Hw1 Sol

MAE 4500 Manufacturing Methods Assignment #1

Solutions Review Questions: a. Define allowance, tolerance, and fit. b. Give examples where the following fits are used: RC5, LC3, LN2, FN3. c. Make a sketch to show a typical surface roughness one would obtain from a stylus instrument traversing perpendicular to the lay of a turned surface. (i)

Indicate relative magnifications;

(ii)

Show the centerline;

(iii)

Show the roughness width;

(iv)

Show the waviness width and height;

(v)

Define Ra , Rq , and Rt (in words or equations).

Solution: a. Allowance is the difference in dimensions required for proper functioning (interference, clearance). Tolerance is the permissible difference between maximum and minimum limits of size [or : difference between the maximum and minimum allowable dimension] [or: the inaccuracy permitted in the dimension of a part]. Fit is the functional difference in the dimension of mating parts [or: the position of the tolerance zone relative to the basic dimensions of mating parts]. b. RC5: class 5 running and sliding fit can be used in Piston and Cylinders LC3: class 3 location clearance fit can be used in shaft and bushing LN2: class 2 location interference fit can be used in key and slot fixture FN3: class 3 force and shrinkage fit can be used in shaft and bearing to transmit motion.

c.

v. Ra = arithmetic average height of aperities [Eq. (3-4)] Rq = root-mean-square average height of asperities [Eq. (3-5)] Rt = maximum peak-to-valley height.

1. A guide shaft and a bushing assembly (Figure 1) are to be used as a part of a device. Positioning accuracy is important for this application, but ease of assembly is also required. The shaft is to have a high accuracy location fit class with the reamed bushing hole into which it fits. Design the assembly: a) Select an appropriate class of fit for this design b) Make a sketch of the shaft and the hole, and dimension them. c) Do an allowance analysis.

Shaft

Bushing 1.2”

FIGURE. 1 Manufacturing Design Solution 1: (a)

Class 1 location fit (LC1) has been selected for this design because of position accuracy and ease of assembly that are needed for this application. On the tolerance chart, the amount of tolerance for LC fits increases with class number. This is an open ended design problem, so choices of LC and LT will be acceptable selections.

(b)

Design of the assembly Class 1 location fit (LC1):

Clearance

Hole

Shaft

0.0 1.0

+0.6 0.0

0.0 -0.4

Basic hole size = minimum acceptable hole size = 1.2000" Basic shaft size = maximum acceptable shaft size = min. hole size – min clearance = 1.2000 – 0.0 = 1.2000" Hole tolerance = maximum – minimum = 0.0006 – 0.0000 = 0.0006 Shaft tolerance = maximum – minimum = (0.0) - (-0.0004) = 0.0004 Largest hole = Smallest hole + hole tolerance = 1.2000 + 0.0006 = 1.2006" Smallest shaft = largest shaft – shaft tolerance = 1.2000 - 0.0004 = 1.1996"

(c)

Allowance Analysis Max. clearance = largest hole – smallest shaft = 1.2006 – 1.1996 = 0.0010" From Table, max clearance = 0.0010" Min. clearance = smallest hole - largest shaft = 1.2000 – 1.2000 = 0.0000" From Table, min clearance = 0.0000" Therefore the design satisfies the standard requirement.

1.2000 +−00..0000 0004

Shaft

Bushing

1.2000 +−00..0006 0000

2. A 1.10” turbine blade pin is to fit a slot in a turbine hub in order to precisely positioning it into a jet engine system. This requires assembly with a slight interference. A class 3 fit is prescribed. (a) Draw the slot and the blade pin and dimension them. (b) Make analysis of the allowances.

Solution: a)

+0.0000

1.1017 − 0.0005

Pin +0.0008

1.1000 −0.0000

Slot

b)

Since the key is to be used for locating a positioning block with slight interference, an LN3 fit will be used. From the tables the following tolerances are found: Limits of Interference Slot +0.4 +0.8 +1.7 0.0

Key +1.7 +1.2

Slot tolerance = 0.0008 – 0.0000 = 0.0008 Key tolerance = 0.0017 – 0.0012 = 0.0005 Basic slot size = Smallest Slot size = 1.1000” Largest Slot = Smallest Slot + Slot tolerance = 1.1000 + 0.0008 = 1.1008” Basic Key size = Largest Key size = Smallest slot size + max. Interference = 1.1000 + (0.0017) = 1.1017” Smallest Key = Largest Key – Key tolerance = 1.1017 – 0.0005 = 1.1012” Allowance Analysis Minimum Interference = Largest Slot – Smallest Key = 1.10008 – 1.1012 = -0.0004” Maximum Interference = Smallest Slot – Largest Key = 1.1000 – 1.1017 = - 0.0017” (Max interference is 0.0017”) +0.0008

Slot

1.1000 −0.0000

Key

1.1017 −0.0005

+0.0000

3. Design a working plug and snap gages to check the bushing and shaft diameters in Problem 1 assuming high production volume is required. Give detail drawings of the gages, indicate the type of material used and describe the surface roughness of the important surfaces. Use a CAD system in your presentation. Manufacturing Design Problem Solution 2: Gage tolerances for the gages will be 10% of the working tolerances of the hole and shaft. Wear allowances will be 5% of the working tolerances. Plug Gage (for checking the hole) GO gage

gage tolerance = 10% of working tolerance =10% (0.0006) =0.00006" wear allowance = 5% of working tolerance =5% (0.0006)=0.00003" gage size = Basic size + wear allowance = 1.2000 + 0.00003 = 1.20003" Plus tolerance = +0.00006" Minus tolerance = -0.00000"

NO GO gage

gage size = maximum acceptable hole size = 1.2006" Plus tolerance = +0.0000" Minus tolerance = -0.00006"

Surface Roughness = (working tolerance/30) = (gage tolerance)/3 = 0.00006"/3 = 20µin Snap Gage (for checking the shaft) GO gage

gage tolerance = 10% of working tolerance = 10% (0.0004)=0.00004" wear allowance = 5% of working tolerance = 5%(0.0004)=0.00002" gage size = maximum acceptable shaft size – wear allowance = 1.20000 – 0.00002 = 1.19998" Plus tolerance = +0.00000" Minus tolerance = -0.00004"

NO GO gage

gage size = minimum acceptable shaft size = 1.19960" Plus tolerance = +0.00004"

Minus tolerance = -0.00000" Surface Roughness = (gage tolerance)/3 = 0.00004"/3 = 13µ in The tolerances can be achieved by finish turning and boring. The gages can be made of aluminum for economic reasons; the high production volume dictates a chromium coating to the important surfaces.

20µ

1.20003

20µ

+ 0.00006 − 0.00000

1.20060

+ 0.00000 − 0.00006

Figure 1 Plug gage. (All dimensions in inches)

1.19998

+ 0.00000 − 0.00004

13µ

Figure 2 Snap gage. (All dimensions in inches)

13µ

1.19960

+ 0.00004 − 0.00000