Flexible Joint - Measurement of the Bending Stiffness

1. Finite Element Model
2. Setup
3. Effect of Bending
4. Computation of the bending stiffness
5. Effect of Shear
6. Effect of Torsion
7. Full stroke measured displacement and applied force as a function of \(H\)
8. Negligible bending of the supporting bar
9. Conclusion

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

1 Finite Element Model

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

2 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 1: Figure caption

3 Effect of 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}

4 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 2, 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%

Figure 2: Error in the estimation of the height of the force sensor and displacement sensor

5 Effect of 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}

Table 1: Height \(H\) to have less than certain amount of error due to shear effects
	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]

6 Effect of 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 3).

Figure 3: 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}

Table 2: Maximum lateral position error \(\epsilon_y\) to have less than certain amount of error due to torsion effects
	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.

7 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 \(H\) \[ F_{\text{max}} \approx \frac{X_b \cdot k_b}{H} \]

Figure 4: 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 3.

Table 3: Maximum displacement and maximum applied force for \(H = 62\,[mm]\)
\(D_b\,[mm]\)	\(F_m\,[N]\)
1.5	1.9

8 Negligible bending of the supporting bar

This should be confirmed with FEM.

9 Conclusion

Table 4: Conclusions in terms of forces and displacement measurements
	Range	Accuracy
Force Measurement	2 N	1% = 0.02 N
Displacement Measurement	1.5 mm	1% = 15 um

Table 5: Conclusions in terms of required positioning accuracy
	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:

Displacement sensors: