# Tolerances in a pin-hole assembly

Hi guys, can someone explain how to do this question? This question is under Geometric Dimensioning and Tolerances and I am clueless on how to tackle problem like this.

EDIT This is the explanation given in textbook regarding MMC and LMC situation.

• figure out the envelopes within which the pins may reside. then place the most extreme case of pin or hole in the most extreme location within the envelope to suit what each of the four ask.
– Abel
Apr 23 at 16:08
• LMC tells us holes are 46.004 and pins are 45.996, but position tolerances vs the 235 basic dimension are not given. Does the question assume 0 position tolerance at MMC (which becomes +-0.004 at LMC) ?? Apr 23 at 19:17

$$L$$ = center line distance of the pin-hole assembly, = 235

$$D_H$$ = diameter of the hole, = 46

$$d_H$$ = tolerance of the hole, = +0.004, -0.000

$$D_P$$ = diameter of the pin, = 46

$$d_P$$ = tolerance of the pin, = +0.000, -0.004

$$A = L + 2(D_H/2 + d_H/2) = L + D_H + d_H = 235 + 46 + 0.004$$

$$B = L - 2(D_H/2 + d_H/2) = L - D_H - d_H = 235 - 46 - 0.000$$

$$C = L = 235$$

$$D = +d_H/2 - d_P/2 = 0.004/2 - (-0.004/2) = 0.004$$ (with respect to the center line, the hole over bored +0.004/2, and the pin shrink -0.004/2)

• For D) the pin would be -0.004 per the LMC Apr 23 at 23:22
• @Jonathan You are correct. See correction made above.
– r13
Apr 24 at 0:01
• I found this explanation on the textbook, please refer to the edited question Apr 24 at 9:24
• @Yuangi Interesting. Thanks for the update.
– r13
Apr 24 at 11:24
• @Yaunqi Chau It would be a good answer to your own question if you can work out the dimensions as asked in the original question and post it to conclude this post. Thanks.
– r13
Sep 21 at 16:49