 Open Access
 Total Downloads : 127
 Authors : Abhishek Kumar Sahu, Dr. Surajit Das
 Paper ID : IJERTV5IS060248
 Volume & Issue : Volume 05, Issue 06 (June 2016)
 DOI : http://dx.doi.org/10.17577/IJERTV5IS060248
 Published (First Online): 09062016
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Identification of Damages in Skeletal Structures using Modal Data
Abhishek Kumar Sahu
M.Tech Scholar Department of Civil Engineering National Institute of Technology
Agartala, India
Dr. Surajit Das
Assistant Professor Department of Civil Engineering National Institute of Technology
Agartala,India
Abstract In the present work the effect of transverse cracks on natural frequencies of a simply supported beam with symmetric overhangs is investigated and an algorithm has been developed for identifying damage in the same. In the numerical example, single and triple cracks are considered in the dynamic analysis. Flexibility matrix of the intact beam and an additional flexibility matrix due to damaged beam is derived and added up to obtain the flexibility matrix of the cracked beam element. Stiffness matrix of cracked beam element is derived by multiplying a transformation matrix with the inverse of the final flexibility matrix of the cracked beam element. The natural frequencies and the corresponding mode shapes of vibration are obtained by solving eigen value problem. It is found that for a simply supported beam with symmetric overhangs, the 1st frequency decreases with an increase in the crack depth, and, it decreases the most at mid span. In case of 2nd and 3rd frequency, these decreases the most at 20% and 80% of the total length from the left support. For triple cracks several important observations are also made.
KeywordsDamage Detection, Structural Health Monitoring, Modal Analysis, Cracks, Beams.

INTRODUCTION
Engineering Structures withstand loads during their service life. Buildings are usually designed on strong column, weak beam concept. So, beams are more vulnerable to cracks. Vibrational measurements are an efficient means of crack detection. Crack leads to reduction in stiffness of beam, thereby reducing its natural frequency. A lot of research work has been done to develop effective methods for crack detection. Pandey and Biswas (1991) have evaluated changes in flexibility matrix in order to locate damage. Pandey et.al. (1990) have used curvature mode shape to detect and locate damage in structure. It is shown that curvature mode shape localizes in the damage region whereas the displacement mode shapes are not localized. Further, MAC and COMAC are not sensitive enough to detect damage in its earlier stage. Morassi and Rovere (1997) have identified localized damage in a multistory steel frame. Vibration tests were performed on a five story steel frame with a notch of fixed position and variable depth. Damage is localized by considering frequencies related to shear type modes only. Rizos et.al. (1989) have used the measurement of flexural vibrations of a cantilever beam with rectangular cross section having transverse surface crack extending uniformly along the width
of the beam to locate crack location and crack depth. The method requires amplitude measurements at two positions of the structure only. The application of this method is limited to moderate cracks only. Liang et.al (1992) have developed theoretical relationship between eigen frequency changes, crack location and crack depth of damaged cantilever and simply supported beam. This theory can be more specifically applied to steel frame structures. Chondros and Dimarogonas (1979) have discussed the influence of crack in a welded joint on the dynamic behavior of a structural member. Local flexibility was used to establish relationship between crack depth to the change of natural frequency for the cases of a cantilever beam with a transverse crack at the welded root of the beam and of a beam welded (clamped) at both ends with a transverse crack at one welded end. This method is applicable to members of simple geometry. It is applied to individual members of large structures where member flexibility is larger than flexibility of supporting members. Mostafa Attar (2012) has used an analytical approach to investigate natural frequencies and mode shapes of a stepped beam with an arbitrary number of cracks and general form of boundary conditions. A simple transfer matrix is used to obtain general form of characteristic equation for the cracked beam. It is a function of crack location, crack depth, frequency, boundary conditions, geometrical and physical parameters of the beam. Boltezar et.al. have shown the crack identification procedure for freefree uniform beams in flexural vibrations. Khiem and Toan (2014) have proposed a novel method for calculating the natural frequencies of a multiple cracked beam and detecting unknown number of multiple cracks from measured natural frequencies. An explicit expression for natural frequencies through crack parameters is derived as modification of Rayleigh quotient for multiple cracked beams. Hu and Liang (1993) have developed two damage modeling techniques. First modeling technique involves use of massless, infinitesimal springs to represent discrete cracks and other employs a continuum damage concept. In spring model, castiglianos theorem and perturbation technique are used to derive crack location, extent of crack and eigen frequency changes. In continuum damage model, effective stress concept together with Hamiltons principle are used to derive similar relationship in continuum form. Antonino Morassi (1993) has shown that frequency sensitivity for any beam like structure can be evaluated on the basis of undamaged system by general
perturbation approach. Frequency sensitivity plays a vital role in crack identification. Freund Herrmann concept of using a spring to represent effect of crack on section is taken into account. Frequency sensitivity is proportional to potential energy stored for relevant mode shape at cross section where crack occurs. Ratios of frequency changes of various orders are independent of crack severity but helps in localizing damage.

PROPOSED ALGORITHM FOR IDENTIFYING TRANSVERSE CRACKS IN SIMPLY SUPPORTED BEAM WITH OVERHANGS

Derivation Of Differential Equation Of Transverse Vibration Of Beam
V
Combining eq. 3 and 4, we have
Using the method of separation of variables:
Substituting eq.6 in 5 and = A , we have,
(5)
(6)
(7)
M
The term on left side of eq. 7 is dependent only on x and the right side only on t. To be equal to each other, both side must be equal to same constant 2. Then the left hand side of eq. 7 can be written as
V
Figure 3.1. Free body diagram of forces acting on beam
Pittman (2004) derived the solution of the differential equation for transverse beam vibration.From the above free body diagram and equation of equilibrium of the vertical forces according to Newtons second law, we have
Where,
And, = Natural frequency in rad/sec.
(8)
or (1)
The sum of moments about any point of the element yields:
or (2)
Substituting eq. 2 into eq. 1, we have
(3)
It is assumed that vibration occurs in symmetric planes of beam. So, the differential equation of the deflected curve is:
(4)
The characteristic equation of the above differential equation is
or (+k)(k)(2+k2)=0 Which gives the eigen values as
k1 = , k2 = , k3 = i, k4 = i which yields the general solution as
X(x) = C1ex + C2ex + C3 cos(x) + C4 sin(x) Or
X(x) = C1 cosh(x) + C2 sinh(x) + C3 cos(x) + C4 sin(x)
C1 , C2 , C3,and C4 are determined from boundary conditions at the ends of a beam.
Since the beam has symmetric overhangs, so we divide it into three sections. Each section will have a separate coordinate system for measuing the distance x with the origin for each section being at the left end of each section.

Proposed damage identification algorithm for simply supported beam with overhangs
General boundary Conditions for Point A in the beam:
A B L
C D At x1 = 0,
x1 x2 x3
S
Figure 2. Geometry of simply supported beam with symmetric overhangs and coordinate system.
From the basic vibration theory, the harmonic motion of the beam and its first, second and third derivative for left overhang (i = 1), centre span (i=2) and right overhang(i=3) are as follows
(14)
(15)
Here, M = 1L for left overhang and 1 for Centre span and right overhang. Subscript L refers to the left side of the crack.
General boundary Conditions for Point D in the beam:
At x3 =
(16)
(17)
General boundary Conditions for crack lying in beam:
(9)
Here, N = 3 for left overhang and Centre span and 3R for right overhang. Subscript R refers to the right side of the crack.
General boundary Conditions for Point B in the beam:
For a single crack located anywhere in the left overhang, Centre span and right overhang, the general boundary conditions are as follows
At xi = Rc ,
At x1= , x2 = 0,
(18)
(19)
(10)
(20)
(11)
(12) (21)
(13)
Here, Rc is the distance of crack from left. i =1,2,3 for left overhang, Centre span and right overhang respectively. Subscript L and R refers to the left and right side of the crack.
Here, M = 1R for left overhang and 1 for centre span and right overhang. N = 2 for left and right overhang and 2L for centre span. Here, Q and S = 1R for left overhang and 1 for centre span and right overhang. Also, R and T = 2 for left and right overhang and 2L for centre span.
General boundary Conditions for Point C in the beam:
axial force and for open cracks above equation can be written as
At x2 = S, x3 = 0,
G = 1 (K K )2 K 2
(28)
E ' I1 I 2
(22)
II1
The expressions for stress intensity factors from earlier studies are given by Uttam Kumar Mishra (2014) as follows
(23)
KI1 = 6P1 LC
bp

6P
F
1 h
1 h
(29)
KI 2
2
bp
F
1 h
1 h
(30)

K
P2
F
(31)
II1 bh
II h
Here, M = 2 for left overhang and right overhang and 2R for centre span. N = 3 for left overhang and centre span and 3L
From definition, the elements of the overall additional flexibility matrix Cij can be
2
for right overhang. Here, Q and S = 2 for left overhang and
C i C
, (I, j=1,2) (32)
right overhang and 2R for centre span. Also, R and T = 3 for
ij P
PP
left overhang and centre span and 3L for right overhang.


PRESENT FINITE ELEMENT FORMULATION FOR FINDING
j i j
Substituting Eq (29),(30),(31) into Eq (28), then into Eq (26) and Eq (32) subsequently we get,
b 2
6PL
6P
2 P
2
NATURAL FREQUENCIES
Cij
E ' PP
1 c
bp
F1 h
1
bp
F1 h
1
bh
F1 h
d
A. Finite Element Formulation
i j
(33)
When crack is induced in a beam, then its flexibility is increased. So, first we calculate the additional flexibility
Substituting i,j (1,2) values, we get
a a
induced in it. Then it is added up with the flexibility matrix of
2 36L2 h
C
2
C
2
h 2
(34)
intact beam element. The inverse of the overall flexibility
C11 E 'b p
xF1 (x)dx xFII (x)dx
matrix thus obtained is multiplied with the transformation matrix to obtain the required stiffness matrix of the cracked
0 0
h
h
a
beam element. This stiffness matrix is assembled along with
72 Lc 2
(35)
the stiffness matrices of the intact beam element and thereafter the natural frequencies are calculated from the equation K 2M=0, where K= Assembled stiffness matrix of the beam,
C12 E 'bp xF1 (x)dx C21
0
0
a
M=Assembled mass matrix and = Natural frequency
72 h 2
(36)
(rad/sec). According to Dimarogonas et.al. (1983) and Tada et.al. (2000) the additional stain energy due to existence of
C22 E 'bp xF1 (x)dx
0
0
2 2
2 2
crack can be expressed as
s
s 4
C =
GdA

F (s)
tan 0.923 0.199 1 sin
(37)
C I
AC
s
2
cos s
2
Where, G = the strain energy release rate, and
AC = the effective cracked area.
1.122 0.561s 0.085s2 0.180s3
FII (s)
(38)
E '
E '
G = 1 (2
n1
KIn
2
2
)2 (
n1
KIIn
2
2
)2 k(
n1
KIIIn
)2
1 s
L3 L2
e C e C
Where, E = E for plane stress
Ctotal
3EI
L2
11 2EI 12
L
(39)
E = E/12 for plane strain
e C
e C
tot
tot
k = 1 +
2EI
21 EI
22
=Poissons ratio
E=Youngs Modulus of elasticity.
KI, KII and KIII = stress intensity factors for sliding, tearing and opening type cracks respectively. Neglecting effect of
The stiffness matrix Kcrack of a cracked beam element can be obtained as Kcrack = LC1 LT , Where, L is the transformation matrix for equilibrium condition
1 0
L 1
beta
rcd=0.0
rcd=0.2
rcd=0.4
rcd=0.6
rcd=0.8
0.1
141.98397
141.8973
141.5952
140.6614
135.8193
0.2
141.98397
141.4341
139.5647
134.254
116.4698
0.4
141.98397
141.785
141.1351
139.4817
135.2091
0.6
141.98397
141.785
141.1351
139.4817
135.2091
0.8
141.98397
141.4341
139.5647
134.254
116.4698
0.9
141.98397
141.8973
141.5952
140.6614
135.8193
tr>
beta
rcd=0.0
rcd=0.2
rcd=0.4
rcd=0.6
rcd=0.8
0.1
141.98397
141.8973
141.5952
140.6614
135.8193
0.2
141.98397
141.4341
139.5647
134.254
116.4698
0.4
141.98397
141.785
141.1351
139.4817
135.2091
0.6
141.98397
141.785
141.1351
139.4817
135.2091
0.8
141.98397
141.4341
139.5647
134.254
116.4698
0.9
141.98397
141.8973
141.5952
140.6614
135.8193
L e
(40)
TABLE II. 2ND NATURAL FREQUENCY OF SIMPLY SUPPORTED BEAM WITH SYMMETRIC OVERHANGS
1 0
0 1
Here, equation 1, 2and 3 are coefficients of additional flexibility matrix, a= crack depth, h= total depth of the beam, Lc= distance of crack from right node of beam element. E=E/(12), where = Poissons ratio, E= modulus of elasticity.


RESULTS BASED ON FORWARD PROCESS FOR FINDING NATURAL FREQUENCIES
Natural Frequencies of simply supported beam with symmetric overhangs having single crack:
Problem Description: A simply supported beam with symmetric overhangs on both ends having various crack depth ratio at various locations is taken. It has following properties and is divided into 4 elements, Length, L = 0.78m, Breadth, b
= 0.04m, Height, h = 0.01m, Mass density, = 7860 kg/m3, Youngs Modulus, E = 210 GPa, Number of elements = 4, Supports located at node 2 and node 4, Elemental length =

m
Figure 1. Convergence Study for single crack simply supported beam with symmetric overhangs.
TABLE I. 1ST NATURAL FREQUENCY OF SIMPLY SUPPORTED BEAM WITH SYMMETRIC OVERHANGS
beta
rcd=0.0
rcd=0.2
rcd=0.4
rcd=0.6
rcd=0.8
0.1
85.650656
85.63637
85.58624
85.42792
84.51325
0.2
85.650656
85.55792
85.22798
84.14173
77.0658
0.4
85.650656
85.26477
83.94458
80.10458
64.98897
0.6
85.650656
85.26477
83.94459
80.10458
64.98898
0.8
85.650656
85.55792
85.22798
84.14173
77.06582
0.9
85.650656
85.63637
85.58624
85.42792
84.51326
TABLE III. 3RD NATURAL FREQUENCY OF SIMPLY SUPPORTED BEAM WITH SYMMETRIC OVERHANGS
beta
rcd=0.0
rcd=0.2
rcd=0.4
rcd=0.6
rcd=0.8
0.1
297.13671
296.9138
296.1418
293.8025
282.7459
0.2
297.13671
295.8371
291.6278
281.176
256.1695
0.4
297.13671
296.9124
296.1733
294.243
288.756
0.6
297.13671
296.9124
296.1733
294.2431
288.7561
0.8
297.13671
295.8371
291.6278
281.176
256.1695
0.9
297.13671
296.9138
296.1418
293.8025
282.746
Figure 2. 1st natural frequency of a simply supported beam with symmetric overhangs.
Figure 3. 2nd natural frequency of a simply supported beam with symmetric overhangs.
TABLE V. 2ND NATURAL FREQUENCY OF SIMPLY SUPPORTED BEAM WITH SYMMETRIC OVERHANGS HAVING TRIPLE CRACK
Beta rcd
beta=0.2,0.4,0.6
beta=0.2,0.4,0.8
beta=0.2,0.6,0.8
beta=0.4,0.6,0.8
0.1
141.7352322
141.6444224
141.6444223
141.7352321
0.3
139.8257202
139.0683819
139.068376
139.8257211
0.5
134.6595384
132.3569261
132.3569234
134.6595553
0.7
119.1343277
114.082506
114.0825185
119.1343434
Beta rcd
beta=0.2,0.4,0.6
beta=0.2,0.4,0.8
beta=0.2,0.6,0.8
beta=0.4,0.6,0.8
0.1
296.6757407
296.3914888
296.3914843
296.6757406
0.3
293.1816274
290.7360878
290.7360562
293.1816314
0.5
284.1079446
276.000332
276.0003583
284.107977
0.7
259.4173901
236.0703659
236.0704891
259.4173403
Beta rcd
beta=0.2,0.4,0.6
beta=0.2,0.4,0.8
beta=0.2,0.6,0.8
beta=0.4,0.6,0.8
0.1
296.6757407
296.3914888
296.3914843
296.6757406
0.3
293.1816274
290.7360878
290.7360562
293.1816314
0.5
284.1079446
276.000332
276.0003583
284.107977
0.7
259.4173901
236.0703659
236.0704891
259.4173403
TABLE VI. 3RD NATURAL FREQUENCY OF SIMPLY SUPPORTED BEAM WITH SYMMETRIC OVERHANGS HAVING TRIPLE CRACK
Figure 4. 3rd natural frequency of a simply supported beam with symmetric overhangs.
From the above figure it can be concluded that for a simply supported beam with overhangs 1st frequency, as the crack depth increases, the frequency decreases. It decreases the most at mid span. In case of 2nd frequency, as the crack depth increases the frequency decreases. It decreases the most near the first support and second support i.e at 20% and 80% of the total length. In case of third frequency, also the frequency decreases with increase in crack depth and decreases the most at 20% and 80% of the total length.
Natural Frequencies of simply supported beam with symmetric overhangs having triple crack:
Problem Description: A simply supported beam with symmetric overhangs on both ends having various crack depth ratio at various locations is taken. It has following properties and is divided into 4 elements, Length, L = 0.78m, Breadth, b
= 0.04m, Height, h = 0.01m, Mass density, = 7860 kg/m3, Youngs Modulus, E = 210 GPa, Number of elements = 4, Supports located at node 2 and node 4, Elemental length =
0.195 m.
TABLE IV. 1ST NATURAL FREQUENCY OF SIMPLY SUPPORTED BEAM WITH SYMMETRIC OVERHANGS HAVING TRIPLE CRACK
TABLE VII. 4TH NATURAL FREQUENCY OF SIMPLY SUPPORTED BEAM WITH SYMMETRIC OVERHANGS HAVING TRIPLE CRACK
Beta rcd
beta=0.2,0.4,0.6
beta=0.2,0.4,0.8
beta=0.2,0.6,0.8
beta=0.4,0.6,0.8
0.1
823.5553966
823.3770847
823.3770825
823.5553952
0.3
807.8733341
806.6115086
806.6115075
807.8733313
0.5
767.7336102
765.62396
765.6239695
767.7336251
0.7
665.2350624
671.2620994
671.2620997
665.2349858
Beta rcd
beta=0.2,0.4,0.6
beta=0.2,0.4,0.8
beta=0.2,0.6,0.8
beta=0.4,0.6,0.8
0.1
1295.374277
1294.538285
1294.538274
1295.374277
0.3
1289.118019
1282.067365
1282.06724
1289.118014
0.5
1274.636156
1252.50435
1252.504407
1274.636196
0.7
1244.677417
1189.939438
1189.94033
1244.677722
Beta rcd
beta=0.2,0.4,0.6
beta=0.2,0.4,0.8
beta=0.2,0.6,0.8
beta=0.4,0.6,0.8
0.1
1295.374277
1294.538285
1294.538274
1295.374277
0.3
1289.118019
1282.067365
1282.06724
1289.118014
0.5
1274.636156
1252.50435
1252.504407
1274.636196
0.7
1244.677417
1189.939438
1189.94033
1244.677722
TABLE VIII. 5TH NATURAL FREQUENCY OF SIMPLY SUPPORTED BEAM WITH SYMMETRIC OVERHANGS HAVING TRIPLE CRACK
Beta rcd
beta=0.2,0.4,0.6
beta=0.2,0.4,0.8
beta=0.2,0.6,0.8
beta=0.4,0.6,0.8
0.1
85.42541912
85.50185911
85.50185897
85.42541911
0.3
83.71291933
84.35523699
84.35523717
83.71291952
0.5
79.22638862
81.22157178
81.22157319
79.22639125
0.7
66.96330109
71.6677455
71.66774656
66.96330167
TABLE IX. 6TH NATURAL FREQUENCY OF SIMPLY SUPPORTED BEAM WITH SYMMETRIC OVERHANGS HAVING TRIPLE CRACK
Beta rcd
beta=0.2,0.4,0.6
beta=0.2,0.4,0.8
beta=0.2,0.6,0.8
beta=0.4,0.6,0.8
0.1
1717.062955
1717.303203
1717.303177
1717.062969
0.3
1710.840064
1712.87886
1712.878668
1710.840161
0.5
1695.381055
1701.879419
1701.879576
1695.381344
0.7
1658.943026
1675.671543
1675.672746
1658.942178
Figure 5. 1st natural frequency of a simply supported beam with symmetric overhangs having triple crack.
Figure 6. 2nd natural frequency of a simply supported beam with symmetric overhangs having triple crack.
Figure 7. 3rd natural frequency of a simply supported beam with symmetric overhangs having triple crack.
Figure 8. 4th natural frequency of a simply supported beam with symmetric overhangs having triple crack.
Figure 9. 5th natural frequency of a simply supported beam with symmetric overhangs having triple crack.
Figure 10. 6th natural frequency of a simply supported beam with symmetric overhangs having triple crack.
Validation of proposed FEM method for triple cracked cantilever beam
Problem Description: A cantilever with crack depth ratio of

for all the three cracks with cracks located at various locations is taken. It has following properties and is divided into 16 elements
Length, L = 0.5m Breadth, b = 0.02m Height, h = 0.02m
Mass density, = 7860 kg/m3. Youngs Modulus, E = 210 GPa Poisons ratio =0.3



RESULTS BASED ON INVERSE PROCESS FOR IDENTIFYING CRACKS FROM FIRST THREE NATURAL
FREQUENCIES
Problem Description: A simply supported beam with symmetric overhangs having cracks located at various locations with various depths is taken. It has following properties. Cracks are identified for single cracked simply supported beam with symmetric overhangs by finding determinant of matrix formed by applying boundary conditions.
Length, L = 10m Breadth, b = 0.2m Height, h = 0.6m
Mass density, = 2350 kg/m3.
Youngs Modulus, E = 2.8*10^10 N/m2. I=0.0036 m4.
TABLE XII. PREDICTED CRACK LOCATION FOR SINGLE CRACK LOCATED AT DIFFERENT LOCATIONS IN CENTRE SPAN OF SIMPLY SUPPORTED BEAM WITH SYMMETRIC OVERHANGS.
CENTRE SPAN
Sl.no.
1st Natural
frequency (1) in rad/sec
2nd Natural
frequency (2) in rad/sec
3rd Natural
frequency (3) in rad/sec
Actual Crack Location, in metres
Predicted Crack Location, in metres
1
126.58
206.27
442.6
0.5
0.5
2
124.092
216.165
436.3544
3
3
3
125.494
210.566
450.17
4
4
4
122.317
198.99
437.47
4.5
4.5
CENTRE SPAN
Sl.no.
1st Natural
frequency (1) in rad/sec
2nd Natural
frequency (2) in rad/sec
3rd Natural
frequency (3) in rad/sec
Actual Crack Location, in metres
Predicted Crack Location, in metres
1
126.58
206.27
442.6
0.5
0.5
2
124.092
216.165
436.3544
3
3
3
125.494
210.566
450.17
4
4
4
122.317
198.99
437.47
4.5
4.5
TABLE X. FIRST THREE NATURAL FREQUENCIES OF A TRIPLE CRACKED CANTILEVER BEAM
td>
7342.7
Case
Crack Location
Method
Natural Frequency
X1/L
X2/L
X3/L
1
2
3
1
0.2
0.4
0.6
M. Attar
416.9159
2612.213
7324.21
Present FEM
417.16
2613.9
7328.4
2
0.2
0.4
0.8
M. Attar
417.0864
2620.455
7318.811
Present FEM
415.07
2618.8
3
0.2
0.6
0.8
M. Attar
417.6464
2617.786
7315.833
Present FEM
415.58
2616.4
7339.6
4
0.4
0.6
0.8
M. Attar
418.7517
2610.361
7311.243
Present FEM
416.57
2609.7
7335.9
TABLE XI. 4TH ,5TH AND 6TH NATURAL FREQUENCY OF A TRIPLE CRACKED CANTILEVER BEAM
Case
Crack Location
Method
Natural Frequency
X1/L
X2/L
X3/L
4
5
6
1
0.2
0.4
0.6
M. Attar
14357.28
23592.02
35604.06
Present FEM
14366.24
23634.15
35634.05
2
0.2
0.4
0.8
M. Attar
14301.02
23602.31
35574
Present FEM
14373.25
23688.01
35549.85
3
0.2
0.6
0.8
M. Attar
14301.53
23603.48
35574.12
Present FEM
14374.84
23688.36
35548.87
4
0.4
0.6
0.8
M. Attar
14338.46
23577.32
35598.16
Present FEM
14408.62
23669.04
35568.63
TABLE XIII. PREDICTED CRACK LOCATION FOR SINGLE CRACK LOCATED AT DIFFERENT LOCATIONS IN LEFT AND RIGHT OVERHANG OF SIMPLY SUPPORTED BEAM WITH SYMMETRIC OVERHANGS.
Discussion:
LEFT OVERHANG
Sl.no.
1st Natural
frequency (1) in rad/sec
2nd Natural
frequency (2) in rad/sec
3rd Natural
frequency (3) in rad/sec
Actual Crack Location, in metres
Predicted Crack Location, in metres
1
130.785
215.276
445.77
1
1
2
125.65
194.23
411.37
2.3
2.3
3
130.43
213.27
441.39
1.5
1.5
RIGHT OVERHANG
1
131.04
216.84
450.23
2
2.1
LEFT OVERHANG
Sl.no.
1st Natural
frequency (1) in rad/sec
2nd Natural
frequency (2) in rad/sec
3rd Natural
frequency (3) in rad/sec
Actual Crack Location, in metres
Predicted Crack Location, in metres
1
130.785
215.276
445.77
1
1
2
125.65
194.23
411.37
2.3
2.3
3
130.43
213.27
441.39
1.5
1.5
RIGHT OVERHANG
1
131.04
216.84
450.23
2
2.1
A new algorithm has been proposed to identify cracks in a simply supported beam with symmetric overhangs. A 16 matrix is formed from the harmonic equation of the beam after applying the boundary conditions. The determinant of the above matrix is equated to zero and therefore graph is plotted between the two unknowns, stiffness (K) and crack location for the first three natural frequencies (rad/sec). The intersection of the curves for corresponding first three natural frequencies gives the crack location and corresponding stiffness.

CONCLUSIONS
8
x 10
10
8
stiffness, K
stiffness, K
6
4
actual stiffness, actual cracklocation=0.5m 0f centre span
omega1(rad/sec)
omega2(rad/sec)
omega3(rad/sec)
X: 0.5
Y: 1.846e+008
X: 0.5
Y: 1.846e+008
Based on the results obtained for simply supported beam having single and multiple cracks following conclusions can be drawn

Frequencies decreases with increase in crack depth.

Frequencies are same for symmetric location of cracks.

For single crack 1st frequency decreases the most at midspan whereas for 2nd and 3rd frequencies it decreases the most near the supports.

For a simply supported beam with overhangs 1st frequency, as the crack depth increases, the frequency decreases. It decreases the most at mid span. In case of 2nd frequency, as the crack depth increases the frequency decreases. It decreases the most near the first support and second support i.e at 20% and 80% of the total length. In
2
0
2
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
crack location, in metres
Figure 11: Location of crack in a single cracked simply supported beam with symmetric overhangs having actual crack location = 0.5m of centre span.
case of third frequency, also the frequency decreases with increase in crack depth and decreases the most at 20% and 80% of the total length.

For the case of simply supported beam with symmetric overhangs having triple cracks, as the crack depth increases, frequency decreases. Frequencies are same for the crack location combination of 0.2L0.4L0.6L and 0.4L0.6L0.8L. It is also same for the crack location combination of 0.2L0.4L0.8L and 0.2L0.6L0.8L. A significant difference among the first six frequencies is
7
x 10
actual stiffness, actual crack location=2.3m of left overhang
that the 1st ,4th and 6th frequencies increases as the crack
omega1(rad/sec)
omega2(rad/sec) omega3(rad/sec)
omega1(rad/sec)
omega2(rad/sec) omega3(rad/sec)
X: 2.3
Y: 1.004e+008
X: 2.3
Y: 1.004e+008
14
12
10
stiffness
stiffness
8
location shifts towards the midspan whereas it is vice versa for the 2nd ,3rd and 5th one.

An algorithm has been developed to identify cracks in simply supported beam with symmetric overhangs. The algorithm can also identify cracks located in simply supported beam with different overhanging lengths.

6
4
2
0
0 0.5 1 1.5 2 2.5
crack location, in metres
Figure 12: Location of crack in a single cracked simply supported beam with symmetric overhangs having actual crack location = 2.3m of left overhang.
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