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Calculating Bond Energy of Steel Using Theoretical Strength Eqn

  • Thread starter FaraDazed
  • Start date
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Homework Statement


Steel has a theoretical strength of 4 x 10^10 Pa a density of 7860 kg/m^3, Youngs Modulus of 2 x 10^11 Pa and the mass number of iron, the main ingredient in steel, is 56.

Calculate the bond energy of the steel using the expression for theoretical strength and surface energy.



Homework Equations


[tex]
S_m = \sqrt{\frac{E\gamma}{r_0}} \\
\gamma = \frac{1}{4}nN_sU_0 \\
N_s ≈ \frac{1}{r_0^2}
[/tex]
where n is the coordination number and N_s is the

The Attempt at a Solution


First before doing anything I rearranged to get the bond energy the subject, by subbing gamma in an rearranging ing as below

[tex]
S_m = \sqrt{\frac{E\frac{1}{4}nN_sU_0}{r_0}} \\
r_0 S_m^2 = E\frac{1}{4}nN_sU_0 \\
U_0 = \frac{r_0 S_m^2}{E\frac{1}{4}nN_s}
[/tex]

The way I did it next is probably the wrong way, I did no use the density given in the question or use the mass number of iron.

I found in a lecture booklet that r_0 of steel is 2 x 10^(-10) and I calculated N_s to be 2.5 x 10^19 by using the approximation given in the relevant equations section. And we have been told to use 10 as an approximation for the coordination number. So with these I did the calculation.

[tex]
U_0 = \frac{(2×10^{-10})(4×10^{10})^2}{(2.1 × 10^{11})\frac{1}{4}10(2.5×10^{19})} \\
U_0 = \frac{3.2×10^{11}}{1.31×10^{31}}=2.44×10^{-20}
[/tex]
And was never sure if the units is joules or eV, if its joules then the answer is at least in the ballpark (I think), if its eV, something has gone terribly wrong. I don't understand how to use the values given for the density of steel and mass number of iron to help in this question.
 
Last edited:

Answers and Replies

  • #2
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338 views and no replies? Please, even if its a little comment it might help me! :)
 

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