Chapter 9.19, Problem 1QP

To determine

The magnitude of the maximum stress at the tip of an internal crack $\left({\sigma }_m\right)$.

Answer to Problem 1QP

The magnitude of the maximum stress at the tip of an internal crack is $\underset{\_}{3959.79\text{MPa}}$.

Explanation of Solution

Given:

The radius of curvature of the crack tip, $p_t=1.9\times {10}^{-4}\text{mm}$

The crack length, $a=3.8\times {10}^{-2}\text{mm}$

The tensile stress, ${\sigma }_0=140\text{MPa}$

Explanation:

Determine the maximum stress that occurs at the crack tip.

${\sigma }_m=2{\sigma }_0{\left(\frac{a}{p_t}\right)}^{1/2}$ (I)

Conclusion:

Substitute $3.8\times {10}^{-2}\text{mm}$ for a, 140 MPa for ${\sigma }_0$, and $1.9\times {10}^{-4}\text{mm}$ for $p_t$ in Equation (I).

$\begin{array}{c}{\sigma }_m=2\left(\text{140}\text{MPa}\right){\left(\frac{3.8\times {10}^{-2}\text{mm}}{1.9\times {10}^{-4}\text{mm}}\right)}^{1/2}\\ =2\left(\text{140}\right)\left(14.142\right)\\ =3959.79\text{MPa}\end{array}$

Thus, the magnitude of the maximum stress at the tip of an internal crack is $\underset{\_}{3959.79\text{MPa}}$.