MP Lecture 03 Moments and Couples

Prepared by Verified by Approved byDr. P. Venkata Reddy Dr. Dinesh Keloth Kaithari Head of the Department

M. Tech, Ph. D (IISc Bangalore) Module Leader Mechanical and Industrial Engineering

CALEDONIAN COLLEGE OF ENGINEERINGMUSCAT - SULTANATE OF OMAN

Module: Mechanical Principles (Code: M1H321458)

Moments and Couples

Moment

A force will cause motion along its direction.

A force may also cause rotation about a fixed point some distance away. This rotation or turning effect of a force iscalled MOMENT.

Moment of a force about a point or axis a measure of the tendency of the force to cause a body to rotate aboutthe point or axis

2Academic Year 2014-15 - semester A Moments and Couples

Moment of a Force

The moment M of a force F about a fixed point A is defined as the product of the magnitude of force F and theperpendicular distance d from point A to the line of action of force F.

MA = F x d

Where force F is in Newtons, N and distance d is in meters, m

Thus moment MA is in Newton-meter, N-m.

Moment Sign Convention

Direction using right hand rule

Anti-clock wise: + VE (Counter-clockwise) Clockwise: - VE


Resultant Moment

Resultant moment, MRo = moments of all the forces = F x d = F1xd1F2xd2+F3xd3

Example 1:

Calculate the moment of the 500 N force about the point A as shown in the diagram.

Solution:

Since the perpendicular distance from the force to the axis point A is 1.5 m, from

MA = F x d = - 500 x 1.5 = - 750 Nm = 750 Nm


Example 2:

Calculate the moment about point A caused by the 500 N force as shown in the diagram.

Solution:

MA = F x d = -500 x AB = -500 x 1.5 sin 60 = -649.5 N-m = 649.5 N-m


Example 3:

Determine the resulting moment about point A of the system of forces on bar ABC as shown in the diagram.

Solution:

AD = (1.5 + 0.5) sin 60 = 2.0 sin 60 = 1.732 m

MA = (-500 x 1.5) + (-800 x 1.732) = - 750 1385.6 = - 2135.6 Nm = 2135.6 Nm


Example 4:

For each case, determine the moment of the force about point O.

Solution:

Line of action is extended as a dashed line to establish moment arm d. Tendency to rotate is indicated and the orbitis shown as a colored curl.

CW clock wise direction CCW counter clock wise direction

)(.200)2)(100()( CWmNmNMa o )(.5.37)75.0)(50()( CWmNmNMb o )(.229)30cos24)(40()( CWmNmmNMc o

)(.4.42)45sin1)(60()( CCWmNmNMd o

)(.0.21)14)(7()( CCWmkNmmkNMe o


Varignons Theorem:

Varignons Theorem states that the moment of a force about any point is equal to the sum of the moments of itscomponents about the same point.

To calculate the moment of any force with a slope or at an angle to the x or y-axis, resolve the force into the Fx andthe Fy components, and calculate the sum of the moment of these two force components about the same point.

Example 5:

Re-Calculate Example 3 Above Employing Varignons Theorem.

Solution:

MA = (-500 x 1.5) + (- 800 sin 60 x 2) =-750 -1385.6 =-2135.6 Nm = 2135.6 Nm* Same answer as in example 3


Couple

A couple consists of a pair of two forces which has thefollowing properties:

-Equal magnitude and opposite in direction-Act along parallel lines of action-Separated by a perpendicular distance d.

A couple causes a body to rotate only without translational motion since the two forces cancels out each othergiving zero resultant. A couple acting in a system of forces will only contribute to the resulting moment but not tothe resulting force.

Magnitude of a Couple:Consider a light bar acted upon by a couple as shown:

The moment of the couple about A isMA = F x d2 - F x d1 = F x (d2 - d1) = F x d

What is the total moment of the couple about point B?MB = F x d/2 + F x d/2 = F x (d/2 + d/2) = F x d

The couples moment about any pivot point is equal to F x dA couple has the same moment about all points on a body.


Procedure for Analysis

1. Establish the coordinate axes with the origin located at a given reference point and the axes having a selectedorientation

2. Force Summation3. Moment Summation

Example 6:

A light bracket ABC is subjected to two forces and two couples as shown. Determine the moment at (a) point Aand (b) point B.

Solution:

(a) MA = (2000 sin 30 x 2.5) (2000 cos 30 x 1.5) + ( 3000 x 0) + 120 80

= 2500 2598 + 0 +120 80 = - 58 Nm = 58 Nm

(b) MB = (2000 sin 30 x 0) (2000 cos 30 x 1.5) + (3000 x 0) + 120 80 = -2558 Nm = 2558 Nm


Force Couple Equivalent

The Force-Couple Equivalent concept will enable us to transfer a force to another location outside its line ofaction.

Consider a force F acting at a point B on a rigid body as shown in diagram (a) below. How do we transfer the forceF from point B to point A?

The above shows how a force can be replaced by a force-couple equivalent.


Example 7:

Determine the force-couple equivalent at point A for the single force of 20 kN acting at point C on the bracketABC.

Solution:MA = (20 sin 40 x 2.3) (20 cos 40 x 1.2) = 11.18 kNm


Example 8:

Determine the force-couple equivalent at point A for Multiple force system of 20 kN acting at point C and another30 kN horizontal force is acting at point B on the bracket ABC.

Solution:

Rx = Fx = 20 cos 40 + 30 = 45.32 kN Ry = Fy = 20 sin 40 = 12.86 kNTherefore R = (45.322 + 12.862) = 47.11 kN

MA = (20 sin 40 x 2.3) (20 cos 40 x 1.2) + (30 x 0) = 11.18 kNm

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MP Lecture 03 Moments and Couples

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