Degarmo's Materials And Processes In Manufacturing
13th Edition
ISBN: 9781119492825
Author: Black, J. Temple, Kohser, Ronald A., Author.
Publisher: Wiley,
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 19, Problem 12RQ
To determine
The basic components of a piercing or blanking die set.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Continuity equation
A
y
x
dx
D
T
معادلة الاستمرارية
Ly
X
Q/Prove that
ди
хе
+ ♥+ ㅇ?
he
me
ze
ོ༞“༠ ?
Q
Derive (continuity equation)?
I want to derive clear mathematics.
motor supplies 200 kW at 6 Hz to flange A of the shaft shown in Figure. Gear B transfers 125 W of power to operating machinery in the factory, and the remaining power in the shaft is mansferred by gear D. Shafts (1) and (2) are solid aluminum (G = 28 GPa) shafts that have the same diameter and an allowable shear stress of t= 40 MPa. Shaft (3) is a solid steel (G = 80 GPa) shaft with an allowable shear stress of t = 55 MPa. Determine:
a) the minimum permissible diameter for aluminum shafts (1) and (2)
b) the minimum permissible diameter for steel shaft (3).
c) the rotation angle of gear D with respect to flange A if the shafts have the minimum permissible diameters as determined in (a) and (b).
Chapter 19 Solutions
Degarmo's Materials And Processes In Manufacturing
Ch. 19 - What distinguishes sheet forming from bulk...Ch. 19 - What is a definition of shearing?Ch. 19 - Prob. 3RQCh. 19 - What measures can be employed to improve the...Ch. 19 - How does fineblanking create shearing in a...Ch. 19 - Prob. 6RQCh. 19 - What types of cuts are made by squaring shears?Ch. 19 - Why might a long shearing cut be made in a...Ch. 19 - What is a slitting operation?Ch. 19 - What is the difference between piercing and...
Ch. 19 - Prob. 11RQCh. 19 - Prob. 12RQCh. 19 - Prob. 13RQCh. 19 - Prob. 14RQCh. 19 - Prob. 15RQCh. 19 - Prob. 16RQCh. 19 - What is the benefit of making dies as a multipiece...Ch. 19 - Prob. 18RQCh. 19 - Prob. 19RQCh. 19 - Prob. 20RQCh. 19 - Prob. 21RQCh. 19 - Prob. 22RQCh. 19 - When making bends in sheet metal, what is the...Ch. 19 - Prob. 24RQCh. 19 - Prob. 25RQCh. 19 - Why does a metal usually become thinner in the...Ch. 19 - Prob. 27RQCh. 19 - Prob. 28RQCh. 19 - Prob. 29RQCh. 19 - What types of operations can be performed on a...Ch. 19 - Prob. 31RQCh. 19 - Prob. 32RQCh. 19 - Prob. 33RQCh. 19 - Prob. 34RQCh. 19 - What is the primary benefit of incorporating a...Ch. 19 - Prob. 36RQCh. 19 - What is the benefit of using a urethane (rubber)...Ch. 19 - What is the objective of the roll bending process?Ch. 19 - What is the role of the form block in draw bending...Ch. 19 - Prob. 40RQCh. 19 - Prob. 41RQCh. 19 - Prob. 42RQCh. 19 - Prob. 43RQCh. 19 - Prob. 44RQCh. 19 - Prob. 45RQCh. 19 - Prob. 46RQCh. 19 - Prob. 47RQCh. 19 - Prob. 48RQCh. 19 - Prob. 49RQCh. 19 - Prob. 50RQCh. 19 - Prob. 51RQCh. 19 - Prob. 52RQCh. 19 - Prob. 53RQCh. 19 - What is the distinction between shallow drawing...Ch. 19 - What is the function of the pressure ring or...Ch. 19 - Prob. 56RQCh. 19 - Prob. 57RQCh. 19 - Prob. 58RQCh. 19 - Prob. 59RQCh. 19 - Prob. 60RQCh. 19 - Prob. 61RQCh. 19 - Prob. 62RQCh. 19 - Prob. 63RQCh. 19 - Prob. 64RQCh. 19 - Prob. 65RQCh. 19 - Prob. 66RQCh. 19 - Prob. 67RQCh. 19 - Prob. 68RQCh. 19 - Prob. 69RQCh. 19 - Prob. 70RQCh. 19 - Prob. 71RQCh. 19 - Prob. 72RQCh. 19 - Prob. 73RQCh. 19 - Prob. 74RQCh. 19 - Prob. 75RQCh. 19 - What are some of the basic methods that have been...Ch. 19 - Prob. 77RQCh. 19 - Prob. 78RQCh. 19 - Prob. 79RQCh. 19 - Prob. 80RQCh. 19 - Prob. 81RQCh. 19 - What properties from a uniaxial tensile test can...Ch. 19 - How is the formability in biaxial tension...Ch. 19 - What is normal anisotropy, R and planar...Ch. 19 - Prob. 85RQCh. 19 - Prob. 86RQCh. 19 - Prob. 87RQCh. 19 - Prob. 88RQCh. 19 - What two hot�forming operations can be used to...Ch. 19 - Prob. 90RQCh. 19 - What are the primary assets and limitations of...Ch. 19 - Prob. 92RQCh. 19 - Prob. 93RQCh. 19 - What are some of the attractive features of...Ch. 19 - What are some of the common types of press frames?Ch. 19 - What are some features that may be included into a...Ch. 19 - Prob. 97RQCh. 19 - Prob. 98RQCh. 19 - Prob. 99RQCh. 19 - Prob. 100RQCh. 19 - The maximum punch force in blanking can be...Ch. 19 - Prob. 2PCh. 19 - Prob. 3PCh. 19 - What are some of the techniques for minimizing the...Ch. 19 - Prob. 5PCh. 19 - Prob. 6PCh. 19 - Prob. 1.1CSCh. 19 - Prob. 1.2CSCh. 19 - Prob. 1.3CSCh. 19 - Polymeric materials have characteristically low...Ch. 19 - If adhesive bonding is specified as a replacement...Ch. 19 - Which of the material/process options do you feel...Ch. 19 - Prob. 1.7CSCh. 19 - Prob. 1.8CSCh. 19 - Prob. 2.2CSCh. 19 - Prob. 2.3CSCh. 19 - Prob. 2.4CSCh. 19 - Prob. 2.5CSCh. 19 - Prob. 2.6CSCh. 19 - Prob. 2.7CSCh. 19 - Prob. 3.1CSCh. 19 - Prob. 3.2CSCh. 19 - Prob. 3.3CSCh. 19 - Both materials will usually require some form of...Ch. 19 - Various design features have been incorporated...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.Similar questions
- First monthly exam Gas dynamics Third stage Q1/Water at 15° C flow through a 300 mm diameter riveted steel pipe, E-3 mm with a head loss of 6 m in 300 m length. Determine the flow rate in pipe. Use moody chart. Q2/ Assume a car's exhaust system can be approximated as 14 ft long and 0.125 ft-diameter cast-iron pipe ( = 0.00085 ft) with the equivalent of (6) regular 90° flanged elbows (KL = 0.3) and a muffler. The muffler acts as a resistor with a loss coefficient of KL= 8.5. Determine the pressure at the beginning of the exhaust system (pl) if the flowrate is 0.10 cfs, and the exhaust has the same properties as air.(p = 1.74 × 10-3 slug/ft³, u= 4.7 x 10-7 lb.s/ft²) Use moody chart (1) MIDAS Kel=0.3 Q3/Liquid ammonia at -20°C is flowing through a 30 m long section of a 5 mm diameter copper tube(e = 1.5 × 10-6 m) at a rate of 0.15 kg/s. Determine the pressure drop and the head losses. .μ= 2.36 × 10-4 kg/m.s)p = 665.1 kg/m³arrow_forward2/Y Y+1 2Cp Q1/ Show that Cda Az x P1 mactual Cdf Af R/T₁ 2pf(P1-P2-zxgxpf) Q2/ A simple jet carburetor has to supply 5 Kg of air per minute. The air is at a pressure of 1.013 bar and a temperature of 27 °C. Calculate the throat diameter of the choke for air flow velocity of 90 m/sec. Take velocity coefficient to be 0.8. Assume isentropic flow and the flow to be compressible. Quiz/ Determine the air-fuel ratio supplied at 5000 m altitude by a carburetor which is adjusted to give an air-fuel ratio of 14:1 at sea level where air temperature is 27 °C and pressure is 1.013 bar. The temperature of air decreases with altitude as given by the expression The air pressure decreases with altitude as per relation h = 19200 log10 (1.013), where P is in bar. State any assumptions made. t = ts P 0.0065harrow_forward36 2) Use the method of MEMBERS to determine the true magnitude and direction of the forces in members1 and 2 of the frame shown below in Fig 3.2. 300lbs/ft member-1 member-2 30° Fig 3.2. https://brightspace.cuny.edu/d21/le/content/433117/viewContent/29873977/Viewarrow_forward
- Can you solve this for me?arrow_forward5670 mm The apartment in the ground floor of three floors building in Fig. in Baghdad city. The details of walls, roof, windows and door are shown. The window is a double glazing and air space thickness is 1.3cm Poorly Fitted-with Storm Sash with wood strip and storm window of 0.6 cm glass thickness. The thickness of door is 2.5 cm. The door is Poor Installation. There are two peoples in each room. The height of room is 280 cm. assume the indoor design conditions are 25°C DBT and 50 RH, and moisture content of 8 gw/kga. The moisture content of outdoor is 10.5 gw/kga. Calculate heat gain for living room : الشقة في الطابق الأرضي من مبنى ثلاثة طوابق في مدينة بغداد يظهر في مخطط الشقة تفاصيل الجدران والسقف والنوافذ والباب. النافذة عبارة عن زجاج مزدوج وسمك الفراغ الهوائي 1.3 سم ضعيف الاحكام مع ساتر حماية مع إطار خشبي والنافذة بسماكة زجاج 0.6 سم سماكة الباب 2.5 سم. الباب هو تركيب ضعيف هناك شخصان في كل غرفة. ارتفاع الغرفة 280 سم. افترض أن ظروف التصميم الداخلي هي DBT25 و R50 ، ومحتوى الرطوبة 8…arrow_forwardHow do i solve this problem?arrow_forward
- Q4/ A compressor is driven motor by mean of a flat belt of thickness 10 mm and a width of 250 mm. The motor pulley is 300 mm diameter and run at 900 rpm and the compressor pulley is 1500 mm diameter. The shaft center distance is 1.5 m. The angle of contact of the smaller pulley is 220° and on the larger pulley is 270°. The coefficient of friction between the belt and the small pulley is 0.3, and between the belt and the large pulley is 0.25. The maximum allowable belt stress is 2 MPa and the belt density is 970 kg/m³. (a) What is the power capacity of the drive and (b) If the small pulley replaced by V-grooved pulley of diameter 300 mm, grooved angle of 34° and the coefficient of friction between belt and grooved pulley is 0.35. What will be the power capacity in this case, assuming that the diameter of the large pulley remain the same of 1500 mm.arrow_forwardYou are tasked with designing a power drive system to transmit power between a motor and a conveyor belt in a manufacturing facility as illustrated in figure. The design must ensure efficient power transmission, reliability, and safety. Given the following specifications and constraints, design drive system for this application: Specifications: Motor Power: The electric motor provides 10 kW of power at 1,500 RPM. Output Speed: The output shaft should rotate at 150 rpm. Design Decisions: Transmission ratio: Determine the necessary drive ratio for the system. Shaft Diameter: Design the shafts for both the motor and the conveyor end. Material Selection: Choose appropriate materials for the gears, shafts. Bearings: Select suitable rolling element bearings. Constraints: Space Limitation: The available space for the gear drive system is limited to a 1-meter-long section. Attribute 4 of CEP Depth of knowledge required Fundamentals-based, first principles analytical approach…arrow_forward- | العنوان In non-continuous dieless drawing process for copper tube as shown in Fig. (1), take the following data: Do-20mm, to=3mm, D=12mm, ti/to=0.6 and v.-15mm/s. Calculate: (1) area reduction RA, (2) drawing velocity v. Knowing that: ti: final thickness V. Fig. (1) ofthrearrow_forward
- A direct extrusion operation produces the cross section shown in Fig. (2) from an aluminum billet whose diameter 160 mm and length - 700 mm. Determine the length of the extruded section at the end of the operation if the die angle -14° 60 X Fig. (2) Note: all dimensions in mm.arrow_forwardFor hot rolling processes, show that the average strain rate can be given as: = (1+5)√RdIn(+1)arrow_forward: +0 usão العنوان on to A vertical true centrifugal casting process is used to produce bushings that are 250 mm long and 200 mm in outside diameter. If the rotational speed during solidification is 500 rev/min, determine the inside radii at the top and bottom of the bushing if R-2R. Take: -9.81 mis ۲/۱ ostrararrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Welding: Principles and Applications (MindTap Cou...Mechanical EngineeringISBN:9781305494695Author:Larry JeffusPublisher:Cengage Learning
Welding: Principles and Applications (MindTap Cou...
Mechanical Engineering
ISBN:9781305494695
Author:Larry Jeffus
Publisher:Cengage Learning
Types of Manufacturing Process | Manufacturing Processes; Author: Magic Marks;https://www.youtube.com/watch?v=koULXptaBTs;License: Standard Youtube License