PSI - Issue 26
Victor Rizov et al. / Procedia Structural Integrity 26 (2020) 75–85 Rizov/ Structural Integrity Procedia 00 (2019) 000 – 000
77
3
in Fig. 1. The left-hand and right-hand crack tips are located in cross-sections of abscissas 3 1 x l = and x l a = + 3 1 , respectively. Here, 3 x is the longitudinal centroidal axis (Fig. 1). The upper and lower crack arms have different heights denoted by 1 h and 2 h , respectively. The beam exhibits continuous material inhomogeneity in the height direction. The material of the beam has non-linear elastic behaviour described by the following stress-strain relation (Kuznetzov and Barzilovitc (1990)):
n m B D H = − − ,
(1)
where is the stress, is the strain, B , D , m and n are material properties. The distribution of material property, B , in the height direction is written as
2 h z
−
+
= + h B B B B K 0
0
,
(2)
3
where
3 h z h −
.
(3)
2
2
In formula (2), 3 z is the vertical centroidal axis of the cross-section of the beam, 0 B and K B are the values of B in the upper and lower surfaces of the beam, respectively. The longitudinal fracture behaviour of the beam is analyzed in terms of the strain energy release rate, G . For this purpose, by assuming a small increase of the crack at the left-hand crack tip, the strain energy release rate can be written as (Rizov (2018))
h
h
h
1
2
2
= − 2 2
2 h
* u dz ,
* h a G u dz 0 1
* a u dz 0
+
−
(4)
0
3
1
2
2
2 h
1
2
−
−
2
* 0 u are the complementary strain energy densities in the upper and lower crack arms and in
where *
* 0 2 a u and
0 1 a u ,
the un-cracked part of the beam ahead of the left-hand crack tip, respectively. In principle, the complementary strain energy density is equal to the area that supplements the area enclosed by the stress-strain curve to a rectangle. Thus, the complementary strain energy density in the upper crack arm is expressed as
* u = −
u
,
(5)
a
a
0
0
1
1
1 0 a u , is equal to the area enclosed by the stress-strain curve. Therefore, by integrating
where the strain energy density,
1 0 a u is obtained as
of (1),
m
n
2
1
1
+
+
n H
2 m u B D a 0 1 = −
−
.
(6)
1
1
+
+
By substituting of (1) and (6) in (5), one derives
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