Download - MP Lecture 03 Moments and Couples
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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
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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
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3Academic Year 2014-15 - semester A Moments and Couples
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
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4Academic Year 2014-15 - semester A Moments and Couples
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
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5Academic Year 2014-15 - semester A Moments and Couples
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
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6Academic Year 2014-15 - semester A Moments and Couples
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
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7Academic Year 2014-15 - semester A Moments and Couples
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
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8Academic Year 2014-15 - semester A Moments and Couples
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.
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9Academic Year 2014-15 - semester A Moments and Couples
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
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10Academic Year 2014-15 - semester A Moments and Couples
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.
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11Academic Year 2014-15 - semester A Moments and Couples
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
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12Academic Year 2014-15 - semester A Moments and Couples
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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