DIMENSIONAL ANALYSIS OF CONSTRUCTIVE UNITS (Links like position deviations of surfaces) Lecturer Egor Efremenkov Tomsk - 2015.

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DIMENSIONAL ANALYSIS OF CONSTRUCTIVE UNITS (Links like position deviations of surfaces) Lecturer Egor Efremenkov Tomsk

Dimension chains may be calculated with links like position deviations of surfaces following types Position deviations  for details  for assemblies

Detail position deviations Identification mark of position tolerances Item of technical requirements Recording form of position deviation and it’s identification mark on dimension chain diagram Tolerance of surface parallelism Б relative A 0,05 mm Perpendicularity tolerance of hole axis A relative hole axis Б 0,04 mm Concentricity tolerance of hole Б relative hole A 0,08 mm

There are two types of dimension chains with closing links А  due to neutral attitude of axis 2 relative axis 1 a) Determination diagram of concentricity deviations Detail position deviations b) So, it is possible that concentricity deviation may be as increasing link as decreasing link for the same dimension chain

Expressions for maximum limit if closing size (dimension) if Е2-1 is decreasing link Detail position deviations if Е2-1 is increasing link

Allow that concentricity deviation of stage 2 relative stage 1 Е2-1 = 0±0,1 mm and concentricity deviation of stage 1 relative axis 3 of center holes Е1-3 = 0±0,05 mm. Detail position deviations For example let’s consider the pin And concentricity deviation of stage 2 relative axis 3 of center holes which is not depicted oh figure. This concentricity deviation is a closing link Е2 - 3 (for dimension chain b)

Tolerance of closing link (previous figure) Detail position deviations Closing link may be wrote like So tolerance of surface (axis) position deviations take into account direction of its deviations. This tolerance more than position tolerance (it is showed on figure) in twice.

It’s show irreversibility of parallelism deviations At the meantime values of deviations are independent of bases choosing. It may be achieved following ways First way is transitioning to specific deviations (tolerance per 1 length); Second way is transitioning to deviations are attribution in common length. Detail position deviations The figure shows that parallelism deviations Р1–2 and Р2–1 is not equal together if these is measured in absolute meanings.

First way After calculation should make back transitioning Second way Detail position deviations For example parallelism deviations are 0 ± 0,02 mm on the length l1 =100 mm; 0±0,05 mm on the length l2 =300 mm; 0±0,03 mm on the length l3 = 200 mm.

Second way Detail position deviations After calculation should make back transitioning to length given l 1, l 2 and l 3

From operational assignment of the machine closing link is parallelism deviation of axis spindle 1 relative table surface 6 [P1-6] = ±0,03 mm. Assembly position deviations Scheme of universal milling machine It should be found position deviations for formative links of machine dimension chain. Closing link is measured like [Р1-6]= ±0,03 мм on the length 300 mm

There are formative links for the machine  squareness deviation of axis spindle 1 relative bad guideways 2 – N1-2  squareness deviation of knee guideways 3 relative bed guideways 2 – N3-2  parallelism deviations of plane 4 relative knee guideways 3 Р4 – 3  parallelism deviations of plane 5 relative plane 4 – Р5 - 4  parallelism deviations of plane 6 relative plane 6 – Р6 - 5 Assembly position deviations

Dimension chain of the machine is showed on the figure Assembly position deviations For simplicity allow that all position deviation should be located on the length given. for example 300 mm. Closing link tolerance of dimension chain is Т 1-6 = Т Т Т Т Т 6-5. Closing link is calculated by Full Interchangeability Method

Take into account hardness of production and mounting for several details then position tolerance are picked Assembly position deviations For formative links position deviations are Т 1-2 = 0,016 (mm); Т 3-2 = 0,014 (mm); Т 4-3 = Т 5-4 = Т 6-5 = 0,01 (mm). N1 - 2 = 0 ± 0,008 mm; N = 0 ± 0,007 mm; Р4-3 = Р5-4 = Р6-5 = 0±0,005 mm.

If Probability Method is used the closing link tolerance is evaluated via tolerances of formative links Assembly position deviations Let’s taking the relative standard deviation like that correspond Simpson’s law with P=1%, t  =2,57 Т 1-2 = 0,03 mm; Т 3-2 = 0,03 mm; Т 4-3 = 0,024 mm; Т 5-4 = 0,024 mm; Т 6-5 = 0,02 mm.

So position deviations for formative links Assembly position deviations N1 - 2 = 0 ± 0,015 (mm); N = 0 ± 0,015 (mm); P4 - 3 = 0 ± 0,012 (mm); P = 0 ± 0,012 (mm); P6 - 5 = 0 ± 0,01 (mm).

Thanks for Your attention