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The problem is shown in Fig.A. 1Two bares where, bar number (2) has a length (2L)
and cross sectional area (A), while the other bar number (1) has length (L) and cross
sectional area (2A).
Fig.A. 1 Two Bar Problem Structure
There are cyclic thermal loading applied on bar (2) such that the specification of the
applied thermal loading will be as shown in Fig.A.2 . While no temperature applied on
the other bar.
Fig.A. 2 Cyclic loading on bar (2)
Elastic Solution:
By applying equilibrium equation β¦
ππ1. οΏ½π΄π΄2οΏ½ + ππ2.π΄π΄ = ππ (1)
By applying compatibility equation β¦
ππ = ππ1. 2πΏπΏ = ππ2. πΏπΏ (2)
From elasticity relation β¦
ππ = ππ/πΈπΈ (3)
Where (E) is the modulus of elasticity, (Ο) are the stresses in bar (1) and (2), (Ξ΅) are the
strains in bar (1) and (2), (q) is the displacement of the two bares, (A) are the
crossectional area and (L) is the length of the bars.
From the last equations we can get the stresses in each bar where:
ππ1 = πππ΄π΄
, ππ2 = ππ2π΄π΄
(4)
Fig.A. 3 Elastic Stresses without Temperature
After applying a cyclic temperature loading on bar (2) where, temperature are vary
through time according to Fig.A. 4, which shows that the temperature applied only on bar
(2) and the other bar do not affected by any temperature gradient.
Fig.A. 4 Temperature variation through time in the two bar structure.
Elastic Solution after applying temperature:
There are no changes happened in the equilibrium and compatibility equations, the only
change happened in the equation of the elasticity relation where β¦
ππ1 = ππ1
πΈπΈ , ππ2 = ππ2
πΈπΈ+ πΌπΌΞππ (5)
So according to that change, the stresses value in each bar will change to β¦
ππ1 = πππ΄π΄
+ πΈπΈπΌπΌΞππ (6)
ππ2 = ππ2π΄π΄β οΏ½1
2οΏ½ πΈπΈπΌπΌΞππ (7)
According to equations (6) and (7) the cyclic thermal loading in each of the two bares
cane be written in the following form:
During the heating half cycle:
Ξππ1th = EΞ±ΞT (8)
Ξππ2th = βοΏ½1
2οΏ½πΈπΈπΌπΌΞππ (9)
During the cooling half cycle:
Ξππ1th = βπΈπΈπΌπΌΞππ (10)
Ξππ2th = οΏ½1
2οΏ½πΈπΈπΌπΌΞππ (11)
Fig.A. 5 Cyclic stresses on the 2 bar structure
Condition for no plasticity due to thermal loading:
Condition # 1:
ππ1 β€ ππy (12)
πππ΄π΄
+ πΈπΈπΌπΌΞππ β€ ππy (13)
πποΏ½ + ΞπποΏ½οΏ½οΏ½οΏ½ β€ 1 (14)
Where β¦
πποΏ½ = πππ΄π΄βππy
(15)
ΞπποΏ½οΏ½οΏ½οΏ½ = πΈπΈπΌπΌΞππππy
(16)
Condition # 2:
ππ2 β₯ ππy (17)
ππ2π΄π΄β οΏ½1
2οΏ½πΈπΈπΌπΌΞππ β₯ ππy (18)
πποΏ½ β ΞπποΏ½οΏ½οΏ½οΏ½ β€ β2 (19)
Equations (14) and (19) explain when the structure will exceeds the elastic limit
and begins to shakedown.
Fig.A. 6 Bree Diagram for the 2 bar problem.
Once the structure exceeds the limit of the elastic limit it means that it will enter the
area that we call it the shakedown area, in a simple way and by explaining what Melan
(Melan, 1938) had explained before in his theory, the shakedown region as indicated in
the Fig.A. is the region where the structure begins to have plastic deformation due to the
load in one of the cycles by exceeding the boundary of elasticity but on the other hand if
we put a cyclic loading value which was less in the value than the last cycle which causes
plasticity in that case the structure will behave elastically again without any new plastic
deformations, and that is why we say on any structure on similar condition that the
structure is shaking down.
In the next step we are trying to know what will happen if the structure exceeds that
shakedown load boundary and explaining a two new phenomenon that we call them
Reversed Plasticity and Ratcheting.
What happened if the structure exceeds the Elastic limit of the first bar (Bar
(1))?
The answer of this question can be according to the following relation, where we
consider that condition # 1. of equation (1.15) was violated:
πποΏ½ + ΞπποΏ½οΏ½οΏ½οΏ½ > 1 (20)
During the first cycle half (Heating of bar (2)):
ππ1 = ππy (21)
By substituting in the equilibrium equation to find (ππ2) β¦
ππ2 = QAβ ππy
2 (22)
By applying the compatibility condition to find the resulting plastic strain after violating
condition # 1. We can get the followingβ¦
ππ = οΏ½ππ1
πΈπΈ+ ππππ1οΏ½ . πΏπΏ = οΏ½ππ2
πΈπΈ+ πΌπΌΞπποΏ½ . 2πΏπΏ (23)
Fig.A. 7 stresses after violation of condition # 1
According to the last equation of the compatibility condition we can get the plastic
strain generating in bar one for this case in the heating step which can be written in the
following form β¦
ππππ1οΏ½οΏ½οΏ½οΏ½ = 2[πποΏ½ + ΞπποΏ½οΏ½οΏ½οΏ½ β 1] (24)
Where,
ππππ1οΏ½οΏ½οΏ½οΏ½ = ππππ1/(πππ¦π¦ .π΄π΄) (25)
By comparing the result of equation (20) by the last analysis we can deduce that the
plastic strain in bar (1) will be zero if equation (14) for condition # 1 was not violated
which is logical.
During the 2nd
Ξππ1th = βπΈπΈπΌπΌΞππ (26)
cycle half (Cooling of bar (2)):
ππ1 = ππy β πΈπΈπΌπΌΞππ (27)
It is obviously appear in equation (26) that the stresses in bar (1) during the cooling that
happened in bar (2) in the second half cycle, the stresses in bar (1) are in a decreasing
manner so we can put anther condition to know what will happened in the structure if
condition # 3 is violated where:
Condition # 3:
For no Yielding β¦. ππ1 = ππy β πΈπΈπΌπΌΞππ > βππy (28)
ΞπποΏ½οΏ½οΏ½οΏ½ β€ 2 (29)
By the same sequence that happened to make condition # 3, there is also anathor
conation for bar (2) to check that the stresses in that bar do not reach the yielding during
the first half cycle, so condition # 4 will be:
Ξππ2th = + οΏ½1
2οΏ½πΈπΈπΌπΌΞππ (30)
ππ2 = QAβ πππ¦π¦
2+ οΏ½1
2οΏ½πΈπΈπΌπΌΞππ β€ πππ¦π¦ (31)
πποΏ½ + οΏ½12οΏ½ΞπποΏ½οΏ½οΏ½οΏ½ β€ 3/2 (32)
After the last analysis steps we answered the last question whish was, what will
happened if the structure exceeded the elastic limit? We knew that there also a limit for
the shakedown area, according to conditions # 3 & 4 we were able to find the borders of
this area, also we proofed what had mentioned before in Melanβs theory (Melan, 1938)
that the limit load for the shakedown area for any structure will be reached for such
perfectly plastic material if the residual stresses reach twice the residual stresses for each
element in the structure which happened in the case of bar (2) in condition # 3.
Limit Load Condition: ππππππ = 2πππ¦π¦ (33)
Also there are anther methods that used to find the limit load in any structure like the
twice elastic slope method.
Reversed plasticity boundary:
Know we are going to investigate the behavior of the structure if condition # 3 is
violated:
ΞπποΏ½οΏ½οΏ½οΏ½ > 2 (34)
This violation will lead us to:
Fig.A. 8 Reversed plasticity
ΞΟ1 = β2Οy (35)
ΞΟ2 = πππ΄π΄
+ Ο2
2β οΏ½ππ
π΄π΄β Ο2
2οΏ½ = Οy (36)
Ξππ = οΏ½Ξππ1
πΈπΈ+ Ξππβππ1οΏ½ . πΏπΏ = οΏ½Ξππ2
πΈπΈβ πΌπΌΞπποΏ½ . 2πΏπΏ (37)
ΞππβΜ ππ1 = Ξππβππ1/(πππ¦π¦ .π΄π΄) (38)
According to Fig.A. 7 and equation (38) we found a new phenomenon that called
reversed plasticity or (Law cycle fatigue area), in this are one of the structure element
begins to reversed the value of its stresses in each cycle with the presence of a significant
amount of plastic strain (ΞππβΜ ππ1) so according to that strain we could be able to guss the
life of any structure according to the corresponding plastic strain obtained in each cycle.
Ratcheting Boundary:
Ratcheting boundary can be obtained by answering the question that say, what will
happen if condition # 4 was violated? In this case structure will behave in a very
dangerous way for any structure exposing to cyclic loading.
Retching means that each element in the structure starts to have a significant
progressive value of plastic strain after each cycle of the cyclic load, leading to large
plastic strain that the structure cannot with stand then the total collapse or failure for the
structure or the mechanical component occurs.
After the violation of condition # 4:
πποΏ½ + οΏ½12οΏ½ ΞπποΏ½οΏ½οΏ½οΏ½ > 3/2 (39)
During cooling:
Ο1 = 2 οΏ½πππ΄π΄β ΟyοΏ½ (40)
ΞΟ1 = 2 οΏ½πππ΄π΄β ΟyοΏ½ β Οy = 2Q
Aβ 3Οy (41)
ΞΟ2 = Οy βQAβ Οy
2= 3Οy β
2QA
(42)
Ξq = L. ΞΟ1
E= 2L. οΏ½ΞΟ2
Eβ Ξππππ1 β πΌπΌΞπποΏ½ (43)
Fig.A. 9 Ratcheting behavior for two bar structure
ΞπποΏ½Μ οΏ½π1 = 2[2πποΏ½ + ΞπποΏ½ β 3] (44)
Where,
ΞπποΏ½Μ οΏ½π1 = ΞΞ΅p 1E
Οy (45)
During 2nd
Ξq = L. οΏ½ΞΟ1
E+ Ξππππ1οΏ½ = 2L. οΏ½ΞΟ2
Eβ πΌπΌΞπποΏ½ (46)
heating of bar (2):
ΞΟ1 = Οy β 2 οΏ½πππ΄π΄β ΟyοΏ½ = 3Οy β
2QA
(47)
ΞΟ2 = 2 οΏ½πππ΄π΄β ΟyοΏ½ β Οy = 2Q
Aβ 3Οy (48)
ΞπποΏ½Μ οΏ½π2 = 2[2πποΏ½ + ΞπποΏ½ β 3] (49)
Where,
ΞπποΏ½Μ οΏ½π2 = ΞΞ΅p 2E
Οy (50)
So according to the last investigation for the ratcheting in the 2 bar structure we would
be able to fine the plastic strain value for each element in the structure corresponds to the
applied cyclic loading, this allows us to find the total plastic strains and estimate the life
time of the structure at the corresponding plastic strain values calculated from equations
such as (50) & (45).