Mechanical Engineering Assignment: Vibration Analysis & Solutions

This document presents a comprehensive solution to a Mechanical Engineering assignment focusing on vibration analysis. The solution addresses several key concepts, including Fourier series approximation, calculation of amplitude in forced vibrations, and application of Duhamel's method to determine system response. The assignment further explores the response of a system to a given forcing function, determining the displacement function. Finally, the document calculates the natural frequencies of a steel beam, considering its physical properties and boundary conditions, and provides a bibliography of relevant sources. This solved assignment is designed to aid students in understanding and solving complex vibration problems in mechanical engineering.

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MECHANICAL ENGINEERING
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Question 2.
A Fourier series approximation to the force show a sequence of approximations.
Given P=
2 3 −1
−3 −4 0
5 6 1
Let A= (v1, v2, v3)
By the given conditions
Pu1= v1
Pu2 = v2
Pu3 = v3
That is matrix equation PA= V = (v1 v2 v3)
PA=V
P-1PA = P-1V
A= P-1V
Now |P| =-1 and adjoint (p) =
−4 3 2
−9 7 3
−4 3 1
^ T
Then P-1 =
4 9 4
−3 −2 −3
−2 −1 −3
Therefore A = P-1V =
4 9 4
−3 −2 −3
−2 −1 −3
−3 −6 −8
4 6 2
5 4 5
=
44 46 6
−34 36 −5
−11 −10 5

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U1=
44
−34
−11
U2=
46
36
−10
U3=
6
−5
5
Question 3.
The amplitude is given by:
A=
Fo/s
√(1¿−( w
wn )n
+ ( 23 w
wn )2
)¿
Wn=√s/m =√3000/6 =10√ 5
A= 80/3000
√ ¿ ¿ ¿ = 80/3000
√ ¿ ¿ ¿ =0.023668m
Therefore A= 2.3668cm
Where by FO=80N
f=8cps
rs =3000N/m
w= 2 π
f

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m= 6kg
But ᵶ =0
Then y =Asin(wt +∅ )
Note: ∅ is the phase difference (not given)
QUESTION 4
By Duhamel's method the response of the system is given by the convolution of the unit
impulse response function and the forcing function. For the undamped case, the unit impulse
response function is:
The unit impulse response function for the undamped motion is:
The forcing function is given by:
The overall response is thus given by:

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And
So the overall response becomes:
Question 5.
Given that;
Mass m =4kg
And k =36 n/m
F (t) ={ 0
−9 sin(4 t)
t <0
t ≥ 0
As per the requirements given in the problem
f (t) = -9 sin(4 t ) t ≥0
m ¨x +kx = f (t)
4 ¨x + 36 x = -9 sin(4 t)
x (t) = −9
4 wn ∫
0
t
sin(4 ᵶ )sin wn(f −ᵶ )dz
x (t)=c1 sin wt+ c2 cos wt + xp

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= c1sin √ R
M f + c2 cos {√ R
M f} + xp
As, √ R
M = √a = 3
C1 sin |√ R
M f | + c2 cos ( √ R
M f ) + C3 sin(4 f )
2(f) = c1sin(3 f ) + c2 cos (3 f ) + c3 sin(4 f )
-4 (4)2 c3 sin 4 f + (36) c3 sin 4 f = -9 sin 4 f
c3 = −9
− ( 4 ) ( 4 ) 2+36
c3 =
9
28
c3 =
27
84 =
9
28
x (f) = c1sin wf + c2cos wf + 27
84 sin wf
c1sin 3 f + c2cos 3 f + 27
84 sin 4 f
f=0
x (0) = 0 → c2 = 0
˙x (0) =0
3c1 + 9. 27
84 (4) = 0

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3c1 + 9/7 =0
C1 = - 3
7
x (t) = - 3
7 sin 3 f + 9
28 sin 4 f
Question 6.
Given that,
The mode of steel with length L = 5m
Cross sectional area A = 50 cm2
And I = 1.2(107) mm4
Psteel =8050kg/m3 = 8.05 × 10-6 kg/mm3
a) The beam is pinned at x=0 and fixed at x= L
Wn =K2√ EI
APL4
Here n=1
K=3.927
W1= (3.927)2
√ ( 2 .1 × ( 1 011 ) × ( 1 .2 ×1 07 )
5000× 8 . 05× 1 0−6 × ( 5000 ) 4 )
=15.4213
√ ¿
=15.4213
√ ¿
=15.4213
√ ¿

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=15.4213 √ ( 2. 52 ×1 012
40250 ×725 )
=15.4213√ ( 2. 52 ×1 012
29181250 )
=15.4213 √ ( 252000000
29181250 )×1 04
=15.4213×293.865
W1=4531.7803
f1=√W1/2π
f1=√ 4531. 7803
6 . 28
f1=√721. 621
f1=26.86 H ᵶ
For f2, f3 , f4,f5, use the K value as 7.069 ,10.210 ,13.362 ,26.7380.
If we repeat the calculations again;
W2= 49.9707× 293.865
W2= 14684.639
f2=√W2/2π
f2=√ 14684 . 639
6 . 28
f2= 48.356 H ᵶ
W3= (10.210^2) ×293.865
W3= 104.24 ×293.865

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W3= 30632.4876
F3=√W3/2π
F3=√ 30632 . 4876
6 . 28
F3=69.84 H ᵶ
W4= 178.54 ×293.865
W4= 52466.65 Hᵶ
F4=√W4/2π
F4=√ 52466 . 65
6 . 28
F4=91.403 H ᵶ

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Question 1

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