In this paper it is numerically analyzed the change in position of mode shape nodes induced by multiple cracks in bar that has been theoretically investigated in previous paper [1] of the authors. The focus is to analyze thoroughly dislocation of node located intermediately between two cracks in dependence upon the crack parameters without assumption on the smallness of crack. | Volume 35 Number 4 4 Vietnam Journal of Mechanics, VAST, Vol. 35, No. 4 (2013), pp. 299 – 311 CHANGE IN MODE SHAPE NODES OF MULTIPLE CRACKED BAR: II. THE NUMERICAL ANALYSIS N. T. Khiem∗ , L. K. Toan, N. T. L. Khue Institute of Mechanics, VAST, 18 Hoang Quoc Viet, Hanoi, Vietnam ∗ E-mail: ntkhiem@ Abstract. In this paper it is numerically analyzed the change in position of mode shape nodes induced by multiple cracks in bar that has been theoretically investigated in previous paper [1] of the authors. The focus is to analyze thoroughly dislocation of node located intermediately between two cracks in dependence upon the crack parameters without assumption on the smallness of crack. Keywords: Multiple cracked bar, crack detection, mode shape nodes, vibration method, modal analysis. 1. INTRODUCTION This paper is devoted to continue the study of change in node position induced by multiple cracks in bar by numerical examples. First, some theoretical aspects that were detailed in the previous paper are summarized to use for numerical computation. The emphasis is on the numerical investigating the case of double cracks and change in position of the node located between cracks. Two cases of boundary conditions: free-free and fixed-free ends are considered. 2. SUMMARIZED THEORETICAL ASPECTS For an uniform bar with Young’s modulus E, density ρ, cross section area A and length L that is assumed to be cracked at the locations e1 , ., en, free longitudinal vibration is described by the equation p Φ00 (x) + λ2 Φ(x) = 0, x ∈ (0, 1), λ = ωL/c0 , c0 = E/ρ (1) with given boundary conditions at both ends x = 0, x = 1 and the compatibility conditions at the crack positions Φ(ej + 0) = Φ(ej − 0) + γj Φ0 (ej ), γj = EA/LKj , j = 1, ., n. (2) It was shown in previous paper [1] that general solution of Eqs. (1), (2) all over the bar length can be represented as 300 N. T. Khiem, L. K. Toan, N. T. L. Khue Φ(x) = Φ0 (x) + n X µj K(x − ej , λ), (3) j=1 where Φ0
