Question

Laboratory activities: Figure 1 shows the drawing and specifications of a simple workpiece with two features of size (the small hole and the larger one) on it. This workpiece is to be produced in large quantities in the FAME area. 3101.003 .00B А в с 4371.002 0.008 А в с q 900.008 1.400.000 8 030 800 2.002 .030 Figure 1
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Answer 1

PAGE-T * Solution..) Considering the conditions with peleneters & figure from the question. 310003 DOM ABC 437 002 9.00 А в с 900.000 1.400 1.008 030 500 :.002 8 030 Figure 1 Table 1: Measurement Data WORKPIECE NUMBER SMALL HOLE Center Coordinates Diameter Х Y LARGE HOLE Center Coordinates Diameter Х Y .604 1 2 3 4 .312 .307 .311 .313 .406 .403 .395 396 .402 403 .402 .395 439 .437 .438 435 .895 .902 .903 .898 .601 599 .598

PAGE-2 these medusement and carry out necessay compilation, Here we study From figure Smell. hole size=4-3103 0.003. = 0.5 •313 T High limit 307 –) Low limit Loige inde size = 8.64374 0.002 =$3 0.435€ High lin't 0.435 low linit Length of wosk piece = 1.400 10.08 Height of jp = 0.900 + 0.008 Constraint Constraint A - work piece SMALL HOLE LAGE HOLE numbel Dierser lositive Dionetes Position Bi Tol Content Gristsent A A A А А. А 2 А A 3 A A 4. A A A

PAGE 3 work piece numbel-1 A Small dienetel Hole = 0.3/2 (Pie constraint A) Lage Hole diameter: 0.438 (Accepted) For position tol. constrant Y co ordinates tol. to.002 y conosdinates tol Smell Hole --> +0.006 Large Hole 0.005 toooos +0.006 +0.001 constein Accepted. Se position toll *WORK Piece No-2 Y diameter А. Smell Hole dia=0.37 +0.003 +0.003 -0.00Z tooool 6.005 Ldge Hole dio =0.437 so position tol Arrepted f 40.oo! & WORK PIECE NO 3: A -0.005 +0.002 SMALL HOLE - CR -0.3V -0.001 LARGE HOUS dio -0.538 A +0.003 0.002 +0001 foi position tol. constraint Accepted.

PAGE 4 WORK PIECENO 4. -Oooos SMALL HOLE diox-313 Accepted -0.00 LARGE HOLE dio = 0.435 Accepted -ovoz -0.002 0.006 0.009 So position toi constraint. Accepted I PLEASE UPVOTE vely * Its mich important for met