Engineering Fundamentals: An Introduction to Engineering (MindTap Course List)
5th Edition
ISBN: 9781305084766
Author: Saeed Moaveni
Publisher: Cengage Learning
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Chapter 20, Problem 54P
To determine
Explain about straight line depreciation and provide examples.
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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 20 Solutions
Engineering Fundamentals: An Introduction to Engineering (MindTap Course List)
Ch. 20.4 - Prob. 1BYGCh. 20.4 - Prob. 2BYGCh. 20.4 - Prob. 3BYGCh. 20.4 - Prob. 4BYGCh. 20.4 - Prob. 5BYGCh. 20.4 - Prob. BYGVCh. 20.5 - Prob. 1BYGCh. 20.5 - Prob. 2BYGCh. 20.5 - Prob. 3BYGCh. 20.5 - Prob. BYGV
Ch. 20.8 - Prob. 1BYGCh. 20.8 - Prob. 2BYGCh. 20.8 - Prob. 3BYGCh. 20 - Prob. 1PCh. 20 - Prob. 2PCh. 20 - Prob. 3PCh. 20 - Prob. 4PCh. 20 - Prob. 5PCh. 20 - Prob. 6PCh. 20 - Prob. 7PCh. 20 - Prob. 8PCh. 20 - Prob. 9PCh. 20 - Prob. 10PCh. 20 - Prob. 11PCh. 20 - Prob. 12PCh. 20 - Prob. 13PCh. 20 - Prob. 14PCh. 20 - Prob. 15PCh. 20 - Prob. 16PCh. 20 - Prob. 17PCh. 20 - Prob. 18PCh. 20 - Prob. 19PCh. 20 - Prob. 20PCh. 20 - Prob. 21PCh. 20 - Prob. 22PCh. 20 - Prob. 23PCh. 20 - Prob. 24PCh. 20 - Prob. 25PCh. 20 - Prob. 26PCh. 20 - Prob. 27PCh. 20 - Prob. 28PCh. 20 - Prob. 29PCh. 20 - Prob. 30PCh. 20 - Prob. 31PCh. 20 - Prob. 32PCh. 20 - Prob. 33PCh. 20 - Prob. 34PCh. 20 - Prob. 35PCh. 20 - Prob. 36PCh. 20 - Prob. 37PCh. 20 - Prob. 38PCh. 20 - Prob. 39PCh. 20 - Prob. 40PCh. 20 - Prob. 41PCh. 20 - Prob. 42PCh. 20 - Prob. 43PCh. 20 - Prob. 44PCh. 20 - Prob. 45PCh. 20 - Prob. 46PCh. 20 - Prob. 47PCh. 20 - Prob. 48PCh. 20 - Prob. 49PCh. 20 - Prob. 50PCh. 20 - Prob. 52PCh. 20 - Prob. 53PCh. 20 - Prob. 54P
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