Please finish Show 3 ways of giving tolerances,using the basic hole system.Limits for hole are: 4.750 and 4.754Limits for shaft are: 4.742 and 4.746Ist method (limits):4.7542nd method (plus and minus):-4.750+004.002nd method (plus or minus):4.752+002Dimension the distance between theholes, using bilateral tolerance (equal ±).B = 1.625T = .006Give limits with unilateral tolerances,using the basic hole system.B 1.875, A003 (positive), TH= TS = .0021.8721.8701.877.1.875Check the correct fit:This is a clearance (), transition ().interference ( ) fit.(5)Give limits with unilateral tolerances,using the ANSI tables and basic hole.system. Class FN 2 fit. Nominal size= 22.1277 2.12576.66172.127 21257 0.022.1262 2.146 APO122.125Check which: This is a clearance ( ).transition (), Interference ( ) fit.Interchangeable () or selective () assy.Max. Int.Min. Int.Give bilateral tolerance dimensions(equal t), using the basic shaft system.B = 1.750, A = .005 (positive), TH= TS = .00661/2Check the correct fit:This is a clearance (), transition ().interference ( ) fit.Give limits with unilateral tolerances,using the ANSI tables and basic hole.system. Class LT 3 fit. Nominal size =1/2Check which: This is a clearance (),transition (), interference ( ) flt.Interchangeable () or selective () assy.Max. Int.Max. Clear.DIMENSIONING

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Answered: Show 3 ways of giving tolerances, using…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Dimension the holes & shafts for the required linear tolerance. Also calculate the hole tolerance, shaft tolerance, maximum clearance, & allowance DIAMETER: 2.00 INCHES A) TYPE OF FIT: RC 3 C) DIAMETER: 10 mm TYPE OF FIT: H9/d9 Hole Tol: Shaft Tol: Max. Clearance: Allowance: Hole Tol: Shaft Tol: Max. Clearance: Allowance: Hole Tol: Shaft Tol: Max. Clearance: Allowance: Hole Tol: Shaft Tol: Max. Clearance: Allowance: DIAMETER: 1.00 INCHES TYPE OF FIT: LT 2 GOD DIAMETER: 12 mm D) TYPE OF FIT: H7/h6
Refer to the figure below. What is the MMC, LMC and geometric tolerance for the Hole and Pin: a. Hole: MMC LMC Geom. Tol. b. Pin: MMC LMC Geom. Tol. C 1.000 1 1.000 Ø.515-.540 Hole +0.010@ABC 1.000 B Ø.495-.500 Pin 0.005MABC A
2. Nominal size is 40mm, the clearance is between 24µm and 92µm, the span of the clearance fit is not bigger than 68µm. To find out: O the upper/lower limit deviation of the hole; the upper/lower limit deviation of the shaft; ® the minimum clearance and the maximum clearance; Designation of fit; Illustration of tolerance interval/zone. The answer is as follows:
A.1.4) Locational interference fits Basic hole system. Limits are in thousandths of an Limits for hole and shaft are applied algebraically to the basic size to obtain Nominal Size Range Inches To Class LN1 Standard Limits Shaft +0.45 +0.25 +0.5 Class LN2 Standard Limits Shaft +0.65 +04 +0.8 Class LN3 Standard Limits Shaft +0.75 +05 +0.9 Hole +0.25 Over Hole 0.4 Hole -0.12 *0.4 0.12 -0.24 +0.3 +0.5 +0.5 +0.3 +0.65 +0.4 +0.8 +0.5 +06 +1.2 0.8 0.24 -0.40 +0.4 +0.6 +1.0 +0.6 +0.4 +0.6 +1.1 0.40 - 0.71 +0.7 +0.7 +14 +0.4 +0.7 +1.3 +0.8 +1.6 +1.0 +10 +1.7 12 2.0 +1.4 +23 +1.6 +2.9 +20 +3.5 +25 0.71 1.19 +0.5 +1.0 +0.8 +0.8 +0.5 1,19 - 1,97 +0.6 +1.1 +1.0 +1.0 +0.6 +1.3 +0.8 +1.6 +1.0 +1.9 +1.2 +22 - 3.15 +1.2 +2.1 +1.2 1.97 +0.7 +14 +1.4 +2.5 +1.6 3.15 -4.73 +0.9 +1.4 +1.6 +2.8 +1.8 +3.2 +2.0 +3.4 +22 +3.9 25 +44 +2.8 - 7.09 +1.0 +1.6 4.73 +1.8 +42 +1.8 +30 +47 7.09 -9.85 +1.2 +1.4 +2.3 +2.0 +2.0 +3.5 +5.9 +45 *6.6 +5.0 +1.2 9.85 - 12.41 +1.4 22 +22 +2.6 +1.6 +2.8 +1.4 12.41 - 15.75 25 +2.5…
The following limits are specified in a limit system, to give a clearance fit between a hole and a shaft: Hole = 10 (+0.05 )/(-0.00 ) mm and Shaft = 10 (-0.003)/(-0.010) mm. Determine the basic size, tolerance and allowance, maximum and minimum clearances, type of fit.
If the pin in the drawing below is produced at Ø.498, what is its total positional tolerance? C 1.000 I 1.000 O.003 .500 O .007 .005 .498 Ø.515-.525 0.010MABC 1.000 K B Ø.498-.500 0.005MABC A
If the pin in the drawing below is produced at Ø.498, what is its total positional tolerance? C 1.000 Y 1.000 .005 .500 .003 O .498 .007 Ø.515-.525 0.010MABC 1.000 B Ø.498-.500 0.005MABC A
Tolerances for a hole and shaft assembly having a nominal size of 50 mm are as follows: Hole = 40 (+0.06 )/(+0.00 ) mm and Shaft = 40 (-0.07)/(-0.015) mm. Determine the basic size, tolerance and allowance, maximum and minimum clearances, type of fit.
7. This problem relates to tolerance stack-up using the absolute tolerance system, as explained in section 1-14. In the figure below, specified dimensions are as follows (all units in inches): a d 1.750 + 0.001 2.000 + 0.003 3.200 + 0.005 ?+0.010
Question 3 The figures (1&2) shows a Hole-Pattern as drawn and as inspected. fill table 1 and calculate the true position tolerance. Show your calculations. Please note: Dimensions shown is relative to the actual size As drawn 4XØ0500 -0.312 Oe.005(M A BC C Pattern-4 holes # 2 1.000 # 3 +) 1.500 +)
Draw the following graphic symbols (a) Total run-out (b) position tolerance (c) symmetry tolerance
A 30 mm diameter hole is made on a turret lathe tothe limits, 30.035 and 30.00. The followingtwo grades of shafts are used to fit in the hole:(a) φ29.955mm and 29.925mm, and (b) φ30.055mm and30.050mm.Calculate the maximum tolerance, clearance and indicatethe type of fit in each case by a sketch.ii) Suggest suitable fits and their letter and tolerancegrades for the components

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Show 3 ways of giving tolerances, using the basic hole system. Limits for hole are: 4.750 and 4.754 Limits for shaft are: 4.742 and 4.746 Ist method (limits): 4.754 2nd method (plus and minus): A -4.750 +004 -.00 2nd method (plus or minus): 4.752 +002