Axial stresses - Free Essay Example

Sample details Pages: 12 Words: 3697 Downloads: 7 Date added: 2017/06/26 Category Physics Essay Type Descriptive essay Tags: Stress Essay Did you like this example? An analysis of the state of stresses generated by torsion in twisted non-circular bars Abstract One of the main problems that must be solved in the design of the machine components stressed in torsion is to establish their optimum shapes and dimensions in order to resist for a given twisting load. Also, in the case of some parts (crankshafts, ornamental bars, blades of screw propeller for torpedoes) manufactured by using plastic torsion as technologic operation the designer must determine the needed deformation energy. An important factor that has a direct implication on the designing methodology and on the resistance to torsion of these components is the state of stress generated by the twisting load application. Don’t waste time! Our writers will create an original "Axial stresses" essay for you Create order Usually, the components that transmit or must support twisting moments are circular or tubular in cross-section, but in certain cases components with other than circular sections can be used. The present paper reviews the main aspects concerning the state of stress generated by elastic and plastic torsion in non-circular bars. Keywords: Torsion; State of stress; Shearing stresses; Axial stresses; Non-circular bars 1. Introduction By considering a non-circular bar stressed in torsion, the following effects generated by the twisting load application are emphasized by the Theory of Elasticity in the twisted bar: the helical orientation of the initially straight fibres, especially, for large angles of twist; the warping of the plane cross-sections; the change in length of the material fibres and differences in length from a deformed fibre to another; the modification of the initially right angle between the deformed fibre and the section normal to it. One of the main factors that generates or influences these effects is the state of stress, which occurs in bars during torsion. Relating to this state of stress generated by torsion, the Theory of Elasticity and Plasticity discusses, generally, about the shearing stresses that act on the cross-sections of the twisted bars. 1.1. Remark The solutions for the problems concerning the shearing stresses magnitude and distribution on the cross-sections of elastic, elastic-plastic and plastic twisted bars, circular or non-circular in cross-section, can be found in Refs.[1],[2],[5],[6],[8],[9],[10],[11],[12],[13],[14],[15],[16],[17]and[18]. Some authors analyzed the possibility that axial stresses to be developed in certain conditions in the twisted bars. Thus, an experimental analysis concerning the development of axial stresses during hot torsion tests, for the case of round samples twisted for different ranges of temperature, is presented in Ref.[7]. According to the obtained results it has been concluded that hot torsion develop in twisted bars axial stresses whose distribution and magnitude depend on temperature range. Some aspects relating to the generation of axial stresses in the twisted non-circular bars are also discussed in Ref.[3]. The knowledge of the values and distribution of the stresses in a twisted bar is of a great importance for the designing and manufacturing processes of different machine components. Thus, by knowing the maximum value and the distribution of the stresses on different sections, the designer can establish an optimum shape of the component section can determine the optimum dimensions of the machine components and can verify certain sections of the bar stressed by torsion. Also, the knowledge of the stresses value and distribution will permit to determine the deformation energy needed to manufacture different parts by using torsion as technologic operation. On the other hand, the knowledge of the state of stress due to torsion will permit to elucidate different phenomena that occur in the twisted bars. Concerning the state of stress generated by torsion in the twisted non-circular bars, the present paper addresses the following questions: 1. What is the real state of stresses generated by torsion in a non-circular bar? 2. What is the variation of the stresses on the cross and longitudinal sections of the twisted non-circular bars? In answer to these questions, the paper performs an analytical analysis concerning the shearing stresses distribution and, based on some theoretical suppositions or experimental investigations, discusses the possibility that the axial stresses to be generated and developed in the twisted non-circular bars. 2. Shearing stresses distribution on different cross-section shapesa review According to the Theory of Elasticity and Plasticity related to torsion, the state of stress on the cross-section of a twisted bar is characterized by the following general properties[2],[3],[6],[10],[13]and[17]: in the case of elastic torsion, the state of stress in an arbitrary point of the section corresponds to the state of pure shear and the shearing stress vector reaches its maximum value on the section contour; in the case of elastic-plastic torsion, the following two regions can be distinguished on the cross-section of bar: an elastic region, located in the close proximity of the axis of bar, where the Hookes law is valid at any point and the stress distribution presents similar aspects like in the case of elastic torsion; a plastic region, which starts when the shearing stress reaches the yield valuekand satisfies the condition of plasticity expressed by the relation: wherexzandyzare the non-zero components of the shearing stresses resultant andmaxis the maximum value of the shearing stress; in the case of fully plastic torsion, by considering that the whole section is in yielding state, the shearing stress vector is constant in magnitude and its direction is perpendicular to the normal on the contour of the plastic region. The distribution and the magnitude of the shearing stresses present different aspects as a function of cross-section shape and dimensions. This paragraph analyses the main aspects concerning the shearing stress distribution on the most frequently used non-circular sections. 2.1. Sections having curvilinear contours In the case of a solid elliptic cross-section, having the semi-axesaandband stressed by the twisting momentT( 1), the distribution of the shearing stresses presents the following aspects[1],[6],[13],[14]and[15]: the resultant of the shearing stresses can be calculated by using the following relation: the two components of the shearing stress vector (xz,yz) have a linear variation along the Oxand Oyaxis, these components and their maximum values being expressed by the following relations: the maximum stress will occur on the elliptic contour in the point closest by the axis of bar atx=a,y=b, its value being given by the relation: whereis the angle of twist per unit length andGis the transverse modulus of elasticity. From the diagram shown in 1, we can observe that the maximum value of the shearing stress is reached in the pointsBandDwhere it is equals toxz. the components of the shear stress vectorxzandyzare equal to zero on the Oyand Oxaxis, respectively; if we consider the ratio between the two components of the shearing stresses vector we obtain the following relation: this relation expresses the fact that on an arbitrary direction OM the shearing stress vector has a constant direction. Plastic yielding will occur at the extremities of the minor axis of the ellipse whenmax=k, the corresponding twisting moment being equal to: In the case of an elliptic section having an elliptic hole ( 2), the shearing stresses variation presents the same aspects like in the case of the solid elliptic sections[11]. By considering a section having a shape in the form of an inverse ellipse ( 3), the components of the shearing stress vector will be given by the following relations: xz=2GcsinuF1,yz=2GccosuF2, whereF1andF2are functions that depend onkand on the real parametersuandv[11]. In the case of a lobe keyed section ( 4), by using a conformable transformation, the shearing stress components can be obtained from the following relation: zbeing a complex variable[11]. In the case of a section in the form of a curvilinear triangle ( 5), the maximum shearing stress occurs in the point located at the middle of the triangle side and has the following value: max=1.15Ga, abeing the triangle side[2]and[9]. 2.2. Sections having contours formed by straight sides In the case of a section in the form of equilateral triangle, having the heighthand the equations of its sides given by:View the MathML source, the distribution of the shearing stresses on the section shown in 6presents the following aspects[1],[2],[6]and[8]: the stress vector components can be calculated using the following relations: the maximum intensity of the shearing stress vector will be reached at the middle of the triangle sides and it can be determined using the following relation: whereais the size of the triangle side. Yielding will start to set in when the shear stress vector reaches the intensitykand the twisting moment becomes equal to: the fully plastic twisting moment can be calculated using the following relation: In the case of a rectangular cross-section, the distribution of the shearing stresses presents the following particularities[5],[12]and[15]: the maximum shearing stress occurs at the middle of the longest side, its value being given by the relation: wherebandaare the dimensions of the cross-section (bbeing the longer side) andmis a factor given in tables as a function ofbanda; the distribution of the shearing stresses on the cross-section is shown in 7; we can observe that at the middle of the smaller side the shearing stress is equal to: xzmax=yzmax, being a coefficient also given in tables as a function of ratiob/a. for a given twisting moment and cross-section area, the maximum shearing stress reaches its smallest value in the case of dimensions leading to the smallest polar second moment of area; for a narrow rectangular cross-section[8],[10]and[18], according to equationxz=2Gx, the shearing stress is linear inx; its maximum value occurs at a/2 ( 8) and is given by the relation: yielding starts at the midpoint of the longer sides when the applied twisting moment becomes equal to: the fully plastic twisting moment is given by the following formula: In the case of a square cross-section[1],[8]and[10], by using a harmonic stress function, the maximum shearing stress will be given by the following relation: abeing the square side andAan expression equal to the yielding starts at the midpoint of the square sides when the stress vector reaches the intensityk; in the case of fully plastic torsion, the twisting moment can be calculated using the following relation: 3. Aspects concerning the possibility that axial stress to be generated by torsion Let us consider the particular case of a rectangular bar stressed in torsion by applying the twisting moment at its both ends. After the twisting load application, we can observe the following effects generated by torsion in the twisted bar: the main effect of the twisting load application will be the helical orientation of the initially straight longitudinal edges of bar ( 9a) and hence the change of their length[4]. Thus, by examining the half of a longitudinal section of bar as non-deformed ( 9b) and deformed after its twisting ( 9c) we can remark the following aspects: the initially straight edge AA of the rectangle OAAO will occupy the new position AA, after its helical orientation around a cylindrical surface of radiusR=OA=OA; the helical orientation of the edge AA will determine the modification of its length from the initial valueL0=LAAto the valueLAA=LAA+LA, where LAis the elongation of the edge of bar; the value of the elongation can be calculated using the following formula: whereRis the radius of the cylindrical surface described by the points of edge during torsion,L0the initially length of edge andnis the number of complete rotations of the ends of bar. 3.1. Remark From relation(21), we can observe that the elongation and the changes in length of the edge depend on the applied angle of twist, the position of the edge in comparison with the neutral axis of bar and the dimensions of bar. in the frame of the same rectangle OAAO, we consider now the side MM which after its twisting will become the helix MM folded around a cylindrical surface of radius OM=OM. From the evident inequality OMOA, we can conclude that the length of the helix MM is smaller than the length of the helix AA (LAALMM). Hence, it results that the changes in length of the rectangle sides or edges of bar depend on their position in comparison with the neutral axis of bar. By this point of view, we can divide the longitudinal section of bar in the following two regions as deformed after twisting: the region from the neutral axis and its neighbouring, where the elongation of the rectangle sides presents the smallest values and becomes equal to zero for the neutral axis; the region from the periphery of section, where the elongation of the rectangle sides presents the greatest values. Hence, we can suppose that the region from the neutral axis of bar, where are registered the smallest deformations, will oppose to the tendency manifested by the peripheral region to be elongated under the action of the twisting load. By examining the internal macrostructure on the etched longitudinal section of a square bar, made in low carbon steel and cold twisted with an angle of twist per unit length equal to 9.42102radmm1( 10), we can observe the presence of two distinct regions as deformed in the axial direction. The first region is located in the close proximity of the axis of bar and represents the rolling core, initially present in the non-twisted bar. In this region, the material will oppose to the tendency of bar to be deformed under the action of the twisting moment and hence minimum torsional deformations and the tendency to contract the material fibres will be registered. The second region is located in the peripheral zones of bar, where maximum torsional deformations and considerable elongation of the material fibres will take place. The above-presented aspects are confirmed by analyzing the variation of the material hardness determined in these two regions after one complete revolution of the ends of bar; thus: in the region from the neutral axis of bar, the hardness increases from a value equal to 196 HB before twisting till a value equal to 212 HB after twisting; in the region from the periphery of bar, the hardness increases from the value equal to 207 HB before twisting till a value equal to 230 HB after twisting. The differences between the values of the material hardness from the periphery and the neutral axis of bar can confirm the presence of these two distinct regions as deformed in the axial direction after twisting. Let us now consider two longitudinal edges of the same outer face of a rectangular bar, the first AA that represents the edge of bar and the second NN placed at the middle of the face ( 11). We can remark that, because these edges will be folded during torsion around cylindrical surfaces of different radii (OA and ON, respectively), the resulted helixes will present different lengths (LNNLAA). The above-presented aspects concerning the effects of torsion in the axial direction of non-circular bars can lead to the following conclusions: The modification of the length of bar edges during torsion and the differences in length resulted between the generated helixes indicate and determine the development of axial deformations and stresses in the twisted non-circular bars. 3.2. Remark The aspects relating to the generation and distribution of the axial stresses in the round bars twisted over the yield limit of material are analyzed in[7]. The analysis proves that the axial stresses can be tensile or compressive depending on temperature and material properties. The main cause that determines the appearance of the compressive stresses is attributed to grains deformation, which has a maximum intensity at low temperatures and at the beginning of the deforming process. The slipping of the grains, which has a maximum intensity at high temperatures and at the end of the deforming process is considered the cause that determines the appearance of the tensile stresses. The opposite tendencies manifested between the outer and inner regions of a non-circular bar indicate the generation of two opposite axial stresses in these two regions: tensile in the peripheral region and compressive in the region from the neutral axis. The differences in length resulted between the deformed longitudinal edges from the outer surface of bar and the variation of their elongation as a function of their position in comparison with the neutral axis of bar can explain the warping of the non-circular sections. These differences can, also, indicate the possibility that the distribution of the axial deformations and stresses to become non-uniform on the longitudinal sections of the twisted non-circular bars. 4. Discussion The distribution of the shearing stresses on the cross-section of the twisted bars is imperative to be known because: the values of these stresses are used in design to determine the shape and dimensions of the section of bar; the shearing stresses distribution and value have an important contribution and influence on different phenomena that occur in bar during torsion. The main aspects concerning the shearing stress variation emphasized by the above-presented analysis are as follows: on the cross-section of twisted non-circular bars only shearing stresses will occur and act; these stresses will determine a state of pure shear. the variation of the shearing stresses depends, generally, on the cross-section shape; thus: in the case of an elliptic section, the shearing stresses resultant has a constant direction that is parallel to the direction of the tangent to the boundary where it is cited by the considered radius; in the points of a given diameter the stresses resultant has a linear increase as we move away from its origin. in the case of rectangular, square and triangular sections the variation of the shearing stresses is non-linear on all the directions and depends on the section shape and dimensions. the shearing stress vector attains its maximum value on the contour of cross-section indifferently of its shape; the maximum intensity of the shearing stress vector will be reached in different points of the contour as a function of cross-section shape and dimensions; thus: in the case of an elliptic section, the maximum shearing stress will occurs at the extremities of the minor axis of the ellipse; in the case of a rectangular bar the maximum shearing stress is reached at the midpoints of the longest sides; in the case of an equilateral triangle the maximum shearing stress will occur at the midpoints of the triangle sides. Generally, the shearing stress decrease in the sense of displacement from the outer contour of cross-section to neutral axis and reach the zero value in the point located on the neutral axis. But, as a function of cross-section shape, the zero value of the shearing stress can be also reached and in other points of the cross-section. Let us consider the corner of a rectangular bar ( 12) and an element of area located in the close proximity of this corner. We can observe that, because on the lateral surface of bar the stress componentsxy=zy=0 are equal to zero, the shearing stress componentsxz,yzwill be, also, equal to zero. Hence, we can conclude that, for a polygonal section, the shearing stresses are equal to zero in the close proximity of the corners. From the above-mentioned effect of torsion concerning the modification of the angle between the longitudinal edge and the cross-section edges of a rectangular bar (initially right angles), we can remark the following aspects: the modification of this angle is smaller in the close proximity of the bar edge and greater near the longitudinal axis of the bar face, and, hence, the modification of the cross-section contour will take place. by examining the internal macrostructure of the material on an etched square cross-section ( 14), we can distinguish two regions as deformed in the transverse plane: the first region is located in the close proximity of the axis of bar and represents the rolling core, initially present in the non-twisted bar; the second region is located in the peripheral zones of bar. The irregular shapes of these regions point out that for large angles of twist the deformation on the cross-section can become non-uniform. By taking into account these aspects concerning the non-uniform deformation of cross-section, we can conclude that, in certain conditions, the shearing stresses can become non-uniform distributed on the cross-section of the twisted non-circular bars. The axial stresses can occur during torsion of the non-circular bars and can be generated by the modification in length of the longitudinal sections. These stresses can be tensile or compressive as a function of position of the longitudinal section in comparison with the neutral axis of bar. Thus, on the peripheral sections of bar the tensile stresses will occur and on the inner sections the compressive stresses will act. In certain conditions, the axial stresses can present a non-uniform distribution on the same longitudinal section or between different sections of bar as a function of their position in comparison with the neutral axis. 5. Conclusions The general theory of elastic or plastic torsion, when discusses about the state of stresses generated by torsion in the twisted bar, take only into consideration the shearing stresses that occur on the cross-section. According to the above presented aspects and suppositions, the complete state of stresses generated by torsion in the twisted non-circular bars must consider not only the shearing stresses on the cross-sections but also axial stresses, compressive or tensile, on the longitudinal ones. The development of axial stresses (tensile or compressive) in the twisted non-circular bars can explain the changes in the length (elongations or contractions) that take place during torsion in the case of bars unconstrained from the axial displacement. Hence, the knowledge of the distribution and values of axial stresses that occur during torsion will permit to determine the precise length of the semi-parts used to manufacture different mechanical components using torsion as technologic operation.

Essay on Gender Roles - 2001 Words

The term gender roles refers to the set of social and behavioral norms that are considered appropriate for individuals of a certain gender. These roles vary between cultures. Gender roles, unlike gender itself, are socially constructed. They may reflect the natural aspirations of the gender, or they may be manipulated, resulting in the oppression of a gender. Historically, gender roles have not always been consistent with those we have today. Though in many ancient societies men have been dominant to women, there are example where women have been considered equal to men, and where women have been dominant over men. In hunting and gathering societies, such as the ones of early humans, males and females were considered equal. Because of†¦show more content†¦Although, men and women were not fully equal, women still had many more rights in Egypt than in other civilizations. 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How effectively did the Soviet Union control Eastern Europe from 1945 to 1968 Free Essays

In 1945, it was very important that Stalin gained control over his sphere of influence; WW2 had come to an end, and the future of the Soviet Union was at an unstable point. From 1945-8 Stalin used a series of ‘salami’ tactics to gain control over the Eastern European countries. This would involve setting up communist parties within a country, winning a coalition government, and ultimately, taking over to run a full communist regime. We will write a custom essay sample on How effectively did the Soviet Union control Eastern Europe from 1945 to 1968? or any similar topic only for you Order Now This was effective in the sense that Stalin, in the space of three years, was able to takeover seven different countries. However, out of these, only one country – Albania – allowed the takeover without any opposition, the other countries gave more of a struggle. There was obviously some backlash against Stalin’s actions, as they did not run smoothly, and this is why he probably had to use more severe tactics when trying to take control over Eastern Europe. In Bulgaria (1945), Stalin had all opposing party members executed, and was only then able to take full control. After this, Stalin had no problems gaining power over such countries – without further opposition, as the same techniques were used in Poland and Czechoslovakia two to three years later. This shows that Stalin did not want to give way to chance, seeing what would happen in each country gradually, by using the minimum force needed. He wanted to be sure of total control, and therefore used the necessary tactics to succeed – it appears that the most extreme were the most effective. Stalin was able to keep such a tight ‘following’ due to the fear and propaganda that followed his infamous Red army. He had placed them within his sphere as a warning to the people. If they opposed, they would be dealt with by the Red Army through means of ‘re-education’. They kept an eye over the people, and crushed any sort of uprising. This would have been a deterrent enough for anybody looking to rebel against communism. This proved to be effective as there are no real incidences where the Red army have been called in to sort out a situation (apart from in Hungary and Czechoslovakia, but that was under different circumstances). When Stalin saw the lure of the Marshall Plan, he definitely felt under pressure to counteract this – coming up with Comecon 1949. This shows he was desperate, as firstly, this goes against communist views, as the idea was to send money to communist countries to aid them in their economic process. This was a bribe, nonetheless; but it worked. No countries already within the control of the Soviet Union left the sphere – if they were tempted by money, they could still have it, and Stalin would still have control. They only probably stayed with Stalin out of the fear of the consequences, and therefore this tactic was successful due to previous attempts of control, mainly because of the Red Army. Stalin finally dies in 1953, and Khrushchev takes over. He introduces a new relaxed policy, and looks to ‘de-Stalinise’ Eastern Europe, hoping to promote better relations with the West. Subsequently, other countries within the sphere wanted a more liberal government, and started to rebel. This shows that Stalin himself, along with his attitude and thinking, was an effective way to keep control over Eastern Europe. he was not a liberal man, he insinuated fear and death throughout the people, who dared not oppose. Now that there is a more liberal leader in office, the people feel this is the opportunity they need to express their feelings. What Khrushchev may have done wrong, was to initiate an ‘instant’ change, which could not be done on a large scale in regards to the whole of Eastern Europe at one given time. Stalin was effective in what he did, as he used gradual changes to implant his policy. 1953 also saw riots in Czechoslovakia, where people where burning Soviet flags and demolishing Stalin statues – this was an obvious attack on Soviet power. However, the riots were quickly put down by the Red army, but the consequences were not great ones, and therefore encouraged strikes in East Germany to occur. East Germans aired their grievances about wage cuts, but the Red army was called in to crush all uprising. 400 people were killed; Khrushchev was reverting back to Stalin’s old methods, however people were not put off, as there are further problems in Hungary in 1956. Does this mean that the Red Army were no longer feared, or was there a change in attitude from the people? A level of both, most likely, but why? This leads back to Stalin and fear that he as an individual imposed on the countries. Khrushchev did not have that same effect, and from then on, people were more open to oppose him, knowing they could probably get away with a lot more. There was also a problem with East Germany, and how the collapse of it could have been crucial, in terms of ‘winning’ the Cold War. Germany had been one of many significant factors in the Cold War, and a collapse in Germany would mean a collapse in Soviet power if the West were to reclaim it. This might have been a short term cause for the Warsaw Pact in 1955 – the Soviet’s version of NATO – this not only gained support against the West, but also helped to unite a dividing sphere. This does not prove to be of any worth (at this time) as in the following year, Hungary has a revolution. There main aim was to get rid of Rakosi – a mini Stalin – through mass demonstrations, which later turned into street fighting. Oddly, Khrushchev introduces Nagy as the new leader of Hungary, instead of sending in the Red army. Was it now that he realised that the Red army did not have so much of an effect that it used to? This is the first time compromising had been used, but quickly backfired. Riots continued, and Nagy had declared Hungary would be leaving the Warsaw Pact. The S.U. was looking at losing control over one country, and sure enough, others would follow. The Soviets again, reverted back to trusted methods, calling in the Red Army to crush all uprising – killing Nagy in the process. The exact same happened with Czechoslovakia in 1968. It now seems that from 1945-68, there were times when the Soviet had complete control over Eastern Europe, and others, quite the opposite. Stalin had realised that in order to have control, he would have to use extreme tactics – the fear of the Red Army, propaganda etc. He also knew that with complete control, there was no edge way for choice or compromise with the people. This is where Khrushchev went wrong, expecting to have complete control when he gave individual countries more power. Knowing of such a relaxed attitude, the people then took advantage of this, and were no longer scared of the repercussions, as there is uprising after uprising under Khrushchev’s rule. By the mid-50’s effective control is on its way to becoming highly non-existent. How to cite How effectively did the Soviet Union control Eastern Europe from 1945 to 1968?, Papers