11X10 1
kN500000EA,
EA
Lu11,
EA
LuS10
2
11
Member L, m S u1 S u1 L (u1)2
L
ab 4 0 1.333 0.000 7.111
dc 4 -16 1.333 -85.333 7.111
ad 3 0 1.000 0.000 3.000
ac 5 20 -1.667 -166.667 13.889
bd 5 0
-1.667 0.000 13.889
bc 3 0
1.000 0.000 3.000
∑ -252 48.000
Members subjected to temperature are: (aB, BC and Cd)
m000504.0252500000
110
m000096.0500000
4811
0X000096.0000504.0 1
TenskN25.5X1
16 kN
(S)
16 kN 12 kN a
d
b
c 4 m
12 kN
3 m
(The Primary Structure, 0, S)
(u1)
a
d
b
c 4 m
3 m
1
1
0 kN
0 kN 0 kN
(The Virtual Structure, 1, u1)
Theory of Structures-1 CH-5: The Method of Consistent Deformations
Q-L (2017-2018) Ch.5-24
6) After finding X1, to find the real values for the bar forces, substitute this value
into the following equation:
11 XuSF
TenskN725.53/40Fab
CompkN925.53/416Fcd
TenskN25.525.510Fad
TenskN25.1125.53/520Fac
CompkN75.825.53/50Fbd
TenskN25.525.510Fbc
16 kN c
16 kN 12 kN a
d
b
4 m
12 kN
3 m