Two-length machine compliance separation ($E=\frac{L_2-L_1}{A_0(q_2-q_1)}$)

Prepare two uniform specimens with the same material, direction and $A_0$, but different deforming lengths $L_1,L_2$. Keep grip engagement, adapters, loading window and measurement configuration equivalent. Determine $q_i=d{x}_i/dF$ separately by linear regression with its own displacement intercept. Subtracting the two series equations eliminates the common machine compliance:This is an inverse problem: one length supplies only one equation for two unknowns, while two distinct lengths separate the length-dependent specimen contribution from the constant contribution. In a graph of compliance against length, the slope is $1/(E{A}_0)$ and the vertical intercept is $C_m$. It is the compliance, not $k_{\mathrm{obs}}$, that is linear in $L$.

The dimensional analysis is revealing: $[q_2-q_1]=m/N$, so $[(L_2-L_1)/(A_0(q_2-q_1))]=N/{\mathrm{m}}^2$. For $L_2>L_1$, the ideal model requires $q_2>q_1$. A negative fitted $C_m$ challenges the common-intercept assumptions or the measurements. Two lengths determine two parameters exactly, so they cannot test the assumed straight line; three or more lengths allow regression residuals to reveal departures. Nearly equal lengths amplify slope errors through the small difference $q_2-q_1$. If the actual apparatus compliances differ by $\Delta C_m$, the inferred length slope becomes $1/(E{A}_0)+\Delta C_m/(L_2-L_1)$.

The following values are stipulated arithmetic inputs, not measurements of a material. Take $A_0=2.00\mathrm{m}{\mathrm{m}}^2$ 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 displacementsThe two different initial offsets deliberately test the use of intercepts. Linear regression of $x_i=a_i+q_{i}F$ gives $a_1=0.005\mathrm{mm}$, $a_2=0.008\mathrm{mm}$ andTheir reciprocal spring constants are $k_{\mathrm{obs},1}=2.85714kN/mm$ and $k_{\mathrm{obs},2}=1.66667kN/mm$. The compliance-versus-length slope isso the recovered Young's modulus and machine compliance areThe corresponding apparatus spring constant is $1/C_m=10.0kN/mm$.

Check the reconstruction at $400\mathrm{N}$. The machine contribution is $C_{m}F=0.040\mathrm{mm}$, 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 $200N/m{\mathrm{m}}^2=200\mathrm{MPa}$. 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.