Axial Tension, Compression, and Deformation

Axial Tension, Compression, and Deformation

For July 2026 preparation, use the axial mechanics relationships in the current NCEES PE Civil Reference Handbook supplied for the test date. This section treats basic linear-elastic response; code member strength, connection design, and compression buckling are separate checks. An axial stress below yield does not prove that a slender compression member is stable.

Equilibrium first, constitutive law second

Cut the member perpendicular to its longitudinal axis and use equilibrium to find internal axial force N. This section takes tension as positive and compression as negative. The average normal stress in a concentric prismatic segment is

σ = N/A.

For linear-elastic material, σ = Eε, and axial strain is ε = δ/L. Combining them gives

δ = NL/(AE).

With the stated sign convention, tensile N produces positive elongation and compressive N produces negative shortening. If a concentrated axial load occurs between two cuts, the axial-force diagram jumps by that load. If distributed axial load acts along the member, N varies with position. Never use the end load automatically for every segment; isolate each region.

The formula assumes the force is concentric, stress is represented adequately by its average away from local disturbances, material is linearly elastic, and deformation is small. Eccentric loading adds bending and belongs in combined-stress analysis. Near holes, connections, or abrupt changes, local stress concentration may make N/A insufficient for detailed design even though it remains useful for basic equilibrium.

Unit discipline

When N is in kips, A is in square inches, and E is in ksi (kip/in²), length must be in inches and the calculated deformation is in inches. Mixing a length in feet with E in ksi makes the answer wrong by a factor of 12. Stress has force-per-area units; strain is dimensionless; deformation has length units. In SI calculations, using newtons, square millimeters, megapascals (N/mm²), and millimeters is similarly self-consistent. Convert temperature units consistently with α; a coefficient per degree Fahrenheit cannot multiply a Celsius temperature change without conversion.

Stepped and nonuniform members

Divide a stepped member wherever N, A, E, or thermal condition changes. For discrete prismatic segments in series,

δtotal = Σ Ni Li/(Ai Ei).

Retain the sign of every Ni. For continuously varying properties, the corresponding form is δ = ∫ N(x)/[A(x)E(x)] dx. Members in series carry forces established by joint equilibrium; members in parallel generally share force according to stiffness and must satisfy compatible end deformation. Do not assume equal force or equal stress without checking the actual arrangement.

Worked stepped tension member

A two-segment steel bar carries a concentric 60-kip tensile force. Segment 1 has L1 = 8 ft = 96 in and A1 = 2.0 in². Segment 2 has L2 = 6 ft = 72 in and A2 = 1.5 in². Take E = 29,000 ksi. Because there is no axial load at the step, equilibrium gives N1 = N2 = +60 kip.

The stresses are

σ1 = 60/2.0 = 30 ksi, and σ2 = 60/1.5 = 40 ksi.

The segment elongations are

δ1 = 60(96)/[2.0(29,000)] = 0.0993 in,

δ2 = 60(72)/[1.5(29,000)] = 0.0993 in.

Thus δtotal = 0.1986 in. The equal elongations are coincidental: segment 2 is shorter but also smaller. If the 60-kip force were compressive and buckling were set aside, the stresses and deformations would have the same magnitudes with negative signs. A compatibility question might prescribe total movement and require solving for the unknown force instead.

Temperature: free movement versus restraint

For a uniform temperature change ΔT, free thermal strain is

εT = αΔT,

and free deformation is δT = αΔT L, where α is the coefficient of thermal expansion. Heating gives positive expansion when α is positive. A freely moving member develops this deformation with no thermal stress. Temperature alone does not imply force.

Restraint changes the problem. Mechanical and thermal strains superpose in the linear range:

εtotal = σ/E + αΔT.

If rigid supports fully prevent the end-to-end movement of a uniformly heated bar, εtotal = 0, so

σ = -EαΔT.

The negative sign means compression. Cooling under full restraint produces tension. The corresponding force is N = σA.

Worked fully restrained bar

A 20-ft steel bar has A = 4.0 in², E = 29,000 ksi, α = 6.5 × 10^-6/°F, and a uniform temperature rise of 80°F. If free, it would expand

δT = (6.5 × 10^-6)(80)(240) = 0.1248 in.

With ideal rigid supports and full restraint,

σ = -29,000(6.5 × 10^-6)(80) = -15.08 ksi,

and N = -15.08(4.0) = -60.3 kip. The stress expression does not contain length because both free expansion and mechanical shortening scale with length. Length still controls the prevented displacement.

Real supports may deform, a gap may close before restraint begins, or connected members may have different temperatures. Then write one compatibility equation rather than using -EαΔT blindly:

Σ[NiLi/(AiEi)] + Σ(αiΔTiLi) = prescribed support movement.

Include signs for support settlement, fabrication error, or gaps. Solve this equation together with equilibrium. A temperature gradient through depth can also cause curvature rather than pure uniform axial deformation; do not reduce that case to a single uniform ΔT without justification.

Final checks

Sketch the axial-force diagram, confirm each jump by joint equilibrium, convert all lengths before substituting, and estimate whether the deformation sign is physically sensible. For temperature, first ask whether movement is free, partial, or fully restrained. Finally separate response from capacity: calculated N, σ, and δ are demands, while yielding, buckling, fracture, and code resistance require their own checks.

Axial Audit Checklist

1. Establish the signed force and actual load path at every segment.
2. Match each segment with its material, area, modulus, length, and restraint.
3. Check both stress or capacity and compatible deformation before closing equilibrium.