Question 16.11: Uniform cantilevered beam of length l carries uniformly dist......

General expression for mode shape of vibration is

y(x)=Y_0 S(k x)+\frac{\theta_0}{k} T(k x)+\frac{M_0}{k^2 E I} U(k x)+\frac{Q_0}{k^3 E I} V(k x)

Since Y_0=0  and \theta_0=0 , then displacement of a beam at free end should be calculated using functions U and V

y(l)=\frac{M_0}{k^2 E I} U(\lambda)+\frac{Q_0}{k^3 E I} V(\lambda), \quad \lambda=k l, \quad k=\sqrt[4]{\frac{m \omega^2}{E I}},        (a)

The bending moment and shear force at free end are known. They are

M(l)=0 \quad \text { and } \quad Q(l)=M y(l) \omega^2=\psi m l \cdot y(l) \omega^2 \quad \psi=M / ml

where y(l)  is displacement of a beam at point where mass M is attached.

The nature of shear force presents the inertial force of lumped mass M. If the acceleration of the mass M is directed in the positive direction (upward), then the inertia force is directed downwards. This means that force acts on the beam in the opposite direction (upward)

\begin{gathered}M(l)=S(\lambda) M_0+\frac{T(\lambda)}{k} Q_0=0        (b)  \\Q(l)=k V(\lambda) M_0+S(\lambda) Q_0=-\psi m l \cdot y(l) \omega^2          (c)\end{gathered}           

Substitution of (a) into (c) leads to the following result

Q(l)=k V(\lambda) M_0+S(\lambda) Q_0=-\psi m l \underbrace{\left[\frac{M_0}{k^2 E I} U(\lambda)+\frac{Q_0}{k^3 E I} V(\lambda)\right]}_{y(l)} \omega^2       (d)

Coefficient \frac{m l \omega^2}{k^3 E I}=\frac{m \omega^2}{E I} \frac{l}{k^3}=k l=\lambda . Relationships (c) after simplification becomes

 [k V(\lambda)+k \psi \lambda \cdot U(\lambda)] M_0+[S(\lambda)+\psi \lambda \cdot V(\lambda)] Q_0=0           (e)

Relationships (b) and (e) present a set of two linear homogeneous algebraic equations with unknowns initial bending moment M_0 and shear force Q_0 . Nontrivial solution is possible if determinant of coefficients is zero.

\left|\begin{array}{cc}k V(\lambda)+k \psi \lambda \cdot U(\lambda) & S(\lambda)+\psi \lambda \cdot V(\lambda) \\S(\lambda) & \frac{T(\lambda)}{k}\end{array}\right|=0

In terms of Krylov-Duncan function the frequency equation is

\frac{S^2(\lambda)-V(\lambda) T(\lambda)}{T(\lambda) U(\lambda)-S(\lambda) V(\lambda)}=\psi \lambda           (f)

This equation in elementary functions becomes

\frac{1+\cosh \lambda \cos \lambda}{\cosh \lambda \sin \lambda-\sinh \lambda \cos \lambda}=\psi \lambda             (g)

Assume the total mass of a beam ml  is twice more than lumped mass M , i.e., \psi=0.5 . Smallest root of this equation is \lambda_1=1.41 , so the smallest frequency vibration \omega_1=\frac{1.41^2}{l^2} \sqrt{\frac{E I}{m}} .

Special Case Let \psi=0 ; it means that M=0 and we have clamped-free beam with uniformly distributed mass m. Corresponding frequency equation becomes \cosh \lambda \cos \lambda=-1 .

Frequencies and shapes of vibrations for different types of beams (one and multispan beams, uniform and nonuniform beams, beams with classical and nonclassical boundary conditions, with lumped and rotational masses, etc.) as well as for frames may be found in handbook (Karnovsky and Lebed 2004a). Also this book contains the general equations of initial parameters method taking into account lumped masses along the beam.