EBK 3N3-EBK: INDUSTRIAL MOTOR CONTROL
7th Edition
ISBN: 9780176919962
Author: Herman
Publisher: VST
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Chapter 14, Problem 13RQ
How can a silicon diode be made to measure temperature?
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2. 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).
Dashed line indicates -
corner of original stock
Intended toolpath-tangent -
arc entry and exit sized to
programmer's judgment
026022 (Slot and Drill Part)
(Setup Instructions.
(UNITS: Inches
(WORKPIECE MAT'L: SAE 1020 STEEL
(Workpiece: 3.25 x 2.00 x0.75 in. Plate
(PRZ Location G54:
( XY 0.0 Upper Left of Fixture
( TOP OF PART 2-0
(Tool List:
)
(
T04
T02 0.500 IN 4 FLUTE FLAT END MILL)
#4 CENTER DRILL
'
T02
0.500 TWIST DRILL
N010 GOO G90 G17 G20 G49 G40 G80 G54
N020 M06 T02 (0.5 IN 4-FLUTE END MILL)
R0.750
N030 S760 M03 G00 x
N040 043 H02 2
Y
(P1)
(RAPID DOWN -TLO)
P4
NO50 MOB (COOLANT ON)
N060 G01 X
R1.000
N070…
6–95. The reaction of the ballast on the railway tie can be assumed uniformly distributed over its length as shown. If the wood has an allowable bending stress of σallow=1.5 ksi, determine the required minimum thickness t of the rectangular cross section of the tie to the nearest 18 in. Please include all steps. Also if you can, please explain how you found Mmax using an equation rather than using just the moment diagram. Thank you!
6–53. If the moment acting on the cross section is M=600 N⋅m, determine the resultant force the bending stress produces on the top board. Please explain each step. Please explain how you got the numbers and where you plugged them in to solve the problem. Thank you!
Chapter 14 Solutions
EBK 3N3-EBK: INDUSTRIAL MOTOR CONTROL
Ch. 14 - Should a metal bar be heated or cooled to make it...Ch. 14 - Prob. 2RQCh. 14 - Prob. 3RQCh. 14 - Why are bimetal strips often formed into a spiral...Ch. 14 - Why should electrical contacts never be permitted...Ch. 14 - What two factors determine the amount of voltage...Ch. 14 - Prob. 7RQCh. 14 - Prob. 8RQCh. 14 - Prob. 9RQCh. 14 - Prob. 10RQ
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- Solving coplanar forcesarrow_forwardComplete the following problems. Show your work/calculations, save as.pdf and upload to the assignment in Blackboard. 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 7118 Zero reference point 1.0005 1.0000 1.252 Bore C' bore 1.250 6.0000 .7118 0.2180 deep (3 holes) 2.6563 1.9445 3.000 diam. slot 0.3000 deep. 0.3000 wide 2.6563 1.9445arrow_forwardComplete 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…arrow_forward
- 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).arrow_forwardWe 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.…arrow_forwardPart 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_forward
- 1.1 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_forwardPlease solve in detail this problem thank youarrow_forward
- 0,5 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_forwardThe 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_forward
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