Thinking Like an Engineer
Thinking Like an Engineer
4th Edition
ISBN: 9781269910989
Author: Elizabeth A. Stephan, David R. Bowman, William J. Park, Benjamin L. Sill, Matthew W. Ohland
Publisher: Pearson Learning Solutions
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Chapter 16.4, Problem 9CC
  1. a. Create the matrix CCM1= [ 18 0.3 4.1 1 0 17 ]   using a single MATLAB command.
  2. b. Create the matrix CCM2= [ 0 0 0 0 0 1 × 10 15 0 0 1 × 10 15 ] using a single MATLAB command.
  3. c. Assume the 3 × 3matrix CCM3 has already been defined as CCM3= [ 3 9 14 6 0.5 4 × 10 2 44 0 1 × 10 3 ]

    Using a single MATLAB command, define a new 2 × 2 matrix named Corners that contains the corner elements of CCM3. Your code should still work correctly if the contents of CCM3 change but it rema1ns a 3 × 3; in other words, DO NOT hard-code the four values.

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An AISI 1018 steel ball with 1.100-in diameter is used as a roller between a flat plate made from 2024 T3 aluminum and a flat table surface made from ASTM No. 30 gray cast iron. Determine the maximum amount of weight that can be stacked on the aluminum plate without exceeding a maximum shear stress of 19.00 kpsi in any of the three pieces. Assume the figure given below, which is based on a typical Poisson's ratio of 0.3, is applicable to estimate the depth at which the maximum shear stress occurs for these materials. 1.0 0.8 Ratio of stress to Pmax 0.4 90 0.6 στ Tmax 0.2 0.5a a 1.5a 2a 2.5a За Distance from contact surface The maximum amount of weight that can be stacked on the aluminum plate is lbf.
A carbon steel ball with 27.00-mm diameter is pressed together with an aluminum ball with a 36.00-mm diameter by a force of 11.00 N. Determine the maximum shear stress and the depth at which it will occur for the aluminum ball. Assume the figure given below, which is based on a typical Poisson's ratio of 0.3, is applicable to estimate the depth at which the maximum shear stress occurs for these materials. 1.0 0.8 Ratio of stress to Pma 9 0.6 στ 24 0.4 Tmax 0.2 0 0.5a a 1.5a Z 2a 2.5a За Distance from contact surface The maximum shear stress is determined to be MPa. The depth in the aluminum ball at which the maximum shear stress will occur is determined to be [ mm.
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