INDUSTRIAL MOTOR CONTROL
7th Edition
ISBN: 9780357670590
Author: Herman
Publisher: CENGAGE L
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Question
Chapter 2, Problem 9RQ
To determine
The reason for using a circle to denote a coil in a schematic of motor control.
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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 2 Solutions
INDUSTRIAL MOTOR CONTROL
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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
- Recall that the CWH equation involves two important assumptions. Let us investigate how these assumptions affect the accuracy of state trajectories under the control inputs optimized in (a) and (b). (c.1): Discuss the assumptions about the chief and deputy orbits that are necessary for deriving CWH.arrow_forwardPROBLEM 2.50 1.8 m The concrete post (E-25 GPa and a = 9.9 x 10°/°C) is reinforced with six steel bars, each of 22-mm diameter (E, = 200 GPa and a, = 11.7 x 10°/°C). Determine the normal stresses induced in the steel and in the concrete by a temperature rise of 35°C. 6c " 0.391 MPa 240 mm 240 mm 6₁ = -9.47 MPaarrow_forwardFor some viscoelastic polymers that are subjected to stress relaxation tests, the stress decays with time according to a(t) = a(0) exp(-4) (15.10) where σ(t) and o(0) represent the time-dependent and initial (i.e., time = 0) stresses, respectively, and t and T denote elapsed time and the relaxation time, respectively; T is a time-independent constant characteristic of the material. A specimen of a viscoelastic polymer whose stress relaxation obeys Equation 15.10 was suddenly pulled in tension to a measured strain of 0.5; the stress necessary to maintain this constant strain was measured as a function of time. Determine E (10) for this material if the initial stress level was 3.5 MPa (500 psi), which dropped to 0.5 MPa (70 psi) after 30 s.arrow_forward
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