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1、Lesson 3 Torsion of a circular bar Let us consider a bar of circular cross section twisted by couples T acting at the ends.A bar loaded in this manner is said to be in pure torsion.It can be shown from considerations of symmetry that cross sections of the circular bar rotate as rigid bodies about th
2、e longitudinal axis,with radii remaining straight and the cross sections remaining circular.Also,if the total angle of twist of the bar is small,neither the length of the bar nor its radius will change.During torsion there will be a rotation about the longitudinal axis of one end of the bar with res
3、pect to the other.For instance,if we consider the left-hand end of the bar as fixed,then the right-hand end will rotate through an angle with respect to the left end.At the same time,a longitudinal line on the surface of the bar,such as line nn,will rotate through a small angle to the position nn.Be
4、cause of this rotation,a rectangular element on the surface of the bar,such as the element shown in the figure between two cross sections distance dx apart,is distorted into a rhomboid.When a shaft is subjected to pure torsion,the rate of change d/dx of the angle of twist is constant along the lengt
5、h of the bar.This constant represents the angle of twist per unit length and will be denoted by.Thus,we see that=/L,where L is the length of the shaft.Then,the shear strain is obtained by.The shear stress which act on the sides of the element have the directions shown in Fig.1.For a linear elastic m
6、aterial,the magnitude of the shear stress is.Equations(1)and(2)relate the strain and stress at the surface of the shaft to the angle of twist per unit length.The state of stress within the interior of the shaft can be determined in a manner similar to that used for the surface of the shaft.Because r
7、adii in the cross sections of the bar remain straight and undistorted during twisting,we see that an interior element situated on the surface of an interior cylinder of radius is also in pure shear with the corresponding shear strain and stress being given by the following expressions.These equation
8、s show that the shear strain and shear stress vary linearly with the radial distance from the center of the shaft and have their maximum values at the outer surface.The shear stresses acting in the plane of the cross section,given by Eq.(3b),are accompanied by equal shear stresses acting on longitud
9、inal planes of the shaft.This result follows from the fact that equal shear stresses always exist on mutually perpendicular planes.If a material is weaker in shear longitudinally than laterally,the first cracks in a twisted shaft will appear on the surface in the longitudinal direction.The state of
10、pure shear stress on the surface ot he shaft is equivalent to equal tensile and compressive stresses on an element rotated through an angle of 450 to the axis of the shaft.If a material that is weaker in tension than in shear is twisted,failure occurs in tension along a helix inclined at 450 the axi
11、s.This type of failure can easily be demonstrated by twisting a piece of chalk.The relationship between the applied torque T and the angle of twist which it produces will now be estabished.The resultant of the shear stresses must be statically equivalent to the total torque T.The shear force acting
12、on an element of are dA is tdA,and the moment of this force about the axis of the bar is t dA.Using Eq.(3b),this moment is also equal to G2dA.The total torque T is the summation over the entire cross-sectional are of such elemental moments,thus where J=2dA is the polar moment of inertia of the circular cross section.From Eq.(4)we obtain=T/GJ which shows that,the angle of twist per unit length,varies directly with the torque T and inversely with the product GJ,known as the torsional rigidity of the shaft.