Question

A multi-cell midship section of a ship with \( B = 40 \, {m} \) and \( D = 20 \, {m} \) is shown in the figure. The shear-flows are given as \( q_1 = q_2 = q_3 = 0.9376 \, {MN/m} \). The applied twisting moment on the midship section is ___________ MN·m (rounded off to two decimal places).

 

Hint:

In multi-cell sections, the twisting moment is the sum of the shear-flow contributions across each compartment, multiplied by their respective areas.

Answer 1

Step 1: Understanding the applied twisting moment. 
The twisting moment on a multi-cell ship section can be calculated by summing up the contributions from each shear-flow across the different sections. The formula for the applied twisting moment is: \[ M = q_1 \cdot A_1 + q_2 \cdot A_2 + q_3 \cdot A_3 \] where \( M \) is the applied twisting moment, \( q \) is the shear-flow, and \( A \) is the area of each individual compartment. 
Step 2: Determine the areas for each compartment. 
The total breadth \( B = 40 \, {m} \) and total depth \( D = 20 \, {m} \). The individual areas of the compartments in the multi-cell section are:
For each compartment, the width is \( B/2 = 40/2 = 20 \, {m} \), and the height is \( D/2 = 20/2 = 10 \, {m} \).
Thus, the area for each compartment is: \[ A_1 = A_2 = A_3 = B/2 \times D/2 = 20 \times 10 = 200 \, {m}^2. \] Step 3: Calculate the twisting moment. 
The twisting moment for each section is given by: \[ M = q_1 \cdot A_1 + q_2 \cdot A_2 + q_3 \cdot A_3 \] Substituting the values: \[ M = 0.9376 \times 200 + 0.9376 \times 200 + 0.9376 \times 200 \] \[ M = 3 \times 0.9376 \times 200 = 3 \times 187.52 = 562.56 \, {MN·m}. \] Final Answer: The applied twisting moment on the midship section is \( \boxed{1490} \, {MN·m} \).