Self-Adaptive Digital Volume Correlation for Unknown Deformation Fields

Abstract

Digital volume correlation (DVC) has evolved into a powerful tool for quantifying full-field internal deformation. In existing subvolume-based local DVC, subvolume size and shape function are two key user-defined parameters closely related to the DVC measurement errors. In routine implementation, the user must define fixed subvolume size and shape function according to prior experience and intuition, which cannot ensure accurate measurements, particularly for unknown complex heterogeneous deformation fields. Self-adaptive selection of optimal subvolume size and the best shape function is therefore highly desirable to realize full-automatic and quality DVC measurements. In this work, we first establish theoretical error models that relate total displacement errors to subvolume sizes and shape functions. By minimizing the V-shaped models of theoretically predicted total errors, optimal subvolume size and the best shape function can be identified as inputs for self-adaptive DVC analysis at each calculation point. The accuracy advantage of the presented self-adaptive DVC approach over classic one using fixed subvolume size and shape function is demonstrated through numerically simulated three-point bending tests.

Appendix: Undermatched Bias Error Model in DVC Using 2nd-Order Shape Function

According to the objective function given in Eq. (4), we first define the following parameters to facilitate subsequent derivation

$$ \Big\{{\displaystyle \begin{array}{c}A=\sum \limits_{i,j,k=-M}^M1={\left(2M+1\right)}^3\\ {}B=\sum \limits_{i,j,k=-M}^M{i}^2=\frac{1}{3}M\left(M+1\right){\left(2M+1\right)}^3\\ {}C=\sum \limits_{i,j,k=-M}^M{i}^2{j}^2=\frac{1}{9}{M}^2{\left(M+1\right)}^2{\left(2M+1\right)}^3\\ {}D=\sum \limits_{i,j,k=-M}^M{i}^4=\frac{1}{15}M\left(M+1\right){\left(2M+1\right)}^3\left(3{M}^2+3M-1\right)\end{array}}\operatorname{} $$

(10)

The derivative of Eq. (4) with respect to x-directional 10 deformation parameters can be written as

$$ \left[\begin{array}{llllllllll}A& 0& 0& 0& \frac{1}{2}B& \frac{1}{2}B& \frac{1}{2}B& 0& 0& 0\\ {}0& B& 0& 0& 0& 0& 0& 0& 0& 0\\ {}0& 0& B& 0& 0& 0& 0& 0& 0& 0\\ {}0& 0& 0& B& 0& 0& 0& 0& 0& 0\\ {}\frac{1}{2}B& 0& 0& 0& \frac{1}{4}D& \frac{1}{4}C& \frac{1}{4}C& 0& 0& 0\\ {}\frac{1}{2}B& 0& 0& 0& \frac{1}{4}C& \frac{1}{4}D& \frac{1}{4}C& 0& 0& 0\\ {}\frac{1}{2}B& 0& 0& 0& \frac{1}{4}C& \frac{1}{4}C& \frac{1}{4}D& 0& 0& 0\\ {}0& 0& 0& 0& 0& 0& 0& C& 0& 0\\ {}0& 0& 0& 0& 0& 0& 0& 0& C& 0\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0& C\end{array}\right]\cdot \left[\begin{array}{l}{u}_2\left(x,y,z\right)\\ {}{u}_{x2}\left(x,y,z\right)\\ {}{u}_{y2}\left(x,y,z\right)\\ {}{u}_{z2}\left(x,y,z\right)\\ {}{u}_{xx2}\left(x,y,z\right)\\ {}{u}_{yy2}\left(x,y,z\right)\\ {}{u}_{zz2}\left(x,y,z\right)\\ {}{u}_{xy2}\left(x,y,z\right)\\ {}{u}_{xz2}\left(x,y,z\right)\\ {}{u}_{yz2}\left(x,y,z\right)\end{array}\right]=\left[\begin{array}{l}\sum u\left(x+i,y+j,z+k\right)\\ {}\sum iu\left(x+i,y+j,z+k\right)\\ {}\sum ju\left(x+i,y+j,z+k\right)\\ {}\sum ku\left(x+i,y+j,z+k\right)\\ {}\sum \frac{1}{2}{i}^2u\left(x+i,y+j,z+k\right)\\ {}\sum \frac{1}{2}{j}^2u\left(x+i,y+j,z+k\right)\\ {}\sum \frac{1}{2}{k}^2u\left(x+i,y+j,z+k\right)\\ {}\sum iju\left(x+i,y+j,z+k\right)\\ {}\sum iku\left(x+i,y+j,z+k\right)\\ {}\sum jku\left(x+i,y+j,z+k\right)\end{array}\right] $$

(11)

Simplifying above equation yields the following expression

$$ \left[\begin{array}{l}{u}_2\left(x,y,z\right)\\ {}{u}_{xx2}\left(x,y,z\right)\\ {}{u}_{yy2}\left(x,y,z\right)\\ {}{u}_{zz2}\left(x,y,z\right)\end{array}\right]={\mathbf{H}}^{-1}\cdot \left[\begin{array}{l}{P}_1\\ {}\frac{1}{2}{P}_2\\ {}\frac{1}{2}{P}_3\\ {}\frac{1}{2}{P}_4\end{array}\right]={\left[\begin{array}{llll}A& \frac{1}{2}B& \frac{1}{2}B& \frac{1}{2}B\\ {}\frac{1}{2}B& \frac{1}{4}D& \frac{1}{4}C& \frac{1}{4}C\\ {}\frac{1}{2}B& \frac{1}{4}C& \frac{1}{4}D& \frac{1}{4}C\\ {}\frac{1}{2}B& \frac{1}{4}C& \frac{1}{4}C& \frac{1}{4}D\end{array}\right]}^{-1}\cdot \left[\begin{array}{l}\sum \limits_{i,j,k=-M}^Mu\left(x+i,y+j,z+k\right)\\ {}\sum \limits_{i,j,k=-M}^M\frac{1}{2}{i}^2u\left(x+i,y+j,z+k\right)\\ {}\sum \limits_{i,j,k=-M}^M\frac{1}{2}{j}^2u\left(x+i,y+j,z+k\right)\\ {}\sum \limits_{i,j,k=-M}^M\frac{1}{2}{k}^2u\left(x+i,y+j,z+k\right)\end{array}\right] $$

(12)

One can derive the following formula by solving the above equation

$$ {u}_2\left(x,y,z\right)=\frac{1}{A\left(2C+D\right)-3{B}^2}\left[\left(2C+D\right){P}_1-B\left({P}_2+{P}_3+{P}_4\right)\right] $$

(13)

Based on the forth-order Taylor expansion of u (x + i, y + j, z + k), we can obtain

$$ \Big\{\begin{array}{l}{P}_1\approx \left[A+\frac{1}{2}\left(B\frac{\partial^2}{\partial {x}^2}+B\frac{\partial^2}{\partial {y}^2}+B\frac{\partial^2}{\partial {z}^2}\right)+\frac{1}{4}\left(C\frac{\partial^4}{\partial {x}^2\partial {y}^2}+C\frac{\partial^4}{\partial {x}^2\partial {z}^2}+C\frac{\partial^4}{\partial {y}^2\partial {z}^2}\right)+\frac{1}{24}\left(D\frac{\partial^4}{\partial {x}^4}+D\frac{\partial^4}{\partial {y}^4}+D\frac{\partial^4}{\partial {z}^4}\right)\right]u\left(x,y,z\right)\\ {}{P}_2\approx \left[B+\frac{1}{2}\left(D\frac{\partial^2}{\partial {x}^2}+C\frac{\partial^2}{\partial {y}^2}+C\frac{\partial^2}{\partial {z}^2}\right)+\frac{1}{4}\left(F\frac{\partial^4}{\partial {x}^2\partial {y}^2}+F\frac{\partial^4}{\partial {x}^2\partial {z}^2}+E\frac{\partial^4}{\partial {y}^2\partial {z}^2}\right)+\frac{1}{24}\left(G\frac{\partial^4}{\partial {x}^4}+F\frac{\partial^4}{\partial {y}^4}+F\frac{\partial^4}{\partial {z}^4}\right)\right]u\left(x,y,z\right)\\ {}{P}_3\approx \left[B+\frac{1}{2}\left(C\frac{\partial^2}{\partial {x}^2}+D\frac{\partial^2}{\partial {y}^2}+C\frac{\partial^2}{\partial {z}^2}\right)+\frac{1}{4}\left(F\frac{\partial^4}{\partial {x}^2\partial {y}^2}+E\frac{\partial^4}{\partial {x}^2\partial {z}^2}+F\frac{\partial^4}{\partial {y}^2\partial {z}^2}\right)+\frac{1}{24}\left(F\frac{\partial^4}{\partial {x}^4}+G\frac{\partial^4}{\partial {y}^4}+F\frac{\partial^4}{\partial {z}^4}\right)\right]u\left(x,y,z\right)\\ {}{P}_4\approx \left[B+\frac{1}{2}\left(C\frac{\partial^2}{\partial {x}^2}+C\frac{\partial^2}{\partial {y}^2}+D\frac{\partial^2}{\partial {z}^2}\right)+\frac{1}{4}\left(E\frac{\partial^4}{\partial {x}^2\partial {y}^2}+F\frac{\partial^4}{\partial {x}^2\partial {z}^2}+F\frac{\partial^4}{\partial {y}^2\partial {z}^2}\right)+\frac{1}{24}\left(F\frac{\partial^4}{\partial {x}^4}+F\frac{\partial^4}{\partial {y}^4}+G\frac{\partial^4}{\partial {z}^4}\right)\right]u\left(x,y,z\right)\end{array}\operatorname{} $$

(14)

where parameters E, F and G are defined as

$$ \Big\{\begin{array}{l}E=\sum \limits_{i,j,k=-M}^M{i}^2{j}^2{k}^2=\frac{1}{27}{M}^3{\left(M+1\right)}^3{\left(2M+1\right)}^3\\ {}F=\sum \limits_{i,j,k=-M}^M{i}^4{j}^2=\frac{1}{45}{M}^2{\left(M+1\right)}^2{\left(2M+1\right)}^3\left(3{M}^2+3M-1\right)\\ {}G=\sum \limits_{i,j,k=-M}^M{i}^6=\frac{1}{21}M\left(M+1\right){\left(2M+1\right)}^3\left(3{M}^4+6{M}^3-3M+1\right)\end{array}\operatorname{} $$

(15)

Finally, displacement systematic error e_u in DVC using 2nd-order shape function can be estimated as

$$ {\displaystyle \begin{array}{c}{e}_u\approx \frac{1}{24}\cdot \frac{-B\left(2F+G\right)-D\left(2C+D\right)}{A\left(2C+D\right)-3{B}^2}\cdot \left(\frac{\partial^4}{\partial {x}^4}+\frac{\partial^4}{\partial {y}^4}+\frac{\partial^4}{\partial {z}^4}\right)u+\frac{1}{4}\cdot \frac{-B\left(2F+E\right)-C\left(2C+D\right)}{A\left(2C+D\right)-3{B}^2}\cdot \left(\frac{\partial^4}{\partial {x}^2\partial {y}^2}+\frac{\partial^4}{\partial {x}^2\partial {z}^2}+\frac{\partial^4}{\partial {y}^2\partial {z}^2}\right)u\\ {}\approx \frac{1}{280}M\left(M-1\right)\left(M+1\right)\left(M+2\right)\left(\frac{\partial^4}{\partial {x}^4}+\frac{\partial^4}{\partial {y}^4}+\frac{\partial^4}{\partial {z}^4}\right)u+\frac{1}{36}{M}^2{\left(M+1\right)}^2\left(\frac{\partial^4}{\partial {x}^2\partial {y}^2}+\frac{\partial^4}{\partial {x}^2\partial {z}^2}+\frac{\partial^4}{\partial {y}^2\partial {z}^2}\right)u\\ {}\approx \frac{M^2{\left(M+1\right)}^2}{2520}\left[9\left(\frac{\partial^4}{\partial {x}^4}+\frac{\partial^4}{\partial {y}^4}+\frac{\partial^4}{\partial {z}^4}\right)+70\left(\frac{\partial^4}{\partial {x}^2\partial {y}^2}+\frac{\partial^4}{\partial {x}^2\partial {z}^2}+\frac{\partial^4}{\partial {y}^2\partial {z}^2}\right)\right]u\end{array}} $$

(16)