Question 17.19: Design diagram of nonsymmetrical portal frame is presented i......

To construct the primary system of the displacement method we need to introduce constraint 1 which prevents angular displacement of fixed joint 1 , and constraint 2 which prevents horizontal displacement of the crossbar 1-2 (Fig. 17.34b).

The canonical equations of the displacement method are:

\begin{aligned}& r_{11} Z_{1}+r_{12} Z_{2}+R_{1 P}=0 \\& r_{21} Z_{1}+r_{22} Z_{2}+R_{2 P}=0\end{aligned}        (a)

For calculation of unit reactions we need to construct the bending moment diagrams in the primary system due the unit angular and linear displacements of the introduced constraints (Fig. 17.34c,d). In both diagrams dotted line means the elastic curve. For computation of ordinates of bending moment diagrams at the specific points we use Table A.28. Argument of correction functions \psi_{i} is \lambda=k l, where k=\sqrt[4]{m_{0} \theta^{2} / E I}, and u=\sqrt[4]{m_{0} \theta^{2} l^{4} / E I}, respectively; in fact \lambda=u. The argument u should be formed for each element separately. Since the rods of the structure can have different parameters m_{0}, l, E I, the argument u for each member should be represented through the argument of one specific element.

For vertical members A-1, B-2 and crossbar 1-2 we have

u_{\mathrm{vert}}=\sqrt[4]{\frac{m_{0} l^{4} \theta^{2}}{E I}}, \quad u_{\mathrm{hor}}=\sqrt[4]{\frac{m_{0} l^{4} \theta^{2}}{4 E I}}  

These arguments are different, so we bring them to the base value. Let u=u_{\text {vert, }}, then

\frac{u_{\text {hor }}}{u}=\sqrt[4]{\frac{1}{4}}=0.707 \rightarrow u_{\text {hor }}=0.707 u         (b)

Expressions for bending moments and shear at the end points of elements in the unit states are as follows.

Unit State 1\left(Z_{1}=1\right). In this case only the elements A-1 and 1-2 are subjected to bending deformations

Vertical member A-1

\begin{aligned}M_{1 A}^{(1)} & =4 i \psi_{2}(u), \quad M_{A 1}^{(1)}=2 i \psi_{3}(u), \text { Table A.28, case } 1, \\Q_{1 A}^{(1)} & =\frac{6 i}{l} \psi_{5}(u), \text { case } 1, \quad i=E I / l\end{aligned}

Horizontal member 1-2

Unit State 2\left(Z_{2}=1\right). In this case only the elements A-1 and B-2 are subjected to bending deformations

\begin{aligned}M_{1 A}^{(2)} & =\frac{6 i}{l} \psi_{5}(u), \quad M_{A 1}^{(2)}=\frac{6 i}{l} \psi_{6}(u), \text { case 2; } \\M_{B 2}^{(2)} & =\frac{3 i}{l} \psi_{7}(u), \quad Q_{2 B}^{(2)}=\frac{3 i}{l^{2}} \psi_{12}(u), \text { case 4; } \\Q_{1 A}^{(2)} & =\frac{12 i}{l^{2}} \psi_{10}(u), \text { case } 2\end{aligned}

Unit reactions

First state

\begin{aligned}& r_{11}=M_{1 A}^{(1)}+M_{12}^{(1)}=4 i \psi_{2}(u)+12 i \psi_{1}(0.707 u) \ & r_{21}=-Q_{1 A}^{(1)}=-\frac{6 i}{l} \psi_{5}(u), \text { case } 1\end{aligned}                   (c)

Second state

r_{12}=-M_{1 A}^{(2)}=-\frac{6 i}{l} \psi_{5}(u), \text { case } 2

r_{22}=Q_{1 A}^{(2)}+Q_{2 B}^{(2)}-J_{12}=\frac{12 i}{l^{2}} \psi_{10}(u)+\frac{3 i}{l^{2}} \psi_{12}(u)-\frac{i}{l^{2}} u^{4}          (d)

 Last term in (d) takes into account the distributed inertial forces of the crossbar in horizontal motion J_{12}=m_{0} l \theta^{2}, where m_{0} l is a total mass of the crossbar. Indeed, u=\sqrt[4]{\frac{m_{0} l^{4} \theta^{2}}{E I}} \rightarrow \theta^{2}=\frac{u^{4} E I}{m_{0} l^{4}}\left(\mathrm{~s}^{-2}\right), and we get the expression J_{12}=i u^{4} / l^{2}.

To determine the frequency of free vibration it is necessary to consider homogeneous equations

\begin{aligned}& r_{11} Z_{1}+r_{12} Z_{2}=0 \\& r_{21} Z_{1}+r_{22} Z_{2}=0\end{aligned}        (e)

A nontrivial solution is achieved if the determinant composed of coefficients for unknowns is zero, i.e.,

D=\left|\begin{array}{ll}r_{11} & r_{12} \\r_{21} & r_{22}\end{array}\right|=r_{11} r_{22}-r_{12}^{2}=0         (f)

In expanded form, the equation D=0 after simplification becomes

\left[4 \psi_{2}(u)+12 \psi_{1}(0.707 u)\right]\left[12 \psi_{10}(u)+3 \psi_{12}(u)-u^{4}\right]-36 \psi_{5}^{2}(u)=0        (f)

This equation is transcendental and it is impossible to obtain its analytical solution. To solve it, the following computational procedure may be recommended. Arbitrary value of dimensionless parameter u should be adopted and corresponding value of r_{11} r_{22}-r_{12}^{2} is computed. For this purpose use the detailed table of the numerical values of the correction functions (0 \leq u \leq 7.0, \quad \Delta u=0.1) (Klein 1972; Smirnov et al. 1984; Karnovsky and Lebed 2001, 2004a).

We perform this procedure for other values of parameter u until the change of the sign of determinant D occurs. Thus, we obtain an interval for u in which the desired root u_{1} is located, such that D\left(u_{1}\right)=0. Root of the frequency equation (f) can be determined numerically using Table A.28 which contains the analytical expressions for reactions.

Omitting cumbersome calculations, we present the final result u_{1}=1.68. Thus, the fundamental frequency of vibration of entire structure becomes

\omega_{1}=\frac{u_{1}^{2}}{l^{2}} \sqrt{\frac{E I}{m_{0}}}=\frac{2.82}{l^{2}} \sqrt{\frac{E I}{m_{0}}}

If the frequency of the excitation coincides with the frequency of free vibration, then a resonance occurs. To avoid resonance, we set \theta=0.8 \omega_{1}=\frac{2.256}{l^{2}} \sqrt{\frac{E I}{m_{0}}}. Therefore, for analysis of steady-state vibration, parameter u_{s}=\sqrt[4]{m_{0} l^{4} \theta^{2} / E I}=\sqrt{2.256} \approx 1.5 should be adopted. Just this parameter corresponds to the given frequency \theta of excitation. Therefore unit reactions r_{i k}\left(u_{s}\right) should be re-evaluated; in particular, r_{11}=4 i \psi_{2}(1.5)+12 i \psi_{1}(0.707 \times 1.5). These reactions to be substituted into nonhomogeneous system of equations (a).

We now turn to the definition of free terms R_{i P} of canonical equations. Since the external load is nodal, there are no bending moments in the loaded state. This force is perceived by the introduced constraint 2; therefore R_{1 P}=0, R_{2 P}=-P. The system of canonical equations is

\begin{aligned}& r_{11}\left(u_{s}\right) Z_{1}+r_{12}\left(u_{s}\right) Z_{2}=0 \\& r_{21}\left(u_{s}\right) Z_{1}+r_{22}\left(u_{s}\right) Z_{2}=P\end{aligned}                  (g)

Solution of these equations

Z_{1}=0.0759 \frac{P l^{2}}{E I}(\mathrm{rad}), \quad Z_{2}=0.210 \frac{P l^{3}}{E I}(m)           (h)

The final bending moment diagram should be constructed according to the formula M_{\text {dyn }}=\bar{M}_{1} Z_{1}+\bar{M}_{2} Z_{2}; it can be seen that the term M_{P}^{0}=0. The bending moment at Section 1 which belongs to the member 1-A in the vicinity of joint 1

\begin{aligned}M_{1 A}=M_{1 A}^{(1)} Z_{1}+M_{1 A}^{(2)} Z_{2} & =4 i \psi_{2}(1.5) Z_{1}+\frac{6 i}{l} \psi_{5}(1.5) Z_{2}= \\& 4 i \times 0.98784 \times 0.0759 \frac{P l^{2}}{E I}-\frac{6 i}{l} 0.95547 \times 0.210 \frac{P l^{3}}{E I}=-0.903 P l \end{aligned}        (i)

The bending moment at any specified section may be calculated similarly. Amplitude bending moment diagram is shown in Fig. 17.34e. The bending moment diagram in case of static loading by the amplitude value of P is shown in Fig. 17.34f. The most dangerous dynamic state occurs in the clamped support of the left column. For this section, the dynamic coefficient is equal to

\mu_{A}=\frac{M_{A}^{\mathrm{dyn}}}{M_{A}^{\text {stat }}}=\frac{1.138}{0.412}=2.76

This example clearly shows the procedural difficulties of the analysis caused by introduction of only one assumption, namely, of accounting for distributed mass. A significant simplification of this exact procedure of dynamic analysis is discussed in the next paragraph.

A further complication of the design diagram of the elements of the primary system of the displacement method can be found in the book (Karnovsky and Lebed 2004a). Table 3.8 of this book contains the reactions of one-span beams with a uniformly distributed mass and attached one lumped mass.