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1、 CHAPTER 14 TRANSVERSE VIBRATION OF EULER BEAM It was recognized by the early researchers that the bending effect is the single most important factor in a transversely vibrating beam. The Euler-Bernoulli model includes the strain energy due to the bending and the kinetic energy due to the lateral di
2、splacement. The Euler-Bernoulli model dates back to the 18th century. Jacob Bernoulli (1654-1705) first discovered that the curvature of an elastic beam at any point is proportional to the bending moment at that point. Daniel Bernoulli (1700-1782), nephew of Jacob, was the first one who formulated t
3、he differential equation of motion of a vibrating beam. Later, Jacob Bernoullis theory was accepted by Leonhard Euler (17071783) in his investigation of the shape of elastic beams under various loading conditions. Many advances on the elastic curves were made by Euler. The Euler-Bernoulli beam theor
4、y, sometimes called the classical beam theory, Euler beam theory, Bernoulli beam theory, or Bernoulli-Euler beam theory, is the most commonly used because it is simple and provides reasonable engineering approximations for many problems. However, the Euler-Bernoulli model tends to slightly overestim
5、ate the natural frequencies. This problem is exacerbated for the natural frequencies of the higher modes. Mathematical formulation: wxdx Figure 1: A beam under transverse vibration Consider a long slender beam as shown in figure 1 subjected to transverse vibration. The free body diagram of an elemen
6、t of the beam is shown in the figure 2. Here, is the bending moment, is the shear force, and is the external force per unit length of the beam. Since the inertia force acting on the element of the beam is ),( txM),( txV ),( txf22),()(ttxwdxxA 247 Figure 2: Freebody diagram of a section of a beam und
7、er transverse vibration balancing the forces in direction gives z22),()(),()(ttxwdxxAVdxtxfdVV=+ , where is the mass density and is the cross-sectional area of the beam. The moment equation about the )(xAy axis leads to 02),()()( =+ MdxdxtxfdxdVVdMM By writing dxxVdV= and dxxMdM= and disregarding te
8、rms involving second powers in , the above equations can be written as dx22),()(),(),(ttxwxAtxfxtxV=+ 0),(),(=txVxtxMBy using the relation xMV= from above two equations 2222),()(),(),(ttxwxAtxfxtxM=+ 248From the elementary theory of bending of beam, the relationship between bending moment and deflec
9、tion can be expressed as 22),()(),(xtxwxEItxM= where E is the Youngs modulus and is the moment of inertia of the beam cross section about the axis. Inserting above two equations, we obtain the equation of the motion for the forced transverse vibration of a non-uniform beam: )(xIy),(),()(),()(222222t
10、xfttxwxAxtxwxEIx=+ For a uniform beam above equation reduces to ),(),()(),()(2244txfttxwxAxtxwxEI =+ For free vibration, , and so the equation of motion becomes 0),( =txf0),(),(22442=+ttxwxtxwc where AEIc= Initial Conditions: Since the equation of the motion involves a second order derivative with r
11、espect to time and a fourth order derivative with respect to , two initial equations and four boundary conditions are needed for finding a unique solution for . Usually, the values of transverse displacement and velocity are specified as and atx),( txw)(0xw )(0.xw 0=t , so that the initial condition
12、s become: )()0,( 0 xwtxw = )()0,(0.xwttxw=Free Vibration: The free vibration solution can be found using the method of separation of variables as )()(),( tTxWtxw = 249Substituting this equation in the final equation of motion and rearranging leads to 222442)()(1)()(= adttTdtTdxxWdxWcwhere is a posit
13、ive constant. Above equation can be written as two equations: 2=a0)()(444= xWdxxWd 0)()(222=+ tTdttTd where, EIAc2224 = The solution to time dependent equation can be expressed as tBtAtT sincos)( += where, A and B are constant that can be found from the initial conditions. For the solution of displa
14、cement dependent equation we assume, sxCexW =)( where C and are constant, and derive the auxiliary equation as: s=2,1s , is =2,1Hence the solution of the equation becomes: xixixxeCeCeCeCxW +=4321)( where , , and are constant. Above equation can also be expressed as: 1C2C3C4CxCxCxCxCxW sinhcoshsincos
15、)(4321+= or, ()()( )( xxCxxCxxCxxCxW ) sinhsinsinhsincoshcoscoshcos)(4321+= The constants , , and can be found from boundary conditions. The natural frequencies of the beam are computed from: 1C2C3C4C()422AlEIlAEI = 250The function is known as normal mode or characteristic function of the beam and )(xW is called the natural frequency of vibration. For any beam, there will be infinite number of normal modes with one natural frequency associated with each normal mode. The unknown c
