中国石油大学工程力学课件tm402ma.ppt

2019/3/30,Kylinsoft,MOM-2-1,Chapter 2 Torsion 第二章 扭转,海盗船长,2019/3/30,Kylinsoft,MOM-2-2,2.1 Introduction,Torsion refers to the twisting of a straight bar when it is loaded by moments (or torques) that tend to produce rotation about the longitudinal axis of the bar.,汽车传动轴,2019/3/30,Kylinsoft,MOM-2-3,2019/3/30,Kylinsoft,MOM-2-4,2.2 Transmission of Power by Circular Shafts,rotary speed: n (rpm) power of the motor: N (kW),If power is expressed in horsepower (hp), the torque will be,1 Transmitted torque: T (Nm),2019/3/30,Kylinsoft,MOM-2-5,2 Internal Torque and Torque Diagram,cross-section method for Internal Torque,Equilibrium Smx = 0,T - m = 0,T = m,sign convention for Internal Torque,by the right-hand rule,2019/3/30,Kylinsoft,MOM-2-6,Example 2-1 A steel shaft is driven at A by a motor that transmits NA=50hp. The gears at B, C and D drive machinery requiring power equal to NB=NC=15hp and ND=20hp, respectively. The shaft is operated at a speed n=300r/min. Draw the diagram to show the twisting moment along this shaft.,Solution (1) Determine the moments applied to the motor and three gears,mB = mC = 351 Nm mD = 468 Nm,2019/3/30,Kylinsoft,MOM-2-7,mA=1170 Nm mB=mC=351 Nm mD=468 Nm,(2) Determine the internal torques,Smx= 0,T1 + MB = 0 T1 = - MB = - 351 Nm,TII + MC + MB = 0,TIII - MD = 0 TIII = MD = 468 Nm,TII = - MC MB = - 702 Nm,2019/3/30,Kylinsoft,MOM-2-8,T1= - 351 Nm TII= - 702 Nm TIII = 468 Nm,如果主动轮A安在D处，如何？,(3) Draw torque diagram,2019/3/30,Kylinsoft,MOM-2-9,2.3 Pure Shear (纯剪切),因为筒壁的厚度很小，认为沿筒壁厚度剪应力不变。 纯剪切:单元体截面上只有剪应力，没有正应力。,g,1 薄壁圆筒扭转,剪应变g ：纯剪切单元体的相对两侧面发生微小的相对错动，使原来互相垂直的两个棱边的夹角改变了一个微量，称为剪切应变或角应变。,2019/3/30,Kylinsoft,MOM-2-10,结构内某一点的变形如图所示，微元原始尺寸为ss ，其线应变和剪应变分别为：,ex= ey= gxy=,讨论题：,ex= ey= gxy=,ex= ey= gxy=,ex= ey= gxy=,2019/3/30,Kylinsoft,MOM-2-11,2 剪应力互等定理,对整个单元体有 S mz = 0 t(dy t )dx = t(dx t )dy 得到： t = t 即：一对相互垂直的平面，剪应力大小相等，垂直交线，方向共同指向或共同背离交线。这是剪应力互等定理。,取单元体,2019/3/30,Kylinsoft,MOM-2-12,2.4 Hooks Low in Shear (剪切虎克定律),扭转试验:,2019/3/30,Kylinsoft,MOM-2-13,弹性模量 E, 泊桑比 m 和剪切弹性模量G 都是材料的弹性常数。,在线弹性范围内，剪应变与剪应力成正比,对各向同性材料有三者的关系,G: 剪切弹性模量 Shear Modulus of Elasticity,单位：GPa,剪切虎克定律:,2019/3/30,Kylinsoft,MOM-2-14,方法与过程,变 形,应变分布,应力分布,应力公式,2.5 Shear Stress in Circular Bars,2019/3/30,Kylinsoft,MOM-2-15,cross sections rotate as rigid bodies about longitudinal axis radii remain straight cross sections remain plane and circular the length of the shaft keeps unchanged.,1 Pure torsion of circular bar (plane assumption),2019/3/30,Kylinsoft,MOM-2-16,2 变形几何关系(Geometry of Deformation),圆截面边缘上a点的剪应变,距圆心 r处的剪应变,在横截面上： df /dx=const,横截面上任意点的剪应变与该点到圆心的距离 r 成正比。即，剪应变线性分布。,2019/3/30,Kylinsoft,MOM-2-17,3 物理关系(Physical Relationship),剪切胡克定律,G ,由剪切虎克定律，得到r 处的剪应力,2019/3/30,Kylinsoft,MOM-2-18,4 静力平衡关系(Static Equilibrium Relationship),横截面上的扭矩,横截面上距圆心 r 处的剪应力,Polar Moment of Inertia (极惯性矩),2019/3/30,Kylinsoft,MOM-2-19,5 圆轴扭转时横截面上距圆心 r 处的剪应力,横截面上最大剪应力,横截面上任意点的剪应力与该点到圆心的距离r 成正比。即，剪应力线性分布。最大剪应力出现在圆周上各点。,抗扭截面模量 Section Modulus of Torsion,2019/3/30,Kylinsoft,MOM-2-20,极惯性矩 Polar Moment of Inertia,面积元,空心轴的极惯性矩和抗扭截面系数,实心轴的极惯性矩和抗扭截面系数,2019/3/30,Kylinsoft,MOM-2-21,6 圆轴扭转的强度条件,根据轴的受力情况作出扭矩图； 根据扭矩图求出最大扭矩Tmax； 按扭转剪应力公式计算最大剪应力tmax ； 阶梯轴，锥形轴的 Wt 不是常量， tmax并不一定发生在Tmax截面上，要综合考虑 T 和Wt, 求得最大的剪应力。,2019/3/30,Kylinsoft,MOM-2-22,Example 2-2,一轴AB传递的功率为N = 7.5kW，转速 n = 360 r/min。轴AC段为实心圆截面，CB段为空心圆截面。已知D = 3cm，d = 2cm。试计算AC段横截面边缘处的剪应力以及CB段横截面上内、外边缘处的剪应力。,解： （1）轴所受的外力偶矩为:,各横截面上的扭矩均为: T = m = 199Nm,（2）AC段和CB段轴横截面的极惯性矩分别为:,2019/3/30,Kylinsoft,MOM-2-23,T = m = 199Nm,(3) AC段轴在横截面边缘处的剪应力为,BC段轴在横截面边缘处内、外边缘的剪应力,2019/3/30,Kylinsoft,MOM-2-24,Example 2-3,解：设实心轴的直径为d，空心轴的外径为D。,设有一实心圆轴与一内外径比为=3/4的空心圆轴，两轴材料及长度相同。承受转矩均为T，已知两轴的最大剪应力 t 相等，试比较两轴的重量。,实心轴,2019/3/30,Kylinsoft,MOM-2-25,实心轴,空心轴,得到,两轴的重量比：,空心轴的直径比实心轴大13%，但重量仅为一半。,2019/3/30,Kylinsoft,MOM-2-26,What will happen if we increase ?,Wrinkling or Buckling of the wall,2019/3/30,Kylinsoft,MOM-2-27,2.6 Torsional Deformations of a Circular Bar,扭转变形的标志为扭转角 (Angle of Twist) f ； 扭转角 f 是指两个横截面绕轴线的相对转角。,GIP: 圆轴的抗扭刚度 Torsional Rigidity,2019/3/30,Kylinsoft,MOM-2-28,阶梯轴两端相对扭转角,内力和尺寸连续变化的轴两端相对扭转角,2019/3/30,Kylinsoft,MOM-2-29,单位长度扭转角(rate of twist),扭转的刚度条件：,(rad/m),(rad/m),2019/3/30,Kylinsoft,MOM-2-30,Example 2-4,传动轴 n = 500 r/min，N1 = 500马力， N2 = 200马力， N3 = 300马力，已知 t = 70MPa，q = 1/m，G = 280GPa。确定AB段，BC段直径。,解： (1) 计算外力偶矩,(Nm),(2)作扭矩图,(Nm),(Nm),2019/3/30,Kylinsoft,MOM-2-31,(3) 计算AB段直径：,由扭转刚度条件,由扭转强度条件:,取d1=85mm,2019/3/30,Kylinsoft,MOM-2-32,(4) 确定BC段直径:,由扭转强度条件得:,由扭转刚度条件得,取d2 = 75mm,2019/3/30,Kylinsoft,MOM-2-33,2.7 Statically Indeterminate Torsion,解： (1)解除B端约束，建立静定基本体系(Released form),Example 2-5 已知M0 = 10kNm，试绘扭矩图。,(3) Force-displacement relations,(2) Equation of compatibility,补充 方程：,解得：,2019/3/30,Kylinsoft,MOM-2-34,(4) 由静力方程，得: mA=mB=m0/3,(5) 作扭矩图，由图可知CD段为危险段。,2019/3/30,Kylinsoft,MOM-2-35,f2,f1,Example 2-6 圆轴上有AB两个凸缘，受力偶矩m作用变形后焊上圆筒，在解除力矩m。设轴和筒的抗扭刚度分别是G1IP1和 G2IP2 ，试求轴内和筒内的扭矩。,轴受扭矩T1，筒受扭矩T2，且 T1-T2=0,解除m后，轴和筒的扭转角分别为f1和f2，有：,补充方程：,解得：,焊接前轴在m作用下