Slope-deflection for frames with sway

In a continuous beam the joints rotate but do not translate. In a frame they may also move sideways. A frame braced by walls or diagonal bracing is prevented from swaying, and slope-deflection proceeds exactly as for beams, with one rotation unknown per rigid joint. An unbraced frame, which relies on the bending stiffness of its beams and columns to resist lateral load, sways, and the sway adds new unknowns and new equations. This lesson analyses the garage end frames, which have moment-resisting connections and fixed bases, under wind and roof load.

When does a frame sway?

A frame sways whenever its joints can translate without stretching any member. Treating members as inextensible (lesson 3), a single-bay portal frame has one independent sway: both corners move sideways by the same amount \(\Delta\) because the beam does not change length. A frame with several stories has one sway per story, and a frame with sloping members (a gable frame) has additional, coupled sways. To count the independent sways, replace every rigid joint and fixed support with a pin and count the degrees of freedom of the resulting linkage: each is an independent sway.

Sway occurs under lateral load, of course, but also under gravity load whenever the frame or its loading is unsymmetric. An unsymmetric roof load, a frame with columns of different heights or stiffness, or a single heavy load placed eccentrically all produce sideways movement. A braced frame has its sways prevented by the bracing; the bracing then carries the force that would otherwise cause sway.

Equations for a swaying frame

The sway enters the slope-deflection equations of the columns through the chord rotation \(\psi = \Delta/h\), clockwise positive. The beam's chord does not rotate, because both its ends move sideways by the same amount. With one more unknown we need one more equation, and it comes from horizontal equilibrium of the whole story: the shears at the tops of the columns must balance the lateral load applied above the column bases.

For a column with no load along its height, the shear follows from a free body of the column: \(V = -(M_{\text{top}} + M_{\text{bottom}})/h\), with the moments clockwise positive and V positive acting to the left at the top. The story shear equation for a single-bay portal with lateral load P to the right at roof level is

If a column carries a lateral load along its height (wind on the wall girts, for example), its free body includes that load, and the shear at its top changes accordingly. If the columns have different heights, each has its own \(\psi = \Delta/h_i\) and its own shear term; the common unknown is \(\Delta\).

The garage end frame under wind

The location of the column inflection points is worth noticing. With an infinitely stiff beam the column tops could not rotate, and the inflection points would sit exactly at mid-height. With a flexible beam they move up toward the beam. Lesson 8 uses this observation to build the portal method, an approximate analysis that assumes inflection points at mid-height of columns and midspan of beams.

Pinned bases: the effect of base fixity

If the same frame had pinned bases, the modified slope-deflection equation would apply to each column, the base moments would be zero, and each column would have to carry its 3.375 kip shear by bending over its full height: the corner moments become \(3.375(20) = 67.5\ \text{kip·ft}\). The sway rises to \(EI_c\psi = 787.5\), a deflection of 0.158 ft or 1.89 in., more than four times the fixed-base value. Base fixity therefore buys a great deal of lateral stiffness, which is why unbraced frames in drift-sensitive buildings often have fixed bases, at the cost of larger footings that can resist moment.

Gravity load and superposition

Under the uniform roof load, a symmetric frame does not sway: \(\theta_C = -\theta_B\) and \(\psi = 0\). The analysis is then a no-sway problem with two rotation unknowns, which symmetry reduces to one.

Superposition is valid because the analysis is linear. In design it is applied with the factors of each load combination: lesson 12 returns to this when the full model is set up. For tall or heavily loaded frames, the sway itself increases the moments (the P-delta effect), and superposition of separate first-order results is no longer exact. Both design standards require second-order effects to be considered for unbraced frames, usually by amplifying the first-order sway moments or by a second-order computer analysis.

Multi-story frames

For a frame of several stories, each story has its own sway \(\Delta_i\) and its own story shear equation: the sum of the column shears just below floor i equals the total lateral load above that level. The office wing, if built as an unbraced two-story, two-bay frame, would have six joint rotations and two sways, eight unknowns in all. That is about the practical limit for hand solution, and it is where moment distribution (lesson 7), approximate methods (lesson 8) and the matrix stiffness method (lessons 9 to 11) take over.

Solving it in code

Sofia's frame program writes the two joint equations and the story shear equation as residual functions, builds the coefficient matrix by evaluating them for unit values of each unknown (valid because they are linear), and solves by Gaussian elimination. Set fixed_bases to False to reproduce the pinned-base results, or change Ic_right to see how an unsymmetric frame sways even under roof load alone.

# Slope-deflection for a single-bay portal frame with sway (bases A and D, corners B and C)
# Unknowns: EI*theta_B, EI*theta_C and EI*psi (column chord rotation = sway / h). Clockwise positive.
h, L = 20.0, 60.0                  # column height and beam span (ft)
Ic_left, Ic_right, Ib = 1.0, 1.0, 2.0   # relative second moments of area
fixed_bases = True

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

def end_moments(v, w, P):
    tB, tC, psi = v
    F = w * L ** 2 / 12
    kl, kr, kb = 2 * Ic_left / h, 2 * Ic_right / h, 2 * Ib / L
    if fixed_bases:
        MAB = kl * (tB - 3 * psi); MBA = kl * (2 * tB - 3 * psi)
        MDC = kr * (tC - 3 * psi); MCD = kr * (2 * tC - 3 * psi)
    else:                          # pinned bases: modified equation, base moments zero
        MAB = 0.0; MBA = 1.5 * kl * (tB - psi)
        MDC = 0.0; MCD = 1.5 * kr * (tC - psi)
    MBC = kb * (2 * tB + tC) - F; MCB = kb * (2 * tC + tB) + F
    return MAB, MBA, MBC, MCB, MCD, MDC

def residuals(v, w, P):            # joint B, joint C, and horizontal equilibrium (shear) of the columns
    MAB, MBA, MBC, MCB, MCD, MDC = end_moments(v, w, P)
    return [MBA + MBC, MCB + MCD, (MAB + MBA) / h + (MCD + MDC) / h + P]

def solve_frame(w, P):
    r0 = residuals([0.0, 0.0, 0.0], w, P)       # equations are linear: build them column by column
    A = [[residuals([1.0 if j == i else 0.0 for j in range(3)], w, P)[r] - r0[r] for i in range(3)] for r in range(3)]
    v = gauss(A, [-x for x in r0])
    return v, end_moments(v, w, P)

z = lambda v: round(v, 6) + 0.0
for name, w, P in (("wind 6.75 kips at roof", 0.0, 6.75), ("roof load 0.82 kip/ft", 0.82, 0.0), ("both together", 0.82, 6.75)):
    v, M = solve_frame(w, P)
    print(name)
    print("  EI theta_B = %.2f  EI theta_C = %.2f  EI psi = %.2f  (sway = psi x h)" % tuple(z(t) for t in v))
    print("  MAB %.2f  MBA %.2f  MBC %.2f  MCB %.2f  MCD %.2f  MDC %.2f kip*ft" % tuple(z(m) for m in M))

wind 6.75 kips at roof
  EI theta_B = 135.00  EI theta_C = 135.00  EI psi = 180.00  (sway = psi x h)
  MAB -40.50  MBA -27.00  MBC 27.00  MCB 27.00  MCD -27.00  MDC -40.50 kip*ft
roof load 0.82 kip/ft
  EI theta_B = 922.50  EI theta_C = -922.50  EI psi = 0.00  (sway = psi x h)
  MAB 92.25  MBA 184.50  MBC -184.50  MCB 184.50  MCD -184.50  MDC -92.25 kip*ft
both together
  EI theta_B = 1057.50  EI theta_C = -787.50  EI psi = 180.00  (sway = psi x h)
  MAB 51.75  MBA 157.50  MBC -157.50  MCB 211.50  MCD -211.50  MDC -132.75 kip*ft

The third case is the sum of the first two, line by line: the program confirms superposition. Building equations by evaluating residuals for unit unknowns is a small version of a powerful idea: the columns of the coefficient matrix are the responses to unit displacements, which is the definition of the stiffness matrix in lesson 9.