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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.


2Q-L ( 2017-2018) Ch.1-

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.


3Q-L ( 2017-2018) Ch.1-

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.

Theory of Structures-1 CH-2: Stability and Determinacy of Structures

Q-L (2017-2018) Ch.2-1

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


Q-L (2017-2018) Ch.2-2

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)


Q-L (2017-2018) Ch.2-3

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


Q-L (2017-2018) Ch.2-4

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)


Q-L (2017-2018) Ch.2-5

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


Q-L (2017-2018) Ch.2-6

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:


degree

(a)

(b)


Q-L (2017-2018) Ch.2-7

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


degree

Vertical Movement

Horizontal

Movement


Q-L (2017-2018) Ch.2-8

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


Q-L (2017-2018) Ch.2-9

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).


internal geometric instability involved, if (b+r = 2j).


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


Q-L (2017-2018) Ch.2-10

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).


internal geometric instability involved, if (3b+r = 3j+c).


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)


Q-L (2017-2018) Ch.2-11

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


Q-L (2017-2018) Ch.2-12

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

Theory of Structures-1 CH-3: Influence Lines for Statically Determinate Structures

Q-L (2017-2018) Ch.3-1

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


Q-L (2017-2018) Ch.3-2

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


Q-L (2017-2018) Ch.3-3

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


Q-L (2017-2018) Ch.3-4

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


Q-L (2017-2018) Ch.3-5

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)


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


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


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


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


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


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


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


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


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


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.


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


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


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


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


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


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


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


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


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)


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):


4

30 yB AM

4

10 yy BF

From sec (1-1)

12

50

4

1

5

30

aaFy

3) UL @ (u2):


2

10 yB AM

2

10 yy BF

From sec (1-1)

6

50

2

1

5

30

aaFy

4) UL @ (u4):


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


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


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


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


20

0

kN

M 30

00

kN

M 10

kN

M


20

0

kN

M 30

00

kN

M 10

kN

M


kN90RMLmaxA

kN90RMLmaxA

Ex: (3-11)


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)


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)


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


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


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


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


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:


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


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


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


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


Q-L (2017-2018) Ch.4-11

4) Tabulate the results to facilitate solution to include the information for each

portion as follows:


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


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


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


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).


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


Q-L (2017-2018) Ch.4-15


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


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


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.


portion as follows:


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


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


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


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:


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


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


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


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)


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


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


Q-L (2017-2018) Ch.4-26

rad5.11245

20

50uL

L

ttuL

L

12

aB

2

aB

aB


(θaB) will be counter clockwise, CCWrad5.112aB

.

2) Due to fabrication error:



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


(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


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:



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


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


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


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


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


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


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


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.


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.


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


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


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


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).


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


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


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


B

L

L/2

A

D

C

P

P/7

Ay=8P/7

MA=PL


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:


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


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).



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


B

D

C

I I

2I

10 kN/m

(The Primary Structure)

A

X3


Q-L (2017-2018) Ch.5-10


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


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


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).


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:


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).



structure.


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


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


Q-L (2017-2018) Ch.5-14



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


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


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.



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


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


Q-L (2017-2018) Ch.5-18



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


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


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.



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.


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


Q-L (2017-2018) Ch.5-21



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


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.



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


Q-L (2017-2018) Ch.5-23

calculate all reactions and (u1), due to that unit load, for each bar of the truss.



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)


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

theory of structures-1 b. e. 3233, 1st term 3rd class ... · text book elementary theory of...

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