2. Benchmark solution

2.1. Calculation method used for the reference solution

The solution is analytical.

Thermal loading is equivalent to a loading defined by a uniform distribution of moments at the edges as shown in the figure.

The value of these moments per unit length is equal to:

.

Either:

. This leads to an even distribution of \(M\) in the plate.

2.2. Benchmark results

So we have \(M=2380.95238N\); the plate being rotated by an angle \(\theta \mathrm{=}53°.1301\), we have components whose absolute value is: \(M\mathrm{\times }\mathrm{cos}\theta \mathrm{=}1428.5715N\) and \(M\mathrm{\times }\mathrm{sin}\theta \mathrm{=}1904.76184N\)

The reactions are defined by a distribution of moments equal to the previous one in absolute value and of opposite sign.

Meshes are squares whose length is equal to \(0.05m\), so the moments in each node must be equal to \({M}_{1}\mathrm{=}M\mathrm{\times }\mathrm{cos}\theta \mathrm{\times }0.05\mathrm{=}71.42857\mathit{N.m}\)

and \({M}_{2}\mathrm{=}M\mathrm{\times }\mathrm{sin}\theta \mathrm{\times }0.05\mathrm{=}95.2381\mathit{N.m}\)

either

2.3. Uncertainty about the solution

Uncertainty is zero.

2.4. Bibliographical references

1. TIMOSHENKO: Theory of plates and shells chapter 2, article 14.