ENGINEERING DESIGN PROCESS
3rd Edition
ISBN: 9781305253285
Author: HAIK
Publisher: CENGAGE L
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Textbook Question
Chapter 9.9, Problem 2P
Why would a designer place items (demos of function) in the morphological chart if he or she knows that they would not be feasible?
Expert Solution & Answer
To determine
The purpose of placing items (demos for function) in the morphological chart if he or she knows that they would not be feasible.
Explanation of Solution
In morphological chart every possible solution is placed even though knowing that they might not be feasible to avail those choices whenever a compatible combination is chosen according to a particular theme such as, economy, environment, ergonomics etc..
A compatible solution might not be correct for a particular theme which makes the designer to take all possible combinations into account.
Also every possible combination is represented in morphological chart to identify the solution space of a design problem.
Thus the above discussion states the purpose of placing items (demos for function) in the morphological chart even if he or she knows that they would not be feasible.
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Find the internal torques for segments AB, BC, and CD (in N-m) by drawing the internal torque diagram, the maximum torsional shear on the shaft in MPa, and the relative rotation of section A with respect to section D in degrees.
Determine the heel and toe stresses and the factor of safeties for sliding and
overturning for the gravity dam section shown in the figure below for the following
loading conditions:
- Horizontal earthquake (Kh) = 0.1
-
Normal uplift pressure with gallery drain working
Silt deposit up to 30 m height
- No wave pressure and no ice pressure
Unit weight of concrete = 2.4 Ton/m³ and unit weight of silty water = 1.4 Ton/m³
- Submerged weight of silt = 0.9 Ton/m³
==
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Solve this question with the presence of gallery and without gallery., discuss
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Solve in table
144 m
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The efficiency for a steel specimen immersed in a phosphating tank is the weight of the phosphate coating divided by the metal loss (both in mg/ft²). An article gave the accompanying data on tank temperature (x) and efficiency ratio (y).
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171
173
174
175
175
176
177
178
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0.82
1.39
1.48
0.95
1.15
1.06
1.00
1.82
Temp.
181
181
181
181
181
182
182
183
Ratio
1.35 1.52
1.55
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2.17
0.92
1.45
0.82
Temp.
183
183
183
185
185
186
187
189
Ratio
1.85
2.04
2.70
1.59
2.54 3.04 1.89
3.12
(a) Determine the equation of the estimated regression line. (Round all numerical values to four decimal places.)
y =
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(183, 0.82)
(183, 1.85)
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(183, 2.70)
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