DHARM

ELEMENTS OF SOIL DYNAMICS AND MACHINE FOUNDATIONS 871

Section

7.5

1.0

1.0

1.0

1.5

0.5 4.5 0.5

1.25

0.5

0.5

1.25

7.5

6.5

All dimensions are in metres

1.0

4.5

Plan

Fig. 20.35 Details of hammer foundation (Example 20.11)

k 2 =

EA

t

p p

p

=

510 55××

060

(^5).

. = 46 × 10

(^5) kN/m
square of the limiting frequency of anvil
ωa^2 =
k
Ma
2 46 10^5 9 81
1000

××.
= 46 × 981 = 45,126/sec^2
Mass of foundation and backfill (Mf) =
3052
981.kN sec
(^2) /m
Frequencies
Square of the limiting frequency of the whole system
ωl^2 = k
MMa f
1
73 2 10^5
()4052 9 81
.

• (/.)
  = × = 17722 sec–2
  Ratio λ 1 =
  M
  M
  W
  W
  a
  f
  a
  f
  ==^1000
  3052 = 0.327654
  The frequency equation is
  ωn^4 – (1 + λ 1 )(ωa^2 + ωl^2 )ωn^2 + (1 + λ 1 )ωa^2 ωl^2 = 0
  Substituting ωa^2 = 45,126, ωl^2 = 17,722, λ 1 = 0.327654,
  ωn^4 = (1 + 0.327654)(45126 + 17722)ωn^2 + (1 + 0.327654)(45126)(17722) = 0
  Solving, ωn 12 = 67.7744 × 10^3 sec–2 and ωn 22 = 15.6660 × 10^3 sec–2
  or ωn 1 = 260.335 rad/s and ωn 2 = 125.164 rad/s