- #1
Soumalya
- 183
- 2
Homework Statement
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.
Homework 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.
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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?