Mechanical Vibrations Test 2

Concept:

The amplitude of vibration,

\(A = \frac{{\frac{{{F_0}}}{k}}}{{\sqrt {{{\left[ {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]}^2} + {{\left( {\frac{{2ξ \omega }}{{{\omega _n}}}} \right)}^2}} }}\)

where, F₀ = Force on the system, k = stiffness of the spring, ξ = damping ratio

At resonance, ω = ω_n

\(A = \frac{{\frac{{{F_0}}}{k}}}{{2ξ }}\)

\(\therefore kA = \frac{{{F_0}}}{{2ξ }} \)

Now,

Spring force at Resonance is given by:

\({F_s} = kA = \frac{{{F_0}}}{{2ξ }} \)

\({F_s} =\frac{{40}}{{2 \times 0.05}} = 400N\)

∴ F_s = 400 N

Concept:

The transverse deflection produced is given by:

\(y = \frac{e}{{{{\left( {\frac{{{\omega _n}}}{\omega }} \right)}^2} - 1}}\)

where, e = eccentricity

Calculation:

Given:

e = 9 mm, ω_n = 50 rad/s and ω = 25 rad/s

Now,

\(y = \frac{{9}}{{{{\left( {\frac{{50}}{{25}}} \right)}^2} - 1}}\)

∴ y = 3 mm 

Concept:

Using Dunkerley’s method

\(\frac{1}{{{f^2}}} = \frac{1}{{f_1^2}} + \frac{1}{{f_2^2}} + \ldots + \frac{1}{{f_n^2}}\)

Calculation:

\(\frac{1}{{{f^2}}} = \frac{1}{{{{100}^2}}} + \frac{1}{{{{200}^2}}}\)

\(\therefore {f^2} = 8000\)

∴ f = 89.44 cycles/sec

∴ N = f × 60

∴ N = 5366.56 rpm

Concept:

The amplitude of whirling is given by:

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

where, e = eccentricity and \(r = \frac{ω }{{{ω _n}}} \)

Calculation:

Given:

m = 40000 gm = 40 kg, e = 0.012 m, k = 320 kN/m = 320 × 10³ N/m and ξ = 0.07

Now,

\(ω = 2\pi \times \frac{{100}}{6} \)

∴ ω = 104.7 rad/s

\({ω _n} = \sqrt {\frac{k}{m}} \)

∴ ω_n = 89.44 rad/s

\(\therefore r = \frac{ω }{{{ω _n}}} = 1.1708\)

Now,

\(A = \frac{{e{r^2}}}{{\sqrt {\left( {1 - {r^2}} \right) + {{\left( {2\xi r} \right)}^2}} }} = 0.047m\)

∴  A = 4.07 cm

Concept:

The amplitude of vibration is given by:

\(x = \frac{{\left( {\frac{{{m_0}e}}{m}} \right){r^2}}}{{\sqrt {{{\left( {1 - {r^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)

where, m = mass of rotor, m₀ = unbalanced mass, r = ω/ωn, ξ = damping ratio

Calculation:

Given: m = 1200 kg, m₀ = 1.5 kg, ξ = 0.30, e = 0.06 m, ω = 1200 rpm

Now,

ω_n = 1550 rpm (∵ (Resonance condition.)

r = ω/ω_n = 0.774

Now,

\(x = \frac{{\left( {\frac{{{m_0}e}}{m}} \right){r^2}}}{{\sqrt {{{\left( {1 - {r^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)

\(x = \frac{{4.49 \times {{10}^{ - 3}}}}{{0.613}}\)

∴ x = 0.073 mm

Concept:

Natural frequency:

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

Damping ratio:

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

Damped natural frequency, \({\omega _d} = {\omega _n}\sqrt {1 - {\xi ^2}}\)

FT is the force transmitted to the foundation. The disturbing force is F. The ratio of FT 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 = 200 kg, k = 35 kN/m = 35000 N/m, F = 100 N, C = 200 N.sec/m

\({\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{{35000}}{{200}}} = 13.228\;rad/s\)

\(\xi = \frac{C}{{2\sqrt {km} }} = \frac{{200}}{{2\sqrt {35000 \times 200} }} = 0.038\)

\({\omega _d} = {\omega _n}\sqrt {1 - {\xi ^2}}\)

\({\omega _d} = 13.288\sqrt {1 - {{\left( {0.038} \right)}^2}} = 13.218\; rad/s\)

The amplitude of force vibration (A) :

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

Here the disturbing force is acting on the mass at an undamped natural frequency so ω = ω_n

ω/ωn = 1

\(A = \frac{{\frac{F}{k}}}{{2\xi }} = \frac{{\frac{{100}}{{35000}}}}{{2 \times 0.038}} = 0.03759\;m\)

A = 37.59 mm

\({\rm{Amplitude\;}}\left( {{A_1}} \right) = \frac{{{F_0}/s}}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {\frac{{2\xi \omega }}{{{\omega _n}}}} \right)}^2}} }}\)

At resonance ω = ω_n

A = 14 mm

Case 1:

\(A = \frac{{{F_0}/s}}{{2\xi }} \Rightarrow 14 = \frac{{{F_0}}}{{2\xi s}}\)

Case 2:

ω/ω_n = 0.75, A = 10 mm

\(10 = \frac{{28\xi }}{{\sqrt {{{\left( {1 - {{0.75}^2}} \right)}^2} + {{\left( {1.5\;\xi } \right)}^2}} }}\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)\({\left( {1 - {{0.75}^2}} \right)^2} + 2.25\;{\xi ^2} = {\left( {2.8} \right)^2}{\xi ^2}\)

\({\xi ^2} = \frac{1}{{5.59}}\left( {0.1914} \right)\)

∴ ξ = 0.185

Concept:

90% isolation ⇒  Transmissibility ratio (TR) = 1 – 0.9 = 0.1

Transmissibility is defined as the ratio of force transmitted to the foundation (F_T) to the disturbing force (F).

\(TR = \frac{{{F_T}}}{F} = \frac{{\sqrt {1 + {{\left( {\frac{{2\xi \omega }}{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {{{\left[ {{{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2} - 1} \right]}^2} + {{\left( {\frac{{2\xi \omega }}{{{\omega _n}}}} \right)}^2}} }}\)

Calculation:

Given: δ = 5 mm, m = 400 kg and isolation = 90 %

For undamped System, ξ = 0

\(T.R = \frac{1}{{{r^2} - 1}} = 0.1 ⇒ r = 3.32\) where r = ω/ω_n

\({\omega _n} = \sqrt {\frac{g}{δ }} = \sqrt {\frac{{9.81}}{{0.005}}} = 44.29\frac{{rad}}{s}\)

ω = r.ω_n = 3.32 × 44.29 = 147.04 rad/s

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

∴ N = 1404.15 rpm

Concept:

\({N_{cr}} = \frac{{{ω _{cr}} \times 60}}{{2\pi }} \) and ωcr = ωn 

where, \({ω _n} = \sqrt {\frac{k}{m}} \)

For simply supported shaft \(k = \frac{{48EI}}{{{l^3}}} \)

where, I = moment of inertia, E = modulus of elasticity, l = length of the shaft

Calculation:

Given: d = 30 mm = 0.03 m, l = 900 mm = 0.9 m, m = 4 kg, e = 0.5 mm = 0.5 × 10^-3m, E = 2 × 10⁵ MPa = 2 × 10¹¹ Pa

Now,

\(I = \frac{\pi }{{64}}{d^4} = 3.976 \times {10^{ - 8}}{m^4}\)

\(k = \frac{{48EI}}{{{l^3}}} = 523588.80\;N\)

\({ω _n} = \sqrt {\frac{k}{m}} \)

∴ ω_n = 361.797 rad/s

Now,

ω_cr = ω_n = 361.797 rad/sec

\( {N_{cr}} = \frac{{{ω _{cr}} \times 60}}{{2\pi }} \)

∴ N_cr = 3454.91 rpm