# Von Mises Truss

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:

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.

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

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

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.