Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
10th Edition
ISBN: 9780073398204
Author: Richard G Budynas, Keith J Nisbett
Publisher: McGraw-Hill Education
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Chapter 10, Problem 2P
It is instructive to examine the question of the units of the parameter A of Eq. (10–14). Show that for U.S. customary units the units for Auscu are kpsi · inm and for SI units are MPa · mmm for ASI. which make the dimensions of both Auscu and ASI different for every material to which Eq. (10–14) applies. Also show that the conversion from Auscu to ASI is given by
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The purpose of this problem is to
show the relationship between material constants typically used in engineering practice. This
is useful because you may often have access to measurements of or tabulated values of some
constants (e.g., Young's modulus and Poisson's ratio) but need another constant (e.g., shear
modulus) for a calculation. Use the expression Cijkl = µ(dildjk + dikdjl) + Ad¿jdkl to derive
the following:
(a) Young's modulus, E = µ(3X+2µ)/(X+μ), from the definition 11 = Ee11 in a unconfined
(022 = 0,033 = 0) uniaxial tension test.
(b) Poisson's ratio, v = \/(2(X + μ)), from the definition v = €22/11 in the same test as
in (a).
(c) Shear modulus, μ = G = 012/(2€12) = E/(2(1 + v)).
Use the results to show that C can also be written Cijkl
Ev/((1+v)(1 − 2v))dij§kl.
=
E/(2(1 + v))(duðjk + dikdjl) +
Henri Darcy, a French engineer, proposed that the pressuredrop ∆ p for fl ow at velocity V through a tube of length Lcould be correlated in the form ∆p
___________ = α LV2 ρ If Darcy’s formulation is consistent, what are the dimensionsof the coefficient α ?
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Chapter 10 Solutions
Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
Ch. 10 - Within the range of recommended values of the...Ch. 10 - It is instructive to examine the question of the...Ch. 10 - A helical compression spring is wound using...Ch. 10 - The spring in Prob. 10-3 is to be used with a...Ch. 10 - A helical compression spring is made with...Ch. 10 - A helical compression spring is to be made of...Ch. 10 - A helical compression spring is made of hard-drawn...Ch. 10 - The spring of Prob. 107 is to be used with a...Ch. 10 - 109 to 1019 Listed in the tables are six springs...Ch. 10 - 109 to 1019 Listed in the tables are six springs...
Ch. 10 - 10-9 to 10-19 Listed in the tables are six springs...Ch. 10 - Prob. 12PCh. 10 - 10-9 to 10-19 Listed in the tables are six springs...Ch. 10 - 10-9 to 10-19 Listed in the tables are six springs...Ch. 10 - 10-9 to 10-19 Listed in the tables are six springs...Ch. 10 - 10-9 to 10-19 Listed in the tables are six springs...Ch. 10 - Prob. 17PCh. 10 - 10-9 to 10-19 Listed in the tables are six springs...Ch. 10 - 10-9 to 10-19 Listed in the tables are six springs...Ch. 10 - Consider the steel spring in the illustration. (a)...Ch. 10 - A static service music wire helical compression...Ch. 10 - Solve Prob. 1021 by iterating with an initial...Ch. 10 - A holding fixture for a workpiece 37.5 mm thick at...Ch. 10 - Solve Prob. 10-23 by iterating with an initial...Ch. 10 - A compression spring is needed to fit over a...Ch. 10 - A compression spring is needed to fit within a...Ch. 10 - A helical compression spring is to be cycled...Ch. 10 - The figure shows a conical compression helical...Ch. 10 - A helical coil compression spring is needed for...Ch. 10 - Solve Prob. 10-30 using the Goodman-Zimmerli...Ch. 10 - Solve Prob. 10-30 using the Sines-Zimmerli...Ch. 10 - Design the spring of Ex. 10-5 using the...Ch. 10 - Solve Prob. 10-33 using the Goodman-Zimmerli...Ch. 10 - A hard-drawn spring steel extension spring is to...Ch. 10 - The extension spring shown in the figure has...Ch. 10 - Design an infinite-life helical coil extension...Ch. 10 - Prove Eq. (10-40). Hint: Using Castigliunos...Ch. 10 - The figure shows a finger exerciser used by...Ch. 10 - The rat trap shown in the figure uses two...Ch. 10 - Prob. 41PCh. 10 - Prob. 42PCh. 10 - Figure 10-13b shows a spring of constant thickness...
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