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Homework Solution 1...

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IE 370-Fall 2020 Homework 1 covering Chapters 1, 2, and 3 in the required textbook (Degarmo’s); Chapters 1 and 2.1 in the lecture slides (Due 11:59 PM EST, September 18)

1. Manufacturing and Production Systems ( 1.5 points) a) Which design does a Toyota manufacturing and assembly factory belong to? (0.5 points) Linked-cell Shop b) The southern hemisphere of the moon has a large quantity of ice, which can be turned into oxygen and hydrogen. Hydrogen is the principal component of rocket fuel. To turn the moon into a cosmic gas station supplying hydrogen fuel for future space travel, which manufacturing system design should we build on the moon? (0.5 points). Continuous Processes If we would like to build spacecraft on the moon using resources transported to the moon, which manufacturing system design should build on the moon? (0.5 points) Project shop

2. A paper clip is made of wire 0.6 mm in diameter. If the original material from which the wire is made is a rod 30 mm in diameter, calculate the longitudinal engineering and true strains that the wire has undergone during processing. (1 point) d  o l o  d f l f  lo e lo lf

2

  30  2     2500    0.6  2500l o  l o   2499 lo

3. A cylindrical specimen of a titanium alloy having an elastic modulus of 110 GPa and an original diameter of 5 mm will experience only elastic deformation when a tensile load of 3000 N is

applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.5 mm. (0.5 point)

 0.5 10 m110 10  3

9

N / m  ( )  5 10 m  2

3

(4)(3000N)

2

 0.35997m  359.97mm

4. A metal tensile specimen has an initial diameter of 5 mm and is 60 mm long. The yield strength is 500 MPa, the elastic modulus is 80 GPa, and the ultimate tensile strength is 600 MPa. (1) Calculate the strain at yield point. ( 1 point); (2) Calculate the maximum load during the test. (1 point)

(1) Calculate the strain at yield point. yield point = yield strength/Elastic modulus = 500 MPa/(80 x 103 MPa) = 6.25 x 103

 Calculate the maximum load during the test. 600 × 106 x (π/4)(0.005m) 2 = 11780.97 N 5. A steel cylinder with 13.8 mm in diameter is tensile-tested to fracture. The engineering stress at fracture is 480 MPa and the cylinder’s diameter at fracture is 10.8 mm. Please calculate (2 points) (1) the percent reduction in area for the cylinder after the test 2

% RA 

(1 point)

2

 13.8mm   10.8mm     2  2     38.75% 2  13.8mm   2  



(2) The true stress at fracture

(1 poi

0.0138m  F   f Ao   480  10 6 N / m 2     2  

 Tf 

F  Ai

71794 N  0.0108m    2  

2

2

 71794N

 783.7MPa

6. If the true stress-strain behavior of a metal is given by true   ∙ true)n, where true is the true stress, true is the true strain,  is a non-zero constant, n is the strain hardening coefficient (non-zero). (1 point) (1) Use the parameters in the above equation to determine the engineering stress (2 points) and engineering strain before necking (0.5 points)

(2) What is the name for the engineering stress at the onset of necking?

Ultimate tensile strength (or ultimate strength, or ultimate tensile stress)

(3) Use the parameters in the above equation to find the engineering stress in (2)....