Harmonically Excited Vibrations

ME-304 – Mechanical Vibrations

𝑚 𝑥 + 𝑘𝑥 = 𝐹 cos𝜔𝑡

Undamped system under Harmonic Excitation:

𝑊𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑒 ℎ𝑜𝑚𝑜𝑔𝑒𝑛𝑜𝑢𝑠 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝐹 = 0 :𝑥ℎ 𝑡 = A1 cos𝜔𝑛𝑡 + 𝐴2 sin𝜔𝑛𝑡

𝑀𝑎𝑛𝑦 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.𝑊𝑒′𝑙𝑙 𝑢𝑠𝑒 𝑡ℎ𝑒 𝑚𝑒𝑡ℎ𝑜𝑑 𝑜𝑓 𝑢𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠𝑥𝑝 𝑡 = 𝑋 cos𝜔𝑡

𝑊𝑒′𝑙𝑙 𝑝𝑙𝑢𝑔 𝑡ℎ𝑖𝑠 𝑏𝑎𝑐𝑘 𝑖𝑛 𝑡𝑜 𝑡ℎ𝑒 𝐸𝑂𝑀 𝑎𝑛𝑑 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑦𝑎𝑓𝑡𝑒𝑟 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑖𝑛𝑔 𝑡𝑤𝑖𝑐𝑒:−𝜔2𝑚𝑋 + 𝑘𝑋 = 𝐹

𝑿 =𝑭

𝒌 −𝒎𝝎𝟐𝒙𝒑 𝒕 =

𝑭

𝒌 −𝒎𝝎𝟐𝒄𝒐𝒔𝝎𝒕

𝑇𝑜𝑡𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 𝑖𝑠𝑥 𝑡 = 𝑥ℎ 𝑡 + 𝑥𝑝 𝑡 = A1 cos𝜔𝑛𝑡 + 𝐴2 sin𝜔𝑛𝑡 + X cos𝜔𝑡

𝑈𝑠𝑖𝑛𝑔 𝐼𝐶𝑠: 𝑥 0 = 𝑥𝑜; 𝑥 0 = 𝑣𝑜𝑤𝑒 𝑐𝑎𝑛 𝑓𝑖𝑛𝑑 𝐴1𝑎𝑛𝑑 𝐴2

𝐴1 = 𝑥𝑜 − 𝑋

𝐴2 =𝑣𝑜𝜔𝑛

𝑇𝑜𝑡𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 𝑖𝑠

𝑥 𝑡 = (𝑥𝑜 − 𝑋) cos𝜔𝑛𝑡 +𝑣𝑜𝜔𝑛

sin𝜔𝑛𝑡 + X cos𝜔𝑡

𝑤ℎ𝑒𝑟𝑒 𝑋 =𝐹

𝑘 − 𝑚𝜔2

𝑋

𝛿𝑠𝑡=

1

1 − 𝑟2

𝐷𝑒𝑓𝑖𝑛𝑒 𝑠𝑡𝑎𝑡𝑖𝑐 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛: 𝛿𝑠𝑡 =𝐹

𝑘

𝑎𝑛𝑑 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑟𝑎𝑡𝑖𝑜: 𝑟 =𝜔

𝜔𝑛

This term is called the amplification factor, amplitude ratio, magnification factor or simply gain

Gain is a function of frequency ratio

In Phase Response(Frequency ratio, r < 1)

Out of Phase Response(Frequency ratio, r > 1)

Resonance(Frequency Ratio, r = 1)

𝐴𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝑧𝑒𝑟𝑜 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠: 𝑥 0 = 0 ; 𝑥 0 = 0

𝑇𝑜𝑡𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 𝑖𝑠𝑥 𝑡 = X (cos𝜔𝑡 − cos𝜔𝑛𝑡)

𝐷𝑒𝑓𝑖𝑛𝑒 𝛿𝑠𝑡 =𝐹

𝑘𝑎𝑛𝑑 𝑟 =

𝜔

𝜔𝑛

𝑤ℎ𝑒𝑟𝑒 𝑋 =𝐹

𝑘 −𝑚𝜔2

𝑥(𝑡) =𝛿𝑠𝑡

1 −𝜔𝜔𝑛

2 (cos𝜔𝑡 − cos𝜔𝑛𝑡 )

𝑥(𝑡) =𝛿𝑠𝑡

1 −𝜔𝜔𝑛

2 (cos𝜔𝑡 − cos𝜔𝑛𝑡 )

𝐴𝑡 𝑟 = 1,𝜔 = 𝜔𝑛; 𝑇ℎ𝑒 𝑟𝑒𝑠𝑝𝑜𝑛𝑠𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑎𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦𝑡ℎ𝑒 𝑏𝑒𝑙𝑜𝑤 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛, 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑

We can use L'Hôpital's rule:

Beating Phenomenon(Frequency Ratio, r close to 1)

We found the response of the system with zero initial conditions to be:

Using trigonometric identities, we can rewrite:

𝑥(𝑡) =𝛿𝑠𝑡

1 −𝜔𝜔𝑛

2 (cos𝜔𝑡 − cos𝜔𝑛𝑡 )

When the difference between the driving frequency and natural frequency is small. We define a small quantity epsilon, ϵ :

Rewriting the response of the system:

𝑥(𝑡) =𝛿𝑠𝑡

1 −𝜔𝜔𝑛

2 (cos𝜔𝑡 − cos𝜔𝑛𝑡 )

𝑥(𝑡) =𝜔𝑛2𝛿 𝑠𝑡

𝜔𝑛2 − 𝜔2

(2 sin(𝜔𝑛 + 𝜔

2𝑡) sin(

𝜔𝑛 − 𝜔

2𝑡) )

𝑥(𝑡) =

𝑘𝑚×𝐹𝑘

4𝜖𝜔(2 sin(

𝜔𝑛 + 𝜔

2𝑡) sin(

𝜔𝑛 −𝜔

2𝑡) )

𝑥(𝑡) =𝐹

2𝑚𝜖𝜔(sin𝜔𝑡 𝑠𝑖𝑛 𝜖𝑡 )

𝑆𝑖𝑛𝑐𝑒 𝜖 𝑖𝑠 𝑟𝑎𝑡ℎ𝑒𝑟 𝑠𝑚𝑎𝑙𝑙: sin 𝜖𝑡 𝑤𝑖𝑙𝑙 ℎ𝑎𝑣𝑒𝑎 𝑙𝑎𝑟𝑔𝑒 𝑡𝑖𝑚𝑒 𝑝𝑒𝑟𝑖𝑜𝑑

𝐴𝑛𝑑 𝜔 𝑖𝑠 𝑚𝑢𝑐ℎ 𝑙𝑎𝑟𝑔𝑒𝑟, sin𝜔𝑡 𝑤𝑖𝑙𝑙 ℎ𝑎𝑣𝑒𝑎 𝑠𝑚𝑎𝑙𝑙 𝑡𝑖𝑚𝑒 𝑝𝑒𝑟𝑖𝑜𝑑

𝑚 𝑥 + 𝑐 𝑥 + 𝑘𝑥 = 𝐹 cos𝜔𝑡

Damped System with Harmonic Force

Many ways to get the particular solution.

1. Using undetermined coefficientsa. Assume a particular solution of the form:

b. Plug it back in to the EOM:

c. Use trigonometric identities to expand cos and sin terms:

d. Equate the coefficients of cos(ωt) and sin(ωt) in above equation

e. Solve for α and X

f. Solution may be expressed in terms of dimensionless numbers

𝜁 𝑎𝑛𝑑 𝑟

2. Using Complex Form

𝑚 𝑥 + 𝑐 𝑥 + 𝑘𝑥 = 𝐹𝑒𝑖𝜔𝑡

𝑊𝑒 𝑐𝑎𝑛 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑎 ℎ𝑎𝑟𝑚𝑜𝑛𝑖𝑐 𝑓𝑜𝑟𝑐𝑖𝑛𝑔 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑢𝑠𝑖𝑛𝑔 𝑐𝑜𝑚𝑝𝑙𝑒𝑥 𝑛𝑜𝑡𝑎𝑡𝑖𝑜𝑛:𝐼𝑓 𝑤𝑒 𝑢𝑠𝑒 𝑎 𝑓𝑜𝑟𝑐𝑖𝑛𝑔 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝐹𝑒𝑖𝜔𝑡 , 𝑡ℎ𝑒 𝑟𝑒𝑎𝑙𝑝𝑎𝑟𝑡 𝑜𝑓 𝐹𝑒𝑖𝜔𝑡 𝑖𝑠 𝐹 cos𝜔𝑡 . 𝑆𝑜 𝑡ℎ𝑒 𝑟𝑒𝑎𝑙 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑤𝑖𝑙𝑙𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑓𝑜𝑟𝑐𝑖𝑛𝑔 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝐹 cos𝜔𝑡

𝐴𝑠𝑠𝑢𝑚𝑒 𝑎 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚:

𝑥 𝑝 = 𝑋𝑒𝑖𝜔𝑡

𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒 𝑡𝑤𝑖𝑐𝑒 𝑎𝑛𝑑 𝑝𝑢𝑡 𝑖𝑡 𝑏𝑎𝑐𝑘 𝑖𝑛 𝑡ℎ𝑒 𝐸𝑂𝑀:

−𝑚𝜔2𝑋𝑒𝑖𝜔𝑡 + 𝑖𝑐𝜔𝑋𝑒𝑖𝜔𝑡 + 𝑘𝑋𝑒𝑖𝜔𝑡 = 𝐹𝑒𝑖𝜔𝑡

𝑋 𝑘 −𝑚𝜔2 + 𝑖𝑐𝜔 = 𝐹

𝑋 𝑘 −𝑚𝜔2 + 𝑖𝑐𝜔 = 𝐹

This is called the “Mechanical impedance” of the system: 𝑍 𝑖𝜔 = 𝑘 −𝑚𝜔2 + 𝑖𝑐𝜔

𝑋 =𝐹

𝑘 −𝑚𝜔2 + 𝑖𝑐𝜔

𝑋 =𝐹

𝑘 −𝑚𝜔2 + 𝑖𝑐𝜔×𝑘 −𝑚𝜔2 − 𝑖𝑐𝜔

𝑘 −𝑚𝜔2 − 𝑖𝑐𝜔

There is an imaginary term in the denominator. We can get rid of that and separate the real and imaginary parts by multiplying and dividing by the complex conjugate of the denominator.

𝑋 = 𝐹𝑘 −𝑚𝜔2

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2− 𝑖

𝑐𝜔

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2

𝑋 𝑖𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝑎 − 𝑖𝑏

𝑁𝑒𝑒𝑑 𝑡𝑜 𝑟𝑒𝑤𝑟𝑖𝑡𝑒 𝑋 𝑖𝑛 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑎 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑎𝑛𝑑 𝑝ℎ𝑎𝑠𝑒 𝑎𝑛𝑔𝑙𝑒:

𝑋 = 𝐴𝑒−𝑖𝜙, 𝑤ℎ𝑒𝑟𝑒 𝐴 = 𝑎2 + 𝑏2 𝑎𝑛𝑑 𝜙 = tan−1𝑏

𝑎

𝐴 = 𝐹𝑘 −𝑚𝜔2

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2

2

+𝑐𝜔

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2

2

𝐴 =𝐹

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2 𝜙 = tan−1𝑐𝜔

𝑘 −𝑚𝜔2

𝑥 𝑝 = 𝑋𝑒𝑖𝜔𝑡 = 𝐴𝑒𝑖(𝜔𝑡−𝜙) =𝐹

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2𝑒𝑖(𝜔𝑡−𝜙)

Use these results to rewrite the particular solution of the system:

𝑥 𝑝 =𝐹

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2cos 𝜔𝑡 − 𝜙

The real part of the particular solution is the same as what we got before:

The equation,

can rewritten in terms of dimensionless numbers:

𝑋

𝛿𝑠𝑡=

1

1 − 𝑟2 + 𝑖2𝜁𝑟≡ 𝐻(𝑖𝜔)

This term is called the “Frequency Response Function”

𝑋 =𝐹

𝑘 −𝑚𝜔2 + 𝑖𝑐𝜔

𝑋 =𝐹/𝑘

1 −𝑚𝜔2

𝑘+𝑖𝑐𝜔𝑘

𝑏𝑦 𝑢𝑠𝑖𝑛𝑔 𝑟 =𝜔

𝜔𝑛; 𝜔𝑛 =

𝑘

𝑚; 𝛿𝑠𝑡 =

𝐹

𝑘𝑎𝑛𝑑 𝜁 =

𝑐

2 𝑘𝑚

Total Solution:

𝛼

Using ICs, we can find unknowns

Note that only the particular solution doesn’t decay exponentially, whereas the homogenous part will decay and die out eventually. So the particular part is often called the ‘Steady State’ solution, and the homogenous part is called the ‘transient’ solution

f. Solution may be expressed in terms of dimensionless numbers

𝑋 =𝐹

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2

𝑋 =𝐹/𝑘

1 −𝑚𝑘𝜔2

2+ 2

𝑐

2 𝑘𝑚𝜔

𝑚𝑘

2

𝑋

𝛿𝑠𝑡=

1

1 − 𝑟2 2 + 2𝜁𝑟 2

𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦 𝜙 = tan−1𝑐𝜔

𝑘 −𝑚𝜔2= tan−1

2𝜁𝑟

1 − 𝑟2

This expression gives us the Gain of the system as a function of ‘frequency ratio’ r and damping ratio ζThis is plotted on the next slide, for different damping ratios

Base Excitation

Lets look at the response of the system when the base of a mass-spring-damper system undergoes harmonic motion.

The Equation of Motion of the system (using the free-body diagram) looks like:

𝑚 𝑥 = −𝑘 𝑥 − 𝑦 − 𝑐( 𝑥 − 𝑦)

𝐹 =𝑚 𝑥

𝑚 𝑥 + 𝑐 𝑥 + 𝑘𝑥 = 𝑐 𝑦 + 𝑘𝑦

𝐼𝑓 𝑦 𝑡 = 𝑌 sin𝜔𝑡

𝑚 𝑥 + 𝑐 𝑥 + 𝑘𝑥 = 𝑐𝜔𝑌 cos𝜔𝑡 + 𝑘𝑌 sin𝜔𝑡

𝑇ℎ𝑒 𝑓𝑜𝑟𝑐𝑖𝑛𝑔 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛: 𝑐𝜔𝑌 cos𝜔𝑡 + 𝑘𝑌 sin𝜔𝑡𝑐𝑎𝑛 𝑏𝑒 rewritten in the fom A sin(𝜔𝑡 − 𝛼)

A sin(𝜔𝑡 − 𝛼) = 𝑐𝜔𝑌 cos𝜔𝑡 + 𝑘𝑌 sin𝜔𝑡𝐴 sin𝜔𝑡 cos 𝛼 − 𝐴 sin 𝛼 cos𝜔𝑡 = 𝑐𝜔𝑌 cos𝜔𝑡 + 𝑘𝑌 sin𝜔𝑡

Must find A and α, by comparing the terms on either side of the equation

𝐴 sin 𝛼 = −𝑐𝜔𝑌𝐴 cos 𝛼 = 𝑘𝑌

𝐴 = 𝑌 𝑐𝜔 2 + 𝑘2

𝛼 = tan−1 −𝑐𝜔

𝑘= tan−1(−2𝜁𝑟)

𝑚 𝑥 + 𝑐 𝑥 + 𝑘𝑥 = 𝐴 sin(𝜔𝑡 − 𝛼)

And the equation of motion is:

This shows that giving excitation of the base is equivalent to applying a harmonic force of magnitude A.

We’ve already solved this for a slightly different forcing function, we don’t really need to solve this again. We can just borrow results we derived before with a few modifications.

One way of doing this, is by comparing with the solution we got using complex notation

𝒎 𝒙 + 𝒄 𝒙 + 𝒌𝒙 = 𝑭𝒆𝒊𝝎𝒕 𝒎 𝒙 + 𝒄 𝒙 + 𝒌𝒙 = 𝑨𝒆𝒊(𝝎𝒕−𝜶)

𝒙 𝒑 = 𝑿𝒆𝒊𝝎𝒕

𝒙 𝒑 =𝑭

𝒌 −𝒎𝝎𝟐 𝟐 + 𝒄𝝎 𝟐𝒆𝒊(𝝎𝒕−𝝓)

Assume particular solution: Assume particular solution:

𝒙 𝒑 = 𝑿𝒆𝒊(𝝎𝒕−𝜶)

𝒙 𝒑 =𝑨

𝒌 −𝒎𝝎𝟐 𝟐 + 𝒄𝝎 𝟐𝒆𝒊(𝝎𝒕−𝜶−𝝓𝟏)

𝑇ℎ𝑒 𝑓𝑜𝑟𝑐𝑖𝑛𝑔 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝐹 cos𝜔𝑡 𝑖𝑠𝑟𝑒𝑝𝑙𝑎𝑐𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑟𝑒𝑎𝑙 𝑝𝑎𝑟𝑡 𝑜𝑓 𝐹 𝑒𝑖𝜔𝑡

𝑇ℎ𝑒 𝑓𝑜𝑟𝑐𝑖𝑛𝑔 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝐴 sin(𝜔𝑡 − 𝛼) 𝑖𝑠𝑟𝑒𝑝𝑙𝑎𝑐𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑖𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑦 𝑝𝑎𝑟𝑡 𝑜𝑓𝐴𝑒𝑖(𝜔𝑡−𝛼)

Solution in complex form: Solution in complex form:

𝒘𝒉𝒆𝒓𝒆 𝝓 = 𝒕𝒂𝒏−𝟏𝒄𝝎

𝒌 −𝒎𝝎𝟐𝒘𝒉𝒆𝒓𝒆 𝝓𝟏 = 𝒕𝒂𝒏−𝟏

𝒄𝝎

𝒌 −𝒎𝝎𝟐

The real part of this solution represents the solution for the forcing function F cos 𝜔𝑡

The imaginary part of this solutionrepresents the solution for the forcingfunction 𝐴 sin(𝜔𝑡 − 𝛼)

Base ExcitationHarmonic Excitation

𝑥 𝑝 =𝐹

𝑘 − 𝑚𝜔2 2 + 𝑐𝜔 2𝑒𝑖(𝜔𝑡−𝜙) 𝑥 𝑝 =

𝐴

𝑘 − 𝑚𝜔2 2 + 𝑐𝜔 2𝑒𝑖(𝜔𝑡−𝛼−𝜙1)

𝑋

𝛿𝑠𝑡=

1

1 − 𝑟2 2 + 2𝜁𝑟 2

Here, we were trying to compare the ‘static displacement’ δst with the magnitude of the particular solution, (or steady state solution), X

𝑥 𝑝 =𝑌 𝑘2 + 𝑐𝜔 2

𝑘 − 𝑚𝜔2 2 + 𝑐𝜔 2𝑒𝑖(𝜔𝑡−𝜙)

We’re trying to compare the magnitude of the input displacement, Y and the magnitude of the particular solution, X

𝑋

𝑌=

𝑘2 + 𝑐𝜔 2

𝑘 − 𝑚𝜔2 2 + 𝑐𝜔 2

𝑋

𝑌=

1 + 2𝜁𝑟 2

1 − 𝑟2 2 + 2𝜁𝑟 2

We can rewrite this in terms of frequency ratio and damping ratio:

𝑤ℎ𝑒𝑟𝑒 𝜙 = 𝛼 + 𝜙1

This term is called displacement transmissibility:

Base Excitation – Force Transmissibility

In base excitation, a force F is transmitted to the base or support due to the reactions from the spring and the dashpot. This force can be determined as:

𝐹 = 𝑘 𝑥 − 𝑦 + 𝑐 𝑥 − 𝑦 = −𝑚 𝑥

𝑥 𝑝 =𝑌 𝑘2 + 𝑐𝜔 2

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2𝑒𝑖(𝜔𝑡−𝜙)

Consider the particular solution:

𝑥 𝑝 = 𝑋 sin(𝜔𝑡 − 𝜙)

𝑤ℎ𝑒𝑟𝑒 𝑋 =𝑌 𝑘2 + 𝑐𝜔 2

𝑘 − 𝑚𝜔2 2 + 𝑐𝜔 2

𝐹 = 𝑚𝜔2X sin 𝜔𝑡 − 𝜙

𝐹 = 𝑚𝜔2𝑌 𝑘2 + 𝑐𝜔 2

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2sin 𝜔𝑡 − 𝜙

The maximum value of this force FT is given by:

𝐹𝑇 = 𝑚𝜔2𝑌 𝑘2 + 𝑐𝜔 2

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2

𝐹𝑇𝑌= 𝑚𝜔2

𝑘2 + 𝑐𝜔 2

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2

𝐹𝑇𝑌= 𝑚𝜔2

1 + 2𝜁𝑟 2

1 − 𝑟2 2 + 2𝜁𝑟 2

Lets try to rewrite this in terms of frequency ratio and damping ratio:

𝐹𝑇𝑘𝑌

=𝑚

𝑘𝜔2

1 + 2𝜁𝑟 2

1 − 𝑟2 2 + 2𝜁𝑟 2

𝐹𝑇𝑘𝑌

= 𝑟21 + 2𝜁𝑟 2

1 − 𝑟2 2 + 2𝜁𝑟 2

This term is called the ‘force transmissibility’

Rotating Unbalance

Rotating Unbalance

Rotating Unbalance

𝑚 𝑥 + 𝑐 𝑥 + 𝑘𝑥 = moe𝜔2 sin𝜔𝑡

𝐹 =𝑚 𝑥

𝑥 𝑝 =𝑚𝑜𝑒𝜔

2

𝑘 − 𝑚𝜔2 2 + 𝑐𝜔 2𝑒𝑖(𝜔𝑡−𝜙)

𝑤ℎ𝑒𝑟𝑒 𝜙 = tan−1𝑐𝜔

𝑘 − 𝑚𝜔2

This is effectively the same problem as before, with F= moeω2

𝑥 𝑝 = 𝑋 sin(𝜔𝑡 − 𝜙)

𝑋 =𝑚𝑜𝑒𝜔

2

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2

Time to make this non-dimensional. Rewrite in terms of zeta and r

𝑋 =𝑚𝑜𝑒𝜔

2

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2 𝑋 =

𝑚𝑜𝑒𝜔2

𝑘

1 − 𝑟2 2 + 2𝜁𝑟 2

𝑋 =

𝑚𝑜𝑒𝜔2

𝑘×𝑚𝑚

1 − 𝑟2 2 + 2𝜁𝑟 2

𝑚𝑋

𝑚𝑜𝑒=

𝜔2 ×𝑚𝑘

1 − 𝑟2 2 + 2𝜁𝑟 2

𝑚𝑋

𝑚𝑜𝑒=

𝑟2

1 − 𝑟2 2 + 2𝜁𝑟 2

This term is the ‘dimensionless displacement magnitude’ of the system.

The force F is transmitted to the foundation due to the rotating unbalanced force is given by:

𝐹 𝑡 = 𝑘𝑥 𝑡 + 𝑐 𝑥(𝑡)

𝐹𝑇 = 𝑚𝑜𝑒𝜔2

1 + 2𝜁𝑟 2

1 − 𝑟2 2 + 2𝜁𝑟 2

The maximum value of this force FT is given by: