Mechanical Engineering Assignment Solutions

This assignment details a comprehensive solution to various mechanical engineering problems, including Fourier series approximation, amplitude calculation, Duhamel's method, and more. The solutions are based on established textbooks in the field of mechanical engineering, providing a thorough understanding of the concepts and methods used.

Mechanical engineering 1
MECHANICAL ENGINEERING
By Name
Course
Instructor
Institution
Location
Date

Mechanical engineering 2
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

Mechanical engineering 3
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

Mechanical engineering 4
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:

Mechanical engineering 5
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

Mechanical engineering 6
= 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

Mechanical engineering 7
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
√ ¿

Mechanical engineering 8
=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

Mechanical engineering 9
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

Mechanical engineering 10
Question 1

Mechanical engineering 11

Mechanical engineering 12

Mechanical engineering 13

Mechanical engineering 14
Bibliography
1. Jorge Angeles, Dynamic Response of Linear Mechanical Systems: Modelling, Analysis and
Simulation (2nd Ed 2011) 154.
2. Daniel J. Inman, Vibration with Control (3rd Ed 2014) 89.
3. Mohammad I. Younis , MEMS Linear and Nonlinear Statics and Dynamics (4th 2014) 138.
4. Reza N. Jazar, Advanced Vibrations: A Modern Approach (1st 2016) 65.
5. Alex.A. Shabana, Theory of Vibration: An Introduction (3rd 2011)45.

Mechanical engineering 15

Mechanical engineering 16