Engineering Mechanics

(^654) „„„„„ A Textbook of Engineering Mechanics
Now taking moments about the centre of gravity (G) of the vehicle,
RF x – RR x=2 Fh
∴ RF – RR= 2
22
2
Fh h I a
xxr
⎛⎞
= ⎜⎟
⎝⎠
... 2
2
Ia
F
r
⎛⎞
⎜⎟=
⎝⎠
Q
= 2
2
hI P
x r I
M
r
××
⎛⎞+
⎜⎟
⎝⎠
2
...
P
a
I
M
r
⎛⎞=
⎜⎟
⎜⎟+
⎝⎠
Q
= 2
hIP
x IMr
×

• ...(ix)
  Adding equations (viii) and (ix),
  RF= 2
  1
  2
  h IP
  Mg
  x IMr
  ⎡⎤⎛⎞
  ⎢⎥+ ⎜⎟
  ⎣⎦⎝⎠+
  and RR= 2
  1

• 2

h IP

Mg

x IMr

⎡⎤⎛⎞

⎢⎥⎜⎟+

⎣⎦⎝⎠

Example 32.1. An industrial truck of total mass 8 tonnes has two pairs of wheels of 400 mm

radius and each pair with axle has a mass of 1 tonne. The radius of gyration of each wheel is

300 mm. The axles are 2·4 m apart and centre of gravity of the truck is mid-way at a height of

1·5 m above the road surface. If the truck is moving with a tractive force of 5·4 kN, acting through its

c.g. Calculate (i) acceleration of the vehicle ; (ii) frictional resistance ; and (iii) reaction of the

wheels.

Solution. Given: Mass of the truck (M) = 8 t; Radius of each wheel (r) = 400 mm = 0·4 m;

Mass of the each pair of wheels with axles (m) = 1 t; Radius of gyration of each wheel (k) = 300 mm

= 0·3 m ; Horizontal distance between the centre of axles (2x) = 2·4 m or x = 1·2 m; Height of the

centre of gravity of the truck above road level (h) = 1·5 m and tractive force (P) = 5·4 kN.

(i) Acceleration of the vehicle

We know that the mass moment of inertia of each pair of wheels,

I=mk^2 = 1(0·3)^2 = 0·09 t-m^2

and acceleration of the vehicle,

a=

2

2 2

5·4

0·63 m/s

0·09

8

(0·4)

P

I

M

r

==

⎛⎞+ +

⎜⎟

⎝⎠

Ans.

(ii) Frictional resistance

We also know that the frictional resistance,

F= 22

0·09 0·63

0·18 kN

22(0·4)

Ia

r

×

== Ans.

(iii) Reactions on the wheels,

We know that the reaction of the front pair of wheels,

RF= 2

1

2

h IP

Mg

x IMr

⎡⎤⎛⎞

⎢⎥+ ⎜⎟

⎣⎦⎝⎠+

= 2

11·50·09 5·4

89·8 39·42kN

21·20·09 8(0·4)

⎡ ⎛⎞× ⎤

⎢⎥×+⎜⎟=

⎣⎦⎝⎠+

Ans.