Linear Units

Deflection Calculations
Examples of calculations of the profile deflection
Example 1 Example 2 Example 3
Type of linear unit:
WH80
Load case:
Case 1 - profile supported in both ends and
fixed at both sides.
Load to be moved by carriage:
mext = 150 kg
Distance between supports:
Lf = 600 mm
Specific unit data:
m’100 = 0,93 kg
mc = 2,75 kg
EAl = 70000 N/mm 2
Iy = 1,93 × 10 6 mm 4
b = 0,0005
Calculated values:
fh = 0,3 mm
fmax = 0,013 mm
Conclusion:
fh > fmax = deflection OK
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Type of linear unit:
M55 (MF06B)
Load case:
Case 2 - profile supported in both ends and
fixed at one side.
Load to be moved by carriage:
mext = 100 kg
Distance between supports:
Lf = 600 mm
Specific unit data:
m’100 = 0,53 kg
mc = 1,2 kg
EAl = 70000 N/mm 2
Iy = 4,59 × 10 5 mm 4
b = 0,0005
Calculated values:
fh = 0,3 mm
fmax = 0,063 mm
Conclusion:
fh > fmax = deflection OK
Type of linear unit:
WM80
Linear Units
Load case:
Case 3 - profile supported and fixed at one
end.
Load to be moved by carriage:
mext = 120 kg
Distance between supports:
Lf = 400 mm
Specific unit data:
m’100 = 1,08 kg
mc = 4,26 kg
EAl = 70000 N/mm 2
Iy = 1,85 × 10 6 mm 4
b = 0,0003
Calculated values:
fh = 0,12 mm
fmax = 0,203 mm
Conclusion:
fh > fmax = deflection not OK
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