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Up to 600 Up to 1000 Up to 600 Up to 1000 Up to 750 Up to 1000 Up to 750 Note) End boss is machined on the flange for E.
Lead wire length symbols: 0.5 mNil (Example) M9N 3 mL (Example) M9NL 5 mZ (Example) M9NZ Normally closed (NC=b contact), solid state switches (Model D-F9G, F9H) are also available. For detail, refer to page 7-9-23 in the Best Pneumatics 2004 Vol. 7 catalog.
Graph (3) 20 x 9.8 x (0.07 + 0.05) x 10/1000 = 0.24 20 x 9.8 x (m1 + m2) x H/1000 0.24 Nm < Effective torque OK < Effective torque (Nm) Find the moment of inertia, "IR" for the load + attachments (2 pcs.) 7 IR = K x (a2 + b2 + 12h2) x (m1 + m2)/(12 x 106) IR = 2 x (202 + 302 + 12 x 102) x (0.07 + 0.05)/(12 x 106) = 0.05 kgm2 (K = 2: Safety factor) Kinetic energy 8 Confirm that the kinetic
(between lead wire and case) 1000 VAC for 1 min.
50 200, 250, 300, 350, 400, 450, 500, 600, 700, 800, 900, 1000 6000 63 200, 250, 300, 350, 400, 450, 500, 600, 700, 800, 900, 1000 Made to Order Specifications (For details, refer to page 10-21-1.)
Mr = 0.2 x 9.8 (20 + 6.8)/1000 = 0.053 Ma = Mmax Find the allowable static moment Ma (Nm).
Confirm that this value is less than 1/20 of the effective torque. 20 x 9.8 x (m1 + m2) x H/1000 20 x 9.8 x (0.07 + 0.05) x 10/1000 = 0.24 < Effective torque (Nm) Graph (3) 0.24 Nm < Effective torque OK Find the moment of inertia, "IR" for the load + attachments (2 pcs.) 7 IR = K x (a2 + b2 + 12h2) x (m1 + m2)/(12 x 106) IR = 2 x (202 + 302 + 12 x 102) x (0.07 + 0.05)/(12 x 106) = 0.05 kgm2
COM. 13 (+) 13 () Voltage limit V, 1 minute, AC 1000 Positive common Negative common Insulation resistance M/km, 20C 5 or more Note) When using a valve with no polarity, either positive common or negative common can be used. Note) The minimum bending radius of the D-sub connector cable is 20 mm.
800 Up to 800 Up to 1000 Up to 1000 Up to 1000 Up to 1000 Up to 1000 37 44 50 56 66 8-25-11 11 Series MGZ With Mounting Bracket Transaxial foot style: (L) 4-LD through LT LH LY X/2 ZZ + Stroke LS + Stroke X X Y LZ LX (mm) Stroke range Bore size (mm) LS ZZ LD Y X LH LT LX LY LZ Up to 800 Up to 800 Up to 1000 Up to 1000 Up to 1000 Up to 1000 Up to 1000 20 25 32 40 50 63 80 86 107 120 138 148
0 200 400 600 800 1000 1200 1400 1600 Stroke [mm] Stroke [mm] (1 Kg = 2.2 lbs) [Graph 7] Allowable load mass by stroke 40 100 y, z + A = 0 y, z + A = 50 Load mass [kg] y, z + A = 100 y, z + A = 200 10 y, z + A = 300 1 0 200 400 600 800 1000 1200 1400 1600 If load center of gravity exceeds the value of y, z + A on the graph, please consult SMC.
75 133 368 Up to 1000 98 345 345 388.5 452 477 368 368 409.5 470 495 8 9 0.2 stroke 98 345 30 to 1000 75 133 368 98 Up to 1000 0.2 stroke 106 388.5 75 141 409.5 106 30 to 1200 Up to 1200 0.2 stroke 111 448 85 153 466 115 30 to 1200 Up to 1200 0.2 stroke 111 468 90 153 486 120 30 to 1200 Up to 998 0.2 stroke 141 579.5 105 176 595.5 30 to 1200 0.17 stroke 9-9-12 13 Series CLS Cylinder with
Stroke (mm) Up to 10 Up to 30 Up to 50 Up to 75 Up to 100 Up to 1000 Up to 800 Up to 1000 Up to 1000 Up to 800 Up to 1000 Up to 1000 Max. speed (mm/s) 20 Selection graph (6) (7) (8) (9) (10) (11) (12) 25 32 Caution If the cylinder is horizontally mounted and the plate end does not reach the load's center of gravity, use the formula below to calculate the imaginary stroke l that includes the
Stroke (mm) Up to 10 Up to 30 Up to 50 Up to 75 Up to 100 Up to 1000 Up to 800 Up to 1000 Up to 1000 Up to 800 Up to 1000 Up to 1000 Max. speed (mm/s) 20 Selection graph (6) (7) (8) (9) (10) (11) (12) 25 32 Caution If the cylinder is horizontally mounted and the plate end does not reach the load's center of gravity, use the formula below to calculate the imaginary stroke l that includes the
Mr = 1 x 9.8 (30 + 10.5)/1000 = 0.39 A6 = 10.5 Mar = 36 Mrmax = 36 K = 1 = 1 '2 = 0.39/36 = 0.011 Examine My. My = 1 x 9.8 (10 + 30)/1000 = 0.39 A3 = 30 May = 1 x 1 x 18 = 18 Mymax = 18 K = 1 = 1 2 = 0.39/18 = 0.022 M = W x 9.8 (Ln + An)/1000 Correction value of moment center position distance An: Table (3) Find the static moment M (Nm).
M = W x 9.8 (Ln + An)/1000 Corrected value of moment center position distance An: Table (1) Examine Mr. Mr = 0.2 x 9.8 (40 + 15.5)/1000 = 0.1 A2 = 15.5 Obtain Mar = 13 from Va = 300 in Graph (3). Find the allowable static moment Ma (Nm). Pitch, Yaw moment: Graph (2) Roll moment: Graph (3) 13 Find the load factor 1 of the static moment.
V = 2gH/1000 x 60 Obtain the total amount of the workpiece. Total weight m = 10 x 0.1 (kg) = 1 (kg) Obtain the intersection of the transfer speed V and the total weight of workpiece m.
V = 2gH/1000 x 60 Obtain the total amount of the workpiece. Total weight m = 10 x 0.1 (kg) = 1 (kg) Obtain the intersection of the transfer speed V and the total weight of workpiece m.
Lateral load Operating pressure: 1 MPa 1000 200 300 500 Stroke Eccentric distance 100 100 Load weight (kg) 10 Courtesy of Steven Engineering, Inc.-230 Ryan Way, South San Francisco, CA 94080-6370-Main Office: (650) 588-9200-Outside Local Area: (800) 258-9200-www.stevenengineering.com 63 Allowable lateral load (N) 50 40 63 1 32 25 20 50 40 10 32 0.1 25 10 100 200 300 500 1000 Maximum piston