Combined loading

Real members rarely carry just one kind of internal action. A column supporting a beam on a bracket carries axial load and bending. A sign post in the wind carries bending, shear and torsion all at once. A retaining wall carries its own weight plus the bending caused by soil pressure. A crane hook, a pre-stressed concrete girder and a footing under an off-centre column all combine axial force with bending moment. This lesson shows how to put the results of the previous lessons together to find the stress at any point in such members.

The principle of superposition

The tool that makes combined loading manageable is superposition: the stress caused by several loads acting together equals the sum of the stresses caused by each load acting alone. Superposition is valid when two conditions hold.

• The material is linear elastic, so stress is proportional to strain.
• Deformations are small, so the geometry, and therefore the internal forces, are not changed appreciably by the loads.

The second condition fails for slender columns, where a small lateral deflection multiplies the bending effect of an axial load. That is the subject of lesson 12. For stocky members and ordinary beams, superposition is accurate and is used throughout structural design.

The procedure is always the same. Find the internal actions at the section of interest (normal force, shear forces, bending moments and torque). Compute the stress each produces at the point of interest, using the formulas from lessons 1 to 6. Add stresses of the same kind acting on the same plane: normal stresses with normal stresses, shear stresses with shear stresses. The result is the state of stress at the point, which lesson 8 then examines on other planes.

Axial force plus bending

An axial force \(P\) and a bending moment \(M\) both produce normal stresses on the cross-section, so they add directly:

With tension positive and \(y\) measured upward from the centroid, a positive moment adds compression at the top and tension at the bottom. The combined distribution is still linear, but the neutral axis, where \(\sigma = 0\), moves away from the centroid. With enough axial compression, the whole section may stay in compression; with enough moment, part of it goes into tension.

This matters a great deal for materials that are weak in tension. Unreinforced masonry walls, stone arches, mass concrete dams and the soil under footings cannot sustain tension, so a designer must keep the resultant force within limits that keep the whole section in compression. Prestressed concrete is designed the other way round: a large compressive force is deliberately applied below the centroid, so that the bending under service loads only reduces, and never reverses, the compression in the bottom of the girder.

Eccentric axial loads

A common source of combined stress is an axial load that does not act through the centroid. A load \(P\) at an eccentricity \(e\) from the centroid is statically equivalent to the same load at the centroid plus a moment \(M = P\,e\). The stress at a fibre a distance \(y\) from the centroid, on the same side as the load, is then

The kern and the middle-third rule

How far can the load move before tension appears? Setting the stress on the far side to zero gives the limiting eccentricity \(e = I/(A\,c)\). For a rectangle of depth \(h\), that is

So a compressive load on a rectangular section causes no tension anywhere if it stays within the middle third of the depth. In two directions, the zone within which the load can act without causing tension is a diamond-shaped region called the kern (or core) of the section. For a solid circle, the kern is a circle of radius \(d/8\).

The middle-third rule is old but still central to the design of footings, gravity retaining walls, masonry piers and arches. For a spread footing, the soil pressure distribution under an eccentric column load follows the same linear formula, and if the resultant falls outside the middle third, part of the footing lifts off the soil. The pressure distribution then becomes triangular over a reduced contact length, and the peak pressure rises sharply.

Biaxial bending

When a member bends about both principal axes at once, as a corner column or a purlin on a sloping roof does, the two bending stresses add to the axial stress:

For a rectangular section the maximum stress occurs at a corner, where both bending terms have their largest magnitude with the same sign. The signs depend on the directions of the moments; drawing a sketch of which corner is pushed hardest is the most reliable check.

Bending, shear and torsion together

Some members carry shear stress from torsion or transverse shear at the same time as normal stress from bending and axial force. A circular sign post or a light standard in the wind is the classic example. The wind on the sign, which is offset from the post, produces a horizontal force (shear), a bending moment that grows toward the base, and a torque about the post's axis.

At the base of such a post, the critical point is usually on the surface where the bending stress is greatest. There the normal stress is \(\sigma = Mc/I\) (plus any axial stress from self-weight), and the shear stress from torsion is \(\tau = Tc/J\). The transverse shear stress \(VQ/(It)\) is zero at that point, because it is an extreme fibre for bending, but it is greatest at the points on the neutral axis, where it adds to the torsional shear. Both locations must be checked. The combination of a normal stress and a shear stress at the same point is not added into one number; it is a state of plane stress, and lesson 8 shows how to find the principal stresses and maximum shear it produces.

Stresses at the corners by code

For a rectangular section under axial load and biaxial bending, the four corner stresses tell you the extremes. The program below computes them for several load cases and reports whether any corner goes into tension, which is the check needed for a masonry pier or a footing.

# Corner stresses in a rectangular section under P, Mx, My (N, mm); compression negative
b, h = 400.0, 600.0                  # width (z-direction) and depth (y-direction)
A = b * h
Iz = b * h**3 / 12                   # bending about z uses y = +-h/2
Iy = h * b**3 / 12                   # bending about y uses z = +-b/2

def corners(P, Mz, My):
    out = {}
    for y in (h / 2, -h / 2):
        for z in (b / 2, -b / 2):
            out[(int(y), int(z))] = P / A - Mz * y / Iz + My * z / Iy
    return out

cases = {"axial only": (-1500e3, 0, 0),
         "plus Mz = 100 kN*m": (-1500e3, 100e6, 0),
         "plus My = 60 kN*m": (-1500e3, 100e6, 60e6),
         "large moments": (-1500e3, 160e6, 80e6)}
for name, (P, Mz, My) in cases.items():
    s = corners(P, Mz, My)
    vals = ", ".join(f"{v:6.2f}" for v in s.values())
    flag = "TENSION at a corner" if max(s.values()) > 0 else "all compression"
    print(f"{name:22s}: {vals} MPa  -> {flag}")

axial only            :  -6.25,  -6.25,  -6.25,  -6.25 MPa  -> all compression
plus Mz = 100 kN*m    : -10.42, -10.42,  -2.08,  -2.08 MPa  -> all compression
plus My = 60 kN*m     :  -6.67, -14.17,   1.67,  -5.83 MPa  -> TENSION at a corner
large moments         :  -7.92, -17.92,   5.42,  -4.58 MPa  -> TENSION at a corner

The moment about one axis alone leaves the section in compression, but adding a moderate moment about the other axis puts a corner into tension. A quick check that matches the code is the biaxial kern condition for a rectangle: no tension occurs if \(e_y/h + e_z/b \le 1/6\), where \(e_y = M_z/P\) and \(e_z = M_y/P\). For the second case, \(0.0667/0.6 = 0.111 \le 0.167\), so no tension. For the third case, \(0.0667/0.6 + 0.040/0.4 = 0.211 > 0.167\), which confirms the tension the code found.