Engineering Mechanics

Chapter 26 : Helical Springs and Pendulums „„„„„ 531

Maximum force in the spring

We know that the angular velocity of the body,

9·8

28·6 rad/s

0·012

g

ω= = =

δ

and maximum acceleration, amax = ω^2 x = (28·6)^2 × 0·025 = 20·4 m/s^2

∴ Maximum inertia force
= Mass × Acceleration = 3 × 20·4 = 61·2 N
We also know that maximum force in the spring occurs when the mass is at its lowest
position. In this position, the force in the spring is equal to the sum of weight of the body and the
inertia force. Therefore maximum force in the spring

= (3 × 9·8) + 61·2 = 90·6 N Ans.
Example 26.5. The weight of an empty railway wagon is 240 kN. On loading it with goods
weighing 320 kN, its spring gets compressed by 80 mm.
(a) Calculate its natural period of vibrations when the railway wagon is (i) empty and
(ii) loaded as above.
(b) It is set into natural vibrations with an amplitude of 100 mm when empty. Calculate the
velocity of the railway wagon when its displacement is 40 mm from statical equilibrium position.

Solution. Given : Weight of empty wagon = 240 kN and when the load (W) is 320 kN,
then deflection (δ) = 80 mm = 0.08 m

(i) Period of vibrations when the wagon is empty

We know that deflection of the spring, when wagon is empty,

1

0·08

240 0·06 m

320

δ= × =

and period of vibrations, 1 1

0·06

22 0·49s

9·8

t

g

δ

=π =π = Ans.

(ii) Period of vibrations when the wagon is loaded

We know that total load on the springs when the wagon is loaded

= 240 + 320 = 560 kN

∴ Deflection of the spring when the wagon is loaded

2

0·08

560 0.14 m

320

δ= × =

and period of vibrations, 2 2

0·14

22 0·75s

9·8

t

g

δ

=π =π =

(b) Velocity of the railway wagon when it is empty

When the wagon is empty, amplitude (r) = 100 mm = 0.1 m and displacement (y) = 40 mm
= 0.04 m

We know that angular velocity of the wagon,

1

22

12·82 rad/s

t 0·49

ππ

ω= = =

and velocity, vry=ω^22 – =12·82 (0·1) – (0·04)^22 = 1·175 m/s Ans.