Chapter 8, Problem 1QAP

To determine

The maximum stress $({\sigma }_m)$ that can be present at the tip of an internal crack.

Answer to Problem 1QAP

Maximum stress $({\sigma }_m)$ at the tip of an internal crack is $2404\text{MPa}$ .

Explanation of Solution

Given:

The Radius of Curvature of the internal crack tip, ${\rho }_t=2.5\times {10}^{-4}\text{mm}$ .

Length of the internal Crack $=2.5\times {10}^{-2}\text{mm}$ .

Tensile Strength Applied, ${\sigma }_0=170\text{MPa}$ .

Formula used:

The maximum stress at the tip of the crack can be calculated by the formula :

  ${\sigma }_m=2{\sigma }_0{\left(\frac{a}{{\rho }_t}\right)}^{\frac{1}{2}}$

where ${\sigma }_m=$ Maximum Stress

  ${\sigma }_0=$ Tensile Strength

  $a=$ Length of the surface Crack

  ${\rho }_t=$ Radius of Curvature of the internal crack tip

Calculation:

To calculate the maximum stress at the tip of the crack, first find out the length of the surface crack :

  $\begin{array}{l}a=\frac{2.5\times {10}^{-2}}{2}\text{mm}\\ a=1.25\times {10}^{-2}\text{mm}\end{array}$

Now, using the formula :

  $\begin{array}{l}{\sigma }_m=2{\sigma }_0{\left(\frac{a}{{\rho }_t}\right)}^{\frac{1}{2}}\\ {\sigma }_m=2\times 170\text{MPa}{\left(\frac{1.25\times {10}^{-2}\text{mm}}{2.5\times {10}^{-4}\text{mm}}\right)}^{\frac{1}{2}}\\ {\sigma }_m=340{\left(50\right)}^{\frac{1}{2}}\text{MPa}\\ {\sigma }_m=340\times 7.07\text{MPa}\\ {\sigma }_m=2403.8\text{MPa}\approx 2404\text{MPa}\end{array}$

Conclusion:

The magnitude of the maximum stress $({\sigma }_m)$ at the tip of an internal crack is $2404\text{MPa}$ .