Question 17.15: Symmetrical portal frame contains three massless rods of len......

Solution This structure has four degrees of freedom, while kinematical indeterminacy of frame is three. The primary system of displacement method contains three introduces constraints 1, 2, 3 as shown in Fig. 17.28b. In case of symmetrical structure it is possible to apply a new type of the primary unknowns, mainly, the group unknowns.

In case of symmetrical vibration the group unknown presents the simultaneous angular harmonic displacements of introduced constraints 1 and 2 in opposite direction. In case of antisymmetrical vibration the group unknown presents the angular harmonic displacements of introduced constraints 1 and 2 in the same direction.

Below we will consider only symmetrical vibration. Corresponding elastic curve and bending moment diagram are shown in Fig. 17.28c. Ordinates of bending moments and shear forces are determined according to Table A.27, rows 1,2, special case a=b=l / 2.

Primary System, Unit State The unit state presents the simultaneous rotation of two introduced constraints Z_{1}=Z_{2}=Z=1. Below are presented the amplitude values of the bending moments at specified points of structure for dimensionless frequency parameter \theta / \omega=0.5, where \omega is a frequency of free vibration.

Vertical member 1-A and 2-B (Fig. 17.28c)

\begin{aligned}& \bar{M}_{1}^{\mathrm{vert}}=\left.\frac{4 E I}{l}\left(1-\frac{7}{4} \frac{\theta^{2}}{\omega^{2}}\right) \frac{1}{1-\theta^{2} / \omega^{2}}\right|_{\theta / \omega=0.5}=4 \frac{E I}{l}\left(1-\frac{7}{4} 0.5^{2}\right) \frac{1}{1-0.5^{2}}=3 \frac{E I}{l} \\& \bar{Q}_{1}^{\mathrm{vert}}=-6 \frac{E I}{l^{2}}\left(1-3 \frac{\theta^{2}}{\omega^{2}}\right) \frac{1}{1-\theta^{2} / \omega^{2}}=-6 \frac{E I}{l^{2}} \cdot 0.333=-\frac{2 E I}{l^{2}} \\& \bar{M}_{m}^{\mathrm{vert}}=\bar{M}_{1}^{\mathrm{vert}}+\bar{Q}_{1}^{\mathrm{vert}} \frac{l}{2}=\frac{E I}{l}\left(1+2 \frac{\theta^{2}}{\omega^{2}}\right) \frac{1}{1-\theta^{2} / \omega^{2}}=2 \frac{E I}{l} \\& \bar{M}_{A}^{\mathrm{vert}}=-\frac{2 E I}{l}\left(1+\frac{1}{2} \frac{\theta^{2}}{\omega^{2}}\right) \frac{1}{1-\theta^{2} / \omega^{2}}=-3 \frac{E I}{l}\end{aligned}

Horizontal member 1- 2, using superposition principle,

\begin{aligned}& \bar{M}_{1}^{\mathrm{hor}}=\frac{4 E I}{l}\left(1-\frac{7}{4} \frac{\theta^{2}}{\omega^{2}}\right) \frac{1}{1-\theta^{2} / \omega^{2}}-\frac{2 E I}{l}\left(1+\frac{1}{2} \frac{\theta^{2}}{\omega^{2}}\right) \frac{1}{1-\theta^{2} / \omega^{2}}=\frac{2 E I}{l}\left(1-4 \frac{\theta^{2}}{\omega^{2}}\right) \frac{1}{1-\theta^{2} / \omega^{2}}=0.0 \\& \bar{Q}_{1}^{\mathrm{hor}}=-6 \frac{E I}{l^{2}}\left(1-3 \frac{\theta^{2}}{\omega^{2}}\right) \frac{1}{1-\theta^{2} / \omega^{2}}+6 \frac{E I}{l^{2}}\left(1+\frac{\theta^{2}}{\omega^{2}}\right) \frac{1}{1-\theta^{2} / \omega^{2}}=24 \frac{E I}{l^{2}} \frac{\theta^{2}}{\omega^{2}} \frac{1}{1-\theta^{2} / \omega^{2}}=\frac{8 E I}{l^{2}} \\& \bar{M}_{m}^{\mathrm{hor}}=\bar{M}_{1}^{\mathrm{hor}}+\bar{Q}_{1}^{\mathrm{hor}} \frac{l}{2}=2 \frac{E I}{l}\left(1+2 \frac{\theta^{2}}{\omega^{2}}\right) \frac{1}{1-\theta^{2} / \omega^{2}}=4 \frac{E I}{l} \\& \bar{M}_{2}^{\mathrm{hor}}=\bar{M}_{1}^{\mathrm{hor}}=0.0 \end{aligned}

Based on the superposition principle, these expressions take into account the simultaneous unit rotation of the first and second introduced constraints (the first and second terms, respectively).

Primary System: Loading State Only horizontal member 1-2 is subjected to bending deformation (Fig. 17.28d)

\begin{aligned}& M_{1 P}^{0}=M_{2 P}^{0}=-\frac{P l}{8} \frac{1}{1-\theta^{2} / \omega^{2}}=-0.1667 P l, \\& M_{m P}^{0}=\frac{P l}{8} \frac{1}{1-\theta^{2} / \omega^{2}}=0.1667 P l, \end{aligned}

Unit reaction and free term of canonical equation  (\theta / \omega=0.5)

\begin{aligned}& \frac{r_{11}}{2}=\bar{M}_{1}^{\text {vert }}+\bar{M}_{1}^{h o r}=\frac{4 E I}{l}\left(1-\frac{7}{4} \frac{\theta^{2}}{\omega^{2}}\right) \frac{1}{1-\theta^{2} / \omega^{2}}+\frac{2 E I}{l}\left(1-4 \frac{\theta^{2}}{\omega^{2}}\right) \frac{1}{1-\theta^{2} / \omega^{2}}= \\& \frac{E I}{l}\left(6-15 \frac{\theta^{2}}{\omega^{2}}\right) \frac{1}{1-\theta^{2} / \omega^{2}}=3 \frac{E I}{l} \\& \frac{R_{1 P}}{2}=M_{1 P}^{0}=-\frac{P l}{8} \frac{1}{1-\theta^{2} / \omega^{2}}=-0.1667 P l \end{aligned}

Primary unknown becomes

Z=-\frac{R_{1 P}}{r_{11}}=\frac{P l^{2}}{8 E I} \frac{1}{6-15 \frac{\theta^{2}}{\omega^{2}}}=\frac{0.1667 P l}{3 i}=0.05556 \frac{P l^{2}}{E I} \quad(\mathrm{rad})

Bending moment diagram may be constructed by formula(M_{P}=\bar{M} Z+M_{1 P}^{0}(Table 17.8)

Shear can be calculated by formula Q=d M / d x. For horizontal member shear forces within portion 1-m and  m -2 are

 Q_{1 \mathrm{~m}}=\frac{0.1667-(-0.3889)}{l / 2} P l=1.1112 P, \quad Q_{m 2}=-1.1112 P 

Inertial force J_{h o r} which acts at mass m of the crossbar may be calculated using the following equation:

\sum Y=0: \quad Q_{1 \mathrm{~m}}-P-J_{\mathrm{hor}}-Q_{m 2}=0 \rightarrow J_{\mathrm{hor}}=2 \times 1.1112 P-P=1.2224 P m

Same approaches should be applied for vertical element

\begin{gathered}Q_{1 \mathrm{~m}}=-\frac{0.1667-0.1111}{l / 2} P l=-0.1112 P, \quad Q_{m A}=-\frac{0.1111-(-0.1667)}{l / 2} P l=-0.5556 P \\\sum X=0: \quad Q_{1 \mathrm{~m}}-J_{\mathrm{ver}}-Q_{m A}=0 \rightarrow J_{\mathrm{vert}}=0.5556 P-0.1112 P=0.4444 P\end{gathered}

All forces which act on the frame are shown in Fig. 17.28f.

Reaction of support are

H_{A}=0.5556 P(\rightarrow), \quad R_{A}=Q_{1 m}^{\text {hor }}=1.1112 P

In fact, all reactions and internal forces contains factor \sin \theta t; therefore ordinates in Fig. 17.28e,f means the amplitude values.

\text { Verification : } \sum Y=R_{A}+R_{B}-P-J=(1.1112+1.1112-1.0-1.2224) P=(2.2224-2.2224) P=0

Table 17.9 contains the bending moments at characteristic sections caused by static action of the load P (Umansky 1973 , Table 8.2.9] and corresponding dynamic coefficients.

In this example, the most dangerous in terms of dynamic coefficients are clamped supports A and B.

Comments

1. In this example, the relative frequency of excitation, i.e., a certain detuning \theta / \omega, had been accepted, while separately the frequency of free vibration \omega and the frequency of excitation \theta are not computed.
2. The symmetry property and the concept of group unknowns allows using only one canonical equation of the displacement method when analyzing symmetric frame vibrations.
3. In the case of symmetric vibration, the horizontal displacement of the crossbar is absent, the diagram of bending moments and longitudinal forces (symmetric internal forces) turns out to be symmetrical, and the diagram of shear forces (antisymmetric internal force) is antisymmetric.
4. In the case of antisymmetric vibrations, the vertical displacement of the point of the horizontal member on the axis of symmetry is absent (frame with lateral shifts); the diagram of bending moments and longitudinal forces turn out to be antisymmetrical, and diagram of shear forces turns out to be symmetrical.

Extensive reference data relating to various structures are presented in the fundamental Handbooks (Blevins 1979; Young et al. 2012; Umansky 1973), etc.