Axial Load

Procedure for Analysis — Axial Displacement

1. Internal forces. Use the method of sections to find \(N\) in each region between concentrated loads or geometry changes (a normal-force diagram helps); for variable loading or area, express \(N(x)\) and \(A(x)\) as functions of position.

2. Displacement. Apply \(\delta = \Sigma NL/AE\) or the integral form, keeping a consistent sign convention. A positive result means elongation (or motion in the assumed positive direction).

Statically Indeterminate

A problem in which the equilibrium equations are not sufficient to determine all the reactions. A bar fixed at both ends and loaded between them has two unknown reactions but only one useful equilibrium equation. The extra equation comes from the geometry of deformation: a compatibility (kinematic) condition — e.g. for two fixed ends, the relative displacement between them must be zero, \(\delta_{A/B} = 0\). The compatibility condition is written in terms of forces through the load–displacement relation \(\delta = NL/AE\).

Procedure for Analysis — Indeterminate Members

1. Equilibrium. Draw the free-body diagram and write the equilibrium equations in terms of the unknown reactions.

2. Compatibility. Express how the displacements of points on the member are constrained by the supports (draw a displacement diagram if helpful).

3. Load–displacement. Substitute \(\delta = NL/AE\) for each segment and solve the equilibrium and compatibility equations simultaneously. A negative result simply reverses the assumed direction of that force.

Force (Flexibility) Method

An alternative way to write the compatibility equation using superposition: choose one support as the redundant and temporarily remove it, making the member statically determinate. The applied load then displaces the freed point by \(\delta_P\); reapplying the unknown redundant reaction alone displaces it back by \(\delta_B\). Since the actual support permits no displacement,

This single equation yields the redundant force directly; the remaining reactions follow from equilibrium. For the classic fixed–fixed bar with a 500-N load applied 2 m from A (total length 5 m), it gives \(F_B = 500(2)/5 = 200\) N and \(F_A = 300\) N — identical to the direct compatibility solution.

Thermal Stress

A statically determinate member expands or contracts freely — no stress develops. In a statically indeterminate member, the supports constrain the thermal displacement, producing thermal stresses that can be large and must be considered in design (hence expansion joints in bridges and pipelines). The analysis combines equilibrium, compatibility, and the load–displacement relations exactly as in Section 4.4, with \(\delta_T\) added to the displacement terms.

Stress Concentration

Complex, non-uniform stress distributions arise not only at points of load application, but at any section where the cross-sectional area changes abruptly — holes, notches, shoulder fillets. The maximum normal stress occurs at the smallest cross section, and equilibrium still requires the stress distribution there to integrate to the internal force: \(N = \int_A \sigma\, dA\) (the volume under the stress diagram, acting through its centroid).

When Stress Concentrations Matter

For brittle materials, or members under fatigue (cyclic) loading, the localized peak stress can initiate a crack and must be checked against the allowable stress. For ductile materials under static load, local yielding at the peak redistributes the stress over the section, so the concentration is usually not critical — one reason ductile materials are preferred. Designers reduce K with generous fillets and relief features (e.g. drilled holes at the ends of grooves).

Elastic Perfectly Plastic Behavior

Members are sometimes designed to yield intentionally (energy absorption) or may be accidentally overloaded. The simplest model is the elastic perfectly plastic (elastoplastic) material: linear elastic up to \(\sigma_Y\), then continued straining at constant stress \(\sigma_Y\) with no strain hardening.

Plastic Load

At a stress concentration, yielding begins when \(\sigma_{max} = K\,N/A\) first reaches \(\sigma_Y\) — i.e. at \(N_Y = \sigma_Y A / K\). Further load does not raise the peak stress; instead the yielded zone spreads across the section while the stress there stays at \(\sigma_Y\). When the entire section has yielded, the stress distribution is uniform at \(\sigma_Y\) and the member carries its largest possible axial force, the plastic load \(N_p = \sigma_Y A\). In indeterminate systems, members can yield one at a time, redistributing load to the still-elastic members until the system reaches its collapse condition.

Residual Stress

If a statically indeterminate member (or system) is loaded into the plastic range and then unloaded, it will not return to a stress-free state: self-equilibrating residual stresses remain locked in. The reason is that plastic strains acquired during loading are permanent, yet the supports still force the displacements to be compatible.

Analysis by superposition: unloading is elastic, so the residual state equals the inelastic stress distribution at full load plus the fully elastic distribution computed for the same load applied in reverse. Members or regions that yielded in tension typically end up in residual compression, and vice versa.