Damage quantification in beam-type structures using modal curvature ratio

Abstract

Modal curvature ratio approach for the localization and quantification of damage in beam-type structures is proposed, demonstrating capability across both dense and sparse measurement grids. This method directly utilizes modal properties (i.e., mode shape) without the requirement of a complete structural model of the structure. This study involves the mathematical derivation of the proposed algorithm, which relies on establishing correlations between modal parameters derived from fundamental theoretical formulations for both pristine and damaged beams. The method’s efficiency and universality are confirmed through numerical case studies that identify and measure damage under various predefined scenarios. Numerical modal analysis is conducted using MATLAB for both damaged and pristine beams. The proposed approach demonstrates a high degree of accuracy in localizing and quantifying the damage. Notably, the approach exhibits a remarkable capability to identify and measure damage with a stiffness loss, as low as 5%. The investigation extends to multiple damage scenarios and considers the influence of noise. The methodology proves effective in discerning the severity of damage, showcasing improvements over prior results. Eventually, an integrated model-free approach for damage locating and quantification is presented for proposal based on the mathematical and numerical method. Considering the versatility of beams as structural elements, the proposed method provides a potential instrument for real-world damage detection in the field of structural health monitoring.

Appendices

Appendix 1

$$\begin{aligned} \left\langle {\eta_{i} (x),w_{k}^{0} (x)} \right\rangle = & \, \frac{1}{L}\int\limits_{{x_{0} - \frac{b}{2}}}^{{x_{0} + \frac{b}{2}}} {\left( {\frac{{i^{2} k^{2} \pi^{4} }}{{L^{4} }}} \right) \times \sin \left( {\frac{i\pi x}{L}} \right)} \times \sin \left( {\frac{k\pi x}{L}} \right){\text{d}}x \\ { = } & \, \frac{{i^{2} k^{2} \pi^{4} }}{{L^{5} }}\int\limits_{{x_{0} - \frac{b}{2}}}^{{x_{0} + \frac{b}{2}}} {\left[ {\cos \frac{(k - i)\pi x}{L} - \cos \frac{(k + i)\pi x}{L}} \right]} {\text{ d}}x \\ { = } & \, \frac{{i^{2} k^{2} \pi^{4} }}{{L^{5} }}\left[ {\frac{L}{(k - i)\pi }\sin \frac{(k - i)\pi x}{L} - \frac{L}{(k + i)\pi }\sin \frac{(k + i)\pi x}{L}} \right]_{{x_{0} - \frac{b}{2}}}^{{x_{0} + \frac{b}{2}}} \\ { = } & \, \frac{{i^{2} k^{2} \pi^{4} }}{{L^{4} }}\left[ \begin{gathered} \frac{1}{(k - i)\pi }\left\{ {\sin \left( {\frac{(k - i)\pi }{L}\left( {x_{0} + \frac{b}{2}} \right)} \right) - \sin \left( {\frac{(k - i)\pi }{L}\left( {x_{0} - \frac{b}{2}} \right)} \right)} \right\} - \hfill \\ \frac{1}{(k + i)\pi }\left\{ {\sin \left( {\frac{(k + i)\pi }{L}\left( {x_{0} + \frac{b}{2}} \right)} \right) - \sin \left( {\frac{(k + i)\pi }{L}\left( {x_{0} - \frac{b}{2}} \right)} \right)} \right\} \hfill \\ \end{gathered} \right] \\ { = } & \, \frac{{i^{2} k^{2} \pi^{4} }}{{L^{4} }}\left[ \begin{gathered} \frac{2}{(k - i)\pi } \times \cos \left\{ {\left( {k - i} \right)\frac{\pi }{L}x_{0} } \right\} \times \sin \left\{ {\left( {k - i} \right)\frac{\pi b}{{2L}}} \right\} - \hfill \\ \frac{2}{(k + i)\pi } \times \cos \left\{ {\left( {k + i} \right)\frac{\pi }{L}x_{0} } \right\} \times \sin \left\{ {\left( {k + i} \right)\frac{\pi b}{{2L}}} \right\} \hfill \\ \end{gathered} \right] \\ \end{aligned}$$

(35)

For limiting case b <  < L, Equation (35) can be written as,

$$\begin{aligned} \left\langle {\eta_{i} (x),w_{k}^{0} (x)} \right\rangle = & \frac{{i^{2} k^{2} \pi^{4} }}{{L^{4} }}\left[ {\frac{b}{L}\left\{ {\cos \left( {(k - i)\frac{\pi }{L}x_{0} } \right) - \cos \left( {(k + i)\frac{\pi }{L}x_{0} } \right)} \right\}} \right] \\ { = } & \, \frac{{2i^{2} k^{2} \pi^{4} b}}{{L^{5} }}\sin \left( {k\pi \frac{{x_{0} }}{L}} \right)\sin \left( {i\pi \frac{{x_{0} }}{L}} \right) \\ \end{aligned}$$

(36)

Appendix 2

$$\begin{aligned} \sum\limits_{k = 2}^{n} {\frac{{4k^{4} b}}{{\left( {1 - k^{4} } \right)L}}\left\{ {\sin \left( {k\pi \frac{{x_{0} }}{L}} \right)} \right\}^{2} } = & - \sum\limits_{k = 2}^{n} {\frac{4b}{L}\left\{ {\sin \left( {k\pi \frac{{x_{0} }}{L}} \right)} \right\}^{2} } \\ = & - \frac{2b}{L}\sum\limits_{k = 2}^{n} {\left( {1 - \cos \left( {2k\pi \frac{{x_{0} }}{L}} \right)} \right)} \, \\ = & - \frac{2b}{L}\left[ {\left\{ {\sum\limits_{k = 0}^{n} {\left( {1 - \cos \left( {2k\pi \frac{{x_{0} }}{L}} \right)} \right)} } \right\} + \left\{ {1 - 1 + 1 - \cos \left( {2\pi \frac{{x_{0} }}{L}} \right)} \right\}} \right] \\ = = & - \frac{2b}{L}\left[ {\left\{ {\sum\limits_{k = 0}^{n} {\left( {1 - \cos \left( {2k\pi \frac{{x_{0} }}{L}} \right)} \right)} } \right\} + \left\{ {1 - \cos \left( {2\pi \frac{{x_{0} }}{L}} \right)} \right\}} \right] \\ \end{aligned}$$

(37)

The term cosine term in the Eq. (37), can be solved as, (where j is the square root of –1)

$$\sum\limits_{k = 0}^{n} {\cos \left( {2k\pi \frac{{x_{0} }}{L}} \right)} = \sum\limits_{k = 0}^{n} {\frac{{e^{{j2k\pi \frac{{x_{0} }}{L}}} - e^{{ - j2k\pi \frac{{x_{0} }}{L}}} }}{2}} = \frac{1}{2}\left[ {\sum\limits_{k = 0}^{n} {e^{{j2k\pi \frac{{x_{0} }}{L}}} } - \sum\limits_{k = 0}^{n} {e^{{ - j2k\pi \frac{{x_{0} }}{L}}} } } \right]$$

(38)

For large n, Eq. (38), can be calculated as,

$$\begin{aligned} \sum\limits_{k = 0}^{n} {\cos \left( {2k\pi \frac{{x_{0} }}{L}} \right)} = & \frac{1}{2}\left[ {\frac{1}{{1 - e^{{j2\pi \frac{{x_{0} }}{L}}} }} - \frac{1}{{1 - e^{{ - j2\pi \frac{{x_{0} }}{L}}} }}} \right] \\ = & \frac{1}{2}\left[ {\frac{{1 - e^{{ - j2\pi \frac{{x_{0} }}{L}}} + 1 - e^{{j2\pi \frac{{x_{0} }}{L}}} }}{{1 - e^{{j2\pi \frac{{x_{0} }}{L}}} - e^{{ - j2\pi \frac{{x_{0} }}{L}}} + 1}}} \right] = \frac{1}{2} \\ \end{aligned}$$

(39)

Substituting Eq. (39) in Eq. (37), it is found that,

$$\sum\limits_{k = 2}^{n} {\frac{{4k^{4} b}}{{\left( {1 - k^{4} } \right)L}}\left\{ {\sin \left( {k\pi \frac{{x_{0} }}{L}} \right)} \right\}^{2} } = - \frac{2b}{L}\left[ {n + 1 - \frac{1}{2} + 1 - \cos \left( {2\pi \frac{{x_{0} }}{L}} \right)} \right]$$

(40)

Appendix 3

A 3% stiffness decrease is applied to four scenarios, each with a different element size, in order to successfully demonstrate damage quantification. Figure 15a–d displays the results of the suggested damage quantification algorithm for a unit-length beam in examples AC1–AC4 with single damage as specified in Table 3. The damage quantification is successfully achieved with an error of 0.664% for case AC1, 0.304% for case AC2, 0.163% for case AC3, and 0.125% for case AC4.

Fig. 15

Comparison of calculated stiffness reduction (CSR) and actual stiffness reduction (ASR) for 3% stiffness loss across varying element sizes and damage region (DR): a 10 elements, DR: 0.20–0.30, b 20 elements, DR: 0.25–0.30, c 50 elements, DR: 0.28–0.30, d 100 elements, DR: 0.29–0.30

Table 3 Cases involving single damage

Using the suggested approach for damage quantification, a 1% drop in stiffness for each of the four examples with different elements is successfully established. Figure 16a–d shows the outcomes of the suggested damage quantification algorithm for a unit-length beam in examples AC5–AC8, each of which involves a single damage location as specified in Table 3. The damage quantification is accomplished, exhibiting an accuracy of 0.199% for case AC5, 0.080% for case AC6, 0.033% for case AC7, and 0.021% for case AC8.

Fig. 16

Comparison of calculated stiffness reduction (CSR) and actual stiffness reduction (ASR) for 1% stiffness loss across varying element sizes and damage region (DR): a 10 elements, DR: 0.20–0.30, b 20 elements, DR: 0.25–0.30, c 50 elements, DR: 0.28–0.30, d 100 elements, DR: 0.29–0.30