Understanding Motor Controls
4th Edition
ISBN: 9781337798686
Author: Stephen L. Herman
Publisher: Delmar Cengage Learning
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Chapter 39, Problem 5RQ
To determine
Identify why the PLC output is used to energize the coil of the motor starter instead of the motor coil.
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Given answers to be: i) 14.65 kN; 6.16 kN; 8.46 kN ii) 8.63 kN; 9.88 kN iii) Bearing 6315 for B1 & B2, or Bearing 6215 for B1
(b)
A steel 'hot rolled structural hollow section' column of length 5.75 m, has
the cross-section shown in Figure Q.5(b) and supports a load of 750 kN.
During service, it is subjected to axial compression loading where one end
of the column is effectively restrained in position and direction (fixed) and
the other is effectively held in position but not in direction (pinned).
i)
Given that the steel has a design strength of 275 MN/m², determine
the load factor for the structural member based upon the BS5950
design approach using Datasheet Q.5(b).
[11]
ii)
Determine the axial load that can be supported by the column
using the Rankine-Gordon formula, given that the yield strength of
the material is 280 MN/m² and the constant *a* is 1/30000.
[6]
300
600
2-300 mm
wide x 5 mm
thick plates.
Figure Q.5(b)
L=5.75m
Pinned
Fixed
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Chapter 39 Solutions
Understanding Motor Controls
Ch. 39 - Why are NEMA symbols such as push buttons, float...Ch. 39 - Prob. 2RQCh. 39 - Why are normally closed components such as STOP...Ch. 39 - Prob. 4RQCh. 39 - Prob. 5RQCh. 39 - List four basic rules for developing a program for...Ch. 39 - A PLC requires that times be programmed in...Ch. 39 - When programming in Boolean, what statement should...Ch. 39 - When programming in Boolean, what statement should...Ch. 39 - In a control circuit, it is imperative that a coil...
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