Portal frames, braced beams and two-pinned arches

A portal frame with fixed bases is statically indeterminate to the third degree, which sounds forbidding. Slope-deflection reduces it to a manageable number of unknown displacements — its kinematic indeterminacy, which is 3: the two joint rotations and the sway.

Sway is the horizontal movement of the top of the frame relative to its base. It enters the slope-deflection equation through ψ, the relative transverse displacement divided by the member length. Two situations rule it out:

• The frame and its loading are both symmetric, so there is no reason for it to lean either way.
• The frame is braced against horizontal movement by something else.

When sway can be ruled out, the analysis collapses. For a symmetric portal under a symmetric load, symmetry also makes the two joint rotations equal and opposite, and three unknowns become one.

A braced beam is a beam propped by a member that is itself elastic — a strut, a tie, a hanger. It is the propped cantilever with one change: the prop gives a little.

Compatibility now says that the deflection of the beam at the brace equals the shortening or stretching of the brace, rather than zero:

δ_beam under load − δ_beam under R = R × L_brace/(A_brace E_brace)

Rearranged, the brace force is the free deflection divided by the sum of the two flexibilities:

R = δ_free / (f_beam + f_brace)

The two limits are worth checking. A rigid brace has zero flexibility, so R = δ_free/f_beam — the fully propped result, 5wL/8 for a UDL. An infinitely flexible brace has infinite flexibility, so R → 0 and the beam is unpropped. Everything real sits in between, and the brace never quite achieves what a rigid prop would.

Finally the two-pinned arch. Removing the crown hinge from a three-pinned arch leaves four unknown reaction components against three equations — indeterminate to degree 1, with the horizontal thrust H as the natural redundant.

Statics alone cannot find H, so compatibility must. The condition is that the horizontal distance between the pins does not change, which by the least-work argument gives:

H = ∫M₀ y ds ÷ ∫y² ds

where M₀ is the bending moment of the equivalent simply supported beam and y the height of the arch axis. Evaluating this for a parabolic arch gives two results worth knowing:

• Under a full UDL: H = wL²/8h — identical to the three-pinned arch. The parabola is funicular for a uniform load, so the arch carries it in pure compression whether or not there is a crown hinge, and the extra restraint does nothing.
• Under a central point load: H = 25WL/128h, against WL/4h for the three-pinned case. The two-pinned arch thrusts about 78% as hard.

The first result is the more instructive. A redundant restraint changes nothing at all when the structure was already carrying the load in the ideal way.