5. Couples and equivalent systems

Fundamentals of Structural Analysis
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Questions and discussion for this lecture live here. Fire away by hitting Reply below :fire:

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Question regarding lecture: 5. Couples and equivalent systems 1:45 min.

If I sum the couple forces as we did in the previous lecture with the moments, I get the same result as summing perpendicular distances 951.6 Nm.

Can you explain why you say “Horizontal forces balance and Vertical Forces Balance”?
The projections are equal and in opposite directions, but each of them have produce the moment.

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Hi, I’m not sure I fully understand your question so please feel free to follow up if you’re still not clear.

Remember that there are three conditions for 2 systems to be equivalent:

  • The sum of the vertical forces must match
  • The sum of the horizontal forces must match
  • The sum of the moments must match

This is why I refer to horizontal and vertical force balance.

Seán

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10000073281965Ă—540 156 KB

To be clear, this is the thing I cannot understand. Second equation says moments in each system must be equal.

If I calculate moment for force A=
40 Ă— 11 + 69.3 x -5 = 93.5 Nm

And B =
40 x (-8) + 69.3 x 17 = 858.1 Nm

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I watched all the video and made hand calculations understood finally this concept…

Last question for this.

On x and y axis there are always 2 solutions when the force is left or right or up and down the origin to generate the same moment?

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Yes, you can generate the required moment using either of the orthogonal force components…

  • sliding the horizontal component up or down the vertical axis, or,
  • sliding the vertical component left or right along the horizontal axis.

Seán

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Thank you. It means there are two solutions for each axis.

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There are two solutions, one for each axis.
S

1 Like
9

Hey , First of all thank you so much for this content which I find very enjoyable to go through . I have a question regarding the last exercice with system C .

While I totally understood the solution , I have to say that before checking it , I tried to solve it myself , and my idea was to just go for calculating the lever arm , not for the individual orthogonal components , but for the Resultant force Rt .

I got d = (1975) / (152,6) = 12,94 m .

I thought that as long as we slide the force in a radius of 12,94 m from the point of rotation , we would get the same moment . However , after checking the solution , It appears we can only do that with the Rx and Ry components . My question is what makes my solution false , and what makes it necessary to work with Rx and Ry only , and not with Rt ?

Thanks in advance .

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Hi @3tmane

I may be interpreting your description of what you did incorrectly (let me know if I am and) - I think the detail you might be missing is that the moment is the product of the force and lever-arm but the lever-arm must be perpendicular to the line of action of the force (the force arrow basically).

So you could use the resultant force, but you would need to calculate the perpendicular distance between the force arrow and the point of rotation for your lever-arm. It’s generally just easier to break the force into its orthogonal components and consider two components of moment with simple (to identify) lever-arms.

Feel free to put your workings on a page a drop an image of it here if you’re still confused or I’ve misinterpreted the problem.

Seán