Linear guide drives Drive concept The linear guides are driven via the hollow shaft in the drive head. The drive head is used to directly install a motor or alternatively (in connection with a center shaft) a hollow shaft gear or a coupling. Depending on the desired speeds of movement for the application, it is necessary to use various motor types in conjunction with intermediate gears. Due to the multitude of drive components available in the market, there are a large number of possible combinations. The MGE modular profile system makes it easy to assemble Bosch Rexroth linear guides through the use of a standard shaft and simple connection dimensions. 00050 Fastening The drive head is directly secured on the profile by means of longitudinal end connectors. The motor, or the intermediate gear flange, can be screwed to the drive head as a complete unit from the left- or righthand side. Synchronous shafts The use of synchronous shafts enables several linear axes to be operated with one drive. Connection dimensions (flange) The connection dimensions for the size LF6, LF and LF0 linear guides are shown on pages -5. 0090 9776
Flange A A flange is generally required for fixing the motor/gear unit in place. The connection dimensions for connection to the drive head are shown below. The connection to the motor/ gear unit depends on the model chosen in each case. See also the connection dimensions for the drive heads LF6S, LF6C, LFC, LFS, LF0C, LF0S 5 6 7 8 9 976 Fastening material flange A Note regarding flange in self-build version. Adapt fastening material to suit differing flange thicknesses. LF6 (b =.5 mm) LF (b = 0 mm) LF0 (b = 0 mm) Drive head flange ISO 76 M8x (x) ISO 76 M8x5 (x) ISO 76 M0x0 (x) ISO 709 8 (x) ISO 709 8 (x) ISO 709 0 (x) 0 LF b Gear / motor Drawing and dimensions for flange in self-build version. Customer-specific adjustment is necessary, depending on the motor and gear unit. The fastening material must be 5 a self-sourced. 6 c Flange 7 Dimension parameter LF6 LF LF0 (mm) (mm) (mm) 8 ø E ø F ø G ø H øl øl a.5 ± 0.5.5 ± 0.5 ± 0.5 b *) min 0.05 * ) * ) * ) 9 0 ø d c *) * ) * ) * ) E A Ø 88-0./-0. Ø -0./-0. Ø 50-0./-0. 0 A ø d 5 F *) min Ø * ) Ø * ) Ø * ) G *) 0.05 A Ø * ) Ø * ) Ø * ) H *) Ø * ) Ø * ) Ø * ) l *) Ø * ) Ø * ) Ø * ) l Ø 78 Ø 00 Ø 5 978 d *) Ø * ) Ø * ) Ø * ) d Ø 0.5 M8 M8 M0 *) Dimension according to customer drive Note: The values must be checked by the user.
Alternative mounting with coupling Instead of direct mounting over the section link, a coupling can be inserted. This allows any user truing errors to be rectified. The coupling must be designed and sourced by the customer. LF Gear / motor 967
Technical data for roller, design, mounting General technical data and calculations Speed v max = 5 m/s Acceleration a max = 5 m/s Temperature resistance -0 C < J < 0 C Lubrication All cam roller guide ball bearings are delivered with built-in lubrication felts. Condition on delivery Ungreased (greasing with Klüber Isoflex Topas NCA 5; see assembly instructions 8 57 6) v Lubricating felts with reservoir are not greased. Guide rods are ungreased. 5 6 7 8 9 Definition of dynamic load rating C Note regarding maximum load F max Definition and calculation of nominal service life Nominal service life at constant speed The radial loading of constant magnitude and direction which a linear rolling bearing can theoretically withstand for a nominal service life of 00 km distance traveled (acc. to ISO 78 Part ). The maximum loads apply to individual loads. Combining loads decreases the expected service life. The calculated service life which an individual linear rolling bearing or a group of apparently identical rolling element bearings operating under the same conditions can attain with a 90% probability, based on the materials of normal manufacturing quality that are generally used today and under usual operating conditions (acc. to ISO 78 Part ) and optimal installation conditions. L = ( ) C 0 5 F 0 5 6 7 L h = ( ) L s n S 60 L = nominal service life (m) L h = nominal service life (h) C = dynamic load capacity (N) F = equivalent load (N) s = length of stroke* (m) = stroke frequency (double stroke) (min - ) n S 8 9 0 *) For s < x L (trolley length) load capacities are reduced. Please consult us.
Fz Load-dependent design Fy Z Mz Fx Y X f The specified values are maximum single loads, which My Mx are reduced when loads are combined B [mm] A [mm] M Note: If cumulative forces and moments arise in your application, please use the Linear Motion Designer LF-MGE linear guide calculation program to recalculate your design 008 0009867 A ) min (mm) F z (N) F y (N) (Nm) M z (Nm) M y (Nm) v max (m/s) M max (Nm) LF6S 75 850 00.6 0.7 x A ) 0. x A ) 5 9.5 / 7.6 LF6C 75 850 00 0. x B ) 0.7 x A ) 0. x A ) 5 5 LFS 90 000 500 78.0.7 x A ).0 x A ) 5 0 LFC 90 000 500.0 x B ).7 x A ).0 x A ) 5 0 LF0S 5 6000 500 0.0 6. x A ).0 x A ) 5 00 LF0C 5 6000 500.0 x B ) 6. x A ).0 x A ) 5 00 ) A min = Length guide bearing ) Note: A (spacing of support wheels, see image above) in mm; see below for calculation of B (track width) Do not exceed the maximum loading of the screwed connections for rails, carriages and fastenings. Take account of the general service life of lubricants Calculating the track width B B = P + x a B = track width (in mm) P = profile width (in mm) a = rod spacing (in mm) LF C Values for a: for LF6C = 9.5 mm (p. -) 90x80 P a for LFC = 9 mm (p. -5) for LF0C = 5 mm (p. -9) LF S Values for a: for LF6S = 0 6 = mm (p. -8) for LFS = 90 = 78 mm (p. -8) for LF0S = 00 0 = 80 mm (p. -) B (mm) 079
M z Z Calculation of load on bearing for a trolley Y X M y Loading of the individual rollers 0987 Roller : Roller : F y M z = -( ) - ( ) A F z B [mm] M y F a = ( ) + ( ) - ( ) B A Roller : Roller : F y M z = +( ) + ( ) A F z A [mm] M y F a = ( ) - ( ) - ( ) B A ff A mm: spacing of the rollers' rotational axes ff B mm: center-to-center distance between the guide rods ff Only compressive forces can be transferred between rollers and guide rods in the radial direction. The following therefore applies in respect of radial forces: 0: = 0 ff Rollers can be loaded equally in both directions. Therefore force F a is sufficient for calculating P and P 0 5 6 7 8 F y M z = -( ) + ( ) A F z M y F a = ( ) + ( ) + ( ) B A F y M z = +( ) - ( ) F z M y F a = ( ) - ( ) + ( ) B A 9 0 Table : Load factors for rollers Load case: F a Load case: < F a x y x 0 y 0 x y x 0 y 0 LF6...5 0.5.6.7 LF.. 5. 0.5.7 5. LF0..9 0.5.5. 5 Table : Rollers - load capacities C for 0 5 m (N) LF6 670 80 LF 800 5000 LF0 00 6600 C o (N) Equivalent dynamic and static loads To calculate the service life of a cam roller guide, the roller with the greatest load has to be considered. The following must be determined: P = max (P,...,P ) P 0 = max (P 0,...,P 0 ) Equivalent dynamic load P P = x + y F a Equivalent static load P 0 P 0 = x 0 + y 0 F a (N): radial load of the roller The following applies: 0: = 0 F a (N): axial load of the roller x, x 0 : radial factor (Table ) y, y 0 : axial factor (Table ) 5 6 7 8 9 0 C: dynamic load capacity (Table ) C 0 : static load capacity (Table ) Static safety values: S 0 recommended C 0 S 0 = ( ) P 0
x Z Size selection Determining the forces and moments which arise Example: F xdyn, F ydyn, F zdyn ) m = 0 kg a =.5 m/s ; dyn, M ydn, M zdyn L = 50 mm L = 5000 mm ) F Z dyn includes the weight of the trolley g = 9.8 m/s All the influences must be considered, e.g.: F z dyn = m g = 9 N M net mass and load acceleration forces and moments M y dyn = m a 70 mm = 5.5 Nm 0009867 process forces and moments loads caused by dampers and/or stops m x a m x g 70 75 00077 00689 Y Fy Y Fz Z Mz X Fx Determining the permissible forces and moments Example of LF 6 S, LF 6 C F z dyn zul = 850 N > F z dyn = 9 N M y dyn zul = 0. A = 0. 75 mm = 0 Nm > M y dyn = 5.5 Nm B [mm] My Mx A [mm] 008 Selection of the suitable size F x, y, z dyn < F x, y, z dyn zul, y, z dyn <, y, z dyn zul If F and M are cumulative: recalculate using linear guide calculation program. Note: When using the program, account must be taken of the directions of the respective forces and moments, i.e. the plus/minus sign, in order for the program to be able to calculate correctly. Example: F z dyn = 9 N < F z dyn zul = 850 N M y dyn = 5,5 Nm < M y dyn zul = 0 Nm You can order the CD directly from our media catalog, number 8 50 900. www.boschrexroth.com/mediadirectory
Calculating the drive m x g m x a 00078 M Peripheral force = m a + m g + F 0 + µ m g zul F 0 : frictional force on return units μ : coefficient of friction of the guide see table below. 5 6 M D 0 m x a m x g Alternative calculation formula = m a + F 0 + µ m g zul Example of LF6C m = 0 kg; a =.5 m/s = (0.5)N + 0 N + (0.05 0 9.8)N = 75 N + 0 N + 0.05 9 N = 9.5 N < zul = 600 N 7 8 9 00078 0 Fy Y Fz Z Mz X Fx Required drive torque M M = ½ D 0 M zul Example: M = ½ 50.9 mm 9.5 N =. Nm M =.5 Nm M zul = 5 Nm B [mm] My Mx A [mm] 008 5 6 v (m/s) zul (N) M zul (Nm) F 0 (N) µ D 0 (mm) D 0 π (mm) LF6S.0 500 9.5 0 0.05 8. 0 LF6S.0... 5.0 00 7.6 0 0.05 8. 0 LF6C 5.0 600 5.0 0 0.05 50.9 60 LFS 5.0 80 0.0 0 0.00 7.0 0 LFC 5.0 80 0.0 0 0.00 7.0 0 LF0S 5.0 000 00.0 5 0.05 0.86 0 LF0C 5.0 000 00.0 5 0.05 0.86 0 F 0 : frictional force on return units; µ : coefficient of friction 7 8 9 0
Checking the selected size Δl Preload force F v of toothed belt 0.5 F v Recommendation: F v = 0.5 Example: = 9.5 N F v = 50 N F v x Fv F v 00078 max F F 00076 Maximum toothed belt force F max F max = F v + 0.5 F zul F min = F v 0.5 > 0 Example of LF6C F max = 50 N + 0.5 9.5 N F max = 96.8 N < F zul = 900 N F min = 50 N 0.5 9.5 N F min =.85 N > 0 F F v F 00079 F v x Fv Δl Required preload length l l = 0.5 F v L / C spez l max Example: L = L L + 00 mm L = 0 50 mm l = 0.5 50 N 0 50 mm / 0 000 N l = 0.8 mm < l max = mm F v 00078 F zul (N) C spez (N) l max (mm) L (mm) Page LF6S 750 5000 L L + 60-6 LF6C 900 0000 L L + 00 - LFS 0 50000 6 L L + 60-6 LFC 0 50000 6 L L + 60 - LF0S 000 870000 L L + 780-0 LF0C 000 870000 L L + 780-7 C spez : Belt stiffness; L: Long toothed belts, see also LF6, LF6C, LFS, LFC, LF0S, LF0C
Dimensioning of synchronous shafts LF6 LF LF0 Support separation SW horizontal 5 000 mm 75 500 mm 5 500 mm vertical 5 000 mm 75 000 mm 5 00 mm Nominal torque max. 0 Nm max. 60 Nm max. 50 Nm Permitted rpm diagram max. 0.7 nk diagram max. 0.7 nk diagram max. 0.7 nk Elasticity axial max. mm max. mm max. mm lateral 0009886 diagram diagram diagram torsion angle diagram 5 diagram 5 diagram 5 5 6 7 8 9 Diagram : Torsion stiffness 0 (Nm/rad) 0000 5000 LF6C/S LFC/S LF0C/S 5 0000 6 7 5000 8 9 0000 0 5000 ) Long synchronous shaft LGW 00056 0 000 000 000 000 5000 6000 LGW ) (mm) LF6 LGW = SW - 7 LF LGW = SW - 0 LF0 LGW = SW - 0
Diagram : Mass inertia Diagram : Critical rpm nk due to bending (0 - kgm ) 0,6 0,5 0, 0, 0, 0, 0,0 9,9 9,8 LF6C/S LFC/S LF0C/S (min - ) 000 000 000 00058 000 000 000 000 5000 6000 LGW (mm) Diagram : Lateral offset 0009886,9,8,7,6,5,,,,,0 0,9 0,8 0,7 0,6 (mm) 0 0 00 80 60 0 0 00059 000 000 000 000 5000 6000 Aligned shafts are recommended. Diagram 5: Torsion angle LGW (mm) 00057 000 000 000 000 5000 6000 LGW (mm) ( ) LF6C/S LFC/S LF0C/S 00060 000 000 000 000 5000 6000 LGW (mm)