Equillibrium of Non Concurrent Coplanar System

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EQUILIBRIUM OF NON-CONCURRENT

COPLANAR FORCE SYSTEM

When a body is in equilibrium, it has neither translatory

nor rotatory motion in any direction. Thus the resultant

force R and the resultant couple M are both zero, and we

have the equilibrium equations for two dimensional forcesystem

Fx = 0;

Fy = 0 Eq(1)

M = 0

These requirements are both necessary and sufficient

conditions for equilibrium.



Supports: A structure is subjected to external forces

and transfers these forces through the supports on to thefoundation. Therefore the support reactions and the

external forces together keep the structure in equilibrium.

There are different types of supports. Some of them are

a) Roller Support b) Hinged or pinned support c) Fixed

or built in support

Some supports are shown in the figure along with the

reactions that can be mobilised.

Types of supports



Types of Supports Action on body

(a) Flexible cable ,belt ,chain,

rope

BODY

BODY

T

Force exerted by cable is

always a tension away from

the body in the direction of

cable

(b) Smooth surfacesContact forces are normal to

the surfaces

FF



(c) Roller

supportContact force is normal to thesurface on which the roller moves.

The reaction will always be

perpendicular to the plane of the

roller . Roller support will offeronly one independent reaction

component.(Whose direction is

known.)



( d )pinned Support / hinged support

This support does not allow any translatory movement of

the rigid body. There will be two independent reaction

components at the support. The resultant reaction can be

resolved into two mutually perpendicular components.

Or it can be shown as resultant reaction inclined at an

angle with respect to a reference direction.

R vR

Rh

θ



R v

MR H

(e) Fixed or Built-in Support

M



(contd .)

This type of support not only prevents the translatory

movement of the rigid body, but also the rotation of the

rigid body. Hence there will be 3 independent reactioncomponents of forces. Hence there will be 3 unknown

components of forces, two mutually perpendicular

reactive force component and a reactive moment asshown in the figure.



TYPES OF BEAMS

A member which is subjected to predominantly transverse loads

and supported in such a way that rigid body motion is preventedis known as beam. It is classified based on the support

conditions. A beam generally supported by a hinge or roller at

the ends having one span(distance between the support) is

called as simply supported beam. A beam which is fixed at oneend and free at another end is called as a cantilever beam.

(a) Simply supported beam

span

MA

VA

B

span

AHA

(b) Cantilever beam



If one end or both ends of the beam project beyond the supportit is known as overhanging beam.

A cantilever with a simple support anywhere along its length is

a propped cantilever.

(c) Overhanging beam

(right overhang)

MA

VA

B

span

A

HA

(d) Propped Cantilever

beam



A beam which is fixed at both ends is called a fixed beam.A beam with more than one span is called continuous beam.

VC

(f) Two Span continuous beam

VA VB

HA HB

MA

VA

span

VB

(e) Fixed beam

MB

HA



Statical ly determinate beam and statical ly indeterminatebeam :

Using the equations of equilibrium given in EQ(1) ,if all the

reaction components can be found out, then the beam is a

statical ly determinate beam ,and if all the reaction

components can not be found out using equations of

equilibrium only, then the beam is a statical ly indeterminate

beam.

In the above fig (a),(b)and ( c ) are statically determinate

beams ,where as (d),(e) and ( f) are statically Indeterminate

beams .



If the number of reaction components is more than thenumber of non-trivial equilibrium equations available then

such a beam is a statically indeterminate beam.

If the number of reaction components is equal to the

number of non-trivial equilibrium equations available then

such a beam is a statically determinate beam

If the number of reaction components is less than the

number of non-trivial equilibrium equations available then

such a beam is an unstable beam.



Determination of Beam reactions

Since three equilibrium equations are available, for a planar

structure a maximum of three unknown independent reactioncomponents can be determined using these equations.

Step I: Draw the free body diagram of the structure showing

the given loadings and the reactions at the supports.

Step 2: Apply the equations Fx = 0, Fy = 0, M = 0.

Assuming some directions and senses for unknown forces

and moments.

Step 3: solve for unknown reactions. If any of them is positive,

it is along the sense initially assumed while drawing the FBD.

If it is negative, it is opposite to the initially assumed sense



4kN/m

12kN/m

(1)Find the reactions at A,B,C and D for the beam loadedas shown in the figure(Ans.R A=R B =34kN;R C=28.84kN;

MC=-140kNm ; θC=-33.69 ˚ )

Problems for practice

12kN/m

4kN/m

20 kN

30kN

1m 2m 1m 1m 2m 1m 1m 2m

AB

C

34

40kNm



(2)A uniform bar AB of weight 50N shown in the figuresupports a load of 200N at its end. Determine the

tension developed in the string and the force supported

by the pin at B.(Ans. T=529.12N;R B=807.15N, θB=64.6˚)

2.5m2.5m200N

2 . 5 m

A

B

60˚

string



(3)Find the position of the hinged support (x),such that thereactions developed at the supports of the beam are equal..

(Ans.x=2m.)

2.0m 1.4m1.0m 3.0m0.6

15kN18kN/m

10kN/m

x



(4)A right angled bar ABC hinged at A as shown in figcarries two loads W and 2W applied at B &C .Neglecting

self weight of the bar find the angle made by AB with

vertical(Ans:θ =18.44˚)

0.5L2W

θ

B

L m

WC

A

Equillibrium of Non Concurrent Coplanar System

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