Question 17.16: The clamped-clamped beam of length l, bending stiffness EI, ......

The origin is placed at the left end of the beam. The positive initial parameters M_{0} and Q_{0} are shown in Fig. 17.30.

At the right support (x=l) two parameters are known. They are the vertical y(l) and angular displacements \varphi(l). So from the set of Eq. (17.44) we will use the general expressions for functions y(x) and \varphi(x)

\begin{aligned}& y(x)=y_0 S_x+\varphi_0 T_x \frac{1}{k}+M_0 U_x \frac{1}{E I k^2}+Q_0 V_x \frac{1}{E I k^3}+\frac{1}{E I k^2}\left[M U_{r M}+\frac{1}{k} P V_{r P}\right] \\& \varphi(x)=y_0 V_x k+\varphi_0 S_x+M_0 T_x \frac{1}{E I k}+Q_0 U_x \frac{1}{E I k^2}+\frac{1}{E I k}\left[M T_{r M}+\frac{1}{k} P U_{r P}\right] \\& M(x)=y_0 U_x E I k^2+\varphi_0 V_x E I k+M_0 S_x+Q_0 T_x \frac{1}{k}+\left[M S_{r M}+\frac{1}{k} P T_{r P}\right] \\& Q(x)=y_0 T_x k+\varphi_0 U_x E I k^2+M_0 V_x k+Q_0 S_x+\left[M V_{r M} k+P S_{r P}\right]\end{aligned}         (17.44)

\begin{aligned}& y(x)=y_{0} S_{x}+\varphi_{0} T_{x} \frac{1}{k}+M_{0} U_{x} \frac{1}{E I k^{2}}+Q_{0} V_{x} \frac{1}{E I k^{3}}, \\& \varphi(x)=y_{0} V_{x} k+\varphi_{0} S_{x}+M_{0} T_{x} \frac{1}{E I k}+Q_{0} U_{x} \frac{1}{E I k^{2}}, \quad k=\sqrt[4]{\frac{m_{0} \theta^{2}}{E I}}\end{aligned}              (a)

Initial parameters are y_{0}=0 and \varphi_{0}=0. At the right support y(l)=y_{l}=0 and \varphi(l)=\varphi_{l}=-1; the negative sign means that support is rotated clockwise direction. Therefore from (a) we get a linear algebraic equations with respect to unknown initial parameters M_{0} and Q_{0}

\begin{aligned}& y(l)=M_{0} U \frac{1}{E I k^{2}}+Q_{0} V \frac{1}{E I k^{3}}=0, \\& \varphi(l)=M_{0} T \frac{1}{E I k}+Q_{0} U \frac{1}{E I k^{2}}=-1\end{aligned}                    (b)

Here index l at the Krylov functions is omitted, i.e., U_{x=l}=U=U(k l), T_{x=l}=T=T(k l), \ldots Solution of this set of equations is

\begin{aligned}Q(0) & =-E I k^{2} \frac{U}{U^{2}-V T} \\M(0) & =E I k \frac{V}{U^{2}-V T}\end{aligned}               (c)

The negative sign for Q(0) means that this shear trends to rotate beam around the opposite end of the beam counterclockwise. Bending moment and shear force at the right support are

\begin{aligned}& M(l)=M_{0} S-Q_{0} T \frac{1}{k}=E k \frac{S V-T U}{U^{2}-V T} \\& Q(l)=M_{0} V k-Q_{0} S=-E k^{2} \frac{U S-V^{2}}{U^{2}-V T}\end{aligned}                (d)

Now all reactions may be presented in terms of trigonometric and hyperbolic functions of dimensionless parameter u=k l. For example,

M(0)=E I k \frac{V}{U^{2}-V T}=E I \frac{u}{l} \frac{2}{2} \frac{V(u)}{U^{2}(u)-V(u) T(u)}=\frac{2 E I}{l} \cdot \frac{u}{2} \cdot \frac{\sinh u-\sin u}{1-\cosh u \cos u}=\frac{2 E I}{l} f(u)               (e)

Formulas (c-e) present the amplitude value of corresponding response functions. Dynamic response involve function \sin \theta t. Coefficient 2 E I / l presents response in case of static rotation of right clamped support through the angle \varphi=1. The complex f(u) presents the correcting function which takes into account nonuniformly distributed inertial forces along entire span. It is easy to show that, if u=0 (static loading), then function f(0)=1.0.

To determine the free vibration frequency of a clamped-clamped beam with a uniformly distributed mass, we need to consider a homogeneous system of equations (b). Nontrivial solution M_{0} \neq 0 and Q_{0} \neq 0 takes place if

D=\left|\begin{array}{cc}U & V / k \\T & U / k \end{array}\right|=0 \rightarrow U^{2}-T V=0

According to (16.26) this equation may be presented in trigonometric and hyperbolic functions as follows:

 \begin{aligned}& S T-U V=\frac{1}{2}(\cosh k x \sin k x+\sinh k x \cos k x) \\& T U-S V=\frac{1}{2}(\cosh k x \sin k x-\sinh k x \cos k x) \\& S^2-U^2=\cosh k x \cos k x ; \quad T^2-V^2=2\left(S U-V^2\right)=\sinh k x \sin k x \\& U^2-T V=\frac{1}{2}(1-\cosh k x \cos k x) ; \quad S^2-T V=\frac{1}{2}(1+\cosh k x \cos k x) \\& T^2-S U=S U-V^2=\frac{1}{2} \sinh k x \sin k x ; \quad 2 S U=T^2+V^2\end{aligned}           (16.26)

U^{2}(k l)-T(k l) V(k l)=\frac{1}{2}(1-\cosh k l \cos k l)=0

Frequency equation for fixed-fixed uniform beam becomes \cosh k l \cos k l=1.

Table A.24 contains the frequency equations, eigenvalues, and nodal points of mode shape for uniform beams with different boundary conditions. Tables A.25 and A.28 contain the analytical expressions for dynamic reactions of uniform one-span beams with distributed masses. These beams are subjected to harmonic kinematic and forced excitation. Reactions in Table A.25 are presented in terms of Krylov-Duncan function, while reactions in Table A.28 are presented in terms of hyperbolic-trigonometric functions.