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Nowadays presence of crack in different engineering structures becomes a serious threat to the performance. Since most of the civil and mechanical structures may be damaged due to material fatigue, mechanical vibration, environmental attack and long-term service. Moreover, dynamical systems of a beam usually possess a non-linear character, which causes practical difficulties on the model-based damage detection techniques. This paper presents a novel approach to detect damage in a simply supported beam. In this study, a numerical simulation using the Finite Element Method (FEM) has been done to determine the frequencies to detect the crack in a concrete beam of length 0.12 m and width 0.015 m. A vibration-based model is employed to simulate the results by using COMSOL Multiphysics. At the tip, by performing the computational analysis it is found that the presence of cracks affects the natural frequencies of the concrete structure. It is observed that after applying load , the frequencies of the cracked beam ha ve been changed with the variation of the location of the crack for all the modes of vibration. It also found that maximum frequency reserved at the cracked point so it will also help us to detect different hidden defect s in any structure. A comparison is also made with the experimental results. It is also found that the effect of crack is more near the fixed end than at the free end.

Identification of structural crack location has gained increasing attentions from the scientific and engineering domains since the unpredicted structural failure may cause catastrophic, economic, and life loss [

Many extensive researches on crack detection in different methods have been performed theoretically and experimentally till today. Kim and Zhao [

The above papers are studied the changes neighborhood in natural frequencies due to the presence of multiple cracks at intervals associate concrete beam. The target of this paper to look out a method for predicting crack parameters (crack depth associated its location) throughout the concrete beam from changes in natural frequencies. Constant studies are disbursed by using COMSOL Multiphysics software to judge modal parameters (natural frequencies and deflections) for various crack position parameters. The technique developed to detect crack by using the FEM software has become popular in recent years.

The vibration-based damage detection has become one of the commonly used tools for crack detection. This approach is mainly based on changes in dynamic characteristics, such as natural frequency and crack position parameter [

The equation of motion in matrix form for vibration of a beam under load is given by,

[ M ] { d 2 q ( x ) d x 2 } + [ [ K ] − P [ K g ] ] { q } = 0 (1)

where,

[M] = Consistent mass matrix.

[K] = Bending stiffness matrix of the beam.

[K_{g}] = Geometric stiffness matrix.

{q} = Displacement vector.

P = External force vector.

For free vibration the forcing function p = 0. So the Equation (1) can be written as,

[ M ] { d 2 q ( x ) d x 2 } + [ K ] { q } = 0 (2)

In-plane, the load P(t) can be expressed in the form as shown below,

P ( t ) = P s + P t cos Ω t (3)

where,

P_{s} = the static portion of P.

P_{t} = the amplitude of the dynamic portion of P and

Ω = the frequency of excitation.

Equation (2) represents an eigen value problem and the roots of the equation give rise to square of the natural frequency given by the equation,

[ K ] − ( ω n ) 2 [ M ] = 0 (4)

Finite Element AnalysisIn this analysis we consider a beam with two degrees of freedom (slope and deflection) per node [

The governing equations of a deformation curved beams can be simplified by determining the force, moment, deflection and twist along the fifth metatarsal. This can be done subjected to both a point wise and a distributed load by using Young’s Modulus and Moments of Inertia given by the following equation,

m ( x ) = d 2 q ( x ) d x 2 E I (5)

V ( x ) = d 3 q ( x ) d x 3 E I (6)

where,

E = Young’s Modulus.

I = Moments of Inertia.

The deflection of the beam due to applied loads on the domain are shown in

Now,

F 1 y = V ( 0 ) = d 3 q ( 0 ) d x 3 E I = E I l 3 ( 12 q 1 + 6 l θ 1 − 12 q 2 + 6 l θ 2 ) (7)

m 1 = − m ( 0 ) = d 2 q ( 0 ) d x 2 E I = E I l 3 ( 6 l q 1 + 4 l 2 θ 1 − 6 l q 2 + 2 l 2 θ 2 ) (8)

F 2 y = − V ( l ) = d 3 q ( l ) d x 3 E I = E I l 3 ( − 12 q 1 − 6 l θ 1 + 12 q 2 − 6 l θ 2 ) (9)

m 2 = − m ( l ) = d 2 q ( l ) d x 2 E I = E I l 3 ( 6 l q 1 + 2 l 2 θ 1 − 6 l q 2 + 4 l 2 θ 2 ) (10)

If we write these values in the matrix form of the above system then we find,

[ F 1 y m 1 F 2 y m 2 ] = [ 12 6 l − 12 6 l 6 l 4 l 2 − 6 l 2 l 2 − 12 − 6 l 12 − 6 l 6 l 2 l 2 − 6 l 4 l 2 ] [ q 1 θ 1 q 2 θ 2 ] (11)

i.e. The Element stiffness matrix due to bending,

[ K ] e = ∫ 0 L [ B ] T [ D ] [ B ] d x

[ K ] e = E I l 3 [ 12 6 l − 12 6 l 6 l 4 l 2 − 6 l 2 l 2 − 12 − 6 l 12 − 6 l 6 l 2 l 2 − 6 l 4 l 2 ] (12)

The stiffness matrix K_{crack} or K_{c} of a cracked beam element: From equilibrium condition as in

Where,

L e = ( − 1 0 L e − 1 1 0 0 1 ) (13)

Hence the stiffness matrix K_{crack} or K_{c} of a cracked beam element can be obtained,

K C = L C t o t a l − 1 L T (14)

The cracked element stiffness matrix becomes,

[ K e ] = 1 C 11 C 22 − C 12 C 21 [ C 22 C 22 L e − C 21 − C 22 C 21 C 22 L e − C 21 C 22 L e 2 − C 21 L e − C 12 L e + C 11 − C 22 L e + C 12 − C 21 L e − C 11 − C 22 − C 22 L e + C 21 C 22 − C 21 C 12 C 12 L e − C 11 − C 12 C 11 ] (15)

Since two end sided of beam are fixed and no deformation will occur so q_{1} = θ_{1} = q_{3} = θ_{3} = 0. At that point there is no bending moment, shear force so the loading condition F_{1} = F_{3} = M_{1} = M_{3} = 0. Only the load applied in the middle portion, F_{2} = 500 N and the bending moment M_{2} and the cracked position of the beam will remain active.

There are three faces present bounding the calculation domain which are thin Elastic Layer (Boundary 6) is shown in

For Symmetry thin Elastic Layer,

n ⋅ u = 0 (16)

A design of computational domain without crack & with crack is shown in

Description | Value |
---|---|

Minimum element quality | 0.2093 |

Average element quality | 0.6135 |

Tetrahedron | 1424 |

Triangle | 863 |

Edge element | 232 |

Vertex element | 32 |

Description | Value |
---|---|

Number of degrees of freedom solved for | 10,093 |

Space dimension | 3 |

Number of domains | 1 |

Number of boundaries | 23 |

Number of edges | 52 |

Number of vertices | 32 |

Space dimension | 3 |

In this study, we have investigated the frequency of the concrete beam containing double crack using finite element method. For our simulation, we construct a solid concrete beam and have used different parameter values according to

In

Description | Value |
---|---|

Length of the beam (L) | 0.12 m |

Width of the beam (h) | 0.015 m |

Thickness of the beam (H) | 0.008 m |

Depth of First crack (D1) | 0.001 m |

Length of First crack (c1l) | 0.001 m |

Height of First crack (c1h) | 0.008 m |

Depth of the second crack (D2) | 0.017 m |

Length of second crack (c2l) | 0.001 m |

Height of second crack (c2h) | 0.001 m |

Description | Value |
---|---|

Density of concrete | 1570 kg/m^{3} |

Young’s modulus | 122.7 [GPa] |

Poisson’s ratio | 0.2 |

Shear modulus | 3.7 [GPa] |

Tensile strength (σ) | 2 - 5 Mpa |

Shear strength (τ) | 6 - 17 MPa |

vibration and differs the frequency. But there is a difference is formed for the (f) that, after the load the beam distorted and deflected. So due to double deflection in this situation, the applied loads distributed into two end sides. Finally we observed that, at the cracked position vibration and frequencies are increased as the load increased and the presence of crack affects the natural frequency of the structure.

position of crack. It is also seen that as long as the crack changes its position (goes to end points) its frequencies are also increased gradually. Finally from the graph it is clear to us that natural frequencies of the beam are directly affected by the location of the cracks.

Validation of the StudyIn

Serial no | Difference between cracks (m) | Crack position of first crack (m) | Crack position of second crack (m) | Present analysis FEA Frequency (Hz) | Priyadarshini A [ | Kisa et al. [ |
---|---|---|---|---|---|---|

1st | 0.05 | 0.01 | 0.06 | 6356.1 | 6465.83 | 6458.34 |

2nd | 0.05 | 0.02 | 0.07 | 6423.6 | 6470.53 | 6457.4 |

3rd | 0.05 | 0.03 | 0.08 | 6485.5 | 6465.81 | 6454.48 |

4th | 0.05 | 0.04 | 0.09 | 6442.3 | 6451.41 | 6448.18 |

5th | 0.05 | 0.05 | 0.10 | 6398.3 | 6397.71 | 6436.01 |

6th | 0.05 | 0.06 | 0.11 | 6233.2 | 6211.68 | 6174.71 |

A numerical simulation is done for a cracked concrete beam with different locations to detect the cracks. The flexibility matrix method is used to calculate the stiffness of the cracked beam by using FEM based vibration model using COMSOL Multiphysics [

The authors gratefully acknowledge for the technical support to the Centre of excellence in Mathematics, Department of Mathematics, Mahidol University, and Bangkok, Thailand. The authors also acknowledge to the Simulation Lab, Department of Mathematics, Chittagong University of Engineering and Technology, Bangladesh, for the support to complete this research.

The authors declare no conflicts of interest regarding the publication of this paper.

Karmaker, R., Deb, U.K. and Das, A. (2020) Modeling and Simulation of a Cracked Beam with Different Location Using FEM. Computational Water, Energy, and Environmental Engineering, 9, 145-158. https://doi.org/10.4236/cweee.2020.94010