K - Moment of a Couple Angel

K - Moment of a Couple Angel - FORCE SYSTEM RESULTANTS...

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Unformatted text preview: FORCE SYSTEM RESULTANTS MOMENT OF A COUPLE LEARNING OBJECTIVES Be able to define a couple Be able to determine the moment of a couple PRE-REQUISITE KNOWLEDGE Units of measurements Trigonometry concepts Vector concepts Rectangular components concepts Moment concept Cross product and dot product DEFINITION OF COUPLE A couple is defined as two parallel forces of the same magnitude acting in opposite direction and separated by a constant distance. F -F Resultant force of a couple = 0 A moment of a couple (also called free vector) or a couple moment has a constant direction and magnitude regardless of the location of the positions of the forces to the reference point MOMENT OF A COUPLE Scalar formulation Magnitude of couple moment : M = ( F )(d ) Where d is the perpendicular distance between the couple forces The direction of the couple moment follows the right-hand rule EXAMPLE - (Scalar Formulation) Determine the magnitude and direction of the couple moment Magnitude of couple moment : d = 6 cos 30 = 5.20 in M = ( F )(d ) = (25)(5.20) = -130 lb - in MOMENT OF A COUPLE Vector formulation M = rF Where r is any position vector projecting from one couple force to another As couple moment is a free vector Also, M = rA ( - F ) + rB F EXAMPLE 1 Determine the couple moment in vector format if F = (25k) N SOLUTION 1 The couple force vector at B: F = ( 25k )N Position vector from A to B: r rAB = {-350i - 200 j}mm The couple moment vector : j - 200 0 k 0 25 i M = rAB F = - 350 0 = ( - 200 25 - 0 ) i - ( - 350 25 - 0 ) j = {-5000i + 8750 j}N - mm EXAMPLE 2 Determine the distance d so that the resultant couple moment has a magnitude of 20 N-m SOLUTION 2 The couple force vectors at B and C : rAB rAC FB = (35k ) N FC = ( - 50i ) N The position vectors from A to B and A to C are: rAB = {-d cos 30j + d sin 30k} mm rAC = {250i - d cos 30 j + d sin 30k} mm SOLUTION 2-Contd. Resultant moment couple : M = rAB FB + rAC FC i =0 0 j 0.866d 0 k i 0.5d + 250 35 - 50 j - 0.866d 0 k 0 .5 d 0 = ( - 0.866d 35 - 0 ) i - ( 0 - ( - 50 ) 0.5d ) j + ( 0 - ( - 50 ) ( - 0.866d ) ) k = d { - 30.31i - 25 j - 43.30k } N - mm SOLUTION 2-Contd. Resultant of couple moments : M = (d / 1000){ - 30.31i - 25 j - 43.30k} N - m The magnitude of the resultant of couple moments : M = ( d / 1000) ( - 30.31) 2 + ( - 25) + ( - 43.30 ) 2 2 = (0.05847)(d ) N - m For M = 20 N-m, d can be calculated as: d = 20/0.05847 = 342 mm ...
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This note was uploaded on 02/03/2012 for the course CE 221 taught by Professor Buch during the Summer '08 term at Michigan State University.

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