Chapter 8, Problem 1RP

If it supports a cable loading of 800 lb, determine the maximum normal stress at section a–a and sketch the stress distribution acting over the cross section. Use the curved-beam formula to calculate the bending stress.

To determine

The maximum tensile stress ${\left({\sigma }_t\right)}_{\max }$ at section a-a.

The maximum compressive stress ${\left({\sigma }_c\right)}_{\max }$.

To sketch:

The stress distribution over the cross section.

Answer to Problem 1RP

The maximum tensile stress ${\left({\sigma }_t\right)}_{\max }$ is 15.8 ksi.

The maximum compressive stress ${\left({\sigma }_c\right)}_{\max }$ is $\underset{\_}{-10.5\text{ksi}}$.

Explanation of Solution

Given information:

The force in the cable is 800 lb.

Diameter of the circular is 1.25 in.

Calculation:

Expression to find the location of neutral $\left(R\right)$ surface from the center of curvature of the hook:

$R=\frac{A}{\sum {\displaystyle \underset{A}{\int }\frac{dA}{r}}}$ (1)

Here, R is the location of neutral axis, A is the cross sectional area of the member, r is the arbitrary position, and $dA$ is the area element on the cross section.

Determine the radius $\left(r\right)$ of the circular cross section as shown below:

$r=\frac{d}{2}$ (2)

Here, d is the diameter of the circular cross section.

Substitute 1.25 in. for d in Equation (2).

$\begin{array}{c}r=\frac{1.25}{2}\\ =0.625\text{in}\text{.}\end{array}$

Determine the area $\left(A\right)$ of circular cross section as shown below:

$A=\pi r^2$ (3)

Here, r is the radius of the circular cross section.

Substitute 0.625 in. for r in Equation (3).

$\begin{array}{c}A=\pi {\left(0.625\right)}^2\\ =1.227184{\text{in}}^{\text{2}}\end{array}$

Determine the value of ${\displaystyle \int \frac{dA}{r}}$ using the relation as shown below:

${\displaystyle \int \frac{dA}{r}}=2\pi \left(\overline{r}-\sqrt{{\overline{r}}^2-c^2}\right)$ (4)

Here, c is the radius of cross section and $\overline{r}$ is the centroid of the section.

Find the distance measured from the center of curvature to the centroid of the cross section $\overline{r}$ value as shown in below:

$\begin{array}{c}\overline{r}=\frac{3.75+2.5}{2}\\ =3.125\text{in}\text{.}\end{array}$

Substitute 0.625 in. for c and 3.125 in. for $\overline{r}$ in Equation (4).

$\begin{array}{c}{\displaystyle \int \frac{dA}{r}}=2\pi \left(3.125-\sqrt{{3.125}^2-{0.625}^2}\right)\\ =2\pi \left(0.0631\right)\\ =0.39670\end{array}$

Substitute $1.227184{\text{in}}^{\text{2}}$ for A and $0.39670$ for ${\displaystyle \int \frac{dA}{r}}$ in Equation (1).

$\begin{array}{c}R=\frac{1.227184}{0.39670}\\ =3.0934\text{in}\text{.}\end{array}$

Sketch the cross section of eye hook as shown in Figure 1.

Let the moment acting at the section be M.

Express to the value of M as shown below:

$M=F\times R$ (5)

Here, F is the load and R is the radius.

$\begin{array}{c}M=800\left(3.09343\right)\\ =2.474\times {10}^3\end{array}$

Determine the bending stress $\left(\sigma \right)$ using the curve beam formula as shown below.

$\sigma =\frac{M\left(R-r\right)}{Ar\left(\overline{r}-R\right)}+\frac{P}{A}$ (6)

Here, M is the applied moment and P is the applied load.

Substitute $2.474\times {10}^3$ for M, $3.0934\text{in}\text{.}$ for R, 0.625 in. for r, $3.125\text{in}\text{.}$ for $\overline{r}$, 2.5 in. for r, $1.227184{\text{in}}^{\text{2}}$ for A, and 800 lb for P in Equation (5).

Determine the maximum tensile stress ${\left({\sigma }_t\right)}_{\max }$ using the curve beam formula:

$\begin{array}{c}{\left({\sigma }_t\right)}_{\max }=\frac{2.475\left({10}^3\right)\left(3.09343-2.5\right)}{1.227184\left(2.5\right)\left(3.125-3.09343\right)}+\frac{800}{1.227184}\\ =\frac{1468.739}{0.096855}+651.89898\\ =15,816\text{psi}\times \frac{1\text{ksi}}{{10}^3\text{psi}}\\ =15.8\text{ksi}\end{array}$

Hence, the maximum tensile stress ${\left({\sigma }_t\right)}_{\max }$ is 15.8 ksi.

Determine the maximum compressive stress ${\left({\sigma }_c\right)}_{\max }$ using the curve beam formula:

Substitute $2.474\times {10}^3$ for M, $3.0934\text{in}\text{.}$ for R, 0.625 in. for r, $3.125\text{in}\text{.}$ for $\overline{r}$, 3.75 in. for r, $1.227184{\text{in}}^{\text{2}}$ for A, and 800 lb for P in Equation (5).

$\begin{array}{c}{\left({\sigma }_c\right)}_{\max }=\frac{2.475\left({10}^3\right)\left(3.09343-3.75\right)}{1.227184\left(3.75\right)\left(3.125-3.09343\right)}+\frac{800}{1.227184}\\ =\frac{-1625.010}{0.14528}+651.89898\\ =-11,185.366+651.89898\\ =-10,533\text{psi}\left(\frac{1\text{ksi}}{{10}^3\text{psi}}\right)\end{array}$

${\left({\sigma }_c\right)}_{\max }=-10.5\text{ksi}$

Hence, the maximum compressive stress ${\left({\sigma }_c\right)}_{\max }$ is $\underset{\_}{-10.5\text{ksi}}$

Sketch the stress distribution (tensile and compressive stress) along the cross section as shown in Figure 2.