Detecting, localizing, and quantifying damage using two-dimensional sensing sheet: lab test and field application

Abstract

Damage to structures in the form of cracks could reduce safety and induce high maintenance cost. Structural health monitoring (SHM) is increasingly employed to detect damage in the structure and inform the stakeholders in a timely manner to allow rehabilitation actions. Reliable crack detection, localization, and quantification are, hence, extremely important. To achieve this goal, a dense network of sensors is often required. Damages even a meter away from sensors are often unable to be detected reliably by a sensing system. Creating a dense network of sensors using the commonly used point sensors (e.g., strain gages) or distributed one-dimensional sensors (e.g., fiber-optic sensors) is expensive and often practically impossible. Sensing sheet is a distributed two-dimensional thin-film sensor comprising of a dense array of resistive strain gage units developed at Princeton University. Based on the principles of large-area electronics (LAE), this thin-film sensor provides an affordable solution to reliably detect and localize damage. This paper derives analytical models for damage detection, localization, and quantification based on sensing sheet. Laboratory experiments are performed by creating artificial damage to verify these models and highlight their uses. Further, the damage quantification algorithm is used to estimate the crack opening in a shrinkage crack on the foundation of the pedestrian bridge at Princeton University. Finally, the results and future research directions are discussed.

Appendices

Appendix A

In this section step-by-step derivation of the change in voltage ratio $\Delta V_{r}$, strain measured by the sensing sheet $\varepsilon$ and the strain ratio $\omega$ for the four cases are detailed.

1.1 Preliminary

Consider the case of a single resistor $R_{3}$ through which a crack passes as shown in Fig. 

Fig. 20

A preliminary case where the crack only passes through resistor $R_{3}$

20. Let the number of serpentines intersected by the crack ${\mathcal{C}}$ be $n_{{\mathcal{C}}}$. Let $\alpha_{{\mathcal{C}}}$ denote the ratio of serpentines intersected by the crack to the total number of serpentines $n$. Let the crack opening be denoted by $w_{{\mathcal{C}}}$ and the ratio $\frac{{w_{{\mathcal{C}}} }}{{l_{r} }} = \psi$.

The total length of the serpentine after crack opening is given by equation A1.

$$l_{f} = (n(1 - \alpha_{c} ) + 1)l_{r} + n\alpha_{c} (l_{r} + w_{{\mathcal{C}}} ) = (n + 1)l_{r} + n\alpha_{c} w_{{\mathcal{C}}}$$

(A1)

The final resistance can then be calculated as

$$R_{3,f} = \rho \frac{{l_{f} }}{{A_{f} }} = \rho \frac{{\left( {(n + 1)l_{r} + n\alpha_{c} w_{{\mathcal{C}}} } \right)}}{{A\left( {1 - \nu \frac{{n\alpha_{c} w_{{\mathcal{C}}} }}{{\left( {n + 1} \right)l_{r} }}} \right)^{2} }}$$

(A2)

On rearranging the terms, we obtain the equation A3.

$$R_{3,f} = \rho \frac{{l_{f} }}{{A_{f} }} = R\frac{{\left( {1 + \frac{{n\alpha_{c} \psi }}{{\left( {n + 1} \right)}}} \right)}}{{\left( {1 - \nu \frac{{n\alpha_{c} \psi }}{{\left( {n + 1} \right)}}} \right)^{2} }} = R\frac{{\left( {1 + \alpha_{c} \psi - \frac{{\alpha_{c} \psi }}{{\left( {n + 1} \right)}}} \right)}}{{\left( {1 - 2\nu \alpha_{c} \psi + \frac{{\nu \alpha_{c} \psi }}{{\left( {n + 1} \right)}} + \nu^{2} \frac{{n^{2} \alpha_{c}^{2} \psi^{2} }}{{\left( {n^{2} + 1} \right)}}} \right)}},$$

(A3)

where $R$ is the initial resistance of the resistor $R_{3}$ and $\psi$ was defined before. Noting that $w_{{\mathcal{C}}} < < l_{r}$, hence $\psi \ll 1$, and $n \ge 50$ we can drop terms such as $\frac{{\alpha_{c} \psi }}{{\left( {n + 1} \right)}}$ and $\nu^{2} \frac{{n^{2} \alpha_{c}^{2} \psi^{2} }}{{\left( {n^{2} + 1} \right)}}$. Further using the binomial expansion for $\frac{1}{1 - x}$ for $x \ll 1$ we get the final resistance in terms of crack opening and proportion of resistor under the influence of crack as given in equation A4.

$$R_{3,f} = R\frac{{\left( {1 + \alpha_{c} \psi } \right)}}{{\left( {1 - 2\nu \alpha_{c} \psi } \right)}} = R\left( {1 + \alpha_{c} \psi } \right)\left( {1 + 2\nu \alpha_{c} \psi } \right)$$

(A4)

Subtracting the original strain from the final strain gives the change in resistance due to the crack opening $w_{{\mathcal{C}}}$.

$$\Delta R_{3} = R_{3} \left( {\left( {1 + \alpha_{c} \psi } \right)\left( {1 + 2\nu \alpha_{c} \psi } \right) - 1} \right)$$

(A5)

Rearranging the terms gives change in resistance with respect to the original resistance of the unit resistor as shown in equation A6.

$$\frac{{\Delta R_{3} }}{{R_{3} }} = \left( {1 + \alpha_{c} \psi } \right)\left( {1 + 2\nu \alpha_{c} \psi } \right) - 1$$

(A6)

This change in resistance results to a voltage drop which is governed by the Eq. 3 in Sect. 2.1. As stated before, the change in resistance in the remaining resistors,$R_{1} ,\;R_{2} ,\;R_{4}$, would be negligible compared to this and hence those changes are approximately 0.

Hence the change in voltage drop for the preliminary scenario would be given by equation A5.

$$\Delta V_{r} = \frac{1}{2} - \frac{{\left( {1 + \alpha_{c} \psi } \right)\left( {1 + 2\nu \alpha_{c} \psi } \right)}}{{1 + \left( {1 + \alpha_{c} \psi } \right)\left( {1 + 2\nu \alpha_{c} \psi } \right)}}$$

(A7)

This change in voltage drop is related to the strain measured by the sensing sheet through equation A6 (For complete derivation, refer [51]).

$$\varepsilon = - \frac{{2\Delta V_{r} }}{{{\text{GF}}\left[ {(1 + \nu ) + \Delta V_{r} \left( {1 - \nu } \right)} \right]}}$$

(A8)

On substituting the drop in voltage in equation A6 we obtain the strain:

$$\varepsilon = - \frac{{2\left( {1 - \left( {1 + \alpha_{c} \psi } \right)\left( {1 + 2\nu \alpha_{c} \psi } \right)} \right)}}{{{\text{GF}}\left[ {(3 + \nu ) + \left( {1 + \alpha_{c} \psi } \right)\left( {1 + 2\nu \alpha_{c} \psi } \right)\left( {3\nu + 1} \right)} \right]}}.$$

(A9)

In the above equation for a given value of $\varepsilon$ and $\alpha_{c}$, $w_{{\mathcal{C}}}$ can be computed by solving the above equations. Finally, using the baseline strain measurement from Eq. 9 in Sect. 2.3 to obtain the voltage ratio.

$$\omega = \frac{{2\left( {1 - \left( {1 + \alpha_{c} \psi } \right)\left( {1 + 2\nu \alpha_{c} \psi } \right)} \right)\left[ {1 + \nu \left( {1 + \psi } \right)\left( {1 + 2\nu \psi } \right)} \right]}}{{\psi \left( {2\nu \left( {\psi + 1} \right) + 1} \right)\left[ {(3 + \nu ) + \left( {1 + \alpha_{c} \psi } \right)\left( {1 + 2\nu \alpha_{c} \psi } \right)\left( {3\nu + 1} \right)} \right]}}$$

(A10)

The above approach holds for the cases where a combination of resistors is intersected by a crack as the change in resistance of individual resistor can be calculated as before and substituted in Eq. 3 of Sect. 2.1 to obtain the change in voltage resistance and compute the strain and strain ratios for various cases in Sect. 2.3.

1.1.1 Case 1: the crack intersects resistors R ₂ and R ₃ only

The case 1 is shown in Fig. 5c. Based on the preliminary calculations shown, the ratio of change of resistances in $R_{2}$ if the proportion of crack being affected in $\alpha_{2}$ is given by:

$$\frac{{\Delta R_{2} }}{{R_{2} }} = \left( {1 + \alpha_{2} \psi } \right)\left( {1 + 2\nu \alpha_{2} \psi } \right) - 1.$$

(A11)

While the values of ratio for $R_{3}$ for a proportion of resistor being affected by crack given by $\alpha_{3}$ is shown in equation A12.

$$\frac{{\Delta R_{3} }}{{R_{3} }} = \left( {1 + \alpha_{3} \psi } \right)\left( {1 + 2\nu \alpha_{3} \psi } \right) - 1$$

(A12)

Substituting the values of these relative change in resistances in Eq. 3 to obtain the voltage drop measured by the sensing system for case 1.

$$\Delta V_{r} = \frac{1}{{1 + \left( {1 + \alpha_{2} \psi } \right)\left( {1 + 2\nu \alpha_{2} \psi } \right)}} - \frac{{\left( {1 + \alpha_{3} \psi } \right)\left( {1 + 2\nu \alpha_{3} \psi } \right)}}{{1 + \left( {1 + \alpha_{3} \psi } \right)\left( {1 + 2\nu \alpha_{3} \psi } \right)}}$$

(A13)

One can substitute this value in equation A8 to obtain the strain recorded by the sensing system:

$$\varepsilon = - \frac{{2\left( {\frac{1}{{\left( {1 + \alpha_{2} \psi } \right)\left( {1 + 2\nu \alpha_{2} \psi } \right) + 1}} - \frac{{\left( {1 + \alpha_{3} \psi } \right)\left( {1 + 2\nu \alpha_{3} \psi } \right)}}{{\left( {1 + \alpha_{3} \psi } \right)\left( {1 + 2\nu \alpha_{3} \psi } \right) + 1}}} \right)}}{{{\text{GF}}\left( {\left( {1 + \nu } \right) + \left( {1 - \nu } \right)\left( {\frac{1}{{\left( {1 + \alpha_{2} \psi } \right)\left( {1 + 2\nu \alpha_{2} \psi } \right) + 1}} - \frac{{\left( {1 + \alpha_{3} \psi } \right)\left( {1 + 2\nu \alpha_{3} \psi } \right)}}{{\left( {1 + \alpha_{3} \psi } \right)\left( {1 + 2\nu \alpha_{3} \psi } \right) + 1}}} \right)} \right)}}.$$

(A14)

1.1.2 Case 2: The crack intersects resistors R ₄ and R ₁ only

The case 2 along with the notation used in shown in Fig. 5d. The ratio of change of resistances in $R_{1}$ if the proportion of crack being affected in $\alpha_{1}$ is given by:

$$\frac{{\Delta R_{1} }}{{R_{1} }} = \left( {1 + \alpha_{1} \psi } \right)\left( {1 + 2\nu \alpha_{1} \psi } \right) - 1.$$

(A15)

The ratios of change of resistances in $R_{4}$ if the proportion of crack being affected in $\alpha_{4}$ is given by:

$$\frac{{\Delta R_{4} }}{{R_{4} }} = \left( {1 + \alpha_{4} \psi } \right)\left( {1 + 2\nu \alpha_{4} \psi } \right) - 1.$$

(A16)

Substituting the values in Eq. 3 to obtain the change in voltage ratio for case 2 in equation A17.

$$\Delta V_{r} = \frac{{\left( {1 + \alpha_{4} \psi } \right)\left( {1 + 2\nu \alpha_{4} \psi } \right)}}{{\left( {1 + \alpha_{4} \psi } \right)\left( {1 + 2\nu \alpha_{4} \psi } \right) + 1}} - \frac{1}{{\left( {1 + \alpha_{1} \psi } \right)\left( {1 + 2\nu \alpha_{1} \psi } \right) + 1}}$$

(A17)

The strain is obtained by substituting the value of $\Delta V_{r}$ in equation A8 and shown in equation A18.

$$\varepsilon = - \frac{{2\left( {\frac{{\left( {1 + \alpha_{4} \psi } \right)\left( {1 + 2\nu \alpha_{4} \psi } \right)}}{{\left( {1 + \alpha_{4} \psi } \right)\left( {1 + 2\nu \alpha_{4} \psi } \right) + 1}} - \frac{1}{{\left( {1 + \alpha_{1} \psi } \right)\left( {1 + 2\nu \alpha_{1} \psi } \right) + 1}}} \right)}}{{{\text{GF}}\left( {\left( {1 + \nu } \right) + \left( {1 - \nu } \right)\left( {\frac{{\left( {1 + \alpha_{4} \psi } \right)\left( {1 + 2\nu \alpha_{4} \psi } \right)}}{{\left( {1 + \alpha_{4} \psi } \right)\left( {1 + 2\nu \alpha_{4} \psi } \right) + 1}} - \frac{1}{{\left( {1 + \alpha_{1} \psi } \right)\left( {1 + 2\nu \alpha_{1} \psi } \right) + 1}}} \right)} \right)}}$$

(A18)

1.1.3 Case 3: The crack intersects resistors R ₃ and R ₄ only

The case 3 along with notation is shown in Fig. 5e. The ratio of change of resistances in $R_{3}$ if the proportion of resistor being affected by crack is $\alpha_{3}$ is given by equation A2 and the ratio of change of resistance for $R_{4}$ when the proportion of resistor being affected by crack is $\alpha_{4}$ is given equation A16. The change in voltage ratio for the present case is given by equation A19.

$$\Delta V_{r} = \frac{{\left( {1 + \alpha_{4} \psi } \right)\left( {1 + 2\nu \alpha_{4} \psi } \right)}}{{\left( {1 + \alpha_{4} \psi } \right)\left( {1 + 2\nu \alpha_{4} \psi } \right) + 1}} - \frac{{\left( {1 + \alpha_{3} \psi } \right)\left( {1 + 2\nu \alpha_{3} \psi } \right)}}{{\left( {1 + \alpha_{3} \psi } \right)\left( {1 + 2\nu \alpha_{3} \psi } \right) + 1}}$$

(A19)

We can rearrange the terms to simplify the above expressions, given in equation A20.

$$\Delta V_{r} = \frac{1}{{\left( {1 + \alpha_{3} \psi } \right)\left( {1 + 2\nu \alpha_{3} \psi } \right) + 1}} - \frac{1}{{\left( {1 + \alpha_{4} \psi } \right)\left( {1 + 2\nu \alpha_{4} \psi } \right) + 1}}$$

(A20)

The measured strain in this case is given by equation A21.

$$\varepsilon = - \frac{{2\left( {\frac{1}{{\left( {1 + \alpha_{3} \psi } \right)\left( {1 + 2\nu \alpha_{3} \psi } \right) + 1}} - \frac{1}{{\left( {1 + \alpha_{4} \psi } \right)\left( {1 + 2\nu \alpha_{4} \psi } \right) + 1}}} \right)}}{{{\text{GF}}\left( {\left( {1 + \nu } \right) + \left( {1 - \nu } \right)\left( {\frac{1}{{\left( {1 + \alpha_{3} \psi } \right)\left( {1 + 2\nu \alpha_{3} \psi } \right) + 1}} - \frac{1}{{\left( {1 + \alpha_{4} \psi } \right)\left( {1 + 2\nu \alpha_{4} \psi } \right) + 1}}} \right)} \right)}}$$

(A21)

The strain ratio can be calculated by dividing this term by baseline strain in Eq. 9.

1.1.4 Case 4: the crack intersects resistors R ₂ and R ₄ only

The case 4 along with notation is shown in Fig. 5f. The ratio of change of resistances in $R_{2}$ if the proportion of resistor being affected by crack is $\alpha_{2}$ is given by equation A11 and the ratio of change of resistance for $R_{4}$ when the proportion of resistor being affected by crack is $\alpha_{4}$ is given by equation A16. The change in voltage ratio for the present case is given by equation A22.

$$\Delta V_{r} = \frac{{\left( {1 + \alpha_{4} \psi } \right)\left( {1 + 2\nu \alpha_{4} \psi } \right)}}{{\left( {1 + \alpha_{2} \psi } \right)\left( {1 + 2\nu \alpha_{2} \psi } \right) + \left( {1 + \alpha_{4} \psi } \right)\left( {1 + 2\nu \alpha_{4} \psi } \right)}} - \frac{1}{2}$$

(A22)

For the cases where $\alpha_{2} = \alpha_{4}$, $\Delta V_{r} = 0$. This would result in the strain and hence the strain ratio to be zero. For other cases, the strain is given by equation A23.

$$\varepsilon = - \frac{{2\left( {\frac{{\left( {1 + \alpha_{4} \psi } \right)\left( {1 + 2\nu \alpha_{4} \psi } \right)}}{{\left( {1 + \alpha_{2} \psi } \right)\left( {1 + 2\nu \alpha_{2} \psi } \right) + \left( {1 + \alpha_{4} \psi } \right)\left( {1 + 2\nu \alpha_{4} \psi } \right)}} - \frac{1}{2}} \right)}}{{{\text{GF}}\left( {\left( {1 + \nu } \right) + \left( {1 - \nu } \right)\left( {\frac{{\left( {1 + \alpha_{4} \psi } \right)\left( {1 + 2\nu \alpha_{4} \psi } \right)}}{{\left( {1 + \alpha_{2} \psi } \right)\left( {1 + 2\nu \alpha_{2} \psi } \right) + \left( {1 + \alpha_{4} \psi } \right)\left( {1 + 2\nu \alpha_{4} \psi } \right)}} - \frac{1}{2}} \right)} \right)}}$$

(A23)

Upon dividing the equation A23 with the equation of baseline strain, we can obtain the strain ratio.

Appendix B

Following steps were performed to establish the equation of the estimated crack path (shown in red dashed line in Fig. 25).

• i. The red solid line in Fig. 17a is used the initial guess for crack orientation obtained by fitting the best line to the centers of the sensors 1, 7, and 8. These sensors were localized using the algorithm presented in Box 2. Let the equation of the line be given by $y = m_{0} x + c_{0}$. Both $m_{0}$ and $c_{0}$ are known.

• ii. Using this initial estimate of the crack path, the initial estimate on the $\omega$ was obtained which is provided in Table 8.

• iii. Using the initial slope value, estimate the values of $\omega$ for various values of $c$ the intercept subject to the constraint that the line must intersect the sensors 1, 7, 8 and compute the $\ell_{1}$-norm between the strain ratio from the estimated line and the experimental strain ratios.

• iv. For each orientation that have $\ell_{1}$-norm better than the initial guess change (increase/decrease) the value of slope at a step size of 0.1 upto 0.5. This corresponds to ~ $5.7^{^\circ } - 22.8^{^\circ }$. This is also subject to the constraint that the line must intersect sensors 1, 7, and 8 and compute the $\ell_{1}$-norm of the difference between the strain ratio from the estimated line and the experimental strain ratios.

• v. The line which has the minimum $\ell_{1}$-norm between the experimental and estimated strain ratios is chosen the crack path.

• vi. That final line is shown in Fig. 17b.