ISO manufacturing tolerancing: three-dimensional transfer with analysis line method

Abstract

Functional tolerancing described in definition drawing of mechanical parts constitutes a contract to be respected by manufacturers. Process engineers have to choose process plans able to manufacture part respecting functional requirements and have to determine manufacturing specifications for each phase. Tolerance zone transfer method offers a three-dimensional algorithm of manufacturing specifications generation with International Organization for Standardization (ISO) standards. Analysis line method, developed in this article, establishes the calculus relation of the results of the tolerance chains according to the tolerances of manufacturing specifications to allow the tolerance synthesis. In this paper, the analysis line method is presented using an example. The aim is to show the hypotheses made during transfers in the context of ISO standards of tolerancing and to define accurately the datum reference frames used and deviations between these frames at particular points named analysis points.

Appendix: Setting-up on three perpendicular planes

This appendix depicts a particular case of general relations established in Section 3.4, with three planes perpendicular each one to the others:

$$ d\left( {F,{\mathbf{f}}} \right) = {\text{Kp}}.d\left( {P.{\mathbf{p}}} \right) + {\text{Ks}}.d\left( {{\text{S}}.{\mathbf{s}}} \right) + {\text{Kt}}.d\left( {T.{\mathbf{t}}} \right);{\text{Kt}} = \frac{\text{fx}}{{\text{tx}}}{\text{; Ks}} = \frac{\text{fy}}{{\text{sy}}}{\text{; Kp}} = \frac{\text{fz}}{{\text{pz}}} $$

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With:

$$ \begin{gathered} {\text{Xs}} = {\text{Xf}}\left( {{\text{Yt}} - {\text{Yf}}} \right).{\text{fx}}/{\text{fy}} \hfill \cr {\text{Ys}} = Y{1} + \left( {Y{2} - Y{1}} \right).\left( {{\text{Xs}} - X{1}} \right)/\left( {X{2} - X{1}} \right) \hfill \cr {\text{Zs}} = Z{1} + \left( {Z{2} - Z{1}} \right).\left( {{\text{Xs}} - X{1}} \right)/\left( {X{2} - X{1}} \right) \hfill \cr {\text{Xp}} = {\text{Xf}} + \left( {{\text{Zt}} - {\text{Zf}}} \right).{\text{fx }}/{\text{fz}} \hfill \cr {\text{Yp}} = {\text{Yf}} + \left( {{\text{Zs}} - {\text{Zf}}} \right).{\text{fy}}/{\text{fz}} \hfill \cr {\text{Zp}} = 0 \hfill \cr \end{gathered} $$

1.1 Analysis line parallel to the secondary plane

If fy = 0 and fz # 0, The analysis line is parallel to the secondary plane, but is nonperpendicular to the tertiary plane. The position of the secondary plane has no effect on F point. Only its orientation around z has influence. P point can be used to void the effect of rotations in x and y.

$$ {\text{Xp}}\,{ = }\,{\text{Xf}}\,{ + }\,\left( {{\text{Zt}}--{\text{Zf}}} \right).{\text{fx/fz;}}\,{\text{Yp}}\,{ = }\,{\text{Yf;}}\,{\text{Zp}}\,{ = }\,{0} $$

Then the relation will be:

$$ \matrix{<!--{*{20}{c}}--> {d\left( {F,{\mathbf{f}}} \right) = {\text{Kp}}{.}d\left( {P{,}{\mathbf{p}}} \right) + \left| {\text{Lz}} \right| \times a\left( {{\text{secondary}}\,{\text{plane,}}\,{\mathbf{r}}} \right) + {\text{Kt}}{.}d\left( {T{,}{\mathbf{t}}} \right)} \hfill \cr {{\text{with: Kt}} = \frac{\text{fx}}{{\text{tx}}}{;}\,{\text{Kp}} = \frac{\text{fz}}{{\text{pz}}}{\text{; Lz}} = \left( {{\text{Yt}} - {\text{Yf}}} \right){\text{.fx;}}} \hfill \cr } $$

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• The vector t = ±x is chosen so that tx will have the same sign as fx.

• The vector p = ±z is chosen so that pz will have the same sing as fz.

• If Lz > 0, r = z. If Lz < 0, r = −z.

1.2 Analysis line parallel to the primary plane

If fz = 0 and fy # 0, The analysis line is parallel to the primary plane. The primary plane only orientates the part.

$$ \begin{gathered} {\text{Xs}}\,{ = }\,{\text{Xf}} + \left( {{\text{Yt}} - {\text{Yf}}} \right){\text{.fx/fy}} \hfill \cr {\text{Ys}} = Y{1} + \left( {Y{2} - Y{1}} \right).\left( {{\text{Xs}} - X{1}} \right)/\left( {X{2} - X{1}} \right) \hfill \cr {\text{Zs}} = Z{1} + \left( {Z{2} - Z{1}} \right).\left( {{\text{Xs}} - X{1}} \right)/\left( {X{2} - X{1}} \right) \hfill \cr \end{gathered} $$

The relation becomes:

$$ \begin{gathered} d\left( {F,{\mathbf{f}}} \right) = {\text{Kt}}.{ }d\left( {T,{\mathbf{t}}} \right) + {\text{ Ks}}.{ }d\left( {S,{\mathbf{s}}} \right) + \left( {a.{\text{Lx}} + b.{\text{Ly}}} \right) \hfill \cr {\text{with: Kt}} = \frac{\text{fx}}{{\text{tx}}}{\text{; Ks}} = \frac{\text{fy}}{{\text{sy}}};{\text{ Lx}} = \left( {{\text{Zs}} - {\text{Zf}}} \right).{\text{fy}};{\text{Ly}} = \left( {{\text{Zf}} - {\text{Zt}}} \right).{\text{fx }} \hfill \cr \end{gathered} $$

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The coefficients Kt and Ks must be positive because the defect influences only add up. This rule allows us to orientate the vectors s and t:

• The vector t = ±x is chosen so that tx will have the same sign as fx.

• The vector s = ±y is chosen so that sy will have the same sign than fy.

α is th rotation around x and β the one around y.

$$ \begin{gathered} d\left( {F,{\mathbf{f}}} \right){ } = {\text{ Kt}}.{ }d\left( {T,{\mathbf{t}}} \right) + {\text{ Ks}}.{ }d\left( {{\text{S}},{\mathbf{s}}} \right){ } + { }\left| {\text{Lx}} \right|.a\left( {{\text{primary plane}},{\mathbf{r1}}} \right){ } + { }\left| {\text{Ly}} \right|.a\left( {{\text{primary plane}},{\mathbf{r2}}} \right) \hfill \cr {\text{If Lx}} > 0,{ }r{1} = x.{\text{ If Lx}} < 0,{ }r{1} = - x. \hfill \cr {\text{If Ly}} > 0,{ }r{2} = y.{\text{ If Ly}} < 0,{ }r{2} = - y. \hfill \cr \end{gathered} $$

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To study the maximum influence of the plane angle, the direction r of the resultant is determined:

$$ {\mathbf{r}} = \frac{{{\text{Lx}}{.}{\mathbf{x}} + {\text{Ly}}{.}{\mathbf{y}}}}{{\left\| {{\text{Lx}}{.}{\mathbf{x}} + {\text{Ly}}{.}{\mathbf{y}}} \right\|}} $$

The angular displacement of this plane around r is noted a (primary plane, r).

$$ d\left( {F,{\mathbf{f}}} \right) = {\text{Kt}}.{ }d\left( {T,{\mathbf{t}}} \right) + {\text{Ks}}.{ }d\left( {S,{\mathbf{s}}} \right) + {\text{Lr}}.a\left( {{\text{primary plane}},{\mathbf{r}}} \right)\,{\text{with }}Lr = \sqrt {{{\text{L}}{{\text{x}}^2} + {\text{L}}{{\text{y}}^2}}} $$

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1.3 Analysis line perpendicular to the tertiary plane

If fy = 0 and fz = 0, the analysis line is perpendicular to the tertiary plane. In this case, the position is given by the tertiary point. The relation is:

$$ \matrix{<!--{*{20}{c}}--> {d\left( {F,{\mathbf{f}}} \right) = \left| {\text{Lz}} \right|.a\left( {{\text{primary plane}},{\mathbf{r}}1} \right) + \left| {\text{Ly}} \right|.a\left( {{\text{primary plane}},{\mathbf{r}}2} \right) + {\text{kt}}{.}d\left( {T{,}{\mathbf{t}}} \right)} \hfill \cr {{\text{with}}\,{\text{Kt}} = {\text{fx}}/{\text{tx}};{\text{Ly}} = \left( {{\text{Yt}} - {\text{Yf}}} \right).{\text{fx}};{\text{Lz}} = \left( {{\text{Zf}} - {\text{Zt}}} \right).{\text{fx}}} \hfill \cr } $$

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• The vector t = ±x is chosen so that tx will have the same sign as fx.

• If Lz > 0, r1 = y. If Lz < 0, r1 = −y.

• If Ly > 0, r2 = z. If Ly < 0, r2 = −z