EBK MANUFACTURING PROCESSES FOR ENGINEE
EBK MANUFACTURING PROCESSES FOR ENGINEE
6th Edition
ISBN: 9780134425115
Author: Schmid
Publisher: YUZU
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Chapter 2, Problem 2.61P

(a)

To determine

The work done in frictionless compression of solid cylinder of 1100-O Aluminum.

(a)

Expert Solution
Check Mark

Answer to Problem 2.61P

The work done in the frictionless compression of the cylinder is W=1562Nm

Explanation of Solution

Calculation:

The work done in frictionless compression of solid cylinder can be calculated by multiplying the specific energy and the volume of the cylinder.

For calculating the value of specific energy, the value of true strain is also required. The reduction in the height of the cylinder is 75%, thus the final height of the cylinder is 10 mm.

The value of true strain can be calculated as

  ε=ln(4010)ε=1.386

The value of bulk modulus and n for the 1100-O Al are

  K=180MPan=0.20

The specific energy can be calculated as

  u=Kεn+1n+1u=(180)(1.386)1.21.2u=222MN/m3

Now, the volume of the cylinder is

  V=πr2lV=π(0.0075)2(0.04)V=7.069×106m3

Now, the work done is

  W=u×VW=222×7.069NmW=1562Nm

Conclusion: Thus, the work done in the frictionless compression of the cylinder is W=1562Nm

(b)

To determine

The work done in frictionless compression of solid cylinder of annealed copper.

(b)

Expert Solution
Check Mark

Answer to Problem 2.61P

The work done in the frictionless compression of the cylinder is W=2391Nm

Explanation of Solution

Calculation:

The work done in frictionless compression of solid cylinder can be calculated by multiplying the specific energy and the volume of the cylinder.

For calculating the value of specific energy, the value of true strain is also required. The reduction in the height of the cylinder is 75%, thus the final height of the cylinder is 10 mm.

The value of true strain can be calculated as

  ε=ln(4010)ε=1.386

The value of bulk modulus and n for the annealed copper are

  K=315MPan=0.54

The specific energy can be calculated as

  u=Kεn+1n+1u=(315)(1.386)1.541.54u=338MN/m3

Now, the volume of the cylinder is

  V=πr2lV=π(0.0075)2(0.04)V=7.069×106m3

Now, the work done is

  W=u×VW=338×7.069NmW=2391Nm

Conclusion: Thus, the work done in the frictionless compression of the cylinder is W=2391Nm

(c)

To determine

The work done in frictionless compression of solid cylinder of annealed 304 stainless steels.

(c)

Expert Solution
Check Mark

Answer to Problem 2.61P

The work done in the frictionless compression of the cylinder is W=10,808Nm

Explanation of Solution

Calculation:

The work done in frictionless compression of solid cylinder can be calculated by multiplying the specific energy and the volume of the cylinder.

For calculating the value of specific energy, the value of true strain is also required. The reduction in the height of the cylinder is 75%, thus the final height of the cylinder is 10 mm.

The value of true strain can be calculated as

  ε=ln(4010)ε=1.386

The value of bulk modulus and n for the annealed 304 stainless steel are

  K=1300MPan=0.30

The specific energy can be calculated as

  u=Kεn+1n+1u=(1300)(1.386)1.31.3u=1529MN/m3

Now, the volume of the cylinder is

  V=πr2lV=π(0.0075)2(0.04)V=7.069×106m3

Now, the work done is

  W=u×VW=1529×7.069NmW=10,808Nm

Conclusion: Thus, the work done in the frictionless compression of the cylinder is W=10,808Nm

(d)

To determine

The work done in frictionless compression of solid cylinder of annealed 70-30 brass.

(d)

Expert Solution
Check Mark

Answer to Problem 2.61P

The work done in the frictionless compression of the cylinder is W=6908Nm

Explanation of Solution

Calculation:

The work done in frictionless compression of solid cylinder can be calculated by multiplying the specific energy and the volume of the cylinder.

For calculating the value of specific energy, the value of true strain is also required. The reduction in the height of the cylinder is 75%, thus the final height of the cylinder is 10 mm.

The value of true strain can be calculated as

  ε=ln(4010)ε=1.386

The value of bulk modulus and n for the annealed 70-30 brass are

  K=895MPan=0.49

The specific energy can be calculated as

  u=Kεn+1n+1u=(895)(1.386)1.491.49u=977MN/m3

Now, the volume of the cylinder is

  V=πr2lV=π(0.0075)2(0.04)V=7.069×106m3

Now, the work done is

  W=u×VW=977×7.069NmW=6908Nm

Conclusion: Thus, the work done in the frictionless compression of the cylinder is W=6908Nm

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part made from AISI 1212 steel undergoes a 20 percent cold-work operation. (a) Obtain the yield strength and ultimate strength before and after the cold-work operation. Determine the percent increase in each strength. (b) Determine the ratios of ultimate strength to yield strength before and after the cold work operation. What does the result indicate about the change of ductility of the part?
Example Problem-2 A cylindrical rod of non cold-worked brass having an initial diameter of 6.4 mm (0.25 in.) is to be cold worked by drawing such that the cross-sectional area is reduced. It is required to have a cold-worked yield strength of at least 345 Mpa (50,000 psi) and a ductility in excess of 20%EL; in addition, a final diameter of 5.1 mm (0.20 in.) is necessary. Describe the solution.
Do both parts

Chapter 2 Solutions

EBK MANUFACTURING PROCESSES FOR ENGINEE

Ch. 2 - Prob. 2.11QCh. 2 - Prob. 2.12QCh. 2 - Prob. 2.13QCh. 2 - Prob. 2.14QCh. 2 - Prob. 2.15QCh. 2 - Prob. 2.16QCh. 2 - Prob. 2.17QCh. 2 - Prob. 2.18QCh. 2 - Prob. 2.19QCh. 2 - Prob. 2.20QCh. 2 - Prob. 2.21QCh. 2 - Prob. 2.22QCh. 2 - Prob. 2.23QCh. 2 - Prob. 2.24QCh. 2 - Prob. 2.25QCh. 2 - Prob. 2.26QCh. 2 - Prob. 2.27QCh. 2 - Prob. 2.28QCh. 2 - Prob. 2.29QCh. 2 - Prob. 2.30QCh. 2 - Prob. 2.31QCh. 2 - Prob. 2.32QCh. 2 - Prob. 2.33QCh. 2 - Prob. 2.34QCh. 2 - Prob. 2.35QCh. 2 - Prob. 2.36QCh. 2 - Prob. 2.37QCh. 2 - Prob. 2.38QCh. 2 - Prob. 2.39QCh. 2 - Prob. 2.40QCh. 2 - Prob. 2.41QCh. 2 - Prob. 2.42QCh. 2 - Prob. 2.43QCh. 2 - Prob. 2.44QCh. 2 - Prob. 2.45QCh. 2 - Prob. 2.46QCh. 2 - Prob. 2.47QCh. 2 - Prob. 2.48QCh. 2 - Prob. 2.49PCh. 2 - Prob. 2.50PCh. 2 - Prob. 2.51PCh. 2 - Prob. 2.52PCh. 2 - Prob. 2.53PCh. 2 - Prob. 2.54PCh. 2 - Prob. 2.55PCh. 2 - Prob. 2.56PCh. 2 - Prob. 2.57PCh. 2 - Prob. 2.58PCh. 2 - Prob. 2.59PCh. 2 - Prob. 2.60PCh. 2 - Prob. 2.61PCh. 2 - Prob. 2.62PCh. 2 - Prob. 2.63PCh. 2 - Prob. 2.64PCh. 2 - Prob. 2.65PCh. 2 - Prob. 2.66PCh. 2 - Prob. 2.67PCh. 2 - Prob. 2.68PCh. 2 - Prob. 2.69PCh. 2 - Prob. 2.70PCh. 2 - Prob. 2.71PCh. 2 - Prob. 2.72PCh. 2 - Prob. 2.73PCh. 2 - Prob. 2.74PCh. 2 - Prob. 2.75PCh. 2 - Prob. 2.76PCh. 2 - Prob. 2.78PCh. 2 - Prob. 2.79PCh. 2 - Prob. 2.80PCh. 2 - Prob. 2.81PCh. 2 - Prob. 2.82PCh. 2 - Prob. 2.83PCh. 2 - Prob. 2.84PCh. 2 - Prob. 2.85PCh. 2 - Prob. 2.86PCh. 2 - Prob. 2.87PCh. 2 - Prob. 2.88PCh. 2 - Prob. 2.89PCh. 2 - Prob. 2.90PCh. 2 - Prob. 2.91PCh. 2 - Prob. 2.92PCh. 2 - Prob. 2.93PCh. 2 - Prob. 2.94PCh. 2 - Prob. 2.95PCh. 2 - Prob. 2.96PCh. 2 - Prob. 2.97PCh. 2 - Prob. 2.98PCh. 2 - Prob. 2.99PCh. 2 - Prob. 2.100PCh. 2 - Prob. 2.101P
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