Vibration isolator Sizing Examples Typical Shock Absorber Applications
Overview
Shock Absorber Sizing Examples EXAMPLE 12: STEP 1: Application Data CASE A Vertical Motor Driven Rotating Arm with Attached Load (W) Weight = 110 lbs.
( STEP 3: Calculate work energy STEP 5: Calculate total energy per hour: not applicable, C=1STEP 6: Calculate impact velocityand confirm selection ω θ ) Starting point of load fromtrue vertical = 20 ) Angular velocity = 2 rad./sec.(T) Torque = 3,100 in-lbs.
( CASE A–Load Aided by Gravity F
D =[T + (W x K x Sin ( θ + Ø))]R V = R
S F
D =[3,100 + (110 x 24 x .77)]16F (Ø) Angle of rotation at impact = 30
Ëš
x ω = 16 x 2 = 32 in./sec.Model LROEM 1.0 is adequate.Needed for higher calculated propelling force.
D =320.8 lbs.E
S (K
Load ) Radius of gyration = 24 in.(R
Ëš
W =F
D x S = 320.8x 1= 320.8 in-lbs.
S ) Mounting radius = 16 in.(C) Cycles/Hr = 1 STEP 4: Calculate total energy per cycle E STEP 2: Calculate kinetic energy
T =E
K + E
W = 328 + 320.8 E I = 1 W
T =648.8 in-lbs./c 386 386 x K
2 = 110x 24
2 I =164 in-lbs-sec
2 2 E
K =I x ω 2 E
2 K =164 x 2 2 E
K = 328 in-lbs.Assume Model OEM 1.0 is adequate(Page 21). STEP 5: Calculate total energy per hour: not applicable, C=1STEP 6: Calculate impact velocityand confirm selection. EXAMPLE 13: STEP 1: Application Data CASE B Vertical Motor Driven Rotating Arm with Attached Load (W) Weight = 110 lbs.
( STEP 3: Calculate work energy ω θ ) Starting point of load fromtrue vertical = 30 ) Angular velocity = 2 rad./sec.(T) Torque = 3,100 in-lbs.
( V = R CASE B–Load Opposing Gravity F
D =[T – (W x K x Sin ( θ – Ø))]R
S F
D =[3,100 – (110 x 24 x .77)]16E (Ø) Angle of rotation at impact = 150
Ëš
S x ω = 16 x 2 = 32 in./sec.Model OEM 1.0 is adequate
W =F
D x S = 67 x 1 = 67in-lbs. (K
Load ) Radius of gyration = 24 in.(R
Ëš
STEP 4: Calculate total energy per cycle E
S ) Mounting radius = 16 in.(C) Cycles/Hr = 1 .
T = E
K + E
W = 328 + 67E STEP 2: Calculate kinetic energy
T = 394.7 in-lbs./c I = 1 W 386 386 x K
2 = 110x 24
2 I =164 in-lbs-sec
2 2 E
K =I x ω 2 E
2 K =164 x 2 2 E
K = 328 in-lbs.Assume Model OEM 1.0 is adequate(Page 21). EXAMPLE 14: STEP1: Application Data (W) Weight = 540 lbs.
( ω θ ) Starting point of load fromtrue vertical = 20 ) Angular velocity = 3.5 rad./sec.(T) Torque = 250 in-lbs.
( STEP 2:Calculate kinetic energy E Vertical Rotating Beam STEP 4: Calculate total energy per cycle K= .289 x √ 4 x L
2 + B
2 K= .289 x √ 4 x 24
2 + 2.5
2 = 13.89
T = E
K + E
W = 1,653 + 730 = 2,383 in-lbs./c I = 5 Wx K
2 386 386 =540x 13.89 STEP 5: Calculate total energy per hour: not applicable, C=1STEP 6: Calculate impact velocity and confirm selection (Ø) Angle of rotation at impact = 50°
(R
Ëš
I =270 in-lbs./sec.
2 E STEP 3: Calculate work energy
2 2 ) Mounting radius = 20 in.(B) Thickness = 2.5 in.
(L) Length = 24 in.
(C) Cycles/Hr = 1
K = I x ω
S 22 =270x 3.5 =1,653in-lbs. V = R Assume Model OEM 1.5M x 2 is adequate (Page 27). F
S x ω = 20 x 3.5 = 70 in./sec.Model OEM 1.5M x 2 is adequate.
D =T + (W x K x Sin ( θ
θ + Ø))R
S F
D =250 + (540 x 13.89 x Sin (20 º + 50 º ))20F
D = 365 lbs.E
W = F
D x S = 365 x 2 = 730 in-lbs.
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