EBK ENGINEERING FUNDAMENTALS: AN INTROD
5th Edition
ISBN: 9780100543409
Author: MOAVENI
Publisher: YUZU
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Question
Chapter 16, Problem 1P
To determine
Draw the top, the front and the right side orthogonal views of the given object and indicate when an object needs only one or two views.
Expert Solution & Answer
Answer to Problem 1P
The top, front and right-side orthogonal view of the given object has been drawn and object which needs one or two views are investigated.
Explanation of Solution
Calculation:
Sketch the given figure as follows,
When the observer observes the object from the front side, the view captured on the front plane is represented below Figure 2.
When the observer observes the object from the top side, the view captured on the top plane is represented below in Figure 3.
When the observer observers the object from the right side, the view captured on the right side plane is represented below in Figure 4.
Refer to Figure 4, the dashed lines are indicates the intersection of two planes that are not visible from the direction of the observer.
Here, the given object is complex geometry. Hence, it requires at least three plane of projected views namely front, top and side views for the full description of object.
Conclusion:
Hence, the top, front and right-side orthogonal views of the given object have been drawn and object which needs one or two views are investigated.
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Students have asked these similar questions
Consider the specifications for an asphaltic concrete mixture and the results of a sieve analysis below.
Coarse aggregates:
Fine aggregates:
Filler:
60%
35%
5%
Percent of Weight of Aggregate or Filler
Passing
Sieve
Designation
Retained on
Sieve
Designation
Coarse
Aggregate
Fine
Aggregate
Mineral
Filler
3/4 in. (19 mm)
1/2 in.
6
1/2 in. (12.5 mm)
3/8 in.
15
3/8 in. (9.5 mm)
No. 4
50
-
No. 4 (4.75 mm)
No. 10
20
1
No. 10 (2 mm)
No. 40 (0.425 mm)
No. 40
9
35
-
No. 80
31
40
No. 80 (0.180 mm)
No. 200 (0.075 mm)
Total
No. 200
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33
-
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25
35
100
100
100
Determine the proportion of different aggregates to obtain the required gradation.
Percent of Total Weight of Mixture
Passing
Sieve
Designation
Retained on
Sieve
Designation
Coarse
Aggregate
Fine
Aggregate
3/4 in. (19 mm)
1/2 in. (12.5 mm)
3/8 in. (9.5 mm)
1/2 in.
3/8 in.
No. 4
No. 4 (4.75 mm)
No. 10
No. 10 (2 mm)
No. 40
No. 40 (0.425 mm)
No. 80
No. 80 (0.180 mm)
No. 200
No. 200 (0.075 mm)
Total
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Mineral
Filler
Total
100
Results obtained from laboratory tests on a sample of RC-250 asphalt cement are given. Determine whether the properties of this material meet the Asphalt Institute specifications for this type of material; if not, note the differences. (For each specification, enter the minimum acceptab
value in the same units as used in the test results.)
• Kinematic viscosity at 140°F (60°C) = 230 centistokes
• Flash point (Tagliabue open cup) = 89°F
• Distillation test where distillate percent by volume of total distillate to 680°F (360ºC)
• To 437°F (225°C) = 27%
• To 500°F (260°C) = 69%
• To 600°F (316°C) = 72%
• Residue from distillation to 680°F (360°C) by volume percentage of sample by difference
• Tests on Residue from Distillation:
• Ductility at 77°F (25°C) = 92 cm
• Absolute viscosity at 140°F (60°C) = 620 poises
⚫ Solubility = 90%
Property
Kinematic Viscosity
=
74%
Specification
Test Results
Were Specifications Met?
centistokes
230 centistokes
---Select--- ✓
Flash Point
°F
89°F…
Problem 2
Two machines produce rivets for a factory job. The number of sub-standard rivets per hour by the two
machines are random variables, denoted by X1 and X2. The bivariate PMF of X1 and X2, Px,x,(x1,x2), is
given in the table below.
X2=0
X2=1
X2=2
X2=3
X₁-0
0.07
0.05
0.02
0.01
X₁ =1
0.05
0.16
0.12
0.02
X₁ =2
0.02
0.12
0.17
0.05
X₁ =3
0.01
0.01
0.05
0.07
Chapter 16 Solutions
EBK ENGINEERING FUNDAMENTALS: AN INTROD
Ch. 16.1 - Prob. 1BYGCh. 16.1 - Prob. 2BYGCh. 16.1 - Prob. 3BYGCh. 16.1 - Prob. 4BYGCh. 16.1 - Prob. BYGVCh. 16.3 - Prob. 1BYGCh. 16.3 - Prob. 2BYGCh. 16.3 - Prob. 3BYGCh. 16.3 - Prob. 4BYGCh. 16.3 - Prob. 5BYG
Ch. 16.3 - Prob. BYGVCh. 16.4 - Prob. 1BYGCh. 16.4 - Prob. 2BYGCh. 16.4 - Give examples of common electrical engineering...Ch. 16.4 - Prob. 4BYGCh. 16.4 - VocabularyState the meaning of the following...Ch. 16 - Prob. 1PCh. 16 - Prob. 2PCh. 16 - Prob. 3PCh. 16 - Prob. 4PCh. 16 - Prob. 5PCh. 16 - Prob. 6PCh. 16 - Prob. 7PCh. 16 - Prob. 8PCh. 16 - Prob. 9PCh. 16 - Prob. 10PCh. 16 - Prob. 11PCh. 16 - Prob. 12PCh. 16 - Prob. 13PCh. 16 - Prob. 14PCh. 16 - Prob. 15PCh. 16 - Prob. 16PCh. 16 - Prob. 17PCh. 16 - Prob. 18PCh. 16 - Prob. 19PCh. 16 - Prob. 20PCh. 16 - Prob. 21PCh. 16 - Prob. 22PCh. 16 - Prob. 23PCh. 16 - Prob. 24PCh. 16 - Prob. 25PCh. 16 - Prob. 26PCh. 16 - Prob. 27PCh. 16 - Prob. 28PCh. 16 - Prob. 29PCh. 16 - Prob. 30PCh. 16 - Prob. 31PCh. 16 - Prob. 32PCh. 16 - Prob. 33PCh. 16 - Prob. 34PCh. 16 - Prob. 35PCh. 16 - Prob. 36PCh. 16 - Prob. 37PCh. 16 - Prob. 38PCh. 16 - Prob. 39PCh. 16 - Prob. 40PCh. 16 - Prob. 41PCh. 16 - Prob. 42PCh. 16 - Prob. 43PCh. 16 - Prob. 44PCh. 16 - Prob. 45PCh. 16 - Prob. 46PCh. 16 - Prob. 47PCh. 16 - Prob. 48PCh. 16 - Prob. 49PCh. 16 - Prob. 50P
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