Equilibrium of System of Coplanar Forces

Equilibrium of

System of

Coplanar forces

Engineering Mechanics Equilibrium of system of coplanar forces

The Automobile Society (India) Page 1

EQUILIBRIUM OF SYSTEM OF COPLANAR FORCES

INTRODUCTION

When a particle is acted upon by a number of forces the resultant force will produce the same

effect as produced by all the given forces. A little consideration will show that if the resultant of

a number of forces, acting on a particle is zero, the particle will be in equilibrium. Such a set of

forces, whose resultant is zero, are called equilibrium forces. The force, which brings the set of

forces in equilibrium, is called an equilibrant.

Condition of equilibrium

Consider a body acted upon by a number of coplanar non-concurrent forces. A little

consideration will show, that as a result of these forces, the body may have any one of the

following states.

1. The body may move in any one direction

2. The body may rotate about itself without moving.

3. The body may move in any one direction and at the same time it may also rotate about

itself.

4. The body may be completely at rest.

Concurrent System

The following set of equation insures equilibrium of a concurrent, spatial system of forces.

1. where Fx, Fy, and Fz = algebraic sum of the x, y, and z components

2. Respectively of the forces of the system.

3.

M = 0 may be used as an alternate to one of the above equation. For example, if it replaces

equation (3), then M must be the algebraic sum of the moments of the forces of the system

about neither an axis neither parallel to nor interesting the z axis

Parallel System

The following set of equation insures equilibrium of a parallel, spatial system of forces.

1. Fy = 0 where Fy = algebraic sum of the forces of the system along the y axis

2. Mx=0 parallel to the system,

3. Mz=0

Mx and Mz = algebraic sums of the moments of the force of the system about the x and axes

respectively.



NONCONCURRENT, NONPARALLEL SYSTEM

The following six equations are necessary and sufficient condition for equilibrium of the most

general force system in three-dimensional space.

1. Fx = 0 where Fx, Fy,, and Fz = algebraic sum of the x, y, and z components

2. Fy = 0 respectively of the forces of the system.

3. Fz = 0

4. Mx=0 Mx, My, and Mz = algebraic sum of the moments of the forces of

5. My=0 the system about x, y, and z axes respectively.

6. Mz=0

All the systems are special cases of this system.

SYSTEM OF FORCES

When two or more forces act on a body, they are called to for a system of forces.

Following systems of forces are important from the subject point of view:

1. Coplanar forces: The forces, whose lines of action lie on the same plane, are known as

coplanar forces.

2. Collinear forces: The forces, whose lines of action lie on the same line, are known as

collinear forces.

3. Concurrent forces: The forces, which meet at one point, are known as concurrent forces.

The concurrent forces may or may not b collinear.

4. Coplanar concurrent forces: The forces, which meet at one point and their line of action

also lay on the same plane, are known as coplanar concurrent forces.

5. Coplanar non-concurrent forces: The forces, which do not meet at one point, but their lines

of action lie on the same, are known as coplanar non-concurrent forces.

6. Non-Coplanar concurrent forces: The forces, which meet at one point, but their lines of

action do not lie on the same plane, are known as non-coplanar concurrent forces.

7. Non-Coplanar non-concurrent forces: The forces, which do not meet at one point and their

lines of action do not lie on the same plane, are called non-coplanar non-concurrent forces.

Couples

A pair of two equal and unlike parallel forces (i.e. forces equal in

magnitude, with lines of action parallel to each other and acting in

opposite directions) is known as a couple.



Classification of Couples

The couples may be, broadly, classified into the following two categories depending upon their

direction, in which the couple tends to rotate the body on which it acts.

1. Clockwise couple, 2. Anticlockwise couple

It is the moment of a force, whose effect is to turn or rotate the body, in

the same direction in which hands of a clock move as shown in fig.(a)

2. It is the moment of a force, whose effect is to turn or rotate the

body, in the opposite direction in which the hands of a clock move as

shown in the fig. (b)

TYPE OF SUPPORTS

The three common types of connections which

join a built structure to its foundation are; roller

pinned and fixed. A fourth type, not often found

in building structures, is known as a simple

support. This is often idealized as a frictionless

surface). All of these supports can be located

anywhere along a structural element. They are

found at the ends, at midpoints, or at any other

intermediate points. The type of support

connection determines the type of load that the

support can resist. The support type also has a

great effect on the load bearing capacity of each element, and therefore the system.

The diagram illustrates the various ways in which each type of support is represented. A single

unified graphical method to represent each of these support types does not exist. Chances are that

one of these representations will be similar to local common practice. However, no matter what

the representation, the forces that the type can resist are indeed standardized.

Note:- The general convention is to take clockwise moment as positive and anticlockwise

moment as negative.



ROLLER SUPPORTS

Roller supports are free to rotate and translate along the surface upon which the roller rests. The

surface can be horizontal, vertical, or sloped at any angle. The resulting reaction force is always

a single force that is perpendicular to, and away from, the surface. Roller supports are commonly

located at one end of long bridges. This allows the bridge structure to expand and contract with

temperature changes. The expansion forces could fracture the supports at the banks if the bridge

structure was locked in place. Roller supports can also take the form of rubber bearing, rockers

or a set of gears which are designed to allow a limited amount of lateral movements.

a roller support cannot provide resistance to a lateral forces. Imagine a structure (perhaps a

person) on roller skates. It would remain in place as long as the structure must only support itself

and perhaps a perfectly vertical load. As soon as a lateral load of any kind pushes on the

structure it will roll away in response to the force. The lateral load could be a shove, a gust of

wind or an earthquake. Since most structures are subjected to lateral loads it follows that a

building must have other types of support in addition to roller supports.

PINNED SUPPORTS

A pinned support can resist both vertical and horizontal

forces but not a moment. They will allow the structural

member to rotate, but not to translate in any direction. Many

connections are assumed to be pinned connections even

though they might resist a small amount of moment in

reality. It is also true that a pinned connection could allow

rotation in only one direction; providing resistance to

rotation in any other direction. The knee can be idealized as

a connection which allows rotation in only one direction and

provides resistance to lateral movement. The design of a

pinned connection is a good example of the idealization of

the reality. A single pinned connection is usually not sufficient to make a structure stable.

Another support must be provided at some point to prevent rotation of the structure. The

representation of a pinned support includes both horizontal and vertical forces.



PINNED CONNECTIONS

These are the typical connection found in almost all trusses. They

can be articulated or hidden from view; they can be very

expressive or subtle.

There is an illustration of one of the elements at the Olympic

Stadium in Munich below. It is a cast steel connector that acts as a

node to resolve a number of tensile forces. Upon closer

examination once can notice that the connection is made of a

number of parts. Each cable is connected to the node by an end

bracket which is connected to a large pin. This is quite literally a

pinned connection. Due to the nature of the geometry of the

bracket and pin, a certain amount of rotation movement would be

permitted around the axis of each pin.

One of the connections from the pyramid of I.M. Pei's Loiuvre

addition follows below. Notice how it too utilized pinned

connections.

Pinned connections are confronted daily. Every time a

hinged door is pushed open a pinned connection has

allowed rotation around a distinct axis; and prevented

translation in two. The door hinge prevents vertical and

horizontal translation. As a matter of fact, if a sufficient

moment is not generated to create rotation the door will

not move at all.



FIXED SUPPORTS

Fixed supports can resist vertical and horizontal forces as

well as a moment. Since they restrain both rotation and

translation, they are also known as rigid supports. This

means that a structure only needs one fixed support in

order to be stable. All three equations of equilibrium can

be satisfied. A flagpole set into a concrete base is a good

example of this kind of support. The representation of

fixed supports always includes two forces (horizontal

and vertical) and a moment.

FIXED CONNECTIONS

Fixed connections are very common. Steel structures of

many sizes are composed of elements which are welded

together. A cast-in-place concrete structure is

automatically monolithic and it becomes a series of rigid connections with the proper placement

of the reinforcing steel. Fixed connections demand greater attention during construction and are

often the source of building failures.

For example: Let this small chair illustrate the way in which two types of "fixed" connections

can be generated. One is welded and the other is comprised to two screws. Both are considered

to be fixed connections due to the fact that both of them can resist vertical and lateral loads as



well as develop a resistance to moment. Thus, it found that not all fixed connections must be

welded or monolithic in nature. Let the hinges at locations A and B be examined in closer detail.

SIMPLE SUPPORTS

Simple supports are idealized by some to be frictionless surface supports. This is correct in as

much as the resulting reaction is always a single force that is perpendicular to, and away from,

the surface. However, are also similar to roller supports in this. They are dissimilar in that a

simple support cannot resist lateral loads of any magnitude. The built reality often depends upon

gravity and friction to develop a minimum amount of frictional resistance to moderate lateral

loading. For example, if a plank is laid across gap to provide a bridge, it is assumed that the

plank will remain in its place. It will do so until a foot kicks it or moves it. At that moment the

plank will move because the simple connection cannot develop any resistance to the lateral load.

A simple support can be found as a type of support for long bridges or roof span. Simple

supports are often found in zones of frequent seismic activity.



Introduction

Whenever some load is attached to a hanging wire, it extends and the load move downwards by

an amount equal to the extension of the wire. When the load moves downwards, it loses its

potential energy. This energy is absorbed (or stored) in the stretched wire, which may be released

by removing the load. On removing the load, the wire will spring back to its original position.

This energy, which is absorbed in a body, when strained within its elastic limit, is known as

Strain energy. It has been experimentally found that this strain energy is always capable of doing

some work. The amount of strain energy, in a body, is found out by the principal of work.

Mathematically

Strain energy = work done

TYPE OF LOAD

Load may act in any of the following three ways:

1. Gradual 2. Sudden 3. With impact

When the Load is Gradually Applied

It is the most common and practical way of loading a body, in which the loading starts from zero

and increases gradually till the body, is fully loaded. E.g. when we lower a body with the help of

a crane, the body first touches the platform on which it is to be placed. On further releasing the

chain, the platform goes on loading till it is fully loaded by the body. This is the case of a

gradually applied load. Now consider a metallic bar subjected to a gradual load.

Let P = Load gradually applied

A = Cross sectional area of the bar,

l = Length of the bar

E = Modulus of elasticity of the bar material, and

l = Deformation of the bar due to load

Since the load applied is gradual, and varies from zero to P, therefore the average load is equal to

P/2

Work done = Force Distance

= Average load Deformation



Since the energy stored is also equal to the work done, therefore strain energy stored

We also know that modulus of resilience

= strain energy per unit volume

Example calculates the strain energy stored in a bar 2 m long, 50 mm wide and 40 mm

thick when it is subjected to a tensile load of 60kN. Take E as 200 GPa.

Solution given: l=2 m = 2103 mm; b=50 mm; t=40 mm; P =60 kN = 6010

3 N and E =

200 GPa = 200103 N/mm

2

We know that stress in the bar

And volume of the bar,

Strain energy stored in the bar,

( )

Ans

When the Load is suddenly applied

Sometimes, in factories and workshops, the load is suddenly applied on a body. E.g. when we

lower a body with the help of a crane, the body is, first of all, just above the platform on which it

is to be placed. If the chain breaks at one at this moment the whole load of the body begins to act

on the platform. This is the case of a suddenly applied load. Now consider a bar subjected to a

sudden load.

P = Load gradually applied







= Stress induced by the application of the sudden load.

Since the load applied is sudden,

the load (P) is constant throughout the process of deformation of the bar.

Work done = Force Distance

= Average load Deformation

We know the strain energy stored.

Since the energy stored is equal to the work, therefore

Or

It means that the stress induced in this case is twice the stress induced, when the same load is

applied gradually. Once the stress () is obtained, the corresponding instantaneous deformation

(l) and the strain energy may be found out as usual.

Example An axial pull of 20 kN is suddenly applied on a steel rod 2.5 m long and 1000

mm2 in cross-section. Calculate the strain energy, which can be absorbed in the rod.

Take E = 200 GPA

Solution Given: P = 20 kN = 20 103 N; l = 2.5 m = 2.510

3 mm; A = 1000 mm

2 and E =

200 GPa = 200 103 N/mm

2.

We know that stress in the rod, when the load is suddenly applied.

And volume of the rod.

Strain energy which can be absorbed in the rod,

Ans.



When the Load is applied with Impact

Sometimes in factories and workshops, the load with impact is applied on a body, e.g. when we

lower a body with the help of a crane, and the chain breaks while the load

is being lowered, the load falls through a distance, before it touches the

platform. This is the case of a load applied with impact.

Now consider a bar subject to a load applied with impact as shown in the

fig.20.1

Let P = Load gradually applied





= Stress induced by the application of the sudden load.

h = Height through which the load will fall, before impact-in on the collar of the

bar

Work done = Load Distance moved

(i)

And energy stored

(ii)

Since energy stored is equal to the work done, therefore

(

) (

)

(

) (

)

Multiplying both sides by

(

)

This is a quadratic equation. We know that

(

) (

) (

)

*

+

Once the stress () is obtained, the corresponding instantaneous deformation (l) or the strain

energy stored may be found out as usual.



Cor. When l is very small as compared to h, then

Work done = Ph

Or

Example A 2 m long alloy bar of 1500 mm2 cross-sectional area hangs vertically and has a

collar securely fixed at its lower end. Find the stress inducted in the bar, when a weight of 2KN

falls from a height 0f 100 mm on the collar. Take E=120Pa. also find the strain energy stored in

the bar.

Solution Given: mm; A=1500 mm2; P=kN=2 103 N;

Gpa .

Stress induced in the bar

We know that in the, extension of the bar will be small and negligible as compared to the

height from where the weight fall on the collar (due to a small value of weight and

a large value of 1 00 mm)

Ans.

Strain energy stored in the bar

We also know that volume of the bar,

Example. A steel bar 3 m long and 2500 mm2 in area hangs vertically, which is securely

fixed on a collar at its lower end. If a weight of 15 kN on the collar from a height of 10 mm,

determine the stress developed in the bar. What will be the strain energy stored in the bar? Take

E as 200 Gpa.

Solution. Given: x x

x



Stress developed in the bar

We know that in this case, extension of the bar will be considerable as compared to the

height from where the weight falls on the collar (due to a large value of weight i.e,. 15 kN

and a small value of Therefore stress developed in the bar,

[

]

*

+

Ans.

Strain energy stored in the bar

We know that volume of the bar,

x x x

And strain energy stored in the bar.

x

x x

Ans.

Example. A copper bar of 12 mm diameter gets stretched by 1 mm under a steady load of

4 kN. What stress would be produced in the bar by a weight of 500 N, if it falls through 80 mm

before striking the collar rigidly fixed to the lower end of the bar? Take youngs modulus for the

bar material as 100 Gpa.

Solution. Given: x

x

Let = Length of the copper bar,

We know that cross-sectional area of the bar,

And stretching of the bar

( )

x x x



We also know that stress produced in the bar by the falling weight.

(

)

(

)

BEAMS

A beam is a structural member which has a length considerable larger than its cross-sectional

dimensions and which carries loads usually perpendicular to the axis of the beam (thus the loads

are at right angles to the length). The loads may be distributed over a very small distance along

the beam, in which case they are called concentrated, or they may be distributed over a

measurable distance, in which case they are called distributed.

Because design criteria usually are concerned with the ability of a beam to withstand shear forces

and bending moments.

Types of Beams

a) Simple-supports are at the ends

b) Cantilever-one end is mounted in a wall and the other end is free (this is the only type

considered here)

c) Overhanging-at least one support is not at the end.



Loading types

Beams are subjected to uniformly distributed loads (UDL), point (concentrated) loads or a

combination of both. The various loading conditions to which a beam may be subjected to are

shown below.



TRUSSES

Assumption

a) Each truss is assumed to be composed of rigid members all lying in one plane. This means

that coplanar force systems are involved.

b) The weights of the members are neglected because they are small in comparison with the

loads.

c) Forces are transmitted from one member, which are called two-force members, will be either

in tension (T) or compression (C)

Solution by the Method of Joints

To use this technique, draw a free body diagram of any pin in this truss, provided no more than

two unknown forces act on that pin. This limitation is imposed because the system of force is a

concurrent one for which, of course, only two equations are available for a solution. Proceed

from one pin to another until all unknowns have been determined.

Solution by the Method of Section

In the method of joints, forces in various members are determined by using free body diagrams

of the pins. In the method of sections, a section of the truss is taken as a free body diagram. This

involves cutting through a number of members, including those members whose forces are

unknown, in order to isolate one part of the truss. The forces in the members cut act as external

forces helping to hold that part of the truss in equilibrium. Since the system is non-concurrent,

nonparallel, three equations are available. Therefore, in any one sectioning no more than three

unknown froes can be found. Be sure to isolate the free body completely and at the same time

have no more than three unknown forces.

Documents

Equilibrium of System of Coplanar Forces