Combined Axial and Flexural Stress

Combined Axial and Flexural Stress

For the July 2026 PE Civil: Structural exam, use the current NCEES PE Civil Reference Handbook, AISC Steel Construction Manual, 15th Edition (2017), and ACI 318-14 when those materials govern. The handbook supplies mechanics relationships; the material standard supplies the applicable strength interaction. Those are related calculations, not interchangeable ones.

Build the elastic stress state

First find the section resultants from a free-body diagram. At a prismatic section under concentric axial force N and bending about one centroidal principal axis, linear-elastic normal stress is

σ = N/A ± Mc/I = N/A ± M/S.

This section takes tension as positive and compression as negative. The axial term is uniform. The bending term varies linearly with distance c from the neutral axis, is zero at that axis, and reaches its extreme magnitude at the farthest fiber. Do not add unsigned magnitudes automatically; identify the physical point and determine whether each contribution is tensile or compressive there.

For biaxial bending about centroidal principal axes, a common signed form is

σ(x,y) = N/A - Mx y/Ix + My x/Iy,

subject to the declared moment and coordinate convention. If axes are not principal, a product-of-inertia term is involved; silently applying the principal-axis expression is then invalid. Evaluate the actual candidate corners. One corner can receive compression from both moments while another receives opposing contributions.

The formula assumes a linear strain distribution, elastic material response, and a section capable of carrying the calculated tension and compression. It describes demand at a point. Local stress concentrations, residual stress, cracking, yielding, and stability can require a different or additional model.

Eccentric force becomes force plus moment

A force P whose line of action is offset by perpendicular eccentricity e from the centroid is statically equivalent at the centroid to the same force plus

M = Pe.

Keep units compatible: kips times inches gives kip-inches. An eccentricity in two directions creates moments about both axes. The offset's sign determines which edge becomes more compressed; drawing the load line is safer than memorizing a plus sign.

Worked eccentric compression example

A rectangular section is 12 in wide and 18 in deep. A 180-kip compressive force acts 3 in from the centroid along the 18-in depth direction. Assume elastic behavior and bending about the width-parallel centroidal axis.

A = 12(18) = 216 in²

I = bh³/12 = 12(18³)/12 = 5,832 in⁴

S = I/c = 5,832/9 = 648 in³

M = Pe = 180(3) = 540 kip-in.

The direct stress is -180/216 = -0.833 ksi. The extreme bending-stress magnitude is M/S = 540/648 = 0.833 ksi. At the near edge, bending adds compression, giving

σnear = -0.833 - 0.833 = -1.667 ksi.

At the far edge, bending opposes compression:

σfar = -0.833 + 0.833 = 0 ksi.

Force equilibrium is preserved because the bending-stress block has zero resultant, and moment equilibrium is preserved because it produces the 540 kip-in couple. Reporting only the maximum stress misses the equally important zero-stress edge.

Kern and no-tension behavior

The kern or core is the region through which a compressive resultant must pass for the linear-elastic stress distribution to remain compressive or zero everywhere. For uniaxial eccentricity about the depth of a rectangle, the kern limit is

|e| ≤ S/A = h/6.

Here h/6 = 18/6 = 3 in, so the worked load lies exactly at the kern boundary and the far edge reaches zero. If |e| < 3 in, both edges remain compressed. If |e| > 3 in, elastic full-section superposition predicts tension at one edge.

That prediction does not mean every material actually carries that tension. Soil bearing interfaces, unbonded foundations, and some unreinforced materials may be modeled as no-tension contacts. Once separation occurs, the compression zone and neutral axis change; P/A ± M/S using the gross area is no longer the final contact distribution. Reinforced concrete can carry tension through reinforcement after concrete cracking, but ACI 318-14 section strength is evaluated by its code model, not by declaring the gross elastic tensile stress acceptable.

Elastic stress is not code interaction

A material-code interaction check compares required strengths with available strengths and incorporates behavior absent from simple elastic addition. For steel, the applicable AISC 15th Edition beam-column provisions use required axial and flexural strengths, available strengths for the chosen LRFD or ASD method, and the prescribed interaction equation. Stability, effective length or direct-analysis requirements, and second-order effects influence those quantities. A ratio such as Pr/Pc is not a stress at one corner.

For reinforced concrete, ACI 318-14 axial-flexural strength comes from strain compatibility, concrete compression behavior, reinforcement forces, nominal strength, and the applicable strength-reduction factor. A point on a Pn-Mn interaction diagram represents section capacity for a strain state; it is not produced by adding Pu/Ag to Mu/Sg.

Use one design framework consistently. Do not compare factored LRFD demand with ASD available strength or insert service-level elastic stress into a strength interaction equation. A good workflow is:

1. obtain N, Mx, and My from the correct load combination and analysis, including required second-order effects;
2. use elastic superposition when the question asks for stress, kern, or an elastic distribution;
3. use the exact AISC 15th or ACI 318-14 interaction procedure when the question asks for member or section strength;
4. confirm axes, signs, units, and the governing edge or interaction ratio.

This separation prevents an accurate mechanics calculation from becoming an invalid code check.