The midpoint deflection

PHYSICS · Multi-part · Fermi University Physics · Computational Physics · Research software practice, verification and validation · Research software practice, verification and validation

Scientific Computing for Physicists, Ch Are they the right equations? Solution verification, benchmarks and validation

Problem

The midpoint deflection w(0.5)w(0.5), in centimetres, of a clamped-clamped elastic beam under a uniform load is found by discretising the bending-energy functional on a finite-difference mesh and minimising it with Newton's method, stopped when the gradient norm satisfies ∣∇E∣=1.0×10−8|\nabla E| = 1.0\times10^{-8} (energy units per cm). At the minimum found, the smallest eigenvalue of the energy Hessian is λmin⁡=15.3\lambda_{\min} = 15.3 (energy units per cm2^2). The minimisation is repeated, to the same stopping criterion, on meshes with N=25N = 25, 5050 and 100100 interior nodes (r=2r=2 between successive meshes), giving converged deflections wN=25=0.0234512 cm,wN=50=0.0235102 cm,wN=100=0.0235247 cm.w_{N=25} = 0.0234512\ \text{cm}, \quad w_{N=50} = 0.0235102\ \text{cm}, \quad w_{N=100} = 0.0235247\ \text{cm}. ((a)) Find the observed order p^\hat p, the extrapolated deflection wextw_{\rm ext} and the grid convergence index at N=100N=100 (Fs=1.25F_s = 1.25). ((b)) Near the minimum, where the energy is quadratic, the distance between the nodal displacements found and the exact minimiser is bounded by ∣∇E∣/λmin⁡|\nabla E|/\lambda_{\min}, in centimetres. Evaluate this bound at the stated stopping criterion and again at a hundredfold looser one (∣∇E∣=1.0×10−6|\nabla E| = 1.0\times10^{-6}), compare both with the absolute discretisation uncertainty from (a), and decide whether a benchmark comparison against a published analytic beam solution at an absolute tolerance of 1×10−4 cm1\times10^{-4}\ \text{cm} could detect that the stopping criterion had been loosened a hundredfold.

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