Four-point shear test

This example deals with two-dimensional analysis of a more complicated concrete specimen--a single-edge-notched beam loaded such that the central portion around the notch is under high shear (Fig. 4a). This so-called Iosipescu geometry was adapted for concrete by Arrea and Ingraffea [1]. For quasibrittle materials such as concrete, the fracture is still locally driven by the maximal tensile stresses. The crack starts propagating from the notch tip in an inclined direction and when it leaves the region of high shear, it approaches the vertical direction and reaches the surface of the specimen to the right of the applied force $F$. Some limited cracking also appears on the opposite side of the loading platen in the region of high tensile stresses due to bending. The final crack trajectory is curved and certain simulations from the literature, especially on coarse meshes, failed to reproduce it correctly. Better results are usually obtained with anisotropic models (such as smeared crack models or anisotropic damage models), but since the present study focuses mainly on the numerical aspects, we will keep using the simple isotropic damage model with a Rankine-type equivalent strain. On the fine mesh with 2135 nodes and 4132 constant-strain elements, shown in Fig. 4b, the essential failure mechanism is captured properly. The constitutive properties are set to: Young's modulus $E=30$ GPa, Poisson's ratio $\nu=0.18$, tensile strength $f_t=3.5$ MPa, strain controling the softening $\varepsilon _f=0.006$, and nonlocal interaction radius $R=5$ mm.

Figure 4: Four-point shear test: (a) specimen geometry and loading, (b) finite element mesh
(a) &
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(b) &
\epsfig{file=4pbt.mesh.eps, width=120mm}%

The formation of the process zone and failure of the specimen are simulated in 40 incremental steps, controled by the increments of the crack mouth sliding displacement. Four solution strategies are compared:

Again, the rate of convergence is linear for the secant or elastic stiffness matrices and quadratic for the tangent stiffness. With the TSM, a fully converged solution is obtained within 4 to 6 iterations at any stage of the analysis; see Fig. 5.

Figure 5: Four-point shear test: convergence characteristics and load-displacement diagram
\begin{figure}\epsfig{file=4pbt.chart.eps, width=120mm}%\end{figure}

For this model, the assembly of a local stiffness matrix (ESM or SSM) takes about 0.11 seconds, while the assembly of the nonlocal stiffness matrix (TSM) takes between 0.11 and 0.28 seconds for the skyline storage scheme (used with the direct solver) or between 0.11 and 0.42 seconds for the compressed row storage (used with the iterative solver). The assembly times vary depending on the stage of analysis, because the number of nonzero entries increases as the process zone evolves. The factorization time for ESM and SSM is 2.19 seconds, while for TSM it varies between 2.2 and 5.9 seconds. The total times needed for the analysis using different solution strategies are given in Table 4. Same as in the previous example, the most efficient technique is SSM for a low accuracy and TSM with a direct solver for a high accuracy. The ESM requires an extremely large number of iterations and is prohibitively expensive. The TSM combined with an iterative solver (GMRES with preconditioning by incomplete $LU$ decomposition) is slightly slower than the TSM with a direct solver. Additional details about the numbers of iterations per step and user times per iteration are provided in Tables 5 and 6.

Table 4: Four-point shear test: total user times
tolerance strategy total time
$\epsilon=10^{-3}$ ESM $>$ 4h
  SSM-10 00h:04m:45s
  TSM 00h:16m:07s
$\epsilon=10^{-6}$ ESM $>$ 10h
  SSM-10 00h:29m:12s
  SSM-5 00h:32m:19s
  TSM 00h:24m:15s
  TSM, GMRES+ILUT 00h:26m:55s

Table 5: Four-point shear test: convergence and user times per iteration (``nite'' is the number of equilibrium iterations) for $\epsilon=10^{-3}$
step nite time[s] nite time[s] nite time[s]
A 499 173 39 28 4 63
B 1590 529 19 14 2 58
C 8411 1489 15 12 3 77

Table 6: Four-point shear test: convergence and user times per iteration (``nite'' is the number of equilibrium iterations) for $\epsilon=10^{-6}$; $\bar\epsilon$ is the actually reached accuracy
step nite t [s] nite t [s] nite t [s] nite ($\bar\epsilon$) t [s] nite ($\varepsilon$) t [s]
A 2191 758 174 123 173 180 6 (1.e-14) 88 6 133
B - - 137 96 136 143 4 (2.e-13) 96 4 48
C - - 110 79 109 112 5 (5.e-12) 115 5 95

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(c) 2016 Borek Patzak, info(at)oofem(dot)org