Moment of a Couple

Moment of a Couple:
Couple: Two forces having the same magnitude, parallel lines of action, and opposite sense.
F & F- form a couple
MA=F . d1- F . d2
=F (d1- d2)
= F . d











M: is called the moment of couple.
• It have the same magnitude and same sense regardless of the location of “A”.
• M is constant =F * distance between the lines of action.
• The sense of “M” is obtained by direct observation.

Equivalent Couples:
Two system of forces are equivalent (have the same effect on a rigid body ) if we can transform one of them into the other by means of one of the following operations (or several):

1- Replacing the forces by their resultant.




2- Resolving into components.




3- Canceling two equal & opposite forces acting on the same particle (on the same line).
4- Attaching to the same particle two equal & opposite forces.
5- Moving a force along its line of action.


Moment about “B”
MB=F d1
& by Varignon’s theorem
MB=Q d2 – P *0
MB=Q d2
? F d1 = Q d2 in magnitude & sense
When a couple acts on a rigid body , it does not matter where the two forces forming the couple act or what magnitude and direction they have. The only thing which counts in the moment of the couple (magnitude & sense ). Couples with the same moment will have the same effect on the rigid body .


Addition of Couples:


Two couples may be replaced by a single couple of moment equal to the algebraic sum of the moments of the given couples.



A given force can be resolved in to a force acting at a given point and a Couple:






3.13 The two couples shown are applied to a 6- by 8- in plate. Knowing that P1=P2=30 lb and Q1=Q2=40 lb, prove that their sum is zero (a) by adding their moments, (b) by combining P1 and Q1 into their resultant R1, combining P2 and Q2 into their resultant R2, and then showing that R1 and R2 are equal and opposite and have the same line of action.

a)

b)

? R1 and R2 are equal and opposite and have the same line of action

3.16 Four pegs are attached to a bord as shown. Two strings are passed around the pegs and pulled with forces of magnitude P= 16 lb and Q= 40 lb. Determine the required diameter of the pegs if the resultant couple applied to the board is to be 300 lb.in. clockwise.






3.18 A crane column supports a 16-kip load as shown. Reduce the load to an axial force along AB and a couple.





3.20 A 65-lb force is applied to a bent plate as shown. Determine an equivalent force-couple system (a) at A (b) at B.
(a) at A

(b) at B




3.22 The force and couple shown are to be replaced by an equivalent single force. Determine the required value of ? so that the line of action of the single equivalent force will pass through point B.








3.23 A hook is held by scews at A and B. (a) Replace the 50-lb load shown by an equivalent force-couple system at B. (b) Find two horizontal forces at A and B which form a couple equivalent to the couple found in part a .

(a) at B

(b)










3.24 A force P is applied to beam AB at point C as shown. Find the vertical forces FA and FB applied respectively , at A and B , which form a system equivalent to P.