山农双语材力3-.ppt

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1、 Mechanics of Materials31 Concept and practicle examples32 External torque of a transmission shaft,internal torque and diagrams33 Pure shear34 Stresses of circular shafts in torsion35 Deformations of circular shaft in torsionCHAPER 3 TORSION 31 Concept and practicle examplesShaft：In engineering the

2、members of which deformations are mainly torsion.Such as transmission shafts in machines,drill rods in oil-drilling rigs etc.Torsion：Resultant of the external forces is a force couple and its acting plane is perpendicular to the axis of the shaft.Under this case the deformation of the rod is torsion

3、.ABOmmOBAThe angle of twist()：The angle of rotation of one section with respect to another.Shearing strain()：The change of a right angle between two straight lines.mmOBAPractical examples in engineeringTransmission shaft Me32 External torque of a transmission shaft,internal torque and diagrams 1、Ext

4、ernal torque of a transmission shaft where：P-power,unit:kilowatt(kW)n rotational speed,unit:r/min 2、Internal torque and its diagram 1)Internal torque：The moment of internal forces acting in arbitrary section of the member in torsion.Designated by“T”.2)Determine the internal torque by the method of s

5、ection:MeMeMeTxnn（+）nn（-）3)Sign conventions for internal torque：“T”will be considered to be positive when the relation between its turning direction and the out normal line obeys the right hand rule,otherwise it will be considered to be negative.4)Internal torque diagram：Sketch that expresses the la

6、w of change of the torque in each cross section along the axis.Purpose the law of change of the torque；Value of|T|max and the position of its section strength calculation（critical section）xT Example 1 A transmission shaft is rotating at n=300r/min.Knowing its input power is PC=500kW，and its output a

7、re PA=150kW，PB=150kW，PD=200kW，Try to plot the internal torque diagram.nA B C D MB MC MDSolution：Calculate the external torqueMADetermine the internal torque（suppose it is positive）nA B C D MB MC MDMAPlot the internal torque diagramEach section in segment BC is a critical section.xT4.789.566.37nA B C

8、 D MB MC MDMA33 Pure shear1)、Experiment：a).Preparation：Plot the longitudinal lines and circumference lines；To act a pair of external torque m。1.Stresses of hollow round shaft with thin wall：（r0:average radius）Thickness of the wall2).After deformation：The circumference lines do not change；The longitu

9、dinal lines are changed into slants。3).Conclusions：Shape,size and distance of the circumference lines on the shaft surface do not change，while rotating with respect to one another along the axis of the shaft。All the longitudinal lines revolve through a same angle 。All squares drawn on the shaft surf

10、ace warp into rhombus with same sizes。acddxbdy No normal stress There are only the shearing stress perpendicular to the radius at each point on the cross section.Magnitude of the shearing stresses are the same on the same section，directions of them coincide with that of the internal torque.4.)Relati

11、on between and ：A small cubic element is shown in the figure：mmOBAL5)、Magnitude of the shearing stress in the hollow shaft with thin wall：A0：Area of the circle with an average radius2、Theorem of conjugate shearing stresses：Formula(a)is called theorem of conjugate shearing stresses.It indicates shear

12、ing stresses always exist on mutually perpendicular plane and occur in equal and opposite pairs and point,perpendicularly,either toward or away from the intersection line of the planes.acddxb dy tz4、Hookes law of shear：There are only shearing stresses and no normal stresses on the four side planes o

13、f the element，which is called pure-shear stressed state。l T=Me Hookes law in shear：The shearing stress is directly proportional to the shearing strain if the shearing stress does not exceed smaller than the proportional sheer limit of the material(p).In above formula G is a elastic constant of mater

14、ial.It is called modulus of elasticity in shearing.as has no dimension,G has the same dimension with .Modulus of elasticity in shearing、Youngs modulus and Possions ratio are three elastic constants.For the isotropic materials there is the following relation between the three constants.（See later cha

15、pters about deduction of the formula）：5、Strain energy and its density acddxb dy dzzxy Infinitesimal work of the element：Strain energy per unit volume：34 Stresses of circular shafts in torsionStress on the cross section of the round shaftGeometric deformationsPhysical relationsStatics 1).After deform

16、ation the cross sections still remain planes;2).No elongation or contraction along the axis；3).Longitudinal lines are still parallel to each other.1、Observation of the torsional test of the straight circular rod：2、Stresses on the cross section of the round shaft in torsion：1).Geometric deformation r

17、elation：The shearing strain at an arbitrary point is directly proportional to the distance from the center of the section to the point.of the variation rate twisting angle along the length.2).Physical relation：Hookes law：Substituting it into the preceding formula we get：3).Static relation：LetAfter s

18、ubstituting into the physical relation we getOpdAFormula to calculate the shearing stress of any point in the section with distance to the center.4).Discussions on the formula：It can be only applied to the isotropic,linear-elastic straight-circular rods with small deformations.In the above formula：T

19、Torque in the cross section.It is determined from the external torques by the method of section.Distance of the point to the center of the circle.IpPolar moment of inertia of section.It is a purely geometric quantity without any physical meaning.unit：mm4，m4。The formula is deduced from the solid circ

20、ular shaft，but it can also be applied to the hollow circular shaft.Values of their Ip are different.a.For a solid circular section：DdOb.For a hollow circular section:dDOd Distribution of the shearing stresses（solid section）（hollow section）In engineering the members with hollow section are widely use

21、d to increase the strength,save materials and decrease the weight of structures.Determine the maximum shearing stress：FromwhenWt section factor in torsion（section factor modulus in torsion),geometric quantity，unit：mm3 or m3 a.For the solid circular section：b.For the hollow circular section：4、Strengt

22、h calculation of the round shaft in torsionStrength condition：For the round shaft with equal sections：(is called permissible shearing stress.)Three aspects of strength calculations：Check the strength：Design the dimension of the section：Calculate the permissible load：Example 2Example 2 Power is 150kW

23、，speed of rotation is 15.4r/s for the rotator of a motor as shown in the figure.Permissible shearing stress is=30MPa.Try to check its strength.TmSolution：Determine its torques and plot the torque diagramCalculate and check the shearing stress strengthStrength of this shaft satisfies request D3=135D2

24、=75 D1=70ABCmmx35 DEFORMATION OF A CIRCULAR SHAFT IN TORSION 1、Deformation in torsionFrom the formulaWe can determine the relative angle of twist between two end sections of the rod with the length l.2、The angle of twist per unit length ：or3、Rigidity conditionorG Ip indicates the capacity that the d

25、imension and material properties of the section resist torsional deformations.It is called the torsional rigidity of the circular shaft.Three aspects of rigidity calculations:Check the rigidity：Design the dimension of the section：Calculate the permissible load：Sometimes,these conditions may be consi

26、dered to select materials.Example 3 Uniformly distributed couple m=20Nm/m acted a circular shaft with length L=2m is shown in the figure.If the ratio of the inside diameter and the outside diameter of the shaft is =0.8,G=80GPa,the permissible shearing stress is=30MPa.Try to design the outside diamet

27、er of the shaft；As =2/m,try to check the rigidity of the shaft and determine the angle of rotation of the right-end section.Solution：Design the outside diameter of the rod40NmxTSubstituting the values we get：D 0.0226m。Check the rigidity according to the rigidity condition in torsion40NmxTAngle of tw

28、ist of the right-end section:Example 4 Knowing：Me1=7.024kNm,Me3=4.21kNm,G=80GPa，=70MPa，=1/m.Try to determine：Diameter d1 of segment AB and diameter d2 of segment BC？If select the same diameter for the whole shaft,how much is the diameter？500400Me1Me3Me2ACBTx7.024 4.21(kNm)From the rigidity condition

29、：500400N1N3N2ACBTx7.0244.21(kNm)Sum up：If select the same diameter for the whole shaft Chapter 3 Exercises 1.In a gear-box,why is the shaft with a slow speed thicker than that with a height speed？2.When the shearing stress and the normal stress coexist in an element,is the theorem of the conjugate s

30、hearing stress amenable?Why?3.For a hollow circular tube of Al,the outside diameter is D100mm，the inside diameter is d80mm,and the length is L2.5m.The Shearing elastic modulus is G28GPa.If a force couple is acted on both ends of the circular tube and results in the pure torsion,try to determine the

31、torsional angle when the maximum shearing stress is 50MPa.For the solid shaft of Al subjected to the same force couple and having the same maximum shearing stress,what is the diameter D2?Determine the ratio of the weight of the hollow tube to that of the solid shaft.the smaller the torque with the m

32、aximum absolute value in the shaft is,the more reasonable it is.So wheel 1 must be replaced with wheel 2.After transposition,torque diagram of the shaft is shown in the figure.Under this case,the maximum diameter of the shaft is 75mm。Tx 4.21(kNm)2.814Solution:As the two have the same material and the same length,the ratio of weights is equal to the ratio of cross-section areas: