Engineering Mechanics

Chapter 26 : Helical Springs and Pendulums „„„„„ 545

∴Total deflection of cantilever at A,

22

1212

Wl W 11 l

W

kb k kb k

⎛⎞ ⎡⎤⎛⎞

δ= ⎜⎟+ = ⎢⎥⎜⎟+

⎝⎠ ⎢⎥⎣⎦⎝⎠

2

21

12

l

kk

b

W

kk

⎡⎤⎛⎞

⎢⎥⎜⎟+

= ⎢⎥⎝⎠

⎢⎥⎣⎦

and natural frequency of vibration of the system,

12

2

12

11

Hz

22

g gk k

n

l

Wk k

b

==

πδ π ⎡⎤⎛⎞

⎢⎥+⎜⎟

⎢⎥⎣⎦⎝⎠

Ans.

Example 26.15. A uniform thin rod as shown in Fig. 26.10 has a mass of 1 kg and carries
a concentrated mass of 2·5 kg at B. The rod is hinged at A, and is maintained in the horizontal
position by a spring of stiffness 1·8 kN/m at C.

Fig. 26.10.

Find the frequency of oscillation. Neglect the effect of the mass of the spring.

Solution. Given : Mass of rod = 1 kg ; Mass at B = 2·5 kg ; Length of rod AB = 300 + 300
= 600 mm = 0·6 m ; Stiffness of spring (s) = 1·8 kN/ m = 1800 N /m.

Let n = Frequency of oscillation.

θ = Small angular displacement of the rod, and

α = Angular acceleration of the rod AB.

We know that mass moment of inertia of the system about A,

IA = M.I. of 1 kg about A + M.I. of 2·5 kg about A

2

1(0·6) 2·5 (0·6) (^22) 1·02 kg-m
3
=+ =
Fig. 26.11.
If the rod AB is given a small angular displacement as shown in Fig. 26.11 and then released,
it will start oscillating about A with simple harmonic motion.