TRANSVERSE VIBRATION ON A BEAM AN ANALYTICAL MODEL

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TRANVERE VIBRATION ON A BEAM AN ANALYTICAL MODEL Derivation of the wave equation Geoffre Kolbe Consider a short section of a beam, length L and having cross-sectional area.

1 TRANVERE VIBRATION ON A BEAM AN ANALYTICAL MODEL Derivation of the wave equation Geoffre Kolbe Consider a short section of a beam, length L and having cross-sectional area. Due to a turning moment applied at x = L, the section is bent in an arc of radius R. The moment at x = L ma be derived b considering an element of the section, area d, situated a distance k from the midline of the section. Due to the applied moment, the element is stretched a distance δx. The force df required to stretch this element is E d δx df = L where E is Young's modulus. δ x L From smmetr, it can be seen that = so k R the equation for df can be written E d k df = R The moment about the midline is k df, so the total moment M applied to the section at x = L will be E M = kdf = k d R E k d is the flexural rigidit of the beam and is usuall written as EI where I = k d and is the second moment of inertia around the midline of the beam. Let K = k d where K is the effective distance from the midline at which the total force F ma be said to act. EK Then the total moment is M = R The radius R can be removed b expressing it in terms of x and. ( ) Expanding, R = x + R R = x + R R + Figure Figure

2 If x is small compared to R, is alwas much smaller than x, so 0 and x = R Differentiating twice with respect to x, The total moment M can now be written as, d M = EK d = R (Note that the second differential of with respect to x is equivalent to the total moment at x.) Let F (x) be the function of the sheer force acting continuousl along the length of the beam but orthogonall to it, such that the total sheer force F x acting on the beam is, ( ) Now, the moment M will be equal and opposite to the sum of the sheer forces acting in the direction, multiplied b their distance x from the left hand end of the segment. Figure o, the total moment M for the segment can be expressed as, ( ) Differentiating both sides with respect to x gives, x F ( x) L d = = x F x M EK d = EK (Note that the third differential of with respect to x is equivalent to the sheer force on the beam over the length x. For a ver short beam where x, this can be approximated to the sheer force at x.) df ( x) d Differentiating once again, F( x) + x = EK Now if the segment length L is ver short, F ( x ) can be considered a constant from x = 0 to x = L. d F x 0 o ( ) and then, F ( x) The sheer force term ( ) ( ) d d = EK F x will be due to the transverse acceleration of beam at x, and can be written as F x = ρ where ρ is the densit and ρ is the mass of the element through the dt segment at x of width.

3 ince F ( x ) is a constant with respect to x over the length of the segment, it is possible to write d ( ) F x written as, = ρ and finall, the equation of motion for transverse vibrations on a beam can be dt olutions to the wave equation EK = t ρ x The partial differentiations in the wave equation above reflect the fact that is an independent x,t = v x ω t function of x and t such that; ( ) ( ) in ubstituting this function into the wave equation gives, v ( x) ω = EK ρ d ( ) v x Let q ω ρ = then the equation can be re-written as q v ( x) EK ( ) d v x = The solutions to this equation are of the exponential tpe such that, v ( x) = c e θ x where c is an arbitrar constant. It will seen that where i = θ = q, so that The general solution then is, ( ) constants. θ = ± q and therefore θ can take on values of ± q or ± iq v x c e c e c e c e qx qx i qx iqx = where... c are arbitrar It is more convenient to express v ( x ) in terms of trigonometric functions, so that the general solution of the x part of the wave equation can be expressed as; v ( x) = A Cos qx + B in qx + C Cosh qx + D inh qx where A, B, C and D are arbitrar constants. The particular solutions of this equation ma be found for a given situation b subjecting it to boundar conditions.

4 The cantilever Consideration of the solutions to the equations for transverse vibrations in the case of a fixed-free beam, the cantilever, serves as a model for the solutions for a free-free beam, pinned-free beam and beams with other boundar conditions applied. Figure shows a beam, cross-sectional area, length l, which has one end fixed (x = 0) and the other end free (x = l). The fixed end is unable to move in the vertical direction, so v(x) = 0 at x = 0 for all time. Too, the dv ( x) angle of the beam does not change at x = 0, so = 0 at x = 0 for all time. Imposing these conditions on the general solution means that A + C = 0. d v ( x) At the free end of the beam, there is necessaril no bending moment, so = 0 at x = l for all time. d v ( x) And there will be no sheer force so = 0 at x = l for all time. Imposing these conditions on the general solution means that B + D = 0. o ( Cosh l Cos l) ( inh l in l ) and ( inh l in l) ( Cosh l Cos l ) Figure C q + q + D q + q = 0 C q q + D q + q = 0 To solve these homogeneous linear equations in C and D, it is necessar that; Cosh ql + Cosq l inh ql + inql inh ql in q l Cosh ql + Cosql = 0 or + Cosh q l Cosql = 0 If ql = Z then CosZ = and this equation ma be solved graphicall b noting the Cosh Z values of Z at which the curves of these two functions intercept.

5 Figure 5 The first few roots are Z =. 875 and Z =. 69. Thereafter, with the function approaching zero asmptoticall, the roots are given to good accurac b, = ( + ) The natural frequencies of a cantilever beam ma now be written as, Z n π n Cosh Z EK Zn EK n ω n = q = ρ l ρ Zn EI = l ρ Z n EI = l m where m is the mass per unit length of the beam. The shape of the standing wave vibrational modes on the beam As shown above, the amplitude ( x ) at x is given b; v ( x) = A Cos qx + B in qx + C Cosh qx + D inh qx ince A + C = 0 and B + D = 0, then;

6 ( ) = ( Cosh Cos ) + ( inh in ) v x C qx qx D qx qx C ql ql + D ql + ql = 0 to express D in terms of C, then dividing v( x) using ( inh in ) ( Cosh Cos ) through b C gives, ( ) ( ) ( in ql inhql) ( ) ( ) Cosh ql + Cosql v x Cosh qx Cosqx + inh qx inqx Figure 6 shows the shapes of the first three modes of natural vibrations. Figure 6

7 Transverse wave propagation velocit While torsional vibrations, longitudinal vibrations and sheer vibrations will propagate along the E beam at the speed of sound, V s =, transverse vibration travels as a phase wave as given b the ρ in π λ equation ( x,t) = a ( x Vt), which is a sinusoidal wave propagating along the beam. EK π EK This equation for satisfies the wave equation = if V = t ρ x λ ρ It should be noted that the transverse vibration velocit is inversel proportional to the wavelength, λ and the velocit is thus dispersive with wavelength, or frequenc. π In the trigonometric expressions above, the term ql ma be replaced b ql = l where λ is the λ wavelength of the particular resonant mode. ubstituting for λ, the expression for velocit ma finall be written as, Z EK V = l ρ

Chapter 16 Mechanical Waves

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