Welding: Principles and Applications (MindTap Course List)
8th Edition
ISBN: 9781305494695
Author: Larry Jeffus
Publisher: Cengage Learning
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Chapter 26, Problem 9R
To determine
The reason for metals not forming into one-large, single crystal.
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Complete the following problems. Show your work/calculations, save as.pdf and upload to the
assignment in Blackboard.
missing information to present a completed program. (Hint: You may have to look up geometry
for the center drill and standard 0.5000 in twist drill to know the required depth to drill).
1. What are the x and y dimensions for the center position of holes 1,2, and 3 in the part shown in
Figure 26.2 (below)?
6.0000
Zero
reference
point
7118
1.0005
1.0000
1.252
Bore
6.0000
.7118
Cbore
0.2180 deep
(3 holes)
2.6563 1.9445
Figure 26.2
026022 (8lot and Drill Part)
(Setup Instructions---
(UNITS: Inches
(WORKPIECE NAT'L SAE 1020 STEEL
(Workpiece: 3.25 x 2.00 x0.75 in. Plate
(PRZ Location 054:
'
XY 0.0 - Upper Left of Fixture
TOP OF PART 2-0
(Tool List
( T02 0.500 IN 4 FLUTE FLAT END MILL
#4 CENTER DRILL
Dashed line indicates-
corner of original stock
( T04
T02
3.000 diam. slot
0.3000 deep.
0.3000 wide
Intended toolpath-tangent-
arc entry and exit sized to
programmer's judgment…
A program to make the part depicted in Figure 26.A has been created, presented in figure 26.B, but some information still needs to be filled in. Compute the tool locations, depths, and other missing information to present a completed program. (Hint: You may have to look up geometry for the center drill and standard 0.5000 in twist drill to know the required depth to drill).
We consider a laminar flow induced by an impulsively started infinite flat
plate. The y-axis is normal to the plate. The x- and z-axes form a plane parallel
to the plate. The plate is defined by y = 0. For time t <0, the plate and the flow
are at rest. For t≥0, the velocity of the plate is parallel to the 2-coordinate;
its value is constant and equal to uw. At infinity, the flow is at rest. The flow
induced by the motion of the plate is independent of z.
(a) From the continuity equation, show that v=0 everywhere in the flow and
the resulting momentum equation is
მu
Ət
Note that this equation has the form of a diffusion equation (the same form as
the heat equation).
(b) We introduce the new variables T, Y and U such that
T=kt, Y=k/2y, U = u
where k is an arbitrary constant. In the new system of variables, the solution
is U(Y,T). The solution U(Y,T) is expressed by a function of Y and T and the
solution u(y, t) is expressed by a function of y and t. Show that the functions are
identical.…
Chapter 26 Solutions
Welding: Principles and Applications (MindTap Course List)
Ch. 26 - What gives metals their desirable properties?Ch. 26 - What is heat?Ch. 26 - What are the basic units of measure for heat?Ch. 26 - What is sensible heat?Ch. 26 - Prob. 5RCh. 26 - What does the color of light given off from a hot...Ch. 26 - Prob. 7RCh. 26 - In steel-making, what is ore combined with in the...Ch. 26 - Prob. 9RCh. 26 - What is an alloy?
Ch. 26 - Using Figure 26-18, answer the following...Ch. 26 - What is a eutectic composition?Ch. 26 - Using Table 26-3, what are the lowest and highest...Ch. 26 - Approximately how many degrees wide is the...Ch. 26 - Referring to Figure 26-20, what color would...Ch. 26 - Referring to Figure 26-20, what is the approximate...Ch. 26 - Prob. 17RCh. 26 - Referring to Figure 26-20, above what temperature...Ch. 26 - Prob. 19RCh. 26 - What is known as the critical temperature of...Ch. 26 - Can a metal have all the mechanical properties at...Ch. 26 - What other properties can a metal's hardness...Ch. 26 - Which property, brittleness or ductility, will let...Ch. 26 - What is toughness?Ch. 26 - What are the common types of strength...Ch. 26 - Prob. 26RCh. 26 - What are the three steps in precipitation...Ch. 26 - How do ferrite and cementite work together to form...Ch. 26 - Why is brine quenching faster than water...Ch. 26 - What can be done to speed up the quenching rate in...Ch. 26 - Why is the formation of martensite a problem when...Ch. 26 - How can the effects of cold working be removed?Ch. 26 - Prob. 33RCh. 26 - Referring to Figure 26-20, what would the preheat...Ch. 26 - Why must the stress-relieving temperature be kept...Ch. 26 - What properties can annealing produce in metals?Ch. 26 - How long does it take the weld metal to go through...Ch. 26 - What are some sources of hydrogen that can...Ch. 26 - How can nitrogen get into an SMA weld?Ch. 26 - What are some of the problems that oxygen can...Ch. 26 - When do cold cracks develop?Ch. 26 - What is carbide precipitation?
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- Part A: Suppose you wanted to drill a 1.5 in diameter hole through a piece of 1020 cold-rolled steel that is 2 in thick, using an HSS twist drill. What values if feed and cutting speed will you specify, along with an appropriate allowance? Part B: How much time will be required to drill the hole in the previous problem using the HSS drill?arrow_forward1.1 m 1.3 m B 60-mm diameter Brass 40-mm diameter Aluminum PROBLEM 2.52 - A rod consisting of two cylindrical portions AB and BC is restrained at both ends. Portion AB is made of brass (E₁ = 105 GPa, α = 20.9×10°/°C) and portion BC is made of aluminum (Ę₁ =72 GPa, α = 23.9×10/°C). Knowing that the rod is initially unstressed, determine (a) the normal stresses induced in portions AB and BC by a temperature rise of 42°C, (b) the corresponding deflection of point B.arrow_forward30 mm D = 40 MPa -30 mm B C 80 MPa PROBLEM 2.69 A 30-mm square was scribed on the side of a large steel pressure vessel. After pressurization, the biaxial stress condition at the square is as shown. For E = 200 GPa and v=0.30, determine the change in length of (a) side AB, (b) side BC, (c) diagnonal AC.arrow_forward
- Please solve in detail this problem thank youarrow_forward0,5 mm 450 mm 350 mm Bronze A = 1500 mm² E = 105 GPa प 21.6 × 10-PC Aluminum A = 1800 mm² £ = 73 GPa = a 23.2 × 10-PC PROBLEM 2.58 Knowing that a 0.5-mm gap exists when the temperature is 24°C, determine (a) the temperature at which the normal stress in the aluminum bar will be equal to -75 MPa, (b) the corresponding exact length of the aluminum bar.arrow_forward0.5 mm 450 mm -350 mm Bronze Aluminum A 1500 mm² A 1800 mm² E 105 GPa E 73 GPa K = 21.6 X 10 G < = 23.2 × 10-G PROBLEM 2.59 Determine (a) the compressive force in the bars shown after a temperature rise of 82°C, (b) the corresponding change in length of the bronze bar.arrow_forward
- The truss shown below sits on a roller at A and a pin at E. Determine the magnitudes of the forces in truss members GH, GB, BC and GC. State whether they are in tension or compression or are zero force members.arrow_forwardA weight (W) hangs from a pulley at B that is part of a support frame. Calculate the maximum possible mass of the weight if the maximum permissible moment reaction at the fixed support is 100 Nm. Note that a frictionless pin in a slot is located at C.arrow_forwardIt is the middle of a winter snowstorm. Sally and Jin take shelter under an overhang. The loading of the snow on top of the overhang is shown in the figure below. The overhang is attached to the wall at points A and B with pin supports. Another pin is at C. Determine the reactions of the pin supports at A and B. Express them in Cartesian vector form.arrow_forward
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