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The vertical reaction of an arch

  1. Nov 1, 2017 #1
    I'm hoping I can find a place to ask these types of questions as a student or layperson. Please no rude flames or responses. I am asking for help because the material is currently a bit out of reach for me. Can I get some help to understand. Thank you.

    From Analysis of Arches by Trupti Sonavane there are figures attached. A point load is said to be established for example in this case it may be said 10Kn for an arch of 12 ft in diameter. There is a defined equation SigMA = BY * 2 * x1 - P(x1+x2) to determine a moment about support A of the arch as defined in the detail. As BY is the vertical reaction and P is defined to be 10Kn is the vertical reaction also 10Kn and if not how is BY to be determined? In addition I understand that there is an output of the function called SigMA (moment about A) however it is unclear to me what that number really actually means physically to the structure for instance is SigMA a vertical moment and if the moment (SigMA) is exceeded what does that mean for the structure?
     

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  2. jcsd
  3. Nov 1, 2017 #2

    BvU

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    Hello Eric, :welcome:
    No reason to expect those, and even then: we have good mentors who weed them out effectively!

    Starting with the first equation on the second picture: that indeed describes the torque (or moment) in the z-direction around point A. In the first picture the symbol is ##A_{z0}##, slightly confusing. If that torque would be positive, the structure would start rotating counterclockwise, and we don't want that. negative - clockwise - don't want that either. So it has to be zero. In other words we have an equation: $$B_{y0} \times 2R \cos\alpha - P \times R\left (\cos\alpha +\cos\theta\right ) = 0$$where ##B_{y0}## is the only unknown. It is easily solved with $$B_{y0} = P\ {\cos\alpha +\cos\theta \over 2\cos\alpha}$$ (I resent the use of the secans in the picture -- needlessly complicating.

    You can rudimentarily check the result by looking at extremes: if ##\theta = \alpha## you expect all the load to press on B, and if ##\theta = \pi - \alpha## it all comes down to A.


    Then:
    The arch also does not accelerate downwards or upwards, but stays at rest. According to Newton ##F = ma## that means the net sum of forces has to be zero. So the story in the second picture continues with a force balance $$A_{y0} + B_{y0} - P = 0 $$ By now, ##B_{y0}## is known so ##A_{y0}##is the only unknown in this equation. Can you understand the result $$A_{y0} = {1\over 2} P\,\left (1 - \cos\theta/\cos\alpha\right ) \ \ ?$$

    upload_2017-11-1_23-30-57.png


    Oh, and: I understand this isn't homework. Nevertheless PF wants such explanations of textbook examples posted in the homework forum -- rightly so.
     
    Last edited: Nov 2, 2017
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