How to find the support reactions of a planar structure
These problems follow one scheme. Choose the body whose equilibrium you consider: the frame or a part of it. Remove the supports in your mind and replace them with reactions: a fixed end gives a force with two components X, Y and a moment, a pin gives X and Y, a roller or a weightless link gives one force along a known line. Then write the three equilibrium equations of a planar force system: ΣX = 0, ΣY = 0, and the sum of moments about a chosen point equals zero.
Choose the moment centre where the lines of two unknowns meet: they drop out of the moment equation and one unknown is found at once. The moment of a force about a point is M = x·Fy − y·Fx, with x, y the coordinates of the point of application measured from the moment centre; counter-clockwise is positive. A couple enters the moment equation as a whole and does not appear in the projections.
Distributed loads
A uniform load is replaced by one force R = q·ℓ at the middle of the span, a triangular one by R = q·ℓ/2 at one third of the length from the side with the larger intensity. If q is given per unit of horizontal projection (snow load), then R = q·ℓ·cos θ, where θ is the angle of the member to the horizontal; a load perpendicular to the member has a resultant along the normal.
Example: an L-shaped frame
Column AB, 3 m high, is rigidly fixed at A, and the beam BC is 4 m long. At C there is a 10 kN force downwards, at B a 5 kN force to the right, the beam carries a vertical load of 2 kN/m per horizontal projection, and at B there is a 6 kN·m counter-clockwise couple. The resultant R = 2·4 = 8 kN acts at the middle of the beam, 2 m from the column.
ΣX = 0: X_A + 5 = 0, so X_A = −5 kN. ΣY = 0: Y_A − 10 − 8 = 0, so Y_A = 18 kN. ΣM_A = 0: M_A − 4·10 − 3·5 − 2·8 + 6 = 0, so M_A = 65 kN·m. Check with moments about point C: ΣM_C = M_A − 4·Y_A + 3·X_A + 2·8 + 6 = 65 − 72 − 15 + 16 + 6 = 0.
Compound structures
If two parts are joined by an internal hinge, there are six unknowns: the support reactions and two components of the hinge force. Write the equations for each part separately (it is convenient to take moments about the hinge, since the hinge force then drops out) and check the answer with the moment equation for the whole structure. The hinge forces on the two parts are equal in magnitude and opposite in direction.
When there is no solution
A single body needs exactly three unknown reactions, and their lines of action must not all be parallel or meet at one point. More unknowns make the problem statically indeterminate (it needs deformations, i.e. strength of materials); fewer, and the structure is a mechanism.