For the dimensioned pin assembly shown below, specify the allowable perpendicularity tolerance for each of the actual pin sizes specified.
ger diameter of the stepped pin is f9.97
0.010 mm. Concentricity can be held to within 0.002 mm for both the stepped pin and the counterbored hole. Assume that concentricity is the only important feature other than the tolerances on the diameters (e.g. negligible straightness, perpendicularity, etc.). For each case below, determine the diameter and tolerance of the smaller diameter of the stepped pin that will permit the parts to mate as specified. Use bilateral, equally divided tolerances on the dimensions. If the required dimensions and tolerances cannot be found, clearly explain why...
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1. At an actual size of 20.07, the pin is at its maximum material condition. The allowable perpendicularity is thus specified as 0.02.
2. At an actual size of 20.05, the pin is 0.02 less than its maximum material condition. Thus this amount may be added to the allowable perpendicularity. The perpendicularity allowance is now 0.04.
3. At an actual size of 20.03, the pin is 0.04 less than its maximum material condition. Thus this amount may be added to the allowable perpendicularity. The perpendicularity allowance is now 0.06.
producing a line-to-line fit. Thus (5.00 - 0.02) = 4.98.
4. The "0.06 MAX" on the drawing specifies that the maximum perpendicularity allowance can be no greater than 0.06. Thus for all actual sizes less than 20.03, the perpendicularity allowance is always 0.06.
table
Actual Size Perpendicularity Allowance
20.07 0.02
20.05 0.04
20.03 0.06
20.01 0.06
19.09 0.06
19.07 0.06
19.05 0.06
19.03 0.06
n and bores to show the above dimensions,
tolerances, and control features so the parts will fit with the desired clearance and interference.
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Example 8.7
Assume that the 8 pin assembly shown below is an existing part. The pins each have a size f8.000
0.010 mm and are located as shown. You are to design the mating part with the same number holes as pins. The holes have a size f8.200
0.020 mm. The parts must fit together, with clearance, in any rotational orientation. Specify the locations of the holes on the mating part, in such a way that the maximum allowable tolerances are produced, which would allow the two parts to fit together. Assume that only size and location tolerances need to be considered.
1. For clearance problems, consider the largest pin, f8.010, and the smallest hole, f8.180. Not yet taking into consideration the location tolerances, the minimum diametrical clearance between the holes and the pins is 0.170. The minimum radial clearance, which is 1/2 the minimum diametrical clearance, is 0.085.
2. Now consider the location tolerances of the pins and holes. To achieve the largest tolerances and still have the parts fit, true position tolerancing and the maximum material conditions will be used.
3. The pin assembly was toleranced using conventional tolerancing. The location of each pin may vary from the nominal location by 0.025. This target area is represented by a 0.050 square.
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4. The greatest distance the pin center can be from the nominal location is at any corner of the square target area. This displacement from the nominal location is 0.035, which is along the diagonal of the square.
5. The maximum radial pin displacement is subtracted from the minimum radial clearance between the pins and the holes. The remainder is the location tolerance of the holes. Thus the location tolerance for the holes about their true positions is (0.085) - (0.035) = 0.050.
6. The appropriate symbols are added to the drawing.m
7. The maximum material condition symbol is added to the hole location tolerance. With this condition, as the actual size of the hole gets larger, the hole location tolerance correspondingly increases.
8. Datums are added to ensure that measurements can be made accurately and consistently.k
step7
step8
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Example 8.3
The size of the pin in the part shown is 14.75
0.005 mm. Determine the size of the hole such that the diametric interference between the two parts is between 0.010 and 0.030 mm. Assume that only the sizes of the parts need to considered. If the parts were to be assembled by thermal conditioning, to what temperature difference would the hole need to be heated, or the pin cooled, to ensure that the parts always assemble with clearance??
1. Consider the case where the diametric interference is the largest. This occurs when the hole is the smallest, and the pin is the largest. Thus (largest pin diameter) - (smallest hole diameter) = (largest diametric interference). Rewritten, the smallest hole diameter is (largest pin diameter) - (largest diametric interference) = (14.755) - (0.030) = (14.725).
2. Next, consider the case where the diametric interference is the smallest. This occurs when the hole is the largest and the pin is the smallest. Thus (smallest pin diameter) - (largest hole diameter) = (smallest diametric interference). Rewritten, the largest hole diameter is (smallest pin diameter) - (smallest diametric interference) = (14.745) - (0.010) = (14.735).
3. The limit sizes of the hole are thus 14.725 to 14.735 mm. This can be rewritten with bilateral and equally distributed tolerances as 14.73
0.005.
backward
audioon
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Example 8.2
The size of the hole in the part shown is 23.50
0.010 mm. Determine the size of the pin such that the diametric clearance between the two parts is between 0.015 and 0.045 mm. Assume that only the sizes of the parts need to considered.
1. Consider the case where the diametric clearance is the smallest. This occurs when the hole is the smallest, and the pin is the largest. Thus (smallest hole diameter) - (largest pin diameter) = (smallest diametric clearance). Rewritten, the largest pin diameter is (smallest hole diameter) - (smallest diametric clearance) = (23.490) - (0.015) = (23.475)
2. Next, consider the case where the diametric clearance is the largest. This occurs when the hole is the largest and the pin is the smallest. Thus (largest hole diameter) - (smallest pin diameter) = (largest diametric clearance). Rewritten, the smallest pin diameter is (largest hole diameter) - (largest diametric clearance) = (23.510) - (0.045) = (23.465)
3. The limit sizes of the pin are thus 23.465 to 23.475 mm. This can be rewritten with bilateral and equally distributed tolerances as 23.47
0.005.der the case where the diametric clearance is the smallest. This occurs when the hole is the smallest, and the pin is the largest. Thus (smallest hole diameter) - (largest pin diameter) = (smallest diametric clearance). Rewritten, the largest pin diameter is (smallest hole diameter) - (smallest diametric clearance) = (23.490) - (0.015) = (23.475)1. Consider the case where the diametric clearance is the smallest. This occurs when the hole is the smallest, and the pin is the largest. Thus (smallest hole diameter) - (largest pin diameter) = (smallest diametric clearance). Rewritten, the largest pin diameter is (smallest hole diameter) - (smallest diametric clearance) = (23.490) - (0.015) = (23.475)1. Consider the case where the diametric clearance is the smallest. This occurs when the hole is the smallest, and the pin is the largest. Thus (smallest hole diameter) - (largest pin diameter) = (smallest diametric clearance). Rewritten, the largest pin diameter is (smallest hole diameter) - (smallest diametric clearance) = (23.490) - (0.015) = (23.475)1. Consider the case where the diametric clearance is the smallest. This occurs when the hole is the smallest, and the pin is the largest. Thus (smallest hole diameter) - (largest pin diameter) = (smallest diametric clearance). Rewritten, the largest pin diameter is (smallest hole diameter) - (smallest diametric clearance) = (23.490) - (0.015) = (23.475)1. Consider the case where the diametric clearance is the smallest. This occurs when the hole is the smallest, and the pin is the largest. Thus (smallest hole diameter) - (largest pin diameter) = (smallest diametric clearance). Rewritten, the largest pin diameter is (smallest hole diameter) - (smallest diametric clearance) = (23.490) - (0.015) = (23.475)1. Consider the case where the diametric clearance is the smallest. This occurs when the hole is the smallest, and the pin is the largest. Thus (smallest hole diameter) - (largest pin diameter) = (smallest diametric clearance). Rewritten, the largest pin diameter is (smallest hole diameter) - (smallest diametric clearance) = (23.490) - (0.015) = (23.475)0.005
backward
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Example 8.1
The part drawing below has been removed from its header. Still there at least five errors or items that are considered poor practice in the dimensioning of the object. Find these items and note, in your mind, how these items should be corrected. Portions of the drawing may be enlarged by clicking on that area.
Tolerance is larger than the feature size on the 5 mm thickness dimension.
Tolerance on radius dimension is missing.
Datum C is not defined for the position of the larger hole.
A location tolerance appears on the larger hole, even though conventional tolerancing is already used.
A location tolerance does not appear on the smaller holes, where true-position tolerancing is used.
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Portions of the drawing may be enlarged by clicking on that area..
Here we show how a shaft is toleranced by conventional methods to produce a clearance fit with a hole of arbitrary size. The shaft has a straightness tolerance which is independent of its diameter.. Notice that as the diameter of the shaft decreases, the clearance, shown in yellow, increases.
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step1
The diameter of the shaft ranges from 4.80 to 5.20.
step2
The tightest fit occurs when the shaft is at its largest allowable diameter. The straightness tolerance of 0.04 is applied to the shaft to ensure a clearance of 0.06..f 0.06.
step3
With conventional tolerancing, the straightness tolerance is independent of the diameter of the shaft. When the shaft diameter is decreased, the clearance increases.
A flatness specification on a surface does not require a reference datum surface.
answer
True.
reference
datums
nopix
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Quiz 8.5
If three different datum planes exist on a part, then they are also perpendicular to each other.
answer
Inconclusive. The plane may or may not be perpendicular to each other, depending on the desired feature orientation..
reference
3datums
nopix
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Quiz 8.4
The pin shown below cannot be dimensioned and toleranced to achieve a 0.005 to 0.030 clearance fit in the hole shown.
answer
True. The total dimensional tolerance of the hole is greater than the total tolerance of the fit.
reference
nopix
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false
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-- These scripts are where you
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It's
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-- setting
inconclusive
display
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8"misc.tbk"
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Quiz 8.3
The parts shown, as toleranced, will produce a clearance fit.
answer
Inconclusive. The fit is a transition fit. The parts will sometimes produce a clearance fit and sometimes an interference fit, depending on the actual size of the parts............
reference
nopix
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Quiz 8.2
The parts shown, as toleranced, will produce a clearance fit.
answer
False. The parts will produce an interference fit.
reference
3datums
Receptacle
Pin Mount
This is an example of non-Maximum Material Condition tolerancing on a pair of mating blocks. The task is to show that the parts will fit together with a clearance fit based on the information given in the engineering drawing. Clearances in the animation are shown as colored bands..). Clearances in teh animation are shown as colored bands.
animationstage
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animerror
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syserrornumber = 0
-- x = 0
-- y = mmendpoint
clip thisanim
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stage "animationstage" hold
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step1
According to the given specifications, the diameters of the holes are allowed to be between 5.06 and 5.10, while the diameters of the pins are within the range 4.98 to 5.02. The longitudinal separation of the pins and the holes is between 19.98 and 20.02.m
step2
To determine if the parts will always fit, we check to see if they fit under the most extreme conditions. With regard to size, we assume the holes have the smallest allowable diameter.
(nominal hole size - lower tolerance)
= minimum diameter
5.80 - 0.02 = 5.06 = 0.066.20 - 0.04 = 0.0666
step3
For the pins, we assume that they have the largest allowable diameter.
(nominal pin size - upper tolerance)
= maximum diameter
5.00 + 0.02 = 5.026 = 0.066.20 - 0.04 = 0.0666
step4
Position also factors into the worst-case scenario. In this case, we assume that the pins are separated by the greatest allowable distance.
(nominal distance - upper tolerance)
= maximum separation
20.00 + 0.02 = 20.02 0.066.20 - 0.04 = 0.0666
step5
We also assume that the holes are separated by the smallest allowable distance.
(nominal distance - lower tolerance)
= minimum separation
20.00 - 0.02 = 19.98 0.066.20 - 0.04 = 0.0666
step6
Finally, we check to see whether or not the points on the holes and pins separated by the longitudinal distance interfere.
The parts will just fit, even under worst-case conditions. Assuming the holes are as far apart as possible and that the pins are as close together as possible gives similar results..
(separation of centers + 2 x radius)
= total separation
Holes: 19.98 + 5.06 = 25.04
Pins: 20.02 + 5.02 = 25.04
hole extents - pin extents = minimum clearance
25.04 - 25.04 = 0.00000
step7
Notice that with conventional tolerancing, the tolerances are absolute. If, for example, the holes have the largest (instead of the smallest) possible diameter, the position tolerances remain the same, while clearance increases.
table
~ G"A
G"` :
table
Actual Diameter Position Tolerance
5.06 0.02
5.07 0.03
5.08 0.04
5.09 0.05
5.10 0.06
5.10 + 0.14 = 5.24
5.00 + 0.24 = 5.24
4.90 + 0.34 = 5.24
4.80 + 0.44 = 5.24
cover5
~ G"A
cover4
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cover1
animationstage
animationstage
thisAnim
animerror
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syserrornumber = 0
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-- x = 150
-- y = x + 30
mmPlay clip
stage "animationstage" hold notify
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The receptacle from the blocks on the preceding page is now toleranced. Now the problem is to determine what position tolerances are allowed for all possible hole diameters in increments of 0.01, filling in the table below. The solution will show that as the diameter of the holes increases, there is a corresponding increase in the position tolerance in order to keep the minimum clearance constant. The increased tolerance facilitates manufacturing and reduces costs.................. manufacturing and reduces costs...
step1
The worst case conditions for the pin mount still correspond to the largest allowable pin diameter
and the greatest possible pin separation.
step2
The size tolerance of the holes remains the same.
Maximum diameter: 5.08 + 0.02 = 5.10
Minimum diameter: 5.08 - 0.02 = 5.0666
step3
The position tolerance of 0.02 is applied at the MMC, where the holes are at the smallest possible diameter.
(nominal pin size - lower tolerance)
= minimum diameter
5.08 - 0.02 = 5.066 = 0.066.20 - 0.04 = 0.0666
(nominal distance - MMC tolerance + (2 x MMC radius)) = extents of holes
20.00 + 0.02 = 20.02
(extents of holes - extents of pins)
= clearance
25.04 - 25.04 = 0.00
(MMC tolerance + hole size variation)
= new position tolerance
0.02 + 0.01 = 0.03980.02 = 19.984 = 0.0666
step7
Completing the table is now straightforward. If the actual diameter of the holes is increased by 0.01, the position tolerance of the holes is also increased by 0.01.
(nominal hole size + upper tolerance)
= maximum diameter
5.00 + 0.02 = 5.02
(nominal distance + upper tolerance)
= maximum separation
20.00 + 0.02 = 20.02
Extents of pins: 20.02 + 5.02 = 25.04k
step4
At the worst-case condition, the holes are at the least possible separation. The parts now fit together with zero clearance, as before.
step5
However, if the actual diameter of the holes is 5.07 (the MMC + 0.01), then the position tolerance on the holes is increased by a corresponding amount.
step6
At worst case condition, where the holes are at the least allowed separation, the parts now fit together with zero clearance.
(nominal separation - new tolerance + 2 x hole radius) = clearance
Extents of Holes: 20.00 - 0.03 + 5.07 = 25.04
extents of holes - extents of pins =
clearance
25.04 - 25.04 = 0.00 = 0.00 0.000 - 25.04 = 0.00
table
table
Actual Diameter Position Tolerance
5.04 0.00
5.05 0.01
5.06 0.02
5.07 0.03
5.08 0.04
5.09 0.05
5.10 0.06= 5.24
4.90 + 0.34 = 5.24
4.80 + 0.44 = 5.24
cover5
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cover2
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cover7
The size tolerance on the holes may be increased by setting the lower limit on the position tolerance to zero. This example shows how to determine what position tolerances are allowed for all possible hole diameters, in increments of 0.01, for the redesigned part.
inst Datum A, then Datum B, and finally Datum C. This procedure is done to prevent any part distortion from causing the part to rock, and possibly altering the measurements.
animationstage
animationstage
thisAnim
animerror
buttonclick
buttonclick
4thisAnim, ref, x, y
syserrornumber = 0
x >= 100
-- x = 100
-- y = x + 30
mmPlay clip
stage "animationstage" hold notify
animerror
Repeat Animation
step1
The worst case conditions for the pin mount correspond to having the pins at the largest allowable diameter, and at the greatest possible separation.
and the greatest possible pin separation.
step2
The receptacle has been redesigned so that the diameter of the holes ranges from 5.04 to 5.10.
step3
At the MMC, the holes are at the least allowable diameter.
(nominal hole size - lower size tolerance)
= MMC diameter
5.07 - 0.03 = 5.0466660.066.20 - 0.04 = 0.0666
(nominal distance - MMC tolerance + (2 x MMC radius)) = extents of holes
20.00 - 0.00 + 5.04 = 25.04
(extents of holes - extents of pins)
= clearance
25.04 - 25.04 = 0.00
(nominal separation - new tolerance
+ 2 x radius) = extents of holes
20.00 - 0.01 + 5.05 = 25.04
(extents of holes - extents of pins) = clearance
25.04 - 25.04 = 0.00
(2 x pin radius + separation)
= extents of pins
Extents of Pins: 5.02 + 20.02 = 25.04
e + upper tolerance)
= maximum separation
20.00 + 0.02 = 20.02
Extents of pins: 20.02 + 5.02 = 25.045.04.04
step4
The parts fit together with zero clearance. The holes cannot be moved, and the position tolerance is zero.
step5
If the actual diameter of the holes is 5.05, then the position tolerance of the holes is increased to 0.01. The parts now fit together with zero clearance as before.
step6
The pattern is now clear. If the actual diameter of the holes increases by 0.01, then the MMC designation on the position tolerance allows this tolerance to be increased by 0.01. The rest of the table is filled in accordingly.
a) A clearance fit is required on the larger diameter of the stepped pin. The smaller diameter of the stepped pin must have 0.000 to 0.015 mm diametric clearance under any condition.
Example 8.5
Consider the mating parts below. The counterbored hole has a hole diameter of f5
0.010 mm and a counterbore diameter of f10
0.010. The larger diameter of the stepped pin is f9.97
0.010 mm. Concentricity can be held to within 0.002 mm for both the stepped pin and the counterbored hole. Assume that concentricity is the only important feature other than the tolerances on the diameters (e.g. negligible straightness, perpendicularity, etc.). For each case below, determine the diameter and tolerance of the smaller diameter of the stepped pin that will permit the parts to mate as specified. Use bilateral, equally divided tolerances on the dimensions. If the required dimensions and tolerances cannot be found, clearly explain why...
suball
a) A clearance fit is required on the larger diameter of the stepped pin. The smaller diameter of the stepped pin must have 0.000 to 0.015 mm diametric clearance under any condition.
b) A clearance fit is required on the larger diameter of the stepped pin. The smaller diameter of the stepped pin must have 0.005 to 0.030 mm diametric interference under any condition.
c) Properly label, on the figure below, both the pin and bores to show the above dimensions,
tolerances, and control features so the parts will fit with the desired clearance and interference.
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1. Notice that the tolerance of the hole is
0.010, or a total tolerance of 0.020. The total tolerance on the clearance fit is only 0.015. The tolerance of the fit is tighter than the tolerance of one of the parts. Thus, no matter how precise the mating part is, the desired fit cannot be achieved. There is no solution for this part of the problem..he problem.
2. For part b), the problem is solved by first temporarily ignoring the concentricity tolerances. They will be added later.
of the larger pin is f10.02
0.01. A = 10.03
3. For the least interference at the hole, the smallest pin must be larger than the largest hole by 0.005. Thus the smallest pin size is (5.000 + 0.010) + 0.005 = 5.015
ame as the smallest hole, producing a line-to-line fit. Thus (5.00 - 0.02) = 4.98.
4. For the most interference at the hole, the largest pin must be larger than the smallest hole by 0.030. Thus the largest pin size is (5.000 - 0.010) + 0.030 = 5.020
5. Using bilateral, equally distributed tolerances, the size of the pin at the hole is f5.0175
0.0025
6. Finally, the effect of concentricity must be checked. At the counterbore, the diametric clearance is calculated by (smallest hole) - (largest pin) = (10.00 - 0.01) - (9.97 + 0.01) = 0.01. The radial clearance is 1/2 the diametric clearance. Thus the radial clearance is 0.005. The concentricites tolerances of the pins and the bore are each 0.002, for a total radial displacement of only 0.004. Thus, even with concentricity tolerances, the part will still fit as desired.
b) A clearance fit is required on the larger diameter of the stepped pin. The smaller diameter of the stepped pin must have 0.005 to 0.030 mm diametric interference under any condition.
"Arial Narrow
TITLE
SCALE
+/- TOLERANCES
UNLESS INDICATED
DRAWN BY
APPROVED BY
DATE
HIGHER ASM
surface
quiz1
datumrel
nopix
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false
false
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False
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correct
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false
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-- These scripts are where you
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? buttons (on
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Dsame message.
It's
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display
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8"misc.tbk"
4thisanim, lastanim, thiswav, lastwav
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Quiz 8.1
For small metal parts, dimensional tolerances of 0.100 mm are easily achievable with standard milling operations.
answer
True.
reference
machine
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thisanim
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-- These scripts are where you
the answer
's quiz question
? buttons (on
/) will always
Dsame message.
It's
-- just a matter
creating
u"correct"
"incorrect"
-- setting
inconclusive
display
F one, since that
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8"misc.tbk"
4thisanim, lastanim, thiswav, lastwav
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Quiz 8.10
For the part shown, the concentricity specification on the pin will be contained within a diameter of 0.025 when the actual diameter of the pin is 4.93.
answer
True.
reference
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-- These scripts are where you
the answer
's quiz question
? buttons (on
/) will always
Dsame message.
It's
-- just a matter
creating
u"correct"
"incorrect"
-- setting
inconclusive
display
F one, since that
be chosen.
8"misc.tbk"
4thisanim, lastanim, thiswav, lastwav
audiodone
animdone
Quiz 8.9
The maximum material condition of a hole occurs when the actual size of the hole is at its largest acceptable dimension.
answer
False. MMC for a hole occurs when the hole is at is smallest acceptable dimension.
reference
nopix
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false
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thisanim
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thiswav
lastwav
enterpage
audiodone
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-- These scripts are where you
the answer
's quiz question
? buttons (on
/) will always
Dsame message.
It's
-- just a matter
creating
u"correct"
"incorrect"
-- setting
inconclusive
display
F one, since that
be chosen.
8"misc.tbk"
4thisanim, lastanim, thiswav, lastwav
audiodone
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Quiz 8.8
For parts fabricated to the tolerances shown, the concentricity specification will not permit all parts to be assembled with clearance on all surfaces on all occasions.
answer
False. Even when the concentricities are at their limit, the clearances are such that the parts will still fit together with clearance..
reference
datumrel
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false
false
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false
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thisanim
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-- These scripts are where you
the answer
's quiz question
? buttons (on
/) will always
Dsame message.
It's
-- just a matter
creating
u"correct"
"incorrect"
-- setting
inconclusive
display
F one, since that
be chosen.
8"misc.tbk"
4thisanim, lastanim, thiswav, lastwav
audiodone
animdone
Quiz 8.7
A Total Indicated Runout (TIR) specification does not require a datum surface.
answer
False. The measured surface must be rotated about a cylindrical datum axis.
True position tolerancing has an allowable error which is independent of direction. Notice that the shape outlined by the allowable location for the center of the hole is a circle, whose perimeter points are equidistant from the center. The circle circumscribes the square and is 57% larger than the square on the previous page. This technique maximizes the allowable error within a given tolerance and still allows the mating parts to fit. The increased tolerance facilitates manufacturing, and reduces costs.osts.
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Datums
Datums are reference surfaces from which measurements are made. They are identified by capital letters between dashes and framed by a rectangle. A datum is usually defined in a view where it appears in edge view. In this case, the datum is the best plane placed on the face of the box.
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This datum is the best cylinder that just fits over the shaft.
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A hole is the inverse of a shaft, and has a corresponding cylindrical datum that just fits inside the hole.
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3datums
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Multiple datums may be used on a single object. In this three plane datum system, the planes are at right angles to each other. The order of the planes is important, since the object contacts the first plane at three points, the second at two, and the third at only one point. This procedure is done to prevent any rocking due to part distortion, which could possibly alter the measurements...g the measurements.es are not required to be perpendicular to each other.....
Datum surfaces are not required to be perpendicular to each other.....
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surface
cylindricitypic
roundnesspic
flatnesspic
0Modern
o9_(o9
$o9v%
o91&o9
o9H#o9
instruct
Surface features may be dimensioned and toleranced independently or relative to datums. Examples of each type of feature are shown above with the standard shorthand symbols used to represent them. The limits are denoted by the green surfaces.
Surface Features
Independent Features
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straightness
queue
buttonclick
mouseenter
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queue "straightness"
= cursor
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Straightness
queue
cylindricity
buttonclick
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queue "cylindricity"
= cursor
= default
Cylindricity
roundness
queue
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queue "roundness"
= cursor
= default
Roundness
flatness
queue
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queue "flatness"
= cursor
= default
Flatness
straightnesspic
*o#U-
0Modern
5'G#5'O(
3G#5'G#
;,G#;,O(
Click one of the above shorthand symbols.for an explanation.
straightness
buttonclick
buttonclick
The designated feature elements must lie within a region bounded by parallel lines that are separated by the given tolerance. For instance, for a straightness tolerance of f0.20 on the diameter of the circle, the axis of the cylinder must lie within a cylindrical region of diameter 0.20.
- click to remove -
F%*%C%
- click to remove -
flatness
buttonclick
buttonclick
- click to remove -
The specified surface must lie between two parallel planes separated by the given tolerance at all points on the surface. The parallel planes do not have to be perpenidcular or parallel to any feature of the given part, but they must be parallel to each other.
roundness
buttonclick
buttonclick
- click to remove -
The boundary of every cross-sectional element of the indicated feature (e.g. a cylinder) must lie within two concentric circles whose radii differ by the specified tolerance. Each cross-section is considered independent of the others.
cylindricity
buttonclick
buttonclick
b-F-_-
- click to remove -
The specified cylindrical surface must lie between two concentric cylinders whose radii differ by the indicated tolerance. The axes of these cylinders do not have to be aligned with any element of the part.
datumrel
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concentricitypic
perpendicularitypic
angularitypic
parallelismpic
perpendicularity
queue
buttonclick
mouseenter
default
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buttonclick
queue "perpendicularity"
= cursor
= default
Perpendicularity
Related to Datums
angularity
queue
buttonclick
mouseenter
default
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buttonclick
queue "angularity"
= cursor
= default
Angularity
parallelism
queue
buttonclick
mouseenter
default
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buttonclick
queue "parallelism"
= cursor
= default
Parallelism
concentricity
queue
buttonclick
mouseenter
default
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buttonclick
queue "concentricity"
= cursor
= default
Concentricity
runout
queue
buttonclick
mouseenter
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buttonclick
queue "runout"
= cursor
= default
Runout
queue
buttonclick
mouseenter
default
mouseleave
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queue "tir"
= cursor
= default
runoutpic
tirpic
Click one of the above shorthand symbols.for an explanation.
instruct
These are features which are toleranced relative to datums, which are shown in purple. The limits are denoted by the green surfaces.
angularity
buttonclick
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The indicated surface must lie between two parallel planes that are separated by the given tolerance and form the desired angle with the referenced datum plane.
lerance of f0.20 on the diameter of the circle, the axis of the cylinder must lie within a cylindrical region of diameter 0.20.
- click to remove -
H&,&E&
- click to remove -
perpendicularity
buttonclick
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For surfaces, the specified surface must lie between two parallel planes that are separated by the given tolerance and are perpendicular to the referenced datum plane. For cylinders, the nominal cylinder axis must lie within a cylindrical region that has a diameter given by the tolerance and is perpendicular to the referenced dtum plane.
- click to remove -
parallelism
buttonclick
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The indicated surface must lie between two parallel planes that are separated by the given tolerance and are parallel to the referenced datum plane.
aightness tolerance of f0.20 on the diameter of the circle, the axis of the cylinder must lie within a cylindrical region of diameter 0.20.
- click to remove -
n-R-k-
- click to remove -
concentricity
buttonclick
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All points of the axis of the specified cylinder must lie within a cylindrical region whose diameter is given by the tolerance. The axis of the cylinder and the axis of the referenced cylindrical datum must coincide.
of the cylinder must lie within a cylindrical region of diameter 0.20.
- click to remove -
- click to remove -
runout
buttonclick
buttonclick
A dial indicator is placed on the part normal to the true geometric shape of the specified surface, and the part is rotated 360 degrees about the axis of the referenced datum. The total indicator movement must be less than the given tolerance. This is done for each section of the surface.
click to remove -
l4P4i4
- click to remove -
buttonclick
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Runout is measured for each section of the surface. The difference between the maximum and minimum indicator readings for the entire procedure must be less than the tolerance. (These readings may occur at different points along the feature.)
ithin a cylindrical region of diameter 0.20.
- click to remove -
- click to remove -
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smmc"
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Maximum Material Condition (MMC)a
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The maximum material condition is the state in which the part, not the feature, is at its greatest volume. For a shaft, this occurs at its largest dimension, while for a hole, it is when the feature is at its smallest.
The shaft from the preceding page may be toleranced using the Maximum Material Condition to keep the clearance constant. The stated straightness tolerance is now applied only at the maximum material condition. As the shaft diameter decreases, the starightness tolerance is increased. The steps in the following example illustrate this by showing how to fill in the tolerance table show below. As the solution progresses, notice that the minimum clearance must remain constant.a
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step3
The rest of the table is completed by the same method. As the actual diameter is decreased by 0.10, the straightness tolerance is increased by 0.10. Using MMC tolerances allows for variable tolerancing. The parts still fit with the proper clearance, and the shaft is easier to manufacture to specifications. to specifications.
If the diameter of the shaft is decreased by 0.10, the straightness is not held constant, but rather increased by an equivalent amount. Notice that the clearance returns to its original size.
step1
The straightness of the shaft is toleranced to 0.04 at the MMC, when the shaft is at its maximum allowable diameter. The minimum clearance is 0.06, as before.
c.wav
a) Determine the minimum nominal size of diameter A, the large pin diameter, e.g. A 0.01 mm..
b) Determine the maximum nominal size of diameter B, the small pin diameter, e.g. B 0.01 mm..
c) Properly label, on the figure below, both the pin and bores to show the above dimensions, tolerances, and control features so the parts will fit with the desired clearance and interference.erence..
Example 8.4
The pin assembly shown below requires a clearance fit on the smaller diameter, and an interference fit on the larger diameter. The nominal diameter of the bore and counterbore are 5 mm and 10 mm, respectively. The tolerances are 0.02 mm on the bore diameters, and 0.01 mm on the pin diameters. Assume that concentricity is the only important feature other than the tolerances on the diameters (e.g. negligible straightness, perpendicularity, etc.). The tightest concentricity that can be held is 0.005 mm for the pins, and 0.01 mm for the bores.
suball
a) Determine the minimum nominal size of diameter A, the large pin diameter, e.g. A 0.01 mm.
b) Determine the maximum nominal size of diameter B, the small pin diameter, e.g. B 0.01 mm.
c) Properly label, on the figure below, both the pin and bores to show the above dimensions,
tolerances, and control features so the parts will fit with the desired clearance and interference.erence......................
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1. The clearance fit on the smaller diameter does affect the interference fit on the larger diameter, so the calculation of the larger diameter should be done first. To guarantee an interference fit at A, its minimum size must be the same as the largest hole, producing a line-to-line fit. Thus (10.00 + 0.02) = 10.02......02
2. Since the tolerance of A is 0.01, the lower limit is added to the minimum size pin to get the nominal size. Thus the size of the larger pin is f10.02
0.01. A = 10.03
3. Now consider the clearance fit at B. Temporarily, ignore the concentricity tolerances. They will be added later. To guarantee clearance at B, its maximum size must be the same as the smallest hole, producing a line- to-line fit. Thus (5.00 - 0.02) = 4.98.
4. Since the tolerance of B is 0.01, the upper limit is subtracted from the maximum size to get the nominal size. Thus the size of the smaller pin, not considering concentricity tolerances, is f4.97
0.01.
5. Now the concentricity tolerances must be added. The centers of the pins can be non-concentric by up to 0.005 radially. To compensate, the diameter of the smaller pin must be reduced by 2 x 0.005 = 0.010 to guarantee a clearance fit.
6. The centers of the bores can also be non-concentric by up to 0.01 radially. To compensate, the diameter of the smaller pin must be reduced by an additional 2 x 0.01 = 0.02 to guarantee a clearance fit.
7. The combination of the concentricity tolerances of the pins and the bore reduce the smaller pin diameter by a total of 0.03. Thus the smaller pin size must be f4.94
0.01. B = 4.94..
8. The drawing is labeled as shown. Note the use of datums for the concentricity specifications...
answer
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Chapter 8: Geometric Tolerancingg
The objectives of this chapter are to:
Define various methods for specifying the tolerances on object dimensions.
Demonstrate proper tolerancing techniques to achieve desired fits on mating objects.
Show examples of basic fabrication techniques to produce guidelines in practical tolerancing.
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Begin Lesson
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tolerancing
step0
machine
datums
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shaft
In an interference fit, the shaft is larger than the hole. To put the two together may require force, or thermal treatment, where the hole is heated to make it expand, or the shaft is cooled to make it contract. The hole below is the same as on the preceding page, 10
0.2 mm. However, the shaft must now be sized to produce an interference fit with an allowance of 1.2 mm and a tolerance of 0.6 mm.
Interference Fitt
step0
Start by clicking the forward arrow below.ow.
step1
The lower limit of the hole is found by subtracting the tolerance from the basic size.
10 - 0.2 = 9.8 mm
>= 9.8 mm= 9.8 mm
step3
The lower limit of the shaft is found by subtracting the tolerance from the upper limit of the shaft.
11.0 - 0.6 = 10.4 mm
Therefore, the diameter of the shaft is described by a bilateral and equal tolerance of 10.7 mm
0.3 mm.
10.7
0.3 mm mmmmmm
step2
The upper limit of the shaft is found by adding the allowance to the lower limit of the hole.e.hole.
9.8 + 1.2 = 11.0 mm
<= 11.0 mm 9.8 mm
1.2 mm mm= 9.8 mm
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shaft
A fit describes the way two or more parts come together. There are three main types of fits. In a clearance fit, one part is always smaller than the other, so there is space between the two. In contrast, for an interference fit, the inner part is larger than the outer one. The minimum clearance or maximum interference between the two parts is called the allowance. When the allowance is equal to zero, the part sizes match exactly and a "line-to-line" fit is produced.
In the example below, there is a hole with a basic size of 10 mm,
0.2 mm. The shaft must be sized with a tolerance of 0.4 mm to provide a clearance fit with an allowance of 0.1 mm.
step0
step0
Start by clicking the forward arrow below.ow.
step1
The lower limit of the hole is found by subtracting the tolerance from the basic size.
10 - 0.2 = 9.8 mm
>= 9.8 mm= 9.8 mm
step3
The lower limit of the shaft is found by subtracting the tolerance from the upper limit of the shaft.
9.7 - 0.4 = 9.3 mm
Therefore, the diameter of the shaft is described by a bilateral and equal tolerance of 9.5 mm,
0.2 mm..................................
0.2 mm.8 mm
step2
The upper limit of the shaft is found by subtracting the allowance from the lower limit of the hole.
9.8 - 0.1 = 9.7 mm
<= 9.7 mm= 9.8 mm
0.1 mm mm= 9.8 mm
shaft
A transition fit describes two parts which are sometimes in interference and sometimes in clearance. The hole below is the same one used on the previous examples. The shaft is to be sized to produce a transition fit with an allowance of 0.1 mm and a tolerance of 0.4 mm.
Transition Fititt
step0
step0
Start by clicking the forward arrow below.ow.
step1
The lower limit of the hole is found by subtracting the tolerance from the basic size.
10 - 0.2 = 9.8 mm
>= 9.8 mm= 9.8 mm
step3
The lower limit of the shaft is found by subtracting the tolerance from the upper limit of the shaft.
9.9 - 0.4 = 9.5 mmmm
Therefore, the diameter of the shaft is described by a bilateral and equal tolerance of 9.7 mm
0.2 mm...bilaterally, is 9.7 mm,
0.2 mm.
0.2 mmm mmmmmm
step2
The upper limit of the shaft is found by adding the allowance to the lower limit of the hole.e.hole.
9.8 + 0.1 = 9.9 mmm
<= 9.9 mmm 9.8 mm
0.1 mm0 mm 9.8 mm
convent
mmsystem.dll
thisAnim
lastAnim
thisWav
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8 = "tola"
"mmsystem.dll"
timegettime()
animdone
audiodone
In conventional tolerancing, the allowable error is dependent on its direction. The shape traced by the allowable location for the center of the hole is a rectangle. Its diagonals are longer than its length and breadth. A rectangular area is traced by the allowable locations of the center of the hole. Note that its diagonals are longer than either its length or breadth.
animationstage
animationstage
thisAnim
animerror
buttonclick
buttonclick
4thisAnim, ref
syserrornumber = 0
mmPlay clip
stage "animationstage" hold
N<> 0
animerror
Repeat Animation
True Position Tolerancing
Compute!
animationstage
Try it out!
Adjust the basic sizes and upper and lower tolerances for a shaft and hole system to create clearance, interference, and transition fits.
shaft
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base1
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4base1
base1
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Nominal Size
Shaft
+txtUY
high2
high2val
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"low2val"
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p + 1
starttime = timegettime()
+ 50
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update
buttonclick
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starttime
low2val
slider1
timegettime
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buttonstilldown
buttonclick
4low2
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"low2val"
Z > 0
h - 1
starttime = timegettime()
+ 50
4temp
= rgbfill
= 196, 0,0
base2
base2
buttonclick
buttonclick
4base2
base2
buttonclick
buttonclick
4base2
= 10
Nominal Size
clearance
Clearance
interference
Interference
transition
Transition
table
step0
Standard fits for cylindrical parts are recommended by ANSI. Sizes, allowances, and tolerances are given in tables. In this example, the table above is used to apply tolerances to a 1/2 inch diameter hole for a given shaft to obtain a class RC-2 (running and clearance) fit. ANSI has separate tables for each type of fit.s separate tables for each type of fit....
step3
0.5000 + 0.0004 = 0.5004
step4
0.5000 + 0.0000 = 0.5000
step5
0.50000 + -0.00025 = 0.49975000
step6
0.50000 + -0.00055 = 0.49945000
step1txt
The hole has a basic diameter of
inch, which is in the 0.40 - 0.71 inch range.
step2txt
The type of clearance fit desired is of class RC 2. The classes vary in the tightness of the fit.
step3txt
To find the upper limit of the hole, the upper tolerance is added to the basic size. Note that values shown in the table are in thousandths of an inch.
step4txt
The lower tolerance is added to the basic size to find the lower limit of the hole. The hole can now be described by a bilateral and equal tolerance of 0.5002
0.0002.
step5txt
The upper limit of the shaft is found by adding the upper tolerance to the basic size.
step6txt
The lower tolerance is added to the basic size to find the lower limit of the shaft.
Tolerance Tables
highlight1
highlight2
highlight4
highlight5
highlight6
highlight2a
highlight3
highlight7
step7txt
The shaft can now be described by a bilateral and equal tolerance of 0.4996
0.00015. Notice that the loosest and the tightest fits are both clearance, and match the maximum and minimum amounts of clearance given on the table.
<BookPath>
:HDMEDIAPATH
pagesys.sbk
=^addToSysBooks
thisAnim
Tolerancing
thisWav
statusBar
menusys.sbk
enterBook
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4thisWav, thisAnim
addToSysBooks "pagesys.sbk"
menusys.
statusBar = "Tolerancing"
nnewBook
System
Arial
Arial
Arial
:CDMEDIAPATH
Arial
Arial
Arial
Arial
TBKWidgets
Arial
New Roman
Arial
Arial
New Roman
Arial
P E:\
Arial
New Roman
Arial
New Roman
Arial
Arial
Symbol
ew Roman
Symbol
Times New Roman
Times New Roman
Arial
New Roman
Symbol
ew Roman
Arial
New Roman
Arial
Graphics Interactive
~(z(z(
machine
instruct
Click one of the above machining processes.r an explanation.
animationstage
animationstage
thisAnim
animerror
buttonclick
buttonclick
4thisAnim, ref
syserrornumber = 0
mmPlay clip
stage "animationstage" hold
N<> 0
animerror
Repeat Animation
Machining Processes
pgrind
queue
buttonclick
buttonclick
queue "pgrind"
Precision
Grinding
Sawing
queue
buttonclick
buttonclick
queue "saw"
Milling
queue
buttonclick
buttonclick
queue "mill"
Extrusion
extrude
queue
buttonclick
buttonclick
queue "extrude"
Drilling
drill
queue
buttonclick
buttonclick
queue "drill"
Lathe
lathe
queue
buttonclick
buttonclick
queue "lathe"
Grinding
queue
grind
buttonclick
buttonclick
queue "grind"
Casting
queue
buttonclick
buttonclick
queue "cast"
EDMSawing
queue
buttonclick
buttonclick
queue "edm"
sawtxt
In sawing, a moving blade with teeth removes material completely through the part along a line. The blade is usually a band or circular in shape. For hand sized parts, this process can easily achieve tolerances of 2 mm. Tolerances of 0.2 mm are considered challenging for this operation.
drilltxt
In drilling, a spinning cylindrical toolbit creates circular holes in a stationary part. For common hole sizes less than 25 mm in diameter, this process is can easily achieve tolerances of 0.1 mm. An additional reaming operation can achieve tolerances of 0.01 mm. Tolerances of 0.001 mm are considered challenging for this operation.
lathetxt
In a turning operation on a lathe or a boring mill, the object is rotated at while a stationary cutting tool is used to remove material. These machines can easily achieve precisions of 0.01 mm for hand sized parts. Precisions of 0.001 mm are considered challenging for this operation.
milltxt
In a milling operation a rotating cutting tool is moved across a surface to smooth and shape it. Milling machines can easily achieve precisions of 0.02 mm for hand sized parts. Precisions of 0.002 mm are considered challenging for this operation.
grindtxt
In rough grinding, the material on a part is removed by a rotating abrasive wheel. Objects can be easily rough ground to a tolerance of 1 mm for hand sized parts. Tolerances of 0.1 wheel. Objects can be easily rough ground to a tolerance of 1 mm for hand sized parts. Tolerances of 0.1 mm are considered challenging for this operation.
edmtxt
In Electrical Discharge Machining, or EDM, a current is passed between a thing conductive wire and a conductive part. The sparks produce charge buildup and thermal stress, causing particles to break off or vaporize. An EDM operation can easily achieve precisions of 0.05 mm for hand sized parts. Precisions of 0.005 mm are considered challenging for this operation.
extrudetxt
Extrusion processes are usually reserved for production quantities. Heated material under pressure is forced through a die. Extrusion can easily achieve precisions of 0.2 mm for hand sized parts. Precisions of 0.02 mm are considered challenging for this operation.
casttxt
Casting and Molding operations are usually reserved for production quantities. Material is melted and poured into a mold before cooling and hardening. As a rule of thumb, casting can easily achieve precisions of 0.5 mm for hand sized parts. Precisions of 0.05 mm are considered challenging for this operation. Actual tolerances would depend on the type of casting operation.
pgrindtxt
Extremely fine precisions and surface finishes are created with precision grinding. A rotating abrasive wheel removes small amounts of material with each pass of the workpiece. The workpiece can be moved linearly, as with a milling machine, or rotated, as with a lathe. With precision grinding, sub-micron flatness, roundness, and surface roughness can be achieved.
picinstruct
currpic
buttonclick
buttonclick
4currpic
picture
"picinstruct"
sawpic
drillpic
edmpic
grindpic
lathepic
millpic
picinstruct
Click the picture to make it disappear.
picture
picinstruct
currpic
buttonclick
buttonclick
4currpic
picture
"picinstruct"
Show Picture
&File
E&xit Alt+F4
Exit the program
&Navigate
navigate
&First Page Ctrl+Home
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&Go to Page... Ctrl+G
Introduction
intro
Go to Introduction chapter
Objectives
intro1
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Sketching
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Objectives
Techniques
Objects
Cartooning
Engineering Drawings
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Objectives
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Working Drawings
Othogonal Projection
ortho
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orth1
Theory
orth2
Standard Views
orth3
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orth4
Common Practices
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orth6
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Go to Pictorials chapter
Objectives
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Sections
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Objectives
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Dimensioning
dimension
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Objectives
Definitions
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Common Shorthand
Tolerancing
tolerance
Go to Tolerancing chapter
Objectives
Definitions
Practical Fabrication Tolerances
True Position
Datums
Surface Features
Descriptive Geometry
descGeom
Go to Descriptive Geometry chapter
Objectives
Basic Principles and Relationships
Line Visibility
Distance Between Lines
Edge Views and True Shapes
Dihedral Angles
Intersection of a Line and a Plane
Intersection of Two Planes
Intersection of a Plane and a Solid
Intersection of Solids
Surface Developments
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&Main Menu Ctrl+Alt+Home
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&Help
Instructions F1
tutor
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About the Authors
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Prof. Dennis K. Lieu
Chris Casey
Su Shien Pang
Paul Krueger
Allison Okamura
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intro
pause
audioOn
paused
thisWav
playing
buttonClick
buttonClick
4thisWav, audioOn
mmStatus
clip
4 = "playing"
mmPause
d = "paused"
mmPlay
notify
repeat
audioOn
thisWav
buttonClick
buttonClick
4thisWav, audioOn
mmPlay clip
notify
Repeat
smpause
smpausedis
backwardsml
backwardDis
forwardsml
forwardDis
pauseup
pauseDis
blank
uppress
downpress
repeat
wwwwp
wwwwp
**""""""**"
**********
*********
********
""
intro
pause
repeat
lastAnim
thisAnim
thisWav
lastWav
enterPage
leavePage
AnimDone
AudioDone
4lastWav, thisWav, lastAnim, thisAnim
4audioOn, startTime
B"repeat"
/"intro"
--switch the
qones
disabled
enabled buttons here
AnimDone
AudioDone
You have reached the end of Tolerancing. To review the material, go to the next page. To continue to Descriptive Geometry, click the button below.e button below.
nextLesson
XdescGeom
buttonClick
buttonClick
descGeom
&Go to next lesson!
intro
pause
audioOn
paused
thisWav
audioerror
playing
buttonClick
buttonClick
4thisWav, audioOn, vol
syserrornumber = 0
mmStatus
clip
W = "playing"
mmPause
= "paused"
mmvolume
Pthiswav =
mmPlay
notify
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audioOn
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audioerror
buttonClick
buttonClick
4thisWav, audioOn, vol
syserrornumber = 0
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clip thiswav =
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audioerror
Repeat
tolerancing
Tolerancing
No specified dimension can be exact, since it is impossible to make a part with perfect precision. Instead, each dimension has an allowable error or tolerance. A smaller or tighter tolerance increases the cost of manufacturing the part. The largest possible tolerance which also ensures the fit and function of the device is usually used to minimize cost..
Limits - the maximum and minimum sizes allowable for the finished part.
Tolerance - the difference between the upper and lower limits.
thisAnim
lastAnim
thiswav
lastWav
enterpage
leavepage
audiodone
animdone
4thisAnim, thiswav, lastAnim, lastWav
= thisWav
= "tol3"
audiodone
animdone
Representing Tolerances
There are several ways to present tolerances, any of which may be used on a given dimension. The following tolerances are equivalent. The basic size, shown in blue, is the chosen size from which tolerances are applied and is shown in decimal form. This differs from the nominal size, which is a whole or fractional number describing the general size of the dimension..