Superposition, indeterminate beams and energy methods

Integrating the elastic curve works, but it is slow for complicated loadings. This lesson offers two faster tools. Superposition builds deflections from a table of standard cases, and it turns statically indeterminate beams, such as propped cantilevers, fixed-end beams and continuous beams, into simple problems of compatibility. Energy methods compute deflections from the strain energy stored in a structure, and they apply just as easily to trusses, frames and curved members as to beams. Both ideas are foundations of the structural analysis courses that follow.

Deflection by superposition

Because \(EIv'' = M\) is linear, the deflection caused by several loads acting together is the sum of the deflections caused by each load alone. Engineers keep tables of slopes and deflections for standard cases, such as the one in lesson 10, and combine them. The skill lies in breaking a loading into cases that appear in the table, and in remembering that a deflection somewhere other than the loaded point may need the slope as well.

For example, a cantilever loaded at a point partway along its length, a distance \(a\) from the fixed end, deflects like a cantilever of length \(a\) up to the load. Beyond the load, the beam carries no moment, so it stays straight, continuing at the slope it had under the load. The tip deflection is therefore the deflection at the load plus the slope at the load times the remaining length:

Statically indeterminate beams

A beam is statically indeterminate when it has more support reactions than equilibrium can determine. A propped cantilever (fixed at one end, simply supported at the other) has four reaction components (one horizontal, two vertical, one moment) and only three equilibrium equations, so it is indeterminate to the first degree. A beam fixed at both ends, or a beam continuous over three supports, has extra unknowns too.

Indeterminate beams are common because they are efficient: the extra restraint reduces moments and deflections, and it provides alternative load paths if one part is damaged. They come at a price. Their internal forces depend on the relative stiffness of their parts, and they develop stresses from support settlement and temperature changes that a determinate beam would simply accommodate.

The method of superposition for indeterminate beams follows the same pattern as the indeterminate axial and torsion problems of lessons 2 and 3.

• Choose redundant reactions: as many as the degree of indeterminacy. Remove them to leave a stable, determinate primary structure.
• Compute the displacement of the primary structure at each redundant due to the applied loads.
• Compute the displacement at each redundant due to the redundant itself, acting alone.
• Write compatibility: the total displacement at each support must match what the real support allows, usually zero. Solve for the redundants, then use equilibrium for everything else.

Two more standard results follow the same way. A beam fixed at both ends under a uniform load has end moments of \(-wL^2/12\) and a midspan moment of \(+wL^2/24\), with a midspan deflection of \(wL^4/(384EI)\), one fifth of the simply supported value. A beam continuous over two equal spans with a uniform load on both has a central reaction of \(5wL/4\), end reactions of \(3wL/8\), and a hogging moment of \(-wL^2/8\) over the central support. These fixed-end moments are the starting values for the moment distribution and stiffness methods taught in structural analysis.

Strain energy

A loaded elastic member stores energy, just as a stretched spring does, and gives it back when unloaded. That stored energy, the strain energy \(U\), provides another route to deflections. For a linear elastic bar, the work done by a load \(P\) that rises gradually from zero while the bar stretches by \(\delta\) is \(P\delta/2\), so

For members where the force or moment varies, the energy is integrated or summed along the length. In a truss, the total strain energy is the sum of \(N_i^2L_i/(2A_iE_i)\) over all the members.

When a single load \(P\) acts on a structure, conservation of energy says the external work equals the strain energy: \(P\Delta/2 = U\), where \(\Delta\) is the deflection under the load in its own direction. For a cantilever with a tip load, \(M = -Px\) measured from the tip, so \(U = \int_0^L P^2x^2/(2EI)\,dx = P^2L^3/(6EI)\). Setting \(P\Delta/2 = P^2L^3/(6EI)\) gives \(\Delta = PL^3/(3EI)\), the same answer as lesson 10 with almost no effort.

Castigliano's theorem

The work–energy method handles only one load and the deflection under it. Castigliano's second theorem removes that limitation: in a linear elastic structure, the deflection at the point of application of a load \(P_i\), in the direction of that load, is the partial derivative of the total strain energy with respect to that load:

For beams and trusses it is easiest to differentiate inside the integral or sum before integrating:

To find a deflection where no load acts, apply an imaginary (dummy) load \(Q\) at that point, carry out the differentiation, and then set \(Q = 0\). To find a rotation, use a dummy couple instead. The same idea, expressed as virtual work with a unit load, is how structural analysis courses compute truss and frame deflections.

Superposition and energy by code

Both methods lend themselves to short programs. The first part of the code below solves the propped cantilever and the two-span beam by compatibility, using the standard deflection formulas. The second part checks the truss result by summing \(N\,\partial N/\partial P\,L/(AE)\), then confirms Castigliano's theorem by differentiating the strain energy numerically.

# Indeterminate beams by superposition, and truss deflection by Castigliano
w, L = 25.0, 7.0                                # kN/m, m

# Propped cantilever: tip deflections of the cantilever under w and under R must cancel
RB_prop = (w * L**4 / 8) / (L**3 / 3)            # EI cancels
print(f"Propped cantilever: R_B = {RB_prop:.2f} kN  (3wL/8 = {3 * w * L / 8:.2f})")

# Two equal spans: remove the middle support of a 2L simple span
RB_cont = (5 * w * (2 * L)**4 / 384) / ((2 * L)**3 / 48)
RA = (w * 2 * L - RB_cont) / 2
MB = RA * L - w * L**2 / 2
print(f"Two spans: R_B = {RB_cont:.2f} kN, R_A = R_C = {RA:.2f} kN, M_B = {MB:.1f} kN*m")

# Truss: members (length mm, force per unit P, area mm^2, E MPa)
P = 30e3
members = {"AB": (2000, 4 / 3, 800, 200e3), "CB": (2500, -5 / 3, 800, 200e3)}
delta = sum((n * P) * n * Lm / (A * E) for Lm, n, A, E in members.values())
print(f"Truss deflection by Castigliano: {delta:.3f} mm")

def U(load):                                      # strain energy in N*mm
    return sum((n * load) ** 2 * Lm / (2 * A * E) for Lm, n, A, E in members.values())
h = 1.0
print(f"Numerical dU/dP: {(U(P + h) - U(P - h)) / (2 * h):.3f} mm")

Propped cantilever: R_B = 65.62 kN  (3wL/8 = 65.62)
Two spans: R_B = 218.75 kN, R_A = R_C = 65.62 kN, M_B = -153.1 kN*m
Truss deflection by Castigliano: 1.969 mm
Numerical dU/dP: 1.969 mm

The numerical derivative of the strain energy matches the hand result, a direct check of Castigliano's theorem. Structural analysis software generalizes exactly these steps into the stiffness method, which handles any number of redundants at once.