Industrial shock absorber Sizing Examples Typical Shock Absorber Applications
Overview
EXAMPLE 3: Vertical Moving Load withPropelling Force Upward STEP 2: Calculate kinetic energy STEP 1: Application Data E STEP 3: Calculate work energy STEP 5: Calculate totalenergy per hour E (W) Weight = 3,400 lbs.
(V) Velocity = 80 in./sec.
(d) 2 Cylinders bore dia. = 6 in.
(P) Operating pressure = 70 psi
(C) Cycles/Hr = 200 F >
D =2 x [ .7854 x d >
2 x P] – W F >
D =2 x [ .7854 x 6 >
2 x 70] – 3,400 F >
T C=E >
T x C E >
D =558 lbs. E >
T C=30,977 x 200 E >
W =F >
D X S E >
T C=6,195,400 in-lbs./hr Model OEM 3.0M x 5 is adequate. >
W =558 x 5 E >
W =2,790 in-lbs. >
2 2 STEP 4: Calculate total energy per cycle E >
K = W W W 772772 x V = 3,400x 80 E >
K =28,187 in-lbs.Assume Model OEM 3.0M x 5 isadequate (Page 31). >
T =E >
K + E >
W E >
T =28,187 + 2,790 E >
T =30,977 in-lbs./c >
Shock Absorber Sizing Examples EXAMPLE 4: Vertical Moving Load with
Propelling Force from Motor STEP 2: Calculate kinetic energy Assume Model OEMXT 2.0M x 2 isadequate (Page 29).
E (W) Weight =200 lbs.
(V) Velocity =60 in./sec.
(Hp) Motor horsepower = 1.5 Hp
(C) Cycles/Hr = 100 E >
W =F >
D X SE >
W =F >
D x SE >
W =295 x 2E >
W =695 x 2E >
W =590 in-lbs. >
W =1,390 in-lbs. STEP 4: Calculate total energy per cycle STEP 4: Calculate total energy per cycle >
K =Wx V >
2 E >
T =E >
K + E >
W 772772 = 200x 60 >
2 E >
T =E >
K + E >
W E >
T =933 + 590E E >
T =933 + 1,390E E >
K =933 in-lbs. >
T =1,523 in-lbs./c >
T =2,323 in-lbs./c STEP 5: Calculate total energy per hour CASEA: UP STEP 5: Calculate total energy per hour STEP 3: Calculate work energy E >
T C=E >
T x CE E 19,800 x Hp– WVF >
T C = E >
T x CE F >
D = >
T C=1,523 x 100E >
T C = 2,323 x 100E C =152,300 in-lbs./hrModel OEM 1.25 x 2 is adequate. 19,800 x 1.5– 20060F >
T T C = 232,300 in-lbs./hrModel OEMXT 2.0M x 2 is adequate. >
D = >
D =295 lbs.
Assume Model OEM 1.25 x 2 is adequate (Page 24).E CASE B: DOWN STEP 3: Calculate work energy F >
D = 19,800 x Hp+ WVF >
D = 19,800 x 1.5+ 20060F >
D =695 lbs. >
(e.g., Load Moving Force Up) EXAMPLE 5: Horizontal Moving Load
STEP 1: Application Data E STEP 1: Application Data STEP 3: Calculate work energy: N/A (W) Weight=1,950 lbs.
(V) Velocity=60 in./sec.
(C) Cycles/Hr=200 STEP 2: Calculate kinetic energy STEP 4: Calculate total energy per cycle E >
T =E >
K = 9,093 in-lbs./c STEP 5: Calculate total energy per hour >
K = W Wx V >
2 E >
T C=E >
T x C E 772 >
T C=9,093 x 200 E E >
K =1950x 60 >
2 T C=1,818,600 in-lbs./hr Model OEMXT 2.0M x 2 is adequate. 772 E >
K =9,093 in-lbs.Assume Model OEMXT 2.0M x 2 isadequate (Page 29). >
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