Engineering Problem Solving...EWI-SP 0 0.2 Production Drawing 11 0B 0.08 A o.os M30xZ 32 32 1+0-0.12...

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Transcript of Engineering Problem Solving...EWI-SP 0 0.2 Production Drawing 11 0B 0.08 A o.os M30xZ 32 32 1+0-0.12...

Page 1: Engineering Problem Solving...EWI-SP 0 0.2 Production Drawing 11 0B 0.08 A o.os M30xZ 32 32 1+0-0.12 A 0.02 - x-x 3 HOLES,Ø6 EQUI-SP 20 32 M30x2.S Machine Drawing (B) FIGURE.1.104
Page 2: Engineering Problem Solving...EWI-SP 0 0.2 Production Drawing 11 0B 0.08 A o.os M30xZ 32 32 1+0-0.12 A 0.02 - x-x 3 HOLES,Ø6 EQUI-SP 20 32 M30x2.S Machine Drawing (B) FIGURE.1.104
Page 3: Engineering Problem Solving...EWI-SP 0 0.2 Production Drawing 11 0B 0.08 A o.os M30xZ 32 32 1+0-0.12 A 0.02 - x-x 3 HOLES,Ø6 EQUI-SP 20 32 M30x2.S Machine Drawing (B) FIGURE.1.104
Page 4: Engineering Problem Solving...EWI-SP 0 0.2 Production Drawing 11 0B 0.08 A o.os M30xZ 32 32 1+0-0.12 A 0.02 - x-x 3 HOLES,Ø6 EQUI-SP 20 32 M30x2.S Machine Drawing (B) FIGURE.1.104

Machine Drawing

Production Drawing

Page 5: Engineering Problem Solving...EWI-SP 0 0.2 Production Drawing 11 0B 0.08 A o.os M30xZ 32 32 1+0-0.12 A 0.02 - x-x 3 HOLES,Ø6 EQUI-SP 20 32 M30x2.S Machine Drawing (B) FIGURE.1.104

Example for tolerance in jigs and fixtures.

Page 6: Engineering Problem Solving...EWI-SP 0 0.2 Production Drawing 11 0B 0.08 A o.os M30xZ 32 32 1+0-0.12 A 0.02 - x-x 3 HOLES,Ø6 EQUI-SP 20 32 M30x2.S Machine Drawing (B) FIGURE.1.104

The fig. (A) shows the fixture to machine a slot in vertical milling machine .Machining requires work plane ‘a’ to be held parallel to locating surface ‘b’ admissible deviation being tolerance

δ1 = 0.1/100mm

The admissible deviation from parallelism of locating plane ‘b’ and supporting plane of the base member of fixture ‘c’. Setting tolerance δm = 0.01/100mm (non parallelism of table surface and horizontal plane) and δelas = 0.03/100mm therefore total tolerance , δ = δ1 – δm- δelas = (0.1-0.01-0.03)/100

Therefore, δ = 0.06/100mm

Page 7: Engineering Problem Solving...EWI-SP 0 0.2 Production Drawing 11 0B 0.08 A o.os M30xZ 32 32 1+0-0.12 A 0.02 - x-x 3 HOLES,Ø6 EQUI-SP 20 32 M30x2.S Machine Drawing (B) FIGURE.1.104

In jig drilling, fig. (B) we may determine half tolerance δ on centre distance between jig bushing by the formula,

Let δ1 – tolerance on the centre distance between the hole in work piece.

S1 & S2 – radial clearance between the change bushing

S3& S4 – maximum radial clearance between the bushings and jobbing tools.

e1 & e2 – admissible eccentricity of change bushing

e3 & e4 – admissible eccentricity of fixed bushing

Therefore total radial clearance, ΣS = S1 + S2 + S3 + S4

And total eccentricity, Σe = e1 + e2 +e3 + e4

Then total tolerance,

δ ≤ [δ1+(ΣS + Σe)]