Coplanar Non-concurrent Force System: This is the force system in which lines of action of individual forces lie in the same plane but act at different

Coplanar Non-concurrent Force System: This is the force system in which lines of action of individual forces lie in the same plane but act at different points of application.

CHAPTER – II RESULTANT OF COPLANAR NON CONCURRENT

FORCE SYSTEM

Fig. 1

F2F1

F3

Fig. 2

F1 F2

F5F4

F3

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TYPES

1. Parallel Force System – Lines of action of individual forces are parallel to each other.

2. Non-Parallel Force System – Lines of action of the forces are not parallel to each other.

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MOMENT OF A FORCE ABOUT AN AXIS

The applied force can also tend to rotate the body aboutan axis, in addition to causing translatory motion. This

rotational tendency is known as moment.

Definition: Moment is the tendency of a force to make arigid body rotate about an axis.

This is a vector quantity.

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Moment Axis: This is the axis about which rotational tendency is determined. It is perpendicular to the plane comprising the moment arm and line of action of the force.

Moment Center (B): This is the point at which the moment axis intersects the plane of the coplanar system. Moment Arm: The perpendicular distance from the moment center to the line of action of the force.

Distance AB = d.

BAd

F

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Consider the example of a pipe wrench.

The force applied which is

perpendicular to the handle of the

wrench tends to rotate the pipe about

its vertical axis. The magnitude of

this tendency depends both on the

magnitude of the force and the

moment arm ‘d’.

EXAMPLE FOR MOMENT5


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MAGNITUDE OF MOMENT

B

F

Ad

It is computed as the product of the the perpendicular distance to the line

of action of the force from the moment center and the magnitude of the force.

MA = d×F

Unit – Unit of Force × Unit of distance kN-m, N-mm etc.

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SENSE OF THE MOMENT

The sense is obtained by the ‘Right Hand Thumb’ rule.

‘If the fingers of the right hand are curled in the direction of rotational tendency of the body, the extended thumb represents the +ve sense of the moment vector’.

M M

For the purpose of additions, the moment direction may be considered by using a suitable sign convention such as +ve for counterclockwise and –ve for clockwise rotations or vice-versa.

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VARIGNON’S THEOREM (PRINCIPLE OF MOMENTS)

Statement: The moment of a force about a moment center (axis) is equal to the algebraic sum of the moments of the component forces about the same moment center (axis).

Proof (by Scalar Formulation):Proof (by Scalar Formulation):

Ry

Py

Q R

A O

p

q r PQy

Let ‘R’ be the given force.‘P’ & ‘Q’ be the component forces of ‘R’. ‘O’ be the moment center. p, r, and q be the moment arms of P, R, and Q respectively from ‘O’., , and be the inclinations of ‘P’, ‘R’, and ‘Q’ respectively w.r.t. the X – axis.

Y

X

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We have,Ry = Py + Qy R Sin = P Sin + Q Sin ----(1)From le AOB, p/AO = Sin From le AOC, r/AO = Sin From le AOD, q/AO = Sin From (1), R ×(r/AO) = P ×(p/AO) + Q ×(q/AO)

i.e., R × r = P × p + Q × q

Moment of the resultant R about ‘O’ = algebraic sum of moments of the component forces P & Q about same moment center ‘O’.

Ry

Py

R

A O

Qy

Q

Pp

qr

Y

X

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VARIGNON’S THEOREM – PROOF BY VECTOR FORMULATION

Consider three forces F1, F2, and F3 concurrent at point ‘A’ as shown in fig. Let r be the position vector of ‘A’ w.r.t ‘O’. The sum of the moments about ‘O’ for these three forces by cross-product is,Mo = ∑(r×F) = (r×F1) + (r×F2) + (r×F3).By the property of cross product,Mo = r × (F1+F2+F3) = r × Rwhere, R is the resultant of the three forces.

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1. It simplifies the computation of moments by judiciously selecting the moment center. The

moment can be determined by resolving a force into X & Y components, because finding x & y distances in many circumstances may be easier than finding the perpendicular distance (d) from the moment center to the line of action.

2. Location of resultant - location of line of action of the resultant in the case of non-concurrent force systems, is an additional information required, which can be worked out using Varignon’s theorem.

APPLICATIONS OF VARIGNON’S THEOREM

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COUPLE

Two parallel, non collinear forces (separated by a certain distance) that are equal in magnitude and opposite in direction form a ‘couple’.

The algebraic summation of the two forces forming the couple is zero. Hence, a couple does not produce any translation, but produces only rotation. d

F

F

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Moment of a Couple: Consider two equal and opposite forces separated by a distance ‘d’. Let ‘O’ be the moment center at a distance ‘a’ from one

of the forces. The sum of moments

of two forces about the point ‘O’ is,

+ ∑ Mo = -F × ( a + d) + F × a = -F× d

Thus, the moment of the couple about ‘O’ is independent of the location, as it is independent of the distance ‘a’.

The moment of a couple about any point is a constant and is equal to the product of one of the forces and the perpendicular distance between them.

O

da

F

F

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RESOLUTION OF A FORCE INTO A FORCE-COUPLE SYSTEM

P

F

=P

FF

Q

F

=Q

M=F × d

d

A given force F applied at a point can be replaced by an equal force applied at another point Q, together with a couple which is equivalent to the original system.

Two equal and opposite forces of same magnitude F

and parallel to the force F at P are introduced at Q.

Fig. (a) Fig. (b) Fig. (c)

F

F

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P

F

=

P

F

Q

F

= Q

M=F × d

d

Fig. (a) Fig. (b) Fig. (c)

F

F

Of these three forces, two forces i.e., one at P and the

other oppositely directed at Q form a couple.

Moment of this couple, M = F × d.

The third force at Q is acting in the same direction as that at P.

The system in Fig. (c) is equivalent to the system in Fig. (a).

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Thus, the force F acting at a point such as P on a rigid

body can be moved to any other given point Q, by adding a

couple M. The moment of the couple is equal to moment of

the force in its original position about Q.

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The resultant of coplanar, non-concurrent force

systems is the one which will produce same rotational and

translational effect as that of the given system. It may be a

force and a moment or a pure moment.

RESULTANT OF COPLANAR NON-CONCURRENTFORCE SYSTEMS

Let F1, F2 and F3 constitutea non concurrent system as shown in the fig.‘O’ – be any convenient referencepoint in the plane.

F1

F3

F2O

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Each force can be replaced by a force of the same magnitude and acting in the same direction at ‘O’ and a moment about ‘O’ as in Fig B. The non concurrent forces can be combined as in the case of concurrent system to get the resultant force R. Thus, the resultant of the system is equal to a force R at ‘O’ and a moment ‘∑Mo’ as shown in fig.C.

F1

F2

OO

∑Mo

O∑Mo

F3

R

Fig. A Fig. BFig. C

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F1

F2

F3


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The single force R and the moment ‘∑Mo’ shown in the fig.C can be replaced by a single force R acting at a distance ‘d’ from ‘O’, which gives the same effect. Thus, the resultant can be reduced to a single force acting at a certain distance from ‘O’. Mathematically,

O∑Mo

R

Fig. C

O

R

Fig. D

d

Rx

Rytan

22 RyRxR

R

Mod

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X and Y intercepts of Resultant:In some problems, it may be required to determine

distances of the resultant R along x-axis and y-axis i.e., X and Y intercepts. Let ‘d’ be the perpendicular distance of the resultant from ‘O’ as shown in the fig. Let Rx=∑Fx and Ry=∑Fy be thecomponents of the resultant in X and Ydirections. By Varignon’s theorem,

R×d= ∑Mo

At B, ∑Mo = Rx×0 + Ry×X R

Y

XO

d

Y-axis

X-axis

Rx

Ry

Rx

Ry

A

B

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Therefore,X= ∑Mo/Ry

Similarly, at A, Ry×0 + Rx×Y = ∑Mo

Therefore,Y= ∑Mo/Rx

RY

XO

d

Y-axis

X-axis

Rx

Ry

Rx

Ry

A

B

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TYPES OF LOADS ON BEAMS

1. Concentrated Loads – This is the load acting for a very small length of the beam.

2. Uniformly distributed load – This is the load acting for a considerable length of the beam with same intensity of W kn/m throughout its spread.

Total intensity = W × L (acts at L/2 from one end of the spread) 3. Uniformly varying load – This load acts for a considerable

length of the beam with intensity varying linearly from ‘0’ at one end to W kN/m to the other representing a triangular distribution. Total intensity of load = area of triangular spread of the load = 1/2× W × L. (acts at 2×L/3 from ‘Zero’ load end)

W kN/m

L

W kN/m

L

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PROBLEMS FOR PRACTICE

1. Determine the resultant of the parallel coplanarforce system shown in fig. Take radius of the circle=1860mm

(Ans. R=2000N towards left, d=626.9mm)

400 N

1000 N

2000 N

600 N

o

60º

60º30º

10º

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2. Four forces of magnitudes 10N, 20N, 30N and 40N acting respectively along the four sides of a square ABCD as shown in the figure. Determine the magnitude, direction and position of resultant w.r.t. A.

(Ans:R=28.28N, θ=45º, X=1.77a)

10N

40N

a

a

30N

20N

A B

CD

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3. Four parallel forces of magnitudes 100N, 150N, 25N and 200N acting at left end, 0.9m, 2.1m and 2.85m respectively from the left end of a horizontal bar of length 2.85m. Determine the magnitude of resultant and also the distance of the resultant from the left end.

(Ans:R=125N, X=3.06m)

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4. Reduce the given forces into a single force and a couple at A. (Ans:F=320kN, θ=14.48º, M=284.8kNm)

100KN80KN

1m

1.5m

70.7kN200kN

A

30º45º

30º

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5. Determine the resultant w.r.t. point A. (Ans:R=450kN, X=7.5kNm)

A

150N

1.5m3m1.5m

150Nm

500N100N

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Documents

Coplanar Non-concurrent Force System: This is the force system in which lines of action of individual forces lie in the same plane but act at different