# Calculating Bond Energy of Steel Using Theoretical Strength Eqn

1. Nov 24, 2013

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
$$S_m = \sqrt{\frac{E\gamma}{r_0}} \\ \gamma = \frac{1}{4}nN_sU_0 \\ N_s ≈ \frac{1}{r_0^2}$$
where n is the coordination number and N_s is the

3. 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

$$S_m = \sqrt{\frac{E\frac{1}{4}nN_sU_0}{r_0}} \\ r_0 S_m^2 = E\frac{1}{4}nN_sU_0 \\ U_0 = \frac{r_0 S_m^2}{E\frac{1}{4}nN_s}$$

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.

$$U_0 = \frac{(2×10^{-10})(4×10^{10})^2}{(2.1 × 10^{11})\frac{1}{4}10(2.5×10^{19})} \\ U_0 = \frac{3.2×10^{11}}{1.31×10^{31}}=2.44×10^{-20}$$
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: Nov 24, 2013
2. Nov 24, 2013