Theory of Structures

1

Theory of Structures by T.A.Majid, K.K.Choong, M.M.Yusof

TOPIC 1

BASIC CONCEPT OF STRUCTURAL ANALYSIS

1.1 INTRODUCTION

A structure can be defined as a form or shape such as buildings, bridges, dams and walls, which are designed to resist any applied loads without losing its own strength capacity and any appreciable deformation. The fundamental purpose of structure is to transmit all applied loads to the point of support systems and ultimately through the foundations to ground.

Structural analysis can be explained as a calculation process carried out on a structure with the purpose of understanding the behavior of the structure under the action of a specified set of loads. Before proceeding to learn about how calculation is carried out, types of structures, their supports and the associated reactions will be first explained in this topic. Furthermore, concepts of planar structures, principle of equilibrium, free body diagram and statical determinacy will be also looked at.

1.2 TOPIC OUTCOMES

a) Able to identify and determine the reaction forces for different types of supports

b) Able to check for equilibrium and solve for reaction forces using the principle of equilibrium for 2D problems by using free body diagram.

1.3 TYPES OF STRUCTURES

Structures can be categorized as those of framed types and mass types. Framed structures resist the applied loads by virtue of their geometry, while mass structures are type of structures which are able to resist the applied loads by virtue of their weight.

The most commonly structural elements are beams, columns, arches, wall, trusses and foundations. Figure 1.1 (a) (f) show examples of these structural elements with typical loading.

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Figure 1.1: Some typical structural elements.

(f) Foundation (e) Truss

(d) Wall (c) Arch

(b) Column

(a) Beam

3


1.4 TYPES OF SUPPORTS

As mentioned earlier, structures could be understood as an assemblage of structural members. A structure must be properly supported and joined together so that the structure is stable and able to carry loads. It is impossible to construct a structure solely from one solid piece of materials. Structures are in almost all cases made from structural members which must be joined using connections. These connections can be used to join member to member as well as member to support. Three types of most commonly used connections are:

(a) Pin connection; (b) Fixed or Rigid connection; and (c) Roller connection.

Figure 1.2 shows two sketches of pinned connections which include connections at (a) support; and (b) between members

Figure 1.2: Non-zero relative rotation of pinned connections:

(a) at support; and (b) between members

One important characteristic of pinned connection is that relative rotation between connecting members or between member and supporting point is permitted. It is noted that relative rotation as depicted in Figure 1.2 has been highly magnified for clarify purpose. In actual structures, rotation is normally very small.

On the other hand, a fixed connection differs from a pinned one is the sense that relative rotation between connecting members or between member and support is not permitted. Figure 1.3 show examples of fixed connection.

(b) (a)

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Figure 1.3: Zero relative rotation of fixed connections:

(a) at support; and (b) between members

In the process of analysis, standard symbols are generally used to represent pin and fixed connections. They are shown in Figure 1.4 where symbols used to denote pin and fixed connections at support are shown in Figure 1.4 (a) and those for connections at joints between members in structures are shown in Figure 1.4 (b).

Figure 1.4: Symbols commonly used in analysis model to represent connections:

(a) at supports and (b) between members

Pinned Fixed

(b) Connections between members

Pinned

Fixed

(a) Connections at supports

(b)

(a)

5


The third type of commonly used connection is roller connection. Symbols used for roller connection are as shown in Figure 1.5.

Figure 1.5: Symbols for roller connections

Comparing with pin connections, rollers are:

(a) similar to pin connections since relative rotation is permitted; and

(b) different from pinned connections since only translation in the direction perpendicular to the supporting plane is not allowed. Translation in the direction parallel to supporting plane is allowed.

1.5 REACTION FORCES

All applied loading on a framed structure will be transferred to the support systems which will then provide the reacting force (or reactions) to maintain equilibrium. Some structures are constrained by supports that do not allow any rigid-body movement. Other support systems resist translational movement but no resistance to rotation. The behavior of the supports can have a critical effect on the structure proper and therefore cannot be ignored at the design stage.

In actual practice, it is necessary to make certain idealized simplifications regarding the nature of supports. The common types of support are fixed, pinned and rollers. Figure 1.6 shows various support conditions with standard symbols.

Type of Support Reactions No. of Unknown

3

Ry

RX

Mz

Fixed

(a)

6


2

1

Figure 1.6: Types of Support Systems and Reactions

At point of roller support, the corresponding structure is not free to translate in the direction perpendicular to the supporting plane. As for the case of pinned supports, the translation is not allowed in directions parallel and perpendicular to the supporting plane. For the third type of rigid supports, apart from no freedom to translate in directions parallel and perpendicular to the supporting plane, the rotation of structure at the point of support is also not allowed.

Due to the above restraints provided by the different types of supports, different types and number of reaction forces will occur at the supports. As shown in Figure 1.6 (a), at the point of rigid support, apart from reaction forces in directions parallel and perpendicular to the supporting plane, moment will also occur at the support point since the rotation of structure at the point of rigid support is restrained.

For the case of pinned support (Figure 1.6 (b)), reaction forces in directions parallel and perpendicular to the supporting plane will occur due to the restraints provided by the support in these two directions.

As shown in Figure 1.6 (c), since at a point of roller support, translation in the direction perpendicular to the supporting plane is restrained, reaction force will occur in that direction.

1.6 EQUATION OF EQUILIBRIUM

A body will move when subjected to a force or moment applied to it. If many forces and moments are applied to such a body, the resultant of all forces ( F) and all moment ( M) will cause the body to move.

A body is said to be in equilibrium if, initially at rest, it remains at rest when subjected to a system of forces and moments. If a body is in equilibrium, then all its members and parts are

Ry

Ry Roller

(c)

RX

Ry

RX

Ry Pinned

(b)

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also in equilibrium. Figure 1.7 shows a general 2D object which is acted upon by a series of loading F1, F2, , F4 and moments M1, M2, , M3. The object is lying in x-y plane.

Figure 1.7: A general planar structure acted upon by a series of loads

For the planar structure shown in Figure 1.7 to be in equilibrium, the following three equations must be satisfied:

= 0xF ; = 0yF ; = 0zM (1.1)

The three equations given in Eq. (1.1) have the following meanings:

= 0xF : Algebraic sum of the x-component of all forces acting on the structure is zero.

= 0yF : Algebraic sum of the y-component of all forces acting on the structure is zero.

= 0zM : Algebraic sum of the moments of all forces about any point in the plane of the structure and the moment of any couple acting on the structure is zero.

In general, the structure mechanics involves determination of unknown forces on the structures. Some of these structures can be completely analyzed by using these equations. These structures are known as statically determinate where it requires the number of independent reaction components to be equal or not exceeding the number of applicable independent equations of equilibrium.

On the other hand, if there are extra or redundant reaction components, then the structure is said to be statically indeterminate, i.e. the number of independent reaction is greater than the number of equilibrium equations available.

1.7 FREE BODY DIAGRAM

When writing down equilibrium equations as given in Eq. (1.1), we need the aid of a useful tool, namely free body diagram. Free body diagram (FBD) could be understood as a

M1

M2

M3

F1

F2 F3

F4

x

y

8


diagram representing the structure (members or the whole structure) where all supports have been removed and replaced with all forces and moment (including reactions) that act on the structure. Figure 1.8 shows a simple example of FBD for a beam subjected to a load P acting at the mid-length of the beam.

Figure 1.8: Example of FBD for a beam

Ax, Ay and By in Figure 1.8 represent reaction forces at supports A and B, respectively. These reaction forces are considered as external forces acting on the beam as shown in the accompanying FBD diagram. FBD could also be usefully employed to show forces which are internal to the structure. Figure 1.9 shows another example of FBD of a beam.

Figure 1.9: Example of FBD for a segment of a beam

In Figure 1.9 (a), the beam is imaginary cut at section a-a. The corresponding FBD diagram for this segment is shown in Figure 1.9 (b). In this figure, Ax and Ay are again reactions at

P

A B

A

Ay

Ax

(a)

(b)

a

a

M N

S

P

P

A B

A B

Ay

Ax

By

(a) The structure

(b) FBD

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support A while N, S and M acting are called internal forces. FBDs as shown in Figure 1.8 and 1.9 will be extensively used later in the calculation of forces.

1.8 STATICAL DETERMINACY

One of the purposes of structural analysis is to determine the unknown forces. In the process of calculating the unknown forces, equilibrium equations are used. In relation to the use of equilibrium equations in the calculation of unknown forces, an important concept called statical determinacy needs to be understood.

Statical determinacy is an important concept in structural analysis. It is a concept which is used to determine whether equilibrium equations alone are sufficient for the purpose of calculating all the unknown forces in a given structure. Statical determinacy could be explained as follows:

If all unknown forces can be determined by using only equilibrium equations, then that structure is called a statically determinate structure. On the other hand, a structure where the number of unknown forces is more than the number of equilibrium equations is called a statically indeterminate structure.

In order to elaborate further the concept of statical determinacy and its determination, example of a beam structure as shown in Figure 1.10 shall be looked at.

Figure 1.10: A cantilever beam subjected to point load.

The corresponding free body diagram for the beam shown in Figure 1.10 is shown in Figure 1.11.

Figure 1.11: Free Body Diagram for the cantilever beam in Figure 1.10

P

Ay

B Ax

MA

P

A B

10


Ax, Ay and MA are the three unknown reactions at support A due to load P applied at the mid span. These three reactions are to be calculated. In order to calculate Ax, Ay and MA, the three equilibrium equations as given in Eq. (1.1) will be used. Using the three equilibrium equations, the three unknown reactions could then be calculated. In this example, it can be seen that the number of equilibrium equations is the same as the number of unknown. And all the unknown forces could be determined.

Lets look at next example shown in Figure 1.12.

Figure 1.12: A beam fixed at both ends subjected to point load.

The difference between example in Figure 1.12 and the one in Figure 1.10 is that the other end of the beam is fixed. Free body diagram for the beam shown is Figure 1.12 could be drawn as shown is Figure 1.13.

Figure 1.13: Free body diagram for the beam shown in Figure 1.12

Again, in order to calculate the unknown forces as shown in Figure 1.13, the set of three equilibrium equations can be used. However, compared to the number of equilibrium equations available which is 3, the number of unknown forces in this case has become 6, i.e. Ax, Ay, MA, Bx, By and MB. There is no other equilibrium equation which can be used. As a consequence, not all the unknown forces can be solved. Three extra equations are needed in order to determine all the six unknown reaction forces.

P

Ay

Ax

MA MB

Bx

By

P

A B

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From the above two examples, it can be seen that:

i. In example shown in Figure 1.10 and 1.11, the set of three equilibrium equations alone is sufficient to determine all the unknown forces. Thus, this is an example of statically determinate structure.

ii. In example shown in Figure 1.12 and 1.13, the set of three equilibrium equations alone is not sufficient to determine all the unknown forces. Thus, this is an example of statically indeterminate structure. Three extra equation (= 6 3) are required in order that all unknown forces in the example shown in Figure 1.12 can be determined. These extra numbers of equations are given the name of degree of indeterminacy.

The procedures to check for statical determinacy of a given structure can be summarized as follows:

1. Draw free body diagram of the structure. 2. Compare the number of unknown forces with the number of equilibrium equations.

Let:

= number of unknown forces = number of equilibrium equations

Then if = : Statically determinate structure > : Statically indeterminate structure .. (1.2)

3. For planar structure, the maximum number of equilibrium equation available is 3. Hence condition as given in Eq. (1.2) could be further simplified as follows:

If 3= : Statically determinate structure 3> : Statically indeterminate structure .. (1.3)

4. If a structure is statically indeterminate, the difference between and as give below:

=i .. (1.4)

is called degree of statical indeterminacy. It could be understood as the number of additional equations required together with the use of equilibrium equations in order that all unknown forces could be determined.

In order to further understanding of the concept of statical determinacy and its determination, two more examples will be looked at.

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Example 1.1

Figure 1.14 shows a two-span continuous beam. Support A is a fixed support, support B is a pinned support whereas support C is a roller support. Check for the statical determinacy of the beam.

Figure 1.14: A two-span continuous beam

Solution:

1. Draw free body diagram.

Figure 1.15: Free body diagram.

2. Count the number of unknown forces in Figure 1.15.

The unknown forces are: Ax, Ay, MA, By, Bx and Cy. Hence the number of unknown forces = 6

3. Compare with the number of equilibrium equation, .

= 3, for planar structure. Hence, > . Beam is a statically indeterminate structure.

4. Degree of statical indeterminacy.

336 === i

P1 P2 P3 P4

Ay

Ax

MA

Bx

By Cy

P1

A C B

P2 P3 P4

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Example 1.2

Figure 1.16 shows a simply supported planar truss structure. Support A is a pinned support and support B is a roller support. Check the statical determinacy of the truss.

Figure 1.16: A simply supported truss

Solution:


Figure 1.17: Free body diagram

2. Count the number of unknown forces .

The unknown forces are: Ax, Ay and By. Hence, = 3

3. Compare with the number of equilibrium equation, .

= 3, for planar structure. Hence, = . Truss is a statically determinate structure.


033 === i

P1 P2 P3 P4 P5

Ay

Ax

By

A B

P1 P2 P3 P4 P5

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Example 1.3

Figure 1.18 shows two beams which are pin jointed at B. The support at A is a fixed support, whereas at C and D are pinned support. Check the statical determinacy of the truss.

Figure 1.18: Two beams with pinned jointed

Solution:



2. Count the number of unknown forces.

The unknown forces are: Ax, Ay, MA, Bx, By, Cx, Cy, Dx and Dy. Hence, = 9

3. Compare with the number of equilibrium equations . Since, there are two beams.

= 2 (3) = 6 Hence, > . Beam is a statically indeterminate structure.


369 === i

P1 P2

Ay

Ax

MA

Cx

Cy

Dx

Dy By

Bx

By

Bx

P1

A C B

P2

D

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1.9 COMPUTATION OF REACTIONS USING EQUATIONS OF EQUILIBRIUM

After reviewing basics about supports such as restraints and corresponding reaction in Section 1.5, how to actually compute reactions will be next looked at. The set of equilibrium equations to be used is as follows:

= 0xF : Algebraic sum of the x-component of all forces acting on the structure is zero.

= 0yF : Algebraic sum of the y-component of all forces acting on the structure is zero.

= 0zM : Algebraic sum of the moments of all forces about any point in the plane of the structure and the moment of any couple acting on the structure is zero.

Since ability to formulae correctly the above set of equilibrium equations is essential for the computation of reaction forces, it is useful at this point to review how to consider forces and moment in equilibrium equations. For this purpose, it will good to start with Example 1.4 and 1.5.

Example 1.4

Compute the reaction forces for the cantilever beam shown in Figure 1.20.

Figure 1.20: Cantilever beam

Solution:


P2

A B C D

P1

M1

L1

L2

L

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2. In Figure 1.21, there are three external loads acting on the beam: i. Concentrated loads P1 and P2 ii. Concentrated moment M1

3. Since the beam is fixed supported at D, there are three reactions at support D, i.e. horizontal reaction force DX, vertical reaction Dy and reaction moment MD, which need to be determined by the use of equilibrium equations.

4. Formulation of equilibrium of forces in horizontal and vertical directions, = 0xF and = 0yF will be first looked at. Referring to Figure 1.21, the corresponding equations will be as follows:

+ = 0xF : 0cos1 = xDP

+ = 0yF : 0sin 21 =+ yDPP

In the above two equations, note that any force whose direction of action is neither in horizontal or vertical directions has to be resolved into components in these two directions, e.g. P1 cos and P1 sin.

5. The next equation to be formulated is equilibrium of moment. The following two formulations of equilibrium of moment will be looked at:

+ M at A=0 : ( ) ( ) ( ) 0sin 22111 =++ LDLPLPMM yD

+ M at D=0 : ( ) ( ) 0sin 22111 =++ LLPLLPMM D

From the above two formulations, it can be seen that: i. For concentrated loads: if their directions of action pass through the point

around which moment is taken, then these forces have no contribution to the

P2

A B C D

P1

M1

L1

L2

L

MD

Dx

Dy

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equilibrium of moment, e.g. Dy, Dx and P1 cos in M at D =0 and Dy, P1 cos in M at A =0.

ii. For concentrated moments: they will always contribute in equilibrium of moment no matter which point is taken as reference for the formulation of the equilibrium equation, e.g. MD and M1 appear in both formulations of M at A =0 and M at D =0. It is important to remember this fact as failure to consider couple will result in wrong values being calculated for other reactions.

Example 1.5

The simply supported beam shown in Figure 1.22 is subjected to distributed loads. Compute the reaction for the beam.

Figure 1.22: Simply supported beam

Solution:



B C D

L1

L2

L3

L4

L

E

w1 w2

w3

Ay

Ax

Fy

A B C D

L1

L2

L3

L4

L

E F

w1 w2

w3

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2. In Figure 1.23, two types of distributed loads are present: i. Uniformly distributed load : acting uniformly at intensity w1 ii. Linearly distributed load : increasing from load intensity w2 to w3

It is noted that the intensity of distributed load is expressed in unit of force/length, e.g. N/m of kN/m.

3. Since the support A is pinned, there are two reactions, i.e. Ax and Ay; support F is a roller support, the reaction is Fy.

4. Formulation of equilibrium equation of forces in horizontal and vertical directions, = 0xF and = 0yF will again looked at.

+ = 0xF : 0=xA

+ = 0yF : ( ) ( )( )3432121 21 LLwwLLwFA yy ++

Since there is no forces with horizontal component, Ax = 0. As for Fy = 0, it can be seen from the corresponding equation that, distributed load is considered as equivalent to a point load with magnitude equal to the area under the distributed load diagram, e.g. i. For uniformly distributed load: magnitude is equal to area of a rectangular

since the corresponding load diagram is a rectangular; and ii. For linearly distributed load: magnitude is equal to the area of a

trapezium since the load diagram is a trapezium.

5. After = 0xF and = 0yF , equilibrium of moment is next looked at. A formulation of = 0M will be studied. Taking moment at support A.

+ M at A=0 :

( ) ( ) ( )( ) ( )

+

+ 3343423112121 3

221

21 LLLLLwwLLLLLw

( ) ( ) ( ) 021

334342 =+

+ LFLLLLLw y

Moment is force X perpendicular distance to the point where moment is taken. Keeping this in mind and by studying the formulation as shown above, it can be seen that contribution by distributed load to moment equilibrium equation is equal to equivalent load corresponding to the distribution load X distance of action of that load

L2-L1

w2 w3 w1

L4-L3

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to the point where moment is taken. Figures below show the individual contribution from the distributed load to the moment equilibrium equation.

( ) ( )

+ 112121 2

1 LLLXLLw

( )( ) ( )

+ 3343423 3

221 LLLXLLww

( ) ( )

+ 334342 2

1 LLLXLLw

D

L3

L4

E F

Ay

Ax Fx

B C

L1

L2

Ay

Ax

20


In the above illustration of how moment due to distributed load is considered, the sign of moment has been omitted. The sign will depend of the positive sense which has been adopted during the formulation of the moment equilibrium equation. It is again emphasized that moment due to distributed load is:

Point load equivalent to area of distributed load diagram X distance of the equivalent point load to point where moment is to be taken.

In the above illustration, point load equivalent to area of distributed load diagram has been denoted using broken arrows.

Example 1.6

Determine the reaction components for the structure shown in Figure 1.24.

Figure 1.24

A B

20 kN

10 kN 3m

3m 6m

D

L3

L4

E F

Ay

Ax Fx

21


Solution:


2. Reactions

There are three reactions to be determined: Ax, Ay and By.

+ = 0xF : 020 =+ xA kNAx 20=

Taking moment at A

+ M at A=0 : ( ) 06103 =+ yB kNBy 5=

+ = 0yF : 010 =+ yy BA 0105 =yA

kNAy 15=

It is noted that in the above calculation, the value By is found to be negative. A negative solution indicates that the direction of reaction assumed at the start of solution process is incorrect.

3. The computed reactions.

A B

20 kN

10 kN 3m

3m 6m

Ax

Ay By

22


A B

20 kN

10 kN 3m

3m 6m

20 kN

15 kN 5 kN

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Exercise 1

For the frame shown in Figure 1.25(a), (b) and (c), check the statical determinacy.

Figure 1.25

Exercise 2

Obtain the reaction components of the beam shown in Figure 1.26.

Figure 1.26

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Exercise 3

Determine the determinacy and the corresponding degree of indeterminacy where applicable for the following structures.

Figure 1.27

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TOPIC 2

ANALYSIS OF PLANE TRUSSES

2.1 INTRODUCTION

A truss is an assemblage of straight members connected at their ends by flexible connections to form a rigid configuration. Because of their light weight and high strength, trusses are widely used, and their applications are supporting bridges and roof of buildings.

In this topic, we will develop the procedures for analyzing statically determinate trusses. The objective of analyzing the trusses is to determine the reactions and member forces. There are many ways to do this. We will be using 2 methods to carry out the analysis where we will consider the equilibrium equations of the whole or parts of the structures. By highlighting or considering only parts of the structure, we will analyze the FBD (free body diagram) and solve for the unknowns. The two methods that will be used:

a) Method of Joints

b) Method of Section

2.2 TOPIC OUTCOMES

a) Able to classify statical determinacy of plane trusses and check geometric stability.

b) Able to determine member forces using joints and section methods.

2.3 COMMON TYPES OF TRUSSES

Trusses are used to carry loads over relatively long spans compared to ordinary beams. They are widely used for roofs over large open areas, such as in warehouses, sport centres, highway, railroad bridges or industrial buildings as shown is Figure 2.1(a) and 2.1(b).

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Figure 2.1(a): Roof truss

Figure 2.1(b): Truss bridge

2.4 CLASSIFICATION OF PLANE TRUSSES

If all the members of a truss and the applied loads lie in a single plane, the truss is called a plane trusses. For plane trusses, the applied loading acts on each truss in its own plane. A typical forms of bridge and roof trusses, many of which have been named after their original designers, are shown in Figure 2.2 below:

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Figure 2.2: Common bridge and roof trusses

2.5 DETERMINANCY AND STABILITY OF PLANE TRUSSES

If all the members of a truss and the applied loads lie in a single plane, the truss is called a plane trusses. Statical determinacy of a plane truss can be checked using equilibrium equations.

A truss is considered to be statically determinate if all of its member forces of reactions can be determined by using the equations of equilibrium. If a plane truss contains m members, j joints, and is supported by r reactions, then if

jrm 2+ the truss is indeterminate truss

28


Statically indeterminate trusses have more members and/or external reactions than the minimum required for stability. The excess members and reactions are called redundants, and the number of excess members and actions is referred to as the degree of static indeterminacy, i, which can be expressed as

( ) jrmi 2+= (2.2)

Example 2.1

Classify each of the plane trusses shown in Figure 2.3 as statically unstable, determinate or indeterminate.

Solution:

The truss shown in Figure 2.3(a) contains 20 members and 12 joints. m = 20 2j = 24 r = 3

Hence, jrm 2

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(b)

The truss shown in Figure 2.3(c) contains 25 members and 12 joints. m = 25 2j = 24 r = 3

Hence, jrm 2>+ the truss is statically indeterminate.

(c)

Figure 2.3

2.6 SIGN CONVENTION AND MEMBER FORCES REPRESENTATION

The sign convention used in the analysis of plane trusses are described in three equations of equilibrium as shown in Figure 2.4:

= 0xF ; the summation of horizontal forces are equal to zero. The horizontal force is assumed positive if it is to the left.

= 0yF . ; the summation of vertical forces are equal to zero. The vertical force is assumed positive if it is upward.

= 0ZM ; the summation of moment at any point are equal to zero. The clockwise moment is positive.

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Figure 2.4: Sign convention.

The internal axial forces at any section of the truss member is equal in magnitude but opposite in direction to the resultant of the components in the direction parallel to the axis of the truss of all the external loads. A tensile member axial force is always represent by an arrows pulling away from the joint and compressive member axial force by an arrows pushing towards the joint as shown in Figure 2.5. It is usually convenient to assume the unknown member forces to be tensile.

Figure 2.5: Internal axial member forces.

2.7 METHOD OF JOINTS

The following step-by-step procedure can be used for the analysis of statically determinate simple plane trusses by the method of joints.

1. Check the trusses for static determinacy, as discussed in preceding chapter. If the truss is found to be statically determinate and stable, proceed to step 2. Otherwise, end the analysis at this stage. (The analysis of statically indeterminate trusses is considered in Part Four of this module) 2. Identify by inspection any zero-force member of trusses. 3. Determine the slopes of the inclined members (except the zero-force member) of the trusses.

y, Positive

x, Positive

Counter-clockwise moment,

Positive

P

P

P

P

Tension (Positive)

Compression (Negative)

31


4. Draw the free-body diagram of the whole truss, showing all external loads and reactions. Write zeros by the members that have been identified as zero-force members. 5. Examine the free-body diagram of the truss to select a joint that has no more than two

unknown force (which must not be collinear) acting on it. If such a joint is found, then go directly to the next step. Otherwise, determine reactions by applying the three equations of equilibrium and equations of condition (if any) to the free body of the whole truss; then select a joint with two or fewer unknowns, and go to the next step.

a. Draw a free-body diagram of the select joint, showing tensile force by arrows pulling away from the joint and compressive force by arrows pushing into the joint. It is usually convenient to assume the unknown member forces to be tensile.

b. Determine the unknown force by applying the two equilibrium equations = 0xF and = 0yF . A positive answer for a member force means that the member is tension, as initially assumed, whereas a negative answer indicates that the member is in compression.

6. If all the desired member forces and reactions have been determine, then go to the next step. Otherwise, select another joint with no more than two unknown, and return to step 6.

7. If the reaction were determine in step 5 by using the equations of equilibrium and condition of the whole truss; then apply the remaining joint equilibrium equations that have not been utilized so far to check the calculations.

IDENTIFICATION OF ZERO-FORCE MEMBERS

Because trusses are usually designed to support several different loading conditions, it is not uncommon to find members with zero forces in them when a truss is being analysed for a particular loading condition. Zero-force members are also added to trusses to brace compression members against buckling and slender tension members against vibrating. The analysis of trusses can be expedited if we can identify the zero-force members by inspection. Two common types of member arrangements that result in zero-force members are the following:

1. If only two non-collinear members are connected to a joint that has no external loads or reactions applied to it, then the force in both members is zero. (Figure 2.6a)

2. If three members, two of which are collinear, are connected to a joint that has no external loads or reactions applied to it, then the force in the member that is not collinear is zero. (Figure 2.6b)

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Figure 2.6: Identification of Zero-Force Member

Example 2.2

Using method of joints, determine the force in each member trusses shown in Figure 2.7.

Figure 2.7

Solution:

B

A E

C

D

L

L

L L

L L L

60 kN

120 kN

y

x

FAB

FAB

(a)

y

x

FAB

FAC

(b)

FAD

33


Free Body Diagram

+ MA = 0 = Dy(2L) - 60 (2L L cos ) - 120 (L) 2Dy = 120 60 cos + 120 Dy = 60 30 cos + 60 But = 60 Dy = 120 30 x 0.5 Dy = 105 kN

+ Fy = 0 = Ay 120 -60 + Dy

Ay = 120 + 60 + Dy Ay = 75 kN

+ Fx = 0 Ax = 0

Joint A

Fy = 0 75 + FAB cos 60 = 0 FAB = -86.6 kN

Fx = 0 FAE + FAB cos 60 = 0 FAE = + 86.6 x 0.5

FAE = + 43.3 kN

B

A

E

C

D

L

L

L L

L L L

60 kN

120 kN

Ay

Ax

Dy

2L L cos

FAB

FAE

75 kN

A

34


FBC

FBE

B

86.6

Joint B

Fy = 0 -FBE sin 60 + 86.6 sin 60 = 0 FBE = 86.6 kN

Fx = 0 FBC + 86.6 cos 60 + 86.6 cos 60 = 0 FBC = -86.6(0.5) x 2 FBC = -86.6 kN

Joint D

Fy = 0 105 + FDC sin 60 = 0 FDC = -121.24 kN

Fx = 0 -FDE + 121.24 cos 60 = 0 FDE = 62.2 kN

Joint C

Fy = 0 -60 + 121.24 sin 60 - FCE sin 60 = 0 FCE = 51.96 kN

FDC

FDE

105 kN

D

121.24 kN FCE

86.6 kN

60 kN

C

35


Checking Joint E

Fx = 0 -43.3 + 60.62 86.6 cos 60 + 51.96 cos 60 = 0 ok

Fy = 0 86.6 sin 60 + 51.96 sin 60 - 120 = 0 ok

Member Force (kN) Condition (+/-) AB 86.6 - Compression BC 86.6 - Compression CD 121.24 - Compression DE 60.62 + Tension EA 43.3 + Tension BE 86.6 + Tension CE 51.96 + Tension

Final Diagram

B

A

E

C

D

86.6

60 kN

120 kN

86.6

51.96 121.24

60.62 43.3

86.6

75 kN 75 kN

120 kN

E 60.62 kN 43.3 kN

86.6 kN 51.96 kN

36


Example 2.3

Using method of joints, determine the force in each member trusses shown in Figure 2.8.

Figure 2.8

Solution:

Free Body Diagram

+ MA = 0 = -10(3) - 15(3) - 30(6) - 15(9) + Ey Ey = 43.33 kN

10 kN

A

E

D C

B

F G

3m 3m

4m

3m

Ax

Ay Ey

15 kN 30 kN 15 kN

53.13

53.13 53.13

10 kN/m

10 kN

A E

D C

B

F G

3m 3m

4m

3m

37


+ Fy = 0 = Ay 10 15 30 15 + 43.33

Ay = 26.67 kN

+ Fx = 0 Ax = 0

Joint A

Fy = 0

26.67 + 54

FAB = 0

FAB = -33.33 kN

Fx = 0

FAG 33.33

53

= 0

FAG = 20 kN

Joint G

Fy = 0 FGB 10 = 0 FGB = 10 kN

Fx = 0 -20 + FGF = 0 FGF = 20 kN

Joint B

Fy = 0

33.33

54

- 15 10 FBF

54

= 0

FBF = 2.08 kN

Fx = 0

33.33

53

+ 2.08

53

FBC = 0

FBC = -21.246 kN

FAB

FAG

26.67 kN

A 0

FGB

FGF

10 kN

G

20 kN

15 kN

FBC

10 kN

B

33.33 kN

FBF

38


Joint D

Fy = 0 -15 - FDE = 0 FDE = -15 kN

Fx = 0 FDC = 0

Joint E

Fy = 0

54 FEC - 15 + 43.33 = 0

FEC = -35.413 kN

Fx = 0

- FEF + 35.413

53

= 0

FEF = 21.248 kN

Joint F

Fy = 0

2.08

54

+ FCF = 0

FCF = -1.664 kN

15 kN

FDE

D FDC

15 kN

43.33 kN

E FEF

FEC

21.248 kN F 20 kN

FCF 2.08 kN

39


Checking Joint C

Fy = 0

-30 + 1.664 + 33.413

54

= 0 OK

Fx = 0

21.246 35.413

53

= 0 OK

Member Force (kN) Condition (+/-) AB 33.33 - Compression BC 21.25 - Compression CD 0 DE 15 - Compression EF 21.25 + Tension FG 20 + Tension GA 20 + Tension BG 10 + Tension BF 2.08 + Tension CF 1.67 - Compression CE 35.42 - Compression

Final Diagram

10 kN

A

E

D C

B

F G

3m 3m

4m

3m

26.67 kN 43.33 kN

10 kN/m

20 20 21.25

33.33 10 1.67 15

2.08 35.42

21.25 0

30 kN

1.664 kN

C 0 kN 21.246 kN

35.413 kN

40


2.8 METHOD OF SECTIONS

This method is an effective method when the forces in some members of a truss are to be determined. Often we need to know the force in just one member with the greatest force in it, for example and the method of sections will yield the force in a particular member without the labour of working out the forces in the rest of truss.

The method of sections involves cutting the truss into two portions by passing an imaginary section through the members whose forces are desired. The desired member forces are then determined by considering the equilibrium equations of one of the two portions of the truss.

There are only three equilibrium equations available, so they cannot be used to determine more than three unknown force. Thus, in general, section should be chosen that do not pass through more than three members with unknown force.

Procedure for Analysis

The following step-by-step procedure can be used for determining the member forces of statically determinate plane trusses by the method of sections.

1. Select the section that passes through as many members as possible whose forces are desired, but not more than three members with unknown forces. The section should cut the truss into two parts.

2. Although either of the two portions of the truss can be used for computing the member forces, select the portion that will require the lease amount of computational effort in determining the unknown forces. To avoid the necessity for the calculation of reactions, if of the two portions of the truss does not have any reactions acting on it, then select this portion for the analysis of members force and go to the next step. If both portions of the truss are attached to external supports, then calculate reaction by applying the equations of equilibrium and condition (if any) to the free body of the entire truss. Next, select the portion of the truss for analysis of member forces that has the least number of external loads and reactions applied to it.

3. Draw the free-body diagram of the portion of the truss selected, showing all external loads and reactions applied to it and the forces in the members that have been cut by the section. The unknown member forces are usually assumed to the tensile and are, therefore, shown on the free-body diagram by arrows pulling away from the joints.

4. Determine the unknown forces by applying the equations of equilibrium. This can be achieved by applying the of equilibrium equations 0,0( == yx FF = )0M at any joint location either of the two portions of the truss.

5. Apply an alternative equilibrium equation, which was not use to compute member forces, to check the calculations. This alternative equation should preferably involve all three member forces determined by the analysis. If the analysis has been perform correctly, and then this alternative equilibrium equation must be satisfied.

41


Example 2.4

Using method of sections, determine the force in members AB, AG, HG, BC, CF and EF for the truss shown in Figure 2.9.

Figure 2.9

Solution:

Free Body Diagram

4m

B

C

D E

F

G

3m

3m

3m

5 kN

5 kN

5 kN

20kN

15 kN

10 kN

Hy

Ax

Ay

a a

b

b

4m

A H

B

C

D E

F

G

3m

3m

3m

5 kN

5 kN

5 kN

20kN

15kN

10 kN

42


+ MA = 0 = -10(3) - 15(6) - 20(9) - 5(3) - 5(6) - 5(9) + 4 Hy

Hy = 97.5 kN

+ Fy = 0 = Ay + 97.5

Ay = -97.5 kN

+ Fx = 0 = Ax + 10 + 15 + 20 + 5 + 5 + 5 Ax = -60 kN

Section a-a

+ MA = 0 = 97.5 (4) + FHG (4) FHG = -97.5 kN

+ Fy = 0 = FAB (4) + 60 (3) 97.5 (4)

FAB = 52.5 kN

+ Fx = 0 = -60 + FAG

54

FAG = 75 kN

A

G

H

B

FAGFAB FHG

60 kN

97.5 kN 97.5 kN

3m

4m

43


Section b-b

+ MF = 0 = -20 (3) - 5 (3) + FCB (4) FCB = 18.75 kN

+ MC = 0 = -20 (3) - 5 (3) - FEF (4) FEF = -18.75 kN

+ MB = 0 = -15 (3) - 20 (6) - 5 (6) - FCF (3) + 18.75 (4) FCF = -40 kN

3m

3m

20kN

15kN

5kN D E

C F

B G

FCB

FCF

FEF

4m

44


2.9 INTRODUCTION TO OTHER TYPES OF PLANAR STRUCTURES

Although a great majority of trusses can be analyzed as plane trusses, there are some truss systems such as transmission towers and lattice domes (Figure 2.10) that cannot be treated as plane trusses because of their shape, arrangement of members, or applied loading.

Figure 2.10: Lattice dome

2.9.1 Planar trusses: Compound and Complex

Compound trusses are constructed by connecting two or more simple plane trusses to form a single rigid body as shown in Figure 2.11. Analysis of compound trusses can usually be expedited by using a combination of the method of joints and method of sections described in the preceding sections.

Figure 2.11: Compound Trusses

Trusses that can be classified neither as simple trusses nor as compound trusses are referred to as complex trusses. Two examples of complex trusses are shown in Figure 2.12. From an

45


analytical viewpoint, the main difference between simple or compound trusses and complex trusses stems from the fact that the methods of joints and sections, as described previously, cannot be used for the analysis of complex trusses.

Figure 2.12: Complex trusses

2.9.2 Space trusses

Space trusses, because of their shape, arrangement of members, or applied loading, cannot be subdivided into plane trusses for the purposes of analysis and must, therefore, be analyzed as three-dimensional structures subjected to three-dimensional force system. To simplify the analysis of space trusses, it is assumed that the truss members are connected at their ends by frictionless ball-and-socket joints, all external loads and reactions are applied only at the joints, and the centriodal axis of each member coincides with the line connecting the centres of the adjacent joints. Because of these simplifying assumptions, the members of spaces trusses can be treated as axial forces members.

m = 17 , j =10, r=3

m + r = 2j

(b)

m = 9 , j =6, r=3

m + r = 2j

(a)

46


Figure 2.13: Space trusses

47


Exercise 1

a) Figure 2.14 (a) shows four different types of plane trusses. Check for the statical determinacy of the trusses.

b) Figure 2.14 (b) shows a plane trusses with pinned supports at A and G. Find the reactions at supports A and G. Identify any zero force member.

c) Determine the force in member BC, Ac and BD for the truss shown in figure 2.14 (b) by using method of sections and classify whether they are in tension or compression.

d) Determine the forces in member GF and FD for the truss shown in Figure 2.14 (b) by using method of joints. Classify whether they are in tension or compression.

Figure 2.14 (a)

48


Figure 2.14 (b)

Exercise 2

a) Figure 2.15 shows a pinned jointed truss with two supports at A and D where A is roller and D is pinned. Check for the statical determinacy of the truss.

b) Determine forces in member BC, BE and DE for the truss shown in Figure 2.15 by using method of sections and classify whether they are in tension or compression.

c) Determine forces in member CF, EF and CE by using method of joints and classify whether they are in tension or compression.

Figure 2.15

49


Exercise 3

a) Describe briefly three characteristics of truss structures and the two methods of analyzing the truss structure.

b) Determine the member forces DF, DE and CE for the truss shown in Figure 2.16 by the method of sections and classify whether they are in tension or compression.

c) Find the member forces of AB, AC and BC using method of joints and classify whether they are in tension or compression.

Figure 2.16

50


TOPIC 3

ANALYSIS OF STATICALLY DETERMINATE BEAMS AND FRAMES

3.1 INTRODUCTION

A beam is a structural element that is capable of withstanding load primarily by resisting bending. The bending force induced into the material of the beam as a result of the external loads, own weight and external reactions to these loads is called a bending moment. A structural frame is complex of columns, and beams, girders, spandrels, and trusses connected to one another and to the columns anchored in a foundation, as well as other components or members necessary for the stability of a structure.

In this topic, the analysis of beams and frames will be looked at. Specifically, how to obtain information about shear force and bending moment in beams and frames will be studied.

3.2 TOPIC OUTCOMES

a) Able to determine statical determinacy of structures. b) Able to draw shear force and bending moment diagrams for statically determinate

beams and frames using equilibrium equations. c) Able to draw shear force and bending moment diagrams for statically determinate

beams and frames using direct integration method.

3.3 INTERNAL FORCES IN BEAMS

Figure 3.1 shows an example of a simply supported beam.

Figure 3.1: A simply supported beam is subjected to point load P

When loads act on truss structures, members in the structure concerned will be in the state of either axial compression or tension. On the other hand, when loads act on beams, members will be bent as illustrated in Figure 3.2.

A B

P

51


Figure 3.2: Bending deformation of a simply supported beam when subjected to point load P

In other words, beams can carry loads by means of bending action. Due to bending deformations, forces will develop within the numbers of beam structures. Such forces are called internal forces. Internal forces developed in members of beam are shear force and bending moment which shall be denoted as S and M, respectively. Figure 3.3 shows the internal forces in a beam.

Figure 3.3: Internal forces in a beam shear force (S) and bending moment (M)

S and M are defined with respect to section of a beam as follows:

i. S is defined to be acting parallel to a section and pointing in upward or downward direction.

ii. M is defined to be acting around an axis perpendicular to the plane of the beams and rotating counter-clockwise or clockwise.

Figure 3.4 shows the definition of shear S and bending moment M over a section beam. Axis x in Figure 3.4 represents the axis of the beam.

P

A

M S M S

A B

P

Bending deformation

52


Figure 3.4: Definition of shear S and bending moment M on a section of a beam

3.4 SIGN CONVENTION AND COMPUTATION OF SHEAR AND BENDING MOMENT

Before proceeding to explain the importance of sign convention, a simple calculation of internal forces in a beam (Figure 3.5) is first looked at. For the purpose of calculation, the concept of free body diagram as explained in Topic 1 is used.

The beam is first cut through section a-a and then separated into two parts as shown in Figure 3.5. Internal force S and M are then indicated on both sections belonging to the parts to the right and left of the cut location. Making use of principle of equilibrium,Fx =0 , Fy =0 and Mz =0 , S and M could be then calculated. There are two choices of free body diagram which could be used: the one on the left or right of the cut section. Results of calculation using each of the free both diagram are shown. Values of S and M are found to be same for both sets of calculation.

Figure 3.5: A simple example showing the computation of internal forces S and M in a beam

A

30 kN

B

a

a

3 m 3 m

(a) A cantilever beam

Ay

M

M M

S

S

(b) Free body diagram

180 kNm

M

S 30 kN

S = 30 kN

M = -90 kNm

30 kN

M

S 30 kN

S = 30 kN

M = -90 kNm

(c) Left section (d) Right section

A B

P

A

M

S

x

53


Before moving to the next section, an example illustrating the use of positive sign convention in the calculation of internal forces in beam is shown in Figure 3.6.

Figure 3.6: Meaning of positive sign convention for S and M

Example 3.1

For the beam shown in Figure 3.7, calculate the internal forces S and M at the sections x-x, y-y and z-z.

Figure 3.7

A B C D

X

X

y

y

z

z

1.5m 4.5m 1m

0.75 kN/m

x-x : just to the right of A

y-y : just to the left of B

z-z : just to the left of D

S

S

M M

Positive S: Portion of the member to the left of the section tends to be pushed upward relative to the portion on the right of the section

Positive M: Beam tends to be bent concave upwards, causing compression on the upper fibres and tension in the lower fibres of the beam at the section

54


Solution:

1. Draw free body diagram and calculate the reaction forces.

+ = 0xF : 0=xB

Take moment at C.

+ M at C=0 : ( )[ ] 0125.55.575.05.45.4 =

+ yB

kNBy 604.1=

+ = 0yF : ( ) 05.575.0 =+ yy CB kNC y 521.2=

2. Section x-x

Section y-y

M = 0

S =0

M

S

A

A M

S

M = 0

S =0

A

By Cy

D

X

X

y

y

z

z

1.5m 4.5m 1m

0.75 kN/m

Bx

55


Section z-z

This example shows that: If appropriate portion after the cut is used, then evaluation of reactions is not necessary. This will result in saving of calculation time.

Example 3.2

For the beam shown in Figure 3.8 below, compute the shear force and bending moment at section x-x which is just to the left of support B. Note that reactions at supports have been calculated and the values are as indicated in Figure 3.8.

Figure 3.8

Solution:

Since the reactions had been calculated, we can directly compute the internal shear force and bending moment using section AB.

S + 10(40) 100 = 0 M + 10(40)(20) 100(40) = 0 S = - 300 kN M = -4000 kNm

From the above two examples, general procedures for the computation of S and M could be summarized as follows:

10 kN/m

A B

Ay=100kN

20 m 20 m

M

S

6 kN/m 10 kN/m

A B C D

Ay=100kN By=482kN Cy=298kN Dy=60kN

20 m 20 m 50 m 20 m 20 m

x

x

M = 0

S =0 M

D

S

56


1. Compute reaction forces. 2. Pass an imaginary cut through the section where S and M are required. Draw the

corresponding free body diagram to be used. Indicate on the free body diagram the positive sign convention for S and M.

3. Apply equilibrium of force in vertical direction and use it to compute S. 4. Apply equilibrium of moment at the cut section to compute M. Although equilibrium

of moment with respect to any other point could also be used, consideration of moment at cut section will eliminate the necessity to consider moment caused by shear S. This is to avoid mistake in value of m due to mistake in S.

3.5 SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR BEAMS

In the design of beams, the information on how S and M change with the change in the location of section is required. Such information is conveniently provided by shear force diagram (SFD) and bending moment diagram (BMD). In this section, computational procedures to produce these two diagrams for beams will be looked at.

The general procedures are as follow: i. Evaluate support reactions.

ii. Make necessary cut through section where equations for S and M are to be evaluated (Depending on the structure and loading condition given, such cutting process might need to be repeated). After choosing suitable free body diagram to be used, apply

a) Equilibrium condition for vertical force in order to obtain equation for S; and b) Equilibrium equation of moment at cut section in order to obtain equation for

M. iii. Making use of the equations for S and M obtained in step (ii) above, draw the

corresponding SFD and BMD.

The following examples will help to illustrate the process of drawing shear force and bending moment diagrams.

Example 3.3

Determine shear force and bending moment at the section which is located distance x from the fixed end A. Draw the shear force and bending moment diagrams.

Figure 3.9: A cantilever beam subjected to point load.

200 kN

A B

x

4 m

57


x 200 kN

800 kNm

M

S

Solution:

1. Draw the free body diagram and determine the reaction forces.

+ = 0xF : 0=xA

+ = 0yF : 0200 =yA kNAy 200=

+ A

M : ( ) 04200 =AM kNmM A 800=

2. Computation of S and M

= 0yF kNS 100=

M at cut section = 0;

0200800 =+ xM 800200 = xM

If the above computational process is examined, it can be seen that S and M at section located at a distance x from point A is expressed in terms of known forces and the distance x. Since this distance x can be anywhere between A and B, equations for S and M are actually equations representing the variation of S and M on sections between A and B. This variation could be seen by changing the value of x between 0 and 4m. If a graph of S (y axis) versus x (x axis) is drawn by using equation S above, the following shear force diagram will be obtained:

200 kN

B

x

4 m

Ay

MA

Ax

58


Similarly, the graph of M versus x is plotted based on equation M above, the following bending moment diagram will be obtained.

Example 3.4

Draw the shear force and bending moment diagrams for the simply supported beam shown in Figure 3.10.

Figure 3.10

Solution:

1. Draw the free body diagram and calculate the reaction forces.

100 kN

A B

2 m 5 m

C

M (kNm)

-800 kNm

0 8 m

BMD (drawn at compression side)

x

S (kN)

-200 kN

0 8 m

SFD

x

59


100 kN

A

2 m AY

x

M

S

x CY

M

S

+ A

M : 0)2(1007 =yC kNC y 571.28=

+ = 0yF : 0100 =+ yy CA kNAy 429.71=

2. Computation of D and M.

-Portion AB

= 0yF yAS = kNS 429.71=

M at cut section = 0; xAM y= xM 429.71=

-Portion BC

= 0yF 100= yAS kNS 571.28=


( )2100 = xxAM y 200571.28 += xM

-Portion BC (Alternative)

= 0yF yCS = kNS 571.28=

M at cut section = 0; xCM y=

xM 571.28=

100 kN

A B

2 m 5 m AY CY

C

A

x AY

M

S

60


100 kN

A B

2 m 5 m AY BY

71.429

142.858

SFD (kN)

BMD(kNm)

-28.571SFD (kN)

142.858

BMD(kNm)

3. SFD and BMD

-Portion AB

S = 71.429 kN M = 71.429x kN

0 x 2m

-Portion BC

S = -28.571 kN M = -28.571x + 200 kNm

2m x 7m

Note that if the results of alternate calculation for portion BC as follow:

S = -28.571 kN M = 28.571x kNm

0m x 5m

Are used, the same diagram will be obtained.

61


The computed SFD and BMD are lastly shown together with the structure analysed. It is important to indicate the title SFD and BMD and the corresponding units used. Note that the BMD is drawn on compression side.

Example 3.5

For the beam shown is Figure 3.11, draw the corresponding shear force and bending moment diagrams.

Figure 3.11

Solution:

1. Draw free body diagram and calculate reactions.

10 kN

A B

2 m 5 m 2 m

5 kN/m

C D

100 kN

A B

2 m 5 m AY BY

71.429

142.858

SFD (kN)

BMD(kNm)

-28.571

62


10 kN

2 m

C D

x

M

S

x

D

M

S

+ B

M : ( ) ( )( ) ( ) 02105.2555 =+ yA kNAy 5.8=

+ = 0yF : ( ) 01055 =+ yy BA kNBy 5.26=

2. Computation of S and M

-Portion AB

= 0yF xAS y 5= xS 55.8 =


02

55.8 =+

+ Mxxx

xxM 5.85.2 2 += -Portion BC

= 0yF kNS 10=

M at cut section = 0; ( )210 = xM 2010 += xM

-Portion CD

= 0yF 0=S


0=M

10 kN

2 m 5 m 2 m

5 kN/m

C D

AY BY

x

5 kN/m

AY

M

S

63


3. Draw SFD and BMD by using equations obtained at above.

3.6 FRAME

Frame composed of straight members connected either rigid (moment-resistant) connections or hinged connection to form stable configurations. Figure 3.12 shows an example of frame.

Figure 3.12: A structural frame subjected to both vertical and lateral loads.

Under the action of external loads, the members of a frame may be, in general subjected to Bending Moment, Shear and Axial (Tension or Compression) forces. Figure 3.13 below

a

a

b

b

10 kN

2 m 5 m 2 m

5 kN/m

C D

AY BY

8.5

-16.5

10

0

SFD (kN)

7.225

-20

BMD (kNm)

64


shows the internal forces of section a-a and b-b in the structural frame as shown in Figure 3.12.

Figure 3.13: Internal forces of a structural frame.

3.7 STATICAL DETERMINACY OF FRAMES

A frame is considered to be statically determinate if the bending moment, shears and axial forces in all its members, as well as all the external reactions, can be determined by using the equations of equilibrium and condition. For better understanding, the example below will give a better idea.

Example 3.6

Check the statical determinacy of the frame shown in Figure 3.14.

Figure 3.14

Solution:

1. Draw the free body diagram.

x

y

A

B

C

D

P1

P2

P3

w

M

V

M V

N N

a

b a

b

65


2. No. of external reactions : Ax, Ay, Dy = 3 No. of equilibrium equations: 3

Statically determinate frame.

Once Ax, Ay and Dy are determined, bending moment, shears and axial forces in all members AB, BC and CD could be determined.

Equation of condition are used whenever there are hinged or roller-type joints in a frame. Figure 3.15 shows frames with hinged and roller joints in a frame.

Figure 3.15: Frames with hinged and roller joints

Hinged joint Roller joint

Rigid joint

x

y

Ax

B

C

Dy

P1

P2

P3

w

Ay

66


For frames with only rigid joints, statical determinacy could be determine by using:

i) the equation:

jrm 33 =+ where m = no. of members r = no. of external reactions

j = no. of rigid joints

eg. m = 3 j = 4 (including the supports) r = 6

( ) 156333 =+=+ rm ( ) 12433 ==j

jrm 33 >+ , statically indeterminate structure Degree of indeterminacy, 31215 ==i

ii) Alternative approach - To cut enough members of the frame by passing imaginary sections and/or to

remove enough supports to render the structure statically determinate. - The total number of internal or external restraints thus removed equals the

DEGREE OF STATICAL INDETRMINACY.

For one imaginary section: Part ABe and DCe become statically determinate. Degree of Statical Indeterminacy, 3=i (= no. of internal restraint removed)

A A

B B C C

D D

e e M

S

M

S

Q Q

(a)

(b) 3 internal restraints:

M, S, Q are removed

67


In order for the frame above to become statically determinate, 3 external restraints are needed to be removed.

External restraints to be removed: Dx, Dy and MD. ABCD become statically determinate.

Degree of statical indeterminacy = no. of external restraint removed 3=i

3.8 ANALYSIS OF PLANE FRAMES

In the design of frames, the information on how S and M change with the change in the location of section is required. Such information is conveniently provided by shear force diagram (SFD) and bending moment diagram (BMD). In this section, determination of member end forces, shears, bending moments and axial forces in plane statically determinate frames will be looked at. The general procedures of computing the shears and bending moment are almost the same as the beams. Example 3.7 gives an idea how to analyze a plane frame.

Example 3.7

Draw the shear, bending moment and axial force diagrams and the quantitative deflected shape for the frame shown in Figure 3.16.

A

B C

Dy (c)

Dx MD

68


Figure 3.16

Solution:

1. Check the statical determinacy

Total no. of internal restraints = 1 x 3 = 3 Total no. of external reactions = 2 + 1 = 3 Total no. of unknown forces = 3 + 3 = 6 Total no. of equilibrium equations = 2 x 3 = 6 Statically Determinate

2. Compute the reactions.

5 kN/m

10m

7 m

10m

10 kN

A

B C

D

69


= 0xF : kNAx 10= , kNAx 10=

+ 0= AM : 010)5)(10(5)10(10 =+ yD kNDy 35=

+ = 0yF : ( ) 0510 =+ yy DA kNAy 15=

3. Shear and bending moment in the members. - Shear and bending moment are defined w.r.t. sections which are perpendicular to

member axis.

- Member AB

5 kN/m

10m

7 m

10m

10

10 kN

B C

15kN

35 kN A

D

a a

b

b

c c

5 kN/m

10m

7 m

10m

10 kN

Ay

Ax

Dy

B C

70


= 0xF : kNS 10= 0=

aaM : xkNmM 10= mx 100

The corresponding SFD and BMD for this section.

- Member BC

10 100

A A

B B

(a) SFD (kN) (b) BMD (kNm)

10 kN

15 kN

A

x

x

S M

5 kN/m

10m

10 kN

10 kN

B

15 kN

A

x

M

S

x

71


= 0Fy : 0515 =+ xS

xS 515 =

0=

bbM : ( ) ( ) 010201525 =

+ x

xxM

200155.2 2 ++= xxM mx 100


- Member CD

B B C C

-35

15

222.5 200

100


5 kN/m

10m

x

10m

10 kN

10 kN

B C

15 kN

A

S M

x

72


= 0xF : 01010 =+S 0=S

0=

ccM : ( )( ) ( ) 01501010105105 =+ xxM 0150101010250 =++ xxM 0=M

Alternative:

= 0xF : 0=S 0=

ccM : 0=M


D D

C C


x

15 kN

D

x

S M

73


The shear force and bending moment diagram.

5 kN/m

10m

7 m

10m

10 kN

10

A

B C

-35

D

B C

222.5 200

100

15

100

A

D

SFD (kN)

BMD (kNm)

74


15

A

B C

D

Axial Force (kN)

35

Quantitative Deflected Shape

(Based on BMD)

75


Example 3.8

Draw the shear, bending moment, and axial force diagrams and the quantitative deflected shape for the frame shown in Figure 3.17.

Figure 3.17

Solution:

1. Statical determinacy

no. of reactions = 3 no. of equilibrium equations = 3 Statically determinate frame.

2. Reactions - If we choose to proceed our calculation from C, there is no need to calculate the

reactions Ax, Ay and MA at A.

3. Shear and bending moment in the members. - Shear and bending moment are defined w.r.t. sections which are perpendicular to

member axis.

A

B C

20 kN/m

80 kN

10m

B C

MA

Ay

Ax

76


- Portion / Member CB

= 0yF : kNS 80= k

= 0aaM : xM 80=

mx 70

80

C B C B

560

(b) BMD (kNm) (a) SFD (kN)

A

B C

20 kN/m

80 kN

10m

a

a

b b

C

80 kN

M

S

x

77


- Portion / Member BA

= 0xF : xS 20=

= 0bbM

( )

=

220780 xxM

210560 xM = mx 100

The shear force and bending moment diagram.

A A 200 -1560

-560


80

C B C B

560

B B

A A 200 -1560

-560


B C

20 kN/m

80 kN

x

M

S

78


A

(c) Axial Force (kN) (d) Qualitative Deflected Shape

-80 A

B C

20 kN/m

80 kN

79


Exercise 1

Figure 3.18 shows a frame with an internal hinge at C. It is supported by two hinge supports at A and E. Draw the corresponding shear force and bending moment diagrams. Sketch also the qualitative deflected shape.

Figure 3.18

Exercise 2

Figure 3.19 shows a rigidly jointed frame with pinned and roller support at A and D, respectively. A horizontal point load of magnitude 40kN acts at joints B and a uniformly distributed load with intensity of 3 kN/m acts along portion C to D of the horizontal member BCD. Draw the corresponding shear and bending moment diagrams.

Figure 3.19

80


Exercise 3

Draw the shear force and bending moment diagrams for the frame shown in Figure 3.20.

Figure 3.20

81


TOPIC 4

ANALYSIS OF CABLES

4.1 INTRODUCTION

Cables constructed of high-strength steel wires are completely flexible and have a tensile strength four to five times greater than structural steel. Because of their strength-to-weight ratio, engineers use cables to construct long-span structures, including suspension bridges and roofs over large arenas and convention halls. Figure 4.1 shows an example of cable stayed bridge, which is the longest bridge in Malaysia Penang Bridge.

Figure 4.1: Penang Bridge a cable stayed bridge.

A cable is a form of structure which is unable to resist bending moments. A structural parabolic arch which is carries a uniformly distributed load is subjected to a very small bending moment at any point on the arch. Therefore, an economical bridge can be constructed as shown in Figure 4.2.

Figure 4.2

Girder on the columns is supported by the structural arch. In this case, the structural arch and the columns are in compression.

Girder

Column

Structural

Arch

82


Since a member in tension requires lesser cross-sectional area if compared to a member in compression, both carrying equal amount of load, therefore it is more economical if the above structure is changed to be a suspended bridge as shown is Figure 4.3. In this case, the cable and hangers are in tension.

Figure 4.3

4.2 TOPIC OUTCOME

a) Able to calculate reactions and tension in cables using equilibrium equations.

4.3 SCOPE OF ANALYSIS

In this analysis, the cable is considered perfectly flexible in bending so that the bending moment on any section of the cable must be zero, and the cable can only transmit load to the supports by means of tension action along its length.

It should be noted that:

1. Maximum tension occurs at maximum reaction point. 2. Minimum tension occurs at the lowest point of the cable.

The bending flexibility of a cable will be intuitively understood by a comparison with a cotton thread from which small weight are hung as shown in Figure 4.3 below.

Girder

Hanger

Cable

83


Figure 4.3: Suspended cable subjected to concentrated loads.

Clearly, the thread has no resistance to bending forces and, since the weight of the thread will be insignificant in comparison with the loads, it will assume a shape consisting of a series of straight line and between the applied loads. It is such as an idealization which has been assumed for the cable in figure 4.3 above.

As the applied loads are all vertical, the horizontal reactions at A and B must be equal and opposite (since they are only horizontal force acting on the structure) and these are denote by Ax and Bx. Simple equilibrium will confirm that the horizontal component of the tensile force at any section will also of value Ax or Bx (Ax = Bx). See Figure 4.4 below.

Figure 4.4

Using = 0BM for the whole structure.

L

C (x,y)

w1 w2

w3

x1

x2

x3

Ay By

Ax Bx

L

C (x,y)

w1 w2

w3

x1

x2

x3

A B

84


( ) ( ) ( ) 0332211 = xLWxLWxLWLAy

( )[ ]L

xLWA nny

=

Consider any typical point C(x,y) and in this case positioned between the concentrated loads W1 and W2. As the bending moment at C is zero, taking moments of all the forces to the left (refer to Figure 4.5) of C gives:

Figure 4.5

( ) 011 = yAxxWxA xy

( )[ ] ( )11 xxWLxLWxyA nnx

=

4.4 TYPES OF CABLES

There are two types of cable that will be considered:

1. Symmetrical cable which is supported at the same level. 2. Unsymmetrical cable which is supported at different levels.

4.5 SYMMETRICAL CABLE SUBJECTED TO UNIFORMLY DISTRIBUTED LOAD

In many practical structures, the suspension bridge for example, the dead load of the suspended superstructure is carried at such numerous positions along the span so that it is sufficiently accurate to consider the load on the cable as uniformly distributed across the span.

Figure 4.6 illustrates the typical problem.

The symmetry of the arrangement demonstrates that

a) The vertical relative forces at the supports are 2L

and

b) The tensile force in the cable at the mid span section is Ax or Bx (Ax = Bx).

C (x,y)

w1

x1

x

Ay

Ax

W2

y

Note: The sum of the moments of all

forces to one side of a point must be zero

as the cable cannot sustain an internal

bending moment.

85


For the whole structure.

wLBA yy =+

For symmetrical cable,

2wLBA yy ==

And, HBA xx ==

Figure 4. 6

Where,

L = horizontal distance (span between two supports) h = vertical distance from the lowest point to supports

Consider a point at mid span (as shown in Figure 4.7).

For equilibrium; = 0cM

HhLLWLwL +

=

4222

84

22 wLwLHh =

8

2wL=

hwLH8

2

=

Figure 4.7

Consider a general point, C (x,y), as shown in Figure 4.8.

+ 0= cM

( ) 022

=

+ x

wLxwxHy

22

2wxwLxHy =

Figure 4.8

A B

h

L

w/unit length

Bx

By

Ax

Ay

A

L/2

w

H

2wL

h

H

2wL

2wL

c

A

x

w

H

2wL

y

H

2wL

v

c(x,y)

86


Substituting the previous value of H and rearranging:-

( ) ( )28

2 xLwxyh

wL =

22)(8

LxLhxy =

2)(4

LxLhxy =

Where y = vertical distance from any point on the cable to supports

This is a parabolic equation and defines the shape of the cable when subjected to a load distributed uniformly with respect to the horizontal projection of the cable.

4.6 FORCES IN CABLE SUBJECTED TO UNIFORMLY DISTRIBUTED LOAD

Refer to the Figure 4.9 below.

The tensile force T, in the cable at any position can be obtained from knowledge of the horizontal and vertical components at the position.

Figure 4.9

For example, at point C (refer to Figure 4.9) in the cable subjected to a uniformly distributed load, where:

Horizontal equilibrium shows that,

hwLH8

2

= and

Vertical equilibrium for the forces to the left of C gives

wxwl

v =2

and acts in the direction shown.

The tensile force T, in the cable C is the resultant of H and v.

Where 222 vHTx +=

A

x

w

H

2wL

y

H

2wL

v

c

T

87


22 vHTx +=

222

28

+

= wx

wLh

wLTx (4.1)

Maximum tensile force in the cable occurs at 0=x , i.e., at supports, and is given by

222

max 28

+

=

wLh

wLT (4.2)

Minimum tensile force in the cable occurs at 2L

x = , substitute into (4.1)

222

min 228

+

=

Lw

wLh

wLT

Hh

wLT ==8

2

min (4.3)

Example 4.1

Figure 4.10 shows a cable subjected to both point load and uniformly distributed load. Determine a) the reactions at A and B b) Tmin and Tmax c) size of the cable if the allowable stress is 660 kN/m2.

Figure 4.10

10kN 10kN

5m

A B

5m 5m 5m

2.5m

5kN/m

88


Solution:

1. Draw free body diagram and calculate reactions.

= 0xF : HBA xx == = 0yF : ( ) 1010205 ++=+ yy BA kNBA yy 120=+ Since it is symmetrical kNBA yy 60==

To prove:

+ 0= AM : ( ) ( ) ( ) ( )( ) 010205151051020 =yB kNBy 60=

kNAy 6060120 ==

2. To determine:

Tmin (Minimum tension)

Consider a point at mid span, x=10m.

10kN 10kN

5m

A B

5m 5m 5m

2.5m

5kN/m

Ax

Ay

Bx

By

10kN

5m

A

5m

2.5m

H

60kN

H

V

C

5kN/m

89


+ 0= CM : ( ) ( )( ) ( ) ( ) 0106051051055.2 =++H kNH 120=

+ = 0yF : ( ) 01051060 = V 0=V

Tmin occurs at mid span = H.

Or

22min VHT +=

22

min 0120 +=T kNT 120min =

To determine Tmax (Maximum tension) which occurs at support A.

22max VHT +=

22

max 60120 +=T kNT 16.134max =

3. To determine the area of cable.

4

2dA pi= and AF

=

Therefore, 660

16.1434

2

=dpi

md 509.0=

90


4.7 UNSYMMETRICAL CABLE SUBJECTED TO UNIFORMLY DISTRIBUTED LOAD

Figure 4.11: Unsymmetrical cable subjected to uniformly distributed load

Figure 4.11 above shows on unsymmetrical cable (supported at different level) where

C = lowest point of the cable. d1 = vertical distance from the lowest point to support A. d2 = vertical distance from the lowest point to support B. l1 = horizontal distance from A to C. l2 = horizontal distance from C to B.

Consider the whole structure

= 0xF where HBA xx == And = 0yF where wLBA yy =+

Consider left hand side (AC), as shown in Figure 4.12.

Figure 4.12 l

A

d1

Ax

Ay

H

V

C

w

A

B

C d1

d2

l1 l2

L

Ax

Ay

Bx

By

w

91


= 0xF where HAx = And = 0yF where 1wlAy =

Also + 0= AM : 021

11 =

lwlHd

1

21

2dwlH =

Consider right hand side (BC), as shown in Figure 4.13.

Figure 4.13

= 0xF where HBx = And = 0yF where 1wlBy = where 12 lLl =

( )1lLwBy =

Also + 0= BM : 02 22

2 =

Hllwl

2

22

2dwlH =

Equate right hand side = left hand side

2

22

1

21

22 dwl

dwl

=

Or 2

1

2

1

dd

ll

= ..(4.4)

B

C

d2

l2

Bx

By

H

w

92


Example 4.2

For the cable shown in Figure 4.14, find H.

Figure: 4.14

Solution:

Given that: mL 40= ; md 51 = ; md 152 =

0= yF ; ( ) kNBA yy 80402 ==+

2

22

1

21

22 dwl

dwlHBA xx ===+

But 31

155

2

12

2

21

===

dd

ll

Therefore 12 3ll = And 4021 =+ ll Hence 403 11 =+ ll ml 641.141 = ml 359.252 =

We know that 1

21

2dwlH =

Therefore ( )( )( ) kNH 87.4252641.142 2

==

To check ( )( )( ) kNdwlH 87.42

152358.252

2

2

2

22

===

A

B

5m

10m

40m

Ax

Ay

Bx

By

2 kN/m

93


4.8 LENGTH OF CABLES (S)

4.8.1 CABLES LOADED WITH POINT LOADS

For the cable loads, it is assumed that point loads, it is assumed that the self weight is negligible. Hence the length is the sum of the length of each segment, as shown in Figure 4.15.

Figure 4.15: Cable loaded with point loads

Length of cable, 4321 ssssS +++=

Each segment is assumed to be straight.

4.8.2 SYMMETRICAL CABLES

For cables loaded with uniformly distributed load we can show that for symmetrical cables:

LdLS

38 2

+=

Figure 4.16

To proof, refer to Figure 4.16,

A B

d

L

w/unit length

Bx

By

Ax

Ay

L

w1 w2

w3

A B

s1

s2 s3

s4

94


If s = length of a small segment

S = total length of cable

222 yxs +=

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