Chapter 1, Problem 59RP

In the marine crane shown, link CD is known to have a uniform cross section of 50 × 150 mm. For the loading shown, determine the normal stress in the central portion of that link.

Fig. P1.59

To determine

The normal stress in the centre portion of the link.

Answer to Problem 59RP

The normal stress in the centre portion of the link is $\underset{\_}{195.3\text{MPa}}$

Explanation of Solution

Given information:

The uniform cross section is $\left(50\times 150\right)\text{mm}$.

Calculation:

Sketch the free body diagram of ABC as shown in Figure 1.

Here, W is the weight of the loading.

Find the weight of the loading as follows:

$W=mg$ (1)

Here, m is the mass of the loading and g is the acceleration due to gravity.

Substitute $80\text{Mg}$ for m and $9.81\text{m}/{\text{s}}^{\text{2}}$ in Equation (1).

$\begin{array}{c}W=80\times 9.81\\ =784.8\text{kN}\end{array}$

Taking moment about A,

$\begin{array}{l}{F}_{CD}\times 15-W\times 28=0\\ 15F_{CD}-28W=0\\ F_{CD}=\frac{28}{15}W\end{array}$ (2)

Substitute $784.8\text{kN}$ for W in Equation (2).

$\begin{array}{c}{F}_{CD}=\frac{28}{15}\left(784.8\text{kN}\right)\\ =1,464.96\\ =1,465\text{kN}\end{array}$

Find the area of the cross section using the relation:

$A=bd$ (3)

Here, b is the width of the cross section and d is the depth of the cross section.

Substitute $50\text{mm}$ for b and $150\text{mm}$ for d in Equation (3).

$\begin{array}{c}A=50\times 150\\ =7,500{\text{mm}}^2{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\\ =7.5\times {10}^{-3}{\text{m}}^{\text{2}}\end{array}$

Determine the normal stress in the central portion of the link as follows:

${\sigma }_{CD}=\frac{F_{CD}}{A}$ (4)

Here, A is the area of the cross section and $F_{CD}$ is force in member CD.

Substitute $7.5\times {10}^{-3}{\text{m}}^{\text{2}}$ for A and $1,465\text{kN}$ for $F_{CD}$ in Equation (4).

$\begin{array}{c}{\sigma }_{CD}=\frac{1,465}{7.5\times {10}^{-3}}\\ =195.3\times {10}^6\text{Pa}\\ \text{=}195.3\times {10}^6\text{Pa}\times \frac{1\text{Mpa}}{{10}^6\text{Pa}}\\ =195.3\text{Mpa}\end{array}$

Thus, the normal stress in the centre portion of the link is $\underset{\_}{195.3\text{MPa}}$.