Stability, determinacy and support reactions

Before computing a single member force, an engineer asks two questions about the model. Is it stable, meaning that it can resist any set of loads without moving as a mechanism? And is it statically determinate, meaning that the equations of equilibrium alone are enough to find all the reactions and internal forces? The answers decide which methods apply. A determinate structure can be solved by statics in a few lines. An indeterminate structure needs extra equations from compatibility of deformations, the subject of lesson 12. An unstable structure cannot be built as modeled, and spotting one on paper is far cheaper than finding out on site.

Equilibrium of a plane structure

A rigid body loaded in its own plane has three independent equations of equilibrium. The usual set is

where O is any point. Equivalent sets use two or three moment equations, provided the moment centers are not all on one line. Choosing moment centers wisely is the main skill in finding reactions: take moments about a point where two unknown forces intersect, and the equation contains only the third unknown. You can solve for each reaction directly instead of solving simultaneous equations.

Every calculation starts with a free-body diagram. Draw the structure, or the part of it you are isolating, remove the supports, and replace each by the reaction components it provides. Show the applied loads with their magnitudes and positions. Assume a direction for each unknown; a negative answer simply means the actual direction is opposite. Replace distributed loads by their resultants for the purpose of finding reactions, but remember that the resultant is only a convenience for whole-body equilibrium.

Equations of condition: internal hinges

Many structures include internal hinges, where two parts are pinned together so that no moment passes between them. Long-span steel roofs sometimes use cantilevered beams with hinges near the points of zero moment, an arrangement called a Gerber system, and three-hinged arches and frames are common for long spans. Each internal hinge provides one extra equation, because the bending moment there is zero. This equation of condition lets you find one extra reaction. A hinge joining more than two members provides more conditions: a hinge joining n members gives \(n - 1\) equations.

Determinacy of beams and frames

Let r be the number of reaction components and c the number of equations of condition from internal hinges. For a structure made of one or more rigid parts in a plane:

The degree of indeterminacy is \(r - (3 + c)\). A propped cantilever (fixed at one end, roller at the other) has four reactions and no hinges, so it is indeterminate to the first degree. A beam fixed at both ends has six reactions, but if all loads are vertical the horizontal reactions are zero and three of them do not matter, leaving it indeterminate to the second degree for bending.

For frames with rigid joints, the counting is easier if you think of cutting the structure into a tree. Each closed loop of rigidly connected members adds three degrees of indeterminacy, because cutting a member in the loop releases an axial force, a shear and a moment. A single-bay portal frame with fixed bases is indeterminate to the third degree; with pinned bases it is indeterminate to the first degree; and with pinned bases and a hinge at the ridge (a three-hinged frame) it is determinate. A more formal count for frames uses \(3b + r\) unknowns (three internal forces per member plus reactions) against \(3j + c\) equations (three per joint plus conditions), where b is the number of members and j the number of joints.

Geometric instability

Counting is not enough. A structure with enough reactions can still be unstable if they are badly arranged. Two arrangements are always unstable.

• All reactions are parallel. Three rollers on a horizontal surface supply three vertical forces but nothing to resist a horizontal load, so the beam slides away.
• All reactions are concurrent, meaning their lines of action pass through one point. They can resist forces but not a moment about that point, so the structure rotates.

Internal mechanisms are the other trap. If hinges are placed so that part of a structure can move without straining any member, the structure is unstable even though its overall count looks fine. Three hinges in a line within one span, or a hinge with nothing to stop one side from rotating, are typical errors. The test is to imagine the structure made of rigid pieces connected by the hinges and supports, and ask whether any piece can move. If it can, the structure is a mechanism.

Finding reactions of determinate structures

With stability and determinacy confirmed, finding reactions is a matter of choosing good equations. Start with a moment equation about a point where as many unknowns as possible pass, then use force equations. For structures with internal hinges, split the structure at the hinge, draw a free body of the simpler part, and use the zero-moment condition there.

Three-hinged frames and arches

A three-hinged frame has pin supports at both bases and a hinge at the crown. It has four reactions (two at each base) and one equation of condition, so it is determinate. It is a popular form for long-span garages and storage buildings because it is determinate, so it does not develop stresses from support settlement or temperature change, and because the horizontal thrusts at the bases reduce the bending moments in the rafters compared with a simple beam.

A determinacy and reaction calculator

For checking, Sofia writes a small program that classifies a structure from its counts and then solves the three equilibrium equations for a determinate beam with any set of point loads. It uses Gaussian elimination, the same method that structural software uses on a much larger scale.

def classify(r, c, kind="beam/frame", b=0, j=0):
    if kind == "truss":
        n = b + r - 2 * j
    else:
        n = r - (3 + c)
    if n < 0:
        return "unstable (too few restraints)"
    return "determinate" if n == 0 else f"indeterminate to degree {n}"

print("pin + 2 rollers:", classify(4, 0))
print("fixed + hinge + roller:", classify(4, 1))
print("fixed-base portal:", classify(6, 0))
print("truss b=13, r=3, j=8:", classify(3, 0, "truss", 13, 8))

def solve(A, b):
    n = len(b); M = [row[:] + [b[i]] for i, row in enumerate(A)]
    for k in range(n):
        p = max(range(k, n), key=lambda i: abs(M[i][k])); M[k], M[p] = M[p], M[k]
        for i in range(k + 1, n):
            f = M[i][k] / M[k][k]
            for jj in range(k, n + 1): M[i][jj] -= f * M[k][jj]
    x = [0.0] * n
    for i in reversed(range(n)):
        x[i] = (M[i][n] - sum(M[i][jj] * x[jj] for jj in range(i + 1, n))) / M[i][i]
    return x

# Beam: pin at A (x=0) gives Ax, Ay; roller at B (x=6) gives By. Loads (x, Fx, Fy).
loads = [(3.0, 0.0, -72.0), (8.0, -10.0, -17.32)]
A = [[1, 0, 0],          # sum Fx: Ax
     [0, 1, 1],          # sum Fy: Ay + By
     [0, 0, 6.0]]        # sum M about A: 6*By
rhs = [-sum(f[1] for f in loads), -sum(f[2] for f in loads), -sum(f[0] * f[2] for f in loads)]
Ax, Ay, By = solve(A, rhs)
print(f"Ax = {Ax:.2f} kN, Ay = {Ay:.2f} kN, By = {By:.2f} kN")

pin + 2 rollers: indeterminate to degree 1
fixed + hinge + roller: determinate
fixed-base portal: indeterminate to degree 3
truss b=13, r=3, j=8: determinate
Ax = 10.00 kN, Ay = 30.23 kN, By = 59.09 kN

The output matches the overhanging beam example. The same three-equation system, generalized to thousands of equations, is what a frame analysis program solves when you press the run button.