In this example we will look at using a couple to determine reactions. The example is taken from Targ (1988) and the reader is urged to look at Sections 18 and 19 for the theory behind this.

The example concerns the problem below. The known forces are P = P’ = 300 N and the dimensions are a = 15 cm, b = 30 cm, and c = 20 cm. We need to determine the reactions at the supports B and C.

Let us begin by noting the following about the supports, a treatment taken in part from the conventions of finite element analysis:

  • Support B has only one degree of freedom (rotation); it cannot transmit moment and cannot move in the x or y directions, but can transmit force in either x or y directions.
  • Support C has two degrees of freedom (rotation and x-axis movement); it cannot transmit moment or force in the x direction but it can transmit force in the y-direction.

This pairing is necessary to maintain static determinacy and is common in elementary statics.

With that out of the way the solution proceeds as follows:

  • The force pair P and P’ are a couple with a moment arm a – c. There is no net force on the system. Only B can be an x-axis reaction, so the x-axis reaction at B is zero.
  • The couple produced by P and P’ is M = (300)(15 – 20) = -1500 N-cm. (clockwise negative moment)
  • We can construct a couple with forces Q and Q’. Their moment arm is 30 cm. The moment of this couple is M = -1500 N-cm = (Q)(30 cm), or Q = Q’ = -50 N which is what is shown.
  • Since the moment of the couples are the same and in the same direction (positive,) the senses of Q and Q’ are arranged accordingly.

It is also possible to solve this problem using straightforward summation of moments and forces. There are no forces to begin with in the y-direction and, as noted, the x-direction forces cancel each other out. Taking moments about B and assuming Q to be positive at C (for a positive moment around B, opposite of what is shown) the summation of moments about the entire system is

aP’ – cP + bQ = 0

– (15)(300) + (20)(300) + (30)Q = 0

Q = -((300)(5))/(30) = -50 N

which is the same result. The force Q’ must be 50 N upward (positive) as shown for summation of forces in the y-direction to be zero.