22-Mec-A2 Kinematics and Dynamics of Machines · December 2014

Question 6 of 6: Three-degree-of-freedom system — modal analysis & forced response

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams, December 2014 — 07-Mec-A2, Kinematics and Dynamics of Machines. 3 hours, OPEN BOOK. Six questions of 20 marks each; candidates answer any FIVE. All SIX are solved in full below (Part A — mechanisms and machine dynamics, Q1–Q4; Part B — mechanical vibration, Q5–Q6).

Reference texts: R. L. Norton, Design of Machinery, 5th ed. (instant centres and velocity analysis, four-bar position/velocity/acceleration, planetary gear kinematics, engine balancing); S. S. Rao, Mechanical Vibrations, 6th ed. (single- and multi-degree-of-freedom free and forced response, modal analysis); J. E. Shigley & C. R. Mischke, Mechanical Engineering Design (epicyclic train-value cross-check).

Question 6: Three-degree-of-freedom system — modal analysis & forced response (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A vertical chain from a fixed support: spring $k$ → mass $2m$ ($x_3$) → spring $k$ → mass $2m$ ($x_2$) → spring $k$ → mass $m$ ($x_1$), with $F(t)$ on the bottom mass. $m=1\ \text{kg}$, $k=10{,}000\ \text{N/m}$, $F=10\sin30t\ \text{N}$ ($\Omega=30\ \text{rad/s}$).

Find. (a) $[M],[K]$ and the EOM; (b) the three modal vectors; (c) the mass-normalised $[\Phi]$; (d) the steady-state $x_1,x_2,x_3$.

Figure 6.1 — three-DOF chain: support–$k$–$2m$($x_3$)–$k$–$2m$($x_2$)–$k$–$m$($x_1$), forced by $F(t)$ on the bottom mass.

Approach. Assemble the mass and stiffness matrices, solve the eigenproblem for the mode shapes, normalise them to the mass matrix, then obtain the forced response by modal superposition (confirmed by direct inversion of the dynamic stiffness).

(a) Equations of motion. Measuring $x_i$ downward from static equilibrium (gravity cancels), $$[M]=\begin{bmatrix}m&0&0\\0&2m&0\\0&0&2m\end{bmatrix},\quad [K]=k\begin{bmatrix}1&-1&0\\-1&2&-1\\0&-1&2\end{bmatrix},\quad [M]\ddot{\mathbf x}+[K]\mathbf x=\begin{Bmatrix}F\\0\\0\end{Bmatrix},$$ with $\mathbf x=(x_1,x_2,x_3)^T$, $m=1$, $k=10^4$, $F=10\sin30t$.
Natural frequencies (check). $\det([K]-\omega^2[M])=0$ yields $\omega_1=36.6,\ \omega_2=100,\ \omega_3=136.6\ \text{rad/s}$, as given.
(b) Modal vectors. Solving $([K]-\omega_r^2[M])\boldsymbol\phi=0$ (setting $\phi_1=1$) gives the clean shapes $$\boldsymbol\phi^{(1)}=\begin{Bmatrix}1\\0.866\\0.500\end{Bmatrix},\quad \boldsymbol\phi^{(2)}=\begin{Bmatrix}1\\0\\-1\end{Bmatrix},\quad \boldsymbol\phi^{(3)}=\begin{Bmatrix}1\\-0.866\\0.500\end{Bmatrix}.$$ Mode 1 all-in-phase (fundamental), mode 2 has the middle mass stationary, mode 3 is the highest.
(c) Mass-normalised modal matrix. Each mode has $\boldsymbol\phi^{(r)T}[M]\boldsymbol\phi^{(r)}=3$, so dividing by $\sqrt3$ gives $$[\Phi]=\frac{1}{\sqrt3}\begin{bmatrix}1&1&1\\0.866&0&-0.866\\0.5&-1&0.5\end{bmatrix}=\begin{bmatrix}0.577&0.577&0.577\\0.500&0&-0.500\\0.289&-0.577&0.289\end{bmatrix},$$ which satisfies $[\Phi]^T[M][\Phi]=[I]$ and $[\Phi]^T[K][\Phi]=\mathrm{diag}(\omega_r^2)$.
(d) Steady-state response. Modal forces $Q_r=\boldsymbol\phi^{(r)T}\mathbf F=0.577\times10 =5.77\ \text{N}$ (all three, since only $x_1$ is loaded and every mode has $\phi_1=1/\sqrt3$). Each modal coordinate is $q_r=\dfrac{Q_r}{\omega_r^2-\Omega^2}\sin\Omega t$ ($\Omega^2=900$), and $\mathbf x=[\Phi]\mathbf q$ gives $$x_1=8.13\sin30t,\ x_2=6.40\sin30t,\ x_3=3.52\sin30t\ \ (\text{mm}).$$ The drive at $30\ \text{rad/s}$ is below the lowest resonance, so all masses move in phase with the force, largest at the driven bottom mass.

Quantity	Value
Natural frequencies	36.6, 100, 136.6 rad/s
Modal vectors ($\phi_1{=}1$)	(1, 0.866, 0.5), (1, 0, −1), (1, −0.866, 0.5)
Normalising factor	$1/\sqrt3$ (each mode)
Steady state $x_1,x_2,x_3$	8.13, 6.40, 3.52 mm (×$\sin30t$)

Check. The modal result equals the direct solution $\mathbf X=([K]-\Omega^2[M])^{-1}\mathbf F$ (8.13, 6.40, 3.52 mm), a full independent check. With no damping and $\Omega<\omega_1$, every modal denominator $\omega_r^2-\Omega^2>0$, so the response is exactly in phase; near any $\omega_r$ the amplitude would blow up.

Topic. Eigen (modal) analysis and modal-superposition forced response of an undamped multi-DOF system.

Key relations. $([K]-\omega_r^2[M])\boldsymbol\phi^{(r)}=0$; mass normalisation $\boldsymbol\phi^{(r)T}[M]\boldsymbol\phi^{(r)}=1$; decoupled modal EOM $\ddot q_r+\omega_r^2q_r=Q_r\sin\Omega t$ with $Q_r=\boldsymbol\phi^{(r)T}\mathbf F$; $\mathbf x=[\Phi]\mathbf q$.

Why this works. The mass-orthonormal modes diagonalise both $[M]$ and $[K]$, so the coupled equations become three independent single-DOF oscillators. Each is solved by the standard harmonic formula and the physical motion is rebuilt by superposing the modes — exact for a linear system.

Common pitfall. Normalising to unit length instead of unit modal mass (then $[\Phi]^T[K][\Phi]\neq\mathrm{diag}(\omega_r^2)$), and forgetting that only the loaded coordinate enters $Q_r$. (Rao, Mechanical Vibrations, ch. 6.)

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