Question

A 'T' section is welded to the flat bottom shell plate of a ship as shown in the following figure (bottom shell longitudinal). The neutral axis of the ship's midship section is 14 m above the bottom shell plate. The distance (X) of neutral axis of the 'T' section from the ship's neutral axis is .................... m (round off to two decimal places) 

 

• A. 12.63
• B. 13.63
• C. 15.24
• D. 11.24

Hint:

When calculating centroids of composite shapes, be careful and consistent with the choice of your reference axis. Choosing the bottom-most or left-most edge as the reference often simplifies calculations by keeping all coordinate values positive.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The neutral axis of a cross-section passes through its centroid. To find the distance X, we first need to calculate the position of the centroid (neutral axis) of the given T-section. The distance X is then the difference between the given position of the ship's neutral axis and the calculated position of the T-section's neutral axis, both measured from the same reference line (the bottom shell plate).
Step 2: Key Formula or Approach:
The formula for the y-coordinate of the centroid (\(\bar{y}\)) of a composite shape is: \[ \bar{y} = \frac{\sum A_i y_i}{\sum A_i} = \frac{A_1 y_1 + A_2 y_2}{A_1 + A_2} \] where \(A_i\) is the area of each component part and \(y_i\) is the distance of the centroid of that part from a reference axis.
Step 3: Detailed Explanation or Calculation:
Let's choose the bottom shell plate as our reference axis (\(y=0\)). The T-section is composed of a vertical web and a horizontal flange. All dimensions must be in consistent units (e.g., mm).
Component 1: Web
- Dimensions: 500 mm (height) \(\times\) 18 mm (width)
- Area: \(A_{web} = 500 \times 18 = 9000\) mm²
- Centroid location from reference axis: \(y_{web} = \frac{500}{2} = 250\) mm
Component 2: Flange
- Dimensions: 300 mm (width) \(\times\) 25 mm (height)
- Area: \(A_{flange} = 300 \times 25 = 7500\) mm²
- Centroid location from reference axis: The flange sits on top of the 500 mm web. Its centroid is at half its own height, measured from the top of the web. \(y_{flange} = 500 + \frac{25}{2} = 500 + 12.5 = 512.5\) mm
Calculate Centroid of the T-section (\(\bar{y}_T\)):
- Total Area: \(A_{total} = A_{web} + A_{flange} = 9000 + 7500 = 16500\) mm²
- Sum of moments of area: \(\sum A_i y_i = (A_{web} \times y_{web}) + (A_{flange} \times y_{flange})\)
\(\sum A_i y_i = (9000 \times 250) + (7500 \times 512.5) = 2,250,000 + 3,843,750 = 6,093,750\) mm³
- Position of T-section's neutral axis: \[ \bar{y}_T = \frac{\sum A_i y_i}{A_{total}} = \frac{6,093,750}{16500} \approx 369.318 \text{ mm} \] Calculate Distance X:
- Position of ship's neutral axis from bottom plate: \(y_{ship} = 14\) m = 14000 mm
- Position of T-section's neutral axis from bottom plate: \(\bar{y}_T = 369.318\) mm
- The distance X is the vertical separation between these two axes: \[ X = y_{ship} - \bar{y}_T = 14000 - 369.318 = 13630.682 \text{ mm} \] Convert X to meters and round to two decimal places: \[ X = \frac{13630.682}{1000} \text{ m} \approx 13.63 \text{ m} \] Step 4: Final Answer:
The distance X is 13.63 m.
Step 5: Why This is Correct:
The step-by-step calculation of the centroid for the T-section and the subsequent subtraction from the ship's neutral axis position yields a value of 13.63 m, which matches option (B).