Question 17.10: Clamped-clamped beam of length l and stiffness EI contains......

Solution The initial kinematical parameters are \varphi(t)=\varphi_{0} \sin \theta t  and y_{0}(t)=0 . The unknown initial forced parameters are M_{0}(t)=M_{0} \sin \theta t  and Q_{0}(t)=Q_{0} \sin \theta . Boundary conditions at the right clamped support are y(l)=0  and \varphi(l)=0 .

Assume that the moment of inertia of mass m  is neglected, while the inertial force J  of mass m  is taken into account. Since y_{0}=0 , then equations of initial parameter method for amplitude values y(x)  and \varphi(x)  should be written as follows

\begin{aligned}& y(x)=\varphi_{0} x-\frac{1}{E I}\left[M_{0} \frac{x^{2}}{2 !}+Q_{0} \frac{x^{3}}{3 !}-\theta^{2} m y_{m} \frac{\left(x-a_{m}\right)^{3}}{3 !}\right] \\& \varphi(x)=\varphi_{0}-\frac{1}{E I}\left[M_{0} x+Q_{0} \frac{x^{2}}{2}-\theta^{2} m y_{m} \frac{\left(x-a_{m}\right)^{2}}{2}\right], \quad a_{m}=a\end{aligned}        (a)

In these equations, a negative sign in front of the bracket means the direction of the y  axis downwards, while in front of the term, which contains \theta^{2}  means that the positive inertial force \theta^{2} m y_{m}  creates a negative moment in the cross section x>a .

Displacement and the force of inertia of mass m  are

\begin{aligned}& y_{m}(x=a)=\varphi_{0} a-\frac{1}{E I}\left[M_{0} \frac{a^{2}}{2}+Q_{0} \frac{a^{3}}{6}\right] \\& \theta^{2} m y_{m}=\theta^{2} m\left(\varphi_{0} a-\frac{M_{0}}{E I} \frac{a^{2}}{2}-\frac{Q_{0}}{E I} \frac{a^{3}}{6}\right)\end{aligned}             (b)

Substitution of the expression for the inertial force into (a) leads to the following formulas for linear and angular displacements

\begin{aligned}& y(x)=\varphi_{0} x-\frac{M_{0}}{E I} \frac{x^{2}}{2}-\frac{Q_{0}}{E I} \frac{x^{3}}{6}+\frac{\theta^{2} m}{E I}\left[\varphi_{0} a-\frac{M_{0}}{E I} \frac{a^{2}}{2}-\frac{Q_{0}}{E I} \frac{a^{3}}{6}\right] \frac{(x-a)^{3}}{6} \\& \varphi(x)=\varphi_{0}-\frac{M_{0}}{E I} x-\frac{Q_{0}}{E I} \frac{x^{2}}{2}+\frac{\theta^{2} m}{E I}\left[\varphi_{0} a-\frac{M_{0}}{E I} \frac{a^{2}}{2}-\frac{Q_{0}}{E I} \frac{a^{3}}{6}\right] \frac{(x-a)^{2}}{2}\end{aligned}              (c)

Considering boundary conditions at the right end (if x=l  then y=0, \varphi=0  ), then relationships (c) become

\begin{aligned}& y(l)=\varphi_{0} l-\frac{M_{0}}{E I} \frac{l^{2}}{2}-\frac{Q_{0}}{E I} \frac{l^{3}}{6}+\frac{\theta^{2} m}{E I}\left[\varphi_{0} a-\frac{M_{0}}{E I} \frac{a^{2}}{2}-\frac{Q_{0}}{E I} \frac{a^{3}}{6}\right] \frac{(l-a)^{3}}{6}=0 \\& \varphi(l)=\varphi_{0}-\frac{M_{0} l}{E I}-\frac{Q_{0}}{E I} \frac{x^{2}}{2}+\frac{\theta^{2} m}{E I}\left[\varphi_{0} a-\frac{M_{0}}{E I} \frac{a^{2}}{2}-\frac{Q_{0}}{E I} \frac{a^{3}}{6}\right] \frac{(l-a)^{2}}{2}=0\end{aligned}           (d)

In the formulas (d) we will isolate members with unknowns M_{0}  and Q_{0} .

\begin{aligned}& \frac{M_{0}}{E I}\left[\frac{l^{2}}{2}+\theta^{2} m \frac{a^{2}}{2} \frac{(l-a)^{3}}{6 E I}\right]+\frac{Q_{0}}{E I}\left[\frac{l^{3}}{6}+\theta^{2} m \frac{a^{3}}{6} \frac{(l-a)^{3}}{6 E I}\right]=\varphi_{0}\left[l+\theta^{2} m a \frac{(l-a)^{3}}{6 E I}\right] \\& \frac{M_{0}}{E I}\left[l+\theta^{2} m \frac{a^{2}}{2} \frac{(l-a)^{2}}{2 E I}\right]+\frac{Q_{0}}{E I}\left[\frac{l^{2}}{2}+\theta^{2} m \frac{a^{3}}{6} \frac{(l-a)^{2}}{2 E I}\right]=\varphi_{0}\left[1+\theta^{2} m a \frac{(l-a)^{2}}{2 E I}\right]\end{aligned}        (e)

Now we need to introduce the dimensionless parameter \theta / \omega . For this purpose the second term in each bracket should be transformed. Unit displacement and frequency of free vibration for clamped-clamped beam are

\delta_{11}=\frac{a^{3} b^{3}}{3 l^{3} E I} \rightarrow \omega=\sqrt{\frac{1}{m \delta_{11}}}=\sqrt{\frac{3 l^{3} E I}{m a^{3} b^{3}}} \rightarrow m=\frac{3 l^{3} E I}{\omega^{2} a^{3} b^{3}}

The second term in the first bracket (e) becomes

\theta^{2} m \frac{a^{2}}{2} \frac{(l-a)^{3}}{6 E I}=\theta^{2} m \frac{a^{2}}{2} \frac{b^{3}}{6 E I}=\frac{\theta^{2}}{\omega^{2}} \frac{l^{3}}{4 a}

Equations (e) in the equivalent form take the form

\begin{aligned}& \frac{M_{0}}{E I}\left[\frac{l^{2}}{2}+\frac{\theta^{2}}{\omega^{2}} \frac{l^{3}}{4 a}\right]+\frac{Q_{0}}{E I}\left[\frac{l^{3}}{6}+\frac{\theta^{2}}{\omega^{2}} \frac{l^{3}}{12}\right]=\varphi_{0}\left[l+\frac{\theta^{2}}{\omega^{2}} \frac{l^{3}}{2 a^{2}}\right] \\& \frac{M_{0}}{E I}\left[l+\frac{\theta^{2}}{\omega^{2}} \frac{3 l^{3}}{4 a b}\right]+\frac{Q_{0}}{E I}\left[\frac{l^{2}}{2}+\frac{\theta^{2}}{\omega^{2}} \frac{l^{3}}{4 b}\right]=\varphi_{0}\left[1+\frac{\theta^{2}}{\omega^{2}} \frac{3 l^{3}}{2 a^{2} b}\right]\end{aligned}         (e)

Solution of this set of equations is (subscript 0 at \varphi  is omitted)

\begin{aligned}& M_{0}=\varphi \frac{4 E I}{l} F, \quad F=\left[1-\frac{\theta^{2}}{\omega^{2}}\left(1+\frac{3}{4} \frac{b}{a}\right)\right] \cdot \frac{1}{1-\theta^{2} / \omega^{2}} \\& Q_{0}=-\varphi \frac{6 E I}{l^{2}} G, \quad G=\left[1-\frac{\theta^{2}}{\omega^{2}}\left(1+\frac{3 a b+b^{2}}{2 a^{2}}\right)\right] \cdot \frac{1}{1-\theta^{2} / \omega^{2}}\end{aligned}              (f)

Special case: If a=b=l / 2 , then the bending moment and shear force at the left clamped supports are

M(0)=\frac{4 E I}{l} \varphi \frac{1-7 \theta^{2} / 4 \omega^{2}}{1-\theta^{2} / \omega^{2}} \sin \theta t, \quad Q(0)=-\frac{6 E I}{l^{2}} \varphi \frac{1-3 \theta^{2} / \omega^{2}}{1-\theta^{2} / \omega^{2}} \sin \theta t             (g)

These formulas present reference data for analysis of the dynamical structures, and are shown in Table A.27, row 1.

Let us substitute these expressions in the first expression (b). Displacement at the middle of the beam caused by a static angular displacement \varphi  becomes

y(x)=\varphi x-\left.\frac{1}{E I}\left(M_{0} \frac{x^{2}}{2 !}+Q_{0} \frac{x^{3}}{3 !}\right)\right|_{x=l / 2}=\varphi \frac{l}{2}-\frac{1}{E I}\left(\frac{4 E I}{l} \frac{l^{2}}{4 \times 2 !}-\frac{6 E I}{l^{2}} \frac{l^{3}}{8 \times 3 !}\right) \varphi=\frac{l}{8} \varphi               (h)

According to (\mathrm{g}) , in case of \theta / \omega=1  resonance occurs. Using the found values of the initial parameters and any parameter \theta / \omega  it is easy to construct a corresponding curve y(x)  of the amplitude values of the dynamic displacements.

In the case of a harmonic force excitation F(t)=F^{*} \sin \theta t  or M(t)=M^{*} \sin \theta t , the corresponding terms should be introduced into the equations of the initial parameters method (17.39). Here, F^{*}  and M^{*}  are the amplitude values of the harmonic force F(t)  and M(t)  the harmonic couple, respectively.

\begin{aligned}\varphi(x) & =\varphi_{0}+\frac{1}{E I}\left[M_{0} x+Q_{0} \frac{x^{2}}{2}+P \frac{\left(x-a_{P}\right)^{2}}{2}+R \frac{\left(x-a_{R}\right)^{2}}{2}+M\left(x-a_{M}\right)\right]+\frac{\theta^{2}}{E I}\left[m y_{m} \frac{\left(x-a_{m}\right)^{2}}{2}+J \varphi_{m}\left(x-a_{m}\right)\right] \\M(x) & =M_{0}+Q_{0} x+\left[M+P\left(x-a_{P}\right)+R\left(x-a_{R}\right)\right]+\theta^{2}\left[m y_{m}\left(x-a_{m}\right)+J \varphi_{m}\right] \\Q(x) & =Q_{0}+(P+R)+\theta^{2} m y_{m}\end{aligned}              ( 17.39)