A beam with a symmetrical T section has a top flange 50 mm wide and 20 mm thick, and a web 40 mm high and 10 mm thick. An additional plate, 10 mm thick and 60 mm wide, is welded above the flange. The moment of inertia of this symmetrical planar cross- section about an axis in its plane normal to the web and in line with the upper face of the 10 mm thick plate works out to 1506,666.66 mm4. The centroidal axis of the combined area is 21.5 mm below this axis, normal to the web. The moment of inertia of this built-up area about the centroidal axis is (in mm4):

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Explanation:

To find the moment of inertia of the given T-section with an additional plate welded on top, we need to follow a systematic approach involving the use of the parallel axis theorem and the properties of composite areas.

Step 1: Determine the individual moments of inertia and areas:

1. Top Plate:

• Width (b<sub>1</sub>) = 60 mm
• Thickness (h<sub>1</sub>) = 10 mm
• Area (A<sub>1</sub>) = b<sub>1</sub> × h<sub>1</sub> = 60 mm × 10 mm = 600 mm²
• Distance from the top of the top plate to its centroid (y<sub>1</sub>) = 5 mm
• Moment of Inertia about its own centroid (I<sub>1</sub>) = (b<sub>1</sub> × h<sub>1</sub>³) / 12 = (60 mm × (10 mm)³) / 12 = 5000 mm⁴

2. Top Flange of T-Section:

• Width (b<sub>2</sub>) = 50 mm
• Thickness (h<sub>2</sub>) = 20 mm
• Area (A<sub>2</sub>) = b<sub>2</sub> × h<sub>2</sub> = 50 mm × 20 mm = 1000 mm²
• Distance from the top of the top plate to the centroid of the flange (y<sub>2</sub>) = 10 mm + 10 mm = 20 mm
• Moment of Inertia about its own centroid (I<sub>2</sub>) = (b<sub>2</sub> × h<sub>2</sub>³) / 12 = (50 mm × (20 mm)³) / 12 = 33333.33 mm⁴

3. Web of T-Section:

• Height (h<sub>3</sub>) = 40 mm
• Thickness (t<sub>3</sub>) = 10 mm
• Area (A<sub>3</sub>) = h<sub>3</sub> × t<sub>3</sub> = 40 mm × 10 mm = 400 mm²
• Distance from the top of the top plate to the centroid of the web (y<sub>3</sub>) = 10 mm + 40 mm/2 + 20 mm = 50 mm
• Moment of Inertia about its own centroid (I<sub>3</sub>) = (t<sub>3</sub> × h<sub>3</sub>³) / 12 = (10 mm × (40 mm)³) / 12 = 53333.33 mm⁴

Step 2: Calculate the moment of inertia about the centroidal axis:

The centroidal axis is 21.5 mm below the upper face of the 10 mm thick plate. So, we need to use the parallel axis theorem to transfer the individual moments of inertia to the centroidal axis.

1. Top Plate:

• Distance to centroidal axis = 5 mm - 21.5 mm = -16.5 mm
• Parallel Axis Theorem: I<sub>1c</sub> = I<sub>1</sub> + A<sub>1</sub> × (distance)²
• I<sub>1c</sub> = 5000 mm⁴ + 600 mm² × (-16.5 mm)²
• I<sub>1c</sub> = 5000 mm⁴ + 600 mm² × 272.25 mm²
• I<sub>1c</sub> = 5000 mm⁴ + 163350 mm⁴ = 168350 mm⁴

2. Top Flange:

• Distance to centroidal axis = 20 mm - 21.5 mm = -1.5 mm
• Parallel Axis Theorem: I<sub>2c</sub> = I<sub>2</sub> + A<sub>2</sub> × (distance)²
• I<sub>2c</sub> = 33333.33 mm⁴ + 1000 mm² × (-1.5 mm)²
• I<sub>2c</sub> = 33333.33 mm⁴ + 1000 mm² × 2.25 mm²
• I<sub>2c</sub> = 33333.33 mm⁴ + 2250 mm⁴ = 35583.33 mm⁴

3. Web:

• Distance to centroidal axis = 50 mm - 21.5 mm = 28.5 mm
• Parallel Axis Theorem: I<sub>3c</sub> = I<sub>3</sub> + A<sub>3</sub> × (distance)²
• I<sub>3c</sub> = 53333.33 mm⁴ + 400 mm² × (28.5 mm)²
• I<sub>3c</sub> = 53333.33 mm⁴ + 400 mm² × 812.25 mm²
• I<sub>3c</sub> = 53333.33 mm⁴ + 324900 mm⁴ = 378233.33 mm⁴

Step 3: Sum the moments of inertia:

Total I<sub>c</sub> = I<sub>1c</sub> + I<sub>2c</sub> + I<sub>3c</sub>

Total I<sub>c</sub> = 168350 mm⁴ + 35583.33 mm⁴ + 378233.33 mm⁴

Total I<sub>c</sub> = 582166.66 mm⁴

Therefore, the moment of inertia of the built-up area about the centroidal axis is 5,82,166.66 mm⁴.

Important Information

To further understand the analysis, let’s evaluate the other options:

Option 1: 2,17,833.34 mm⁴

This value does not consider the correct centroidal distance for the top plate, flange, and web or does not properly sum up the individual moments of inertia using the parallel axis theorem.

Option 2: 70077.52 mm⁴

This value is significantly lower than the correct value and indicates a miscalculation in the individual moments of inertia or incorrect application of the parallel axis theorem.

Option 4: 1.33 × 10⁵ mm⁴

This value also does not align with the correct calculation and indicates errors in the intermediate steps or misapplication of the principles of moment of inertia calculation.

Conclusion:

The correct calculation involves proper use of the parallel axis theorem and accurate summation of the individual moments of inertia for the composite area. The correct moment of inertia about the centroidal axis is 5,82,166.66 mm⁴.

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