Compute the elastic modulus for the following metal alloys, whose stress-strain behaviors may be observed in the "Tensile Tests" module of Virtual Materials Science and Engineering (VMSE): (a) titanium, (b) tempered steel, (c) aluminum, and (d) carbon steel. How do these values compare with those presented in Table 6.1 for the same metals?

Answers

Answer 1

The elastic modulus for steel given in the table is 207 GPa, which is in reasonable agreement with this value E = 200.75 GPa

The elastic modulus is the slope in the linear elastic region:

б2 = б1 / Э2 - Э1

Since stress-strain curves for all of the metals/alloys pass through the o

rigin, and if we take σ1 = 0 then ε1 = 0. Determinations of σ2 and ε2 are possible by moving the cursor to some arbitrary point in the linear region of the curve and then reading corresponding values in the “Stress” and “Strain” windows that are located below the plot.

(a) A screenshot for the titanium alloy in the elastic region is shown below

Here the cursor point resides in the elastic region at a stress of 492.4 MPA (which is the value of σ2) at a strain of 0.0049 (which is the value of ε2). Thus, the elastic modulus is equal to

| 492.4-0

0.0049-0

E=100489.79 MPA

E=100.5 GPA

The elastic modulus for titanium given in the table is 107 GPA, which is in reasonably good agreement with this value.

(b) A screenshot for the tempered steel alloy in the elastic region is shown below

Here the cursor point resides in the elastic region at a stress of 916.7 MPA (which is the value of σ2) at a strain of 0.0045 (which is the value of ε2). Thus, the elastic modulus is equal to :

1916.7-0

0.0045 -0

E = 203711.11 MPA

E = 203.7 GPA

The elastic modulus for steel given in the table is 207 GPA, which is in good agreement with this value.

(c) A screenshot for the aluminum alloy in the elastic region is shown below.

Here the cursor point resides in the elastic region at a stress of 193.6 MPa (which is the value of σ2) at a strain of 0.0028 (which is the value of ε2). Thus, the elastic modulus is equal to:

193.6 – 0

0.0028 - 0

E = 69142.85 MPa

E = 69.14 GPa

The elastic modulus for aluminum given in the table is 69 GPa, which is in excellent agreement with this value.

(d) A screenshot for the carbon steel alloy in the elastic region is shown below

Here the cursor point resides in the elastic region at a stress of 160.6 MPa (which is the value of σ2) at a strain of 0.0008 (which is the value of ε2). Thus, the elastic modulus is equal to:

160.6 – 0

0.0008 - 0

E = 200750 MPa

E = 200.75 GPa

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Compute The Elastic Modulus For The Following Metal Alloys, Whose Stress-strain Behaviors May Be Observed

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Writting the code:

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You are given a data stream that has been compressed to a length of 100,000 bits, and told that it is the result of running an "ideal" entropy coder on a sequence of data. You are also told that the original data consists of samples of a continuous waveform, quantized to 2 bits per sample. The probabilities of the uncompressed values are as follows: 00 1/2 01 3/8 10 1/16 11 1/16. What (approximately) was the length of the uncompressed signal?

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