Machine compliance ($C_m$)

Machine compliance is the extra load-path displacement per applied force attributed to the apparatus in a specified measurement model. Its units are $\mathrm{m}{\mathrm{N}}^{-1}$, the reciprocal units of a spring constant. Here $C_m$ includes the frame, load cell, adapters and grips insofar as their deformation contributes to the recorded crosshead displacement $x$. It is a property of that configuration and readout, rather than a universal constant of the testing machine.

Assume a uniform specimen with initial length $L$, constant $A_0$ and Young's modulus $E$, small axial elastic deformation, quasistatic loading, no slip, and a linear, length-independent apparatus response over the chosen force window. Equilibrium makes the same force act through specimen and apparatus, while their extensions add. The two springs in series therefore giveThe offset $x_0$ describes the displacement origin after seating; the slope $q$ describes total compliance. Fitting an intercept avoids forcing an arbitrary zero through the origin, but cannot repair slip or a curved response.

If all of $x-x_0$ is incorrectly attributed to specimen extension, the inferred modulus isFor positive $C_m$, this is too small. The fractional bias is most serious for a short, thick, stiff specimen because its own compliance $L/(E{A}_0)$ is small. A local extensometer instead has $d(\Delta L_g)/dF=L_g/(E{A}_0)$ in this ideal model, excluding deformation outside its contacts.

A shaped specimen requires more care. With varying initial cross-sectional area $A(s)$ and uniform $E$, its axial compliance is $\int ds/[EA(s)]$ in the one-dimensional approximation. Shoulder and end deformation must be included explicitly or remain identical between tests and be absorbed into an effective intercept. Changing only the extensometer gauge length on an unchanged specimen does not change its total compliance and cannot perform a two-length machine compliance separation.

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.