Mechanical Vibration Problems: Spring Constants, Deflection, and DOF

This assignment provides detailed solutions to several mechanical vibration problems. It begins by determining the equivalent spring constants for various spring configurations and cantilever beams made of aluminum and steel. Next, it calculates the spring stiffness required to reduce the deflection of a cantilever beam supporting a machine, considering 25%, 50%, and 75% reductions. The assignment also determines the equivalent mass and natural frequencies for different vibration systems. Finally, it analyzes the degrees of freedom in various mechanical systems, identifying associated motions like oscillating and linear motion. Desklib offers a wide range of solved assignments and past papers to aid students in their studies.

Running head: MECHANICAL VIBRATION
MECHANICAL VIBRATION
Name of Student
Institution Affiliation

MECHANICAL VIBRATION 2
QUESTION ONE
Part a
The spring constant in series is given by
K= 1
K 1 + 1
K 2
In parallel it is given by
K = k1 +k2
K1 and k2 are in parallel hence
K = k1 +k2
k1 +k2 is in series with k3, therefore
K= 1
k 1+ K 2 + 1
K 3
K= k 2+ k 1+ k 3
K 3 (k 1+ K 2)

MECHANICAL VIBRATION 3
k 2+ k 1+ k 3
K 3 (k 1+ K 2) is in parallel with k4 therefore ;
k 2+ k 1+ k 3
K 3 (k 1+ K 2) + k4 = k 2+ k 1+ k 3+k 3 k 4 (k 1+k 2)
K 3(k 1+K 2)
k 2+ k 1+ k 3+k 3 k 4 (k 1+k 2)
K 3(k 1+ K 2) is series with k5 therefore,
k 3(k 1+k 2)
k 2+k 1+ k 3+k 3 k 4 ¿ ¿+ 1
K 5
K= K 5[k 1+ K 2+ k 3+k 3 k 4 ( k 1+k 2 )]¿ ¿
k 5 k 3 [ k 1+k 2 ]+k 1+k 2+k 3+k 3 k 4 ¿ ¿
Part b ,
Spring constant for a cantilever is given by
K= E∗W t3
4 l3
Where E is the young’s modulus w is the width, t is the thickness and l is the length (Scott,
2009).
For aluminum

MECHANICAL VIBRATION 4
K= 8.3 ×1 010 × 0.015× 0.0053
4 ×0.83
K= 155.625
2.048 = 7 5.9887
For steel
K= E∗W t3
4 l3
K= 2.07 ×1 010 × 0.015× 0.013
4 ×0.83
K= 310.5
2.048
K= 151.611
The arrangement of the aluminum and the steel are in series, therefore the resultant is given by;
K= 1
K al + 1
K st
K= 1
75.9887 + 1
151.611
K = 50.18356
QUESTIOON TWO
M= 300kg

MECHANICAL VIBRATION 5
L= 1.4 m
W= 0.8 m
D(t) = 0.15 m
E = 1.06×1011 N/m2
The moment of inertia of the beam is given by
I = d3 W
12
I = 0.153 × 0.8
12 = 0.0027
12
I = 0.000225
For load at the mid span is simply supported beam
K1= 48 EI
l3
K1= 48 ×1.06 ×1 011 ×0.000225
1.43
K1= 1144800000
2.744
K1= 417201166.18
The deflection of the beam due to load at mid span is

MECHANICAL VIBRATION 6
δ 1= W
K 1 = 300× 9.81
417201166.18
δ 1= 2943
417201166.18
δ 1= 7.0541×1 0−6
Part a
Final deflection is 25% of the original, therefore
δf=0. 25 δ 1
δf= W
Keq
0.25 δ 1= 300× 9.81
K 1+ Ks
0.25×7.0541×1 0−6= 2943
k 1+ks
735.499016+1.763575×1 0−6= 2943
Ks = 2208
1.763575× 10−6
Ks = 1252.0289 MN/m

MECHANICAL VIBRATION 7
Part B
Final deflection is 50% of the original, therefore
δf=0. 5 δ 1
δf= W
Keq
0.5 δ 1= 300× 9.81
K 1+ Ks
0.5×7.0541×1 0−6= 2943
k 1+ ks
1471.499016+3.527075 ×1 0−6= 2943
Ks = 1471.501
3.527075× 10−6
Ks = 417.201 MN/m
Part C
Final deflection is 75% of the original, therefore
δf=0. 75 δ 1
δf= W
Keq
0.75 δ 1= 300× 9.81
K 1+ Ks

MECHANICAL VIBRATION 8
0.75×7.0541×1 0−6= 2943
k 1+ks
2207.2497+5.29075×1 0−6= 2943
Ks = 735.7503
5.29075× 10−6
Ks = 139.068 MN/m
QUESTION THREE
The masses of the arms are negligible therefore; at point O , equivalent mass will be ;
M1 amd m2 are in parallel.

MECHANICAL VIBRATION 9
Eq mass= m1 m2
m1+m 2 and this is in series to m3,
Resultant mass at point O will be
m1 m2
m1+m 2 +m3
Mass = m1+m 2(m1+m 2)
m1 m2
Fn= √ k
m
2 π
K= k 1 k 2
k 1+ k 2+k3
K= k 1 k 2+ k 3 (k 1+k 2)
k 1+ k 2
Therefore natural frequency,
Fn= √ k 1 k 2+k 3 ( k 1 k 2 ) ×(m 1+ m2)
k 1+ k 2¿ ¿ ¿ ¿
Part b

MECHANICAL VIBRATION 10
Equivalent mass = ⍴v+m
Equivalent mass = ⍴v+m
Natural frequency fn = √ k
m
2 π
K1 is parallel K2
Resultant k = k 1 k 2
k 1+ k 2
Fn = k 1 k 2
k 1+ k 2 × 1
ρv+ m
Fn = √ k 1 k 2
( k 1+k 2 ) + ρv +m
2 π

MECHANICAL VIBRATION 11
QUESTION FOUR
Part a
The degree of freedom = (DOF)
Where n is the number of link, l is the number of lower pair, h is the number of higher pair for
this system the links are 3
DOF= 3(3-1) - 2×2 -1
DOF = 6-4-1
DOF= 1
And from this system the types of motions associated with this DOF are oscillating and linear
motion (Dickner, 2017).
Part b

MECHANICAL VIBRATION 12
The number of link n = 8
DOF = 3(n-1) – 2 l – h
DOF= 3(8-1) – (2×7) – 1
DOF = 21-14-1
DOF = 6
The motion which are associated with these DOF are rotatory motion and linear motion.
Part c

MECHANICAL VIBRATION 13
The number of link is 4
DOF = 3(n-1) – 2l –h
DOF = 3(3-1) – 2*2 – 1
DOF = 6-4-1
DOF = 1
The motion associating with this is oscillating.
Part d
The number of links is 11
DOF = 3( n-1) – 2l –h

MECHANICAL VIBRATION 14
= 3 (11-1) -2*11 -1
DOF = 30-22-1
DOF= 7
The linear motion and the oscillating motions are the motions which are associated with these
DOF.

MECHANICAL VIBRATION 15
References
Dickner, N. (2017). Six Degrees of Freedom. Toronto: Knopf Canada.
Scott, R. J. (2009). Degrees of Freedom: Louisiana and Cuba after Slavery. Harvad: Harvard University
Press.