{ "cells": [ { "cell_type": "markdown", "id": "markdown-000", "metadata": {}, "source": [ "# Numerical solutions of elastic systems\n", "\n", "[Read the lesson](https://tirta.co/a-byte-of-physics/numerical-solutions-elastic-systems). This notebook includes the lesson text and its Python examples in reading order. Run the setup cell, then run the remaining cells from top to bottom. In Colab, use File → Save a copy in Drive to keep your edits. You can run and edit this view without changing the course source. Examples use illustrative data unless stated otherwise." ] }, { "cell_type": "code", "id": "code-001", "metadata": {}, "source": [ "import importlib.metadata\n", "import subprocess\n", "import sys\n", "\n", "requirements = {\"numpy\": (2, 0), \"sympy\": (1, 13), \"matplotlib\": (3, 8)}\n", "to_install = []\n", "for package, minimum in requirements.items():\n", " try:\n", " installed = importlib.metadata.version(package)\n", " version = tuple(int(part) for part in installed.split(\".\")[:2])\n", " except importlib.metadata.PackageNotFoundError:\n", " version = (0, 0)\n", " if version < minimum:\n", " to_install.append(f\"{package}>={minimum[0]}.{minimum[1]}\")\n", "if to_install:\n", " subprocess.check_call([sys.executable, \"-m\", \"pip\", \"install\", *to_install])\n", "\n", "import numpy as np\n", "import sympy as sp\n", "import matplotlib\n", "import matplotlib.pyplot as plt\n", "\n", "print(\"Python\", sys.version.split()[0])\n", "print(\"NumPy\", np.__version__)\n", "print(\"SymPy\", sp.__version__)\n", "print(\"Matplotlib\", matplotlib.__version__)\n" ], "execution_count": null, "outputs": [] }, { "cell_type": "markdown", "id": "markdown-002", "metadata": {}, "source": [ "An elastic model becomes a numerical problem when it contains many connected degrees of\n", "freedom, a complicated shape, or loads that change with time. The numerical method must\n", "respect the same force, energy, and boundary-condition statements as the physical model.\n", "A calculation that finishes without an error message is not automatically a correct\n", "calculation.\n", "\n", "## From elastic equilibrium to a linear system\n", "\n", "For a small-deformation, linearly elastic system, equilibrium is commonly written\n", "\n", "\n", "$$\n", "\\mathbf K\\mathbf u=\\mathbf f\n", "$$\n", "\n", "\n", "Here $\\mathbf K$ describes how\n", "the structure resists deformation,\n", "$\\mathbf u$ contains unknown\n", "nodal displacements, and\n", "$\\mathbf f$ contains applied nodal\n", "loads. This equation is the discrete counterpart of force and torque equilibrium.\n", "\n", "The matrix is assembled from elements. Each element contributes forces involving only\n", "its own nodes, so its local stiffness is added into the corresponding rows and columns of\n", "the global matrix. Shared nodes receive contributions from every attached element." ] }, { "cell_type": "code", "id": "code-003", "metadata": {}, "source": [ "import numpy as np\n", "\n", "stiffness = np.zeros((3, 3))\n", "for i, j in [(0, 1), (1, 2)]:\n", " stiffness[np.ix_([i, j], [i, j])] += 100 * np.array([[1, -1], [-1, 1]])\n", "loads = np.array([0.0, 0.0, 10.0])\n", "print(\"Assembled stiffness (N/m):\", stiffness)\n" ], "execution_count": null, "outputs": [] }, { "cell_type": "markdown", "id": "markdown-004", "metadata": {}, "source": [ "For conservative linear elasticity, the correctly constrained stiffness matrix is\n", "usually symmetric. A negative quadratic energy direction indicates either an unstable\n", "configuration, a sign error, or a model outside the assumptions of ordinary linear\n", "elasticity.\n", "\n", "## A one-dimensional bar example\n", "\n", "A uniform axial bar element with length\n", "$L$, cross-sectional area\n", "$A$, and Young's modulus\n", "$E$ has element matrix\n", "\n", "\n", "$$\n", "\\mathbf K_e=\\frac{EA}{L}\\begin{bmatrix}1&-1\\\\-1&1\\end{bmatrix}\n", "$$\n", "\n", "\n", "If both endpoints move by the same amount, their separation is unchanged. Multiplying\n", "the matrix by that rigid-translation displacement produces zero force. This zero mode is\n", "physical for a free bar. Fixing one endpoint removes the rigid mode and makes the other\n", "displacement solvable under an axial load.\n", "\n", "The extension under load\n", "$F$ is\n", "\n", "\n", "$$\n", "\\Delta L=\\frac{FL}{EA}\n", "$$\n", "\n", "\n", "This analytic result is a valuable unit test for matrix assembly. The numerical answer\n", "should reproduce it, including the facts that doubling length doubles extension and\n", "doubling area or modulus halves extension.\n", "\n", "## Boundary conditions are part of the physics\n", "\n", "A free elastic body can translate and rotate without strain, so its stiffness matrix has\n", "rigid-body zero modes. Attempting to invert that matrix does not reveal a bad solver; it\n", "reveals missing support information. Prescribed displacements remove selected unknowns\n", "and generate reaction forces. Prescribed forces belong in the load vector.\n", "\n", "One robust approach partitions degrees of freedom into free and constrained sets. Solve\n", "only for the free displacements, then recover reactions from the complete equilibrium\n", "equation. Replacing constrained rows with identity rows is also possible, but reactions\n", "must still be computed from the original system." ] }, { "cell_type": "code", "id": "code-005", "metadata": {}, "source": [ "displacements = np.zeros(3) # Node 0 is fixed at zero displacement.\n", "free = [1, 2]\n", "displacements[free] = np.linalg.solve(stiffness[np.ix_(free, free)], loads[free])\n", "reactions = stiffness @ displacements - loads\n", "assert np.allclose(reactions[free], 0.0)\n", "print(\"Displacements (m):\", displacements)\n", "print(\"Reactions (N):\", reactions)\n" ], "execution_count": null, "outputs": [] }, { "cell_type": "markdown", "id": "markdown-006", "metadata": {}, "source": [ "The sum of applied forces and support reactions should vanish in static equilibrium.\n", "Their moments should vanish as well. These global checks often expose an incorrect node\n", "index, omitted load, or accidentally constrained direction.\n", "\n", "## Residual and conditioning\n", "\n", "After solving, calculate the residual\n", "\n", "\n", "$$\n", "\\mathbf r=\\mathbf K\\mathbf u-\\mathbf f\n", "$$\n", "\n", "\n", "on unconstrained equations. A small residual says that the computed vector solves the\n", "discrete equations. It does not prove that the mesh, material law, or loads represent the\n", "real system.\n", "\n", "Conditioning describes how strongly small input or roundoff changes can affect the\n", "solution. Very different stiffness scales, extreme element aspect ratios, nearly\n", "unconstrained motion, or inconsistent units can make a system ill-conditioned. Scaling\n", "and preconditioning can help the numerical solver, but they cannot repair a physically\n", "incorrect model.\n", "\n", "Never form a matrix inverse merely to solve the system. Use a linear solver appropriate\n", "to the matrix. A direct factorization is convenient for moderate problems. Iterative\n", "methods can handle large sparse systems, but need convergence tolerances and often a\n", "preconditioner. Report both the residual criterion and the displacement or energy change\n", "used to decide convergence.\n", "\n", "## Nonlinear elastic equilibrium\n", "\n", "Large deformation, contact, and nonlinear material response produce internal forces that\n", "are not a constant matrix times displacement. Static equilibrium becomes\n", "\n", "\n", "$$\n", "\\mathbf R(\\mathbf u)=\\mathbf f_{\\mathrm{int}}(\\mathbf u)-\\mathbf f_{\\mathrm{ext}}=\\mathbf 0\n", "$$\n", "\n", "\n", "Newton's method repeatedly linearizes this residual using a tangent stiffness matrix.\n", "Each step solves for a displacement correction, but a full correction may be too large.\n", "Load stepping and line searches improve robustness." ] }, { "cell_type": "code", "id": "code-007", "metadata": {}, "source": [ "# Nonlinear spring: internal force = 100 u + 1000 u^3.\n", "u = 0.0\n", "for load in np.linspace(0, 10, 6):\n", " for iteration in range(30):\n", " residual = 100 * u + 1000 * u**3 - load\n", " if abs(residual) < 1e-10:\n", " break\n", " u -= residual / (100 + 3000 * u**2)\n", " else:\n", " raise RuntimeError(\"Newton iteration did not converge.\")\n", " print(f\"Load {load:.1f} N: displacement {u:.6f} m\")\n" ], "execution_count": null, "outputs": [] }, { "cell_type": "markdown", "id": "markdown-008", "metadata": {}, "source": [ "A converged nonlinear solution can depend on the load path and initial guess. Multiple\n", "equilibria, buckling, contact changes, plasticity, or damage make that dependence\n", "physical rather than merely numerical.\n", "\n", "## Dynamic elastic systems\n", "\n", "When inertia matters, the semidiscrete equation is\n", "\n", "\n", "$$\n", "\\mathbf M\\ddot{\\mathbf u}+\\mathbf C\\dot{\\mathbf u}+\\mathbf K\\mathbf u=\\mathbf f(t)\n", "$$\n", "\n", "\n", "The mass matrix $\\mathbf M$ stores\n", "inertia, and the damping matrix\n", "$\\mathbf C$ represents specified\n", "dissipation. With no damping or external work, a correct continuous model conserves the\n", "sum of kinetic and elastic energy.\n", "\n", "For one undamped mass and spring, angular frequency is\n", "\n", "\n", "$$\n", "\\omega=\\sqrt{\\frac{k}{m}}\n", "$$\n", "\n", "\n", "For the kick-then-drift form of symplectic Euler, linear stability of this particular\n", "oscillator requires\n", "$\\omega\\Delta t<2$.\n", "This is a stability limit, not an accuracy guarantee. Ordinary forward Euler is unstable\n", "for the undamped oscillator for every positive step size because it systematically adds\n", "energy." ] }, { "cell_type": "code", "id": "code-009", "metadata": {}, "source": [ "position, velocity, mass, dt = 0.1, 0.0, 1.0, 0.001\n", "for _ in range(100):\n", " acceleration = -100 * position / mass\n", " velocity += dt * acceleration\n", " position += dt * velocity\n", "print(f\"Dynamic displacement after 0.1 s: {position:.6f} m\")\n" ], "execution_count": null, "outputs": [] }, { "cell_type": "markdown", "id": "markdown-010", "metadata": {}, "source": [ "In a multi-degree system, the highest natural frequency normally imposes the tightest\n", "explicit time-step restriction. A finer mesh often raises that frequency, so spatial\n", "refinement without time-step refinement may destabilize an otherwise unchanged model.\n", "Implicit methods can remain stable at larger steps, but large steps can still erase phase\n", "and high-frequency motion.\n", "\n", "## Energy and modal checks\n", "\n", "For a linear system, elastic energy is\n", "\n", "\n", "$$\n", "U=\\frac12\\mathbf u^{\\mathsf T}\\mathbf K\\mathbf u\n", "$$\n", "\n", "\n", "and kinetic energy is\n", "\n", "\n", "$$\n", "T=\\frac12\\dot{\\mathbf u}^{\\mathsf T}\\mathbf M\\dot{\\mathbf u}\n", "$$\n", "\n", "\n", "Free-vibration modes satisfy the generalized eigenvalue problem\n", "\n", "\n", "$$\n", "\\mathbf K\\boldsymbol\\phi=\\omega^2\\mathbf M\\boldsymbol\\phi\n", "$$\n", "\n", "\n", "Negative values of $\\omega^2$\n", "signal an unstable direction under the stated linearization. Zero values correspond to\n", "rigid motion or a mechanism unless supports should have removed them. Compare the lowest\n", "frequencies with analytic limits or a more refined mesh.\n", "\n", "## A verification ladder\n", "\n", "Begin with cases whose answers are known: a single spring, a uniform bar, or a symmetric\n", "beam. Then test assembly invariants. Element forces should appear as equal-and-opposite\n", "pairs, a free body's energy should be unchanged by rigid motion, and a supported static\n", "solution should balance forces and moments.\n", "\n", "Next refine the mesh and time step separately. Compare displacement at fixed physical\n", "locations, reaction forces, total elastic energy, and important natural frequencies.\n", "Convergence toward the same result is stronger evidence than an attractive deformed\n", "shape. Finally, vary uncertain material values and boundary assumptions. If the answer is\n", "very sensitive to a poorly known input, report that sensitivity rather than extra decimal\n", "places.\n", "\n", "Numerical elasticity is trustworthy when the discrete equations reflect the intended\n", "physics, the solver satisfies those equations, and independent tests show that the answer\n", "does not depend materially on arbitrary resolution choices.\n", "\n", "## A worked Python notebook\n", "\n", "Two springs of stiffness 100 newtons per metre are connected in series with one end fixed. The question for this calculation is precise: how does **end displacement**\n", "change when we change **applied end load**, while keeping the other choices fixed?\n", "The earlier sections explain the physical relationship. Here we turn one part of that\n", "relationship into a calculation that you can run, inspect, and change in Google Colab.\n", "\n", "The linear system assumes small displacement and unchanged spring stiffness. Keep that sentence beside the results. A formula can return a number outside\n", "its useful range, and a plotting library will usually draw that number without asking\n", "whether the physical assumptions still hold. The selected interval is from\n", "0 to 20 N. It gives us a specific domain to discuss\n", "instead of leaving the graph's range to a software default.\n", "\n", "Use the notebook link above the lesson, run the setup cell, and then run the cells in\n", "order. The code below is also included there. NumPy handles the arrays of numerical\n", "values, SymPy keeps the equation symbolic, and Matplotlib draws the results. If these\n", "libraries or notebook cells are unfamiliar, start with\n", "[Your first physics notebook](https://tirta.co/a-byte-of-physics/your-first-physics-notebook).\n", "\n", "### State the relationship before choosing a graph\n", "\n", "In this small example, `q` stands for applied end load, expressed in N. The output is\n", "end displacement, expressed in m. The equation below uses the numerical values in those\n", "units. Any fixed coefficients therefore carry the units needed to make the relation\n", "consistent. A value supplied in another unit must be converted before it enters this\n", "expression.\n", "\n", "\n", "$$\n", "f(q) = \\frac{q}{50}\n", "$$\n", "\n", "\n", "This is a function of one changing input. It is not a statement that the physical\n", "system has only one variable. We deliberately hold the other inputs fixed so that one\n", "cause of variation is visible. Changing several inputs at once would make it harder\n", "to tell which change produced the output difference. Write down the held-fixed choices\n", "before trying a different experiment.\n", "\n", "At 10 N, the expression gives 0.2 m. This is our\n", "reference calculation. It is a prediction of the stated example, not a new measurement.\n", "Its displayed digits help us compare implementations. They should not be read as a\n", "claim that physical inputs are known to six significant figures." ] }, { "cell_type": "code", "id": "code-011", "metadata": {}, "source": [ "import numpy as np\n", "import sympy as sp\n", "import matplotlib.pyplot as plt\n", "\n", "q = sp.Symbol(\"q\", real=True)\n", "expression = sp.sympify(\"q/50\", locals={\"q\": q})\n", "reference_input = 10.0\n", "reference_output = float(expression.subs(q, reference_input))\n", "print(\"Input:\", reference_input, \"N\")\n", "print(\"Predicted end displacement:\", reference_output, \"m\")\n" ], "execution_count": null, "outputs": [] }, { "cell_type": "markdown", "id": "markdown-012", "metadata": {}, "source": [ "`sp.Symbol` declares a mathematical symbol. It does not assign a numerical measurement\n", "to that symbol. `subs` then replaces the symbol with our chosen input, and `float`\n", "converts the resulting symbolic number into a standard Python numerical value.\n", "Keeping these steps separate lets us inspect the equation before numerical substitution.\n", "It also prevents a later plot from becoming the only surviving record of the model.\n", "\n", "### Compare a few cases you can explain\n", "\n", "Before drawing hundreds of points, calculate three. The lower endpoint gives\n", "0 m. The midpoint, at 10 N, gives\n", "0.2 m. The upper endpoint gives 0.4 m.\n", "Read these together with the assumptions: they describe different choices of applied end load\n", "within the same example, with the other inputs held fixed.\n", "\n", "These three values are a useful first check, but they do not establish the shape between\n", "them. Two endpoints can be equal even when the curve bends or oscillates in between.\n", "Likewise, a large change between endpoints does not tell us whether the relationship is\n", "linear. We need the equation and intermediate samples to answer those questions." ] }, { "cell_type": "code", "id": "code-013", "metadata": {}, "source": [ "evaluate = sp.lambdify(q, expression, modules=\"numpy\")\n", "inputs = np.array([0.0, 10.0, 20.0])\n", "outputs = np.asarray(evaluate(inputs), dtype=float)\n", "outputs = np.broadcast_to(outputs, inputs.shape)\n", "for input_value, output_value in zip(inputs, outputs):\n", " print(f\"{input_value:.6g} N -> {output_value:.6g} m\")\n", "assert np.all(np.isfinite(outputs))\n" ], "execution_count": null, "outputs": [] }, { "cell_type": "markdown", "id": "markdown-014", "metadata": {}, "source": [ "`lambdify` turns the symbolic expression into a numerical function. We explicitly select\n", "NumPy as its numerical backend so it can accept an array. The same expression now has\n", "two useful forms: one for algebra and one for calculating many cases. The assertion\n", "checks that the chosen inputs produce finite outputs. It would catch an infinity or an\n", "undefined numerical result, but a finite value can still be physically inappropriate.\n", "\n", "### Make a graph that answers the question\n", "\n", "The horizontal axis shows applied end load; the vertical axis shows end displacement. Every plotted\n", "point is a prediction of the same expression. We mark the reference case separately\n", "so that a reader can connect the earlier printed value to its place on the curve.\n", "The axis labels include units because a shape without a scale cannot tell us the size\n", "of a physical effect." ] }, { "cell_type": "code", "id": "code-015", "metadata": {}, "source": [ "input_grid = np.linspace(0.0, 20.0, 301)\n", "output_grid = np.asarray(evaluate(input_grid), dtype=float)\n", "output_grid = np.broadcast_to(output_grid, input_grid.shape)\n", "fig, ax = plt.subplots(figsize=(7, 4))\n", "ax.plot(input_grid, output_grid, label=\"Stated model\")\n", "ax.scatter([reference_input], [reference_output], label=\"Reference case\", zorder=3)\n", "ax.set_xlabel(\"Applied end load (N)\")\n", "ax.set_ylabel(\"End displacement (m)\")\n", "ax.legend()\n", "ax.grid(alpha=0.25)\n", "fig.tight_layout()\n", "plt.show()\n" ], "execution_count": null, "outputs": [] }, { "cell_type": "markdown", "id": "markdown-016", "metadata": {}, "source": [ "Look for a sign change, a turning point, or a region where the curve becomes steep.\n", "Connect each feature to the expression rather than to the colour of the line. If the\n", "curve has a minimum or maximum, changing the input in opposite directions may have\n", "similar effects on the output near that point. If it is nearly straight over a short\n", "interval, a local linear approximation may be useful even when the complete expression\n", "is nonlinear.\n", "\n", "Increasing the number of samples makes the line smoother and may reveal features that\n", "a coarse grid missed. It does not improve the physical inputs or enlarge the model's\n", "validity range. For quantities that can take only discrete values, use the line as a\n", "guide to the relationship and interpret only the allowed input values. A plotting\n", "function does not enforce such restrictions for you.\n", "\n", "### Measure the local sensitivity\n", "\n", "A derivative asks a more specific question than the whole graph: how quickly does\n", "end displacement change near the reference input? Its units are the output unit divided by\n", "the input unit. For this calculation, the derivative at 10 N is\n", "0.02 in those ratio units. Locally, the output increases as the\n", "input increases. This statement is local; the sign or size of the slope can be different\n", "elsewhere on the curve.\n", "\n", "\n", "$$\n", "f'(q) = \\frac{1}{50}\n", "$$\n", "\n", "\n", "For a small input change, multiply the derivative by that change to estimate the output\n", "change. Then compare that estimate with a fresh evaluation of the full expression.\n", "The distinction between these two calculations matters: the derivative gives a tangent\n", "approximation, while substitution evaluates the chosen expression at the new point." ] }, { "cell_type": "code", "id": "code-017", "metadata": {}, "source": [ "slope_expression = sp.diff(expression, q)\n", "slope_at_reference = float(slope_expression.subs(q, reference_input))\n", "changed_input = 10.1\n", "linear_prediction = reference_output + slope_at_reference * (\n", " changed_input - reference_input\n", ")\n", "full_prediction = float(evaluate(changed_input))\n", "print(\"Symbolic derivative:\", slope_expression)\n", "print(\"Tangent prediction:\", linear_prediction, \"m\")\n", "print(\"Full expression:\", full_prediction, \"m\")\n", "print(\"Approximation difference:\", full_prediction - linear_prediction, \"m\")\n" ], "execution_count": null, "outputs": [] }, { "cell_type": "markdown", "id": "markdown-018", "metadata": {}, "source": [ "Here the changed input is 10.1 N. The tangent estimate is\n", "0.202 m, while the full expression gives\n", "0.202 m. If the two values are close, it means the chosen\n", "input change is small enough for this local approximation at this point. It does not\n", "mean the entire model is linear. If the slope is zero, a quadratic or higher-order term\n", "may supply the leading change instead.\n", "\n", "This is also the beginning of uncertainty propagation. If the input has a small stated\n", "standard uncertainty, the magnitude of the slope times that uncertainty estimates one\n", "contribution to output uncertainty. Do not automatically use the same rule for a wide\n", "interval, a discontinuity, a turning point with zero first derivative, or several\n", "correlated uncertain inputs. Those situations need more of the model than one slope.\n", "\n", "### Check the derivative with numerical differences\n", "\n", "We can estimate the same slope from values on either side of the reference point.\n", "Subtract the lower output from the upper output and divide by twice the input step.\n", "Parentheses are essential: the entire output difference must be divided by the entire\n", "input interval. Without them, Python can execute a different mathematical expression\n", "while still producing a plausible number.\n", "\n", "The calculation below uses three decreasing step sizes. Each estimate is compared with\n", "the symbolic derivative, so the reference is independent of the numerical difference\n", "formula. Near an endpoint, it is acceptable here to evaluate a small extension only\n", "because the displayed mathematical expression is smooth there; such a check does not\n", "assert that extrapolated values describe additional physical cases." ] }, { "cell_type": "code", "id": "code-019", "metadata": {}, "source": [ "input_span = 20.0\n", "for fraction in [1e-2, 1e-3, 1e-4]:\n", " step = input_span * fraction\n", " upper = float(evaluate(reference_input + step))\n", " lower = float(evaluate(reference_input - step))\n", " numerical_slope = (upper - lower) / (2 * step)\n", " difference = numerical_slope - slope_at_reference\n", " print(f\"step={step:.6g}: slope={numerical_slope:.6g}, difference={difference:.6g}\")\n" ], "execution_count": null, "outputs": [] }, { "cell_type": "markdown", "id": "markdown-020", "metadata": {}, "source": [ "For a smooth function, reducing the step often reduces the approximation error at first.\n", "Very small steps can eventually make subtraction of nearly equal floating-point values\n", "less reliable. Therefore “smaller” is not an unlimited guarantee of “better.” Compare\n", "the observed behaviour over a useful range of steps instead of choosing the smallest\n", "number the computer can represent.\n", "\n", "An absolute slope difference is useful when the expected slope is near zero. A relative\n", "difference would divide by a very small reference and could look enormous even when the\n", "absolute discrepancy is tiny. Away from zero, a relative difference can help compare\n", "quantities with very different scales. The tolerance should reflect the question and\n", "the numerical method, rather than a universal decimal threshold copied into every test.\n", "\n", "### Change the example and preserve the reasoning\n", "\n", "Make one copy of the reference calculation before editing it. Choose a second value of\n", "applied end load inside the stated interval, predict whether end displacement will rise or fall,\n", "and then evaluate the expression. Place the prediction and the result in a text cell.\n", "If they disagree, identify whether the mistake was in the sign, the unit conversion,\n", "the algebra, or the assumption about how the curve behaves.\n", "\n", "Next choose a shorter plotting interval around the reference point. Compare the local\n", "curve with its tangent approximation. Then widen the interval again. The difference\n", "shows why a useful local estimate need not describe the whole range. Keep the axis\n", "labels and numerical scale visible when comparing the plots; automatic axis limits can\n", "make two different effects appear equally large.\n", "\n", "Finally, return to the physical limit stated at the start: The linear system assumes small displacement and unchanged spring stiffness. Describe one change\n", "to the setup that would violate that limit. State what new input, equation, or observation\n", "would be needed before trusting an answer in the changed situation. This last step\n", "connects the Python experiment to the physics instead of treating successful execution\n", "as the only criterion for a useful result.\n", "\n", "Before saving, restart the runtime and run all cells in order. A result that depends on\n", "an earlier, now-deleted input cell is not reproducible from the saved notebook. Keep\n", "unrounded values for later calculations, round only the reported output, and label any\n", "new data as measured, assumed, or generated. The graph and printed values should remain\n", "traceable to the same expression and the same choices of inputs.\n", "\n", "### Recover the output change by integrating the slope\n", "\n", "There is another way to check that the expression and its derivative describe the same\n", "relationship. Add up the small output changes predicted by the slope across the chosen\n", "input interval. In the continuous limit, this integral equals the output at the upper\n", "endpoint minus the output at the lower endpoint. The result has units of m:\n", "the input unit in the integration cancels the input unit in the denominator of the slope.\n", "\n", "This statement is about the mathematics of this particular relationship. It does not\n", "assign a new physical meaning to an arbitrary area under a graph. For example, an area\n", "under force versus distance represents work, while an area under a different pair of\n", "quantities may represent something else or have no standard name. Always carry the\n", "axis units through the calculation before naming an integral." ] }, { "cell_type": "code", "id": "code-021", "metadata": {}, "source": [ "slope_function = sp.lambdify(q, slope_expression, modules=\"numpy\")\n", "endpoint_change = float(evaluate(20.0) - evaluate(0.0))\n", "for points in [101, 201, 401]:\n", " grid = np.linspace(0.0, 20.0, points)\n", " slopes = np.asarray(slope_function(grid), dtype=float)\n", " slopes = np.broadcast_to(slopes, grid.shape)\n", " integrated_change = np.trapezoid(slopes, grid)\n", " print(\n", " points,\n", " \"integrated change:\",\n", " integrated_change,\n", " \"endpoint change:\",\n", " endpoint_change,\n", " )\n" ], "execution_count": null, "outputs": [] }, { "cell_type": "markdown", "id": "markdown-022", "metadata": {}, "source": [ "The trapezoidal rule joins neighbouring slope samples with straight segments and adds\n", "their signed areas. A negative contribution remains negative; replacing every slope\n", "by its magnitude would calculate a different quantity. Increasing the sample count\n", "provides a practical refinement check. If the values do not settle toward the endpoint\n", "difference, first inspect the interval, singularities, and array ordering.\n", "\n", "Do not conclude that two agreeing calculations prove the model is physically complete.\n", "Both use the same selected relationship, so an omitted effect would be omitted from\n", "both. Their agreement is evidence that the numerical tools implement that relationship\n", "consistently. Comparing with an independent experiment or a more appropriate physical\n", "model is a separate test.\n", "\n", "Keep this comparison in the notebook when you change the expression. It is particularly\n", "useful after adding a coefficient, changing a sign convention, or converting an input\n", "unit. A code cell that produces a smooth graph can conceal these mistakes; a second\n", "calculation of the same output change gives you another place to notice them.\n", "\n", "## Sources\n", "\n", "- [OpenStax, *University Physics Volume 1*, Section 12.3, \"Stress, Strain, and Elastic Modulus\"](https://openstax.org/books/university-physics-volume-1/pages/12-3-stress-strain-and-elastic-modulus).\n", "- [OpenStax, *University Physics Volume 1*, Section 15.1, \"Simple Harmonic Motion\"](https://openstax.org/books/university-physics-volume-1/pages/15-1-simple-harmonic-motion).\n", "- [MIT OpenCourseWare, *8.01SC Classical Mechanics*, Chapter 26, \"Elastic Properties of Materials\"](https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/resources/mit8_01scs22_chapter26/).\n", "- [Benjamin Crowell, *Light and Matter*, force, energy, and spring-model discussions](https://open.umn.edu/opentextbooks/textbooks/light-and-matter).\n", "- [Physics LibreTexts, \"Finite Element Method\"](https://phys.libretexts.org/Bookshelves/Mathematical_Physics_and_Pedagogy/Computational_Physics_(Chong)/12%3A_Finite_Element_Methods)." ] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "name": "python", "version": "3.12" }, "colab": { "name": "numerical-solutions-elastic-systems.ipynb", "provenance": [] }, "tirta": { "lesson": "https://tirta.co/a-byte-of-physics/numerical-solutions-elastic-systems", "source_sha256": "a7329b349c12e7394f08447340299e68191d6735b09e99e02a46296e00c7bcc8" } }, "nbformat": 4, "nbformat_minor": 5 }