2 Bar problem detailed analysis.pdf

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 …

𝜎𝜎1. �𝐴𝐴2� + 𝜎𝜎2.𝐴𝐴 = 𝑄𝑄 (1)

By applying compatibility equation …

𝑞𝑞 = 𝜀𝜀1. 2𝐿𝐿 = 𝜀𝜀2. 𝐿𝐿 (2)

From elasticity relation …

𝜀𝜀 = 𝜎𝜎/𝐸𝐸 (3)

Where (E) is the modulus of elasticity, (σ) are the stresses in bar (1) and (2), (ε) 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:

𝜎𝜎1 = 𝑄𝑄𝐴𝐴

, 𝜎𝜎2 = 𝑄𝑄2𝐴𝐴

(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 …

𝜀𝜀1 = 𝜎𝜎1

𝐸𝐸 , 𝜀𝜀2 = 𝜎𝜎2

𝐸𝐸+ 𝛼𝛼Δ𝑇𝑇 (5)

So according to that change, the stresses value in each bar will change to …

𝜎𝜎1 = 𝑄𝑄𝐴𝐴

+ 𝐸𝐸𝛼𝛼Δ𝑇𝑇 (6)

𝜎𝜎2 = 𝑄𝑄2𝐴𝐴− �1

2� 𝐸𝐸𝛼𝛼Δ𝑇𝑇 (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:

Δ𝜎𝜎1th = EαΔT (8)

Δ𝜎𝜎2th = −�1

2�𝐸𝐸𝛼𝛼Δ𝑇𝑇 (9)

During the cooling half cycle:

Δ𝜎𝜎1th = −𝐸𝐸𝛼𝛼Δ𝑇𝑇 (10)

Δ𝜎𝜎2th = �1

2�𝐸𝐸𝛼𝛼Δ𝑇𝑇 (11)

Fig.A. 5 Cyclic stresses on the 2 bar structure

Condition for no plasticity due to thermal loading:

Condition # 1:

𝜎𝜎1 ≤ 𝜎𝜎y (12)

𝑄𝑄𝐴𝐴

+ 𝐸𝐸𝛼𝛼Δ𝑇𝑇 ≤ 𝜎𝜎y (13)

𝑄𝑄� + Δ𝑇𝑇���� ≤ 1 (14)

Where …

𝑄𝑄� = 𝑄𝑄𝐴𝐴∗𝜎𝜎y

(15)

Δ𝑇𝑇���� = 𝐸𝐸𝛼𝛼Δ𝑇𝑇𝜎𝜎y

(16)

Condition # 2:

𝜎𝜎2 ≥ 𝜎𝜎y (17)

𝑄𝑄2𝐴𝐴− �1

2�𝐸𝐸𝛼𝛼Δ𝑇𝑇 ≥ 𝜎𝜎y (18)

𝑄𝑄� − Δ𝑇𝑇���� ≤ −2 (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:

𝑄𝑄� + Δ𝑇𝑇���� > 1 (20)

During the first cycle half (Heating of bar (2)):

𝜎𝜎1 = 𝜎𝜎y (21)

By substituting in the equilibrium equation to find (𝜎𝜎2) …

𝜎𝜎2 = QA− 𝜎𝜎y

2 (22)

By applying the compatibility condition to find the resulting plastic strain after violating

condition # 1. We can get the following…

𝑞𝑞 = �𝜎𝜎1

𝐸𝐸+ 𝜀𝜀𝑝𝑝1� . 𝐿𝐿 = �𝜎𝜎2

𝐸𝐸+ 𝛼𝛼Δ𝑇𝑇� . 2𝐿𝐿 (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 …

𝜀𝜀𝑝𝑝1���� = 2[𝑄𝑄� + Δ𝑇𝑇���� − 1] (24)

Where,

𝜀𝜀𝑝𝑝1���� = 𝜀𝜀𝑝𝑝1/(𝜎𝜎𝑦𝑦 .𝐴𝐴) (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

Δ𝜎𝜎1th = −𝐸𝐸𝛼𝛼Δ𝑇𝑇 (26)

cycle half (Cooling of bar (2)):

𝜎𝜎1 = 𝜎𝜎y − 𝐸𝐸𝛼𝛼Δ𝑇𝑇 (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 …. 𝜎𝜎1 = 𝜎𝜎y − 𝐸𝐸𝛼𝛼Δ𝑇𝑇 > −𝜎𝜎y (28)

Δ𝑇𝑇���� ≤ 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:

Δ𝜎𝜎2th = + �1

2�𝐸𝐸𝛼𝛼Δ𝑇𝑇 (30)

𝜎𝜎2 = QA− 𝜎𝜎𝑦𝑦

2+ �1

2�𝐸𝐸𝛼𝛼Δ𝑇𝑇 ≤ 𝜎𝜎𝑦𝑦 (31)

𝑄𝑄� + �12�Δ𝑇𝑇���� ≤ 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 Melan’s 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: 𝜎𝜎𝑟𝑟𝑟𝑟 = 2𝜎𝜎𝑦𝑦 (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:

Δ𝑇𝑇���� > 2 (34)

This violation will lead us to:

Fig.A. 8 Reversed plasticity

Δσ1 = −2σy (35)

Δσ2 = 𝑄𝑄𝐴𝐴

+ σ2

2− �𝑄𝑄

𝐴𝐴− σ2

2� = σy (36)

Δ𝑞𝑞 = �Δ𝜎𝜎1

𝐸𝐸+ Δ𝜀𝜀∗𝑝𝑝1� . 𝐿𝐿 = �Δ𝜎𝜎2

𝐸𝐸− 𝛼𝛼Δ𝑇𝑇� . 2𝐿𝐿 (37)

Δ𝜀𝜀∗̅𝑝𝑝1 = Δ𝜀𝜀∗𝑝𝑝1/(𝜎𝜎𝑦𝑦 .𝐴𝐴) (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 (Δ𝜀𝜀∗̅𝑝𝑝1) 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:

𝑄𝑄� + �12� Δ𝑇𝑇���� > 3/2 (39)

During cooling:

σ1 = 2 �𝑄𝑄𝐴𝐴− σy� (40)

Δσ1 = 2 �𝑄𝑄𝐴𝐴− σy� − σy = 2Q

A− 3σy (41)

Δσ2 = σy −QA− σy

2= 3σy −

2QA

(42)

Δq = L. Δσ1

E= 2L. �Δσ2

E− Δ𝜀𝜀𝑝𝑝1 − 𝛼𝛼Δ𝑇𝑇� (43)

Fig.A. 9 Ratcheting behavior for two bar structure

Δ𝜀𝜀�̅�𝑝1 = 2[2𝑄𝑄� + Δ𝑇𝑇� − 3] (44)

Where,

Δ𝜀𝜀�̅�𝑝1 = Δεp 1E

σy (45)

During 2nd

Δq = L. �Δσ1

E+ Δ𝜀𝜀𝑝𝑝1� = 2L. �Δσ2

E− 𝛼𝛼Δ𝑇𝑇� (46)

heating of bar (2):

Δσ1 = σy − 2 �𝑄𝑄𝐴𝐴− σy� = 3σy −

2QA

(47)

Δσ2 = 2 �𝑄𝑄𝐴𝐴− σy� − σy = 2Q

A− 3σy (48)

Δ𝜀𝜀�̅�𝑝2 = 2[2𝑄𝑄� + Δ𝑇𝑇� − 3] (49)

Where,

Δ𝜀𝜀�̅�𝑝2 = Δεp 2E

σy (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).