Essentials of Materials Science and Engineering, SI Edition
4th Edition
ISBN: 9781337672078
Author: ASKELAND, Donald R., WRIGHT, Wendelin J.
Publisher: Cengage Learning
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Question
Chapter 16, Problem K16.1KP
Interpretation Introduction
Interpretation:
An example of a thermoplastic elastomer based on an interpenetrating polymer network is to be given along with the method of its preparation.
Concept Introduction:
Thermoplastic elastomers are the polymers which belong to the class of copolymers and have both the properties of thermoplastics and elastomers. Due to this they can be substantially stretched to moderate elongation on application of force and upon its removal tends to go back to their near original shape.
Interpenetrating polymer networks are the class of polymers which contains two or more networks of polymers that are partially interlaced to each other but are not bonded covalently to each other. These polymers cannot be separated from each other without the chemical bonds being broken.
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Please find Vo using Mesh analysis
(a) Determine the Nataf model for the joint PDF fxx, (xx) of the basic (physical) random variables X₁
and X, with marginal PDF's
fx(x)=e, 0≤x
(Exponential distribution)
fx₁ (x2)=x2e-0.5x, 0≤x (Rayleigh distribution)
and correlation coefficient
Pxx=0.50
Note: Use Table 6 of paper by Liu and Der Kiureghian, 1986.
(b) Generate a 3D surface plot and contour plot of the joint PDF fxx, (x,x) using Matlab or any other
software of your choice.
(c) What is the standard deviation of X2?
(d) Construct a transformation from the physical X space (defined by random variables X, and X,) to the
standard normal U space (defined by the statistically independent standard normal random variables
(U, and U₂), i.e., U=T(X). Also describe the inverse transform X=T(U) and the Jacobian
matrices J
=
ди
θα
and Ju
Ox
ди
(e) According to the inverse transformation X = T¹ (U) and using Matlab, generate 1,000 samples from
the Nataf joint PDF fxx, (x1,x2) derived in part (a). Start by generating samples of U using a…
Based on the results obtained, comment on the relative importance of the body and the tails of thedistributions of R and S on the probability of failure with increasing central safety factor CSF .
Chapter 16 Solutions
Essentials of Materials Science and Engineering, SI Edition
Ch. 16 - Prob. 16.1PCh. 16 - Prob. 16.2PCh. 16 - Prob. 16.3PCh. 16 - Prob. 16.4PCh. 16 - Prob. 16.5PCh. 16 - Prob. 16.6PCh. 16 - Prob. 16.7PCh. 16 - Prob. 16.8PCh. 16 - Prob. 16.9PCh. 16 - Prob. 16.10P
Ch. 16 - Prob. 16.11PCh. 16 - Prob. 16.12PCh. 16 - Prob. 16.13PCh. 16 - Prob. 16.14PCh. 16 - Prob. 16.15PCh. 16 - Prob. 16.16PCh. 16 - Prob. 16.17PCh. 16 - Prob. 16.18PCh. 16 - Prob. 16.19PCh. 16 - Prob. 16.20PCh. 16 - Prob. 16.21PCh. 16 - Prob. 16.22PCh. 16 - Prob. 16.23PCh. 16 - Prob. 16.24PCh. 16 - Prob. 16.25PCh. 16 - Prob. 16.26PCh. 16 - Prob. 16.27PCh. 16 - Prob. 16.28PCh. 16 - Prob. 16.29PCh. 16 - Prob. 16.30PCh. 16 - Prob. 16.31PCh. 16 - Prob. 16.32PCh. 16 - Prob. 16.33PCh. 16 - Prob. 16.34PCh. 16 - Prob. 16.35PCh. 16 - Prob. 16.36PCh. 16 - Prob. 16.37PCh. 16 - Prob. 16.38PCh. 16 - Prob. 16.39PCh. 16 - Prob. 16.41DPCh. 16 - Prob. 16.42DPCh. 16 - Prob. 16.43DPCh. 16 - Prob. 16.44CPCh. 16 - Prob. K16.1KPCh. 16 - Prob. K16.2KPCh. 16 - Prob. K16.3KP
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