Question 17.18: Design diagram of two-span continuous uniform beam A−1−B wit......
Design diagram of two-span continuous uniform beam A-1-B with equal spans is shown in Fig. 17.33a. The beam is subjected to excitation force P(t)=P \sin \theta t at the middle point of span A-1. Discuss the procedure for determining the primary unknown. Use the expressions for dynamic reactions Krylov-Duncan functions (Table A.25). The uniformly distributed mass m_{0} per unit length should be taken into account.
Solution The primary system contains additional constraint 1 which prevents angular displacement of the beam at support 1 (Fig. 17.33b). The primary unknown is angle of rotation Z_{1} of the introduced constraint 1. Bending moment diagrams in primary system caused by harmonic angular displacement with unit amplitude Z_{1}=1 (the unit state) and by given force P(t) (loaded state) are shown in Fig. 17.33c,d, respectively; both diagrams are curvilinear.
Ordinates at specified points for unit and loading states, according to Table A.25, cases 1 and 3, respectively, are:
Unit state (all Krylov-Duncan functions has subscript k l )
\begin{aligned}& \bar{M}_{1 A}=\frac{4 E I}{l} f_{3}(\lambda), \quad f_{3}=\frac{\lambda}{4} \frac{T U-S V}{U^{2}-T V} \\& \bar{M}_{1 B}=\frac{3 E I}{l} f_{10}(\lambda), \quad f_{10}=\frac{\lambda}{3} \frac{T^{2}-V^{2}}{T U-S V}\end{aligned}
Loaded state (Table A.25, case 5)
M_{1 A-P}^{0}=-\frac{P}{k} \frac{U_{k l} V_{k a}-V_{k l} U_{k a}}{U_{k l}^{2}-T_{k l} V_{k l}}
Equilibrium of joint 1 in the unit and loaded states leads to the following reactions:
\begin{aligned}r_{11}=\bar{M}_{1 A}+\bar{M}_{1 B} & =\frac{4 E I}{l} \frac{\lambda}{4} \frac{T U-S V}{U^{2}-T V}+\frac{3 E I}{l} \frac{\lambda}{3} \frac{T^{2}-V^{2}}{T U-S V} \\& =k E I \frac{(T U-S V)^{2}+\left(U^{2}-T V\right)\left(T^{2}-V^{2}\right)}{\left(U^{2}-T V\right)(T U-S V)} \\R_{1 P}=M_{1 A P}^{0} & =\frac{P}{k} \frac{U_{k l} V_{k a}-V_{k l} U_{k a}}{U_{k l}^{2}-T_{k l} V_{k l}}\end{aligned}
For given data l, m_{0}, \theta, E I we get the following parameters: k=\sqrt[4]{m_{0} \theta^{2} / E I}\left(\right. length \left.{ }^{-1}\right), k a, \lambda=k l, and after that using Table of Krylov functions we can calculate all functions of the argument k a and k l. They are
\begin{aligned}& U_{k a}=U(k a), \quad V_{k a}=V(k a), \\& S=S_{k l}=S(k l), \quad T=T_{k l}=T(k l), \cdots\end{aligned}
Canonical equation of displacement method is r_{11} Z_{1}+R_{1 P}=0, so primary unknown becomes Z_{1}=-\frac{R_{1 P}}{r_{11}}.
Final amplitude of bending moments and shear at specified sections are
\begin{aligned}& M_{1 B}=\bar{M}_{1 B} Z_{1}, \quad M_{1 A}=\bar{M}_{1 A} Z+M_{1 A P}^{0}, \cdots \\& Q_{B}=\bar{Q}_{B} Z_{1}, \quad Q_{1 B}=\bar{Q}_{1 B} Z+M_{1 B P}^{0}, \cdots\end{aligned}
| Table A.25 Amplitude values of dynamical reactions of uniform beams with distributed mass m_0 ; \lambda=k l , \varphi(t)=\varphi \sin \theta t ; \Delta(t)=\Delta \sin \theta t ; \varphi,\Delta are amplitudes of angular and linear displacements, \varphi=1, \Delta=1(\mathrm{Kiselev} 1980) | ||||
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Bending moment M(0)=M_0 | Bending moment M(l) | Shear force Q(0) | Shear Force Q(l) |
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\begin{gathered}\frac{4 E I}{l} f_3(\lambda) \\f_3=\frac{\lambda}{4} \frac{T U-S V}{U^2-T V}\end{gathered} | \begin{gathered}\frac{2 E I}{l} f_5(\lambda) \\f_5=\frac{\lambda}{2} \frac{V}{U^2-T V}\end{gathered} | \begin{gathered}-\frac{6 E I}{l^2} f_4(\lambda) \\f_4=\frac{\lambda}{4} \frac{S U-T^2}{U^2-T V}\end{gathered} | \begin{gathered}-\frac{6 E I}{l^2} f_6(\lambda) \\f_6=\frac{\lambda^2}{6} \frac{U}{U^2-T V}\end{gathered} |
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\begin{gathered}-\frac{6 E I}{l^2} f_4(\lambda) \\f_4=\frac{\lambda}{4} \frac{S U-T^2}{U^2-T V}\end{gathered} | \begin{gathered}-\frac{6 E I}{l^2} f_8(\lambda) \\f_8=\frac{\lambda^2}{6} \frac{U}{U^2-T V}\end{gathered} | \begin{gathered}\frac{12 E I}{l^3} f_7(\lambda) \\f_7=\frac{\lambda^3}{12} \frac{S T-U V}{U^2-T V}\end{gathered} | \begin{gathered}\frac{12 E I}{l^3} f_9(\lambda) \\f_9=\frac{\lambda^3}{12} \frac{U}{U^2-T V}\end{gathered} |
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\begin{gathered}\frac{3 E I}{l} f_{10}(\lambda) \\f_{10}=\frac{\lambda}{3} \frac{T^2-V^2}{T U-S V}\end{gathered} | 0 | \begin{gathered}-\frac{3 E I}{l^2} f_{11}(\lambda) \\f_{11}=\frac{\lambda^2}{3} \frac{U V-S T}{T U-S V}\end{gathered} | \begin{gathered}-\frac{3 E I}{l^2} f_{12}(\lambda) \\f_{12}=\frac{\lambda^2}{3} \frac{T}{T U-S V}\end{gathered} |
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\begin{gathered}-\frac{3 E I}{l^2} f_{11}(\lambda) \\f_{11}=\frac{\lambda^2}{3} \frac{U V-S T}{T U-S V}\end{gathered} | 0 | \begin{gathered}-\frac{3 E I}{l^3} f_{13}(\lambda) \\f_{13}=\frac{\lambda^2}{3} \frac{U^2-S^2}{T U-S V}\end{gathered} | \begin{gathered}-\frac{3 E I}{l^3} f_{14}(\lambda) \\f_{14}=-\frac{\lambda^2}{3} \frac{S}{T U-S V}\end{gathered} |
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\frac{P}{k} \frac{U_{k l} V_{k a}-V_{k l} U_{k a}}{U_{k l}^2-T_{k l} V_{k l}} | -\frac{P}{k} \frac{U_{k l} V_{k a}-V_{k l} U_{k a}}{U_{k l}^2-T_{k l} V_{k l}} | P \frac{U_{k l} U_{k a}-T_{k l} V_{k a}}{U_{k l}^2-T_{k l} V_{k l}} | P \frac{U_{k l} U_{k a}-T_{k l} V_{k a}}{U_{k l}^2-T_{k l} V_{k l}} |