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  1. #1
    Senior Member dleccord's Avatar
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    Any engineers in here?

    Pretend we have a cantilever I-beam that is subjected to an inclined load at the end of the beam and we have to take into account of the self-weight of the beam.

    If we're solving for the bending moment at the end of the beam and we need to consider the self weight of the beam, are we suppose to find the weight=mg a distance (in this case half the length of the beam) from the end of the beam and consider that into your moment in the z and y axis?

    Cheers!

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    Administrator CbadRider's Avatar
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    Moved from Road Cycling to Foo.
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    So Tom only hires people that are nutty? Is part of the requirement to be a moderator on this site is that you have to be nuts??
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    Quote Originally Posted by dleccord View Post
    If we're solving for the bending moment at the end of the beam and we need to consider the self weight of the beam, are we suppose to find the weight=mg a distance (in this case half the length of the beam) from the end of the beam and consider that into your moment in the z and y axis?
    no

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    Perhaps you could draw a diagram?
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  5. #5
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    Yes, the weight of an object can be represented as a force in the free body diagram by its mass x gravity located at the object's center of mass and in the direction of gravity.

  6. #6
    You Know!? For Kids! jsharr's Avatar
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    One on't cross beams gone owt askew on treadle.
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    Quote Originally Posted by colorider View Post
    Phobias are for irrational fears. Fear of junk ripping badgers is perfectly rational. Those things are nasty.

  7. #7
    derailleurs are overrated bigbenaugust's Avatar
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    uggh, stop making me flash back to statics.
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    Senior Member no motor?'s Avatar
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    Quote Originally Posted by jsharr View Post
    One on't cross beams gone owt askew on treadle.
    I don't think he was expecting the Spanish Inquisition....

  9. #9
    Super Moderator BillyD's Avatar
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    wut
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    You Know!? For Kids! jsharr's Avatar
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    Quote Originally Posted by no motor? View Post
    I don't think he was expecting the Spanish Inquisition....
    Does anyone?
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    Quote Originally Posted by colorider View Post
    Phobias are for irrational fears. Fear of junk ripping badgers is perfectly rational. Those things are nasty.

  11. #11
    derailleurs are overrated bigbenaugust's Avatar
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    But don't the modulus of the material and also its cross section enter into this at some point?

    (I took statics a very long time ago, didn't do so well, and didn't keep the book.)
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    Senior Member dleccord's Avatar
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    The problem gives you the density, the cross sectional area A of the I-beam, and the length L of the entire beam. It's a cantilever beam so it is fixed to a wall at one end. The other free end of the beam is subjected to an inclined force P at an angle alpha below the z-axis through the center of the beam from a y-z axis view. Without considering the weight of the beam and the bending moment it produces, moments My and Mz are produced. Since we take into account of the weight of the beam, that weight also causes a moment at the center of the cantilever with respect to the wall. I basically moved that moment vector to the end of the beam where the inclinded force occurs and then add a moment there, thus changing the resultant M in the y-z plane. The part I am not so clear on is, I have the mass of the beam because I have density, length, and cross sectional area; these three values give me the mass of the beam. Now the question is where do I place this load, or is it a distributed load along the entire beam. If not, do I just put a concetrated load smack down the center of the beam and find a moment with respect to the end of the beam and add moment components?

    Am I right?

  13. #13
    I ain't no newbie redirekib's Avatar
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    Absolutely.
    "Never send a monkey to do a man's job." ~ Captain Leo Davidson ~

  14. #14
    long time visiter Alfster's Avatar
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    Is this approx what you're looking at? Note - the UDL is converted to a point load (W) located at the center of the beam. Ma, fx and fy are resultant moments and forces.

    diagram.jpg

    Sum of Moments = W x 1/2(Dim X) + P (Dim Y)

    Note - this generates a moment about only one axis, not two. Since you indicated 2 moments are generated, does that mean the inclined point load at the end of the beam is not as I have it shown (ie: located in 2 planes rather than just 1)?

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    I'm an engineer and I can do these in my sleep. Throw us a more complicated one.

    Moment due to beam weight is W*L^2/2. W is weight per unit length, L is length of the cantilever. Total beam weight is WL. Moment arm is L/2 since the weight of the beam acts at the middle point of the beam. Therefore M=W*L^2/2.

    Alf's diagram is my understanding of the statement of the problem as well. The moment due to the inclined load can be calculated using the distance measured from the end of the beam perpendicular to a line acting through the load, as shown in the diagram. OR you can use trigonometry to break the inclined load into horizontal and vertical components. Then the moment is equal to the vertical component times the length of the cantilever. The horizontal component has no affect on the bending moment.

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    Quote Originally Posted by bigbenaugust View Post
    But don't the modulus of the material and also its cross section enter into this at some point?

    (I took statics a very long time ago, didn't do so well, and didn't keep the book.)

    Youngs modulus of the material and the properties of the cross section have no affect on the static forces, but would have an affect on deflections and internal stresses.

  17. #17
    Senior Member dleccord's Avatar
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    Actually, the problem I have describe was not clear but this is a biaxial bending moment case since the force P is at an angle alpha, say 80 degrees, below the z-axis whereas the diagram above shows a force acting down primarily the y-axis and we don't consider the force in the x direction since it's along the line of action. We have forces acting on the end of the beam on the z-axis and on the y-axis. Therefore, this will cause two moments respectively. Now the sum of these moment components are affected by the weight of the beam but I believe it's just the y-component force that is actually being affected.

  18. #18
    The Site Administrator: Currently at home recovering from a couple of strokes,please contact my assistnt admins for forum issues Tom Stormcrowe's Avatar
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    Quote Originally Posted by bjtesch View Post
    I'm an engineer and I can do these in my sleep. Throw us a more complicated one.

    Moment due to beam weight is W*L^2/2. W is weight per unit length, L is length of the cantilever. Total beam weight is WL. Moment arm is L/2 since the weight of the beam acts at the middle point of the beam. Therefore M=W*L^2/2.

    Alf's diagram is my understanding of the statement of the problem as well. The moment due to the inclined load can be calculated using the distance measured from the end of the beam perpendicular to a line acting through the load, as shown in the diagram. OR you can use trigonometry to break the inclined load into horizontal and vertical components. Then the moment is equal to the vertical component times the length of the cantilever. The horizontal component has no affect on the bending moment.

    Clarify the formula for me....

    Is that better represented as W*(L^2)/2, or [(WL)^2]/2 or a compound exponent (^2/2)? If the third, then wouldn't that translate out to a 1st power expression?
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  19. #19
    long time visiter Alfster's Avatar
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    Well I'm not going to take another guess at another diagram. Working with forces in different planes is not that difficult. Just remember, moments in the same plane are additive. You can then find a resultant moment from the two different planes (about two separate axes). However, when sizing an i-beam you typically work with separate minor and major axis bending. The resultant is determined when you compare the ratio of stresses to ensure you don't exceed a fully stressed beam ratio of 1.0.

  20. #20
    Senior Member RollCNY's Avatar
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    Apparently things have changed since my Strength of Materials class. There were no I-beams, only H-beams and W-beams. And the times they are a changin..

  21. #21
    long time visiter Alfster's Avatar
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    Quote Originally Posted by RollCNY View Post
    Apparently things have changed since my Strength of Materials class. There were no I-beams, only H-beams and W-beams. And the times they are a changin..
    I-beams are actually different than W-beams. A W-beam is a wide-flanged beam, used for major support beams or columns. I-beams are narrow-flanged beams used for lower stress applications. An I-beam is less stable under bending stresses, requiring additional lateral support on the compression flange.

  22. #22
    Look! My Spine! RubenX's Avatar
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    Oh well... this is not electric, electronic, or computer engineering so i'm out.
    "Hoy es un dia normal, pero yo voy a hacerlo intenso" ~ Juanes

  23. #23
    Senior Member dleccord's Avatar
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    Here's a picture of what it looks like.



  24. #24
    long time visiter Alfster's Avatar
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    So how would you solve this problem?

  25. #25
    Senior Member dleccord's Avatar
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    This problem without the weight would be cake.

    Density, cross sectional area, and the length of the beam is given, which means we can solve for the mass of the beam. Which also means that if the problem doesnt say neglect the weight of the beam, we should include the weight of the beam into the problem.

    The problem here is that I don't know if we're suppose to place the weight of the beam in the middle and from there calculate for the moment vector the weight of the beam produces and add to moment components at the end of the beam.

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