LM Guide THK - Industrial Technologies

Индастриал Технолоджис www.industrial‐technologies.com.ua +38 044 498 21 94Point of SelectionCalculating the Applied Load[Example of calculation]When one LM block is usedModel No.: SSR20XV1Gravitational acceleration g=9.8 (m/s 2 )Mass m=10 (kg)l 1=200mml 2=100mmFig.6 When One LM Block is UsedNo.1 P1=mg+KAR1•mg•l 1+KCR•mg•l 2=98+0.27598200+0.12998100=6752 (N)No.2 P2=mg– KAL1•mg•l 1+KCR•mg•l 2=98– 0.13798200+0.12998100=– 1323 (N)No.3 P3=mg– KAL1•mg•l 1– KCL•mg•l 2=98– 0.13798200– 0.064498100=– 3218 (N)No.4 P4=mg+KAR1•mg•l 1– KCL•mg•l 2=98+0.27598200– 0.064498100=4857 (N)When two LM blocks are used in close contact with each otherModel No.: SNS30R2Gravitational acceleration g=9.8 (m/s 2 )Mass m=5 (kg)l 1=200mml 2=150mmNo.1 P1=No.2 P2=No.3 P3=No.4 P4=mg2mg2mg2mg2No.3No.2No.3No.2No.4No.1l 1 l 2Fig.7 When Two LM Blocks are Used in Close Contact with Each Other+KAR2•mg•l 1+KCR•–KAL2•mg•l 1+KCR•–KAL2•mg•l 1–KCL•+KAR2•mg•l 1–KCL•mg•l 22mg•l 22mg•l 22mg•l 22====492492492492Note1) Since an LM Guide used in vertical installation receives only a moment load, there is no need to apply a load force(mg).Note2) In some models, load ratings differ depending on the direction of the applied load. With such a model, calculate anequivalent load in the direction of the smallest load rating.No.4No.1mmml 1 l 2+0.01849200+0.0842 491502–0.015149200+0.0842 491502–0.015149200–0.0707 491502+0.01849200–0.0707 491502m=510.3 (N)=186 (N)=–383.3 (N)=–58.9 (N)LM GuideIndustrial TechnologiesA1-63