Geotechnical Engineering

DHARM

866 GEOTECHNICAL ENGINEERING

Maximum amplitude at resonance

Amax =

(/)Pk

DD

0

21 −^2

Substituting P 0 = 5 kN

k = 32,194 kN/m

Amax = 1 mm = 1 × 10–3 m

1 × 10–3 =

(/ )

()

5 32194

21 DD−^2

(^21) DD−^2 =
(/ )
–3
5 32194
110 ×
= 0.1553
Squaring,
4 D^2 (1 – D^2 ) = 0.024
4 D^4 – 4D^2 + 0.024 = 0
Solving for D^2 ,
and hence D = 0.997 or 0.078. The plausible value is 0.078.
Example 20.8: A single cylinder engine with the following particulars is to be placed on a
concrete foundation. Find the maximum unbalanced force generated by the engine:
crank radius = 80 mm
Length of connecting rod = 280 mm
operating frequency = 1800 rpm
Weight of reciprocating parts = 49 N
Pz = (Mrec + Mrot)Rω^2 cos ωt + Mrec
R
L
(^22) ω
cos 2ωt
The maximum value is
Pz = MrecRω^2 + MrecR
L
(^22) ω
, approximately
= MrecRω^21 +
F
HG
I
KJ
R
L
fn = 1800 rpm =
1800
60 cps = 30 cps
ωn = 2πfn = 2π × 30 rad/s
Substituting this and R = 80 × 10–3 m, L = 280 × 10–3 m, and
Mrec =
49
981.
kN sec^2 /m
we have Pz =
49 80
981 10
1 80
280
3 2302
×
×
F +
HG
I
KJ
×
.
()π N
= 18,254 N = 18.25 kN.