Question 19.4: The uniform beam shown in Figure E19.4 is supported by three...
The uniform beam shown in Figure E19.4 is supported by three spring supports as indicated. The cross sectional moment of inertia of the beam is I, the length of each of the four equal segments shown in the figure is l, and the modulus of elasticity is E. The spring stiffnesses are k_{1}=20 k, k_{2}=30 k, and k_{3}=25 k, where k=E I / l^{3}. During an experiment three independent displacement patterns are produced by applying the following set of forces: (1) F_{1}=50 k, F_{2}=100 k, F_{3}=75 k; (2) F_{1}=100 k, F_{2}=150 k, F_{3}=90 k; (3) F_{1}=80 k, F_{2}=90 k, F_{3}=70 k. Using computer simulation generate the resulting displacements along the free and support d.o.f. Hence, obtain the three constrained mode shapes. Verify that they are identical to those computed from Equation 19.6 . Next obtain the support reactions by computer simulation and hence determine the effective constrained coordinate stiffness matrix \tilde{\mathbf{K}}_{s s}. Compare the result with that obtained directly from Equation 19.11b.
The stiffness matrices for the beam are given by
\begin{aligned}\mathbf{K}_{f f} =&k\left[\begin{array}{rrrrrrr}4 & -6 & 2 & 0 & 0 & 0 & 0 \\-6 & 24 & 0 & 6 & 0 & 0 & 0 \\2 & 0 & 8 & 2 & 0 & 0 & 0 \\0 & 6 & 2 & 8 & -6 & 2 & 0 \\0 & 0 & 0 & -6 & 24 & 0 & 6 \\0 & 0 & 0 & 2 & 0 & 8 & 2 \\0 & 0 & 0 & 0 & 6 & 2 & 4\end{array}\right] \text { (a) } \\\mathbf{K}_{s f}=&k\left[\begin{array}{rrrrrrr}6 & -12 & 6 & 0 & 0 & 0 & 0 \\0 & -12 & -6 & 0 & -12 & 6 & 0 \\0 & 0 & 0 & 0 & -12 & -6 & -6\end{array}\right]=\mathbf{K}_{f s}^{T} \quad (b) \\\mathbf{K}_{s s} =&k\left[\begin{array}{rrr}32 & 0 & 0 \\0 & 54 & 0 \\0 & 0 & 37\end{array}\right] \text { (c) }\end{aligned}
where the rotations \theta are represented by l \theta.
The assembled beam stiffness matrix is given by
\mathbf{K}_{b}=\left[\begin{array}{ll}\mathbf{K}_{f f} & \mathbf{K}_{f s} \\\mathbf{K}_{s f} & \mathbf{K}_{s s}\end{array}\right] \text { (d) }
The matrix of support spring stiffness is
\mathbf{k}_{s}=\left[\begin{array}{ccc}k_{1} & 0 & 0 \\0 & k_{2} & 0 \\0 & 0 & k_{3}\end{array}\right]=k\left[\begin{array}{rrr}20 & 0 & 0 \\0 & 30 & 0 \\0 & 0 & 25\end{array}\right]
The stiffness matrix for the assembly of beam and the supporting spring is obtained from
\mathbf{K}=\left[\begin{array}{cc}\mathbf{K}_{f f} & \mathbf{K}_{f s} \\\mathbf{K}_{s f} & \mathbf{K}_{s s}+\mathbf{k}_{s}\end{array}\right] (f)
The vector of forces applied on the beam is obtained from
\mathrm{F}=\left[\begin{array}{c}0 \\\mathrm{~F}_{s}\end{array}\right](\mathrm{g})
where 0 is a null matrix of size 7 \times 3 and \mathrm{F}_{s} is given by
\mathrm{F}_{s}=k\left[\begin{array}{rrr}50 & 150 & 80 \\100 & 150 & 90 \\75 & 90 & 70\end{array}\right] \text { (h) }
The displacements resulting from the application of forces \mathrm{F} is
\mathbf{u}=\mathbf{K}^{-1} \mathbf{F}=\left\{\begin{array}{c}\Delta_{f} \\\Delta_{s}\end{array}\right\}
which gives
The displacements of the supports are
\boldsymbol{\Delta}_{s}=\left[\begin{array}{lll}2.5105 & 5.0126 & 3.9928 \\3.3193 & 4.9832 & 3.0096 \\3.0084 & 3.6101 & 2.7942\end{array}\right](\mathrm{k})
In practical application the data presented in Equations (\mathrm{j}) and (\mathrm{k}) will be obtained through measurements.
The constraint modes are now obtained from
\boldsymbol{\Psi}_{s}=\left\{\begin{array}{c}\boldsymbol{\Psi}_{f s} \\\mathbf{I}\end{array}\right\}
where
\boldsymbol{\Psi}_{f s}=\boldsymbol{\Delta}_{f} \boldsymbol{\Delta}_{s}^{-1}=\left[\begin{array}{rrr}-0.6250 & 0.7500 & -0.1250 \\0.4063 & 0.6875 & -0.0937 \\-0.5313 & 0.5625 & -0.0313 \\-0.2500 & -0.0000 & 0.2500 \\-0.0937 & 0.6875 & 0.4063 \\0.0312 & -0.5625 & 0.5313 \\0.1250 & -0.7500 & 0.6250\end{array}\right] \quad(\mathrm{m})
The modes obtained in Equation (\mathrm{m}) are identical to those obtained from Equation 19.6. The support reactions are given by
\mathbf{R}_{s}=-\mathbf{k}_{s} \boldsymbol{\Delta}_{s}=-k\left[\begin{array}{rrr}50.21 & 100.25 & 79.86 \\99.58 & 149.50 & 90.29 \\75.21 & 90.25 & 69.86\end{array}\right] \text { (n) }
Again, in practical application the support reactions will be obtained from measurements The constrained coordinate stiffness matrix is now given by
The \tilde{\mathbf{K}}_{s s} from Equation (o) is identical to that calculated from Equation 19.11b.