Moment distribution

Slope-deflection turns a continuous beam or frame into a set of simultaneous equations. In 1930, before electronic computers, Hardy Cross published a way to avoid solving them: lock every joint against rotation, then release the joints one at a time and let them rotate, distributing and carrying over moments until the joints are in balance. Each step needs only arithmetic, the process converges quickly, and the engineer can watch moments flow through the structure. Moment distribution was the dominant method of frame analysis for forty years, and it remains the fastest way to check a continuous beam or a small frame by hand.

Three ideas: stiffness, distribution and carryover

The method rests on three quantities, all of which follow from the slope-deflection equation of lesson 5.

• Member stiffness. The moment needed at the near end of a member to rotate it through one radian is its stiffness K. With the far end fixed, \(K = 4EI/L\). With the far end pinned, \(K = 3EI/L\), which is the modified stiffness.
• Distribution factor. When a joint with several members is rotated, each member resists in proportion to its stiffness. An unbalanced moment applied to the joint is shared among the members by the distribution factors

which add to 1.0 at each joint. A fixed support has a distribution factor of zero (it absorbs any moment without rotating), and a pin with no other member has a distribution factor of 1.0.

• Carryover factor. Rotating the near end of a member with a fixed far end induces a moment at the far end equal to one half of the near-end moment, with the same sign. The carryover factor is therefore \(+\tfrac{1}{2}\) for prismatic members toward a fixed or locked end, and zero toward a pin.

The procedure

• Compute the stiffness of every member end at the joints that can rotate, and the distribution factors at each joint.
• Lock all joints and compute the fixed-end moments, with the slope-deflection sign convention (clockwise positive on the member end).
• Release a joint. Its unbalanced moment is the sum of the member end moments there. Distribute the negative of this sum to the members by their distribution factors, so the joint balances.
• Carry over one half of each distributed moment to the far end of its member (unless the far end is a pin).
• Repeat for the other joints, cycling until the carried-over moments are negligible. Add each column of the table to obtain the final end moments.

Joints can be released one at a time or all together in each cycle; both converge to the same answer. Releasing all joints together and carrying over afterwards is easier to tabulate and is the scheme the program below uses.

The corridor beam under pattern load

Convergence is fast when the carryover pathways are short and the distribution factors are moderate. Each cycle here reduces the unbalance by a factor of \(0.571 \times 0.5 = 0.29\). Joints connected through very stiff members converge more slowly, and in that case releasing the joints one at a time (Gauss-Seidel style) speeds things up.

Checking the result

Moment distribution is a sequence of small additions, and arithmetic slips are the main risk. Three checks catch nearly all of them. First, every joint must balance in the final sums. Second, the change in moment at each end of a member (final minus fixed-end) can be checked against the slope-deflection equation: the rotations it implies must be the same for every member meeting at a joint. Third, the moments obtained should match the patterns you expect: the fully loaded spans either side of B should produce a larger hogging moment over B than the all-spans case, and they do (59.4 compared with 54.0).

Frames without sway

For a braced frame, or a symmetric frame under symmetric load, moment distribution proceeds exactly as for beams. Every rigid joint gets distribution factors from all the members that meet there, columns included, and columns with fixed bases receive carryovers to their bases. A column with a pinned base uses the modified stiffness \(3EI/L\), just like the end span of a beam.

For the garage end frame of lesson 6 (fixed bases, columns 20 ft, beam 60 ft with twice the column I), the stiffnesses at corner B are \(K_{BA} = 4EI_c/20 = 0.20EI_c\) and \(K_{BC} = 4(2EI_c)/60 = 0.133EI_c\), giving distribution factors of 0.60 to the column and 0.40 to the beam. Under the symmetric roof load the beam's far end rotates equally and oppositely, which halves its effective stiffness; using that symmetric stiffness \(2EI/L\) for the beam, one distribution gives the exact corner moment of 184.5 kip·ft found in lesson 6.

Frames with sway

Moment distribution cannot let joints translate, because every step holds the joints in position. Sway is handled by superposition in two stages.

• Stage 1, no sway. Add an imaginary horizontal restraint at the beam level to stop sway, and distribute the fixed-end moments of the loads. Compute the column shears and the force R that the restraint must provide.
• Stage 2, sway. Remove the loads, impose an arbitrary sway \(\Delta'\) with the joints locked (which induces fixed-end moments \(-6EI\Delta'/h^2\) at both ends of every column), distribute these, and compute the column shears and the restraint force \(R'\) they correspond to.
• Combine: final moments = stage 1 + \((R/R')\) times stage 2, chosen so the imaginary restraint force is canceled.

For the garage frame under 6.75 kip wind, stage 1 has no fixed-end moments at all, so the restraint carries the whole 6.75 kips. In stage 2, choose a sway giving fixed-end moments of \(-100\) kip·ft at every column end. Distribution with factors 0.60 and 0.40, and carryover to the bases, converges to \(-50.0\) kip·ft at the column tops and \(-75.0\) kip·ft at the bases. The column shears are \(2(50.0 + 75.0)/20 = 12.5\) kips, so the scale factor is \(6.75/12.5 = 0.54\), and the final moments are \(-27.0\) and \(-40.5\) kip·ft, matching lesson 6.

A frame with n independent sways needs n separate sway stages and n simultaneous equations for the scale factors, which removes much of the method's appeal. For multi-story unbraced frames, the matrix stiffness method is the practical choice.

Solving it in code

Iteration is what computers do best. The program below runs the same table automatically, releasing both joints in each cycle and stopping when the carryovers fall below 0.005. Change the span loads to try the other patterns from lesson 2, or change one EI to see how a stiffer span attracts moment.

# Moment distribution (Hardy Cross) for the three-span corridor beam A-B-C-D
# Ends A and D are pins: the end spans use the modified stiffness 3EI/L and no carryover to the pin.
L = [20.0, 20.0, 20.0]             # spans (ft)
EI = [1.0, 1.0, 1.0]               # relative EI
w = [1.0, 1.0, 0.4]                # factored load on each span (kip/ft): live load on AB and BC only

# member ends at the two interior joints: (span, end) with end 0 = left, 1 = right
K = [3 * EI[0] / L[0], 4 * EI[1] / L[1], 4 * EI[1] / L[1], 3 * EI[2] / L[2]]   # BA, BC, CB, CD
DF = [K[0] / (K[0] + K[1]), K[1] / (K[0] + K[1]), K[2] / (K[2] + K[3]), K[3] / (K[2] + K[3])]
# fixed-end moments, clockwise positive; end spans already released at the pin (wL^2/8)
M = [w[0] * L[0] ** 2 / 8, -w[1] * L[1] ** 2 / 12, w[1] * L[1] ** 2 / 12, -w[2] * L[2] ** 2 / 8]
print("distribution factors BA %.3f BC %.3f | CB %.3f CD %.3f" % tuple(DF))
print("%-10s %9s %9s %9s %9s" % ("", "BA", "BC", "CB", "CD"))
print("%-10s %9.2f %9.2f %9.2f %9.2f" % ("FEM", M[0], M[1], M[2], M[3]))

for cycle in range(1, 30):
    # balance joint B and joint C at the same time, then carry over half to the far ends
    uB = M[0] + M[1]; uC = M[2] + M[3]
    d = [-DF[0] * uB, -DF[1] * uB, -DF[2] * uC, -DF[3] * uC]
    co = [0.0, 0.5 * d[2], 0.5 * d[1], 0.0]          # BC <- half of CB, CB <- half of BC
    M = [M[i] + d[i] + co[i] for i in range(4)]
    if cycle <= 4:
        print("%-10s %9.2f %9.2f %9.2f %9.2f" % ("cycle %d" % cycle, M[0], M[1], M[2], M[3]))
    if max(abs(x) for x in co) < 0.005:
        break
uB = M[0] + M[1]; uC = M[2] + M[3]                    # final balance, no carryover
M = [M[0] - DF[0] * uB, M[1] - DF[1] * uB, M[2] - DF[2] * uC, M[3] - DF[3] * uC]
print("%-10s %9.2f %9.2f %9.2f %9.2f   (after %d cycles)" % ("final", M[0], M[1], M[2], M[3], cycle))
print("support moments: B = %.2f kip*ft, C = %.2f kip*ft (hogging)" % (M[0], M[2]))

distribution factors BA 0.429 BC 0.571 | CB 0.571 CD 0.429
                  BA        BC        CB        CD
FEM            50.00    -33.33     33.33    -20.00
cycle 1        42.86    -46.67     20.95    -25.71
cycle 2        44.49    -43.13     24.76    -23.67
cycle 3        43.91    -44.22     23.75    -24.14
cycle 4        44.04    -43.93     24.06    -23.97
final          44.00    -44.00     24.00    -24.00   (after 7 cycles)
support moments: B = 44.00 kip*ft, C = 24.00 kip*ft (hogging)

The first four printed cycles reproduce the rows of the hand table (each printed row includes that cycle's distribution and carryover), and the final line shows the converged values. Seven cycles reach full precision; hand calculation rarely needs more than three or four for design accuracy.