The plastic strips propagation in an ideal elastic-plastic body with colinear system of shear cracks has been studied. The cracks are opposed to the stable stresses on the faces. Stressand-strain state of the body, dependencies of the strips length on the loading for random distance between cracks and the level of their faces friction have been found. The value of critical loading has been found where the plastic strips are merging and some plastic fracture occurs.
1. Introduction 2. Problem statement and formalization. centers), is loaded by the infinite shear efforts ∞ = ∞, ∞ = 0 (Figure ). Under extra compression by normal stresses conditions on infinities ∞ = − the crack is closed and it can’t excite any normal homogeneous field of stresses in the environment: ( , )= ∞ = − = const. The interaction of cracks faces is accompanied by their friction resulted in some extra tangent stresses on the cracks faces which can oppose the shear and whose values are supposed to be stable and the same
= 0 = 0 ( 0 < ), where 0 - static coefficient of the sliding friction. When ∞ > 0 the cracks faces are shifted, some extra tangent stresses are acting on their faces = 0 = 0 and some plastic strips are developing on the cracks continuation due to the concentration of stresses − − ≤ | + 2 | ≤ + , = 0
( ∈ ), −∞ < < ∞, whose length should be found. In these plastic strips the yield criterion must be fulfilled: 2 + 2 = 2, where – shear limit of liquidity.
Unlike linear case, due to the nonlinear character of the problem it cannot be reduced to the study of similar problems only in case of the given stress on infinities ∞ = ∞ − 0. stress some anti flat stress-and-strain state arises in the body which can be found by the shear ( , ) . Two nonzero components of the stresses tensor are given by the formulae = ⁄ and = = ⁄
( – shear modulus of the material). The shear ( , ) is symmetric referred to the lines ( ∈ )and is antisymmetric about X-axis. That is why it can be determined only in a halfstrip 0 ≤
≤ , 0 ≤ < ∞(area ).
Due to the balance conditions and Hooke’s law, the function ( )= _ ( , )+ ( , ) is analytical one in the elastic part of the body. So, to determine stress-and-strain state of the body we will define a boundary problem for function ( ) in the area , consisting in the necessity to fulfill four conditions.
1. Due to the symmetry we have obtained
Im ( )= 0
(( = , ≥ 0)⋃( = , + ≤ ≤ )⋃( = + , ≥ 0)). 2. On cracks faces the stress
= 0 = const, so 3. In the area ≤ ≤ + the yield criterion has been fulfilled, so 4. Stress-and-strain state on infinities is defined by the formula
Re ( )= 0
(
= , 0 ≤ ≤ ).
| ( )| =
(
= , ≤ ≤ + ).
→∞
lim ( )= _∞.
(1)
(2)
(3)
(
As the function ( )in the zone is analytical and one-sheet, it conformally maps to the part of circle | | ≤ , Re ≥ 0, Im ≥ 0 (zone Figure 2). In this case the following points match the areas boundaries
and : = ∞+ ℎ → = ∞, = 0 → = 0, = → = 0 − √ 2 − 02, = + → = . Section ( = , 0 ≤ < ∞) within the zone is mapped to the interval (Im = 0, 0 ≤ Re ≤ ∞); interval ( = , 0 ≤ ≤ ) – to Re = 0 , section ( = , +
≤ 〰 ≤ )⋃( = + , ≥ 0) – to the interval (Im = 0, ∞ ≤ Re ≤ ). The interval ( = , ≤ ≤ + ), corresponding to the plastic strip, is mapped to the circle − √ 2 − 02 ≤ Im ≤ 0, arch (| | = , −arccos( 0⁄ )≤ arg ≤ 0 ). 3. Study of plastic strips propagation
The solution of the boundary problem (1)-(
= {Im ≥ 0} (see Figure 2) and we will find the function ( ) in parametric form ( )= 6( )exp( 0)+exp(− 0),
6( )+1 3( )+ 3( )−
, 3( )= √1−( (++1)1 ) , − )), 0 = arccos 0, = − 2
. 2+1 6( )= 5( ) 0⁄ , 5( )= = −tg ( 2 0 (2arctg ∞− 0 √ 2− 02 = ( ), = ( )
( ∈ ) Function ( )is given as a composition of elementary mappings:
As (0 < < 1) we consider the analytical in the upper half plane function receiving actual added values at the same values of .
In the finishing point of the strip = . So, from the formula (
The function determined by composition of elementary mappings ( )looks like:
( )= 2 arcsin (√ sin ).
2 As ( )= + , the length of plastic strips can be obtained from the last formula = 2 arcsin ( 1
sin )− . √ +1
2
The length of plastic strips as functions of stresses have been calculated for different stresses and
distance between cracks and for different levels of the faces friction and are shown Figure 3.
(
The strips length can’t exceed the half distance between the tops of neighboring cracks. Stress ∞ = ∞∗ when = − can cause some ductile fracture of the body. From the formula (8) it is clear that the value of critical loading ∞ = ∞∗ has satisfied the equation sin( /(2 ))= √ + 1. Hence, 0∗ = 0 + √ 2 + 02tg ( 0 (1 − 2 )).
(9)
Under fixed levels of the friction of faces conditions ( 0 = const) the last par has given the dependence of critical loading
on the distance between cracks (Figure 4).
The interaction of faces has very strong impact on the value of critical loading for the cracks locating very close to each other. The bigger distance between cracks the more weaken is the interaction impact. formulae
(8), dependencies [9]:
Without the interaction of faces ( 0 = 0) the (9) have given the known = 2 arcsin ( 22+− ∞∞22 sin )− , ∞∗ = tg ( (1 − 2 4 )).
In case of large distance between cracks the function stresses has been expressed by the formulae
(
The conducted study has proved a considerable impact of cracks interaction on plastic strips development near their tops. Therefore, the important problem requiring a special attention is the study of cracks interaction located in neighbourhood in an elastic-plastic body resulted in possible plastic strips coalescence and plastic fracture. 4. References