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Materials Science and Engineering
9th Edition
ISBN: 9781118324578
Author: Jr. William D. Callister
Publisher: WILEY
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Question
Chapter 2.10, Problem 22QP
To determine
(a). The main difference among ionic, covalent and metallic bonding.
(b). The Pauli Exclusion Principle.
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Students have asked these similar questions
The beam ABC has a mass of 79.0 kg and is supported by
the rope BDC that runs through the frictionless pulley at D
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rope acts on the beam at points B and C and counteracts
the moments due to the beam's weight (acting vertically at
the midpoint of its length) and the weight of the winch
(acting vertically at point C) such that the resultant moment
about point A is equal to zero. Assume that rope segment
CD is vertical and note that rope segment BD is NOT
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b.) Model the two forces the rope exerts on the beam as a
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Enter your answer in Cartesian components.
c.) Model the two forces the rope exerts on the beam as a
single equivalent force (no couple) and determine the
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equivalent force acts (measured parallel to the beam from A
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w1
Three distributed loads act on a beam as shown. The load
between A and B increases linearly from 0 to a maximum
intensity of w₁ = 12.8 lb/ft at point B. The load then varies
linearly with a different slope to an intensity of w₂ = 17.1
lb/ft at C. The load intensity in section CD of the beam is
constant at w3 10.2 lb/ft. For each load region, determine
the resultant force and the location of its line of action
(distance to the right of A for all cases).
cc 10
BY NC SA
2016 Eric Davishahl
=
WI
W2
W3
-b-
C
Values for dimensions on the figure are given in the following
table. Note the figure may not be to scale.
Variable Value
a
4.50 ft
b
5.85 ft
с
4.28 ft
The resultant load in region AB is FR₁ =
lb and acts
ft to the right of A.
The resultant load in region BC is FR2
lb and acts
=
ft to the right of A.
The resultant load in region CD is FR3 =
lb and acts
ft to the right of A.
The T-shaped structure is embedded in a concrete wall at A
and subjected to the force F₁ and the force-couple system
F2 1650 N and M = 1,800 N-m at the locations shown.
Neglect the weight of the structure in your calculations for
this problem.
=
a.) Compute the allowable range of magnitudes for F₁ in the
direction shown if the connection at A will fail when
subjected to a resultant moment with a magnitude of 920 N-
m or higher.
b.) Focusing on the forces and igonoring given M for now.
Using the value for F1, min that you calculated in (a), replace
the two forces F₁ and F2 with a single force that has
equivalent effect on the structure. Specify the equivalent
→>
force Feq in Cartesian components and indicate the
horizontal distance from point A to its line of action (note
this line of action may not intersect the structure).
c.) Now, model the entire force system (F1,min, F2, and M)
as a single force and couple acting at the junction of the
horizontal and vertical sections of the…
Chapter 2 Solutions
Materials Science and Engineering
Ch. 2.10 - Prob. 1QPCh. 2.10 - Prob. 2QPCh. 2.10 - Prob. 3QPCh. 2.10 - Prob. 4QPCh. 2.10 - Prob. 5QPCh. 2.10 - Prob. 6QPCh. 2.10 - Prob. 7QPCh. 2.10 - Prob. 8QPCh. 2.10 - Prob. 9QPCh. 2.10 - Prob. 10QP
Ch. 2.10 - Prob. 11QPCh. 2.10 - Prob. 12QPCh. 2.10 - Prob. 13QPCh. 2.10 - Prob. 14QPCh. 2.10 - Prob. 15QPCh. 2.10 - Prob. 16QPCh. 2.10 - Prob. 17QPCh. 2.10 - Prob. 18QPCh. 2.10 - Prob. 19QPCh. 2.10 - Prob. 20QPCh. 2.10 - Prob. 21QPCh. 2.10 - Prob. 22QPCh. 2.10 - Prob. 23QPCh. 2.10 - Prob. 24QPCh. 2.10 - Prob. 25QPCh. 2.10 - Prob. 26QPCh. 2.10 - Prob. 27QPCh. 2.10 - Prob. 1SSPCh. 2.10 - Prob. 2SSPCh. 2.10 - Prob. 1FEQPCh. 2.10 - Prob. 2FEQPCh. 2.10 - Prob. 3FEQPCh. 2.10 - Prob. 4FEQP
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