9.19 Proble ms
Example 9.1:Determine the natural frequency of a machine foundation which has a base area of
2.5 m ×2.5 m and a weight of 150 kN including the weight of the machine. Take the value of the
coefficient of elastic uniform compression as 4.35 × 10 4 kN/m3 .
Solution:
As we know,
M=150/9.81 kN.sec2 /m
ω n = √(4.35 104 2.52 9.81/150)=133.34 rad/sec
Natural Frequency, fn = ω n /2π = 21.22 Hz
Example 9.2:The exciting force in a constant force-amplitude excitation is 80 kN. The natural
frequency of the machine foundation is 2 Hz. The damping factor is 0.25. Determine the
magnification factor and the transmitted force at an operating frequency o f 6 Hz.
Solution:
Frequency ratio ξ = 6/2 =3, Damping factor D=0.25
Substituting the values of D and ξ, η1 = 0.122
Substituting the values of D and ξ, T= 0.190
Transmitted force = 0.190 80=15.24 kN
Example 9.3: Design a suitable block foundation for a two-cylinder vertical compressor for the
following data:
Crank angles: 0 and π/2
Weight of compressor = 220 kN
Operating speed: 650 rpm
Total weight of rotating mass: 0.08 kN
Total weight of reciprocating mass: 0.30 kN
Radius of crank: 0.42 m
Safe bearing capacity of soil under static condition: 125 kN/m2
Co-efficient of elastic uniform compression: 45,000 kN/m3
Solution:
Calculation of unbalanced forces
Pz = (Mrec + Mrot ) Rω 2 cos (ωt+π/4)
Mrec = 0.30/9.81 kN.sec2 /m
Mrot = 0.08/9.81 kN.sec2 /m
Pz.max = √2*(0.38/9.81)*0.42*(650*2π/60)2 kN = 106.6 kN
Similarly Px can be calculated as;
Px = Mrot Rω 2 sin (ωt+π/4)
Substituting the known values, we obtain P x.max = 22.44 kN
Size of Foundation
Let us assume a block of size 7m X 4m X 1m high for compressor foundation. The compressor
& motor should be arranged so that; the eccentricity of the resultant forces due to the weight of
the machine and foundation block do not exceed 5%
Weight of the foundation block= 7 4 1 24= 672 kN
Pressure on Foundation Soil
Total weight coming on soil = 220+ 672 = 892 kN
Area of Base = 7 4 = 28 m2
Therefore, stress on soil = 31.85 kN/m2
Safe bearing capacity = 125 kN/m2
Assuming reduction factor of 0.5, dynamic bearing capacity = 62.5 kN/m2 [Hence OK]
Amplitudes of Motion
ω n2 = CuA/M= 13587.17 / sec2
ω 2 = (650*2π/60)2 = 4633 / sec2
Amplitude of vibration
= P0 / M(ω n2 - ω2 ) = 0.1309 mm
Since the value is less than 0.150 mm, is acceptable.