Geotechnical Engineering

DHARM

ELEMENTS OF SOIL DYNAMICS AND MACHINE FOUNDATIONS 865

But fn =

1

2 π

CA

M

ut.

Substituting all known values,

18 =

1

2

1 125

π 1 927

Cu×.
.
whence Cu, the coefficient of elastic uniform compression = 2.191 × 10^4 kN/m^3

Next, Weight of machine = 90 kN

Weight of block = 3 × 4 × 2 × 24 = 567 kN

Total Weight = 666 kN

Corresponding Mass =

666

981.

kN sec^2 /m = 67.89 kN sec^2 /m

Base area, Ap = 12 m^2

If we denote Cu for the test block as Cut and that for the prototype block for the machine

as Cup the relationship between these two is

C

C

A

A

u

u

t

p

p

t

=

∴ Cup = 2.191 × 10^4 ×

1 125

10

.

= 1.37 × 10^3 kN/m^3

(Since Ap is to be limited to 10 m^2 )

Hence the natural frequency of the machine foundation

=

1

2

735 10 12

67 89

3

π

.

.

××

= 5.74 Hz.

Example 20.6: What will be the percentage variation in the value of the coefficient of elastic
uniform compression if the diameter of the plate is halved?

C

C

A

A

D

D

D

D

u

u

2

1

1

2

1

2

1

1 2

===

(/)

= 2

∴ CCuu 21 =^2
This means that the coefficient of elastic uniform compression increases by cent per cent
if the diameter of the plate is halved.

Example 20.7: The resonant frequency of a block foundation, excited by an oscillator is ob-
served as 20 Hz. The amplitude of vibration at resonance is 1 mm. The magnitude of the
dynamic force at 20 Hz is 5 kN. If the total weight of the block and oscillator is 20 kN, calculate
the damping factor associated with it.

fn =

1

2 π

k

M

20 =

1

220981 π

k
(/.)
whence k = 32, 194 kN/m