ME 328.3 M4‐M7 Solution 2015 THE LAB REPORT IS TO INCLUDE THE FOLLOWING TASKS IN SEQUENCE: 1. Band Saw & Chucking for CNC Lathe: Given: Bar Stock Diameter = 52 mm Material = 6061‐T6 Bar stock length = 20 feet Bar stock cost = $335.28 per bar Bar stock density = 2.70 g/cm 3 a) Find the set‐up time (tr) and total cutting time (te) Table 7, ME 328 Lab Manual: tr = 5 min te = 0.75 min b) How many workpieces can we machine out of one bar? How much is required per piece? ܮൌ െ ܮ௪ ܮ௨ ܮ ܮௗ௦௪ ൌ 56 െ 47.3 25 70 2 5 ൌ 110.7 → 111 Number per bar = ? # ݎ ݎൌ 20 ݐ111 ൌ 54.92 → 54 ݏ ݎ .ݎc) How many bars, and how many cuts do we need to rough cut all workpieces? # ݏݎൌ 9000 54 ൌ 166.67 → 167 ݏݎ# ݏݐݑൌ 9000 ሺ1 ݐݑ ݎሻ d) What is the remaining length of each barstock after the last workpiece is cut? For bars 1‐166 there is the same length left: (NOTE: 20 feet = 6096 mm) 20 ݐെ 111 ∗ 54 ൌ 102 ݎ
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ME 328.3 M4 M7 IN SEQUENCE 1. Saw for CNC Labs/ME... · ME 328.3 M4‐M7 Solution 2015 THE LAB REPORT IS TO INCLUDE THE FOLLOWING TASKS IN SEQUENCE: 1. Band Saw & Chucking for CNC
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ME 328.3 M4‐M7 Solution 2015
THE LAB REPORT IS TO INCLUDE THE FOLLOWING TASKS IN SEQUENCE:
1. Band Saw & Chucking for CNC Lathe: Given:
Bar Stock Diameter = 52 mm
Material = 6061‐T6
Bar stock length = 20 feet
Bar stock cost = $335.28 per bar
Bar stock density = 2.70 g/cm3
a) Find the set‐up time (tr) and total cutting time (te)
Table 7, ME 328 Lab Manual:
tr = 5 min
te = 0.75 min
b) How many workpieces can we machine out of one bar?
How much is required per piece?
56 47.3 25 70 2 5110.7 → 111
Number per bar = ?
# 20111
54.92 → 54 .
c) How many bars, and how many cuts do we need to rough cut all workpieces?
# 900054
166.67 → 167
# 9000 1
d) What is the remaining length of each barstock after the last workpiece is cut?
For bars 1‐166 there is the same length left: (NOTE: 20 feet = 6096 mm)
20 111 ∗ 54 102
For the final bar:
20 111 ∗ 36 2100 .
e) Calculate the material cost. Keep in mind that the barstock is only available in 20 foot lengths.
167 ∗ $335.28 $55,991.76
f) Determine the chucking time (ttb) for the CNC lathe for each workpiece.
From table 9 of the ME 328 lab Manual, the chucking time depends on the mass of the bar stock:
∗ ∗4
2.70 ∗52
4∗ 106 607.8 0.6078
ttb = 0.3 min
2. Create a CNC‐set‐up sheet, which can be downloaded from the ME328 WEB site, showing all necessary information to set up the Baxter Entrepreneur CNC lathe (refer to the lab manual for working range details).
ZW=56.0
barstock
3. Calculate the workpiece datum and the reference points for tool change (G50’s) for the CNC‐lathe. The safety distance for the tool change position u=25.0 mm.
Note: X home, Z home, reference tool information (ID#5) and equations are all given on page 53 of the lab manual. Work Piece Datum: X = Xhome – XsetRT X = 522.727 – 127.000 = 395.727 mm Z = Zhome – ZW – ZsetRT Z = 573.629 – 56.000 – 50.800 = 466.829 mm Reference point for the Reference Tool (ID#5) RP XRT = 300.000 mm (given in lab manual) RP ZRT = LW + (Lmax – ZsetRT) + u RP ZRT = 85.000 + (135.800-50.800) + 25.000 = 180.000 mm Reference points for the other tools, tool 2 as an example: RP X = XRT+ (XsetRT – Xset) RP X = 300.000 + (127.000 – 127.000) = 300.000 mm RP Z = ZRT + (ZsetRT – Zset) RPZ = 180.000 + (50.800-135.800) = 95.000 mm
CALCULATION of REFERENCE POINT (RP) for TOOL CHANGE
REFERENCE TOOL RP X = 300.000 mm
RP Z = 180.000 mm
CALCULATION of REFERENCE POINT (RP) for TOOL CHANGE
T01 RP X = 300.000 mm
RP Z = 180.000 mm
T02 RP X = 300.000 mm
RP Z = 95.000 mm
T03 RP X = 320.800 mm
RP Z = 141.300 mm
T04 RP X = 297.000 mm
RP Z = 180.000 mm
T05 RP X = 248.300 mm
RP Z = 179.700 mm
T06 RP X = no tool mm
RP Z = no tool mm
T07 RP X = no tool mm
RP Z = no tool mm
T08 RP X = 299.700 mm
RP Z = 172.200 mm
4. T01 is used for a facing pass, leaving 0.5mm for finishing, after the facing pass the tool is moved rapidly away at a 45° angle using the following block of code: G00 U4.0 W2.0. Calculate the points required (3) for this tool path with tool nose radius compensation, the safety distance for the start of the tool path is 2 mm.
Given:
Barstock diameter = 52.0mm
Safety distance, u, = 2.0 mm
Finishing allowance, ap = 0.5 mm
For movement from P2 to P3 use: G00 U4.0 W2.0
P1: 56, 70.5
1 2 52 2 ∗ 2 56
1 70 0.5 70.5
P2: ‐4.0, 70.5
2 2 ∗ 2 4.0
2 1 70.5
P3: 0.0, 72.5
3 2 4.0 0.0
3 2 2.0 72.5
2.0
5. Calculate all points required (23) for external roughing with T01 with tool nose radius compensation. Show all unique geometries for the points. Leave 0.5 mm allowance for the finishing pass, maximum allowable depth of cut = 2.2 mm. The roughing pass overcuts by 2.0 mm for point 5, and the safety distance is 2.0 mm.
Given:
Start diameter = 52.0 mm
Start length after facing = 72.5 mm
Ap = 0.5 mm
DoCmax = 2.2 mm
a.) DoC and NoC = ?
Smallest Diameter for roughing = ?
Below is an enlarged drawing of the 2.2x45° chamfer on the end of the part:
252 32.01
22.2
4.54 → 5
252 32.01
25
2.0
P1: 48, 72.5
1 2 ∗ 52 2 ∗ 2 48.0
1 70 0.5 2 72.5 3
P2: 48, 30.53
2 1 48.0
ap=0.5 mm
Dfinal = 36 mm
Smallest
Diameter for
Roughing = Dmin
∆xα = 45°
2.2 mm
(drawing)
Dcorner
= 36 – 2.2(2) = 31.6
∆ tan2
∗ 0.2071
2 ∗ ∆ 32.01
P3: 49.0, 27.8
3 ∗ 2 48 0.5 ∗ 2 49.0
Z3 is similar to Z2, just at a different point on the arc, therefore using the same equations as in P2:
∆ 7.7
∆ 0.0
3 29 0.0 1.2 27.8
P4: 49.0, ‐2.0
X4=X3=49.0
Z4=‐2.0 (overcut amount given)
P5: 53.0, ‐2.0
5 4 2 49 2 ∗ 2 53
Z5=Z4=‐2.0
P6: 53.0, 72.5
X6=X5=53.0
Z=29
drawing
Z2
∆x
∆z
X1 = X2 = 48
R = r+ap+Rs
R = 6.0+0.5+1.2
R = 7.7
∆22
362
482
362
1.2 7.2
∆ ∆ 7.7 7.2 2.73
2 29 ∆ 29 2.73 1.2 30.53
X=36.0
drawing
Z6=Z1=72.5
P7: 44.0, 72.5
7 1 ∗ 2 48 2 ∗ 2 44.0
Z7=Z1=72.5
P8: 44.0, 33.48
Similar to Point 2:
∆ 5.2
∆ 5.6789
8 33.48
X8=X7=44.0
P9 to P15 are all similar to previous points.
P9: 48.0, 30.53
P9=P2
P10: 48, 72.5
P10=P1
P11: 40, 72.5
11 7 2 ∗ 44 2 ∗ 2 40
11 7 72.5
P12: 40, 34.80
Similar to Point 2:
∆ 3.2
∆ 7.0035
12 34.80
X12=X11
P13: 44, 33.48
P13=P8
P14: 44, 72.5
P14=P7
P15: 36, 72.5
15 11 2 ∗ 40 2 ∗ 2 36
Z15=Z14=72.5
P16,23: 36.0, 67.80
X16=X15=36.0
P17: 37.0, 67.30
See GEO‐Figure 5 ME 328 Lab Manual Page 61, this is Point 2 in that image.
∆ 0.4958
Z17=70‐2.2‐∆z = 70‐2.2‐0.4958 = 67.30
17 2 36 2 ∗ 0.5 37.0
P18 to P22 have similar geometry or calculations as points previously covered.
P18: 37, 35.31
X18=X17=37.0
Z18 is similar to Z2
∆ 1.7
∆ 7.5099
70‐2.2=67.8
0.5 mm
45°
X16=36
C D
β= 45°
ε= 22.5°
0.5cos 45°
0.707
∗ 1.2 ∗ 22.5°0.497056
16 67.8 67.80
18 35.31
P19: 40, 34.80
P19=P12
P20: 40, 72.5
P20=P11
P21: 32, 72.5
21 15 2 ∗ 36 2 ∗ 2 32
Z21=Z20=72.5
P22: 32, 69.80
Similart o P16
2.0
0.7071
0.497056
22 69.80
X22=X21=32
P23: 36, 67.80
P23=P16
6. Calculate all points required (5) for internal finishing with T03 with tool nose radius compensation. Show all unique geometries for the points, for point 1 the safety distance in Z is 2.0 mm from the faced bar stock (which still has 0.5 mm on for finishing).
10.0
NOTE: Drilling has already been completed at this point, so the 24 mm hole is in place.
P1: 26.8, 72.5
1 70 0.5 2 72.5
X1=X2 (calculate X2 as shown below)=26.8
P2: 26.8, 70.0
2 70.0
P3: 24, 68.6
X3 = from drawing = 24.0
3 70 1 0.4 68.6
P4: 20, 68.6
X4=20 (given in drawing, 10x2=20)
Z4=Z3=68.6
P5: 20, 72.5
X5=X4=20.0
Z5=Z1=72.5
Z=70.0 (drawing)
Hole Diameter = 24.0
r=1.0 (drawing)
2 ∗ 2 ∗ 224 1 ∗ 2 0.4 ∗ 2 26.8
7. Calculate all points required (9) for external roughing with T04 with tool nose radius compensation. Show all unique geometries for the points. Leave 0.5 mm allowance for the finishing pass, maximum allowable depth of cut = 1.6 mm. The roughing pass overcuts by 1.5 mm for point 8, safety distance = 2.0 mm.
First the NoC and DoC need to be determined:
Known:
Start Diameter = 49 mm (left from T01)
Final Diameter = 42.0 mm (drawing)
ap = 0.5
Final roughed diameter = 42+0.5*2=43.0 mm
DoCmax = 1.6 mm
,2
49 4321.6
1.875 → 2
,2
49 4322
1.5
P1: 53, 25.316
1 2 ∗ ∗ 2 ∗ 48 0.5 ∗ 2 2 ∗ 2 53.0
Degree of the angle = ?
Using the dimensions from the drawing:
482
422
23 17.8→ 30°
P2: 46, 19.25
2 2 ∗ 49.0 2 ∗ 1.5 46.0
Z2: geometry is similar to Z1:
∆
462
422
30°230°
3.4641
30°0.530°
1.0
30°2
∗ 0.214359
2 17.8 ∆ 17.8 3.4641 1 0.214359 0.8 19.25
P3: 46, ‐1.5
3 2 46
3 1.5
P4: 53, ‐1.5
4 1 53
Diameter = 48.0Diameter = 49.0
Diameter = 48.0+2ap+2u
D=48+2*0.5+2*2=53.0
30°
g f
∝0.530°
1.0
∝2
∗30°2
∗ 0.8 0.214359
23 ∆ 23 4.330127.33
∆∝
0.5 230°
4.3301
. . .
.
Z = 23.0 ∆z = 4.3301
Z1
4 3 1.5
P5: 53, 19.25
5 4 53.0
5 2 19.25
P6: 46, 19.25
P6 = P2
P7: 43, 16.65
7 ∗ 2 42 0.5 ∗ 2 43.0
Z7: similar to P2:
∆0.530°
0.866
1.0
0.214359
7 17.8 ∆ 17.8 0.866 1 0.214359 0.8 16.65
P8: 43, ‐1.5
8 7 43
8 3 1.5
P9: 53, ‐1.5
P9 = P4
8. Calculate all points required (11) for external finishing with T05 with tool nose radius compensation. Show all unique geometries for the points, after the finish facing pass too tool moves to position 3 with the following block of code: G00 U4.0 W2.0. The finishing pass overcuts by 1.0 mm for point 10, safety distance = 2.0 mm, for P11, the X safety is from the largest remaining stock diameter on the part.
P1: 40, 70
1 2 ∗ 36 2 ∗ 2 40
1 70.0
P2: 20, 70
2 20.0
2 1 70.0
P3: 24, 72
3 2 2 ∗ 2 20 2 ∗ 2 24.0
3 2 2 70 2 72.0
P4: 27.13, 72
10.0
P5: 36, 67.566
This geometry is shown in the Figure on page 61 of the ME 328 Lab Manual for P2
5 36.0
∆ ∗2
0.4 0.4 ∗45°2
0.2343
5 ∆ 70 2.2 0.2343 67.566
P6: 36, 34.987
6 5 36.0
X = 31.6
Z=70.0
(part geometry
bottom corner
of chamfer)
∆z = u = 2.0
∆x = ∆z = 2.0
f sin0.445°
0.56568
0.56568 0.4 0.16568
45°0.16568
45°0.2343
4 31.6 2 ∗ ∆ 2 ∗ 31.6 2 ∗ 2 2 ∗ 0.2343 27.13
4 70.0 2.0 72.0
P7: 48, 28.6
7 48.0
7 29 29 0.4 28.6
For Z7: the 29 is from the drawing geometry.
P8: 48, 22.50
8 48.0
Centre Point of Arc
(from drawing)
X = 36.0
Z=29.0
6 0.4 6.4
∆ 6.4 0.4 6.38748
6 29 ∆ 29 6.38748 0.4 34.987
Finished Geometry
corner
X=48.0
Z=23.0
α = 30° (given geometry)
β = α/2 = 15°
∆ ∗0.4 ∗ 15°0.103527
∆z
8 23 ∆ 23 0.103527 0.4 22.496
P9: 42, 17.30
Similar to P8:
9 42.0
9 17.8 ∆ 17.8 0.103527 0.4 17.296
P10: 42, ‐1.0
10 9 42.0
10 1.0
P11: 52, ‐1.0
11 , 2 ∗ 48 2 ∗ 2 52.0
11 10 1.0
9. Calculate all required points (7) for T08 (part off tool), with tool nose radius compensation. Show the geometry for all points, initial safety distance in X (P1) is 2.0 mm from the initial barstock diameter, all other safety distances = 2.0 mm. Initial plunge cut (P2) has an X coordinate = 37.
P1: 56, 0.0
1 2 ∗ 52 2 ∗ 2 56.0
1 0.0
P2: 37, 0.0
2 37.0
2 1 0.0
P3: 46.0, 0.0
3 2 ∗ 42 2 ∗ 2 46.0
3 2 0.0
P4: 46, 3.617
4 3 46.0
P5: 38.766, 0.0
Finished Corner
X=42.0
Z=1.5
u=2.0
β = α/2 = 22.5° y
∗ 0.2 ∗ 22.5°0.08284
4 1.5 1.5 2 0.08284 0.2 3.617
P6: 0.0, 0.0
6 0.0
6 5 0.0
P7: 56, 0.0
7 1
Bottom of Chamfer
X=42‐2*1.5=39.0
Z=0.0
∆x
∆ ∗ 0.2 ∗ 22.5°0.0828
5 39.0 2 ∗ ∆ 2 ∗ 39 2 ∗ 0.0828 2 ∗ 0.2 38.766
T01 Point Summary:
Point X Z
Face 1 56.0 70.05
Face 2 ‐4.0 70.5
Face 3 0.0 72.5
1 48.0 72.5
2 48.0 30.53
3 49.0 27.8
4 49.0 ‐2.0
5 53.0 ‐2.0
6 53.0 72.5
7 44.0 72.5
8 44.0 33.48
9 48.0 30.53
10 48.0 72.5
11 40.0 72.5
12 40.0 34.8
13 44.0 33.48
14 44.0 72.5
15 36.0 72.5
16 36.0 67.80
17 37.0 67.30
18 37.0 35.31
19 40.0 34.80
20 40.0 72.5
21 32.0 72.5
22 32.0 69.80
23 36.0 67.80
T03 Point Summary:
Point X Z
1 26.8 72.5
2 26.8 70.0
3 24.0 68.6
4 20.0 68.6
5 20.0 72.5
T04 Point Summary:
Point X Z
1 53.0 25.316
2 46.0 19.25
3 46.0 ‐1.5
4 53.0 ‐1.5
5 53.0 19.25
6 46.0 19.25
7 43.0 16.65
8 43.0 ‐1.5
9 53.0 ‐1.5
T05 Point Summary:
Point X Z
1 40.0 70.0
2 20.0 70.0
3 24.0 72.0
4 27.13 72.0
5 36.0 67.566
6 36.0 34.987
7 48.0 28.6
8 48.0 22.5
9 42.0 17.30
10 42.0 ‐1.0
11 52.0 ‐1.0
T08 Point Summary:
Point X Z
1 56.0 0.0
2 37.0 0.0
3 46.0 0.0
4 46.0 3.617
5 38.766 0.0
6 0.0 0.0
7 56.0 0.0
10. Calculate the expected tool life for T01 and the maximum power used for this tool, . μ Assume the following values for the Taylor Equation: C= 198, F= ‐0.12, E= ‐0.34, G= ‐0.14.
a.) Determine Feed rate:
8 ∗→ ∗ ∗ 8 1.2 ∗ 0.0116 ∗ 8 0.1136
0.337 → 25% → 0.25
b.) Determine Vc:
From the supplied Table 8, with f=0.25 mm/rev and DoC=2.0 mm (from Q5):
200
c.) Determine Tool life:
∗ ∗200
198 ∗ 2 . ∗ 0.25 .
.14.89
d.) Determine VBa:
10.040.1
10.040.1
0.3 1.12
e.) Determine Maximum Power for this tool:
∗ ∗ ∗60 ∗ 10 ∗
2.0 ∗ 0.25 ∗ 640 ∗ 20060 ∗ 10 ∗ 0.85
1.12 1.40
11. Calculate the maximum power and rpm used for the drill on the CNC lathe.
Given:
f=0.12 mm/rev
Vc=120 m/min
a.) RPM:
∗ 1000∗
120 ∗ 100024 ∗
1591
b.) Power:
∗ ∗ ∗12 ∗ 10000 ∗
24 ∗ 0.12 ∗ 640 ∗ 12012 ∗ 10000 ∗ 0.85
1.12 2.43
12. Determine the surface speed for T04.
Known:
f=0.4 mm/rev (given)
DoC=1.5 mm, Q7
From Given Table 8 and linear interpolation:
178
13. Calculate the feed rate for the external finishing tool T05.
From the drawing: Rth = 16 µm
8 ∗→ ∗ ∗ 8 0.4 ∗ 0.016 ∗ 8 0.0512
0.226 → 10% → 0.20
14. Write a Program for the Baxter CNC lathe, using the information determined in this report. Use SPACE.exe to format the text. Use PLOT.exe to check your program for movement errors.