Chapter 7.5, Problem 76P

To determine

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

Answer to Problem 76P

The values of ${\sigma }_y$ are $\underset{\_}{60\text{MPa and }-122\text{MPa}}$.

Explanation of Solution

Given information:

The components of stress ${\sigma }_x=-50\text{MPa}$ and ${\tau }_{xy}=48\text{MPa}$.

The maximum shear stress ${\tau }_{\max }=73\text{MPa}$.

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 $73\text{MPa}$ for ${\tau }_{\max }$ and $48\text{MPa}$ for ${\tau }_{xy}$ in Equation (4).

$\begin{array}{l}73=\sqrt{u^2+{48}^2}\\ 5,329=u^2+2,304\\ u^2=3,025\\ u=\pm 55\text{MPa}\end{array}$

Case 1:

For $u=55\text{MPa}$:

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

Substitute $-50\text{MPa}$ for ${\sigma }_x$ and $55\text{MPa}$ for u in Equation (2).

$\begin{array}{c}{\sigma }_y=-50+2\times 55\\ =60\text{MPa}\end{array}$

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

Substitute $-50\text{MPa}$ for ${\sigma }_x$ and $55\text{MPa}$ for u in Equation (3).

$\begin{array}{c}{\sigma }_{\text{avg}}=-50+55\\ =5\text{MPa}\end{array}$

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 $5\text{MPa}$ for ${\sigma }_{\text{avg}}$ and $73\text{MPa}$ for R in Equation (5).

$\begin{array}{l}{\sigma }_{\max ,\min }=5\pm 73\\ {\sigma }_{\max }=78\text{MPa}\\ {\sigma }_{\min }=-68\text{MPa}\end{array}$

Calculate the maximum shearing stress as shown below.

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

Substitute $78\text{MPa}$ for ${\sigma }_{\max }$ and $-68\text{MPa}$ for ${\sigma }_{\min }$ in Equation (6).

$\begin{array}{c}{\tau }_{\max }=\frac{1}{2}\left(78-\left(-68\right)\right)\\ =\frac{1}{2}\times 146\\ =73\text{MPa }\left(\text{OK}\right)\end{array}$

For $u=-55\text{MPa}$:

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

Substitute $-50\text{MPa}$ for ${\sigma }_x$ and $-55\text{MPa}$ for u in Equation (2).

$\begin{array}{c}{\sigma }_y=-50+2\times \left(-55\right)\\ =-160\text{MPa}\end{array}$

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

Substitute $-50\text{MPa}$ for ${\sigma }_x$ and $-55\text{MPa}$ for u in Equation (3).

$\begin{array}{c}{\sigma }_{\text{avg}}=-50-55\\ =-105\text{MPa}\end{array}$

Calculate the principal stresses $\left({\sigma }_a\text{and }{\sigma }_b\right)$ as shown below.

Substitute $-105\text{MPa}$ for ${\sigma }_{\text{avg}}$ and $73\text{MPa}$ for R in Equation (5).

$\begin{array}{l}{\sigma }_{a,b}=-105\pm 73\\ {\sigma }_a=-32\text{MPa}\\ {\sigma }_b=-178\text{MPa}\end{array}$

Hence, the principal stresses ${\sigma }_{\max }=0$ and ${\sigma }_{\min }=-178\text{MPa}$.

Calculate the maximum shearing stress as shown below.

Substitute $0$ for ${\sigma }_{\max }$ and $-178\text{MPa}$ for ${\sigma }_{\min }$ in Equation (6).

$\begin{array}{c}{\tau }_{\max }=\frac{1}{2}\left(0-\left(-178\right)\right)\\ =89\text{MPa}>73\text{MPa}\end{array}$

Hence, the value of $\underset{\_}{{\sigma }_y=60\text{MPa}}$.

Case 2:

Assume the maximum principal stress ${\sigma }_{\max }=0$.

Calculate the minimum principal stress as shown below.

Substitute $0$ for ${\sigma }_{\max }$ and $73\text{MPa}$ for ${\tau }_{\max }$ in Equation (6).

$\begin{array}{l}73=\frac{1}{2}\left(0-{\sigma }_{\min }\right)\\ {\sigma }_{\min }=-146\text{MPa}\end{array}$

The minimum principal stress ${\sigma }_b={\sigma }_{\text{avg}}-R$.

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

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

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

Substitute $-50\text{MPa}$ for ${\sigma }_x$, $-146\text{MPa}$ for ${\sigma }_b$, and $48\text{MPa}$ for ${\tau }_{xy}$.

$\begin{array}{c}u=\frac{{48}^2-{\left(-50-\left(-146\right)\right)}^2}{2\times \left(-50-\left(-146\right)\right)}\\ =\frac{-6,912}{192}\\ =-36\text{MPa}\end{array}$

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

Substitute $-50\text{MPa}$ for ${\sigma }_x$ and $-36\text{MPa}$ for u in Equation (2).

$\begin{array}{c}{\sigma }_y=-50+2\times \left(-36\right)\\ =-122\text{MPa}\end{array}$

Calculate the value of R as shown below.

Substitute $-36\text{MPa}$ for $u$ and $48\text{MPa}$ for ${\tau }_{xy}$ in Equation (4).

$\begin{array}{c}R=\sqrt{{\left(-36\right)}^2+{48}^2}\\ =\sqrt{3,600}\\ =60\text{MPa}\end{array}$

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

${\sigma }_a={\sigma }_b+2R$

Substitute $-146\text{MPa}$ for ${\sigma }_b$ and $60\text{MPa}$ for R.

$\begin{array}{c}{\sigma }_a=-146+2\times 60\\ =-26\text{MPa}\left(\text{OK}\right)\end{array}$

Hence, the principal stresses ${\sigma }_{\max }=0\text{, }{\sigma }_{\min }=-146\text{MPa,}$ and ${\tau }_{\max }=73\text{MPa}$.

Therefore, the value of $\underset{\_}{{\sigma }_y=-122\text{MPa}}$.