Plastic collapse will take place whenever the strain is localized and not uniformly distributed. Considère’s criterion [5] is used and sketched in Figure 2. That is the reason why the stresses versus strain plots, in Figure 1, are limited to different maximum strains, for the different hardening exponent. From Figure 1 it is clear that plastic instability take place for large plastic strains (for the exponent being considered). If the elastic contribution is neglected, the maximum uniform strain is limited to 1/n.

For a given bending moment, M, the stresses in the longitudinal direction, s, distribute in the section replicating the stress versus strain plots (shown in Figure 1). Bending moment is

Or, because the identical material behaviour assumed for tension and compression

For a non-linear material, it is still valid to assume that

the originally flat cross-sections will remain flat after bending (Euler-Bernoulli’s assumption on plane-sections [6-8], and neglecting the effect of the shear stresses in wrapping the cross-sections [7]), then the longitudinal strain distributes linearly for the neutral plane to the top and bottom of the section [1].

e_max is the maximum strain produced at the cross section. In most cases, it will not be valid to consider that the neutral plane still passes through the cross-section centroid (an iterative procedure to locate its position will be required), but, if the cross section has a double symmetry along the horizontal and vertical axis, then the neutral plane will not change its position.

Recasting Equation (3) into strain terms

But in Equation (1) the strain is given as a function of the stress (and not vice versa), then Equation (5) can be expressed in stress terms as

After making the integral of Equation (6), the follow ing equation for M is obtained

Equation (7) allows computing de maximum stress in a section once the applied bending moment at this section is known. The applied bending moment depends on the geometry of the beam, loads… but not of the material that the beam is made of. This Equation (7) is not linear, so an iterative, numerical procedure should be used. If, for example, a Newton-Raphson is to be used, it is convenient to have an explicit expression for its derivative with respect to the maximum stress in the section [9, Press et al.]

where

and

Figure 3 shows a 3-dimensional plot of the evolution of the maximum bending moment (vertical axis) for a square cross section (12.5 × 12.5 mm) in steel (E = 210 GPa, n = 10) with a yield strength (e_p = 0.2%) of 600 MPa. Square symbols are plotted at increments of the maximum stress of 10 MPa until fulfilling Considère’s condition (plastic instability, 890 MPa).

For a horizontal cantilever beam of length L, that support a vertical load at its free end F, the applied bending moment varies along its length as

x represents the position along its length. For example, the maximum moment occurs at its fixed end, x = 0, M_max = FL. For a particular beam of length L = 1 m and an applied load F = 10 kN, the maximum moment is 10 kN×m. For a rectangular cross section of width b = 40 mm and depth h = 40 mm, it is possible to solve iteratively Equation (7) for the maximum stress at this section. The initial guess might be obtained neglecting the elastic contribution in Equation (7) (introducing E = ¥):

In the example case, it results in 656 MPa. Introducing this first guess into Equation (7) a bending moment is obtained (9.367 kN×m, smaller than the imposed one). Now the derivative in Equation (8) (dM/ds = 2.243 ×10^-5 m³) can be used to refine the solution (s _max = 685.7 MPa).

Introducing this maximum stress in the material constitutive Equation (1), the maximum strain at this section (at the top or its bottom is obtained) is obtained: e_max = 0.01086.

Temporarily we shall introduce the moment that will produce the same maximum strain for an elastic material of identical cross section is

We shall call it elastic-equivalent-bending moment. For the example, M_eqe = 24.3 kN×m.

Summarizing, at the fixed end of the beam the actual bending moment is 10 kN×m. This moment produces a maximum strain of 0.01086 at the top (or bottom) of this cross section. Identical maximum strain is produced for an elastic beam (of the same geometry and modulus of elasticity), but with a larger moment: M_eqe = 24.3 kN×m.

The bending moment, M, is a function of the position along the cantilever beam, x (see Equation (11)). So, the elastic-equivalent moment, M_eqe, can be derived for all along the cantilever beam sections. Figure 4 compares the actual and the elastic-equivalent moments for the proposed cantilever beam.

If the strains all along the top and bottom of both beams: the one built of a Ramberg-Osgood material and the elastic one, are identical, so are displacements, angles, maximum deflection… From this point all the computations can be carried out for the elastic beam under the M_eqe distribution. For example, the slope (derivative of the vertical displacements, v, with respect to the longitudinal position, x)

For the example, at the free end it results in an angle of 0.149 rad.

The beam vertical deflections

Any numeric integration procedure can be used to compute Equation (15). In the proposed example, the maximum deflection (at the loaded free end, x = L) is v_max = 108 mm. From Equations (14,15) it is clear that the equivalent elastic moment was a convenient conceptual scaffold, but are not actually required; only the maximum strains are needed.

A similar analysis can be carried out for a circular solid cross-section. The bending moment is

Recasting the vertical position into strains, that dis-

tributes linearly from top to bottom of the section (Euler-Bernoulli’s assumption on plane-sections),

But Ramberg-Osgood equation provides the strain as a function of stress, so, in stress terms (see Equation (18)).

That should be solved numerically. Figure 5 shows the evolution of the maximum strain and maximum stress at the solid circular section and the resulting bending moment. For comparison with the previous square section, the diameter is chosen to obtain the same area in the cross section as used in Figure 3 for the square cross section. If both figures are compared (3 and 5, for the square and circular cross sections), the square section supports slightly larger bending moment than the circular one (413 vs. 389 N×m), for a given area and the specified hardening material exponent (n = 10).

If a rod is used as a cantilever beam, as explained for the rectangular beam, the applied bending moment varies along its length x, see Equation (11). To compute the maximum stress in the different positions along its length, a first numeric estimation can be obtained neglecting the elastic contribution [1].

For the proposed example, F = 10 kN, L = 1 m, the maximum moment is 10 kN×m, at the fixed cantilever end. If the same cross section, used for the square beam, is distributed in a solid circular geometry, it result in a radius, R = 22.57 mm. Using Equation (19) the first guess for the maximum stress is s_max = 694.7 MPa. Using Equation (1) the maximum strain is e_max = 0.01197. If these estimations are introduced in the Equation (18) for the moment produced by the longitudinal stresses within the section is 9.469 kN×m, smaller than the applied one (10 kN×m). Any numerical procedure should be used to solve Equation (18). For example, a Newton-Raphson’s; the derivative of the bending moment with respect to the maximum stress is numerically estimated in 2.072 × 10^-5 m³. Per forming iterations it results in a maximum stress s_max = 721.5 MPa and the corresponding maximum strain is e_max= 0.01607.

In the same way, Equation (18) can be solved for the maximum strains along the different cross sections of the beam, as shown in Figure 6.

The beam slopes along its length is computed from

And deflections

Figure 7 represents the beam vertical deflections for the proposed cantilever beam. Note that if the load is positive (upwards), so are the deflections. (Most frequently loads are downwards because of gravity.)

Figure 7 also represents one half of a three point

bending test-piece loaded at its centre with a double load, 20 kN, of identical circular rod (45.14 mm f). These deflections correspond to the displacement observed at the mid-plane and they do not account for probable roll indentations. Note that these indentations are not elastic (but in the case of n = 1) and Hertz’s equations [10,11] are strictly not applicable at the contacts with the rolls.