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Axial traction of a beam on support

  1. Sep 29, 2012 #1

    My question is rather off the usual queries concerning beams, which is maybe why i'm struggling to find a solution. If we suspend a totally rigid beam at its extremities with springs to measure forces, the springs will both indicate q x L /2.

    If the beam is supple, it will sag and exert a traction force on springs placed axially at each extremity (see diagram). How will this traction be related to q, L and E ??

  2. jcsd
  3. Sep 29, 2012 #2
    If you understand the principles of horizontal and vertical equilibrium you can calculate the effect.

    The result of the beam deflection is a rotation of the end so that the axial force is neither vertical nor horizontal, but it will be perpendicular to the rotated end face of the beam.

    This force will have vertical and horizontal components equal to the vertical and horizontal forces in the springs.

    Please note that you can post the diagram directly here. Accessing your link introduces an unwanted and unwelcome advertisement window.
  4. Sep 30, 2012 #3
    yes, the forces in the two springs at each end can be represented by a single force F acting at a cerain angle alpha. So two unknowns which can be resolved by the two equations coming from force equilibrium and moment equilibium for pure flexion in small deflection theory. Second order or more differential equations I imagine. Sunday morning, the sun is shining, I'll tacle the maths tonight.

    Sorry for the image, when I clicked I got a request for a URL link, I suppose there's a help for posting images which i didn't take time to read.
  5. Sep 30, 2012 #4
    There are no reaction moments for the spring support situation you describe.

    The two equations arise by resolving that force into vertical and horizontal components, one in each spring.

    Another simple way to attack this is to consider a prestressing analysis.

    Normally the prestressing force is fixed, but in your case is depends upon the spring constants for the horizontal springs.

    You can obtain an expression for this and then consider the effect on the maximum bending moment and thefore deflection in the centre of the beam. Prestressing reduces the horizontal shear and moves the position of the neutral axis, thus reducing the bending moment.

    Remember that, although this is a statically indeterminate problem symmetry cuts your work in half.
    Last edited: Sep 30, 2012
  6. Sep 30, 2012 #5
    Unless i'm mistaken I shouldn't need to calculate the beam deformation, which should alleviate me of any second order maths.

    i'm only interested in horozontal pull at the extremitires.
  7. Sep 30, 2012 #6
    Have you tried what I suggested?

    You have not supplied all the beam properties. These are needed to complete the problem.

    Attached Files:

  8. Sep 30, 2012 #7

    My using springs to measure forces was an error, the spring properties just complicate things. So I'll rephrase the question :

    a homogenous rectangular beam b, h, L and E, charge q (newtons per unit length), suspended by a weightless non-extensible string of length p at the centre point of each extremity face, the suspension points being distance L apart, what will be the tension and the angle in the strings?
    Last edited: Sep 30, 2012
  9. Sep 30, 2012 #8
    OK the answer to your simpler problem is simpler.

    The normal theory states that plane section remain plane after bending and rotate as a plane about a point on the neutral axis.

    This means that since your strings are attached to the centre or neutral axis they will not be displaced sideways and if the strings were originally vertical, they will remain vertical.

    I have sketched the end plane face of the beam, greatly exaggerated. It has rotated an angle θ about the suspension point on the neutral axis.
    Simple geometry tells us that this is also the angle the plane face now makes with the string.

    The sketch also includes the standard formula for this angle. Note that the result is in radians.
    Note also this formula includes reference to the moment of inertia, I, of the beam and I have included a formula for this in terms of dimensions, for convenience.

    I should also note that my earlier comment about an axial prestress decreasing the moment was not quite right. What actually happens is that the axial prestress increases the stiffness, reducing the deflection or rotation, thus making the apparent or effective moment less. The real applied moment is, of course, still determined by the loads.

    Attached Files:

  10. Oct 1, 2012 #9
    First of all thanks for staying with me on this problem. As you said the neutral fibre doesn't move and forces remain vertical at the extremities, so the simple small deflection bending approach cannot estimate axial traction forces exerted by the extremities on whatever is supporting the beam. This is in away very helpful for me, a sort of confirmation.

    This image (sorry I haven't found how to post images correctly) illustrates what I want to know. A metal beam is fixed to a wall with a bolt on each side. By considering the bolt is in a cantilver situation, it's easy to calculate the forces the bolt has to withstand.

    What I'm trying to find is what will be the force trying to pull the bolt out of the wall, there surely must be one as soon as the beam starts to flex.

    Last edited: Oct 1, 2012
  11. Oct 1, 2012 #10
    The last time I did this calculation was to bolt slab formwork into RC walls to cast a floor.

    I was more concerned that the shear strenght would be adequate since the deflection required to pull out an anchor would entail pretty excessive movement.

    Anyway the pull out force is quite easy.

    Look again at my second sketch and imagine that the end of the beam is bolted at the midpoint by two brackets (you need two one each side for a rectangular section to avoid twisting).

    This arrangement then converts the beam to a restrained beam. As a result of the restraint the is now and end moment at each end.

    This moment can only be provided by the a couple made from the pull out force and the reaction at the bottom corner of the beam pushing against the wall. These forces will be equal and opposite, say Rh.

    The end moment for a restrained beam such as yours is M = qL2/12. This equals the couple force times their separation (half the beams depth) = 0.5 *Rhh

    Equating these will provide the pull out force.


    Be careful with this it can get you into deep water.

    See here

    Last edited: Oct 1, 2012
  12. Oct 2, 2012 #11

    Establishing a couple for the fixation on the wall and equating it to the beam's end moment is the link I didn't know. Thanks so much. This allows me to go further in evaluatings ways of fixing joists to an existing maisonry wall, where I see everything from diam 10 every metre to a 16 every 30 cm, at least on French forums.

    By the way i did post this question, albeat badly formulated, on a French forum. No reply.

    Thanks for your link to the ultra short beam discussion. My ignorance is frightening!

    I could submit you a thousand questions but this is not a housebuilding forum, although I've got a niggling feeling that you would have an answer to most of them.

  13. Oct 4, 2012 #12

    I'm come back on this question of fixing floor joists to masonry walls just to ask you to check my calculations.
    Let's assume a 4 metre wide floor and everything and everybody on it including the cat weighs 300 kg/m2, as a round figure. Joists are 50cm apart so the load of each joist is 1500N per meter. We can therefore calculate that the joists exert a 2000 N-m couple on each wall fixation.

    Simpson SAE380/76/2 metal fixations are used, with four 10mm through bolts. These fasteners are 15.2cm high and let's say the bottom bolts are 7 cm and the upper bolts 14 cm above the base pivot point. Fh and Fl (high and low) in newtons are the forces in the through bolts. We can say, and I again thank you for this ;

    2 x Fh x 0.14 + 2 x Fl x 0.07 = 2000
    also, as angles are small, we can say that Fh / Fl = 14/7=2
    and all that gives Fh = 6000 N.

    Are my calculations correct, that is I've not missed out on a zero or forgotten to multiply or divide by txo or something like that? I've seen a 10mm galvanised expansion through bolt in quality concrete as being able to resist a 7600 kN traction force.

    I'm aware that we have a horizontal traction force, a vertical shear force plus a bending moment along the bolt, all acting together, heaven knows how to assess all that.

    Am I right in saying that the non-rigidity of connections (nails in wood, metal fasteners not totally rigiid) will absorb a substantial part of the couple, thus alleviating the traction on the through bolt?
  14. Oct 5, 2012 #13
    Please confirm the section size of the timber beams, along with some indication of timber type/grade.

    This is very important since the limiting factor is likely to be the bearing stress/and or pull out stress of the through bolts on the timbers. The steel can take a great deal more concentrated stress than timber.

    As a matter of interest, this is why we normally use joist hangers or sit the timbers on brackets rather than bolt through the sides as I think you are doing.
    I would advise these if at all possible. Joist hangers can be bolted to the wall in much the same way as your side brackets, but provide unde support for the timber. They are readily and cheaply available from builders merchants.
  15. Oct 5, 2012 #14
    Last edited: Oct 5, 2012
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