Chapter 7.10, Problem 45SEP

(a)

To determine

The critical length of crack if the vessel made of $\text{Al}7178\text{-T}651$.

Answer to Problem 45SEP

The critical length of crack is $0.00216\text{in}$.

Explanation of Solution

Write the expression for the circumferential stress or hoop stress in cylindrical vessel.

    $\sigma =\frac{Pr}{t}$                                                                                          (I)

Here, the internal pressure is $P$, the radius of vessel is $r$ and the thickness of vessel is $t$.

Write the expression of typical fracture toughness is,

    $K_{IC}=Y\sigma \sqrt{\pi a}$

    $a={\left(\frac{K_{IC}}{Y\sigma \sqrt{\pi }}\right)}^2$                                                                              (II)

Here, fracture toughness is $K_{IC}$, yield strength is $Y$, circumferential stress or hoop stress is $\sigma$ and critical thickness of the vessel is $a$.

Write the expression of critical length of crack for the vessel is,

    $C=2a$                                                                                      (III)

Here, critical thickness of the vessel is $a$.

Conclusion:

Substitute $5000\text{Psi}$ for $P$, $18\text{in}$ for $r$ and $0.25\text{in}$ for $t$ in Equation (I).

    $\begin{array}{c}\sigma =\frac{\left(5000\text{Psi}\right)\left(18\text{in}\right)}{\left(0.25\text{in}\right)}\\ =\frac{\left(5000\text{Psi}\right)\left[\frac{1\text{ksi}}{1000\text{Psi}}\right]\left(18\text{in}\right)}{\left(0.25\text{in}\right)}\\ =360\text{ksi}\end{array}$

Refer table 7.1 “Typical fracture toughness values for selected engineering alloys”, to obtain the value of $K_{IC}$ as $21\text{ksi}\sqrt{\text{in}}$ corresponding to the material $\text{Al}7178\text{-T}651$.

Substitute, $21\text{ksi}\sqrt{\text{in}}$ for $K_{IC}$, $1$ for $Y$ and $360\text{ksi}$ for $\sigma$ in Equation (II).

    $\begin{array}{c}a={\left(\frac{21\text{ksi}\sqrt{\text{in}}}{1\times 360\left(\text{ksi}\right)\times \sqrt{\pi }}\right)}^2\\ =\left(\frac{\left({\left(21\right)}^2\text{in}\right)}{{\left(360\right)}^2\times \pi }\right)\\ =0.00108\text{in}\end{array}$

Substitute $0.00108\text{in}$ for $a$ in Equation (III).

    $\begin{array}{c}C=2\times 0.00108\text{in}\\ =0.00216\text{in}\end{array}$

Thus, the critical length of crack is $0.00216\text{in}$.

(b)

To determine

The critical crack length if the vessel made of alloy steel $\left(17-7\text{pH}\right)$.

Answer to Problem 45SEP

The critical length of crack is $0.024\text{in}$.

Explanation of Solution

Conclusion:

Refer table 7.1 “Typical fracture toughness values for selected engineering alloys”, to obtain the value of $K_{IC}$ as $70\text{ksi}\sqrt{\text{in}}$ corresponding to the material alloy steel $\left(17-7\text{pH}\right)$.

Substitute, $70\text{ksi}\sqrt{\text{in}}$ for $K_{IC}$, $1$ for $Y$ and $360\text{ksi}$ for $\sigma$ in Equation (II).

    $\begin{array}{c}a={\left(\frac{70\text{ksi}\sqrt{\text{in}}}{1\times 360\left(\text{ksi}\right)\times \sqrt{\pi }}\right)}^2\\ =\left(\frac{\left({\left(70\right)}^2\text{in}\right)}{{\left(360\right)}^2\times \pi }\right)\\ =0.012\text{in}\end{array}$

Substitute $0.00108\text{in}$ for $a$ in Equation (III).

    $\begin{array}{c}C=2\times 0.012\text{in}\\ =0.024\text{in}\end{array}$

Thus, the critical length of crack is $0.024\text{in}$.