capítulo 1. sistema de la fuerza resultantes y equilibrio

8/10/2019 Captulo 1. Sistema de La Fuerza Resultantes y Equilibrio

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IStatics1 Force-System Resultants and Equilibrium Russell C. Hibbeler

Force-System ResultantsEquilibrium

2 Centroids and Distributed Forces Walter D. Pilkey and L. Kitis

Centroid of a Plane Area Centroid of a Volume Surface Forces Line Forces

Calculation of Surface Area and Volume of a Body with Rotational Symmetry

Determination of Centroids

3 Moments of Inertia J. L. Meriam

Area Moments of Inertia Mass Moments of Inertia


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1Force-SystemResultantsand Equilibrium

1.1 Force-System ResultantsConcurrent Force SystemsMoment of a Force Couple

Resultants of a Force and Couple SystemDistributed Loadings

1.2 EquilibriumEquations of EquilibriumFree-Body DiagramSupport

ReactionsFrictionConstraintsInternal Loadings

Numerical Applications

Statics is a branch of mechanics that deals with the equilibrium of bodies, that is, those that are either

at rest or move with constant velocity. In order to apply the laws of statics, it is first necessary to understand

how to simplify force systems and compute the moment of a force. In this chapter these topics will be

discussed, and some examples will be presented to show how the laws of statics are applied.

1.1 Force-System Resultants

Concurrent Force Systems

Force is a vector quantity that is characterized by its magnitude, direction, and point of application.

When two forces F1and F2are concurrentthey can be added together to form a resultant FR=F1+F2using the parallelogram law,Figure 1.1.Here F1and F2are referred to as components of FR. Successiveapplications of the parallelogram law can also be applied when several concurrent forces are to be added;

however, it is generally simpler to first determine the two components of each force along the axes of a

coordinate system and then add the respective components. For example, the x, y, z (or Cartesian)

components of Fare shown in Figure 1.2.Here, i,j,kare unit vectors used to define the direction of the

positivex, y, zaxes, and Fx, Fy, Fzare the magnitudes of each component. By vector addition, F=Fx i+Fyj + Fzk. When each force in a concurrent system of forces is expressed by its Cartesian components,the resultant force is therefore

(1.1)

where SFx, SFy, SFz represent the scalar additions of thex, y, zcomponents, respectively.

F i j kR x y z F F F= + +

Russell C. HibbelerUniversity of Louisiana at Lafayette

2005 by CRC Press LLC


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Sometimes the moment arm d is geometrically hard to determine. To make the calculation easier, the

force is first resolved into its Cartesian components and then the moment about point O is determined

using the principle of moments,which states that the moment of the force about Ois equal to the sum of

the moments of the forces components about O.Thus, as shown inFigure 1.5,we haveMO=Fd=Fxy+Fyx.The moment about point Ocan also be expressed as a vector cross product of the position vector r,

directed from O to any point on the line of action of the force and the force F, as shown inFigure 1.6.Here,

MO=rF (1.3)

If r and F are expressed in terms of their Cartesian components, then as in Figure 1.7 the Cartesian

components for the moment about O are

(1.4)

Couple

A coupleis defined as two parallel forces that have the same magnitude and opposite directions and are

separated by a perpendicular distance d, as in Figure 1.8.The moment of a couple about the arbitrary

point Ois

FIGURE 1.5 FIGURE 1.6

FIGURE 1.7

y

xMOO

Fx

y

FyF

d

XMO

O

r F

FFx

Fz

Fy

y

z

zr

xO

x

y

MO

M r F i j k i j k

i j k

i j k

o x y z F F F

yF zF zF xF xF yF

x y z

F F F

x y z

z y x z y x

x y z

= = + + +

= - + - + -

=

(

( ) ( ) ( )

) ( + )



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(1.5)

Here the couple moment MCis independent of the location of the moment point O. Instead, it dependsonly on the distance between the forces; that is, r in the above equation is directed from any point on

the line of action of one of the forces (-F) to any point on the line of action of the other force F. Theexternal effect of a couple causes rotation of the body with no translation, since the resultant force of a

couple is zero.

Resultants of a Force and Couple System

A general force and couple-moment system can always be replaced by a single resultant force and couple

moment acting at any point O. As shown in Figure 1.9(a) and Figure 1.9(b), these resultants are

(1.6)

(1.7)

where SF=F1+F2+F3is the vector addition of all the forces in the system, and SMO=(r1F1) +(r2F2) +(r3F3) +M1+M2is the vector sum of the moments of all the forces about point Oplus the sumof all the couple moments. This system may be further simplified by first resolving the couple moment

into two components one parallel and the other perpendicular to the force FR, as in Figure

FIGURE 1.8

FIGURE 1.9

O

F

r

d

r1 r2

F

M r F r F r r F

r F

C= =1 2 1 2 -

=

+ ( ) ( )

F FR =

M MR OO=

MRO

==

M

O

d

FR

(c)

O

r1

r3r2

F3

F2

F1

M1

M2

(a)

MRO

M

M

FR

O

(b)



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1.9(b).By moving the line of action of FRin the plane perpendicular to M^a distance d = M /FR, so that

FR creates the moment M^ about O, the system can then be represented by a wrench, that is, a single

force FRand collinear momentM ||, Figure 1.9(c).

Note that in the special case of q =90, Figure 1.9(b), M||=0and the system then reduces to a singleresultant force FRhaving a specified line of action. This will always be the case if the force system is either

concurrent, parallel, or coplanar.

Distributed Loadings

When a body contacts another body, the loads produced are always distributed over the surface area of

each body. If the area on one of the bodies is small compared to the entire surface area of the body, the

loading can be represented by a single concentrated force acting at a point on the body. However, if the

loading occurs over a large surface area such as that caused by wind or a fluid the distribution of

load must be taken into account. The intensity of this surface loading at each point is represented as a

pressure and its variation is defined by a load-intensity diagram. On a flat surface, the load intensity

diagram is described by the loading functionp=p(x,y), which consists of an infinite number of parallelforces, as in Figure 1.10. Applying Equation (1.6) and Equation (1.7), the resultant of this loading and

its point of application ( ) can be determined from

(1.8)

(1.9)

Geometrically, FR is equivalent to the volume under the loading diagram, and its location passes

through the centroid or geometric center of this volume. Often in engineering practice, the surface loading

is symmetric about an axis, in which case the loading is a function of only one coordinate, w =w(x).Here the resultant is geometrically equivalent to the area under the loading curve, and the line of action

of the resultant passes through the centroid of this area.

Besides surface forces as discussed above, loadings can also be transmitted to another body without

direct physical contact. These body forces are distributed throughout the volume of the body. A common

FIGURE 1.10

x y,

F p x y dAR = ( , )

xx p x y dA

p x y dAy

y p x y dA

p x y dA= =

( , )

( , )

( , )

( , )

x

y

p = p(x, y)

P

FR

y

x



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example is the force of gravity. The resultant of this force is termed theweight; it acts through the bodys

center of gravity and is directed towards the center of the earth.

1.2 Equilibrium

Equations of Equilibrium

A body is said to be in equilibrium when it either is at rest or moves with constant velocity. For purposes

of analysis, it is assumed that the body is perfectly rigid, meaning that the particles composing the body

remain at fixed distances from one another both before and after applying the load. Most engineering

materials deform only slightly under load, so that moment arms and the orientation of the loading

remain essentially constant. For these cases, therefore, the rigid-body model is appropriate for analysis.

The necessary and sufficient conditions to maintain equilibrium of a rigid body require the resultant

external force and moment acting on the body to be equal to zero. From Equation (1.6) and Equation

(1.7), this can be expressed mathematically as

(1.10)

(1.11)

If the forces acting on the body are resolved into their x, y, z components, these equations can be

written in the form of six scalar equations, namely,

(1.12)

Actually, any set of three nonorthogonal, nonparallel axes will be suitable references for either of these

force or moment summations.

If the forces on the body can be represented by a system of coplanar forces, then only three equations

of equilibrium must be satisfied, namely,

(1.13)

Here thexandyaxes lie in the plane of the forces and point Ocan be located either on or off the body.

Free-Body Diagram

Application of the equations of equilibrium requires accountability for allthe forces that act on the body.

The best way to do this is to draw the bodys free-body diagram.This diagram is a sketch showing an

outlined shape of the body and so represents it as being isolated or free from its surroundings. On this

sketch it is necessary to show all the forces and couples that act on the body. Those generally encountered

are due to applied loadings, reactions that occur at the supports and at points of contact with other

bodies, and the weight of the body. Also one should indicate the dimensions of the body necessary for

F 0=

M 0O =

F M

F M

F M

x Ox

y Oy

z Oz

= == =

= =

0 0

0 0

0 0

F

F

M

x

y

O

=

=

=

0

0

0



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computing the moments of forces. Once the free-body diagram has been drawn and the coordinate axes

established, application of the equations of equilibrium becomes a straightforward procedure.

Support Reactions

Various types of supports can be used to prevent a body from moving. Table 1.1 shows some of the most

common types, along with the reactions each exerts on the body at the connection. As a general rule, if

a support prevents translation in a given direction, then a force is developed on the body in that direction,

whereas if rotation is prevented, a couple moment is exerted on the body.

Friction

When a body is in contact with a rough surface, a force of resistance called friction is exerted on the

body by the surface in order to prevent or retard slipping of the body. This force always acts tangent to

the surface at points of contact with the surface and is directed so as to oppose the possible or existing

motion of the body. If the surface is dry, the frictional force acting on the body must satisfy the equation

(1.14)

TABLE 1.1 Force Systems

Connection

cable

Reaction Connection

smooth surface

Reaction

roller ball and socket

pin single pin

fixed supportfixed support

F

Fy

Fx

Fy

Fy

Fx

M

Fz

Fy

Fx Fz

My

Mz

FzFx

Fy

Fx

Fz

Fy

Mx

Mz

My

F Ns< m



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states that if a body is in equilibrium, then so is any segment

of the body. For example, if an imaginary section is passed

through the body in Figure 1.14(a), separating it into two parts,

the free-body diagram of the left part is shown in Figure

1.14(b). Here the six internal resultant components are

exposed and can be determined from the six equations of

equilibrium given by Equation (1.12). These six components

are referred to as the normal force, Ny, the shear-force com-

ponents, Vx and Vz, the torque or twisting moment, Ty, andthe bending-moment components,Mx andMz.

If only coplanar loads act on the body [Figure 1.15(a)], then

only three internal resultant loads occur [Figure 1.15(b)],

namely, the normal force,N,the shear force, V, and the bending

moment, M.Each of these loadings can be determined from

Equation (1.13). Once these internal resultants have been com-

puted, the actual load distribution over the sectioned surface,

called stress, involves application of the theory related to

mechanics of materials.

Numerical Applications

The following examples illustrate application of most of the

principles discussed above. Solution of any problem generally

requires first establishing a coordinate system, then representing the data on a diagram, and finally

applying the necessary equations for solution.

FIGURE 1.13 FIGURE 1.14

A

B

F

BNB

F

Ay

AxA

Mx

F2

F1

F2

F1

F4

F3

Mz

Ty

x

z

(a)

(b)

Vz

Vx Nyy

FIGURE 1.15

y

M

V

Nx

F1

F1

F2

F2

F3

F4

(b)

(a)



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Example 1.1

Simplify the system of three parallel forces acting on the plate in Figure 1.16 to a single resultantforce and specify where the force acts on the plate.

Solution

First, Equation (1.6) and Equation (1.7) are applied in order to replace the force system by a single

resultant force and couple moment at point O.

Ans.

Since the forces are parallel, note that as expected FRis perpendicular to .

The two components of can be eliminated by moving FR along the respectiveyandxaxes an

amount:

Ans.

Ans.

Both coordinates are positive since FR, acting at r= {0.556i+2.17j} m, will produce the requiredmoment =rFR.

Example 1.2

Determine the reactions at the supports for the beam shown inFigure 1.17(a).

Solution

Using Table 1.1,the free-body diagram for the beam is shown inFigure 1.17(b).The problem is

statically determinate. The reaction NB can be found by using the principle of moments and

FIGURE 1.16

1.5 mx

1 m

0.5 m

2 my

300 N

200 N

z

A

rA

400 N

C

O

B

rC

rB

FR

y x

F F F k k k k NR R= = - - - = - 300 400 200 900{ }

M M M r k r k r k

i j j j k

i j k

i j

( ) ( ) ( )

( . ) ( ) ( . ) ( )

( . . ) ( )

{ }

R O R A B C O O= = - + - + -

= + - + -

+ - + -

= - +

300 400 2002 1 5 300 2 5 400

0 5 2 5 200

1950 500 N m

MRO

MRO

x M FOy R= = =/ / .500 900 0 556N m N m

y M FOx R= = =/ / .1950 900 2 17N m N m

MRO



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summing moments about pointAto eliminate Axand Ay. Applying Equation (1.13) with referenceto the coordinate system shown gives

Ans.

Ans.

Ans.

Since the answers are all positive, the assumed sense of direction of the reactive forces is shown

correctly on the free-body diagram.

Example 1.3

The compound beam shown in Figure 1.18(a)consists of two segments, AB and BC,which arepinned together at B. Determine the reactions on the beam at the supports.

Solution

The free-body diagrams of both segments of the beam are shown in Figure 1.18(b). Notice how the

principle of action equal but opposite reaction, Newtons third law applies to the two force

FIGURE 1.17

60

500 kN

5 34

2m

B

60A

2 m 2 m

(a)

500 kN

5 34

2m 60

2 m 2 m

(b)

Ax

Ay

NB

x

y

M N

N

N

A B

B

B

= - + +

+ =

=

0 500 3 5 2 60 4 2 6060 2 60 0

150

( / )( ) cos ( cos )

cos ( sin )

N m m m

m

N

F A

A

x x

x

= - - =

= 0 500 4 5 150 60 0

530

( / ) sinN N

N

F Ay y

y

= - + =

= 0 500 3 5 150 60 0

225

( / ) cosN N

NA



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Defining Terms

Concurrent forces Forces that act through the same point.

Couple Two forces that have the same magnitude and opposite directions and do not have the same

line of action. A couple produces rotation with no translation.

Free-body diagram A diagram that shows the body free from its surroundings. All possible loadsand relevant dimensions are labeled on it.

Friction A force of resistance caused by one surface on another.

Method of sections This method states that if a body is in equilibrium, any sectioned part of it is also

in equilibrium. It is used for drawing the free-body diagram to determine the internal loadings

in any region of a body.

Parallelogram law The method of vector addition whereby two vectors, called components, are joined

at their tails; parallel lines are then drawn from the head of each vector so that they intersect

at a common point forming the adjacent sides of a parallelogram. The resultant vector is the

diagonal that extends from the tails of the component vectors to the intersection of the lines.

Principle of moments This concept states that the moment of the force about a point is equal to thesum of the moments of the forces components about the point.

Principle of transmissibility A property of a force that allows the force to act at any point along its

line of action and produce the same external effects on a body.

Weight The gravitational attraction of the earth on the mass of a body, usually measured at sea level

and 45latitude.Wrench A force and collinear moment. The effect is to produce both a push and simultaneous twist.

Reference

Hibbeler, R. C. 2004. Engineering Mechanics: Statics,10th ed. Prentice Hall, Englewood Cliffs, NJ.

Further Information

Many textbooks are available for the study of statics; they can be found in any engineering library.

capítulo 1. sistema de la fuerza resultantes y equilibrio

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