# Consider two materials - Aluminum and Steel. For each of these two materials, consider the...

## Question:

Consider two materials ? Aluminum and Steel. For each of these two materials, consider the surface energy term as follows: **{eq}\gamma_s{/eq} = 1 J/m{eq}^2{/eq}** For each of the two materials, consider two cases where the defects size are**1 um** and **1 mm**.

Calculate the fracture strength as predicted by Models for ideal and real materials, as discussed in class. Discuss the trends that you observe in the model predictions for the fracture process.

## Strength of Materials

The material property with the help of which the amount of load that a material can oppose or resist without getting deformed is known as the strength of the material.

## Answer and Explanation:

**Given Data:**

- The surface energy of the material is: {eq}{\gamma _s} = 1\;{\rm{J/}}{{\rm{m}}^2} {/eq}.

- The defect of the size is: {eq}{s_1} =...

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**Given Data:**

- The surface energy of the material is: {eq}{\gamma _s} = 1\;{\rm{J/}}{{\rm{m}}^2} {/eq}.

- The defect of the size is: {eq}{s_1} = 1\;{\rm{\mu m}} {/eq}.

- The defect of the size is: {eq}{s_2} = 1\;{\rm{mm}} {/eq}.

**For Aluminium**

The expression for the fracture strength when the defect size is {eq}{s_1} {/eq}.

{eq}{\sigma _f} = \sqrt {\dfrac{{2{E_{Al}}\left( {{\gamma _s} + {\gamma _P}} \right)}}{{{A_{{s_1}}}}}} {/eq}

Here, the modulus of elasticity of aluminium is {eq}{E_{Al}} {/eq}, the plastic strength of aluminium is {eq}{\gamma _P} {/eq}, the area of the defect is {eq}{A_{{s_1}}} {/eq}

The expression for the fracture strength when the defect size is {eq}{s_2} {/eq}.

{eq}{\sigma _f} = \sqrt {\dfrac{{2{E_{Al}}\left( {{\gamma _s} + {\gamma _P}} \right)}}{{{A_{{s_2}}}}}} {/eq}

**For Steel**

The expression for the fracture strength when the defect size is {eq}{s_1} {/eq}.

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from Introduction to Engineering

Chapter 2 / Lesson 6