Chapter 7.5, Problem 77P

For the state of stress shown, determine two values of σ_y for which the maximum shearing stress is 10 ksi.

Fig. P7.77

To determine

The two values of ${\sigma }_y$ for the state of stress.

Answer to Problem 77P

The values of ${\sigma }_y$ are $\underset{\_}{2\text{ksi and 9}\text{.33}\text{ksi}}$.

Explanation of Solution

Given information:

The components of stress ${\sigma }_x=14\text{ksi}$, and ${\tau }_{xy}=8\text{ksi}$.

The maximum shear stress ${\tau }_{\max }=10\text{ksi}$.

Calculation:

Consider $u=\frac{{\sigma }_y-{\sigma }_x}{2}$ (1)

Modify Equation (1) as shown below.

$\begin{array}{l}2u={\sigma }_y-{\sigma }_x\\ {\sigma }_y={\sigma }_x+2u\end{array}$ (2)

Calculate the average normal stress $\left({\sigma }_{\text{avg}}\right)$ as shown below.

${\sigma }_{\text{avg}}=\frac{{\sigma }_x+{\sigma }_y}{2}$

Substitute ${\sigma }_x+2u$ for ${\sigma }_y$.

$\begin{array}{c}{\sigma }_{\text{avg}}=\frac{{\sigma }_x+{\sigma }_x+2u}{2}\\ =\frac{2{\sigma }_x+2u}{2}\\ ={\sigma }_x+u\end{array}$ (3)

Calculate the value u as shown below.

${\tau }_{\max }=\sqrt{u^2+{\tau }_{xy}^2}$ (4)

Substitute $10\text{ksi}$ for ${\tau }_{\max }$ and $8\text{ksi}$ for ${\tau }_{xy}$ in Equation (4).

$\begin{array}{l}10=\sqrt{u^2+8^2}\\ 100=u^2+64\\ u^2=36\\ u=\pm 6\text{ksi}\end{array}$

Case 1:

For $u=6\text{ksi}$:

Calculate the value of ${\sigma }_y$ as shown below.

Substitute $14\text{ksi}$ for ${\sigma }_x$ and $6\text{ksi}$ for u in Equation (2).

$\begin{array}{c}{\sigma }_y=14+2\times 6\\ =26\text{ksi}\end{array}$

Calculate the average normal stress $\left({\sigma }_{\text{avg}}\right)$ as shown below.

Substitute $14\text{ksi}$ for ${\sigma }_x$ and $6\text{ksi}$ for u in Equation (3).

$\begin{array}{c}{\sigma }_{\text{avg}}=14+6\\ =20\text{ksi}\end{array}$

Calculate the principal stresses $\left({\sigma }_{\max }\text{and }{\sigma }_{\min }\right)$ as shown below.

${\sigma }_{\max ,\min }={\sigma }_{\text{avg}}\pm R$ (5)

Substitute $20\text{ksi}$ for ${\sigma }_{\text{avg}}$ and $10\text{ksi}$ for R in Equation (5).

$\begin{array}{l}{\sigma }_{a,b}=20\pm 10\\ {\sigma }_a=30\text{ksi}\\ {\sigma }_b=10\text{ksi}\end{array}$

Hence, the principal stresses are ${\sigma }_{\max }={\sigma }_a=30\text{ksi}$ and ${\sigma }_{\min }=0$.

Calculate the maximum shearing stress as shown below.

${\tau }_{\max }=\frac{1}{2}\left({\sigma }_{\max }-{\sigma }_{\min }\right)$ (6)

Substitute $30\text{ksi}$ for ${\sigma }_{\max }$ and $0$ for ${\sigma }_{\min }$ in Equation (6).

$\begin{array}{c}{\tau }_{\max }=\frac{1}{2}\left(30-0\right)\\ =15\text{ksi}\ne \text{7}\text{.5 ksi}\end{array}$

For $u=-6\text{ksi}$:

Calculate the value of ${\sigma }_y$ as shown below.

Substitute $14\text{ksi}$ for ${\sigma }_x$ and $-6\text{ksi}$ for u in Equation (2).

$\begin{array}{c}{\sigma }_y=14+2\times \left(-6\right)\\ =2\text{ksi}\end{array}$

Calculate the average normal stress $\left({\sigma }_{\text{avg}}\right)$ as shown below.

Substitute $14\text{ksi}$ for ${\sigma }_x$ and $-6\text{ksi}$ for u in Equation (3).

$\begin{array}{c}{\sigma }_{\text{avg}}=14+\left(-6\right)\\ =8\text{ksi}\end{array}$

Calculate the principal stresses $\left({\sigma }_{\max }\text{and }{\sigma }_{\min }\right)$ as shown below.

Substitute $8\text{ksi}$ for ${\sigma }_{\text{avg}}$ and $10\text{ksi}$ for R in Equation (5).

$\begin{array}{l}{\sigma }_{a,b}=8\pm 10\\ {\sigma }_a=18\text{ksi}\\ {\sigma }_b=-2\text{ksi}\end{array}$

Hence, the principal stresses are ${\sigma }_{\max }={\sigma }_a=18\text{ksi}$ and ${\sigma }_{\min }={\sigma }_a=-2\text{ksi}$.

Calculate the maximum shearing stress as shown below.

Substitute $18\text{ksi}$ for ${\sigma }_{\max }$ and $-2\text{ksi}$ for ${\sigma }_{\min }$ in Equation (6).

$\begin{array}{c}{\tau }_{\max }=\frac{1}{2}\left(18-\left(-2\right)\right)\\ =10\text{ksi}\left(\text{OK}\right)\end{array}$

Hence, the value of $\underset{\_}{{\sigma }_y=2\text{ksi}}$.

Case 2:

Assume the minimum principal stress ${\sigma }_{\min }=0$.

Calculate the maximum principal stress as shown below.

Substitute $0$ for ${\sigma }_{\min }$ and $10\text{ksi}$ for ${\tau }_{\max }$ in Equation (6).

$\begin{array}{l}10=\frac{1}{2}\left({\sigma }_{\max }-0\right)\\ {\sigma }_{\max }=20\text{ksi}\end{array}$

The maximum principal stress ${\sigma }_a={\sigma }_{\text{avg}}+R$.

Substitute ${\sigma }_x+u$ for ${\sigma }_{\text{avg}}$ and $\sqrt{u^2+{\tau }_{xy}^2}$ for R.

$\begin{array}{l}{\sigma }_a={\sigma }_x+u+\sqrt{u^2+{\tau }_{xy}^2}\\ \sqrt{u^2+{\tau }_{xy}^2}={\sigma }_a-u-{\sigma }_x\\ u^2+{\tau }_{xy}^2={\left({\sigma }_a-u-{\sigma }_x\right)}^2\\ u^2+{\tau }_{xy}^2={\sigma }_a^2+u^2+{\sigma }_x^2+2{\sigma }_a\left(-u\right)+2\left(-u\right)\left(-{\sigma }_x\right)+2\left(-{\sigma }_x\right){\sigma }_a\end{array}$

$\begin{array}{l}{u}^2+{\tau }_{xy}^2={\sigma }_x^2+{\sigma }_a^2+2u\left({\sigma }_x-{\sigma }_a\right)-2{\sigma }_x{\sigma }_a+u^2\\ {\tau }_{xy}^2={\left({\sigma }_a-{\sigma }_x\right)}^2-2u\left({\sigma }_a-{\sigma }_x\right)\\ 2u\left({\sigma }_a-{\sigma }_x\right)={\left({\sigma }_a-{\sigma }_x\right)}^2-{\tau }_{xy}^2\\ u=\frac{{\left({\sigma }_a-{\sigma }_x\right)}^2-{\tau }_{xy}^2}{2\left({\sigma }_a-{\sigma }_u\right)}\end{array}$

Substitute $14\text{ksi}$ for ${\sigma }_x$, $20\text{ksi}$ for ${\sigma }_a$, and $8\text{ksi}$ for ${\tau }_{xy}$.

$\begin{array}{c}u=\frac{{\left(20-14\right)}^2-8^2}{2\left(20-14\right)}\\ =\frac{-28}{12}\\ =-2.3333\text{ksi}\end{array}$

Calculate the value of ${\sigma }_y$ as shown below.

Substitute $14\text{ksi}$ for ${\sigma }_x$ and $-2.3333\text{ksi}$ for u in Equation (2).

$\begin{array}{c}{\sigma }_y=14+2\times \left(-2.3333\right)\\ =9.3334\text{ksi}\end{array}$

Calculate the value of R as shown below.

Substitute $-2.3333\text{ksi}$ for $u$ and $8\text{ksi}$ for ${\tau }_{xy}$ in Equation (4).

$\begin{array}{c}R=\sqrt{{\left(-2.3333\right)}^2+8^2}\\ =\sqrt{69.4443}\\ =8.3333\text{ksi}\end{array}$

Calculate the average normal stress $\left({\sigma }_{\text{avg}}\right)$ as shown below.

Substitute $14\text{ksi}$ for ${\sigma }_x$ and $-2.3333\text{ksi}$ for u in Equation (3).

$\begin{array}{c}{\sigma }_{\text{avg}}=14+\left(-2.3333\right)\\ =11.6667\text{ksi}\end{array}$

Calculate the principal stress ${\sigma }_b$ as shown below.

${\sigma }_b={\sigma }_{\text{avg}}-R$

Substitute $11.6667\text{ksi}$ for ${\sigma }_{\text{avg}}$ and $8.3333\text{ksi}$ for R.

$\begin{array}{c}{\sigma }_b=11.6667-8.3333\\ =3.3334\text{ksi}\end{array}$

Hence, the principal stresses ${\sigma }_{\max }=20\text{ksi, }{\sigma }_{\min }=0\text{,}$ and ${\tau }_{\max }=10\text{ksi}$.

Therefore, the value of $\underset{\_}{{\sigma }_y=9.33\text{ksi}}$.