= Worked two-length tensile compliance example
The following values are stipulated arithmetic inputs, not measurements of a material. Take $A_0=2.00\ \mathrm{mm^2}$ and lengths $L_1=50.0\ \mathrm{mm}$, $L_2=100.0\ \mathrm{mm}$. At <forces> $F=(0,100,200,300,400)\ \mathrm{N}$, stipulate <crosshead displacements>
$$
\begin{aligned}
x_1&=(0.005,0.040,0.075,0.110,0.145)\ \mathrm{mm},\\
x_2&=(0.008,0.068,0.128,0.188,0.248)\ \mathrm{mm}.
\end{aligned}
$$
The two different initial offsets deliberately test the use of intercepts. <Linear regression> of $x_i=a_i+q_iF$ gives $a_1=0.005\ \mathrm{mm}$, $a_2=0.008\ \mathrm{mm}$ and
$$
q_1=3.50\times10^{-4}\ \mathrm{mm/N},\qquad
q_2=6.00\times10^{-4}\ \mathrm{mm/N}.
$$
Their reciprocal <spring constants> are $k_{\mathrm{obs},1}=2.85714\ \mathrm{kN/mm}$ and $k_{\mathrm{obs},2}=1.66667\ \mathrm{kN/mm}$. The compliance-versus-length slope is
$$
m=\frac{6.00\times10^{-4}-3.50\times10^{-4}}{100.0-50.0}\ \mathrm{N^{-1}}
=5.00\times10^{-6}\ \mathrm{N^{-1}},
$$
so the recovered <Young's modulus> and <machine compliance> are
$$
E=\frac1{A_0m}=100000\ \mathrm{N/mm^2}=100\ \mathrm{GPa},\qquad
C_m=q_1-mL_1=1.00\times10^{-4}\ \mathrm{mm/N}.
$$
The corresponding apparatus <spring constant> is $1/C_m=10.0\ \mathrm{kN/mm}$.
\Image[/tensile-two-length-compliance-fit.png]
{title=Stipulated displacement fits separate length-dependent compliance from apparatus compliance}
{description=Original calculated example, not experimental material data. Left: independent linear displacement fits at lengths 50 and 100 millimetres retain their different offsets. Right: the line through the two compliance values has slope 1/(EA0) and intercept Cm. The dashed segment extrapolates the model to zero length; it is not a third observation.}
Check the reconstruction at $400\ \mathrm{N}$. The machine contribution is $C_mF=0.040\ \mathrm{mm}$, and the two specimen contributions are $0.100\ \mathrm{mm}$ and $0.200\ \mathrm{mm}$. Adding the respective offsets gives the stipulated $0.145\ \mathrm{mm}$ and $0.248\ \mathrm{mm}$. An <extensometer> spanning $L_g=25.0\ \mathrm{mm}$ on either ideal uniform specimen would instead read $0.050\ \mathrm{mm}$ extension: <engineering strain> $0.00200$, at <engineering stress> $200\ \mathrm{N/mm^2}=200\ \mathrm{MPa}$. Their ratio is again $100\ \mathrm{GPa}$. Ignoring <machine compliance> gives apparent moduli $71.43\ \mathrm{GPa}$ and $83.33\ \mathrm{GPa}$ even after removing the offsets. No <yield point> or real material identity is implied by these stipulated linear inputs.
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