# Verification by Simple Geometric Model

## Verification by Simple-Shaped Model

### Elastic static analysis

This verification was performed with a mesh-divided cantilever, as shown in Fig. 9.1.1. The analysis was performed with seven load conditions, exA–exG, as illustrated in Fig. 9.1.2. Please note that exG has the same load conditions as those of exA using the direct method solver.

The verification result of each load condition is presented in Tables 9.1.1–9.1.7.

Fig. 9.1.1: Example of Mesh Partitioned Cantilever Beam (Hexahedral Element)

(b) exD: Gravitation
(d) exE: Centrifugal force
Item Value
Young's Modulus $E = 4000.0\ kgf/mm^2$
Length $L = 10.0\ mm$
Poisson's Ratio $\nu = 0.3$
Sectional area $A = 1.0\ mm^2$
Mass density $\rho = 8.0102 \times 10^{-10}\ kg\,s^2/mm^4$
Second moment of area $I = 1.0/12.0\ mm^4$
Gravitational acceleration $g = 9800.0\ mm/s^2$
Linear coefficient of thermal expansion $\alpha = 1.0 \times 10^{-5}$
Fig. 9.1.2: Verification conditions of the cantilever model
Table 9.1.1: exA: Verification results of the concentrated load problem
Case Name Number of elements Predicated Value : $\delta_{max}= -1.000$ Remarks
NASTRAN ABAQUS FrontISTR
A231 40 -0.338 -0.371 -0.371 33 nodes / plane stress status problem
A232 40 -0.942 -1.002 -1.002 105 nodes / plane stress status problem
A241 20 -0.720 -0.711 -0.711 33 nodes / plane stress status problem
A242 20 -0.910 -1.002 -1.002 85 nodes / plane stress status problem
A341 240 -0.384 -0.384 -0.386 99 nodes
A342 240 -0.990 -0.990 -0.999 525 nodes
A351 80 -0.353 -0.355 -0.351 99 nodes
A352 80 -0.993 -0.993 -0.992 381 nodes
A361 40 -0.954 -0.985 -0.984 99 nodes
A362 40 -0.994 -0.993 -0.993 220 nodes
A731 40 - - -0.991 33 nodes / direct method
A741 20 - - -0.996 33 nodes / direct method
Table 9.1.2: exB: Verification results of the surface-distributed load problem
Case name Number of elements Predicated value : $\delta_{max}= -3.750$ Remarks
NASTRAN ABAQUS FrontISTR
B231 40 -1.281 -1.403 -1.403 33 nodes / plane stress status problem
B232 40 -3.579 -3.763 -3.763 105 nodes / plane stress status problem
B241 20 -3.198 -2.680 -2.680 33 nodes / plane stress status problem
B242 20 -3.426 -3.765 -3.765 85 nodes / plane stress status problem
B341 240 -1.088 -1.449 -1.454 99 nodes
B342 240 -3.704 -3.704 -3.748 525 nodes
B351 80 -3.547 -1.338 -1.325 99 nodes
B352 80 -0.3717 -3.716 -3.713 381 nodes
B361 40 -3.557 -3.691 -3.688 99 nodes
B362 40 -3.726 -3.717 -3.717 220 nodes
B731 40 - - -3.722 33 nodes / direct method
B741 20 - - -3.743 33 nodes / direct method
Table 9.1.3: exC: Verification results of the volume load problem
Case Name Number of elements Predicated Value : $\delta_{max}= -2.944^{-5}$ Remarks
NASTRAN ABAQUS FrontISTR
C231 40 - -1.101e-5 -1.101e-5 33 nodes / plane stress problem
C232 40 - -2.951e-5 -2.951e-5 105 nodes / plane stress problem
C241 20 - -2.102e-5 -2.102e-5 33 nodes / plane stress problem
C242 20 - -2.953e-5 -2.953e-5 85 nodes / plane stress problem
C341 240 - -1.136e-5 -1.140e-5 99 nodes
C342 240 - -2.905e-5 -2.937e-5 525 nodes
C351 80 - -1.050e-5 -1.039e-5 99 nodes
C352 80 - -2.914e-5 -2.911e-5 381 nodes
C361 40 - -2.895e-5 -2.893e-5 99 nodes
C362 40 - -2.915e-5 -2.915e-5 220 nodes
C731 40 - - -2.922e-5 33 nodes / direct method
C741 20 - - -2.938e-5 33 nodes / direct method
Table 9.1.4: exD: Verification results of the gravitation problem
Case name Number of elements Predicated Value : $\delta_{max}= -2.944^{-5}$ Remarks
NASTRAN ABAQUS FrontISTR
D231 40 -1.101e-5 -1.101e-5 -1.101e-5 33 nodes / plane stress status problem
D232 40 -2.805e-5 -2.951e-5 -2.951e-5 105 nodes / plane stress status problem
D241 20 -2.508e-5 -2.102e-5 -2.102e-5 33 nodes / plane stress status problem
D242 20 -2.684e-5 -2.953e-5 -2.953e-5 85 nodes / plane stress status problem
D341 240 -1.172e-5 -1.136e-5 -1.140e-5 99 nodes
D342 240 -2.906e-5 -2.905e-5 -2.937e-5 525 nodes
D351 80 -1.046e-5 -1.050e-5 -1.039e-5 99 nodes
D352 80 -2.917e-5 -2.914e-5 -2.911e-5 381 nodes
D361 40 -2.800e-5 -2.895e-5 -2.893e-5 99 nodes
D362 40 -2.919e-5 -2.915e-5 -2.915e-5 220 nodes
D731 40 - - -2.922e-5 33 nodes / direct method
D741 20 - - -2.938e-5 33 nodes / direct method
Table 9.1.5: exE: Verification results of the centrifugal force problem
Case name Number of elements Predicated value : $\delta_{max}= 2.635^{-3}$ Remarks
NASTRAN ABAQUS FrontISTR
E231 40 2.410e-3 2.616e-3 2.650e-3 33 nodes / plane stress status problem
E232 40 2.447e-3 2.627e-3 2.628e-3 105 nodes / plane stress status problem
E241 20 2.386e-3 2.622e-3 2.624e-3 33 nodes / plane stress status problem
E242 20 2.387e-3 2.627e-3 2.629e-3 85 nodes / plane stress status problem
E341 240 2.708e-3 2.579e-3 2.625e-3 99 nodes
E342 240 2.639e-3 2.614e-3 2.638e-3 525 nodes
E351 80 2.642e-3 2.598e-3 2.625e-3 99 nodes
E352 80 2.664e-3 2.617e-3 2.616e-3 381 nodes
E361 40 2.611e-3 2.603e-3 2.603e-3 99 nodes
E362 40 2.623e-3 2.616e-3 2.616e-3 220 nodes
E731 40 - - 2.619e-3 33 nodes / direct method
E741 20 - - 2.622e-3 33 nodes / direct method
Table 9.1.6: exF: Verification results of the thermal stress load problem
Case name Number of elements Predicated Value : $\delta_{max}= 1.000^{-2}$ Remarks
NASTRAN ABAQUS FrontISTR
F231 40 - 1.016e-2 1.007e-2 33 nodes / plane stress status problem
F232 40 - 1.007e-2 1.007e-2 105 nodes / plane stress status problem
F241 20 - 1.010e-2 1.010e-2 33 nodes / plane stress status problem
F242 20 - 1.006e-2 1.006e-2 85 nodes / plane stress status problem
F341 240 - 1.047e-2 1.083e-2 99 nodes
F342 240 - 1.018e-2 1.022e-2 525 nodes
F351 80 - 1.031e-2 1.062e-2 99 nodes
F352 80 - 1.015e-2 1.017e-2 381 nodes
F361 40 - 1.026e-2 1.026e-2 99 nodes
F362 40 - 1.016e-2 1.016e-2 220 nodes
Table 9.1.7: exG: Verification results of the direct method (concentrated load problem)
Case name Number of elements Predicated value : $\delta_{max}= -1.000$ Remarks
NASTRAN ABAQUS FrontISTR
G231 40 -0.338 -0.371 -0.371 33 nodes / plane stress status problem
G232 40 -0.942 -1.002 -1.002 105 nodes / plane stress status problem
G241 20 -0.720 -0.711 -0.711 33 nodes / plane stress status problem
G242 20 -0.910 -1.002 -1.002 85 nodes / plane stress status problem
G341 240 -0.384 -0.384 -0.386 99 nodes
G342 240 -0.990 -0.990 -0.999 52 nodes
G351 80 -0.353 -0.355 -0.351 99 nodes
G352 80 -0.993 -0.993 -0.992 381 nodes
G361 40 -0.954 -0.985 -0.984 99 nodes
G362 40 -0.994 -0.993 -0.993 220 nodes
G731 40 - - -0.991 33 nodes / direct method
G741 20 - - -0.996 33 nodes / dierct method

### Non-linear static analysis

#### (2-1) exnl1: Geometrical non-linear analysis

The verification model of exI is the same as those of exA–G. A schematic diagram of the verification model is shown in Fig. 9.1.3. A geometric non-linear analysis was performed on this model. The verification results are presented in Table 9.1.8.

The non-linear calculation is a ten-step process with an increment value of 0.1 P and a final load of 1.0 P.

Fig. 9.1.3: Verification model

Table 9.1.8: exI: Verification results (maximum deflection amount log)
Case Name 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Linear Solution
I231 - - - - - - - - - - -
I232 - - - - - - - - - - -
I241 - - - - - - - - - - -
I242 - - - - - - - - - - -
I341 0.039 0.077 0.116 0.154 0.193 0.232 0.270 0.309 0.348 0.386 0.386
I342 0.099 0.200 0.300 0.400 0.499 0.599 0.698 0.797 0.896 0.995 0.999
I351 0.035 0.070 0.105 0.141 0.176 0.211 0.246 0.281 0.316 0.351 0.351
I352 0.099 0.198 0.298 0.397 0.496 0.595 0.693 0.792 0.890 0.987 0.992
I361 0.070 0.139 0.209 0.278 0.348 0.417 0.487 0.556 0.625 0.694 0.984
I362 0.099 0.197 0.298 0.397 0.496 0.595 0.694 0.793 0.891 0.988 0.993

#### (2-2) 　exnl2: Elastoplasticity deformation analysis

Based on the test conducted at National Agency for Finite Element Methods and Standards (NAFEMS; U.K.): Test NL1, this verification problem was verified by elastoplastic deformation analysis that incorporated geometric non-linearity and multiple hardening rules. The elastoplastic deformation analysis model is shown in Fig. 9.1.4.

Fig. 9.1.4: Elastoplasticity deformation analysis Model

(1) Verification conditions:

Item Value
Material Mises elastoplastic material
Young's Modulus $E = 250 GPa$
Poisson's Ratio $\nu = 0.25$
Initial yield stress $5 MPa$
Initial yield strain $0.25\times 10^{-4}$
Isotropic hardening factor $H_i = 0$ or $62.5 GPa$

(2) Boundary conditions

Item Boundary conditions Value
step 1 Forced displacement in nodes 2 and 3 $u_x = 0.2500031251 *10^{-4}$
step 2 Forced displacement in nodes 2 and 3 $u_x = 0.25000937518*10^{-4}$
step 3 Forced displacement in nodes 3 and 4 $u_y = 0.2500031251 *10^{-4}$
step 4 Forced displacement in nodes 3 and 4 $u_y = 0.25000937518*10^{-4}$
step 5 Forced displacement in nodes 2 and 3 $u_x = -0.25000937518*10^{-4}$
step 6 Forced displacement in nodes 2 and 3 $u_x = -0.2500031251 *10^{-4}$
step 7 Forced displacement in nodes 3 and 4 $u_y = -0.25000937518*10^{-4}$
step 8 Forced displacement in nodes 3 and 4 $u_y = -0.2500031251 *10^{-4}$

All the nodes that are not mentioned here are fully constrained. The theoretical solution for this problem is presented as follows:

Strain ($\times10^{-4}$)
[$\varepsilon_x$, $\varepsilon_y$, $\varepsilon_z$]
Equivalent Stress ($MPa$)
[$H_i=0\ H_k=0$; $H_i=62.5\ H_k=0$]
0.25, 0, 0 5.0; 5.0
0.50, 0, 0 5.0; 5.862
0.50, 0.25, 0 5.0; 5.482
0.50, 0.50, 0 5.0; 6.362
0.25, 0.50, 0 5.0; 6.640
0, 0.50, 0 5.0; 7.322
0, 0.25, 0 3.917; 4.230
0, 0, 0 5.0; 5.673

The results of the calculations are as follows:

Strain ($\times10^{-4}$)
[$\varepsilon_x$, $\varepsilon_y$, $\varepsilon_z$]
Equivalent Stress ($MPa$)
[$H_i=0\ H_k=0$; $H_i=62.5\ H_k=0$]
0.25, 0, 0 5.0 (0.0%); 5.0 (0.0%)
0.50, 0, 0 5.0 (0.0%); 5.862 (0.0%)
0.50, 0.25, 0 5.0 (0.0%); 5.482 (0.0%)
0.50, 0.50, 0 5.0 (0.0%); 6.362 (-0.05%)
0.25, 0.50, 0 5.0 (0.0%); 6.640 (-0.21%)
0, 0.50, 0 5.0 (0.0%); 7.322 (-0.34%)
0, 0.25, 0 3.824 (-2.4%); 4.230 (-2.70%)
0, 0, 0 5.0 (0.0%); 5.673 (5.673 (-2.50%)

### Contact analysis (1)

Based on the contact patch test problem (CGS-4) from NAFEMS (U.K.), this verification problem tests the finite sliding contact problem function with friction. The contact analysis model is shown in Fig. 9.1.5

Fig. 9.1.5: Contact analysis model

The equilibrium condition of this problem is as follows:

In the adhesive friction stage, the frictional force is as follows:

In the sliding friction stage, the frictional force is as follows:

The comparison between the calculation results and the analysis solution is as follows.

$\mu$ $F/G$ Analysis Solution $F/G$ Calculation Results
0.0 0.1 0.1
0.1 0.202 0.202
0.2 0.306 0.306
0.3 0.412 0.412

### Contact analysis (2): Hertz contact problem

In this verification, the Hertz contact problem between a cylinder of infinite length and an infinite plane was analyzed. The cylinder’s radius was R=8 mm, and the deformable body’s Young’s modulus E and Poisson’s ratio µ were 1100 MPa and 0.0, respectively. Moreover, assuming that the contact area was much smaller than the cylinder’s radius, and also considering the symmetry of the problem, the analysis was performed with a quarter model of the cylinder.

Fig. 9.1.6: Hertz contact problem analysis model

#### (1) Verification results of contact radius

The theoretical formula to calculate the contact radius is as follows:

where

With the actual calculation, when the pressure is $F=100$, the contact radius is $a=1.36$.

Fig. 9.1.7 shows the equivalent nodal force of the contact point. The contact radius is obtained by extrapolating this nodal force distribution.

Fig. 9.1.7: Equivalent nodal force distribution of the contact point

#### (2) Verification results of maximum shear stress

With the theoretical solution, when the contact position is $z=0.78a$, the maximum shear stress is as follows:

With the actual calculation conditions, it becomes

The actual result obtained was

Fig. 9.1.8: Shear stress distribution (maximum value = 15.6)

### (3) Eigenvalue analysis

The verification models of exJ and exK are the same as those of exA–G. A schematic diagram of the verification model is shown in Fig. 9.1.9. An eigenvalue analysis was performed for this model to determine the primary, secondary, and tertiary eigenvalues. The iteration and direct method solvers were used for exJ and exK, respectively. Furthermore, the verification results are presented in Tables 9.1.9–9.1.12.

Fig. 9.1.9: Verification Model

The vibration eigenvalue of the cantilever is determined by the following equations:

Primary :

Secondary :

Tertiary :

The property values of the verification model are:

Item Value
$I$ $10.0 mm$
$E$ $4000.0 kgf /mm^2$
$l$ $1.0/12.0 mm^4$
$\omega$ $7.85 * 10^{-6} kgf/mm^3$
$g$ $9800.0 mm/sec^2$

Therefore, the primary to tertiary eigenvalue are as follows:

Mode number Value
$n_1$ 3.609e3
$n_2$ 2.262e4
$n_3$ 6.335e4
Table 9.1.9: exJ: Iteration method verification results with the primary eigenvalue
Case Name Number of elements Predicated value : n1=3.609e3 Remarks
NASTRAN FrontISTR
J231 40 5.861e3 5.861e3 33 nodes / plane stress status problem
J232 40 3.596e3 3.593e3 105 nodes / plane stress status problem
J241 20 3.586e3 4.245e3 33 nodes / plane stress status problem
J242 20 3.590e3 3.587e3 85 nodes / plane stress status problem
J341 240 5.442e3 5.429e3 99 nodes
J342 240 3.621e3 3.595e3 525 nodes
J351 80 3.695e3 4.298e3 99 nodes
J352 80 3.610e3 3.609e3 381 nodes
J361 40 3.679e3 3.619e3 99 nodes
J362 40 3.611e3 3.606e3 220 nodes
Table 9.1.10: Iteration method verification results of exJ with the secondary eigenvalue
Case name Number of elements Predicated value : n2=2.262e4 Remarks
NASTRAN FrontISTR
J231 40 3.350e4 3.351e4 33 nodes / plane stress status problem
J232 40 2.163e4 2.156e4 105 nodes / plane stress status problem
J241 20 2.149e4 2.516e4 33 nodes / plane stress status problem
J242 20 2.149e4 2.143e4 85 nodes / plane stress status problem
J341 240 3.145e4 3.138e4 99 nodes
J342 240 2.171e4 2.155e4 525 nodes
J351 80 2.208e4 2.546e4 99 nodes
J352 80 2.156e4 2.149e4 381 nodes
J361 40 2.202e4 2.168e4 99 nodes
J362 40 2.154e4 2.144e4 220 nodes

Note: In the three-dimensional (3D) models, the primary and secondary values have equal roots. Therefore, the secondary value in the table represents the tertiary calculation value.

Table 9.1.11: Direct method verification results of exK with the primary eigenvalue
Case name Number of elements Predicated Value : n1=3.609e3 Remarks
NASTRAN FrontISTR
J231 40 5.861e3 5.861e3 33 nodes / plane stress status problem
J232 40 3.596e3 3.593e3 105 nodes / plane stress status problem
J241 20 3.586e3 4.245e3 33 nodes / plane stress status problem
J242 20 3.590e3 3.587e3 85 nodes / plane stress status problem
J341 240 5.442e3 5.429e3 99 nodes
J342 240 3.621e3 3.595e3 525 nodes
J351 80 3.695e3 4.298e3 99 nodes
J352 80 3.610e3 3.609e3 381 nodes
J361 40 3.679e3 3.619e3 99 nodes
J362 40 3.611e3 3.606e3 220 nodes
J731 40 - 3.606e3 220 nodes
J741 20 - 3.594e3 220 nodes
Table 9.1.12: Direct method verification results of exK with the secondary eigenvalue
Case name Number of elements Predicated value : n2=2.262e4 Remarks
NASTRAN FrontISTR
J231 40 3.350e4 3.351e4 33 nodes / plane stress status problem
J232 40 2.163e4 2.156e4 105 nodes / plane stress status problem
J241 20 2.149e4 2.516e4 33 nodes / plane stress status problem
J242 20 2.149e4 2.143e4 85 nodes / plane stress status problem
J341 240 3.145e4 3.138e4 99 nodes
J342 240 2.171e4 2.155e4 525 nodes
J351 80 2.208e4 2.546e4 99 nodes
J352 80 2.156e4 2.149e4 381 nodes
J361 40 2.202e4 2.168e4 99 nodes
J362 40 2.154e4 2.144e4 220 nodes
J731 40 - 2.156e4 220 nodes
J741 20 - 2.153e4 220 nodes

Note: In the 3D models, the primary and secondary values have equal roots. Therefore, the secondary value in the table represents the tertiary calculation value.

### (4) Heat conduction analysis

The common conditions for the steady-state heat conduction analysis are shown in Fig. 9.1.10. The individual conditions for the verification cases exM–exT are shown in Fig. 9.1.11. The mesh division used here is equivalent to that of exA.

The verification results (temperature distribution table) of each case are presented in Tables 9.1.13–9.1.20.

Length between AB $L = 10.0m$
Cross-sectional area $A = 1.0 mm^2$

Temperature dependency of thermal conductivity

Thermal conductivity $\lambda(W/mK)$ Temperature $(^\circ C)$
50.0 0.0
35.0 500.0
20.0 1000.0
Fig. 9.1.10: Verification conditions of steady-heat conduction analysis
exM: Linear material
exN: Specified tempreature problem
exO: Concentrated heat flux problem
exP: Distributed heat flux problem
exQ: Convective heat transfer problem
exS: Volume heat generation problem
exT: Internal gap problem
Fig. 9.1.11: Analysis conditions for each verification case
Table 9.1.13: Verification results of exM (steady-state calculation of linear material)
Case name Element type Number of elements/nodes Distance from edge A (m)
Edge A 2.0 4.0 6.0 8,0 Edge B
M361A 361 40／33 0.0 100.0 200.0 300.0 400.0 500.0
M361B 361 40／105 0.0 100.0 200.0 300.0 400.0 500.0
M361C 361 20／33 0.0 100.0 200.0 300.0 400.0 500.0
M361D 361 20／85 0.0 100.0 200.0 300.0 400.0 500.0
M361E 361 240／99 0.0 100.0 200.0 300.0 400.0 500.0
M361F 361 24／525 0.0 100.0 200.0 300.0 400.0 500.0
M361G 361 80／99 0.0 100.0 200.0 300.0 400.0 500.0
Table 9.1.14: Verification results of exN (specified temperature problem)
Case name Element type Number of elements/nodes Distance from edge A (m)
Edge A 2.0 4.0 6.0 8,0 Edge B
ABAQUS 361 40／99 0.0 87.3 179.7 278.2 384.3 500.0
N231 231 40／33 0.0 87.2 179.5 278.0 384.1 500.0
N232 232 40／105 0.0 86.0 178.3 276.8 382.9 500.0
N241 241 20／33 0.0 87.3 179.7 278.2 384.3 500.0
N242 242 20／85 0.0 87.3 179.7 278.2 384.3 500.0
N341 341 240／99 0.0 87.3 179.7 278.2 384.3 500.0
N342 342 24／525 0.0 87.9 179.9 278.0 383.6 500.0
N351 351 80／99 0.0 87.3 179.7 278.2 384.3 500.0
N352 352 80／381 0.0 87.3 179.7 278.2 384.3 500.0
N361 361 40／99 0.0 87.3 179.7 278.2 384.3 500.0
N362 362 40／330 0.0 87.3 179.7 278.2 384.3 500.0
N731 731 40／33 0.0 87.3 179.7 278.2 384.3 500.0
N741 741 20／33 0.0 87.3 179.7 278.2 384.3 500.0
Table 9.1.15: Verification results of exO (concentrated heat flux problem)
Case name Element type Numbewr of elements/nodes Distance from edge A (m)
Edge A 2.0 4.0 6.0 8,0 Edge B
ABAQUS 361 40／99 0.0 103.2 213.7 333.3 464.8 612.6
O231 231 40／33 0.0 103.2 213.7 333.3 464.8 612.6
O232 232 40／105 0.0 103.2 213.7 333.3 464.8 612.6
O241 241 20／33 0.0 103.2 213.7 333.3 464.8 612.6
O242 242 20／85 0.0 103.2 213.7 333.4 465.2 618.0
O341 341 240／99 - - - - - -
O342 342 24／525 0.0 104.4 214.9 334.7 466.3 614.6
O351 351 80／99 - - - - - -
O352 352 80／381 0.0 103.2 213.7 333.3 465.0 624.2
O361 361 40／99 0.0 103.2 213.7 333.3 464.8 612.6
O362 362 40／330 0.0 103.2 213.7 333.4 465.5 623.5
O731 731 40／33 0.0 103.2 213.7 333.3 464.8 612.5
O741 741 20／33 0.0 103.2 213.7 333.3 464.8 612.6
Table 9.1.16: Verification results of exP (distribution heat flux problem)
Case name Element type Number of elements/nodes Distance from edge A (m)
Edge A 2.0 4.0 6.0 8,0 Edge B
ABAQUS 361 40／99 0.0 103.2 213.7 333.3 464.8 612.6
P231 231 40／33 0.0 103.2 213.7 333.3 464.8 612.6
P232 232 40／105 0.0 103.2 213.7 333.3 464.8 612.6
P241 241 20／33 0.0 103.2 213.7 333.3 464.8 612.6
P242 242 20／85 0.0 103.2 213.7 333.3 464.8 612.6
P341 341 240／99 - - - - - -
P342 342 24／525 0.0 103.2 213.7 333.3 464.8 612.6
P351 351 80／99 - - - - - -
P352 352 80／381 0.0 103.2 213.7 333.3 464.8 612.6
P361 361 40／99 0.0 103.2 213.7 333.3 464.8 612.6
P362 362 40／330 0.0 103.2 213.7 333.4 465.5 612.6
P731 731 40／33 0.0 103.2 213.7 333.3 464.8 612.5
P741 741 20／33 0.0 103.2 213.7 333.3 464.8 612.6
Table 9.1.17: Verification results of exQ (convective heat transfer problem)
Case name Element type Number of elements/nodes Distance from edge A (m)
Edge A 2.0 4.0 6.0 8,0 Edge B
ABAQUS 361 40／99 0.0 89.2 183.8 284.8 393.9 513.2
Q231 231 40／33 0.0 89.2 183.8 284.8 393.9 513.2
Q232 232 40／105 0.0 89.2 183.8 284.8 393.9 513.2
Q241 241 20／33 0.0 89.2 183.8 284.8 393.9 513.2
Q242 242 20／85 0.0 89.2 183.8 284.8 393.9 513.2
Q341 341 240／99 - - - - - -
Q342 342 24／525 0.0 89.2 183.8 284.8 393.9 513.2
Q351 351 80／99 - - - - - -
Q352 352 80／381 0.0 89.2 183.8 284.8 393.9 513.2
Q361 361 40／99 0.0 89.2 183.8 284.8 393.9 513.2
Q362 362 40／330 0.0 89.2 183.8 284.8 393.9 513.2
Q731 731 40／33 0.0 89.2 183.8 284.8 393.9 513.2
Q741 741 20／33 0.0 89.2 183.8 284.8 393.9 513.2
Table 9.1.18: Verification results of exR (radiation heat transfer problem)
Case name Element type Number of elements/nodes Distance from edge A (m)
Edge A 2.0 4.0 6.0 8,0 Edge B
ABAQUS 361 40／99 0.0 89.5 184.4 285.8 395.3 515.2
R231 231 40／33 0.0 89.5 184.4 285.8 395.3 515.2
R232 232 40／105 0.0 89.5 184.4 285.8 395.3 515.2
R241 241 20／33 0.0 89.5 184.4 285.8 395.3 515.2
R242 242 20／85 0.0 89.5 184.4 285.8 395.3 515.2
R341 341 240／99 - - - - - -
R342 342 24／525 0.0 89.5 184.4 285.8 395.3 515.2
R351 351 80／99 - - - - - -
R352 352 80／381 0.0 89.5 184.4 285.8 395.3 515.2
R361 361 40／99 0.0 89.5 184.4 285.8 395.3 515.2
R362 362 40／330 0.0 89.5 184.4 285.8 395.3 515.2
R731 731 40／33 0.0 89.5 184.4 285.8 395.3 515.2
R741 741 20／33 0.0 89.5 184.4 285.8 395.3 515.2
Table 9.1.19: Verification results of exS (volume heat generation problem)
Case name Element type Number of elements/nodes Distance from edge A (m)
Edge A 2.0 4.0 6.0 8,0 Edge B
ABAQUS 361 40／99 0.0 103.2 213.7 333.3 464.8 612.6
S231 231 40／33 0.0 103.2 213.7 333.3 464.8 612.6
S232 232 40／105 0.0 103.2 213.7 333.3 464.8 612.6
S241 241 20／33 0.0 103.2 213.7 333.3 464.8 612.6
S242 242 20／85 0.0 103.2 213.7 333.3 464.8 612.6
S341 341 240／99 - - - - - -
S342 342 24／525 0.0 103.2 213.7 333.3 464.8 612.6
S351 351 80／99 - - - - - -
S352 352 80／381 0.0 103.2 213.7 333.3 464.8 612.6
S361 361 40／99 0.0 103.2 213.7 333.3 464.8 612.6
S362 362 40／330 0.0 103.2 213.7 333.3 464.8 612.6
S731 731 40／33 0.0 103.2 213.7 333.3 464.8 612.6
S741 741 20／33 0.0 103.2 213.7 333.3 464.8 612.6
Table 9.1.20: Verification results of exT (internal gap problem)
Case name Element type Number of elements/nodes Distance from edge A (m)
Edge A 2.0 4.0 6.0 8,0 Edge B
ABAQUS 361 40／99 0.0 88.6 182.4 282.6 387.7 500.0
S231 231 40／33 0.0 88.6 182.4 282.6 387.7 500.0
S232 232 40／105 0.0 88.6 182.4 282.6 387.7 500.0
S241 241 20／33 0.0 88.6 182.4 282.6 387.7 500.0
S242 242 20／85 0.0 88.6 182.4 282.6 387.7 500.0
S341 341 240／99 - - - - - -
S342 342 24／525 0.0 88.6 182.4 282.6 387.7 500.0
S351 351 80／99 - - - - - -
S352 352 80／381 0.0 88.6 182.4 282.6 387.7 500.0
S361 361 40／99 0.0 88.6 182.4 282.6 387.7 500.0
S362 362 40／330 0.0 88.6 182.4 282.6 387.7 500.0
S731 731 40／33 0.0 88.6 182.4 282.6 387.7 500.0
S741 741 20／33 0.0 88.6 182.4 282.6 387.7 500.0

### (5) Linear dynamic analysis

In exW, a linear dynamic analysis was performed on the same mesh-divided cantilever that was discussed earlier in the Elastic static analysis subsection, with the verification conditions shown in Fig. 9.1.12. The objective was to verify the impact of time increment on the results with the same mesh division. Both the implicit and explicit methods of dynamic analysis and the element types 361 and 342 were used. The verification results are presented in Table 9.1.22 and shown in Figs. 9.1.13–9.1.15.

Analysis Model

Analysis Model

Time history of external force $F$

The theoretical solution for vibration point displacement is as follows:

where

Fig. 9.1.12: Verification conditions of linear dynamic analysis

Verification conditions:

Length $L$ $10.0\ mm$
Cross-sectional width $a$ $1.0\ mm$
Cross-sectional height $b$ $1.0\ mm$
Young's modulus $E$ $4000.0\ kgf/mm^2$
Poisson's ratio $\nu$ $0.3$
Density $\rho$ $1.0E-09\ kgf\,s^2/mm^3$
Gravitational acceleration $g$ $9800.0\ mm/s^2$
External force $F_0$ $1.0\ kgf$
Element Hexahedral linear element
Tetrahedral secondary element
Solution Implicit method
Parameter $\gamma$ of Newmark-$\beta$ method 1/2
Parameter $\beta$ of Newmark-$\beta$ method 1/4
Explicit method
Damping N/A
Table 9.1.21: Verification conditions of the linear dynamic analysis (continuation)
Case Name Element Type No. of Nodes No. of Elements Solution Time Increment $\Delta t$ [sec]
W361_c0_im_m2_t1 361 99 40 Implicit method 1.0E-06
W361_c0_im_m2_t2 361 99 40 Implicit method 1.0E-05
W361_c0_im_m2_t3 361 99 40 Implicit method 1.0E-04
W361_c0_ex_m2_t1 361 99 40 Explicit method 1.0E-08
W361_c0_ex_m2_t2 361 99 40 Explicit method 1.0E-07
W361_c0_ex_m2_t3 361 99 40 Explicit method 1.0E-06
W342_c0_im_m2_t1 342 525 240 Implicit method 1.0E-06
W342_c0_im_m2_t2 342 525 240 Implicit method 1.0E-05
W342_c0_im_m2_t3 342 525 240 Implicit method 1.0E-04
W342_c0_ex_m2_t1 342 525 240 Explicit method 1.0E-08
W342_c0_ex_m2_t2 342 525 240 Explicit method 5.0E-08
W342_c0_ex_m2_t3 342 525 240 Explicit method 1.0E-07
Table 9.1.22: Verification results of linear dynamic analysis of exW (cantilever)
Case name Element type Number of nodes Number of elements Solution z-direction displacement: $u_z(mm)$ when time $t = 0.002(s)$
Theorical solution repeated to sextic equation FrontISTR
W361_c0_im_m2_t1 361 99 40 Implicit method 1.9753 1.9302
W361_c0_im_m2_t2 361 99 40 Implicit method 1.9753 1.8686
W361_c0_im_m2_t3 361 99 40 Implicit method 1.9753 0.3794
W361_c0_ex_m2_t1 361 99 40 Explicit method 1.9753 1.9302
W361_c0_ex_m2_t2 361 99 40 Explicit method 1.9753 1.9247
W361_c0_ex_m2_t3 361 99 40 Explicit method 1.9753 Divergence
W342_c0_im_m2_t1 342 525 240 Implicit method 1.9753 1.9431
W342_c0_im_m2_t2 342 525 240 Implicit method 1.9753 1.8719
W342_c0_im_m2_t3 342 525 240 Implicit method 1.9753 0.3873
W342_c0_ex_m2_t1 342 525 240 Explicit method 1.9753 1.9359
W342_c0_ex_m2_t2 342 525 240 Explicit method 1.9753 1.9358
W342_c0_ex_m2_t3 342 525 240 Explicit method 1.9753 Divergence

Fig. 9.1.13: Deformation diagram and equivalent stress distribution of the cantilever (W361_c0_im_m2_t2)

(a) Element Type 361 : Implicit method

(b) Element Type 361 : Explicit method

Fig. 9.1.14: Time history of vibration point displacement $$u_z$$

(a) Element Type 342 : Implicit method

(b) Element Type 342 : Explicit method

### (6) Frequency response analysis

In this verification, a frequency response analysis was performed on a cantilever, and the results were compared with those obtained using the ABAQUS software. The analysis model and verification conditions are presented below:

Analysis conditions:

Young's modulus $E$ $210000\ N/mm^2$
Poisson's ratio $\nu$ $0.3$
Density $\rho$ $7.89E-09\ t/mm^3$
Gravitational acceleration $g$ $9800.0\ mm/s^2$
Load $F_0$ $1.0\ N$
Parameter of Rayleigh damping $R_m$ $0.0$
Parameter of Rayleigh damping $R_k$ $7.2E-07$
Fig. 9.1.15 : Analysis model (tetrahedral primary element (126 elements and 55 nodes))

The eigenvalues up to the fifth order and the frequency response of the vibration points obtained from eigenvalue analysis are as follows:

mode FrontISTR ABAQUS
1 14952 14952
2 15002 15003
3 84604 84539
4 84771 84697
5 127054 126852

Fig. 9.1.16 : Frequency dependence of displacement strength of vibration points