The workflow

The Golden Quad prescribes a two-step method that separates the part of the work that requires an engineer from the part that does not.

The first step is to write down every relation in Figure 8.9.1 for the problem at hand, one equation at a time, without eliminating anything. Choosing which cuts to make, which supports constrain what, and which material model applies is the mechanics, and it is where modelling judgment lives.

The second step is to solve the resulting system. This is not mechanics. It is algebra, it is mechanical, it is where hand calculations go wrong, and it belongs to the computer. We therefore never eliminate unknowns by substitution on paper. We collect the equations in a list and hand them to SymPy.

Applied to a rod in simple tension, the three relations are one line each,

\[ \left\lbrace \begin{array}{l} \sigma = F/A\\ \sigma = E\varepsilon\\ \delta = L\varepsilon \end{array}\right. \qquad\Longrightarrow\qquad \boxed{\delta = \frac{FL}{EA}} \tag{8.9.1}\]

which is the elementary formula for the elongation of a bar. The group \(EA\) is the axial stiffness, the counterpart of the bending stiffness \(EI\) of 8.14.6, and \(EA/L\) is the spring constant of the bar.

For bodies in series the same force passes through each and the deformations add,

\[ \delta = \sum_i \delta_i = \sum_i \frac{N_i L_i}{E_i A_i} \tag{8.9.2}\]

Example 1: A heated rod between rigid walls

A linear elastic rod is fastened between two rigid walls and then heated by \(\Delta T\). Determine the stress in the rod and the force it exerts on the walls.

No external load is applied, which makes this a problem that pure statics cannot touch: equilibrium alone gives \(N = P\) and stops. The extra information comes from the deformation relation, because the walls impose \(\delta = 0\) while the material wants to expand.

Writing down all five relations without eliminating anything gives

\[ \left\lbrace \begin{array}{ll} N - P = 0,\quad \sigma = N/A & \text{Equilibrium}\\ \varepsilon_{\text{tot}} = \delta/L, \quad \delta = 0 & \text{Deformation and boundary condition}\\ \sigma = E\varepsilon_{\text{elast}} & \text{Material, elastic relation}\\ \varepsilon_T = \alpha\Delta T & \text{Thermal strain}\\ \varepsilon_{\text{tot}} = \varepsilon_{\text{elast}} + \varepsilon_T & \text{Total strain} \end{array}\right. \]

which is seven equations in the seven unknowns \(N\), \(P\), \(\sigma\), \(\delta\), \(\varepsilon_{\text{tot}}\), \(\varepsilon_{\text{elast}}\) and \(\varepsilon_T\).

N, P, sigma, delta = sp.symbols('N P sigma delta', real=True)
eps_tot, eps_el, eps_T = sp.symbols('varepsilon_tot varepsilon_elast varepsilon_T', real=True)
E, A, L, alpha, dT = sp.symbols('E A L alpha Delta_T', real=True)

eqs = [sp.Eq(N - P, 0),
       sp.Eq(sigma, N/A),
       sp.Eq(eps_tot, delta/L),
       sp.Eq(delta, 0),
       sp.Eq(sigma, E*eps_el),
       sp.Eq(eps_T, alpha*dT),
       sp.Eq(eps_tot, eps_el + eps_T)]

sol = sp.solve(eqs, [N, P, sigma, delta, eps_tot, eps_el, eps_T], dict=True)[0]
sol | la

The rod pushes on the walls with \(P = -EA\alpha\Delta T\) and carries a stress \(\sigma = -E\alpha\Delta T\). Both are negative for a positive temperature change, which is right: heating a rod that cannot expand puts it in compression. A dimensional check passes, since \(E\alpha\Delta T\) has the units of \(E\).

The magnitude deserves a moment. For steel, \(E\alpha \approx 210\,000 \cdot 12\cdot 10^{-6} = 2.52~\text{MPa}\) per degree, so a forty degree summer swing produces \(100~\text{MPa}\) in a fully restrained member, a third of the way to yield with no load applied whatsoever. This is why rails buckle, why bridges have expansion joints, and why a pipe run between two fixed anchors needs a bend in it.

Observe also that the answer is independent of the length. A short restrained rod and a long one develop the same stress, because both the free expansion and the elastic strain needed to cancel it scale with \(L\).

Example 2: Two rods of different materials

The same walls, but now two rods of different materials and cross sections joined in series between them, heated by \(\Delta T\).

Equilibrium of the three free bodies shows that the same normal force passes through both rods, since nothing is applied at the junction. The deformation relation is where the problem is decided: each rod is free to change length, but their sum must be zero because the walls have not moved.

N1, N2, S1, S2 = sp.symbols('N_1 N_2 S_1 S_2', real=True)
sig1, sig2, d1, d2 = sp.symbols('sigma_1 sigma_2 delta_1 delta_2', real=True)
eE1, eE2, eT1, eT2, et1, et2 = sp.symbols(
    'varepsilon_E1 varepsilon_E2 varepsilon_T1 varepsilon_T2 varepsilon_tot1 varepsilon_tot2', real=True)
E1, E2, A1, A2, L1, L2, a1, a2 = sp.symbols('E_1 E_2 A_1 A_2 L_1 L_2 alpha_1 alpha_2', positive=True)

eqs = [sp.Eq(-S1 + N1, 0), sp.Eq(-N1 + N2, 0), sp.Eq(-N2 + S2, 0),   # equilibrium
       sp.Eq(sig1, N1/A1), sp.Eq(sig2, N2/A2),
       sp.Eq(d1, L1*et1), sp.Eq(d2, L2*et2), sp.Eq(d1 + d2, 0),      # deformation and BC
       sp.Eq(sig1, E1*eE1), sp.Eq(sig2, E2*eE2),                      # material
       sp.Eq(eT1, a1*dT), sp.Eq(eT2, a2*dT),                          # thermal strain
       sp.Eq(et1, eE1 + eT1), sp.Eq(et2, eE2 + eT2)]                  # total strain

unknowns = [N1, N2, S1, S2, sig1, sig2, d1, d2, eE1, eE2, eT1, eT2, et1, et2]
sol = sp.solve(eqs, unknowns, dict=True)[0]

ltx(r"N &=", sp.simplify(sol[N1]),
    r"\\ \sigma_1 &=", sp.simplify(sol[sig1]),
    r"\\ \sigma_2 &=", sp.simplify(sol[sig2]), aligned=True)

Fourteen equations, fourteen unknowns, and not one of them eliminated by hand. The force is

\[ N = -\frac{\Delta T\left(L_1\alpha_1 + L_2\alpha_2\right)} {\dfrac{L_1}{E_1A_1} + \dfrac{L_2}{E_2A_2}} \]

with the numerator collecting how much the assembly wants to expand and the denominator collecting how easily it gives way. Two rods of the same material reproduce Example 1, and a very compliant second rod, either long or thin or soft, drives \(N\) towards zero because it absorbs the expansion of the first. That is the principle behind an expansion loop.

Where the diagram comes from

The English name used here is not standard. In Swedish teaching the figure is den gyllene kvadraten, and students who have met it elsewhere will recognise the four corners under that name, although its origin is hard to pin down: Swedish discussions of it attribute the term to textbook authors without naming one.

The diagram itself is older and more general than solid mechanics. Tonti [1] observed that the same square recurs throughout physics, with a configuration variable and a source variable joined by a chain of operators and a constitutive law across the middle, and he used it to classify a large number of field theories. The 1972 paper is hard to obtain; the construction is given book length treatment in [2]. What we have drawn for a bar is one cell of that classification.

Under Tonti’s name the diagram is standard in finite element teaching. Felippa [3] presents the governing equations of exactly our bar as a strong-form Tonti diagram, with unknown quantities in boxes of one colour and given data in another, and the field equations and boundary conditions drawn as the lines between them. He also uses a discrete version for what he calls mechanics-of-materials elements, which is the same square with matrices in place of derivatives. That is the version we meet in Chapter 8.12, where \(\bm A\), \(\bm S^e\) and \(\bm A^\mathsf{T}\) occupy the three sides. The three-relation grouping appears without the square in Ottosen and Petersson [4], who organise the entire finite element method around it.

What survives every renaming is the asymmetry between the three sides. Equilibrium and kinematics are statements about the world that hold whatever the bar is made of, and only the middle link changes when the material does. Replacing Hooke’s law by a plastic or a viscoelastic model redraws one side and leaves the other three untouched, which is why a single finite element code can handle all of them.

The Golden Quad and the continuum

Nothing in Figure 8.9.1 is restricted to one dimension. The four corners and three arrows are the same in the continuum, with each scalar replaced by its tensor counterpart, as we set out at the start of the chapter. A one-dimensional problem allows us to write the relations as scalar equations and solve them algebraically; a general problem forces the same relations into partial differential form and we solve them with the finite element method. The bookkeeping changes, the structure does not.

What we gain from the picture is a habit. Faced with an unfamiliar problem, ask which of the three relations are missing and write them down. A problem that statics alone cannot solve, meaning a statically indeterminate one, is always a problem in which the deformation relation carries information that equilibrium does not, and the missing equation is a statement about compatibility of displacements. Both examples above were of exactly this kind, and both were solved the moment the geometric condition was written down alongside the equilibrium equations.

The next chapter takes the same three relations and, instead of applying them to a finite bar, applies them to an infinitesimal slice. The result is the differential equation the bar formula 8.9.1 is a special solution of.