Introduction to indeterminate structures: the force method

Most real structures are statically indeterminate. Continuous beams run over several supports, building frames have rigid joints and fixed or pinned bases on both sides, and trusses often have extra members. Indeterminate structures have more unknown forces than equilibrium equations, so statics alone cannot find them. The missing equations come from compatibility: the requirement that the deformed structure fits together and respects its supports. This lesson introduces the oldest systematic approach, the force method, which uses exactly the deflection tools of lessons 10 and 11. CIV-321 continues with the slope-deflection, moment distribution and matrix stiffness methods.

Why build indeterminate structures?

Redundancy brings real advantages. For the same loads and spans, indeterminate structures generally have smaller maximum moments and deflections, because continuity lets adjacent spans help each other. A two-span continuous beam under uniform load has a maximum moment of \(wL^2/8\) over the middle support, the same as a simple span, but its span moments drop to about 56 per cent of the simple-span value, and its deflections drop by more than half. Indeterminate structures also have alternative load paths. If one member is damaged by a vehicle impact, fire or corrosion, the load can redistribute to others instead of causing collapse, a property engineers call robustness.

There are costs. Because internal forces depend on relative stiffness, the analysis is more involved and must be repeated if member sizes change. Indeterminate structures also develop internal forces from effects that cause no forces in determinate ones: support settlement, temperature change, shrinkage, and fabrication misfit. A continuous beam whose middle support settles redistributes its moments, and a fixed arch heated by the sun develops bending. The designer must account for these effects, or choose a determinate form, such as the three-hinged frames of the salt shed, where foundations are uncertain.

The idea of the force method

The force method, also called the flexibility or consistent deformation method, works by turning an indeterminate structure into a determinate one and then restoring compatibility.

• Determine the degree of indeterminacy n and choose n redundants: reactions or internal forces whose removal leaves a stable determinate structure, called the primary or released structure.
• Compute the displacements of the primary structure, at the location and in the direction of each redundant, caused by the actual loads.
• Compute the displacements of the primary structure caused by a unit value of each redundant acting alone. These are flexibility coefficients: \(f_{ij}\) is the displacement at redundant i caused by a unit value of redundant j.
• Write one compatibility equation for each redundant, stating that the total displacement at that location equals its known value (usually zero at a support): \(\Delta_{i0} + \sum_j f_{ij}X_j = \Delta_i\).
• Solve for the redundants, then find all other forces by equilibrium, which is now possible because the redundants are known.

One redundant: the propped cantilever

Continuous beams

A beam continuous over two spans has one redundant reaction. Removing the middle support leaves a simple beam spanning both spans, whose midspan deflection under the load is easy to compute. A unit upward force at the middle support of that simple beam produces the flexibility coefficient, and compatibility gives the middle reaction.

The moment over the middle support is the same as each simple span's moment, so the continuous beam needs the same section, but its deflection is much smaller: for one span of a two-span beam under full load, the maximum deflection is about \(wL^4/(185EI)\) instead of \(5wL^4/(384EI) = wL^4/(77EI)\). That gain in stiffness is why continuous construction is standard for floor beams, bridge girders and slabs.

Maxwell's reciprocal theorem

With two or more redundants, the compatibility equations form a system \(\sum_j f_{ij}X_j = \Delta_i - \Delta_{i0}\), and you need all the flexibility coefficients. James Clerk Maxwell proved in 1864 that for a linearly elastic structure \(f_{ij} = f_{ji}\): the displacement at point i due to a unit load at j equals the displacement at j due to a unit load at i. The flexibility matrix is therefore symmetric, which halves the work and provides a check on the arithmetic. Betti later generalized the theorem to any two load systems: the work done by the first set of loads acting through the displacements caused by the second equals the work done by the second set through the displacements caused by the first.

The reciprocal theorem also explains the Müller-Breslau principle of lesson 8. The influence line for a reaction is the deflected shape when that reaction is released and displaced by one unit, because, by reciprocity, the reaction caused by a unit load at x equals the deflection at x caused by a unit displacement of the released support, scaled appropriately.

Support settlement and temperature

The compatibility equation also handles known support movements. If the middle support of the two-span beam settles by \(\delta\), compatibility becomes \(-\Delta_{B0} + R_B f_{BB} = -\delta\), so the middle reaction drops by \(\delta/f_{BB} = 6EI\delta/L^3\), and the end reactions increase to keep equilibrium. Stiff beams on short spans are the most sensitive, because \(6EI/L^3\) is large.

Temperature effects enter the same way. A uniform temperature change in a beam on rollers causes no force, but if both ends are restrained, the free expansion \(\alpha\,\Delta T\,L\) must be cancelled by an axial force, which compatibility gives as \(EA\,\alpha\,\Delta T\). A temperature difference between the top and bottom of a continuous beam, as under strong sun, causes curvature that the intermediate supports resist, producing moments that a determinate beam would not have.

Choosing redundants and looking ahead

Any set of redundants that leaves a stable primary structure works, but a good choice simplifies the computation. Removing interior supports of a continuous beam leaves a simple beam whose deflections are standard formulas. Inserting hinges over the supports instead, so that the redundants are the support moments, leaves a series of simple spans whose flexibility coefficients involve only adjacent spans, which gives a banded system of equations, the three-moment equation of Clapeyron. For a frame, cutting each closed loop releases three internal forces.

The force method becomes cumbersome for structures with many redundants, because each needs its own displacement calculations and the equations are coupled. The displacement (stiffness) methods take the opposite approach: they treat the joint displacements as unknowns and write equilibrium equations at the joints. That is the basis of moment distribution, slope-deflection, and the matrix stiffness method used by every structural analysis program today. You will study them in CIV-321, but the ideas of this course (equilibrium, compatibility, and the link between forces and deformations) remain the foundation.

The force method by computer

Sofia's program applies the force method to a two-span beam with any uniform loads on the spans, computing the deflections of the primary structure by numerical integration of \(mM/(EI)\), exactly as in lesson 11, rather than from formulas. It then checks the classic results and adds the settlement case.

# Two-span continuous beam A-B-C by the force method; redundant = reaction at B
L1, L2 = 7.5, 7.5                  # spans (m)
w1, w2 = 19.0, 19.0                # factored uniform loads (kN/m)
EI = 70200.0                       # kN*m^2
settle = 0.0                       # settlement of B (m), downward positive
Lt = L1 + L2

def M0(x):                         # primary structure (simple beam A-C) moment from loads
    RA = (w1 * L1 * (Lt - L1 / 2) + w2 * L2 * (L2 / 2)) / Lt
    m = RA * x - w1 * min(x, L1) * (x - min(x, L1) / 2)
    if x > L1: m -= w2 * (x - L1) ** 2 / 2
    return m

def m1(x):                         # moment from a unit upward force at B
    RA = -L2 / Lt
    return RA * x + (x - L1 if x > L1 else 0.0)

n = 6000; h = Lt / n
xs = [i * h for i in range(n + 1)]
def integ(f):
    return sum((f(xs[i]) + f(xs[i + 1])) / 2 * h for i in range(n))
D0 = integ(lambda x: m1(x) * M0(x)) / EI          # upward displacement at B due to loads
f11 = integ(lambda x: m1(x) ** 2) / EI            # upward displacement due to unit upward force
for s in (0.0, 0.010):
    RB = (-s - D0) / f11
    RA = ((w1 * L1 * (Lt - L1 / 2) + w2 * L2 * L2 / 2) - RB * L2) / Lt
    MB = RA * L1 - w1 * L1 ** 2 / 2
    print(f"settlement {s*1000:4.0f} mm: RB = {RB:6.1f} kN, RA = {RA:5.1f} kN, MB = {MB:7.1f} kN*m")
print(f"check 5wL/4 = {1.25 * w1 * L1:.1f} kN, -wL^2/8 = {-w1 * L1**2 / 8:.1f} kN*m")

settlement    0 mm: RB =  178.1 kN, RA =  53.4 kN, MB =  -133.6 kN*m
settlement   10 mm: RB =  168.1 kN, RA =  58.4 kN, MB =   -96.2 kN*m
check 5wL/4 = 178.1 kN, -wL^2/8 = -133.6 kN*m

Changing one span's load to zero gives the pattern-loading case for maximum positive moment in the other span, which lesson 8's Müller-Breslau sketch predicted. With unequal spans or loads, the program still works, because nothing in it assumes symmetry.