Approximate methods and influence lines for indeterminate structures

Exact methods need member sizes, because indeterminate structures share load by stiffness. At the start of a design the sizes are not yet known. Engineers break this circle with approximate methods: assumptions about where the moment is zero, or how the shear divides, that make the structure determinate. The results are good enough to size members, which can then be checked by an exact analysis. The second half of this lesson turns to moving and variable loads. Influence lines for indeterminate structures are curved, but the Müller-Breslau principle from CIV-224 still gives their shape at a glance, and that shape tells you where to put the load.

Approximate analysis for gravity load

Each assumption about a point of zero moment in a beam removes one unknown. For a beam continuous at both ends under uniform load, the exact inflection points of a fixed-fixed beam lie at \(0.211L\) from each end (lesson 2). Real beams in frames have ends that are neither pinned nor fixed, so a common approximation places the inflection points at \(0.1L\) from each end. The central \(0.8L\) is then a simple beam and the end portions are cantilevers carrying its reactions.

These bracket the true values. A fixed-fixed beam has \(0.042wL^2\) at midspan and \(-0.083wL^2\) at its ends; a simple beam has \(0.125wL^2\) at midspan and nothing at its ends. Concrete design standards formalize a similar idea through tabulated moment coefficients for continuous beams and slabs with regular spans, such as \(wL^2/14\) for the positive moment in an end span built integrally with its support and \(wL^2/9\) for the negative moment at the first interior support of a two-span beam. They are calibrated to cover pattern loading, so they apply only within stated limits on span ratios and live-to-dead load ratios.

The portal method for lateral load

Lesson 6 showed that, in an unbraced frame under lateral load, columns and beams bend in double curvature with inflection points near their middles. The portal method, developed for the early steel skyscrapers, makes three assumptions:

• There is an inflection point at mid-height of every column.
• There is an inflection point at midspan of every beam.
• In each storey, each interior column carries twice the shear of each exterior column, because an interior column acts as part of two adjacent portals.

With these assumptions the frame becomes determinate. The procedure is mechanical: find the shear in each storey from the loads above it, divide it among the columns, multiply each column shear by half its height to get its end moments, then use joint equilibrium from left to right to find the beam end moments, and beam shears to find the column axial forces.

The cantilever method

For tall, slender frames, where overall bending of the building dominates, the cantilever method gives better results. It keeps the inflection-point assumptions but replaces the shear rule with an axial rule: the column axial forces vary linearly with distance from the centroid of the column areas, as the fibre stresses do in a cantilever beam. Taking moments of the storey loads about the inflection points of each level then gives the axial forces, from which beam shears and moments follow.

For the office wing frame, with equal column areas, the centroid is at the interior column, which therefore carries no axial force. At mid-height of the ground storey the overturning moment is \(20.0(6.0) + 36.0(2.0) = 192\ \text{kN·m}\), giving exterior column forces of \(192/(2 \times 7.5) = 12.8\) kN, identical to the portal result. For this symmetric low-rise frame the two methods agree; they diverge for tall frames and for unequal bays.

Influence lines for indeterminate structures

An influence line shows how a response at one point (a reaction, shear or moment) varies as a unit load moves across the structure. In CIV-224 the Müller-Breslau principle gave influence lines for determinate structures as straight-line mechanisms. The principle holds equally for indeterminate structures: the influence line for a response is the deflected shape of the structure when the restraint corresponding to that response is removed and a unit displacement is imposed in its place. The difference is that the remaining structure is still indeterminate, so the deflected shape is a smooth curve rather than a set of straight lines.

For the reaction at an interior support B of a two-span continuous beam with equal spans L, removing the support and pushing B up by one unit gives the curve

and its mirror image in the second span. At midspan of the first span \(a = L/2\) the ordinate is \(11/16 = 0.6875\): a load there sends more than two thirds of itself to B. For the moment at B, inserting a hinge and rotating the two sides by a unit relative angle gives

which is negative everywhere: load anywhere on either span causes hogging at B. Its largest magnitude is \(0.0962L\) at \(a = L/\sqrt{3} = 0.577L\).

Qualitative influence lines and pattern loading

Often only the shape matters. Sketching the Müller-Breslau curve for a response tells you which spans to load for its maximum, without computing a single ordinate. For the sagging moment at the middle of a span, insert a hinge there and bend the beam: the curve is positive in that span, negative in the adjacent spans and positive again in the spans beyond. Hence the rule from lesson 2: load the span and every second span from it. For the hogging moment over a support, the curve is negative in the two adjacent spans and positive in the next ones. The same sketches applied to a multi-storey frame give the checkerboard live load patterns used for the maximum beam moments in buildings.

Solving it in code

The program applies the portal method to any regular frame: give it the bay widths, the tier heights and the lateral loads. Its column, beam and axial results reproduce the hand calculation, and adding a third bay or a third level shows how quickly the method scales.

# Portal method for a regular unbraced frame under lateral load
# Assumptions: inflection points at mid-height of columns and midspan of beams;
# interior columns carry twice the shear of exterior columns in each tier of columns.
bays = [7.5, 7.5]                  # bay widths (m)
heights = [4.0, 4.0]               # column tier heights from the top down (m)
F = [20.0, 36.0]                   # lateral loads at each level from the roof down (kN)

ncol = len(bays) + 1
shares = [1.0 if c in (0, ncol - 1) else 2.0 for c in range(ncol)]
col_moment = []                    # column end moments, tier by tier
tier_shear = 0.0
for s, h in enumerate(heights):
    tier_shear += F[s]
    V = [tier_shear * k / sum(shares) for k in shares]
    Mc = [v * h / 2 for v in V]
    col_moment.append(Mc)
    print("tier %d (from top): shear %.1f kN, column shears %s kN, column moments %s kN*m"
          % (s + 1, tier_shear, [round(v, 2) for v in V], [round(m, 2) for m in Mc]))

axial = [0.0] * ncol
for s in range(len(heights)):
    # joint equilibrium from left to right: beam moment = column moments above and below minus beam on the left
    above = col_moment[s - 1] if s > 0 else [0.0] * ncol
    below = col_moment[s]
    left = 0.0; beamM = []
    for c in range(ncol - 1):
        m = above[c] + below[c] - left
        beamM.append(m); left = m
    beamV = [2 * m / b for m, b in zip(beamM, bays)]
    # beam shears accumulate as column axial forces (tension on the windward side)
    for c in range(ncol):
        if c > 0: axial[c] -= beamV[c - 1]
        if c < ncol - 1: axial[c] += beamV[c]
    print("level %d beams: end moments %s kN*m, shears %s kN; column axial below %s kN"
          % (s + 1, [round(m, 2) for m in beamM], [round(v, 2) for v in beamV], [round(a, 2) + 0.0 for a in axial]))

tier 1 (from top): shear 20.0 kN, column shears [5.0, 10.0, 5.0] kN, column moments [10.0, 20.0, 10.0] kN*m
tier 2 (from top): shear 56.0 kN, column shears [14.0, 28.0, 14.0] kN, column moments [28.0, 56.0, 28.0] kN*m
level 1 beams: end moments [10.0, 10.0] kN*m, shears [2.67, 2.67] kN; column axial below [2.67, 0.0, -2.67] kN
level 2 beams: end moments [38.0, 38.0] kN*m, shears [10.13, 10.13] kN; column axial below [12.8, 0.0, -12.8] kN

Approximate methods earn their place in modern practice as checks: if a computer model's column moments under wind differ from the portal estimate by more than about 30 per cent for a regular frame, it is worth finding out why before trusting either.