Degarmo's Materials And Processes In Manufacturing
13th Edition
ISBN: 9781119492825
Author: Black, J. Temple, Kohser, Ronald A., Author.
Publisher: Wiley,
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Chapter 24, Problem 18RQ
To determine
The basic principle of a universal dividing head.
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λ=
te average
Vc =
MDN
1000
T = S sin
Where λ is Chip Reduction Coefficient
Vc is cutting speed
te is thickness of cut chip
T is thickness of non-cut chip
S is feed rate
Ⓒ is cutting angle = 45⁰
D is diameter of workpiece = 40mm
Results
m/min
Depth of
cut (a)
mm
22222
Cutting
speed
(Vc)
m/min
500
500
500
500
500
Feed rate
(S)
mm/rev
Required:
1.Complete the solution to the table
with the given rules
2-Draw chart between feed rate (S) and chip
reduction coefficient
2
tc:
tc₂
tcs
0.11
0.39 0.58 0.43
0.14
0.42 0.62 0.5
0.20
0.34 0.42 0.32
0.24
0.9 0.96 0.95
0.28 0.92 0.85 0.85
te average
(mm)
T
(mm)
with discuss it.
Chip
Reduction
Coefficient
(2)
High accuracy cutting is very important in modern machining processes. This study is focus in
modern machining process of EDM wire cutting with diameter wire 0.25 mm (Brass), material
aluminum with thickness 10 mm and initial spark gap recommended is 50 microns. By applying the
assumption value of size desired 60 mm (X, Y) and output size 59.984 (X axis) and 60.012 (Y axis).
Complete the value of (A,B,C,D,E and F) in design of experimental (DOE) as in table1 and explain
the result outcome methods.
Table 1: Design of experimental (DOE) table
Size
Trial
output size
size of
% error
True or actual
desired
spark
(mm)
measurement
of
spark or gap
gap
error (mm) / side
overall
(mm)
(mm)
size
A
B
D
E
F
Determine machining time to produce the final product shown in Fig. 2 from the blank shown in Fig.
1 using face milling process. Diameter of the cutter is 50 m, dp = 1mm; N= 200 rpm, S-8 mm; and
f= 0.7 mm/ rev.
- 50
48 -
150
30
29
144
100
98
Fig. 2
Fig. 1
Chapter 24 Solutions
Degarmo's Materials And Processes In Manufacturing
Ch. 24 - Suppose that you wanted to machine cast iron with...Ch. 24 - How is the feed per tooth related to the feed rate...Ch. 24 - Why must the number of teeth on the cutter be...Ch. 24 - Why is the question of up or down milling more...Ch. 24 - For producing flat surfaces in mass-production...Ch. 24 - Milling has a higher metal removal rate than...Ch. 24 - Which type of milling (up or down) is being done...Ch. 24 - Why does down milling dull the cutter more rapidly...Ch. 24 - What parameters do you need to specify in order to...Ch. 24 - In Figure 24.2b, the tool material is carbide....
Ch. 24 - What is the advantage of a helical-tooth cutter...Ch. 24 - What would the cutting force diagram for Fc look...Ch. 24 - Could the stub arbor-mounted face mill shown in...Ch. 24 - In a typical solid arbor milling cutter shown in...Ch. 24 - Make some sketches to show how you would you set...Ch. 24 - Make some sketches to show how you would set up a...Ch. 24 - Explain how controlled movements of the work m...Ch. 24 - Prob. 18RQCh. 24 - What is the purpose of the hole-circle plate on a...Ch. 24 - You have selected a feed per tooth and a cutting...Ch. 24 - How much time will be required for a milling...Ch. 24 - If the depth of cut is 0.35 in., what is the metal...Ch. 24 - Prob. 4PCh. 24 - Calculate the spindle rpm and table feed (ipm) for...Ch. 24 - A gray cast iron surface 6 in. wide and 18 in....Ch. 24 - Prob. 7PCh. 24 - In Figure 24.12, the feed is 0.006 in. per tooth....Ch. 24 - Suppose you want to do the job described in...Ch. 24 - The Bridgeport vertical-spindle milling machine is...Ch. 24 - The KC Machine Works, which does job shop...
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- =te average Vc = TDN 1000 TS sin e Where is Chip Reduction Coefficient ve is cutting speed te is thickness of cut chip T is thickness of non-cut chip S is feed rate is cutting angle = 45⁰ D is diameter of workpiece = 40mm Results Depth of cut (a) mm 2 NNNN 2 2 m/min 2 Cutting speed (Vc) m/min 500 500 500 500 500 Feed rate (S) mm/rev 0.11 0.14 0.20 0.24 0.28 tc; tc₂ tc3 0.39 0.58 0.43 0.42 0.62 0.5 0.34 0.42 0.32 0.9 0.96 0.95. 0.92 0.85 0.85 te average (mm) H (mm) Chip Reduction Coefficient (2) Discussion 1- Why the thickness of chip after cutting process is larger than chip thickness before cutting process? 2- What is the reason of variety of the chips types with cutting conditions? 3- What is the difference in the feed rate between turning machining and milling machining? 4- Draw chart between feed rate (S) and chip reduction coefficient (A) with discuss it.arrow_forwardQuestion 2 Face milling, using an Ø 75 mm cutter with 10 inserts rotating at 280 rpm, is being carried out on a copper alloy block as shown in Figure 1. Direction of cut Cutter Workpiece 37,5 200 37,5 Figure 1: Illustration of the face milling operation on a copper alloy material The feed rate during the machining process is 560 mm/min for a 1.5 mm depth of cut. 2.1 If the cutter is rotating clockwise, in which type of milling operation does it result? What is the advantages and disadvantages of this milling operation? 2.2 Calculate the cutting time required for the machining process shown in Figure 1. 2.3 Calculate the power required for the machining operation. 2.4 If the number of inserts is reduced to 6, how would it influence the manufacturing process? Provide reasons for your answer.arrow_forwardA vertical boring mill is used to bore the inside diameter of a large batch of tube-shaped parts. The diameter = 28.0 in and the length of the bore = 14.0 in. Current cutting conditions are: speed = 200 ft/min, feed = 0.015 in/rev, and depth = 0.125 in. The parameters of the Taylor equation for the cutting tool in the operation are: n = 0.23 and C = 850 (ft/min). Tool change time = 3.0 min, and tooling cost = $3.50 per cutting edge. The time required to load and unload the parts = 12.0 min, and the cost of machine time on this boring mill = $42.00/hr. Management has decreed that the production rate must be increased by 25%. Is that possible? Assume that feed must remain unchanged in order to achieve the required surface finish. What is the current production rate and the maximum possible production rate for this job?arrow_forward
- Solve this early but correctlyarrow_forwardPneumatics Bending Tool Problem: Sheet metal is to be bended on a pneumatically actuated bending rig. After it has been clamped the piece is initially bent by a double-acting cylinder A and subsequently bent into the finished form by cylinder B. The sequence is initiated manually by a push button. The circuit must be so constructed that each time a start signal is given a full cycle is completed. Positional Sketch: 1.0 2,0 Electro-pneumatics Sequential Circuit Problem: Initial position all cylinder are retracted. Pressing the start pushbutton makes 5 cycles. After making 5 cycles lamp is energized for 3 secs. Cycle Operation: Cylinder A extend and hold the work piece. Cylinder B extend for 5 secs. Then retract. Cylinder A retract.arrow_forward...Which of the following is the surface velocity of a pulley 18 inches in diameter mounted on a shaft running at 350 rpm?Group of answer choices 122 fpm 1240 fpm 1649 fpm 1187 fpmarrow_forward
- HW3: A block of steel has a length I. = 600 mm and a width w = 450 mm, The thickness is Machined 10 mm, once by Peripheral Milling and once by Planing Machine. Calculate Machining time for each operation if you know the following: 1. For peripheral milling: cutting speed V = 80 m/min, Feed f 0.15 mm/rev, the cutter diameter D- 50 mm with five teeth and depth of cut d- 3 mm. 2. For Planing: clearance on both sides e = 25 mm, cutting speed V = 36.56 m'min, return to cutting ratio m - 0.25, feed f- 5 mm'stroke and depth of cut d- 5 mm. 3. Calculate Material removal rate in both Processes. Solution:arrow_forwardQ1:- Put a mark of True or false for below:- (a):- The milling cutting tool has a single cutting edge? (b):- At the break-even point, the total revenue equals the total cost? (c):- The right-hand rule is used to define machine axes? (d):- Design error is a simple mistake? (e):- The quality of products is poor in mass production? (f):- Allowances must be left before turning? (g):- Up-ward milling is better than down-ward milling?arrow_forwardAn end milling operation is carried out along a straight line path that is 325 mm long. The cut is in a direction parallel to the x‑axis on a CNC machining center. Cutting speed = 30 m/min, and chip load = 0.06 mm. The end milling cutter has two teeth and its diameter = 16.0 mm. The x‑axis uses a DC servomotor connected directly to a leadscrew whose pitch = 6.0 mm. The optical encoder emits 400 pulses per revolution of the screw. Determine (a) feed rate during the cut, (b) rotational speed of the motor, and (c) pulse rate of the encoder at the feed rate indicated.arrow_forward
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