Firestone EMDG198 manual Airmount Vibration Isolation, Selection and Isolation Formula

AIRMOUNT® VIBRATION ISOLATION

SELECTION AND ISOLATION FORMULA

Refer to the selection guide on page 33 for Airmount load and isolation capabilities. Follow this procedure:

1. LOAD CAPACITY

Select one or two Airmounts that can support the load at each mounting point. It is normally best to design for pressures in the 60 to 80 psig range. Consider only the 1M1A-0 and the single and double convoluted types at first. Please notice that in the range of 210 to 63,890 pounds you will, in most cases, find both a single and double convoluted style part which will support the load.

2. DETERMINE ISOLATION EFFECTIVENESS

Select the disturbing frequency that is closest to the actual forced frequency (400, 800, or 1500 cpm). Then check the per- centage of isolation for the parts that were selected in 1 above.

3. DETERMINE DESIGN HEIGHT

THE AIR SPRING SHOULD BE USED AT THE DESIGN HEIGHT GIVEN. The double convoluted part is used at a design height somewhat higher than its single convolution equivalent. Make sure that the design height falls within the height restrictions. Also, the double convoluted part will show a higher percentage of isolation (less transmitted vibration) than the single convoluted air spring. The reason for this is that the double convoluted part has a greater internal volume of air than the single convoluted version of the same size. At disturb- ing frequencies in the 400 to 800 cpm range, the double convoluted part is a significantly better vibration isolator than the single convoluted part. At disturbing frequencies of 800 to 1500 cpm, the gap closes considerably. At frequencies of 1500 cpm and above, the difference is negligible.

4. DETERMINE EXACT INTERNAL PRESSURE AND ISOLATION EFFECTIVENESS

The chances are that your specific vibration problem does not fall neatly into the load and disturbing frequency criteria as pre- sented in the selection guide.

Therefore, once a preliminary part selection has been made, turn to the individual data page for that part in order to deter- mine the specific internal pressure required and the percentage of isolation attainable.

CONSIDER THIS EXAMPLE:

Isolate a vibrating screen which weighs a total of 16,400 pounds, preferably with one isolator at each corner. The vibrat- ing mechanism is rotating at a speed of 850 rpm (cpm) with a total stroke of 5/16 inch.

a. Determine the load at each mounting point:

16,400 = 4,100 lbs.

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Scan down the 80 psig load column in the selection guide. It appears that either a #19 or a #22 will support the load at a pressure between 60 and 80 psig.

b. Determine Isolation Effectiveness.

Read the % of Isolation at 800 cpm for the #19 and #22 (since 800 is closest to our machine speed of 850 cpm). A #19 is at 96.0% and a #22 is at 98.2%. Looking at isolation effec- tiveness in terms of % TRANSMISSION, the #19 will transmit 100 – 96.0, or 4.0% of the vibrations. A #22 will transmit 100 – 98.2, or 1.8% of the vibrations. So, even though there does not seem to be much difference between 96.0% and 98.2% isolation, the #22 is in fact a better isolator by approxi- mately a factor of two when comparing transmitted vibration.

c. Determine Design Height.

Let’s say we have chosen the #22 because 96.0% isolation for a #19 is considered to be too low. A #22 should be used at 9.5 inches as shown in the second column on page 33.

d. Determine Exact Internal Pressure and Isolation Percentage. Turn to page 61 for detailed information on the #22.

a)What exact pressure will be required to support the load of 4,100 Ibs? Refer to the information in the block entitled “Dynamic Characteristics at 9.5 in Design Height.”

4,280 lbs. = 53.5 in2 = effective area @ 9.5 inches

80 lbs/in2@ 80 psig

Divide the actual load by the effective area:

4,100 lbs. = 76.6 psig required to support

53.5 in24,100 lbs. at 9.5 inches

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