Statics and Mechanics of Materials
2nd Edition
ISBN: 9780073398167
Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David Mazurek
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 3.3, Problem 49P
Two parallel 60-N forces are applied to a lever as shown. Determine the moment of the couple formed by the two forces (a) by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples, (b) by using the perpendicular distance between the two forces, (c) by summing the moments of the two forces about point A.
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Problem (17): water flowing in an open channel of a rectangular cross-section with width (b) transitions from a
mild slope to a steep slope (i.e., from subcritical to supercritical flow) with normal water depths of (y₁) and
(y2), respectively.
Given the values of y₁ [m], y₂ [m], and b [m], calculate the discharge in the channel (Q) in [Lit/s].
Givens:
y1 = 4.112 m
y2 =
0.387 m
b = 0.942 m
Answers:
( 1 ) 1880.186 lit/s
( 2 ) 4042.945 lit/s
( 3 ) 2553.11 lit/s
( 4 ) 3130.448 lit/s
Problem (14): A pump is being used to lift water from an underground
tank through a pipe of diameter (d) at discharge (Q). The total head
loss until the pump entrance can be calculated as (h₁ = K[V²/2g]), h
where (V) is the flow velocity in the pipe. The elevation difference
between the pump and tank surface is (h).
Given the values of h [cm], d [cm], and K [-], calculate the maximum
discharge Q [Lit/s] beyond which cavitation would take place at the
pump entrance. Assume Turbulent flow conditions.
Givens:
h = 120.31 cm
d = 14.455 cm
K = 8.976
Q
Answers:
(1) 94.917 lit/s
(2) 49.048 lit/s
( 3 ) 80.722 lit/s
68.588 lit/s
4
Problem (13): A pump is being used to lift water from the bottom
tank to the top tank in a galvanized iron pipe at a discharge (Q).
The length and diameter of the pipe section from the bottom tank
to the pump are (L₁) and (d₁), respectively. The length and
diameter of the pipe section from the pump to the top tank are
(L2) and (d2), respectively.
Given the values of Q [L/s], L₁ [m], d₁ [m], L₂ [m], d₂ [m],
calculate total head loss due to friction (i.e., major loss) in the
pipe (hmajor-loss) in [cm].
Givens:
L₁,d₁
Pump
L₂,d2
오
0.533 lit/s
L1 =
6920.729 m
d1 =
1.065 m
L2 =
70.946 m
d2
0.072 m
Answers:
(1)
3.069 cm
(2) 3.914 cm
( 3 ) 2.519 cm
( 4 ) 1.855 cm
TABLE 8.1
Equivalent Roughness for New Pipes
Pipe
Riveted steel
Concrete
Wood stave
Cast iron
Galvanized iron
Equivalent Roughness, &
Feet
Millimeters
0.003-0.03 0.9-9.0
0.001-0.01 0.3-3.0
0.0006-0.003 0.18-0.9
0.00085
0.26
0.0005
0.15
0.045
0.000005
0.0015
0.0 (smooth) 0.0 (smooth)
Commercial steel or wrought iron 0.00015
Drawn…
Chapter 3 Solutions
Statics and Mechanics of Materials
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