Chapter 2, Problem 2.85P

(a)

To determine

The value of principal stress $\left({\sigma }_1\right)$ under the given tri-axial state of stress condition.

Answer to Problem 2.85P

The value of the principal stress $\left({\sigma }_1\right)$ under the given tri-axial state of stress is $56.694\text{MPa}$

Explanation of Solution

Given:

Yield strength of the material is $75\text{MPa}$.

Principal(normal) stress in x-direction is ${\sigma }_1$ .

Principal(normal) stress in y-direction is ${\sigma }_2=0$ .

Principal(normal) stress in z-direction is ${\sigma }_3=-\frac{{\sigma }_1}{2}$ .

Concept used:

Here ${\sigma }_2$ = 0. Hence the above problem crumbles down from tri-axial state of stress problem to a biaxial state of stress problem.

Write the Maximum distortion energy theory (Von mises Theory) as,

  ${\left({\sigma }_1-{\sigma }_2\right)}^2+{\left({\sigma }_2-{\sigma }_3\right)}^2+{\left({\sigma }_3-{\sigma }_1\right)}^2\le 2{\left(S_y\right)}^2$.......(1)

Here, ${\sigma }_1$ is principal stress in x-direction, ${\sigma }_2$ is principal stress in y-direction, ${\sigma }_3$ is principal stress is z-direction, $S_y$ is yield strength of the material.

Calculation:

Substitute $0$ for ${\sigma }_1$ , $0$ for ${\sigma }_2$ , $-\frac{{\sigma }_1}{2}$ for ${\sigma }_3$ and $75\text{MPa}$ for $S_y$ in equation (1).

  $\begin{array}{c}{\left({\sigma }_1-0\right)}^2+{\left(0-\left(-\frac{{\sigma }_1}{2}\right)\right)}^2+{\left(\left(-\frac{{\sigma }_1}{2}\right)-{\sigma }_1\right)}^2\le 2{\left(75\right)}^2\\ {\left({\sigma }_1\right)}^2+{\left(\frac{{\sigma }_1}{2}\right)}^2+{\left(-\frac{3{\sigma }_1}{2}\right)}^2\le 2{\left(75\right)}^2\\ \frac{14}{4}{\sigma }_1^2\le 2{\left(75\right)}^2\\ {\sigma }_1=56.694\text{MPa}\end{array}$

Conclusion:

Thus the value ofthe principal stress $\left({\sigma }_1\right)$ under the given tri-axial state of stress is $56.694\text{MPa}$ .

(b)

To determine

The value of principal stress $\left({\sigma }_1\right)$ under the given tri-axial state of stress condition.

Answer to Problem 2.85P

The value of principal stress $\left({\sigma }_1\right)$ under the given tri-axial state of stress is $57.61\text{MPa}$ .

Explanation of Solution

Given:

Yield strength of the material is $75\text{MPa}$.

Principal stress in x-direction is ${\sigma }_1$ .

Principal stress in y-direction is ${\sigma }_2=\frac{{\sigma }_1}{3}$ .

Principal stress in z-direction is ${\sigma }_3=-\frac{{\sigma }_1}{2}$ .

Concept used:

Here ${\sigma }_2$ = $\frac{{\sigma }_1}{3}$ . Hence the problem is a tri-axial state of stress problem.

Write the Maximum distortion energy theory (Von mises Theory) as;

  ${\left({\sigma }_1-{\sigma }_2\right)}^2+{\left({\sigma }_2-{\sigma }_3\right)}^2+{\left({\sigma }_3-{\sigma }_1\right)}^2\le 2{\left(S_y\right)}^2$.......(1)

Here, ${\sigma }_1$ is principal stress in x-direction, ${\sigma }_2$ is principal stress in y-direction, ${\sigma }_3$ is principal stress is z-direction, $S_y$ is yield strength of the material.

Calculation:

Substitute $0$ for ${\sigma }_1$ , $0$ for ${\sigma }_2$ , $-\frac{{\sigma }_1}{2}$ for ${\sigma }_3$ and $75\text{MPa}$ for $S_y$ in equation (1).

  $\begin{array}{c}{\left({\sigma }_1-\frac{{\sigma }_1}{3}\right)}^2+{\left(\frac{{\sigma }_1}{3}+\frac{{\sigma }_1}{2}\right)}^2+{\left(-\frac{{\sigma }_1}{2}-{\sigma }_1\right)}^2\le 2{\left(75\right)}^2\\ \left(\frac{4{\sigma }_1^2}{9}\right)+\left(\frac{25{\sigma }_1^2}{36}\right)+\left(\frac{9{\sigma }_1^2}{4}\right)\le 2{\left(75\right)}^2\\ \frac{122{\sigma }_1^2}{36}\le 2{\left(75\right)}^2\\ {\sigma }_1=57.61\text{MPa}\end{array}$

Conclusion:

Thus the value ofthe principal stress $\left({\sigma }_1\right)$ under the given tri-axial state of stress is $57.61\text{MPa}$