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Beams in equilibrium

  1. Feb 12, 2015 #1

    I don't understand, what is the equation that we use in beams

    is it that forces 1,2,3 equal with the reactions at the two supporting points?
    or the torques are 1,2,3 equal with the reactions at the two supporting points?
    and how do we find the reactions at the supporting points?

  2. jcsd
  3. Feb 12, 2015 #2


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    At equilibrium, the beam will not be moving. So total vertical force is zero. And total torque is zero around any point.

    Recall your equation for torque. Pick a convenient point and work out the torque as a function of the force provided by 1, 2, 3, and each fulcrum.
  4. Feb 12, 2015 #3
    well, I see two problems

    total vertical forces may be zero, but it's impossible to calculate them, because each of the forces is distributed to the two supporting points, unequally, since they don't act at the same distance from each of the supporting points

    second, saying total moments is easier, but the problem is which axis should I take as reference? and if I take the axis that passes from the left supporting point, how am I going to calculate the reaction on that point, where moment is zero, since distance from the axis is zero? do you understand what I mean?
  5. Feb 12, 2015 #4


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    Not going to help you any more until I see you make some effort to solve the problem yourself.
  6. Feb 12, 2015 #5


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    The magnitude of the forces marked 1, 2, and 3 and their locations along the beam must be known already. This beam has only two supports, hence two unknown reactions occur at these locations.

    This particular beam is statically determinant, since you can write two equations of equilibrium, which when solved, will provide the unknown reactions.

    The sum of the moments about a particular point must equal zero.

    If you take the left support as the reference point for calculating moments, this sum will include the moment due to the unknown reaction at the opposite end of the beam, and also the moments created by each of the forces 1, 2, and 3 about the reference point.

    By selecting one of the supports as the reference point for the moment, we can write a moment equation with one unknown quantity: the reaction opposite from the reference point. By solving for this reaction, the only unknown remaining is the reaction at the moment reference, which can be easily solved by using the sum of the forces equation.
  7. Feb 18, 2015 #6
    excuse me, I still don't know how to do this

    I only know that sum of forces and moments is zero, since there is equilibrium

    but that is one equation, with TWO unknowns, the reaction to each of the two supporting points

    so how do I solve this?
  8. Feb 18, 2015 #7


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    No, you are confused. There are two equations of static equilibrium:

    ∑F = 0 and
    ∑M = 0

    You don't combine these two equations into one big equation, just because they are equal to one another.

    The following simple example illustrates this procedure:

  9. Feb 18, 2015 #8
    this webpage is very bad
    it introduces a term (Ay) in an equation without explaining what it stands for

    1) can you tell me please why sum of forces and sum of moments are two different equations? aren't all the forces produce moments and aren't all moments products of forces?

    2) we take the equations sum of forces and sum of moments equal to zero for any of the supporting points? or one for each of the points? WHAT ARE EXACTLY the equations we take? explain please with words, not with symbols and maths, say for example that we calculate the sum of forces in any of the supporting points and then we calculate the sum of moments in the VERY same supporting point, or whatever

    but if we calculate the sum of moments on a supporting point, the reacting forces have zero distance from the axis of rotation (which is the supporting point) so are they zero?

  10. Feb 18, 2015 #9


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    Ay is what that site is calling the reaction at point A.

    There are two equations of equilibrium because we are trying to say:

    1.) There is no net force on the beam, which would tend to accelerate it off into space, and
    2.) there is no net torque which would tend to cause the beam to rotate about an arbitrary axis.

    Sure, the net moment on the beam is produced by the forces acting on it, but there could also be isolated couples applied to the beam, which couples have no net force associated with them.

    Again, you are misinterpreting what the condition of equilibrium is. Equilibrium applies to the whole beam, not just the supports.

    We are trying to analyze the forces and moment acting on the beam, as if it were separated from the rest of the world by using a concept known as the free body.

    Analyzing your diagram in the first post, and assuming the unknown reactions are RA for the left support and RB for the right support, the first equilibrium equation is:

    ∑F = F1 + F2 + F3 + RA + RB = 0

    where F1, etc. are the known forces applied as shown.

    The second equation of equilibrium is:

    ∑M = (x1 * F1) + (x2 * F2) + (x3 * F3) + (xB * RB) = 0

    Here, we have taken the location of RA as the reference point for measuring the moment arms xA, etc., used in calculating the moments due to each force or reaction. In the equation above, the only unknown quantity should be the reaction, RB.

    In a beam with two supports, we typically write a moment equation using only one of the supports as a reference. The moment produced by the reaction at that support will be zero, but there is a moment produced by the unknown reaction at the other support. When the moment equation is written out, there should be only one unknown quantity, the reaction at the second support. Since the sum of the moments must be zero, we can solve for the reaction which makes this equation true. That gives the value of one of the reactions.

    Since there are only two unknown reactions to start with, and we have found one of those unknown reactions using the sum of the moments, likewise, we can use the sum of the force equation to find the value of the remaining reaction, the one which occurs at the reference location we originally selected in order to calculate the moments acting on the beam.

    I selected the beam problem at the link in Post #7 because it appeared to be pretty simple. However, you won't be able to grasp these concepts involving equilibrium fully until you have tried to solve some simple problems on your own. There are many more examples available on the Web, and books on Statics or Strength of Materials can also be used to find additional such problems.
  11. Feb 19, 2015 #10
    thanks, I get these equations now

    but what if for example, F3 is not a force, but a moment?

    how the equations would be?
  12. Feb 19, 2015 #11


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    If F3 were a force, instead of being included in the sum of the forces equation, it would be included in the sum of the moments equation.

    ∑F = F1 + F2 + RA + RB = 0


    ∑M = (x1 * F1) + (x2 * F2) + (F3) + (xB * RB) = 0

    Notice that the location of the moment F3 is immaterial to writing the moment equation.
  13. Feb 19, 2015 #12
    now I get the whole process to solve these problems
    but I have an objection
    in ∑F you take into account only forces and not moments: why that? why you completely ignore the F3 moment that requires a force to be exerted?
    in ∑M I agree with it, as you take into account all the moments (F3) AND the moments created by all the acting forces
    but why in ∑F you ignore the F3 moment? isn't it actually a force (let's name it F3f) that acts in reference with a distance from some axis? or because there is another moment acting reversely (as a couple) so we ignore it?
    So basically we use ∑F to the whole beam and ∑M to a single supporting point?

    To be honest, I would come up with these equations:

    ∑M the same as yours but the ∑F would be:
    ∑F = F1 + F2 + F3/distance(from where really?) + RA + RB = 0
  14. Feb 19, 2015 #13


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    Generally, moments applied to beams in the manner we have discussed are also known as couples. A couple consists of a pair of forces of equal magnitude, but oriented in the opposite direction from one another and separated by a distance d. If the separation is d and the magnitude of each force is F, then the couple is equivalent to a net force of zero and a moment equal to F*d. This is why applied moments are included in the moment equation and not the force equation. Couples have the advantage in that their application to the beam does not have to occur at any particular point on the beam. Statically, couples can be applied anywhere on the beam and the reactions are not affected.

    In writing the sum of the moments equation for equilibrium, we choose one of the supports as the reference about which the moments are calculated merely for convenience. In fact, we could choose any point about which to write a moment equation, and this arbitrary point is not confined to being located on the beam.

    If we write a moment equation using an arbitrary point, not located at one of the supports, then the moment equation must contain moments due to each of the unknown reactions at the supports. This is OK, but you will also have the two unknown reactions appearing simultaneously in the sum of the forces equation and the sum of the moments equation. In order to find the reactions then, you must solve simultaneously two equations in two unknowns. There is no problem mathematically in doing so, but why create the additional calculation work when you can write the moment equation referenced to one of the supports, which eliminates having to solve for that reaction?

    The moment equation written about one support determines the value of the reaction at the support opposite of the reference point, which value can be calculated using simple algebra and arithmetic. Once the value of one reaction is determined from the moment equation, the reaction of the other support, the support used as a reference for the moment calculation, also can be calculated from the force equation using simple algebra and arithmetic.
  15. Feb 19, 2015 #14
    I still not get why and how to distinguish forces from moments/couples
    can't we replace moments/couples by simple forces?
    what is the actual difference between forces (that also can generate a moment and that also generate a reaction in the support point, so they could simply considered couples?) and couples? I suppose both forces and couples are two vectors. What is their difference? Their magnitude? their point of action? their direction? Can't they simply be interchanged each other?
  16. Feb 19, 2015 #15


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    Sure. But even a beam loaded entirely with forces will have an equivalent free body model where the moments due to these forces must be balanced so that the beam doesn't rotate.

    Maybe this article can explain and illustrate the concepts of a force and the moment of a force better than I can:

  17. Feb 19, 2015 #16
  18. Feb 19, 2015 #17


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    Well one superficial reason forces and moments are handled separately is due to the units of their magnitudes, since moments are inherently 'force times distance' instead of just force.

    I think the major distinction between the two falls into the effect of each when applied to an arbitrary body. According to Newton's Second Law, the application of a force F to a body of mass m results in that body accelerating its motion, or F = ma, in a linear fashion. The application of a moment M to a body results in causing that body tending to accelerate its rotation about an arbitrary point, or M = Iα, where I is the mass moment of inertia of the body and α is the angular acceleration of the body.

    Learning to distinguish between what a force is and what a moment is and how a force can produce a moment is key to understanding how to solve the problem in the OP. I've tried to describe this distinction as best I can, but you will have to take additional steps on your own to cement this knowledge in your own mind. Unfortunately, I lack the skill and the means to do so.
    Last edited: Feb 20, 2015
  19. Feb 20, 2015 #18
    if in my diagram, one of the vertical arrows is not a force, but a moment/couple
    let's say 10Nm
    shouldn't we be given also which axis is this value taken in reference to?
    a 10Nm moment, could be calculated in regards to any axis! shouldn't we know which axis in order to solve the problem?
  20. Feb 20, 2015 #19


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    [Capitalization and punctuation repaired]
    It does not matter what axis is chosen. The moment associated with a couple is independent of axis. You can freely choose which axis to use. Try to choose one that makes the resulting equations easier to solve.
  21. Feb 20, 2015 #20

    it start making sense now

    moments, are simply couple of forces acting with a distance between them, so that try to rotate the beam

    1) if we know a moment that it's eg. 5Nm and we know the point it acts, can we find the exact value of forces in distance from that point? eg 1m far from the center of the moment, the forces are 5N? and 2 meters far forces are 2.5N?

    2) if we know a force, can we find how that force is distributed at different points from its point of application? eg. if a force 10N is applied to the center of the beam, we can surely deduct that 5N of force is equally exerted at its ends? but what if the point of application of the force is not exactly in the center? is there a method to calculate the distribution? thanks!
    Last edited: Feb 20, 2015
  22. Feb 27, 2015 #21
    anyone please?
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