## Ideal Strength in Terms of Physical Parameters

9 We shall first derive an expression for the ideal strength in terms of physical parameters, and in the next section the strength will be expressed in terms of engineering ones.

Solution I: Force being the derivative of energy, we have F = ^, thus F = 0 at a = a0, Fig. 11.4, and is maximum at the inflection point of the U0 — a curve. Hence, the slope of the force displacement curve is the stiffness of the atomic spring and should be related to E. If we let x = a — a0, then the strain would be equal to e = —. Furthermore, if we define the stress as a = F, then the a — e n ao ' aj'

curve will be as shown in Fig. 11.5.

From this diagram, it would appear that the sine curve would be an adequate approximation to this relationship. Hence, a = am— sin2nX- (11.5)

and the maximum stress a^hO— would occur at x = J. The energy required to separate two atoms is thus given by the area under the sine curve, and from Eq. 11.4, we would have

1 From watching raindrops and bubbles it is obvious that liquid water has surface tension. When the surface of a liquid is extended (soap bubble, insect walking on liquid) work is done against this tension, and energy is stored in the new surface. When insects walk on water it sinks until the surface energy just balances the decrease in its potential energy. For solids, the chemical bonds are stronger than for liquids, hence the surface energy is stronger. The reason why we do not notice it is that solids are too rigid to be distorted by it. Surface energy 7 is expressed in J/m? and the surface energies of water, most solids, and diamonds are approximately .077, 1.0, and 5.14 respectively.

Force

Force

a0 |
Interatomic Distance | |

/ |
Interatomic | |

^ Distance |

Figure 11.4: Energy and Force Binding Two Adjacent Atoms

ytheor

Also for very small displacements (small x) sin x « x, thus Eq. 11.5 reduces to

, 2nx Ex elliminating x,

Substituting for A from Eq. 11.9, we get ytheor '

ytheor

ao 2n

Solution II: For two layers of atoms a0 apart, the strain energy per unit area due to a (for linear elastic systems) is

If y is the surface energy of the solid per unit area, then the total surface energy of two new fracture surfaces is 2y.

For our theoretical strength, U = 2y ^ ma2E °° = 2y or ahl? = 2J

max max

Note that here we have assumed that the material obeys Hooke's Law up to failure, since this is seldom the case, we can simplify this approximation to:

which is the same as Equation 11.12

Example: As an example, let us consider steel which has the following properties: y = 1 mJi; E 2 x 1011 m~2; and a0 « 2 x 10~10 m. Thus from Eq. 11.12 we would have:

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