Shock absorber Sizing Examples Typical Shock Absorber Applications

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Overview

Shock Absorber Sizing Examples EXAMPLE 6: Horizontal Moving Load with Propelling Force STEP 1: Application Data STEP 2: Calculate kinetic energy STEP 1: Application Data STEP 3: Calculate work energy STEP 5: Calculate total energy per hour E (W) Weight = 1,950 lbs. (V) Velocity = 60 in./sec. (d) Cylinder bore dia. = 3 in. (P) Operating pressure = 70 psi (C) Cycles/Hr = 200 F
D = .7854 x d
2 x P F
D = .7854 x 3
2 x 70 F
T C=E
T x C E
D =495 lbs. E
T C=10,083 x 200 E
W =F
D x S E
T C =2,016,600 in-lbs./hr Model OEMXT 2.0M x 2 is adequate.
W =495 x 2 E
W =99O in-lbs. W W x V
2 K = STEP 4: Calculate total energy per cycle E 772 E
2 K =1,950x 60 =E + E 772
T K W E =9,093 + 990 E E
T K =9,093 in-lbs. Assume Model OEMXT 2.0M x 2 isadequate (Page 29).
T =10,083 in-lbs./c EXAMPLE 7: Horizontal Moving Load, Motor Driven STEP 2: Calculate kinetic energy (W) Weight = 2,200 lbs. (V) Velocity = 60 in./sec. (Hp) Motor horsepower = 1.5 Hp (C) Cycles/Hr = 120 E STEP 3: Calculate work energy STEP 5: Calculate total energy per hour F 19,800
o x HpVF
D = E
T C=E
T x C E 19,800
o x 1.560F
T C=11,249 x 120 E
D =
T C=1,349,880 in-lbs./hr Model OEMXT 2.0M x 2 is adequate.
D =495 lbs.E
W =F
D x S E
K = W W x V
2 772
W =495 x 2E
W =990 in-lbs. E
2 K =2,200x 60 772 STEP 4: Calculate total energy per cycle E
K =10,259 in-lbsAssume Model OEMXT 2.0M x 2 isadequate (Page 29). E
T =E
K + E
W E
T =10,259 + 990 E
T =11,249 in-lbs./c EXAMPLE 8: Free Moving Load Down an Inclined Plane STEP 1: Application Data (W) Weight=550 lbs. (H) Height=8 in. ( STEP 3: Calculate work energy STEP 5: Calculate total energy per hour E F
D =W x Sin
α F
D =550 x .5F
T C=E
T x C E
α ) Angle of incline = 30
D =275 lbs.E
T C = 5,225 x 250 E (C) Cycles/Hr = 250

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W =F
D x SE
T C = 1,306,250 in-lbs./hr STEP 2: Calculate kinetic energy
W =275 x 3E STEP 6: Calculate impact velocityand confirm selection V= √ 772 E
K =W x H E
W =825 in-lbs.
K =550 x 8 E STEP 4: Calculate total energy per cycle x x HV=
K =4,400 in-lbs. Assume Model OEMXT 1.5M x 3 isadequate (Page 27). √ 772 x x 8= 79 in./sec.Model OEMXT 1.5M x 3 is adequate. E
T =E
K + E
W E
T =4,400 + 825E
T =5,225 in-lbs./c
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