4. Results and Discussions
Figure 4.1 and Table 4.1 shows the obtained results from the Multiphysics simulation carried out on the 50 samples.
Figure 4.1.
Displacement in FCC-based a) structural steel, b) AA6061, c) AA7075, d) Ti-6Al-4V , and e) Inconel 718 lattice structure beam at edge load of 10 KN/m.
Figure 4.1.
Displacement in FCC-based a) structural steel, b) AA6061, c) AA7075, d) Ti-6Al-4V , and e) Inconel 718 lattice structure beam at edge load of 10 KN/m.
Table 4.1.
Obtained displacement magnitude from the Multiphysics based modeling.
Table 4.1.
Obtained displacement magnitude from the Multiphysics based modeling.
| Alloy type |
Alloy Strength (MPa) |
Applied Load (N/m) |
Displacement magnitude (mm) |
| Structural Steel |
250 |
1000 |
0.003518 |
| Structural Steel |
250 |
2000 |
0.0070361 |
| Structural Steel |
250 |
3000 |
0.010554 |
| Structural Steel |
250 |
4000 |
0.014072 |
| Structural Steel |
250 |
5000 |
0.01759 |
| Structural Steel |
250 |
6000 |
0.021108 |
| Structural Steel |
250 |
7000 |
0.024626 |
| Structural Steel |
250 |
8000 |
0.028144 |
| Structural Steel |
250 |
9000 |
0.031662 |
| Structural Steel |
250 |
10000 |
0.03518 |
| AA6061 |
276 |
1000 |
0.01014 |
| AA6061 |
276 |
2000 |
0.02028 |
| AA6061 |
276 |
3000 |
0.030419 |
| AA6061 |
276 |
4000 |
0.040559 |
| AA6061 |
276 |
5000 |
0.050699 |
| AA6061 |
276 |
6000 |
0.060839 |
| AA6061 |
276 |
7000 |
0.070978 |
| AA6061 |
276 |
8000 |
0.081118 |
| AA6061 |
276 |
9000 |
0.091258 |
| AA6061 |
276 |
10000 |
0.1014 |
| AA7075 |
503 |
1000 |
0.0098541 |
| AA7075 |
503 |
2000 |
0.019708 |
| AA7075 |
503 |
3000 |
0.029562 |
| AA7075 |
503 |
4000 |
0.039417 |
| AA7075 |
503 |
5000 |
0.049271 |
| AA7075 |
503 |
6000 |
0.059125 |
| AA7075 |
503 |
7000 |
0.068979 |
| AA7075 |
503 |
8000 |
0.078833 |
| AA7075 |
503 |
9000 |
0.088687 |
| AA7075 |
503 |
10000 |
0.098541 |
| Ti6Al4V |
880 |
1000 |
0.0066571 |
| Ti6Al4V |
880 |
2000 |
0.013314 |
| Ti6Al4V |
880 |
3000 |
0.019971 |
| Ti6Al4V |
880 |
4000 |
0.026628 |
| Ti6Al4V |
880 |
5000 |
0.033286 |
| Ti6Al4V |
880 |
6000 |
0.039943 |
| Ti6Al4V |
880 |
7000 |
0.0466 |
| Ti6Al4V |
880 |
8000 |
0.053257 |
| Ti6Al4V |
880 |
9000 |
0.059914 |
| Ti6Al4V |
880 |
10000 |
0.066571 |
| Inconel718 |
1034 |
1000 |
0.0034893 |
| Inconel718 |
1034 |
2000 |
0.0069786 |
| Inconel718 |
1034 |
3000 |
0.010468 |
| Inconel718 |
1034 |
4000 |
0.013957 |
| Inconel718 |
1034 |
5000 |
0.017446 |
| Inconel718 |
1034 |
6000 |
0.020936 |
| Inconel718 |
1034 |
7000 |
0.024425 |
| Inconel718 |
1034 |
8000 |
0.027914 |
| Inconel718 |
1034 |
9000 |
0.031404 |
| Inconel718 |
1034 |
10000 |
0.034893 |
The heatmap depicted in Figure 4.2 shows varied displacement patterns for various alloy kinds under different loads. AA6061 has the highest displacement sensitivity, reaching 0.1014 mm at maximum load, followed by AA7075 at 0.0985 mm. In comparison, Inconel718 and Structural Steel exhibit strikingly similar, smaller displacement patterns (approximately 0.035 mm at maximum load), indicating improved structural stability. Ti6Al4V has a moderate displacement characteristic (0.0666 mm at maximum load), placing it between aluminum alloys and more rigid materials. Each alloy's linear color transition from dark to light as load increases suggests consistent, predictable material behavior under stress.
Figure 4.2.
Displacement heat maps across loads and alloys.
Figure 4.2.
Displacement heat maps across loads and alloys.
The plot shown in Figure 4.3 depicts the linear relationship between applied load and displacement for several alloys, with the behavior clearly related to the strength values. AA6061 and AA7075 have the sharpest gradients, suggesting the maximum displacement sensitivity, whereas Inconel718 (1034 MPa) has the least displacement under load. The parallel lines indicate consistent elastic behavior across all materials, with displacement magnitudes inversely proportional to strength values. This linear reaction suggests that deformation will occur predictably within the tested load range.
Figure 4.3.
Load vs displacement of different alloys.
Figure 4.3.
Load vs displacement of different alloys.
In this study, the PINN developed using python programming predicts the displacement magnitude of materials under applied edge loads by combining physics-based constraints with data-driven learning at 1000 number of epochs. The PINN makes use of a unique loss function that blends a penalty term based on physics with a data-fitting term (mean squared error). The implemented PINN architecture takes two inputs i.e. alloy strength (
and applied edge load
and yields the output value as displacement magnitude
. The data were divided in 80-20 ratio i.e. 80 percent data were used for training purpose and 20 percent data were used for testing purpose. The architecture of the neural network is defined by equation 4.1.
Where
is the input feature vector represented as
. Equation 4.2 represents the transformation at the
layer using a ReLU activation function.
Where
and
are the weights and bias matrices for the
layer and the function
applied the ReLU activation function. The neural network architecture uses a feedforward approach to process material deformation characteristics across various layers of increasing abstraction. The network begins with an input layer that accepts two features: alloy strength and applied load. These inputs are then routed through a network of dense hidden layers. The first two hidden layers have 64 neurons each and use ReLU (Rectified Linear Unit) activation functions, which add nonlinearity and help the network to learn complicated patterns in the data. Following this, a third hidden layer with 32 neurons uses ReLU activation to further compress the characteristics into a more compact representation. Finally, the network closes with an output layer made up of a single neuron that generates the projected displacement value.
The total loss function depicted in equation 4.3 is given by summing up the two loss components i.e. data-driven loss which is the mean squared error (MSE) between the predicted displacement
and the true displacement
as depicted in equation 4.4 and other component is physics-based loss which incorporates a physics-based constraint which relates the applied load and alloy strength as represented in equation 4.5.
Where
is the number of datapoints,
is a hyperparameter controlling the weight of the physics-based loss term, and
is the physics-based term calculated using equation 4.6.
Where
is a simple representation of the physical relationship between stress and strain, assuming linear behavior under applied load and
is a small constant for numerical stability (
in the present study). The physics term is further normalized depicted in equation 4.7 for the numerical stability and to match the scale of displacement values.
Where
is the mean of
across the batch, and
is the standard deviation across the batch. This normalization ensures that the physics-based loss functions at the same scale as the data-driven loss.
The model is trained using the Adam optimizer with a learning rate of 0.001. A custom training step is constructed using TensorFlow's GradientTape, which simplifies the computation of gradients for the loss function in relation to the model's parameters. These gradients are then used to update the model weights via the optimizer, ensuring efficient convergence throughout the optimization process.
Figure 4.4 a) depicts a continuous displacement prediction across alloy strength and applied load combinations, with larger displacements (red) occurring in low strength-high load regions and smaller displacements (blue) in high strength-low load regions. Figure 4.4 b) shows the actual data points utilized for training, indicating that the model's predictions are consistent with experimental values. The discrete color-coded dots provide clear strength-dependent displacement behavior. Figure 4.4 b) shows rapid early convergence followed by steady refinement, with loss stabilizing at approximately 0.3 after 600 epochs, indicating effective model training.
Figure 4.4.
Comprehensive visualization of PINN model performance showing a) predicted displacement surface across material strength and load ranges, b) actual training data distribution, and c) model convergence through training epochs. The color gradients represent displacement magnitude in millimeters.
Figure 4.4.
Comprehensive visualization of PINN model performance showing a) predicted displacement surface across material strength and load ranges, b) actual training data distribution, and c) model convergence through training epochs. The color gradients represent displacement magnitude in millimeters.
The models performance is evaluated using three metric features i.e. R
2 score, Mean Square error (MSE), and Mean Absolute Error (MAE) shown in Figure 4.5. R
2 score depicted in equation 4.8 determines how well the model's predictions match the actual values. MSE calculates the average squared difference between the predicted and actual values as depicted in equation 4.9. MAE calculates the average absolute difference between predicted and actual values as depicted in equation 4.10.
Figure 4.5.
Comparison of metrics features obtained for linear regression model and PINN model.
Figure 4.5.
Comparison of metrics features obtained for linear regression model and PINN model.
A comparison of performance metrics between the PINN and Linear Regression models indicates considerable variations in predicting ability as shown in Figure 4.1. The PINN model outperforms all important measures. The PINN model has a higher R² Score (0.7923) than Linear Regression (0.5686), indicating that it explains more of the dependent variable's variance. This significant difference of about 0.22 points shows that the PINN model reflects the data's underlying patterns more well. In terms of error metrics, the PINN model likewise performs significantly better. The PINN model's Mean Squared Error (MSE) is 0.00017417, which is less than half of the Linear Regression's MSE of 0.00036187. This lower MSE shows that the PINN model's predictions have lower average squared deviations from the true values. Similarly, the Mean Absolute Error (MAE) shows the same pattern, with PINN obtaining 0.00767965 versus Linear Regression's 0.01624120. This suggests that the PINN model's predictions depart less from true values in absolute terms, with an MAE about 52% lower than the Linear Regression model.
The distribution plots shown in Figure 4.6 of prediction errors for the Linear Regression and PINN models show major differences in their predictive tendencies. The Linear Regression model has a more symmetrical, bell-shaped error distribution centered at 0.01 mm, indicating a consistent but minor overestimation bias in its predictions. This symmetrical pattern suggests that the model's mistakes are uniformly distributed on both sides of the mean, as is common for linear models dealing with complex interactions. In comparison, the PINN model has a much different error distribution pattern, with a noticeable right-skewed shape. Its peak is closer to 0.00 mm, showing more accuracy in most predictions. The PINN distribution has a higher maximum density of around 37.0 than Linear Regression's 19.0, indicating that a greater proportion of its predictions cluster around true values. However, the PINN model's distribution has a broader right tail that extends to around 0.04 mm, showing that while it generally performs better, it may occasionally yield bigger errors in specific instances. The error ranges are also different across the two models, with Linear Regression covering from -0.02 mm to 0.03 mm in a more uniform spread, whilst the PINN model's faults range from about -0.01 mm to 0.04 mm. The PINN's sharper peak and concentrated distribution around zero error illustrate its superior predictive performance in the majority of cases, despite the presence of infrequent outliers. These distributional properties are consistent with and complement the preceding performance measures, demonstrating the PINN model's overall improved prediction accuracy over the standard Linear Regression technique.
Figure 4.6.
Error distribution plot for a)Linear regression model, and b) PINN model.
Figure 4.6.
Error distribution plot for a)Linear regression model, and b) PINN model.
A comparison of actual versus predicted displacement plots for PINN and Linear Regression models, Figure 4.7a) and Figure 4.7b), respectively, indicates large differences in their predictive performances. The dashed line in the plots reflects perfect prediction, where actual and predicted values are equal, and colors for data points correspond to their magnitude of absolute error, ranging from blue-low error-to red-higher error. The PINN model gives a very good prediction accuracy and has most of the points tightly grouped around the perfect prediction line over the entire range of displacements between 0.00 and 0.10 mm. Most of the predictions are of relatively lower magnitudes of absolute errors represented by mostly blue-colored points but for a few at large values of displacements. This consistent clustering along the diagonal line indicates that the PINN model has successfully grasped the linearity and nonlinearity of the displacement relationship.
On the other side, the Linear Regression model gives more scattered predictions with a lot of deviation from the perfect prediction line, particularly for the middle range between 0.04 and 0.08 mm. The color gradient of the points shows a trend in increasing prediction errors with more points showing lighter blue to red colors, which indicates bigger absolute errors compared to the PINN model. A systematic deviation of points from the diagonal line, especially for the mid-range values, would mean that the Linear Regression model is not able to capture the underlying complexity of the relationship between displacements. It can also be seen from the visualization that both models have problems with the extreme values, mainly around the biggest displacement measures of 0.10 mm, where the prediction errors of both models increase, given by the red points.
Figure 4.7.
Actual vs Predicted displacement magnitude plots for a) PINN model, and b) Linear regression model.
Figure 4.7.
Actual vs Predicted displacement magnitude plots for a) PINN model, and b) Linear regression model.
In particular, residual plots for Linear Regression shown in Figure 4.8 a) and PINN depicted in Figure 4.8 b) show dramatically different patterns in their prediction errors for different displacement values. The residuals are the differences between predicted and actual values; the dashed line at zero represents a perfect prediction, while point colors show absolute magnitude errors ranging from blue (low) to red (high).
The residual plot of the Linear Regression model does show a pattern in the residuals that is somewhat troubling: it shows residuals increasing with larger actual values. Indeed, the residuals for smaller displacements—starting from 0.00 to 0.04 mm—stay relatively small and close to zero; this can be seen by the blue dots. However, there is an upward trend in residuals as the actual values increase, and the largest residuals (about 0.034) are at the highest level of displacement values (at 0.10 mm). Such a pattern, in the progression from blue to red of the color scheme, indicative of underpredicting larger displacement values, could be seen as an implication of a biased Linear Regression model. On the other hand, the residual plot for the PINN model shows more homogeneous and controlled error patterns. Residuals are generally smaller in magnitude, with most points clustered closer to the zero line and showing predominantly blue coloring, which indicates lower absolute errors. While there is still one noticeable outlier at the highest displacement value of 0.10 mm, the overall spread of residuals is more uniform across the range of actual values. The PINN model keeps the prediction accuracy relatively stable for different values of displacement without showing a systematic bias as was seen in the Linear Regression model. It also shows better balance in the residuals distribution above and below zero, which may hint at less biased predictions.
Figure 4.8.
Residual plots for plot for a)Linear regression model, and b) PINN model.
Figure 4.8.
Residual plots for plot for a)Linear regression model, and b) PINN model.
The 3D surface plots for the PINN shown in Figure 4.9 a) and Linear Regression shown in Figure 4.9 b) models show that there is a stark difference on how both models estimate displacement from alloy strength and applied load . Both of the visualizations map the predicted displacement values to a blue to red color scale to reflect the displacement level. The surface plot depicted for the PINN model reveals a more curved three-dimensional profile of the variables. It illustrates a curved shaped surface with different gradients, especially areas with larger applied load and smaller alloys. It appears that the PINN model has succeeded in capturing nuanced relationships between the strength of the alloy used and the applied load due to surface features such as gentle waves and a higher degree of rounding on the outer surface. These displacement values have an approximate order of 0.00 and 0.08 mm, and the maximum displacements (depicted in red) are observed at high loads and low alloy strength. However, the Linear Regression model gives a significantly more straightforward plane with equal slopes in the entirety of the prediction space. This means that this linear surface represents a direct relationship meaning displacement rises uniformly as load goes up and as alloy strength goes down. The displacement range is smaller: 0.01-0.06 mm and the blending from a blue to a red colour is gradual to highly predictable. This linear behavior is an inherent problem with the model since it can only account for first order effects between the variables.
Figure 4.9.
3D surface visualization of predicted displacement for a) PINN model, and b) Linear regression model.
Figure 4.9.
3D surface visualization of predicted displacement for a) PINN model, and b) Linear regression model.