Front flange FLV. Max. [knm/stroke]
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1 SDN 45 Safety shock absorbers SDN are a low cost alternative to industrial shock absorbers appropriate to customers requirements. Typical applications: cranes, storage and retrieval unit for highbay warehouse, heavy machinery, etc. 0,9 4,5 m/s. Brake force max.: 80 kn (max. energy capacity). Spring force: N. Temperature range: - 10 C to + 80 C. Front flange FLV Rear flange Foot mounting FB Dimensions: Technical data: Type A B C Energy capacity Allowed angular deviation [mm] Max. [knm/stroke] FLV + FB [ ] Weight [ ] [kg] SDN , SDN , SDN , SDN , SDN ,0 4,5 3,5 21 SDN , SDN ,2 3,5 2,5 25 SDN , SDN ,0 2,5 1,5 31 SDN , SDN ,4 1,5 0,
2 SDN 60 Safety shock absorbers SDN are a low cost alternative to industrial shock absorbers appropriate to customers requirements. Typical applications: cranes, storage and retrieval unit for highbay warehouse, heavy machinery, etc. 0,5 4,5 m/s. Brake force max.: 160 kn (max. energy capacity). Spring force: N. Temperature range: - 10 C to + 80 C. Front flange FLV Rear flange Foot mounting FB Dimensions: Technical data: Type A B C Energy capacity Allowed angular deviation [mm] Max. [knm/stroke] FLV + FB [ ] Weight [ ] [kg] SDN SDN SDN SDN ,5 3,5 31 SDN SDN ,5 2,5 37 SDN SDN ,5 1,5 45 SDN SDN ,5 0,5 57 SDN
3 SDN 75 Safety shock absorbers SDN are a low cost alternative to industrial shock absorbers appropriate to customers requirements. Typical applications: cranes, storage and retrieval unit for highbay warehouse, heavy machinery, etc. 0,5 4,5 m/s. Brake force max.: 210 kn (max. energy capacity). Spring force: N. Temperature range: - 10 C to + 80 C. Front flange FLV Rear flange Foot mounting FB Dimensions: Technical data: Type A B C Energy capacity Allowed angular deviation [mm] Max. [knm/stroke] FLV + FB [ ] Weight [ ] [kg] SDN SDN SDN SDN ,5 3,5 39 SDN SDN ,5 2,5 45 SDN SDN ,5 1,5 56 SDN SDN ,5 0,5 70 SDN SDN ,5 0,5 90 SDN
![Capacity charts: The following parameters will be needed in the energy absorption calculation: 1. m [kg] 2. v [m/s] 3. Propelling force F [N] 4.](/docs-images/92/109382839/images/4-0.jpg "C [1/h] The load range is calculated with those parameters. Pre-determine a stroke length and verify the calculation. 1. Total energy/stroke 2. Total energy/hour 3.")
![Effective mass E T E TC m e [Nm] [Nm/h] [kg] Case 1: without propelling force m = 50 kg v = 1,5 m/s C = 100 1/h E K/E T = ½ m v 2 = ½ 50 kg (1,5 m/s) 2 = 56 Nm E TC = E T C = 56 Nm 100 1/h = 5600](/docs-images/92/109382839/images/4-1.jpg "Nm/h m e = 2 E T / v 2 = 2 56 Nm / (1,5 m/s) 2 = 50 kg SES 11 x 25 B selected Case 2: with propelling force m = 100 kg v = 1,5 m/s F D = 1000 N C = 200 1/h s = 0,025 m Propelling force E K = ½ m v 2")
4 Capacity charts: The following parameters will be needed in the energy absorption calculation: 1. m [kg] 2. v [m/s] 3. Propelling force F [N] 4. C [1/h] The load range is calculated with those parameters. Pre-determine a stroke length and verify the calculation. 1. Total energy/stroke 2. Total energy/hour 3. Effective mass E T E TC m e [Nm] [Nm/h] [kg] Case 1: without propelling force m = 50 kg v = 1,5 m/s C = 100 1/h E K/E T = ½ m v 2 = ½ 50 kg (1,5 m/s) 2 = 56 Nm E TC = E T C = 56 Nm 100 1/h = 5600 Nm/h m e = 2 E T / v 2 = 2 56 Nm / (1,5 m/s) 2 = 50 kg SES 11 x 25 B selected Case 2: with propelling force m = 100 kg v = 1,5 m/s F D = 1000 N C = 200 1/h s = 0,025 m Propelling force E K = ½ m v 2 = ½ 100 kg (1,5 m/s) 2 = 112,5 Nm E W = F D s = 1000 N 0,025 m = 25 Nm E T = E K + E W = 112,5 Nm + 25 Nm = 137,5 Nm E TC = E T C = 137,5 Nm 200 1/h = Nm/h m e = 2 E T / v 2 = 2 137,5 Nm / (1,5 m/s) 2 = 122 kg SES 1.1 M x 1 B selected Case 3: on driven rollers m = 900 kg v = 1,0 m/s C = 200 1/h s = 0,05 m µ = 0,3 Coefficient of friction steel/steel E K = ½ m v 2 = ½ 900 kg (1,0 m/s) 2 = 450 Nm E W = m µ g s = 900 kg 0,3 9,81 m/s 2 0,05 m = 132 Nm E T = E K + E W = 450 Nm + 137,5 Nm = 582 Nm E TC = E T C = 582 Nm 200 1/h = Nm/h m e = 2 E T / v 2 = Nm / (1,0 m/s) 2 = 1164 kg STD 2.0 M x 2 selected 32
5 Case 4: with motor drive m = 3000 kg v = 1,4 m/s HM = 2,5 P = 3 kw C = 1/h s = 0,125 m Arresting torque factor for motors Drive power E K = ½ m v 2 = ½ 3000 kg (1,4 m/s) 2 = 2940 Nm E W = 1000 P s HM / v = kw 0,125 m 2,5 / 1,4 m/s = 670 Nm E T = E K + E W = 2940 Nm Nm = 3610 Nm E TC = E T C = 3610 Nm 1 1/h = 3610 Nm/h m e = 2 E T / v 2 = Nm / (1,4 m/s) 2 = 3684 kg STD 3.0 M x 5 selected Case 5: Free falling mass m = 50 kg h = 0,5 m C = 300 1/h s = 0,05 m Height of fall v = 2 g h = 2 9,81 m/s 2 0,5 m = 3,1 m/s E K = m g h = 50 kg 9,81 m/s 2 0,5 m = 245 Nm E W = m g s = 50 kg 9,81 m/s 2 0,05 m = 24,5 Nm E T = E K + E W = 245 Nm + 24,5 Nm = 269,5 Nm E TC = E T C = 269,5 Nm 300 1/h = Nm/h m e = 2 E T / v 2 = 2 269,5 Nm / (3,1 m/s) 2 = 55 kg STD 1.5 M x 2 selected Case 6: Rotating mass/rotary table with driving torque J = 60 kgm 2 ω = 1,2 1/s r = 0,5 m M = 200 Nm C = /h s = 0,025 m Moment of inertia Angular velocity Radius (shock absorber) Driving torque v = ω r = 1,2 1/s 0,5 m = 0,6 m/s E K = ½ J ω 2 = ½ 60 kgm 2 (1,2 1/s) 2 = 43,2 Nm E W = M s / r = 200 Nm 0,025 m / 0,5 m = 10 Nm E T = E K + E W = 43,2 Nm + 10 Nm = 53,2 Nm E TC = E T C = 53,2 Nm /h = Nm/h m e = 2 E T / v 2 = 2 53,2 Nm / (0,6 m/s) 2 = 296 kg STD 1.0 M selected 33
6 Case 7: Swivelling mass with driving torque m = 30 kg v m = 1,0 m/s r = 0,4 m R m = 0,6 m M = 40 Nm C = 1500/h s = 0,02 m Radius (shock absorber) Radius (mass) Driving torque E K = ½ m v 2 = ½ 30 kg (1,0 m/s) 2 = 15 Nm E W = M s / r = 40 Nm 0,02 m / 0,4 m/s = 2 Nm E T = E K + E W = 15 Nm + 2 Nm = 17 Nm E TC = E T C = 17 Nm /h = Nm/h v = v m r / R m = 1,0 m/s 0,4 m / 0,6 m = 0,67 m/s m e = 2 E T / v 2 = 2 17 Nm / (0,67 m/s) 2 = 76 kg SES 10 x 20 A selected Case 8: Swivelling mass with driving force m = 3000 kg v m = 1,5 m/s r = 1,0 m R m = 1,3 m R F = 0,5 m F D = 4000 N C = 150/h S = 0,1 m Radius (shock absorber) Radius (mass) Radius (force) Driving force E K = ½ m v 2 = ½ 3000 kg (1,5 m/s) 2 = 3375 Nm E W = F D s R F / r = 4000 N 0,1 m 0,5 m/1,0 m= 200 Nm E T = E K + E W = 3375 Nm Nm = 3575 Nm E TC = E T C = 3575 Nm 150 1/h = 536,25 knm/h v = v m r / R m = 1,5 m/s 1,0 m/1,3 m = 1,15 m/s m e = 2 E T / v 2 = Nm / (1,15 m/s) 2 = 1352 kg STD 4.0 M x 4 selected Case 9: on incline m = 10 kg h = 0,2 m α = 20 C = 500 1/h s = 0,016 m Height Angle of inclination E K = m g h = 10 kg 9,81 m/s 2 0,2 m = 19,62 Nm E W = m g s sin α = 10 kg 9,81 m/s 2 0,016 m sin 20 = 0,54 Nm E T = E K + E W = 19,62 Nm + 0,54 Nm = 20,16 Nm E TC = E T C = 20,16 Nm 500 1/h = Nm/h v = 2 g h = 2 9,81 m/s 2 0,2 m = 1,98 m/s m e = 2 E T / v 2 = 2 20,16 Nm / (1,98 m/s) 2 = 10,3 kg SES 14 S selected 34
7 Additional sizing formulas and calculations: Effective mass m e [kg] Deceleration a [m/s 2 ] m e = 2 E T / v 2 a = 0,6 v 2 / s Brake force F B [N] Deceleration time t B [s] F B = 1,2 E T / s t B = 2,5 s / v The above formulas apply to correctly selected and adjusted shock absorbers. Please take more precautions than may be necessary to be on the safe side. Special versions are available on request: Description Application Shock absorber with swivelling fixing Clevis mounting Shock absorber with special characteristic line Very high impact velocity Very low impact velocity Shock absorber in stainless steel Hostile environment Outdoor application Shock absorber with alternative seals Hostile environment Deviating ambient temperatures Shock absorber with special stroke length Shock absorber with nickel plated outside parts Hostile environment Outdoor application Shock absorber with air/oil-tank High frequencies requiring an increased energy capacity/h Controlled return stroke of piston rod Shock absorber with special fastening thread Pre-determined fastening elements 35
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