Chapter 7.10, Problem 52SEP

(a)

To determine

The critical crack length for alloy steel plate for the applied stress.

Answer to Problem 52SEP

The critical crack length for alloy steel plate for the applied stress is $0.04476\text{m}$.

Explanation of Solution

Write the expression for the critical length.

    $a=\frac{1}{\pi }\times {\left(\frac{K_C}{\gamma \sigma }\right)}^2$                                                           (I)

Here, the fracture toughness of the material is $K_C$, the tensile stress is $\sigma$ and the constant is $\gamma$.

Conclusion:

Substitute $45\text{MPa}\cdot {\text{m}}^{1/2}$ for $K_C$, $120\text{MPa}$ for $\sigma$ and $1$ for $\gamma$ in Equation (I).

    $\begin{array}{c}a=\frac{1}{\pi }\times {\left(\frac{45\text{MPa}\cdot {\text{m}}^{1/2}}{1\left(120\text{MPa}\right)}\right)}^2\\ =0.3183\times \left(0.140625\text{m}\right)\\ =0.04476\text{m}\end{array}$

Thus, the critical crack length for alloy steel plate for the applied stress is $0.04476\text{m}$.

(b)

To determine

The crack length.

The largest allowable initial crack length.

Answer to Problem 52SEP

The crack length is $0.02238\text{m}$.

The largest allowable initial crack length is $0.792\text{mm}$.

Explanation of Solution

Write the expression for the crack length of the alloy steel plate.

    $a_f=0.5a$                                                                               (II)

Write the expression for the stress range for the steel.

    ${\sigma }_r={\sigma }_{\max }-{\sigma }_{\min }$                                                                     (III)

Here, the maximum compressive stress is ${\sigma }_{\max }$ and the minimum compressive stress is ${\sigma }_{\min }$.

Write the expression for the fatigue life in cycles.

    $N_f=\frac{a_f^{-\left(m/2\right)+1}-a_o^{-\left(m/2\right)+1}}{\left[-\left(m/2\right)+1\right]A{\sigma }_r^m{\pi }^{m/2}{\gamma }^m}$                                              (IV)

Here, the initial crack length is $a_o$ and the fatigue constant are $A$ and $m$.

Conclusion:

Substitute $0.04476\text{m}$ for $a$ in Equation (II).

    $\begin{array}{c}{a}_f=0.5\left(0.04476\text{m}\right)\\ =0.02238\text{m}\end{array}$

Here, the minimum tensile stress is $0\text{MPa}$.

Substitute $120\text{MPa}$ for ${\sigma }_{\max }$ and $0\text{MPa}$ for ${\sigma }_{\min }$ in Equation (III).

    $\begin{array}{c}{\sigma }_r=120\text{MPa}-0\text{MPa}\\ =120\text{MPa}\end{array}$

Here, the fatigue constant are $A=2\times {10}^{-12}$ and $m=3$.

Substitute $120\text{MPa}$ for ${\sigma }_r$, $3$ for $m$, $2\times {10}^{-12}$ for $A$, $1$ for $\gamma$, $0.02238\text{m}$ and $3\times {10}^6\text{cycles}$ for $N_f$ in the Equation (IV).

    $\begin{array}{l}3\times {10}^6\text{cycles}=\frac{{\left(0.02238\text{m}\right)}^{-\left(3/2+1\right)}-a_o^{-\left(3/2\right)+1}}{\left[-\left(3/2\right)+1\right]\left(2\times {10}^{-12}\right){\left(120\text{MPa}\right)}^3{\left(\pi \right)}^{3/2}{\left(1\right)}^3}\\ 3\times {10}^6\text{cycles}=\frac{6.6845{\text{m}}^{-0.5}-a_o^{-0.5}}{\left[-0.5\right]\left(2\times {10}^{-12}\right)\left(9622.02\times {10}^3\text{MPa}\right)}\\ a_o=0.000792\text{m}\left(\frac{{10}^3\text{mm}}{1\text{m}}\right)\\ a_o=0.792\text{mm}\end{array}$

The crack length is $0.02238\text{m}$.

The largest allowable initial crack length is $0.792\text{mm}$.