Thermo-Mechanical Analysis of Elasto-Plastic Cyclic Torsion of a Tubular Element

Journal of Thermal Stresses, 34: 893–910, 2011Copyright © Taylor & Francis Group, LLCISSN: 0149-5739 print/1521-074X onlineDOI: 10.1080/01495739.2011.601255

THERMO-MECHANICAL ANALYSIS OF ELASTO-PLASTICCYCLIC TORSION OF A TUBULAR ELEMENT

Joze Stropnik and Franc KoselFaculty of Mechanical Engineering, University of Ljubljana,Ljubljana, Slovenia

The paper treats a tubular element which presents a cantilever beam loaded by atorque moment on the free end. The value of torque is chosen so that the stressesrange in the elasto-plastic domain. The rheological properties of the material followthe elastic-linear hardening model. The tubular element is subjected to a cyclic torsionof such frequency that heat is generated in the tubular element. The quantity of thegenerated heat depends on the amplitude of the torque, frequency of the load andmechanical properties of the tubular element material. The paper compares the resultsof the temperature change at characteristic points along the tubular element obtainedvia numerical analysis as well as experimentally.

Keywords: Elasto and plastomechanics; Heat generation; Heat transfer; Thermomechanics; Torsion

In general, simultaneous effects of mechanical and thermal loads applied to astructural element in the elasto-plastic domain present an elasto-plastic thermomechanical problem. In treating this problem all necessary material propertiesof the continuum have to be considered. Furthermore, the loads applied are tobe within such a range as to avoid elasto-plastic deformations. They perform acertain amount of deformation work in the elasto-plastic domain, and when theunloading has been completed the equilibrium state between elastic and plastictorque moments in the tubular element is restored. The tubular element releasesthe elastic part of the deformation, while the plastic part induces permanentdeformation. During the process of plastic deformation crystals movement along socalled sliding planes generate heat due to internal friction, which can be noted bythe increased temperature of the tubular element.

Thermo-elastic problems have been discussed by Mendelsen [9], Nowacki [8],Parkus [10] and Boley and Weiner [2]. Kapoor [6] examined thermo-elastic stressesin a thick-walled cylinder and spherical shell, the curved surfaces of which wereaxially symmetric exposed to a known heat generation. Cooper and Pilkey [5]explored the problem of distribution of normal and tangential stresses along theaxis of a straight beam, with a given quasi-static temperature load of the beam.So far, thermo-elastic problems have been solved by employing numerical methods

Received 8 May 2009; accepted 11 June 2011.Address correspondence to Joze Stropnik, Faculty of Mechanical Engineering, University of

Ljubljana, 1000 Ljubljana, Slovenia. E-mail: [email protected]

893

894 J. STROPNIK AND F. KOSEL

Zienkiewicz [13], Srikanth and Zabaras [12] and experimental methods Brustengaet al. [3], Loung [7].

The paper discusses the problem of cyclic torsion with torque moment beingapplied to a tubular element, taking into consideration the elastic linear hardeningproperties of the material. When the tubular element is loaded with a cyclic torqueof such a value that the maximum stress is in the elastic domain solely elastic shearstress appears in the tubular element and there is no heat generation. In the casewhen the cyclic torque load is grows, the shear stress exceeds the yield point andin the cross-section area, elastic as well as plastic shear strain occurs. The materialthat has already been plastically deformed responds elastically when the load isremoved (the case of elastic unloading). When the load is completely removed,elastic deformations disappear and plastic deformations remain. When the load isreapplied, in the reversed direction, the material initially responds in a linear plasticmanner. In the next phase of loading, the yield point increases by a certain degree,which means that hardening of the material has occurred.

ELASTO-PLASTIC CYCLIC TORSION

The paper deals with a tubular element under torsional load, where thematerial has an elasto-linear hardening rheological model. The function � �� for thechosen material is seen from Figure 1(a). The cross-section is in the elastic domainuntil the stresses in the outer boundary �r = b� exceed the yield point �0. Withincreasing torque moment, the plastic domain is growing from the outer boundarytowards the inside until the radius r0 is reached. With increasing torque moment,the stresses in the plastic domain are growing as well until they reach the innerboundary �r = a�.

Figure 1 Diagram �− � and the cross-section of the beam.

ELASTO-PLASTIC CYCLIC TORSION OF TUBULAR ELEMENT 895

The total shearing strain in the elasto-plastic domain is a combination ofelastic and plastic strains, as follows:

��r� = �0G

+ ��r�− �0Gt

(1)

where ��r� is the stress at the cross-section of the circle with the radius r (Figures 1and 2), G is a shear modulus in the elastic domain and Gt is a tangential shearmodulus of the material in the plastic domain (Figure 1(b)).

The tubular element is subjected to torque moment Mt, which causes stressgeneration �max in the surface �r = b�. The entire cross-section is in the elasticdomain at the point when the stress on the outer surface �r = b� of the tubularelement reaches a value �0. This occurs with the torque being

M0el =

��0�b4 − a4�

2b(2)

If the torque Mt exceeds the value expressed in Eq. (2), a part of the cross-sectiona ≤ r < r0 is in the elastic domain �Mt > M0

el�, while the remaining part r0 ≤ r ≤ b

of the cross-section is in the plastic domain [10], Figures 2 and 3.In this case when the plastic domain of the cross-section is in the range: r0 ≤

r ≤ b, the torque of the elastic part of the domain Mr0el and the torque of the plastic

part of the domain Mr0pl are

Mr0el =

��0�r40 − a4�

2r0� M

r0pl =

�

6

[4�0�b

3 − r30 �+ ��maks − �0�3b4 − 4r0 · b3 + r40

b − r0

](3)

Figure 2 Elasto-plastic state at the cross-section when moment Mt is applied.


Figure 3 Stress state after unloading.

The total torque is the sum of torques Mr0el and M

r0pl :

Mt = Mr0el +M

r0pl (4)

Due to the effects of torque Mt, a distortion occurs, which may be represented by aspecific angle of twist per unit length �, and is given for each individual part of thecross section, i.e., either in the elastic or in plastic domain, Eqs. (5) and (6).

�el =2Mr0

el

� ·G · �r40 − a4�(5)

�pl =2

� ·Gt · �b4 − r40 �

[M

r0pl +

2�3

· �0 · �b3 − r30 � ·(Gt

G− 1

)](6)

Specific angles of twist must be identical, therefore �el = �pl, which yields themaximum shearing strain �max = �0

br0, the shearing strain �0 being at a yield point �0.

The maximum tangential stress occurring in the outer surface of the tubularelement is:

�max = �0 +Gt��max − �0� (7)

The value of the torque, the size of the plastic domain of the cross-section and thespecific angle of twist are interdependent quantities. In other words, if one quantityis known, the other two can be determined as well.

DISTORTION AND DILATATION ENERGY IN THE TUBULAR ELEMENTSUBJECTED TO TORQUE AND RESIDUAL STRESSES

The distortion energy is:

W = 12

∫�V�

�ij · ij · dV (8)


In the case of pure torsion of the tubular element all normal stresses as well as stress�yz are zero. The resultant of stresses � = ��r�, composed of the components �xy and�xz, equals torsion stress at the cross-section of the circle with the radius r, and canbe written as:

� = ��r� =√�2xy + �2xz (9)

Total energy necessary for distortion and dilatation of the tubular element subjectedto torque is obtained from

Wcel = Wel +Wpl

Wcel =� · l · �2012G · r20

{3�r40 − a4�+

[8r0 · �b3 − r30 � ·

(1− Gt

G

)+ 6

Gt

G· �b4 − r40 �

]}(10)

At the moment when the tubular element, which is partially in the plastic domain,is unloaded, the equilibrium state is disrupted, preventing the recovery to the initialstate. At the same time residual stresses (Ress 2005) [11] occur, making only part ofthe input energy reversible. The irreversible part of the energy is transformed intoheat energy.

When the tubular element is in the elasto-plastic state (Figure 2), it is subjectedto torque Mt, Eq. (4). Such a state is a result of the input energy stated in Eq. (10).After unloading the resultant torque equals zero. The resulting stress state afterunloading is illustrated in Figure 3.

Maximum stress after unloading �rmax is given in the following equation

�rmax =Mt

Wt

= 2Mt · b��b4 − a4�

(11)

Stress distribution �r = �r�r� can be expressed in the following equation

�r =�max

b· r = 2Mt

��b4 − a4�· r (12)

The tubular element which is subjected to torque Mt, Eq. (4), enters partially in theplastic domain at its outer boundary, while the core remains in the elastic domain.When the tubular element is unloaded, the following residual tangential stresses �z,are present:

�z = �ep − �r (13)

In the above Eq. (13), �ep stands for stresses that result from the elasto-plasticstate of the tubular element, whereas �r stands for stresses after unloading, given inEq. (12). On the outer edge of the cross-section �r = b� residual stress �zb occurs.

�zb = �rmax − �max =2Mt

��b4 − a4�· b − �max (14)


Figure 3 clearly shows the radius rx at which residual stresses equal zero. It can becalculated from the condition �r = �ep as follows:

rx =�0C

·(1− Gt

G

)(15)

where the constant C is:

C = 2Mt

��b4 − a4�− �0 ·Gt

r0 ·G

ENERGY DUE TO RESIDUAL STRESSES AND ENERGY TRANSFORMEDINTO HEAT

Energy of residual stresses can be determined by the following Eq. (16).

Wzi =12

∫�V�

� · � · dV� i = 1� 2� 3 (16)

Individual energy quantities due to residual stresses are determined for particularareas of the cross-section of the tubular element as follows:

for the part of the cross-section rx < r < b

Wz1 =� · l · �2zb

G · �b − rx�2·∫ b

rx

r · �r − rx�2 · dr

= � · l · �2zb12G · �b − rx�

2

[3�b4 − r4x �− 8rx�b

3 − r3x �+ 6r2x �b2 − r2x �

](17)

for the part of the cross-section r0 < r < rx

Wz2 =� · l · �2z0

G · �rx − r0�2·∫ rx

r0

r · �rx − r�2 · dr

= � · l · �2z012G · �rx − r0�

2

[3�r4x − r40 �− 8rx�r

3x − r30 �+ 6r2x �r

2x − r20 �

](18)

for the part of the cross-section a < r < r0

Wz3 =� · lG

·∫ r0

a

[�za + ��z0 − �za� ·

r − a

r0 − a

]2

· r · dr = � · l12G

· B (19)

In the Eq. (19) B stands for

B =

6�2za · �r20 − a2�+ 4�za ·

(�z0 − �zar0 − a

)· [2�r30 − a3�− 3a · �r20 − a2�

]+(�z0 − �zar0 − a

)2

· [3�r40 − a4�− 8a · �r30 − a3�+ 6a2 · �r20 − a2�]

(20)


After the tubular element is unloaded, part of the energy W ∗el, is released:

W ∗el =

� · l · �∗2max

4G · b2 �b4 − a4�� where �∗max =2Mt · b

��b4 − a4�(21)

The total amount of energy due to residual stresses in the tubular element is

Wz = Wz1 +Wz2 +Wz3 (22)

Part of the energy which is transformed into heat is therefore:

�W = Wcel −W ∗el −Wz (23)

Eq. (23) states the amount of energy when the torque acts upon the tubularelement Mt in the clockwise direction, the tubular element at the same time beingrotated at the angle of twist � and shear strain � = r · �, which equals the sum ofelastic �e and plastic �p shear deformation. These conditions are shown in ��− ��diagram in Figure 4 by the function 0AB. This is followed by the elastic unloadinguntil point C. In the next phase, the tubular element undergoes the torque Mt inthe counter-clockwise direction in such a way that the angle of twist � and shearstrain � remain the same as in the first phase. This can be seen in Figure 4, functionCDE. Next, the elastic unloading until point F is carried out. During this procedure

Figure 4 Diagram �− � under torsional cyclic loading with torque moment.


slight hardening of the material occurs, therefore the yield point �02 is slightlyhigher than the yield point �01. In the case of the chosen material, the yield pointafter the hardening effect increased minimally, therefore in this phase the amount

Figure 5 Testing machine. 1: framework; 2: electric motor with a brake; 3: reduction gear; 4: eccentriccrank; 5: slotted link; 6: shaft; 7: fixed support; 8 crank; 9: test specimen; 10: support (color figureavailable online).


of generated heat equals the amount of the heat generated in the previous phase(stated in the Eq. (23)). In the following cycle the procedure is repeated at the sameangle of twist �. Due to the hardening of the material, the yield point increasesslightly, however with the angle of twist remaining the same, the specimen assumesadditional plastic properties after each cycle (Figure 4). After a couple of initialloading cycles, the material reaches the final yield point.

EXPERIMENTAL MONITORING OF HEAT GENERATION IN THEDYNAMICALLY LOADED TUBULAR ELEMENT

For the purposes of experimental monitoring of heat generation in the tubularelement to which an alternating cyclic torque was applied, a special testing machine(Figure 5) was designed and built. In designing the machine, attention was paidto creating a machine with proper dimensions that would enable us to conducttests on the given specimens and would allow recording the temperature changewhen the cyclic torque is applied to the tubular element. An aluminium tubularelement, as seen in Figure 8, was chosen as the test specimen. The machine enabledchanges of frequency and amplitude of the torque applied to the tubular element.Time and temperature were measured on 3 segments of the test specimen by meansof a computerized monitoring and Spyder 8 recording instrument (Figure 6). Thedata were collected, processed and graphically presented via a computer programCatman 3.1. The temperature was measured in 0.5 seconds or 1 seconds – intervals,with 0.1�C allowable deviation. Spyder 8 has 16 channels, 5 of which were usedduring the experiment (4 to measure temperature and 1 for time). For each channel2048 data were set.

The test specimen was supported with two supports, one of them fixed and theother simply supported which you can see from Figures 5 and 8. The torque alteredharmonically, its frequency being changed by altering the rotational frequency ofthe electric motor with an in-built frequency transformer.

The amplitude of the torque depended on the initial setting of the eccentriccrank. The aluminium test specimen was thermally insulated from the inside as well

Figure 6 Monitoring and recording instrument Spyder 8 (color figure available online).


as from the outside, except for the places where the tabular element is connectedwith the fixed support and on the other side with the crank. These places presentthe locations of the most intense heat conduction. Temperatures were measured onthe inner side of the wall of the tabular element in three points 1, 2 and 3, this areT1, T2 and T3 (Figure 8).

NUMERICAL ANALYSIS

Quantities of residual stresses and the effect of the generated heat werenumerically evaluated on aluminium tubular elements of 200mm length, thedimensions of the annular cross-section being b = 11mm, a = 95mm. Furthermore,experimental monitoring of the temperature change in the tubular element wascarried out as well. Tubular elements upon which the heat generation was monitoredwere made from aluminium alloy AlMgSi0,5F22 (according to DIN 9107), whichhas, according to DIN 40501/1 the yield stress �0 = 160 240MPa. To determinethe actual yield stress and in order to set the hardening model, tension testswere performed on the ZWICK 50 testing machine. Figure 7 shows the resultsfor four test specimens, the average yield stress being �0 = 227MPa, Young’smodulus E = 68566MPa, the average tangential modulus of elasticity Et = 312MPaand Poisson’s ratio 0.29. Shear yield point was determined to be �0 = 131MPa(according to v. Mises theory) and 113.5MPa (according to Tresca theory). Theshear modulus of elasticity obtained from Young’s modulus and Poisson’s ratio wasG = 26575MPa. The torsion test was carried out on the AMSLER testing machine,which resulted in a yield point �0 = 120MPa and shear modulus of elasticityG = 26320MPa. These two values were referred to in the subsequent calculations.Tangential shear modulus Gt = 104MPa follows from Et, taking into considerationPoisson’s ratio 0.5 in the plastic domain.

Figure 7 Stress-strain �� − � function for aluminium alloy AlMgSi0,5F22.


Figure 8 Geometric computational model of the test specimen.

Table 1 Amounts of generated energy

r0 Mt rx Wcel W ∗el Wz �W

No. mm N m mm N mm N mm N mm N mm

1 1100 11131 11000 460464 460464 000 0002 10501 11556 10596 689192 496231 099 1928603 1011 11773 10401 866611 515091 423 3510954 975 11882 10306 1028663 524642 806 5032165 950 11904 10288 1140507 526600 934 612961

r0: elastic-plastic radius; Mt : torque; rx: radius at which residual stressesequal zero; Wcel: total distorsion and dilatation energy; W ∗

el: released energy;Wz: energy due to residual stresses; �W : into heat transformed energy.

Table 1 shows amounts of generated energy for different depths r0 of thecross-section of the tubular element in the plastic domain. In example 1, theentire cross-section lies within the elastic range �r0 = b�, therefore there is noheat generation ��W = 0�. Subsequent examples witness increased plastification ofthe cross-section, transforming consequently greater amounts of energy into heat,reaching the maximum value at r0 = a = 950mm, when the entire cross-sectionacquires plastic properties. The calculated energy values apply to one loading cycle,with an increasing number of cycles, the generation of the heat in the tubularelement increased as well.

The temperature was calculated in the same points 1, 2 and 3, on the inner sideof the wall of the tabular element with professional software ANSYS (1994) [1] fora geometric computational model, see Figure 8. Numerical analysis was based onreal geometrical data for each individual test specimen and mechanical properties ofthe material as established during laboratory tests. When dealing with heat transfer,


Table 2 Heat properties of materials

Quantity Symbol Unit Steel Aluminium Insulation

Density kg/m3 7850 2830 33Specific heat capacity c J/�kg ·K� 460 896 1600Thermal conductivity � W/�m ·K� 57.25 229 0.032Heat transfer coefficient � W/�m2 ·K� 7.23�1� – 9.31�2�

�1�In case of heat transfer from the steel surface into the surroundings.�2�In case of heat transfer from the insulation surface into the

surroundings.

heat properties of the material listed in Table 2 were taken into account. The heatgeneration (following Eq. (23)) calculated per unit time was considered as the loadupon the test specimen. It should be pointed out that the energy in question dependson the frequency and amplitude of the load applied to the test specimen. The inputdata �A is the amount of energy upon the unit of the middle surface of the tubularelement cross-section expressed as:

�A = 2f · �W� · �a+ b� · l (24)

where f stands for the frequency of the load, �W stands for the energy at the semicycle of load, a in b indicate the radius of the tubular element, and l stands for thedistance between the two supports.

HEAT TRANSFER FROM THE TEST SPECIMEN

The heat generated in the tubular element subjected to alternating cyclictorsion is conducted along the supports and insulation and is subsequentlytransferred and radiated into its surroundings. The amount of the radiated heat isnegligible in comparison to the overall heat transfer. Table 2 shows heat propertiesof individual materials taken into consideration during numerical analysis.

Numerical values for density, specific heat capacity and thermal conductivityof individual materials were obtained from thermodynamic reference tables andfrom the insulation manufacturer, whereas the heat transfer coefficient wascalculated according to specific test conditions. The heat was transferred from thecylindrical surface of the insulation and from the steel surfaces of the supports.Considering the heat transfer, however, it must be borne in mind that thesurrounding air was not stationary due to the movements of individual elements ofthe testing machine. Air movement velocity was estimated as v� = 05m/s.

The average heat transfer coefficient from the insulation surface into the air is�m = 931W/�m2 ·K� and is obtained from Eq. (25) [4].

�m = �

d·Num (25)


The average Nusselt number is given in Eq. (26).

Num = C ·Rem · Prn ·(Pr�Prs

)025

(26)

where the Reynolds number is Re = v�·d�

and the Prandtl number is Pr = �· ·cp�

.The preceding equations are based upon the following values of air at 25�C:

thermal conductivity � = 00263W/�m ·K�; kinematic viscosity � = 1553mm2/s;mass density = 1185 kg/m3; specific heat capacity cp = 1013 kJ/�kg ·K�;diameter of the outer surface of the insulation d = 60mm, exponents n = 037 (forPr < 10) and m = 06; and constant C = 026 (for Re = 103 2× 105).

Values of Prandtl numbers for air on the surface of the insulation Prs as wellas at a greater distance from the boundary layer Pr� depend on the temperature onthe surface of the insulation and on the temperature of the surrounding air. Sincethe two temperatures do not differ considerably, also Pr� equals Prs and the ratiois �Pr� /Prs� � 1.

As far as conditions for convection heat transfer from a plane steel surface areconcerned, the perimeter of the plane front wall d = 150mm and the air movementvelocity v� = 05m/s are taken into account. The calculated Reynolds number is4829, whereas the Prandtl number is 0.713. The average Nusselt number 41.23 isobtained from Eq. (27); (the Eq. being valid for 06 < Pr < 50 and Re < 105).

Num = 0664Re05 Pr0333 (27)

The average heat transfer coefficient along a plane surface is �m = 723W/�m2 ·K�,and it results from Eq. (25).

EVALUATION OF THE RESULTS OBTAINED VIA NUMERICAL ANALYSISAND EXPERIMENTALLY GATHERED RESULTS

There were 17 tests conducted during the experiment. On each specimen analternating torque of various amplitudes and frequencies was applied. The results oftwo different specimens are here analyzed.

Figure 9 shows the time-temperature change for test specimen P17. The datafrom numerical analysis and experimental results are compared. The dimensionsof the test specimen are b = 1075mm and a = 9175mm. The entire cross-sectionof the specimen is within the elastic domain until the torque reaches the value ofM0

el = 1099Nm. During the test, the testing machine was set to such a degree ofeccentricity as to create the entire angle of twist of the specimen at the value of 4.96�,which caused plastification of the cross-section up to the radius r0 = 1050mm. Thetorque of Mr0

el = 9101Nm was applied to the part of the cross-section in the elasticdomain, whereas the part of the cross section in the plastic domain was subjectedto M

r0pl = 2127Nm torque, hence in total Mt = 11228Nm. After one loading cycle

�W = 1864Nm of energy was transformed into heat. During the entire test thefrequency of the load was 2Hz, the total number of cycles being 1200 in 600seconds. From the temperature curve it is evident that the maximum temperaturevalue T1 is reached at point 1, whereas the minimum value of the temperature T3 is


Figure 9 Comparison of the results for specimen P17 obtained experimentally and via numericalanalysis.

at point 3. Such results may be explained by the heat conduction along the specimeninto the steel supports and in the subsequent convection heat transfer into thesurroundings. The difference between the measured and the calculated temperaturemay be explained by thermal properties of materials, which do not prove to be

Figure 10 Comparison of the results for specimen D8 obtained experimentally and via numericalanalysis.


exactly the same in real circumstances as in the adopted rheological model of thematerial of the specimen.

Figure 10 illustrates temperature change for specimen D8 with b = 1111mm,a = 898mm, whose wall thickness was by 0.555mm thicker than that of testspecimen P17. The entire cross-section of the specimen is in the elastic domain untilthe torque exceeds the value of M0

el = 14816Nm. During the test the total angleof twist was 4.77�, which caused plastification of the cross-section up to the radiusr0 = 1092mm. A torque of the value M

r0el = 13320Nm was applied to the part of

the cross-section which was in the elastic range, whereas the part in the plasticrange was subjected to M

r0pl = 1738Nm torque, the total torque therefore being

Mt = 15058Nm. After one loading cycle �W = 146Nm of energy was transformedinto heat.

During the initial 2700 s of the test, the specimen was subjected to a 2Hzfrequency, which was subsequently changed to 3Hz. The frequency change of theload provoked a considerable increase in the generated heat, which was manifestedin the substantial temperature rise at all 3 test points.

KINEMATIC HARDENING OF THE MATERIAL

Tensile-compression tests along one axis were carried out by means of theInstron 8802 universal testing machine on three test specimens. The test specimenswere made from AlMgSi0,5F22 material. The tests were performed in the 20.5�Cenvironment, the temperature of the specimen, however, did not exceed 35�C andthe temperature change did not influence significantly the mechanical propertiesof the material. During the experiment the yielding processes as well as kinematichardening of the material were under scrutiny. Each experiment was carried outwith the constant amplitude for normal deformation being 1.2 % and with the samespeed of deformation for each of the 6 cycles of the loading. Deformations weremeasured by means of extensometer Instron 2620-63.

The results manifest minimal differences among the test specimens, thereforethe diagram � − shown in Figure 11 represents the results obtained from one testspecimen only. The diagram reveals that the mechanical properties of the materialhave been stabilized already after the first cycle of loading, which is evident inthe neatly matching curves of the subsequent loading cycles. Consequently, similarbehaviour of the material may be predicted for the cyclic torsion loading, thereforethe diagram �− �, taking into consideration v. Mises stress theory, would provesimilar to the diagram shown in Figure 12. Furthermore, Figure 12 shows an explicitlinear increase �� during loading and unloading as well as relatively high yieldingpoint �0 in the first half of the first loading cycle.

In the second half of the first loading cycle, as well as in all subsequent cycles,however, tangential stresses �� are no longer linear and the yield point is no longerclearly manifested. Furthermore, the highest tangential stress �max, �max for the samemaximal tangential strain, �max is a little lower than the one in the initial part ofthe first loading cycle. All these differences result from the kinematic hardening ofthe material which creates a greater hysteresical loop. A bigger hysteresical loopindicates more energy which is transformed into heat.


Figure 11 Tensile-compression cyclic test �� − � for aluminium alloy AlMgSi0,5F22.

Figure 12 Stress-strain ��− �� function for torsional cyclic loading.


CONCLUSIONS

The paper discusses a quasi-static thermoplastic problem of the continuum.For the solution of the problem, a rheological model for elasto-plastic material wasused. The results of tensile testing of the specimen material matched well with theadopted model and provided evidence confirming its choice.

The primary goal of the paper was to analyse the effects of cyclic torqueapplied to a tubular element, the value of which caused the tangential stresses in aparticular part of the cross-section to exceed the yield point, inducing a shift into theelasto-plastic domain. Considering the adopted rheological model of the materialand the shape of the cross-section to which the torque was applied, equations forthe values of the torque and tubular element deformation in the elasto-plastic regionwere formulated. The equations make it possible to calculate the value of the totaltorque and the angle of twist at the given size of the part of the cross-section whichacquires plastic properties and vice versa.

Equations for elastic and plastic work in the elasto-plastic domain were given.Alternating torque was applied to the tubular element in the elasto-plastic domain.On locations where the cross-section tangential stresses exceeded the yield stress,the outer surface �r = b� of the cross section became plastic, while the rest of thecross-section formed the elastic core.

One part of the input energy is transformed into residual stresses �Wz�, andanother part of the energy �W ∗

el� is released after unloading. The part of energy,however, which equals the subtraction of the sums of energies �Wz +W ∗

el� fromthe total energy �Wcel� necessary to reach the elasto-plastic state (Eq. (23)) istransformed into heat.

The paper presents a numerical example of how to determine the extent of theplastic domain and the value of the torque applied to a tubular element at a pre-setdeformation (angle of twist). To determine the quantity of generated heat, whichdepends on the thermodynamic and mechanical properties of the material, testspecimen dimensions, amplitude and frequency of the torque, a numerical analysisof the temperature change was carried out. Considering at the same time the heattransfer the analysis was conducted by means of ANSYS program on 17 tubularelements. The comparison of the experimental results and numerical analysis showson average a 4.9% deviation. The paper presents numerical and experimental resultsfor 2 tubular elements.

The results obtained form the tensile-compression cyclic test for materialsof the same quality as the test specimens for cyclic torsional tests show thatdeviations from the presupposed elastic-linear hardening model occur mostly due tothe kinematic hardening of the material. This is the reason why the generated heat inthe case of real material response differs from the calculated one for the presupposedmodel of the material. The work in the future will be directed into research of theeffects of the kinematic hardening of the material upon the generated heat at cyclictorsion.

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