Chapter 14, Problem 1RP

The steel pipe has an inner diameter of 2.75 in. and an outer diameter of 3 in. If it is fixed at C and subjected to the horizontal 20-lb force acting on the handle of the pipe wrench, determine the principal stresses in the pipe at point A, which is located on the surface of the pipe.

To determine

Find 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:

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

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:

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

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.

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

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:

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

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

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

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

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