Applied Fluid Mechanics (7th Edition)
7th Edition
ISBN: 9780132558921
Author: Robert L. Mott, Joseph A. Untener
Publisher: PEARSON
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Chapter 6, Problem 6.15PP
To determine
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Given value into required form by using related equation.
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2. Express the following complex numbers in rectangular form.
(a) z₁ = 2еjл/6
(b) Z2=-3e-jπ/4
(c) Z3 =
√√√3e-j³/4
(d) z4 = − j³
A prismatic beam is built into a structure. You can consider the boundary conditions at A and B to be
fixed supports. The beam was originally designed to withstand a triangular distributed load, however,
the loading condition has been revised and can be approximated by a cosine function as shown in the
figure below. You have been tasked with analysing the structure. As the beam is prismatic, you can
assume that the bending rigidity (El) is constant.
wwo cos
2L
x
A
B
Figure 3: Built in beam with a varying distributed load
In order to do this, you will:
a. Solve the reaction forces and moments at point A and B.
Hint: you may find it convenient to use the principal of superposition.
(2%)
b. Plot the shear force and bending moment diagrams and identify the maximum shear force
and bending moment.
(2%)
c. Develop an expression for the vertical deflection. Clearly state your expression in terms of x.
(1%)
Question 1: Beam Analysis
Two beams (ABC and CD) are connected using a pin immediately to the left of Point C. The pin acts
as a moment release, i.e. no moments are transferred through this pinned connection. Shear forces
can be transferred through the pinned connection. Beam ABC has a pinned support at point A and a
roller support at Point C. Beam CD has a roller support at Point D. A concentrated load, P, is applied
to the mid span of beam CD, and acts at an angle as shown below. Two concentrated moments, MB
and Mc act in the directions shown at Point B and Point C respectively. The magnitude of these
moments is PL.
Moment Release
A
B
с
°
MB = PL
Mc=
= PL
-L/2-
-L/2-
→
P
D
Figure 1: Two beam arrangement for question 1.
To analyse this structure, you will:
a) Construct the free body diagrams for the structure shown above. When constructing your
FBD's you must make section cuts at point B and C. You can represent the structure as three
separate beams. Following this, construct the…
Chapter 6 Solutions
Applied Fluid Mechanics (7th Edition)
Ch. 6 - Convert a volume flow rate of 3.0 gal/min to...Ch. 6 - Convert 459 gal/min to rrP/s.Ch. 6 - Convert 3720 gal/min to mJ/sCh. 6 - Convert 34.3 gal/min to mJ/sCh. 6 - Convert a volume flow rate of 125 L/min to m3/s.Ch. 6 - Convert 4500 L/min to m5/s.Ch. 6 - Convert 15 000 L/minto m3/s.Ch. 6 - Convert 459 gal/min to L/mninCh. 6 - Convert 3720 gal/min to L/minCh. 6 - Convert 23.5cm2/stom3/s.
Ch. 6 - '6.11 Convert 0.296cm5/stom3/s.Ch. 6 - Convert 0.105 cm3/s to L/minCh. 6 - Convert 3.53103m3/s to L/min.Ch. 6 - Convert 5.26106m3stoL/min.Ch. 6 - Prob. 6.15PPCh. 6 - Convert 20 gal/min to ft'/s.Ch. 6 - Convert 2500 gal/min to ft5/s.Ch. 6 - Convert 2.50 gal/min to ft3/s.Ch. 6 - Convert 125 ft3/s to gal/minCh. 6 - Convert 0.060 ft3/s to gal/min.Ch. 6 - Convert 0.03 ft5/s to gal/minCh. 6 - Convert ft5/s sto gal/minCh. 6 - Table 6.21 lists the range of typical volume flow...Ch. 6 - Table 6.2 lists the range of typical volume flow...Ch. 6 - A certain deep-well pump for a residence is rated...Ch. 6 - A small pump delivers 0.85 gal/h of liquid...Ch. 6 - A small metering pump delivers 11.4 gal of a water...Ch. 6 - A small metering pump delivers 19.5 mL/min of...Ch. 6 - Water at 10 C is flowing at 0.075 m3/s Calculate...Ch. 6 - Oil for a hydraulic system (sg =0.90 ) is flowing...Ch. 6 - A liquid refrigerant (sg = 1.08) is flowing at a...Ch. 6 - After the refrigerant from Problem 6.31 flashes...Ch. 6 - A fan delivers 640ft3/min (CFM) of air. 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