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Statics: Statically Indeterminate Unstable Plane Truss

  1. Jul 28, 2016 #1
    1. The problem statement, all variables and given/known data

    The signboard truss is designed to support a horizontal wind load of 800 lb. If the resultant of this load passes through point C, calculate the forces in members BG and BF.

    Untitled.jpg



    2. Relevant equations

    For a solution using the method of sections for plane trusses,any three independent equations of static equilibrium i.e,

    ∑Fx=0 ∑Fy=0 and ∑MO=0

    where x- and y- are any two mutually perpendicular directions and O is any point on the plane of the truss members.

    OR

    ∑MA=0 ∑MB=0 and ∑MC=0

    where A,B and C are any three points on the plane of the truss members not lying along the same straight line.

    OR

    ∑Fx=0 ∑MA=0 and ∑MB=0

    where where A and B are any two points on the plane of the truss members such that line AB is not perpendicular to the x- direction.


    3. The attempt at a solution

    I was able to calculate the forces in members AB, BG and FG by passing a plane through the three members and considering the FBD of the portion of the truss above the section.Having determined the force in member FG I determined the force in member BF by considering the equilibrium of joint F(method of joints).

    Now if we consider the equilibrium of the entire truss as a whole the horizontal components of the support reactions at A and G are statically indeterminate i.e, the truss is statically indeterminate externally.Again counting the number of members and joints for the truss we find the truss has 10 members(10 unknown member forces) and 7 joints(14 available independent equations of equilibrium for 7 joints).So the 7 joints are collectively under equilibrium under the action of 10 unknown member forces and 4 unknown external support reactions(a total of 14 unknowns).So we have the equal number of independent equations and unknowns to calculate all the unknowns including all the components of the supports reaction at A and G.We might also notice that the truss is unstable(deficient of internal members) and likely to collapse under load when removed form its external supports(member AB free to rotate about B and not completely fixed).Therefore,my question is that since we were able to determine the support reactions at A and G completely using the equations of equilibrium for all the joints(even though the truss seems to be have an extra constraint in the horizontal direction and is statically indeterminate as a whole) should we still call the truss statically indeterminate externally?

     
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  3. Jul 28, 2016 #2

    haruspex

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    Are they? Seems to me that you know the direction of the force at A, and taking moments about G tells you the magnitude, so both components are known.
     
  4. Jul 28, 2016 #3
    Oh yes! I missed that completely :headbang:

    But I have a new alteration of the truss after making a few adjustments to it which is troubling me again.What if the truss was like the one in the figure below and we wished to calculate all the member forces?
    Truss.jpg

    What I did is replace the member BF with two equal,opposite and collinear forces at B and F such that the truss is under equilibrium and place the member AG which was not there before.In doing so the number of members(unknown member forces) and the number of joints remain unchanged.We now have 14 unknowns( 10 unknown member forces and 4 unknown external support reactions) and 14 independent joint equations of equilibrium.Theoretically we should be able to solve for the 14 unkowns from the same number of independent equations.But starting at joint D and evaluating joints C,E,F and B in order by the method of joints we obtain forces in members CD,DE,BC,CE,BE,EF,FG,AB and BG until we arrive at joints A and G where we are left with 5 unknowns(member force AG,Ax,Ay,Gx and Gy) and 4 joint equilibrium equations.So the truss seems to be statically indeterminate even if we have the same number of unknowns and independent equations to solve for them!

    Why aren't we able to solve for all the member forces even if we have the adequate number of equations in hand?
     
  5. Jul 28, 2016 #4

    haruspex

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    Your equal and opposite forces at B and F are completely equivalent to a truss. Since the force is not given, it's another unknown.
    So the only change is the extra member AG. The force in that is clearly indeterminate since it is parallel with the force in the ground between the same two points.
     
  6. Jul 28, 2016 #5
    Yes my intention is to replace the member BF with the two equal,opposite and collinear forces at B and F such that the magnitudes of the forces are known.In that case they are not unknown quantities.

    Assuming that we know the magnitudes of the applied forces at B and F and the member BF is not physically present there but its effect is replaced by the two forces which are known,we have the same number of independent equations of equilibrium from all the joints as the number of unknown forces(10 unknown member forces and 4 unknown external support reactions).But that doesn't seem to guarantee that we be able to determine all the unknowns if we have an equal number of independent equations in hand as the number of unknowns.So can we conclude that equating the number of equations with the number of unknowns is not a sufficient condition to verify whether all the unknowns could be determined or not?

    Another point that I wanted to clarify is that is it absolutely necessary for a truss to be statically determinate externally in order to be statically determinate internally as well?
    Is there an example of a truss for which the unknown external support reactions(assuming all other applied loads to be known quantities) could not be completely determined initially from the equations of equilibrium for the entire truss as a whole but could well be determined from the 2j number of joint equlibrium equations along with all the unknown member forces?

    What I mean to imagine is a plane truss which has 'm' members and 'j' joints and is in equilibrium under the action of 'r' number of unknown external support reactions and other externally applied known forces such that it satisfies the relation m+r=2j where r>3(hence statically indeterminate externally).Assuming that we have such a truss which is obviously unstable and would collapse when removed from its external supports but is in a stable configuration under the externally applied loads and the 'r' external support reactions, we still have the opportunity determine all the member forces and the unknown external support reactions with the help of those '2j' independent equations.So such a truss if it exists could be statically indeterminate externally but statically determinate internally.
     
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