test-bench-flexible-joint-adv/bending.org

Flexible Joint - Measurement of the Bending Stiffness

• Introduction
• Finite Element Model
• Setup
• Effect of Bending
• Computation of the bending stiffness
• Effect of Shear
• Effect of Torsion
• Full stroke measured displacement and applied force as a function of $H$
• Negligible bending of the supporting bar
• Conclusion

Introduction   ignore

The goal is to design a test bench to measure the bending stiffness of a flexible joint with 1% accuracy.

Finite Element Model

<<sec:fem>>

From the Finite Element Model, the stiffnesses and strokes of the flexible joint have been computed.

ka = 94e6; % Axial Stiffness [N/m]
ks = 13e6; % Shear Stiffness [N/m]
kb = 5;    % Bending Stiffness [Nm/rad]
kt = 260;  % Torsional Stiffness [Nm/rad]

Fa = 469;    % Axial Force before yield [N]
Fs = 242;    % Shear Force before yield [N]
Fb = 0.118;  % Bending Force before yield [Nm]
Ft = 1.508; % Torsional Force before yield [Nm]

Xa = Fa/ka; % Axial Stroke before yield [m]
Xs = Fs/ks; % Shear Stroke before yield [m]
Xb = Fb/kb; % Bending Stroke before yield [rad]
Xt = Ft/kt; % Torsional Stroke before yield [rad]

	Stiffness [N/um]	Max Force [N]	Stroke [um]
Axial	94	469	5
Shear	13	242	19

	Stiffness [Nm/rad]	Max Torque [Nmm]	Stroke [mrad]
Bending	5	118	24
Torsional	260	1508	6

Setup

<<sec:setup>>

Let's say a force is applied on top of the flexible joint with a distance $H_F$ with the joint's center.

The displacement of the flexible joint is also measure at an height $H_D$ from the joint's center.

Figure caption

Effect of Bending

<<sec:bending>>

The torque applied is:

\begin{equation} M_b = F \cdot H_F \end{equation}

The flexible joint is experiencing a rotation $R_b$ due to the torque $M_b$:

\begin{equation} R_b = \frac{M_b}{k_b} = \frac{F \cdot H_F}{k_b} \end{equation}

This rotation is then measured by the displacement sensor. The measured displacement is:

\begin{equation} D_b = H_D \tan(R_b) = H_D \tan\left( \frac{F \cdot H_F}{k_b} \right) \label{eq:bending_meaured_disp} \end{equation}

Computation of the bending stiffness

From equation eqref:eq:bending_meaured_disp, we can compute the bending stiffness:

\begin{equation} k_b = \frac{\tan^{-1}\left( \frac{D_b}{H_D} \right)}{F \cdot H_F} \end{equation}

And therefore, to precisely measure $k_b$, we need to:

• precisely measure the motion $D_b$
• precisely measure the applied force $F$
• precisely know the height from the flexible joint's center to the measurement point $H_D$
• precisely know the height from the flexible joint's center to the force application point $H_F$

If there are estimation errors for $H_D$ or $H_F$ as shown in Figure fig:bending_effect_error_vertical, this will induce an error for the estimation of the stiffness.

For 1% accuracy estimation of $k_b$, we can write the following approximate requirements:

	Accuracy
Force Measurement	1%
Displacement Measurement	1%
$H_D$	1%
$H_F$	1%

Error in the estimation of the height of the force sensor and displacement sensor

Effect of Shear

<<sec:shear>>

The effect of Shear on the measured displacement is simply:

\begin{equation} D_s = \frac{F}{k_s} \end{equation}

We would like to have this displacement much smaller than the displacement induced by the bending effects:

\begin{equation} D_b \gg D_s \end{equation}

Which is equivalent as to have:

\begin{equation} H_D \tan\left( \frac{F \cdot H_F}{k_b} \right) \gg \frac{F}{k_s} \end{equation}

Here to simplify, we suppose $FH_F/k_b \ll 1$ (which is the case in practice), and we suppose $H_D = H_F = H$.

The obtained condition is then:

\begin{equation} H \gg \frac{k_b}{k_s} \end{equation}

	10% error	1% error	0.1% error
$H\,[mm]$	6	62	620

In order to limit the effect of shear of less than 1%, the height from the joint's center to the force application point and to the measurement point should be larger than 62mm.

H = 62e-3; % [m]

Effect of Torsion

<<sec:torsion>>

If the application force is not aligned with the vertical axis of the flexible joint, this will induce a torsion motion that will induce a measurement error (Figure fig:bending_effect_torsion).

Horizontal position error of the force sensor and displacement sensor

Let's note the offset of the force sensor $\epsilon_{F,y}$ and the offset of the measurement point $\epsilon_{D,y}$. The vertical torque (torsion) will be equal to:

\begin{equation} M_t = F \cdot \epsilon_{F,y} \end{equation}

And the induced torsion:

\begin{equation} R_t = \frac{M_t}{k_t} = \frac{F \cdot \epsilon_{F,y}}{k_t} \end{equation}

The effect on the measured displacement is:

\begin{equation} D_t = \epsilon_{D,y} \tan \left( R_t \right) = \epsilon_{D,y} \tan\left( \frac{F \cdot \epsilon_{F,y}}{k_t} \right) \end{equation}

And we would like to have:

\begin{equation} D_b \gg D_t \end{equation}

Which is equivalent as to have:

\begin{equation} H_D \tan\left( \frac{F \cdot H_F}{k_b} \right) \gg \epsilon_{D,y} \tan\left( \frac{F \cdot \epsilon_{F,y}}{k_t} \right) \end{equation}

Supposing $H_F = H_D = H$ and $\epsilon_{F,y} = \epsilon_{D,y} = \epsilon_{y}$, the condition becomes:

\begin{equation} \epsilon_{y} \ll H \sqrt{\frac{k_t}{k_b}} \end{equation}

	10% error	1% error	0.1% error
$\epsilon_y\,[mm]$	44.7	4.5	0.4

For 1% error, the lateral positioning errors $\epsilon_y$ for both the force sensor and the displacement sensor should be less than 4.5mm.

Full stroke measured displacement and applied force as a function of $H$

Applying a force with a large height $H$ means the induced rotation (for constant force) will be larger. This also means that the measured displacement $D_b$ will also be larger.

Note that we here suppose the force axis is co-linear with the measurement axis ($H_F = H_D = H$).

Let's compute:

• $D_b$ as a function of $H$ \[ D_b \approx H \tan (R_b) \]
• the applied force $F_{\text{max}}$ to induce the maximum rotation as a function of $H$ \[ F_{\text{max}} \approx \frac{X_b \cdot k_b}{H} \]

Applied force $F_{\text{max}}$ and measured displacement $D_b$ as a function of $H$

With an offset of 62mm, we obtained values shown in Table tab:disp_force_range.

$D_b\,[mm]$	$F_m\,[N]$
1.5	1.9

Negligible bending of the supporting bar

This should be confirmed with FEM.

Conclusion

	Range	Accuracy
Force Measurement	2 N	1% = 0.02 N
Displacement Measurement	1.5 mm	1% = 15 um

	Value	Precision	Comment
$H_D$	62mm		For negligible Shear
$H_F$	62mm		Same
$\epsilon_y$	0	4.5mm	For negligible Torsion
$\epsilon_z$	0	1% = 0.6mm	For torque estimation precision

Load cells:

• https://www.te.com/usa-en/product-CAT-FLS0020.html?q=&n=510872&d=681051%20564247&type=products&samples=N&inStoreWithoutPL=false&instock=N#mdp-tabs-content
• https://www.digikey.com/en/products/detail/honeywell-sensing-and-productivity-solutions/FSS005WNSB/6056404

Displacement sensors:

• http://www.vaco-france.com/outillage-de-metrologie/comparateur-a-cadran-de-precision.htm
• https://fr.rs-online.com/web/p/comparateurs/1940101/