how do you calculate the load on support points of a triangular frame (bracket) mounted to a wall?

Question

The setup: the bracket is a right triangle mounted perpendicular to a wall so that one leg (AB) is against the wall, one leg (BC) extends out from wall (these form 90degree angle) and hypotenuse (CA) is in compression (extending down and back to the bottom of leg AB) when a downward point load (1000lb.) is applied to point C. Points A and B are fixed to the wall with mechanical connectors. Legs AB and BC are equal length (2 feet). Assume frame and connectors are of adequate strength. Questions: What is the required shear strength of connectors at A and B? What is the pullout resistance or tension on connector at B? What is the compressive force on member BC?

assume the frame connections are fixed and the frame is more than adequate to support the loads. i am only interested in the forces at the two connections to the wall--points A and B.
think of it as determining the shear strength of bolts at these two points and the pull-out resistance of the bolt anchor at point B.

Answer 1

points a and b take 1/2 load in shear (500 lb each)

tension at b = 1000 (2ft)/(2ft) =1000 lb
thus a is 1000 in the other direction (sum of F in hor. = 0)

Answer 2

The type of calculation required for the bracket you described varies greatly based on the type of connections between the structural members. If we had rigid connections we would need to know the moments of inertia of the structural members, and their cross-sectional area. So lets say we are assuming the connections are hinged. There's a force couple including the downward point load of 1000 pounds and the 1000 pound force at the wall generated by the mechanical connectors resisting that load. These two forces are two feet apart and generate a moment of (1000lb)(2ft) = 2000ft.lb.
The pull-out at point B would be calculated as follows:
1000(BC)/(AB)
Fortunately it's a 45 degree right triangle and all calculations are easy.
Remember all horizontal forces have to balance and all vertical forces have to balance. For example, there's a 1000 pound force pushing into the wall at A to balance the pull-out at B.
You said there's a member AB. It's length is fixed so we can only assume the shear force is distributed evenly between A and B. BC is not in compression, it's in tension. 1000lbs. Same as the pull-out at B.
AC is in compression, and since it's hinged at both ends, it's incapable of carrying a bending moment. Since it's at 45 degrees the horizontal force is equal to the vertical force 1000lbs. The magnitude of the compression is the resultant of the horizontal and vertical forces:
1000/sin45 = 1414lbs