Mechanical Vibrations Test 1

Concept:

\( {ω _d} = {ω _n}\sqrt {1 - {ξ ^2}} \)

where, ω_n = natural frequency, ξ = damping ratio

Calculation:

Given: C = 0.2 C_c , mass (m) = 50 kg, stiffness (k) = 30 kN/m

\(ξ = \frac{C}{{{C_c}}} = 0.2\)

\({ω _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{{30 \times {{10}^3}}}{{50}}} = 24.5\;rad/sec\)

Concept:

\(\xi = \frac{C}{{{C_c}}}\)

Equation of motion

\(m\ddot x + c\dot x + kx = 0\)

Calculation:

Comparing, we get

m = 3 kg, C = 9 Ns/m,  k = 27 N/m

\({C_c} = \sqrt {{km}} = 2\sqrt {27 \times 3} = 18\;\;Ns/m\)

\(\therefore \xi = \frac{C}{{{C_c}}} = \frac{9}{{18}}\)

∴ ξ = 0.5

Concept:

\({\rm{Phase\;difference\;}}\left( \phi \right) = {\tan ^{ - 1}}\left( {\frac{{2\xi \left( {\frac{\omega }{{{\omega _n}}}} \right)}}{{1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}}}} \right)\)

At resonance ω  = ω_n

∴ ϕ = tan^-1 (∞) = 90°

Concept:

Logarithmic decreament \(\left( \delta \right) = \frac{{2\pi \xi }}{{\sqrt {1 - {\xi ^2}} }}\)

Calculation:

ξ = 0.3

\(\therefore \delta = \frac{{2\pi\; \times\; 0.3}}{{\sqrt {1\; - \;{{0.3}^2}} }}\)

∴ δ = 1.976

Ratio of 2 successive amplitude is always constant and given as,

\(\frac{{{X_n}}}{{{X_{n + 1}}}} = {e^\delta } = {e^{1.976}}\)

\(\therefore \frac{{{X_n}}}{{{X_{n + 1}}}} = 7.21\)

Equation of motion will be:

\(0.5\;\ddot x + 300\;x = 0\) 

\(0.5\;\ddot x + 300\;x = 0\) 

Natural frequency:

\(\omega = \sqrt {\frac{k}{m}} = \sqrt {\frac{{300}}{{0.5}}} = 24.5\;rad/s\) 

\(f = \frac{\omega }{{2\pi }} = 3.9\;Hz\)

Note:

\(\omega = \sqrt {\frac{k}{m}} = \sqrt {\frac{g}{{{\delta _{st}}}}} \) 

Concept:

Damping ratio \(= \frac{C}{{{C_c}}} \)

where C_c = 2√ km and c = damping coefficient.

Calculation:

\(16\ddot x + 5\dot x + 4x = 0\). 

By comparing this equation with \(m\ddot x + c\dot x + kx = 0\)

\(\begin{array}{l} m\; = 16,\;c = 5,k = 4\\ {C_c} = 2√ {km} \\ {C_c} = 2√ {4 \times 16} = 16 \end{array}\)

∴ Damping ratio \(= \frac{C}{{{C_c}}} = \frac{5}{{16}}\)

Concept:

\(X = \frac{{\frac{{{m_0}r{\omega ^2}}}{k}}}{{\sqrt {{{\left( {1 - {r^2}} \right)}^2} + {{\left( {2\xi r} \right)}^2}} }}{\rm{\;and\;}}{\omega _n} = \sqrt {\frac{k}{{M\;}}\;} \)

For large ω, the ratio ω/ω_n ≫ 1

\(\therefore X = \frac{{{m_0}r}}{M}\)

Calculation:

\(X = \frac{{{m_0}r}}{M}\)

\(X = \frac{{1\; \times \;1}}{{100}}\)

∴ X = 0.01 cm   

Concept:

\(\omega = \sqrt {\frac{g}{\delta }} \;\)

For cantilever beam:

\(\begin{array}{l} \delta = \frac{{P{L^3}}}{{3EI}}\\ \omega = \sqrt {\frac{g}{\delta }} = \sqrt {\frac{{3EIg}}{{P{L^3}}}} \Rightarrow \omega \propto \sqrt {\frac{1}{{{L^3}}}} \\ \frac{{{\omega _2}}}{{{\omega _1}}} = \sqrt {\frac{{L_1^3}}{{L_2^3}}} = \sqrt {\frac{{{L^3}}}{{{{\left( {\frac{L}{2}} \right)}^3}}}} = 2\sqrt2 \\ {\omega _2} > {\omega _1} \end{array}\)

Concept:

\({\rm{Transmissibility\;}} = \frac{{{{\left[ {1 + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]}^{1/2}}}}{{{{\left[ {{{\left( {1 - \frac{{{\omega ^2}}}{{\omega _n^2}}} \right)}^2} + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]}^{1/2}}}}\)

Calculation:

ω  = √2 ω_n

\(\therefore Transmissibility = \frac{{{{\left( {1 + 8{\xi ^2}} \right)}^{\frac{1}{2}}}}}{{{{\left( {1 + 8{\xi ^2}} \right)}^{\frac{1}{2}}}}} = 1.0\)

Concept:

F_T is the force transmitted to the foundation. The disturbing force is F. The ratio of F_T to F is called transmissibility.

\(T.R = \frac{{{F_T}}}{F} = \frac{{\sqrt {1 + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

The steady state amplitude for the system:

\(A = \frac{{\frac{F}{k}}}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

Calculation:

Given: m = 10 kg, k = 4000 N/m, C = 40 Ns/m, F = 60 N

ω = 40 rad/sec

\({\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{{4000}}{{10}}} = 20\;rad/sec\)

Frequency Ratio:

\(r = \frac{\omega }{{{\omega _n}}} = \frac{{40}}{{20}} = 2\)

The critical damping coefficient:

\({c_c} = 2m\;{\omega _n} = 2\sqrt {mk} = 2\sqrt {10 \times 4000} = 400\;N.s/m\)

The damping factor, ξ:

\(\xi = \frac{C}{{{C_c}}} = \frac{{40}}{{400}} = 0.1\)

Transmissibility:

\(T.R = \frac{{{F_T}}}{F} = \frac{{\sqrt {1 + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

\(T = \frac{{\sqrt {1 + {{\left( {2 \times 0.1 \times 2} \right)}^2}} }}{{\sqrt {{{\left( {1 - {2^2}} \right)}^2} + \left( {2 \times 0.1 \times 2} \right)^2} }} = \frac{{1.07703}}{{3.02654}} = 0.3559\)

The amplitude of the force transmitted:

F_T = (T.R) F = (0.3379) × 60

∴ FT = 21.35 N

Concept:

To calculate lowest critical speed (w)

\(\omega = 2\pi f = 2\pi \frac{N}{{60}}\) 

f = frequency, N = rpm, ∴ f = N/60

⇒ N = 60 f

Dunkerley’s empirical formula

\(\frac{1}{f} = \frac{1}{{{f^2}}} + \frac{1}{{f_2^2}}\)

 Calculation:

\(\frac{1}{{{f^2}}} = \frac{1}{{{{200}^2}}} + \frac{1}{{{{310}^2}}}\)

\(\Rightarrow f = 168.059\) 

∴ N = 60 f = 10083.55 rpm

Concept:

For long bearing: fixed-fixed end condition

\(k = \frac{{192\;EI}}{{{\ell ^3}}}\)

\({\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{{192\;EI}}{{m{\ell ^3}}}} \)

Calculation:

\(I = \frac{{\pi {d^4}}}{{64}} = \frac{{\pi\; \times \;{{\left( {0.02} \right)}^4}}}{{64}} = 7.854 \times {10^{ - 9}}{m^4}\)

\({\omega _n} = \sqrt {\frac{{192\; \times \;2\; \times \;{{10}^{11}}\; \times \;7.854\; \times \;{{10}^{ - 9}}}}{{19.2\; \times \;{{\left( {0.5} \right)}^3}}}} = 354.5\;rad/s\)

\({f_n} = \frac{{{\omega _n}}}{{2\pi }} = \frac{{354.5}}{{2\pi }} = 56.42\;\;Hz\)

N = 60 f_n = 60 × 56.42

∴ N = 3385.13 rpm

Fluid medium ⇒ viscous damping

m = 4 kg, F₀ = 50 N

Resonance amplitude (X) = 20 mm

\(X = \frac{{{F_0}}}{{2\xi k}},\;T = \frac{1}{f} = \frac{1}{{12}}s\)

\(\omega = \frac{{2\pi }}{T} = 2\pi \times 12\)

∴ ω = 24π rad/s

Now,

\(\xi = \frac{C}{{2\sqrt {mk} }}\)

\(\xi k = \frac{C}{2}\sqrt {\frac{k}{m}} = \frac{C}{2}\omega \)

\(\therefore X = \frac{{2{F_0}}}{{2C\;\omega }} = \frac{{{F_0}}}{{C\omega }}\)

\(C = \frac{{{F_0}}}{{X\omega }}\)

\(C = \frac{{50}}{{24\pi }} \times \frac{1}{{0.020}}\)

∴ C = 33.157 Ns/m

Concept:

\(X = \frac{{{F_0}/k}}{{\sqrt {{{\left( {1 - {r^2}} \right)}^2} + {{\left( {2\xi r} \right)}^2}} }}\)

Calculation:

\({\omega _n} = \sqrt {\frac{{{k_{eq}}}}{M}}\)

k_eq = 4 × k = 4 × 0.2 × 10⁶ = 0.8 × 10⁶ N/m

M = 45 kg

\(\therefore {\omega _n} = \sqrt {\frac{{0.8\; \times \;{{10}^6}}}{{45}}} = 133.33\;rad/s\)

ω = 2πn = 2π × 32 = 201.06 rad/s

\(r = \frac{\omega }{{{\omega _n}}} = \frac{{201.06}}{{133.33}}\)

∴ r = 1.508

Now,

Magnification factor of an undamped system is given as,

\(MF = \frac{1}{{\sqrt {{{\left( {1 - {r^2}} \right)}^2} + {{\left( {2\xi r} \right)}^2}} }}\)

For Undamped, ζ = 0

\(\therefore MF = \frac{1}{{\left| {\left( {1 - {r^2}} \right)} \right|}}\)

\(MF = \frac{1}{{\left| {1 - {{\left( {1.51} \right)}^2}} \right|}} = 0.7812\)

Using equation of amplitude

\(X = \frac{{{F_0}/{k_{eq}}}}{{\sqrt {{{\left( {1 - {r^2}} \right)}^2} + {{\left( {2\xi r} \right)}^2}} }}\)

\(\therefore {F_0} = \frac{{m\omega _n^2X}}{{MF}}\)

\({F_0} = \frac{{45\; \times \;{{\left( {133.3} \right)}^2}\; \times \;0.0015}}{{0.7812}}\)

∴ F­₀ = 1.536 kN ≈ 1.54 kN

Concept:

Natural frequency:

\({\omega _n} = \sqrt {\frac{k}{m}}\)

2ξω_n = c/m

Damping factor:

\(\xi = \frac{c}{{{c_c}}} = \frac{c}{{2\sqrt {km} }} = \frac{c}{{2m{\omega _n}}}\)

Damped frequency:

\({{\rm{\omega }}_{\rm{d}}} = \sqrt {1 - {\xi ^2}} {\omega _n}\)

Calculation:

Given: m = 0.2 kg, k = 1 kN/m = 1000 N/m, c = 8.5 Ns/m

\({\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{{1000}}{{0.2}}} = 70.71\;rad/s\)

\(\xi = \frac{c}{{2\sqrt {km} }} = \frac{{8.5}}{{2\sqrt {1000 \times 0.2} }} = 0.3\)

\({{\rm{\omega }}_{\rm{d}}} = \sqrt {1 - {\xi ^2}} {\omega _n} = \sqrt {1 - {{\left( {0.3} \right)}^2}} \times 70.71 = 67.45\;rad/s\)

% change in frequency of oscillation = (ω_d – ω_n)/ω_n

= (67.45 - 70.71)/70.71 = -0.0461 = -4.6%