Crane Hook Design Calculation

Problem Statement: Design a crane hook to lift an average automobile weighing 1500 kg with a factor of safety of 5. The hook is made of structural steel, having a yield strength of 250 MPa. Calculate the dimensions of the hook’s cross-section assuming two types of sections: rectangular and circular.

Solution:

Step 1: Calculate the load (\(P\)) due to the weight of the automobile:

The weight of the car (\(m\)) is assumed to be 1500 kg. Using the formula \(P = m \cdot g\), where \(g = 9.81 \, \text{m/s}^2\), we get:

\( P = 1500 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 14,715 \, \text{N} \)

Step 2: Determine the allowable stress:

Given the yield strength of the steel, \( \sigma_y = 250 \, \text{MPa} \), and a factor of safety (FOS) of 5, the allowable stress is:

\( \sigma_{allowed} = \frac{\sigma_y}{FOS} = \frac{250 \, \text{MPa}}{5} = 50 \, \text{MPa} \)

Step 3: Calculate the required cross-sectional area:

The required cross-sectional area can be calculated using the formula \( A = \frac{P}{\sigma_{allowed}} \):

\( A = \frac{14,715 \, \text{N}}{50 \times 10^6 \, \text{N/m}^2} = 2.943 \times 10^{-4} \, \text{m}^2 = 294.3 \, \text{mm}^2 \)

Rectangular Cross-Section:

Assuming the width-to-height ratio \( b:h = 1:2 \), we can express the area as:

\( A = b \cdot h = \frac{h}{2} \cdot h = \frac{h^2}{2} \)

Solving for \(h\):

\( 294.3 = \frac{h^2}{2} \quad \Rightarrow \quad h^2 = 588.6 \quad \Rightarrow \quad h \approx 24.26 \, \text{mm} \)

Thus, the height \(h \approx 24.26 \, \text{mm}\), and the width \(b = \frac{h}{2} \approx 12.13 \, \text{mm}\).

Circular Cross-Section:

For a circular cross-section, the area is given by:

\( A = \frac{\pi d^2}{4} \)

Solving for \(d\):

\( 294.3 = \frac{\pi d^2}{4} \quad \Rightarrow \quad d^2 = \frac{4 \cdot 294.3}{\pi} \approx 374.52 \quad \Rightarrow \quad d \approx 19.36 \, \text{mm} \)

Summary of Dimensions: