Issue 41
V. Rizov, Frattura ed Integrità Strutturale, 41 (2017) 491-503; DOI: 10.3221/IGF-ESIS.41.61
m
m
m
2
B b
1
B b
2 B br
m
m
1
1
0 1
1 1
1
2
N
1
2
1 3 1
m
m h m
1
z
m
B br
2
1
n
1
m
m
m
m
2
2
1
1
2 1
1
(18)
1
2
1
2
2
m
m
2
1
h
f
q
f
q
f
q
f
q
3
3
2
2
z
2 1 B b q m
2
1
n
1
q
q
q
q
1
1
2
1
2
2
f
q
f
q
3
2
h
f q
f q
2 1 1 n z f q
q
q
1
2
m
m
2
1 1
1
B b
2 B br
1
m
m
m
2
2
1
2
M B b
y
0 1
2
1
2 m h
3
f
q
f
q
3
3
z
m
2 1 B br q
2
1
n
1
q
q
m
m
1
1
1
1
2
1
2
2
m
f
q
1
3
h
f q
f q
f
q
f
q
2
2
2
z
z
2
n
n
1
1
q
q
q
q
1
1
1
2
1
2
(19)
f q
f
q
2
f
q
f
q
f
q
f
q
4
4
3
3
z
2 1 B b q m
3
1
n
1
q
q
q
q
1
1
2
1
2
2
f
q
f
q
4
3
h
f q
f q
f
q
f
q
2
2
2
3
z
z
3
n
n
1
1
q
q
q
q
1
1
1
2
1
2
f q
f
q
2
/ 2
/ 2 h
/ f q m ( f and q are positive integers). In (19), 1 N and 1 y M are
where 1 1 h
z
z
, 2
and
1 1 n
1 1 n
1
determined by (15). It is obvious that at m =1 the non-linear stress-strain relation (1) transforms into the Hooke’s law, assuming that B = E (here E is the modulus of elasticity). This means that at m =1 Eq. (19) should transform into the formula for curvature of linear-elastic beam. Indeed, at m =1, 0 B E and 1 2 0 B B Eq. (19) yields
M Ebh
12
y
1
(20)
1
3
1
which is exact match of the formula for curvature of a linear-elastic homogeneous beam of width, b , and height, 1 h . Eqs. (18), and (19) should be solved with respect to 1 1 n z and 1 by using the MatLab computer program. By substituting of (12) and (16) in (11) the distribution of complementary strain energy density in the lower crack arm is written as
1
m m
1
2
m z z
2
r z
y
4
n
1 1
1
1
1
* 0
1
B
u
B
B
(21)
L
0
1
2
2
2
m
1
b
h
496
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