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MindTap Engineering for Garber/Hoel's Traffic and Highway Engineering, 5th Edition, [Instant Access], 1 term (6 months)
5th Edition
ISBN: 9781305581159
Author: Nicholas J. Garber; Lester A. Hoel
Publisher: Cengage Learning US
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Question
Chapter 12, Problem 1P
To determine
The description of the two basic demand forecasting situations in transportation planning.
Expert Solution & Answer
Explanation of Solution
Introduction:
The number of vehicles or persons per unit time under a set of given land-use, socioeconomic, and environmental conditions that are expected to travel on a given transportation system are expressed as the travel demand.
In transportation planning, there are two basic demand-forecasting situations.
- One situation involves travel demand studies for urban areas
- Second situation deals with intercity travel demand.
The first one forecasts, was developed in the 1950s and 1960s, and it requires extensive data which can be prepared by
Data gathered about travel characteristic can be further aggregated by the zone for formulating the relationships between the variables and then for the calibration of various models.
The second demand forecasting situations gathers data on a larger extent than for the first one. The intercity travel collects data such as city population, average city income, and travel time or the travel cost between different cities.
Conclusion:
Thus, there are two basic demand-forecasting situations in transportation planning. The first one deals with the travel demand studies for urban areas, and the second one involves intercity travel demands.
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Students have asked these similar questions
The given beam has continuous lateral support. If the live load is twice the dead load, what is the
maximum total service load, in kips / ft, that can be supported? A992 steel is used: Fy = 50 ksi,
Fu=65 ksi. Take L = 30 ft.
bf
For W40 x 149:
2tf
=
7.11,
=
=
54.3, Z 598 in.³
tw
W
W40 X 149
L
(Express your answers to three significant figures.)
a. Use LRFD.
Wtotal =
kips/ft
b. Use ASD.
Wtotal
kips/ft
The beam shown in the figure below is a W16 × 31 of A992 steel and has continuous lateral support.
The two concentrated loads are service live loads. Neglect the weight of the beam and determine
whether the beam is adequate. Suppose that P = 52 k.
For W16 × 31: d = 15.9 in., tw
=
0.275 in., h/tw = 51.6,
and M = M₁ = 203 ft-kip, Mn/₁ = Mp/α = 135 ft-kip.
P
Р
W16 x 31
a. Use LRFD.
Calculate the required moment strength, the allowable shear strength, and the maximum shear.
(Express your answers to three significant figures.)
Mu
=
OvVn
=
ft-kip
kips
kips
Vu =
Beam is -Select-
b. Use ASD.
Calculate the required moment strength, the allowable shear strength, and the maximum shear.
(Express your answers to three significant figures.)
Ma =
Vn/b
-
Va =
Beam is -Select-
ft-kip
kips
kips
Determine the smallest value of yield stress Fy, for which a W-, M-, or S-shape from the list below will
become slender.
bf/2tfh/tw
Shape
W12 × 72
8.99
22.6
W12 × 26
8.54
47.2
M4 × 6
11.9
22.0
M12 x 11.8
6.81
62.5
M6 × 4.4
5.39
47.0
S24 × 80
4.02
41.4
S10 × 35
5.03
13.4
(Express your answer to three significant figures.)
Fy
=
ksi
To which shape does this value apply?
-Select-
✓
Chapter 12 Solutions
MindTap Engineering for Garber/Hoel's Traffic and Highway Engineering, 5th Edition, [Instant Access], 1 term (6 months)
Ch. 12 - Prob. 1PCh. 12 - Prob. 2PCh. 12 - Prob. 3PCh. 12 - Prob. 4PCh. 12 - Prob. 5PCh. 12 - Prob. 6PCh. 12 - Prob. 7PCh. 12 - Prob. 8PCh. 12 - Prob. 9PCh. 12 - Given a table with production and attraction data,...
Ch. 12 - Given a table with production and attraction data,...Ch. 12 - Prob. 12PCh. 12 - Prob. 13PCh. 12 - Prob. 14PCh. 12 - Prob. 15PCh. 12 - Prob. 16PCh. 12 - Prob. 17PCh. 12 - Prob. 18PCh. 12 - Prob. 19PCh. 12 - Prob. 20PCh. 12 - Prob. 21PCh. 12 - Prob. 22PCh. 12 - Prob. 23PCh. 12 - Prob. 24PCh. 12 - Prob. 25PCh. 12 - Prob. 26PCh. 12 - Prob. 27PCh. 12 - Prob. 28P
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