Chapter 9, Problem 1RP

To determine

The principal stresses at point A.

Answer to Problem 1RP

The principal stresses at point A $\left({\sigma }_1\right)$ and $\left({\sigma }_2\right)$ are $\underset{\_}{119\text{psi}}$ and $\underset{\_}{-119\text{psi}}$.

Explanation of Solution

Calculate the normal stress $\left({\sigma }_A\right)$ acting at point A using the relation:

${\sigma }_A=\frac{M_{y}z}{I_y}$ (1)

Here, $M_y$ is the moment in y-direction, z is the distance in z-direction from centroid to point A, and $I_y$ is the moment of inertia.

Sketch the internal forces and moment in free body diagram at point A as shown in Figure 1.

Apply Equilibrium equations to find the value of moment at point A.

Sum of moments in y-direction is equal to 0.

$\begin{array}{l}\Sigma M_y=0\\ M_y-\left(20\times 10\right)=0\\ M_y=200\text{lb}\cdot \text{in}\end{array}$

Sum of moments in x-direction is equal to 0.

$\begin{array}{l}\Sigma M_x=0\\ T_x+\left(20\times 12\right)=0\\ T_x=240\text{lb}\cdot \text{in}\end{array}$

Sum of forces in z-direction is equal to 0.

$\begin{array}{l}\Sigma V_z=0\\ V_z-20=0\\ V_z=20\text{lb}\end{array}$

Find the moment of inertia of the section $\left(I\right)$:

Outer radius of the pipe is 1.5 in. and the inner radius of the pipe is 1.375 in.

$\begin{array}{c}I=\frac{\pi }{4}\left[{\left(1.5\text{in}\right)}^4-{\left(1.375\text{in}\right)}^4\right]\\ =1.1687{\text{in}}^4\end{array}$

Find the polar moment of inertia of the section $\left(J\right)$:

$\begin{array}{c}J=\frac{\pi }{2}\left[{\left(1.5\text{in}\right)}^4-{\left(1.375\text{in}\right)}^4\right]\\ =2.3374{\text{in}}^4\end{array}$

Sketch the cross section at point A as shown in Figure 2.

Find the first moment of area at point A ${\left(Q_A\right)}_z$ using the relation:

${\left(Q_A\right)}_z=\Sigma \overline{y}\text{'}A\text{'}$ (2)

Here, $\overline{y}\text{'}$ is the centroid distance to point A and $A\text{'}$ is the area from centroid to point A.

Refer to Figure 2.

$\begin{array}{c}{\left(Q_A\right)}_z=\frac{4\times 1.5\text{in}.}{3\pi }\left[\frac{1}{2}\times \pi \times {\left(1.5\text{in}.\right)}^2\right]-\frac{4\times 1.375\text{in}.}{3\pi }\left[\frac{1}{2}\times \pi \times {\left(1.375\text{in}.\right)}^2\right]\\ =0.51693{\text{in}}^3\end{array}$.

Substitute $200\text{lb}\cdot \text{in}$ for $M_y$, $0$ for z, and $1.1687{\text{in}}^4$ for $I_y$ in Equation (1).

$\begin{array}{c}{\sigma }_A=-\frac{\left(200\text{lb}\cdot \text{in}\right)\left(0\right)}{1.1687{\text{in}}^4}\\ =0\end{array}$

Find the shear stress $\left({\tau }_A\right)$ at point A using the relation:

${\tau }_A={\left(\frac{VQ}{It}\right)}_z-\frac{T\rho }{J}$ (3)

Substitute $20\text{lb}$ for $V_z$, $0.51693{\text{in}}^3$ for ${\left(Q_A\right)}_z$, $1.1687{\text{in}}^4$ for $I$, $2\times 0.125\text{in}\text{.}$ for t, $240\text{lb}\cdot \text{in}$ for $T$, $1.5\text{in}.$ for $\rho$, and $2.3374{\text{in}}^4$ for J in Equation (3).

$\begin{array}{c}{\tau }_A=\frac{\left(20\text{lb}\right)\left(0.51693{\text{in}}^3\right)}{\left(1.1687{\text{in}}^4\right)2\left(0.125\text{in}.\right)}-\frac{\left(240\text{lb}\cdot \text{in}\right)\left(1.5\text{in}.\right)}{\left(2.3374{\text{in}}^4\right)}\\ =-118.6\text{psi}\end{array}$

Sketch the state of stress at point A as shown in Figure 3.

Refer to Figure 3.

The value of normal stresses are ${\sigma }_x=0$ and ${\sigma }_z=0$.

The value of shear stress is ${\tau }_{xz}=-118.6\text{psi}$.

Find the principal stresses $\left({\sigma }_1\right)$ and $\left({\sigma }_2\right)$ at point A:

${\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_z}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_z}{2}\right)}^2+{\tau }_{xz}^2}$ (4)

Substitute $0$ for ${\sigma }_x$, $0$ for ${\sigma }_z$, and $-118.6\text{psi}$ for ${\tau }_{xz}$ in Equation (4).

$\begin{array}{c}{\sigma }_{1,2}=0\pm \sqrt{0+{\left(-118.6\right)}^2}\\ =0\pm 118.6\\ {\sigma }_1=0+118.6\\ =118.6\text{psi}\end{array}$

$\begin{array}{c}{\sigma }_1\approx 119\text{psi}\\ {\sigma }_2=0-118.6\\ =-118.6\text{psi}\\ \approx -119\text{psi}\end{array}$

Therefore, the principal stresses $\left({\sigma }_1\right)$ and $\left({\sigma }_2\right)$ at point A are $\underset{\_}{119\text{psi}}$ and $\underset{\_}{-119\text{psi}}$.