structure knowledge

7/29/2019 structure knowledge

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CHAPTER 4.0: STRUCTURE

4.1 Understand plane trusses

4.1.1 Define a truss

4.1.2 Explain plane trusses

4.2 Analyze truss and force using related method

4.2.1 Analyze a truss using the method of joints

4.2.2 Analyze the force in the members of a truss using the method of sections

4.3 Understand frames and machines

4.3.1 Explain frames and machines

4.3.2 Determine the force acting at the joints and supports of a frame or machine


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4.1 INTRODUCTION

1. This chapter concerned the equilibrium of a single rigid body, and all forces involvedwere external to the rigid body.

2. The structure related to equilibrium of structure which made of several connected parts.

3. It involves external forces and internal forces which hold together on that structure

Example: Crane which carries a load W(Fig. 1)

Figure 1

1) B pointForce used at B by member BE on member AD has been represent as equal and opposite to the

force that used at the same point by member CF on BE.

2) E pointForce used at E by member BE on CF is equal and opposite to the force that used at the same

point by member AD on BE.

3) C point

Force used at C by CF on AD is equal and opposite to the force that used by AD on BE.

So, the B,E, and C point can be related with Newton 3rd

LawForce of action and reaction

between bodies in contact have same magnitude, same line of action and opposite sense.

4.1.1 ENGINEERING STRUCTURES

a) Truss (framework supporting a roof, bridge, building, etc.)Designed to support loads and usually stationary, fully constrained structure. Consist ofstraight member connected at joint located at the ends of each member. Hence, the

member of truss is two force members, i.e., member acted upon by two equal andopposite forces directed along the member. Can be divided into two trusses which issimple truss and planar truss.

b) FramesWhich are also designed to support loads and are also usually stationary, fully

constrained structures. However, frames always contain at least one multi-force member,


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i.e., a member acted upon by three or more forces which, in general, are not directed

along the members.

c) MachinesWhich are designed to transmit and modify forces and are structures containing moving

parts. Machines like frames always contain at least one multi-force member.

1. 4.2 TRUSS2. Composed of slender members joined together at their end points.3. Truss member are connected at their extremities only: thus no member is continuous

through a joint.

4. The members commonly used in construction consist of wooden struts or metal bars. Thejoint connections are usually formed by bolting or welding the ends of the members to a

common plate, called a gusset plate(Fig 2(a)), or by simply passing a large bolt or pin

through each of the members (Fig. 2(b)).

a. Figure 25. When concentrated load is to be applied between two joints, or when a distributed load is

to be supported by the truss, as in the case of a bridge truss, a floor system must be

provided which, through the use of stringers and floor beams, transmits the load to thejoint (Fig.3)

6. Figure 3

7. Although the members are actually joined, it is customary to assume that the membersare pinned together; therefore, the forces acting at each end of a member reduce to a

single force and no couple.


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8. Thus, the only forces assumed to be applied to a truss member are a single force at eachend of the member. Each member can then be treated as a two-force member, and the

entire truss can be considered as a group of pins and two-force member.

Figure 4

4.2.1 SIMPLE TRUSSES

1. To prevent collapse, the form of a truss must be rigid. For example, the four-bar shapeABCD(Fig. 5 (a)) will collapse unless a diagonal, such asAC, is added for support.

2. The simplest form that is rigid or stable is a triangle. Consequently, a simple truss isconstructed by starting with a basic triangular element, such as ABC (Fig.5. (b))

connecting two members (AD and BD) to form an additional element.3. As each additional element consisting of two members and a joint is placed on the truss,

it is possible to construct a simple truss.

(a)

(b)Figure 5


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4.2.2 PLANAR TRUSSES

1. The trussABCDE(Fig. 6 (a)) is an example of a typical roof-supporting truss. The roofload is transmitted to the truss at the joints by means of a series ofpurlins, such asDD.

2. Since the imposed loading acts in the same plane as the truss (Fig. 6 (b)), the analysis ofthe forces developed in the truss member is two dimensional.

Figure 6

3. In the case of a bridge (Fig. 7 (a)), the load on the deck is first transmitted to stringers,

then to floor beams and finally to the joint B, C, and D of the two supporting side trusses.

Like the roof truss, the bridge truss loading is also coplanar(Fig. 7 (b))

Figure 7

4. When bridge of roof trusses extend over large distances, a rocker or roller is commonly

used for supporting one end (Fig. 6(a),7(a)). This type of support allows freedom for

expansion or contraction of the members due to temperature or application loads.

4.2.3 Assumption for design

To design both the members and the connections of a truss, it is first necessary to determine the

force developed in each member when the truss is subjected to a given loading. In this regard,

two important assumptions will be made:

1) All loadings are applied at the joints.In most situations, such as for bridge and roof trusses, this assumption is true. Frequently

in the force analysis the weight of the members is neglected since the forces supported bythe members are usually large in comparison with their weight. If the member weight is

(b)


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to be included in the analysis, it is generally satisfactory to apply it as a vertical force,

half of its magnitude applied at each end of the member.

2) The members are joined together by smooth pins.In cases where bolted or welded joint connections are used, this assumptions is

satisfactory provided the center lines of the joining members are concurrent, ( Fig 2(a))

Each truss member acts as a two-force member, and therefore the forces at the ends of the

member must be directed along the axis of the member(Fig 8):

Tensile force (T): force tends to elongate the member

Compressive force (C): force tends to shorten the member

Figure 8

4.3 THE METHOD OF JOINT

To analyze or design a truss, the forces must be obtained in each of its members.If we were to consider a FBD of the entire truss, then the forces in the members would beinternal forces, and they could not be obtained from an equilibrium analysis.

Instead, if we consider the equilibrium of a joint of the truss then a member force

becomes an external force on the joint FBD, and the equation of equilibrium can beapplied to obtain its magnitude. This forms the basis for the method of joints.

As an example, consider the pin at joint B of the truss (Fig 9).Three forces act on the pin,

namely, the 500-N force and the forces exerted by membersBA and BC. The FBD is shown

(Fig. 9 (a)), FBA is pulling on the pin, which means that memberBA is in tension; whereasFBCis pushing on the pin, and consequently memberBCis in compression. These effects

are clearly demonstrated by isolating the joint with small segments of the memberconnected to the pin. The pushing or pulling on these small segments indicates the effect of

the member being either in compression or tension.


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Figure 9

Procedure for analysis:

1. Draw the FBD of a joint having at least one known force and at most two unknown

forces. (If this joint is at one of the supports, it generally will be necessary to know theexternal reactions at the truss support)

2. Use one of the two methods described above for establishing the sense of an unknownforce.

3. Orient the x and y axis such that the force on the FBD can be easily resolved into their xand y components and then apply the two force equilibrium equations Fx = 0 andFy =

0. Solve for the two unknown member forces and verify their correct sense.

4. Continue to analyze each of the other joints, where again it is necessary to choose a jointhaving at most two unknowns and at least one known force.

5. Once the force in a member is found from the analysis of a joint at one of its ends, theresult can be used to analyze the force acting on the joint at its other end. Remember that

a member in compression pushes on the joint and a member in tension pulls on thejoint.


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4.4 THE METHOD OF SECTIONS

1. The method of sections is used to determine the loading acting within a body and basedon the principle that if a body is in equilibrium then any part of the body is also in

equilibrium.

2. The method of sections can also be used to cut or section the members of an entiretruss. If the section passes through the truss and the FBD of either of its two parts is

drawn, the equation of equilibrium then can be applied to that part to determine the

member forces at the cut section.3. Since only three independent equilibrium equations (Fx = 0, Fy = 0, Mo = 0) can be

applied to the isolated part of the truss try to select a section that, in general, passes

through not more than three members in which the forces are unknown.4. For example (Fig 10 (a)), if the force in member GC is to be determined, section aa

would be appropriate. FBD of the two parts are shown (Fig 10 (b),(c)). In particular, note

that the line of action of each member force is specified from thegeometry of the truss,

since the force in a member passes along its axis. Also, the member forces acting on one

part of the truss are equal but opposite of those acting on the other partNewton 3rd

law.

Figure 10

5. Member assumed to be in tension (BC and GC) are subjected to a pull, whereas themember in compression (GF) is subjected to a push(Fig. 11)

Figure 11


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Procedure for analysis

Free-Body Diagram (FBD)1. Make a decision as to how to cut orsection the truss through the members where forces

are to be determined.

2. Before isolating the appropriate section, it may first be necessary to determine the trusssexternal reactions. Then three equilibrium equations are available to solve for member

forces at the cut section.

3. Draw FBD of that part of the sectioned truss which has the least number of forces actingon it.

4. Use one of the two methods described above for establishing the sense of an unknownmember force.

Equation of equilibrium1. Moments should be summed about a point that lies at the intersection of the lines of

action of two unknown forces, so that the third unknown force is determined directlyfrom the moment equation.

2. If two of the unknown forces are parallel, forces may be summed perpendicular to thedirection of these unknowns to determine directly the third unknown force.


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4.5 SPACE TRUSSES

1. A space truss consists of members joined together at their ends to form a stable three-dimensional structure. The simplest element of a space truss is a tetrahedron, formed by

connecting six members together(Fig 12).

Figure 12

2. Any additional members added to this basic element would be redundant in supportingthe force P.

3. Simple space truss can be built from this basic tetrahedral element by adding threeadditional members and a joint, forming a system on multi-connected tetrahedrons.

4.5.1 Assumptions for design

1. The members of a space truss may be treated as a two-force member provided theexternal loading is applies at the joints and the joints consist of ball-and-socketconnections.

2. These assumptions are justified if the welded or bolted connections of the joinedmembers intersect at a common point and the weight of the members can be neglected.

3. In cases where the weight of a member is to be included in the analysis, it is generallysatisfactory to apply it as a vertical force, half of its magnitude applied at each end of the

member.

4.5.2 Procedure for analysis

Either the method of joints or the method of sections can be used to determine the forces

developed in the members of a simple space truss.

Method of joints1. Basically, if the forces in all the members of the truss must be determined, the method ofjoints is most suitable for the analysis.

2. When using method of joints, it is necessary to solve the three scalar equilibriumequations Fx = 0, Fy = 0, Fz= 0 at each joint.

3. The solution of simultaneous equations can be avoided if the force analysis begins at thejoint is hard to visualize, it is recommended that a Cartesian vector analysis be used for

the solution.


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Method of sections1. If only the few member forces are to be determined, the method of sections may be used.

When an imaginary section is passed through a truss and the truss is separated into twoparts, the force system acting on one of the parts must satisfy thesix scalar equilibrium

equations: Fx = 0, Fy = 0, Fz= 0, Mx = 0, My = 0, Mz= 0.

2.

By proper choice of the section and the axes for summing forces and moments, many ofthe unknown member forces in a space truss can be computed directly, using a single

equilibrium equation.

Example:


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4.6 FRAMES AND MACHINES

1. Common types of structures which are often composed of pin-connected multi-forcemembers (members that are subjected to more than two forces)

2. Provided a frame or machine is properly constrained and contains no more supports ormembers than are necessary to prevent collapse, the forces acting at the joints andsupports can be determined by applying the equations of equilibrium to each member.

3. Once the forces at the joints are obtained, it is then possible to design the size of themembers, connections, and supports using the theory of mechanics of material and anappropriate engineering design code.

Free-Body Diagram

In order to determine the forces acting at the joints and supports of a frame or machine,

the structure must be disassembled and the FBD of its parts must be drawn.

1. Isolate each part by drawing its outlined shape. Then show all the forces and/or couplemoments that act on the part. Make sure to label or identify each known and unknown

force and couple moment with reference to an establishedx,y coordinate system. Also,

indicate any dimensions used for taking moments. Most often the equations ofequilibrium are easier to apply if the forces are represented by their rectangular

components. As usual, the sense of an unknown force or couple moment can be assumed.

2. Identify all the two-force members in the structure and represent their FBD as having twoequal but opposite collinear forces acting at their points of application. By recognizing

the two force members, we can avoid solving and unnecessary number of equilibrium

equations.

3. Forces common to any two contacting members act with equal magnitudes but oppositesense on the respective members. If the two members are treated as a system ofconnected members, then these forces are internal and are not shown on the FBD of

the system; however, if the FBD of each member is drawn, the forces are external and

must be shown on each of the FBD.


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Example:


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Equations Of Equilibrium

1. Provided the structure (frame or machine) is properly supported and contains no moresupports or members than are necessary to prevent its collapse, then the unknown forces

at the supports and the connections can be determined from the equations of equilibrium.

2. If the structure lies in thex-yplane, then the each FBD drawn the loading must satisfyFx = 0, Fy = 0, Mo = 0. The selection of the FBD used for the analysis is completelyarbitrary. They may represent each of the members of the structure, a portion of the

structure, or its entirety.

3. Example (Fig 12 (a)): Finding the six components of the pin reactions at A,B and c forthe frame. If the frame is dismembered (Fig 12 (b)), these unknowns can be determined

by applying the three equations of equilibrium to each of the two members (total of six

equations).the FBD of the entire frame can also be used for part of the analysis (Fig 12

(c)).4. Hence, if so desired, all six unknowns can be determined by applying, the three

equilibrium equations to the entire frame (Fig 12 (c)) and also to either one of its

members.5. Furthermore, the answers can be checked in part by applying the three equations ofequilibrium to the remaining second member. In general, then, thus problem can be

solved by writing at most six equilibrium equations using FBD of the members and/or the

combination of connected members.6. Any more than six equations written would not be unique from the original six and would

only serve to check the result.

Figure 13

4.6.1 Procedure for analysis

The joint reactions on frames or machines (structures) composed of multiforce members can be

determined using the following procedure:

Free-Body Diagram1. Draw the FBD of the entire structure, a portion of the structure, or each of its members.

The choice should be made so that it leads to the most direct solution of the problem.

2. When the FBD of a group of members of a structure is drawn, the forces at theconnected parts of this group are internal forces and are not shown on the FBD of the

group.

3. Forces common to two members which are in contact act with equal magnitude butopposite sense on the respective FBD of the members.


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4. Two-force members, regardless of their shape, have equal but opposite collinear forcesacting at the ends of the member.

5. In many cases it is possible to tell by inspection the proper sense of the unknown forcesacting on a member; however, if it seems difficult, the sense can be assumed.

6. A couple moments is a free vector and can act at any point on the FBD. Also, a force is

a sliding vector and can act at any point along its line of action.

Equations of equilibrium1. Count the number of unknowns and compare it to the total number of equilibrium

equations that are available. In two dimensions, there are three equilibrium equations thatcan be written for each member.

2. Sum moments about a point that lies at the intersection of the lines of action of as manyunknown forces as possible.

3. If the solution of a force or couple moment magnitude is found to be negative, it meansthe sense of the force is the reverse of that shown on the FBD.


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