SMC Corporation of America
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Search Results "CDM2E20-100Z-N-M9BV"

4, 6, 8) Note 3) 145 + 50 (n 4) 2 (n = 4, 8, 12, 16) Note 4) 115 + 50 (n 4) 2 (n = 4, 8, 12, 16) Note 4) 125 + 50 (n 4) 2 (n = 4, 8, 12, 16) Note 4) 135 + 50 (n 4) 2 110 + 50 (n = 4, 8, 12, 16) Note 4) Different surfaces (n = 4, 8, 12, 16) Note 4) n 75 + 50 (n 2) (n = 2, 3, 4) 90 + 50 (n 2) (n = 2, 4, 6, 8) Note 3) 115 + 50 (n 2) (n = 2, 4, 6, 8) Note 3) 125 + 50 (n 2) (n = 2, 4,

4, 6, 8 ) 80 + 30 (n 2) (n = 2, 4, 6, 8 ) 35 + 30 (n 2) (n = 2, 3, 4 ) 90 + 30 (n 2) (n = 2, 4, 6, 8 ) 100 + 100 (n 2) (n = 2, 4, 6, 8 ) D-A3 D-G39 D-K39 Different surfaces 100 + 100 (n 2) (n = 2, 3, 4 ) Same surface 1 pc. 75 80 90 10 75 75 80 80 Different surfaces 35 55 90 90 90 + 30 (n 2) (n = 2, 4, 6, 8 ) Same surface 75 + 30 (n 2) (n = 2, 4, 6, 8 ) 80 + 30 (n 2) (n = 2, 4, 6,

(n:: 4, 8, 12, 16 ) (n ::: 4.8, 12, 16) Wrlh2 pes jDifferenl sllriaces. 11 5 120 20 Samesuriace), Wrlh 1 pc D-M90AL ---.-~~-, 115 +40(n 2 4) '~O + 40(12 4; 20+40In ;2) Wllh n pes. (n = 2. 4. 6. S .. ) (n = 4 . 8, 12, 16...),_-+_ ~4. 8. 12.16.. -) D~ F5D1.J5 CJ Same surface) With 1 pc.

D-A9 30 Note 1) 15 5 (n = 2, 3, 4, 5) (n = 2, 4, 6) Note 3) (n 2) 2 35 + 35 (n 2) 20 + 35 D-M9V 35 20 5 (n = 2, 3, 4, 5) (n = 2, 4, 6) Note 3) (n 2) 2 25 + 35 (n 2) 15 + 35 D-A9V 25 15 5 (n = 2, 3, 4, 5) (n = 2, 4, 6) Note 3) (n 2) 2 D-M9WV D-M9AV 35 + 35 (n 2) 20 + 35 35 20 10 (n = 2, 3, 4, 5) (n = 2, 4, 6) Note 3) (n 2) 2 D-C7 D-C80 60 + 45 (n 2) 20 + 45 60 20 5 (n = 2, 3, 4, 5

How to Order MXW 16 100 B M9N S Number of auto switches Nil S n 2 pcs. 1 pc. n pcs.

0.6 MPa Pressure 0.6 MPa Pressure 0.6 MPa 40 40 Gripping force (N) Gripping force (N) 0.5 MPa 0.5 MPa 30 30 0.4 MPa 0.4 MPa L L 20 20 0.3 MPa 0.3 MPa 10 10 0.2 MPa 0.2 MPa 0 0 10 20 30 40 50 60 10 20 30 40 50 60 Gripping point L (mm) Gripping point L (mm) MHK2-20D MHK2-20D 80 60 70 Pressure 0.6 MPa Pressure 0.6 MPa Pressure 0.6 MPa Pressure 0.6 MPa 50 Gripping force (N) Gripping force (N)

MHKL2-20D MHKL2-20D D80 60 70 Pressure 0.6 MPa 2050 Gripping force (N) 60 Holding force (N) Pressure 0.6 MPa Pressure 0.6 MPa 0.5 MPa 50 40 0.5 MPa 40 0.4 MPa 30 0.4 MPa 30 0.3 MPa 20 0.3 MPa 20 0.2 MPa 10 10 0.2 MPa 0 0 20 40 60 80 20 40 60 80 Gripping point L (mm) Gripping point L (mm) MHKL2-25D MHKL2-25D 100 120 Pressure 0.6 MPa Pressure 0.6 MPa 100 80 Gripping force (N) Gripping force

io n ge io at of at ot ot R an R e n r ng tio ra ta n Ro io at ot R Unit Used as Flange Mount The L dimensions of this unit are shown in the table below.

O1 thread N through-hole These 5 figures show the piston rod extended.

20 30 40 50 60 70 0 10 20 30 40 50 60 70 0 0 20 40 60 80 100 120 0 20 40 60 80 100 120 Gripping point distance R [mm] Gripping point distance R [mm] Gripping point distance R [mm] Gripping point distance R [mm] 1 2 1 2 MHL2-20DZ MHL2-25DZ MHL2-25D Z MHL2-20D Z 200 200 120 120 160 160 Gripping force [N] Gripping force [N] Gripping force [N] Gripping force [N] Pressure 0.6 MPa Pressure 0.6

Lead 6: EQY25B 300 Force [N] Select a model based on the pushing force set value and force while referencing the force conversion graph. Lead 12: EQY25A 200 Selection example) Based on the graph shown on the right side, Pushing force: 100 [N] Pushing force set value: 40 [%] The EQY25DHB can be temporarily selected as a possible candidate.

Number of auto switches [mm] Number of auto switches 1 2 n Auto switch model Different surfaces Same surface Different surfaces Same surface D-M9m 25 25 40 20 + 35 (n 2) 2 (n = 2, 4, 6)*1 55 + 35 (n 2) (n = 2, 3, 4, 5) D-M9mW 25 25 40 20 + 35 (n 2) 2 (n = 2, 4, 6)*1 55 + 35 (n 2) (n = 2, 3, 4, 5) D-M9mA 25 25 40 25 + 35 (n 2) 2 (n = 2, 4, 6)*1 60 + 35 (n 2) (n = 2, 3, 4, 5) D-M9mV

M3 M3E m x g FE L2 DFrom above, n = 1 + 2 + 3A + 3B = 0.1 + 0.082 + 0.35 + 0.28 = 0.812 From n = 0.812 1, it is applicable.

Thus, F= 5 x 0.05 x 9.8 = 2.5 (N) 1. Calculation of applied lateral load F F = N m g (N) = 0.2 x 10 x 0.1 x 9.8 = 2.1 (N) CC 70 Lateral load F (N) RB 60 0.4 MPa 50 J 40 2. Confirmation of allowable lateral load From the graph, the allowable lateral load at L = 50 mm and P = 0.4 MPa is 18 N. Because 2.1 N < 18 N, it is applicable. 30 20 D2.

Thus, F= 5 x 0.05 x 9.8 = 2.5 (N) 1. Calculation of applied lateral load F F = N m g (N) = 0.2 x 10 x 0.1 x 9.8 = 2.1 (N) CC 70 Lateral load F (N) RB 60 0.4 MPa 50 J 40 2. Confirmation of allowable lateral load From the graph, the allowable lateral load at L = 50 mm and P = 0.4 MPa is 18 N. Because 2.1 N < 18 N, it is applicable. 30 20 D2.

Bore size (mm) Allowable kinetic energy for intermediate stop Es (J) 10 0.03 15 0.13 25 0.45 8-16-8 8 Series CY1F Model Selection 2 Selection Calculation The selection calculation finds the load factors (n) of the items below, where the total (n) does not exceed 1. n = 1 + 2 + 3 1 Load factor n 1 = m/mmax Item Note MX Review m m max is the maximum load mass at a 1.

Load [N] Load [N] MXZ16 MXZ16 MXZ16 0.15 0.15 0.15 Table displacement [mm] Table displacement [mm] Table displacement [mm] ST 25, 30 ST 25, 30 ST 25, 30 ST 15, 20 0.10 0.10 0.10 ST 15, 20 ST 15, 20 ST 5, 10 0.05 0.05 0.05 ST 5, 10 ST 5, 10 0.00 0 5 10 15 20 25 0.00 0 5 10 15 20 25 0.00 0 5 10 15 20 25 Load [N] Load [N] Load [N] MXZ20 MXZ20 MXZ20 0.25 0.40 0.25 ST 35, 40 ST 45, 50 ST 45,