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75 PSF Equipment on 50 PSF Floor

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Subject: 75 PSF Equipment on 50 PSF Floor

Dear Chris,

Analysis:

1)area of equipment= 2.5 ft x 6.167 ft = 15.417 sq ft
2)Load = 1200 pounds
3)Load intensity = 1200 / 15.417 = 77.8 psf
4)Design load originally considered = 50 psf
5)You need therefore an area of 24 sq feet (minimum)

Suggestion:

6)Proposed revised base area = 3.5 ft x 7 ft =
24.5sqft
7)Fabricate  a grid frame of steel with these outer
dimensions
8)Place the equipment and fix it to this grid.
9)The grid should have sufficient depth to permit
dispersion of the load to this calculated expanded 
area.

Will this  do?
Regards
G Vishwanath
Bangalore, India


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