Elastic Deflection Methods

Elastic Deflection Methods

For July 2026 problems, use the beam formulas and mechanics relationships in the current NCEES PE Civil Reference Handbook supplied for the test date. A deflection formula is not identified only by its load. Support conditions, evaluation point, stiffness, and sign are part of the formula.

Slope and displacement are different responses

Let transverse deflection be v(x). For small rotations, beam slope is

θ(x) = dv/dx.

Slope is a rotation, measured in radians and dimensionless in calculations. Deflection is a translation with length units. At the center of a symmetrically loaded simple beam, slope is zero by symmetry while deflection is usually maximum, not zero. At a simple support, deflection is zero but slope can be nonzero. Confusing these conditions corrupts integration constants.

For an Euler-Bernoulli beam under a consistent sign convention, curvature is related to bending moment by a form of

d²v/dx² = M(x)/(EI).

Some sign conventions place a minus sign on one side. Either is workable when the load, moment, curvature, and deflection signs are consistent. E is elastic modulus and I is the section moment of inertia about the bending axis. Use the stiffness specified for the problem; do not silently alternate among gross, cracked, transformed, or effective inertia.

Method 1: integrate the moment-curvature equation

Find reactions, write M(x) for every interval, then integrate:

EI θ(x) = ∫M(x) dx + C1

EI v(x) = ∫∫M(x) dx dx + C1x + C2.

Determine constants from actual boundary and continuity conditions. Typical geometric conditions are:

An internal hinge has zero bending moment but generally not zero deflection. At a free end, deflection and slope are not prescribed as zero; end shear and moment conditions follow from loading. Piecewise moment functions require enough constants and continuity or compatibility equations.

Method 2: handbook cases and superposition

Tabulated formulas save time when the diagram exactly matches. Confirm whether the listed quantity is maximum deflection, deflection at a named point, or end slope, and confirm whether the load is total force or force per length. Powers of span are a strong clue: a point-load beam deflection commonly scales with PL³/(EI), while uniform-load deflection scales with wL⁴/(EI).

Linear-elastic responses can be superposed:

vtotal(x) = Σvi(x) and θtotal(x) = Σθi(x).

Add signed responses at the same coordinate and in the same direction. Every component case must use the same structure, supports, span, coordinate system, and stiffness model. One cannot add a simply supported formula to a cantilever formula and claim the result describes either beam. Superposition also becomes unreliable after yielding, contact changes, large-displacement behavior, or other nonlinear response.

Worked combined loading

A simply supported steel beam spans 16 ft and has constant E = 29,000 ksi and I = 400 in⁴. It carries a full-span uniform load w = 1.0 kip/ft and a 12-kip point load at midspan. Find elastic midspan deflection. Convert L = 192 in and w = 1/12 kip/in. Both handbook cases have the same beam and boundary conditions, so superposition is valid.

For the uniform load,

δw = 5wL⁴/(384EI)

= 5(1/12)(192⁴)/[384(29,000)(400)] = 0.1271 in downward.

For the midspan point load,

δP = PL³/(48EI)

= 12(192³)/[48(29,000)(400)] = 0.1525 in downward.

The signed responses act in the same direction, so

δtotal = 0.1271 + 0.1525 = 0.2796 in downward.

By symmetry, the midspan slope is zero even though the deflection is the largest downward displacement. At each simple support, deflection returns to zero. An answer of zero midspan deflection obtained from θ = 0 would confuse rotation with translation.

Method 3: virtual work and a unit load

Virtual work is efficient when only one displacement is required. For bending deformation of a linear-elastic beam,

δ = ∫ M(x)m(x)/(EI) dx,

where M is the moment from real loads and m is the moment from a unit force applied at the desired displacement location in the desired direction. The integral covers every member and interval contributing to response. Its sign reports displacement relative to the chosen unit-force direction.

To calculate a rotation instead, apply a unit moment at the desired point and direction; the corresponding virtual moment diagram replaces m. A unit force extracts translation, while a unit moment extracts rotation. For a cantilever of length L with a real tip load P, measured from the fixed end, M = -P(L-x) and the same-direction unit-load diagram is m = -(L-x). Then

δtip = ∫0^L P(L-x)²/(EI) dx = PL³/(3EI),

which agrees with the standard case. If axial deformation matters, a corresponding ∫Nn/(AE) dx term may be added; include only deformation modes required by the model.

Choose and verify

Use integration when the moment function and boundary conditions are straightforward, tables when the case matches exactly, superposition when several compatible linear cases combine, and virtual work when one response in a larger system is sought. Then perform four checks:

1. dimensions reduce to length for deflection and to a dimensionless value for slope;
2. supports and symmetry satisfy their geometric conditions;
3. curvature sense follows the bending-moment sign and the deflected shape is physically plausible;
4. load scaling is sensible—doubling a load doubles linear-elastic response, while doubling span can increase deflection by a cubic or fourth-power factor.

These checks distinguish a correct formula lookup from a correct structural solution.