Answer: Find Moments & Reactions for Force Couple on Frame

AI Thread Summary
A force couple is acting on a fixed frame, requiring the calculation of moments and reactions at the base. The discussion highlights that the reactions are assumed to equal the acting forces of 1430 lbs, with moments calculated at 18232 in-lbs for both legs of the frame. Concerns are raised about the presence of torsional reactions and the implications of forces acting perpendicular to the frame's members. The assumption of a point of zero moment halfway up the columns is questioned due to differing conditions at the top and bottom, suggesting that a more credible zero moment point exists along the beam. The conversation emphasizes the importance of careful calculations to avoid errors and enhance understanding of the mechanics involved.
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Homework Statement


A force couple acts on the frame, which is fixed at the base. Find the moment(s) and reaction(s) at the fixed locations.

Homework Equations


ΣMa = 0
ΣMb = 0
ΣFy = 0
ΣFx = 0

The Attempt at a Solution


Since it would be statically indeterminate, assumptions would be that the reactions are equivalent to the acting forces (1430 lbs), and points of inflection exist halfway along the 25.5" length of the legs? This results in a theoretical "pinned" connection at the center thus the FBD for one leg cut in half:

ΣMa = -(12.75)*1430 + Ma = 0
Ma = 18232 in-lbs

Obviously both moments have to be equal, so Mb = 18232 in-lbs.

Am I on the right track? Wouldn't there be a Torsional reaction as well? I'm just a little lost because all the examples of frames I can find don't have forces acting perpendicular to the plane of the members (acting "into the page").
 

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Your ΣMa ignores the 1430 lbf in the other direction (at the corner above B)

What if the horizontal member were cut in half ?
 
So if it were cut in half what would be the internal shear force acting on it? My guess would be 1430, but then if you did a force balance it wouldn't make sense because you already have the reaction at the base as 1430, and then the applied 1430 in the opposite direction. That would mean the force in the horizontal member would have to be zero?
 
The assumption that a point of zero moment occurs half-way up the columns is unconvincing, because the conditions at the top and the bottom of the columns are not the same; at the bottom it is encastre and no rotation is assumed to occur, but at the top of the columns, the fixity is provided by a connection to a flexible member (the beam) where rotations about the x,y and z axes are possible. the assumption that a point of zero bending moment occurs half-way along the beam is more credible. However that would be accompanied by a torsional moment there (constant along the beam length). I am thinking qualitatively here, and am aware that careful calculations done properly can reveal errors in thinking at the same time as teaching us something.
 
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