(1.15.9)--chapter12-9 Principal Axes and P材料力学材料力学.ppt

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1、Chapter 12 Review of Centroids and Moments of Inertia Mechanics of MaterialsPrincipal Axes and Principal Moments of inertiaWelcome to mechanics of materials,in this video,we are going to discuss Principal Axes and Principal Moments of inertia.Review:Transformation equations Showing how Ix1,Iy1,Ix1y1

2、 vary as the angle of rotation varies.Questions:What are the maximum and minimum values of the moment of inertia?At which that makes the moment of inertia a maximum or a minimum?In the last section,*the transformation equations for moments and products of inertia were discussed,thoese equations*show

3、 how the moments and products of inertia vary as the angle of rotation varies.*Questions arise:*what are the maximum and minimum values of the moment of inertia,and at which angle that makes the moment of inertia a maximum or a minimum.You will find answers to these questions in this section.Princip

4、al Axes and Principal Moments of inertiaPrincipal Axes 1.Principal AxesPrincipal Moments of inertia:-the maximum and minimum values of the moment of inertia.(P:principal angle)First,introduce two*properties,principal moment of inertia and principal axes.*The maximum and minimum values of moment of i

5、nertia of an area about an origin O are known as the principal moments of inertia,and*the corresponding axes are known as pricinpal axes,which are*defined by the angle that makes the moments of inertia a maximum or a minimum.*For an arbitrary origin O,as shown in the figure,to find the values of ang

6、le that make the moment of inertia*Ix1 or Iy1 a maximum or a minimum,*take the derivative with respect to of Ix1 or Iy1,and set it equal to zero*.Solving for from this equation gives tan2p*.in which p denotes the principal angle defining a principal axis.Note here the same result can be obtained by

7、taking the derivative of Iy1.This equation yields*two values of p differing by 90 degree,note that for this euqation,there are two values of the angle 2p in the range from 0 to 360 degree,which differ by 180 degree.These two angles,define the two perpendicular principal axes.One of these axes corres

8、ponds to the maximum moment of inertia,and the other corresponds to the minimum moment of inertia.This equation yields*two values of differing by 90 degree,note that for this euqation,there are two values of the angle 2p in the range from 0 to 360 degree,which differ by 180 degree.These two angles p

9、,define the two perpendicular principal axes.One of these axes corresponds to the maximum moment of inertia,and the other corresponds to the minimum moment of inertia.The product of inertia is zero for the principal axes.(P:principal angle)1.Principal AxesConclusions:(1)principal axes through an ori

10、gin O are a pair of orthogonal axes for which the moments of inertia are a maximum and a minimum;(2)the orientation of the principal axes is given by the angle P;(3)the product of inertia is zero for principal axes;(4)an axis of symmetry is always a principal axis.When the rotation angle is p,at thi

11、s orientation,x1 and y1 become a set of principal axes at point O,now examine the product of inertia*Ix1y1,substite p into the corresponding transformation equation,by using the*trigonometric identities,it is found that*Ix1y1=0.This means that*the product of inertia is zero for the principal axes.*T

12、he preceding observations may be summarized as follows*:(1)for any arbitrary origin O,there is always at least one pair of principal axes.Principal axes are orthogonal axes for which the moments of inertia are a maximum and a minimum;(2)the orientation of the principal axes defined by the angle p;(3

13、)the product of inertia is zero for principal axes;and(4)an axis of symmetry of an area is always a principal axis of this area.-every axis through the point is a principal axis;-the moments of inertia are constant for all axes through the point.The principle points for the rectangle2.Principal poin

14、tQuestion:Do this rectangle have other principal points?If yes,where are they?x-y axes are a pair of principal axesx-y axes are another pair of principal axesNow consider an arbitrary origin at a given point O,it is known that one pair of principal axes always exists.If there exists a different pair

15、 of principal axes at origin O,which means a second maximum exists,then the only possibility is that the moment of inertia remains constant,which means that*every axis through the point is a principal axis,every pair of axes is a set of principal axes,and all moments of inertia are the same.Such poi

16、nt O is known as a principal point.An illustration of this situation is*a rectangle of width 2b and height b,at the origin O,*the x-y axes are principal axes,because the y axis is an axis of symmetry.*The x-y axes,with the same origin,are also principal axes,because the product of inertia Ixy=0,beca

17、use the triangles are symmetrically located with respect to the x and y axes.It follows that every pair of axes through O is a set of principal axes and every moment of inertia is the same,equal to 2b4/3.Therefore,point O is a principal point for the rectangle.Question is do this rectangle have othe

18、r principal points,try to locate them.Remarks(2)In general,every plane area has two principal points.(1)The centroid is a principal point.-for areas having two different axes of symmetry not perpendicular to each other;-for areas having three or more different axes of symmetry.-lie equidistant from

19、the centroid on the principal centroidal axis having the larger principal moment of inertia.Here are some discussions about principal points.First,through the centroid of an area,*if there exist two different axes of symmetry that are not perpendicular to each other,or*there exist three or more diff

20、erent axes of symmetry,the centroid of such area is a principal point.These conditions are fulfilled for a circle,for all regular polygons(*equilateral triangle,*square,regular pentagon,regular hexagon,and so on,and for many other symmetric shapes.*In general,every plane area has two principal point

21、s.These points lie equidistant from the centroid on the principal centroidal axis having the larger principal moment of inertia.3.Principal Moments of inertia(1)First methodprincipal anglesP Principeal moments of inertia I1,I2Now determine the principal moments of inertia,assuming that Ix,I y,and Ix

22、yare known.There are two methods.One method is to*determine the two values of p and then*substitute these values of p into the transformation equation for Ix1 or I y1.It yields*two principal moments of inertia,denoted by I1 and I2.The advantage of this method is that it gives which of the two princi

23、pal angles up corresponds to I1 or I2.Maximum principal moments of inertia:1+2=+3.Principal Moments of inertia(2)geometric methodThe second method isgeometric method,by finding general formulas for the principal moments of inertia.Here is the*figure which is a geometric representation for the equati

24、on*.From this trigangle,it can be obtained*sin2p,*cos2p,and*the hypotenuse R.Now substitute the expressions for cos 2p and sin 2p into the transformation equation for Ix1,and obtain the algebraically larger of the two principal,moments of inertia,denoted by the symbol*I1.Consisering this*relationshi

25、p,and substitute I1 into this equation and solving for*I2.These two equations for I1 and I2 provide a convenient way to calculate the principal moments of inertia.Example 1:Determine the orientations of the principal centroidal axes and the magnitudes of the principal centroidal moments of inertia for the cross-sectional area of the Z-section shown in Figure.Use the following numerical data:height h=200 mm,width b=90 mm,and constant thickness t=15 mm.This is the end of this video.End