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The moment of two equally opposing force would create a couple-moment. Since the couple-moment value is invariant to the point considered to calculate the couple-moment, I can replace the two equally opposing force by a couple moment. I can place the couple moment anywhere, but the position will change the bending moment diagram (BMD). Since the bending moment diagram is different for two equivalent force system I think I'm doing something wrong. Can anyone point out to me what is it?

Example: Say a 10m long cantilever beam has a couple-moment of 10 Nm at 8m distance from fixed position. Now it has a specific BMD. Now I can replace the couple-moment by two opposite forces of 2.5 N each 4m apart (one at 10m from fixed distance, and one at 6m from fixed distance). This new force system is equivalent to the first force system. But the second force system has different bending moment diagram. In fact I can move the couple-moment anywhere and still have an equivalent force system; but with different BMDs.

Summary: Since moment of a couple is always the same about any point I can move around the couple position having equivalent force systems. But these equivalent force systems will have different BMDs. Why this discrepancy?

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  • $\begingroup$ How many forces are acting on the object under consideration? You cannot have an equilibrium situation with just one couple. $\endgroup$
    – Farcher
    Commented Oct 9, 2016 at 23:05
  • $\begingroup$ Have added an example in the question to answer yours. Thanks $\endgroup$
    – Naveed
    Commented Oct 10, 2016 at 19:21

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I think I found my mistake.

I started with the fact that 'the moment of a couple is same about any point'(Fact -1). And stemming from this fact I assumed that I can move around this couple and still have equivalent force system. This assumption is wrong. When I move around a couple 'Fact -1' is still true but the force systems are not equivalent. The definition of equivalent force system is that 'Two or more systems of forces and moments are said to be equivalent iff they have the same resultant force and resultant moment about any (and all) points (i.e., BMD, shear force diagram, axial force diagram are all same). So the definition of equivalent force system is set in the way precisely so that we don't get the discrepancy I was seeing earlier!

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  • $\begingroup$ You can move the pure torque around only for a rigid body. But for a flexible body it would make a difference, since you can think of a flexible body as a many rigid bodies elastically connected. So the point of application changes which infinitesimal rigid body receives the torque. $\endgroup$
    – John Alexiou
    Commented May 7, 2018 at 0:27
  • $\begingroup$ You can move a couple anywhere if you are only determining static equilibrium (reaction forces and moments) but you cannot move it when determining bending moments for mechanics of materials analysis. $\endgroup$
    – Bob D
    Commented Dec 16, 2018 at 18:04
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I think, in static equilibrium condition, we can move the bending couple at a point to any other point.Because in static equilibrium, the effect of the couple will be nullified (there will not be any rotations) by the reaction forces.

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You can move a moment (couple) around and not change its value. But you can't move a force around (off its line of action anyway) without changing the associated/added moment. In a traditional beam-bending diagram (which I guess you are referring to) we cut the end of the beam and replace it with a force and moment. But there is a reaction force at the other end (the other half of the moment) that doesn't move. So we are only moving one of the two forces, thus need to adjust the associated moment.

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