On numerical moment-curvature relationship of a beam

Abstract

In complex bending problems involving material and geometric non-linearity, quite often moment-curvature based approach is preferred over stress-strain based methods. For such an approach, available uniaxial stress-strain test data or models are required to be converted into moment-curvature relationship. The process of con-version of uniaxial stress-strain relationship into a moment-curvature relationship is non-unique. And hence, complete moment-curvature law can be modelled suiting any of the several hardening laws. Such modelling will be very important when abeam is under cyclic load producing reverse plastic deformation. In this paper, an approach is presented to obtain a unique moment-curvature relationship from any given stress-strain law. Standard elasto-plastic models viz.elastic-perfectly plastic, isotropic and kinematic hardening are considered to produce corresponding unique moment-curvature relationships. The results indicate that an isotropic curvature hardening model, corresponding to an elastic perfectly plastic stress-strain model, would be erroneous. Additionally, step by step procedure of using the approach in solving a large deflection elasto-plastic beam problem, is demonstrated here.

Appendix I. Derivation of moment-curvature relationship for a circular cross-section following elastic-perfectly plastic law

In figure 8, the cross-section of area A is assumed to be elastic in the range $-a\le Y \le a$ and elasto-plastic outside this range. The total moment M is given by:

$$\begin{aligned} {M}=\int _{A} {\sigma }\,\,Y\, \,dA \end{aligned}$$

(A1)

Figure 8

The stress and strain state of a typical solid circular cross-section following elastic-perfectly plastic law.

The above equation corresponding to a circular cross-section reads:

$$\begin{aligned} {M}=4\int _{0}^{\frac{h}{2}} {\sigma }\,\,Y\sqrt{\frac{h^2}{4}-Y^2}\, \,dY \end{aligned}$$

(A2)

Using $\sigma =E\epsilon$ and $\epsilon =\kappa Y$ at $Y=a$, a is determined to be:

$$\begin{aligned} a=\frac{\sigma _0}{E\kappa } \end{aligned}$$

(A3)

To perform the integration of Eq. A2, it is divided into elasic and elasto-plastic regions as given by:

$$\begin{aligned} {M}=4E\kappa \int _{0}^{a}Y^2\sqrt{\frac{h^2}{4}-Y^2}\, \,dY+4\sigma _0\int _{a}^{\frac{h}{2}} \,\,Y\sqrt{\frac{h^2}{4}-Y^2}\, \,dY \end{aligned}$$

(A4)

Using the following substitution: $Y=\frac{h}{2}\sin {\theta }$, Eq.A3 and pertinent normalization, the above equation simplifies into Eq. 15.