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Method of Joints

  1. Oct 21, 2015 #1
    1. The problem statement, all variables and given/known data
    IMG_1340.JPG
    Determine the axial forces in the members of the truss and indicate whether they are in tension or compression.

    2. Relevant equations
    Fx = 0
    Fy = 0
    Mp = 0

    3. The attempt at a solution
    IMG_1339.JPG
    I attempted to find the reactions at the supports first. So:
    Fx = 0 = Bx + 800cos(250)N
    Fy = 0 = By + Cy + 800sin(250)N
    Mb = 0 = Cy(1.4m) + 800cos(250)(0.4)Nm + 800sin(250)(0.7)Nm

    These give
    Cy = 454N
    Bx = 273.6N
    By = 297.8N

    I found Alpha at A, that was 60.25, which gives an angle of 29.75 at the joint B.
    (sorry the photo doesn't line up with the text, this was discovered after taking).

    This would give Fy = Tab = 297.8/sin(29.75)
    However, this is very much incorrect...

    I'm legitimately lost and, if anyone helps me, please point out which assumptions i made incorrectly and at least a general idea of why they're incorrect. I won't respond well to vagueness because in my mind i've covered a lot of what I already know with some sense of logic as i've come to understand it.
     
  2. jcsd
  3. Oct 21, 2015 #2

    SteamKing

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    I think what you need to do, after calculating the reactions at B and C, is to analyze the members in the truss by drawing a separate free body diagram for each.

    Because the truss as a whole is in equilibrium, each member must also be in equilibrium. You can write the equations of statics for each member and determine the axial force in each, along with whether it is in tension or compression.
     
  4. Oct 21, 2015 #3
    Are you sure? I've been a little confused on how finding them works. I know that if you rotate your triangle so the adjacent side goes up then you have to take cosines, but I was intending to continue the angle forward so that I wouldn't have to rotate the triangle. It would instead face the way it would've in quadrant 1 but in quadrant 3 (thus resolving sign and trig errors naturally)
     
  5. Oct 21, 2015 #4
    That's what I was attempting to do at the bottom. Is my understanding of the entire structure correct? I know that if the sins and cosines are incorrect then I need to go back and review my understanding of the unit circles, but I was attempting that by starting with B and building from there:
    So Tab = 297/sin(29.75)
    I shouldn't need to do the other joint to solve this one should I? It sounds like my error is in finding the reactions at the support because once you do that simple joints like this should just fall into place.
     
  6. Oct 21, 2015 #5

    SteamKing

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    Rather than trying to figure out what angle to use on a 360° circle, it's often less confusing to take the angle as indicated on the diagram, in this case 20°, and use that to figure out the components of the applied load. By using an angle of 250°, its easy to get an extra negative sign figured into the components.

    In your calculation of the reactions at B and C, your moment equation is incorrect, due to having the wrong signs on the moment components produced by the applied load at A. Here, the moment due to the horizontal component of the load acts in the opposite direction of the moment produced by the vertical component. Your moment equation has both moment components acting in the same direction.
     
  7. Oct 21, 2015 #6
    Okay, I accept your wisdom and will use it if that's what's been proven to work most accurately, I just have a few questions so that I know why to favor one over the other. Currently, I'm hearing that using the 360° circle is a lazy way that can lead to carelessness with directions, however why is this the case?
    It seems to make sense mathematically that if you plot a line through the origin and create similar triangles through the components then in quadrant 1 you have 2 triangles: The one above and below the line. The one below the line will use normal sin/cos conventions and the one above will use the opposite. So it seems easier and less prone to errors to pick a side of this line always refer to it. In my case, i'm choosing the 'under' side, so that if you continue the angle until you are on the other side of the line then the sin and cos will remain the same. Drawing the triangles, this is what I found and if we're supposed to merely assume which direction these forces are going (because the math will take care of us if we set it up right) then it seems most beneficial to have a standardized method that will leave the least amount of room for subjective reasoning.

    This is why in my equilibrium equations I have Fy = 0 = By + Cy + 800sin(250)N. because I do not know which direction these things go until I compute them.

    Sorry if this is off-topic, but it's a fundamental question I've been struggling with since the beginning and i've found that forgetting the fundamentals are the bane of learning math.
     
  8. Oct 21, 2015 #7
    Ugh. I saw that as soon as I read it. I get careless with these equations trying not to assume too much with them that I forget moments aren't relative to the origin of the "plane" but the origin of their action. Thank you, how embarassing.
    I redid the moment equation, and strangely enough it switched my BY and CY forces (roughly)
    When I sought out TAB = 454 / sin(29.75) = 917, which is 17N too large, but I wasn't rounding. I was just seeing if that got me in the ballpark.
     
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