Von Mises Truss

Figure 1: Von Mises Truss
Figure 1: Von Mises Truss
Figure 2: Animation showing snap-through behaviour as load increases
Figure 2: Animation showing snap-through behaviour as load increases

The Von Mises Truss is an example of a structure that is susceptible to snap-through buckling, as such is not correctly modelled by a linear eigenvalue buckling analysis, and we must instead use the more general non-linear eigenvalue buckling analysis (what is the difference?).

In this example we’ll model the truss in Figure 1 with the dimensions \(H = 1.0\textrm{ m}\) and \(L = 5.0\textrm{ m,}\) and an axial stiffness \(EA = 1280\textrm{ MN}\). In the table below, we compare the predicted buckling load from both the linear and non-linear eigenvalue analyses:

Method Critical Load
Linear 19111 kN
Non-linear 3790 kN

We can see from the table that if we’d naively used the linear buckling analysis we’d’ve overestimated the capacity by a factor of 5!

The critical load is found by running a displacement controlled non-linear analysis and looking for the peak load before the snap-through (see Figure 3). The dashed line shows where the snap-through occurs.

Figure 3: Load-displacement curve
Figure 3: Load-displacement curve

We can then investigate the non-linear eigenvalues around the critical load to find the point they go to zero (see Figure 4).

Figure 4: Eigenvalues against applied load
Figure 4: Eigenvalues against applied load

We can compare these numerical results to the theoretical results described by Bazant in Stability of Structures (Equation 4.4.5).

Figure 5: Von Mises Truss at critical deflection
Figure 5: Von Mises Truss at critical deflection

Bazant showed that the critical rotation can be described by the equation:

\[ \cos q_{0} = (\cos \alpha)^{\frac{1}{3}} \]

where \(q_{0}\) and \(\alpha\) are defined on Figures 1 and 5. This can be transformed in terms of the critical deflection (see derivation):

\[ \delta_{crit} = H - L \sqrt{ \Biggl( \frac{H^{2}}{L^{2}} + 1 \Biggr)^{\frac{1}{3}} - 1} \]

where \(\delta_{crit}\), \(H\) and \(L\) are as defined on Figures 1 and 5. Plugging the numbers into this equation for our model, \(H = 1.0\textrm{ m}\) and \(L = 5.0\textrm{ m}\), we find the predicted critical displacement to be \(\delta_{crit} = 426.4\textrm{ mm}\), which matches the peak of our load-displacement curve in Figure 3.