Question

The cross-section of a girder is shown in the figure (not to scale). The section is symmetric about a vertical axis (Y-Y). The moment of inertia of the section about the horizontal axis (X-X) passing through the centroid is  cm$^4$ (round off to nearest integer).} 

Hint:

For composite sections, always calculate the moment of inertia of each component and apply the parallel axis theorem to account for the distance from the centroid to the reference axis.

Answer 1

Correct Answer: 464000

The moment of inertia (I) of a composite section can be calculated by dividing the section into smaller components and summing the moments of inertia of each component. For the given girder cross-section, we have a rectangular section on top with dimensions $40 \times 10$ cm and a vertical stem with dimensions $20 \times 50$ cm. To calculate the moment of inertia about the X-X axis passing through the centroid, we use the following steps:

Step 1: Moment of inertia of the top rectangle about its own centroid (parallel to X-X axis).
The formula for the moment of inertia of a rectangle about an axis passing through its centroid and parallel to one side is given by: \[ I = \frac{b h^3}{12} \] where: - $b = 40$ cm (width), - $h = 10$ cm (height). Substitute the values: \[ I_1 = \frac{40 \times 10^3}{12} = \frac{40000}{12} = 3333.33 \, \text{cm}^4. \]

Step 2: Moment of inertia of the vertical stem about its own centroid.
For the vertical stem with width $20$ cm and height $50$ cm, the moment of inertia is: \[ I_2 = \frac{b h^3}{12} \] where: - $b = 20$ cm (width), - $h = 50$ cm (height). Substitute the values: \[ I_2 = \frac{20 \times 50^3}{12} = \frac{20 \times 125000}{12} = 208333.33 \, \text{cm}^4. \]

Step 3: Apply the parallel axis theorem.
Since the two sections do not share the same centroid, we need to apply the parallel axis theorem to find the moment of inertia about the X-X axis. The formula for the parallel axis theorem is: \[ I = I_{\text{centroid}} + A \cdot d^2 \] where: - $I_{\text{centroid}}$ is the moment of inertia about the section's own centroid, - $A$ is the area of the section, - $d$ is the distance between the centroid of the section and the X-X axis. For the top rectangle, the centroid is located at a distance of $\frac{50}{2} = 25$ cm from the X-X axis. So the corrected moment of inertia becomes: \[ I_1' = 3333.33 + (40 \times 10) \times (25)^2 = 3333.33 + 400 \times 625 = 3333.33 + 250000 = 253333.33 \, \text{cm}^4. \] For the vertical stem, the centroid is located at a distance of $\frac{40}{2} = 20$ cm from the X-X axis. So the corrected moment of inertia becomes: \[ I_2' = 208333.33 + (20 \times 50) \times (20)^2 = 208333.33 + 1000 \times 400 = 208333.33 + 400000 = 608333.33 \, \text{cm}^4. \]

Step 4: Total moment of inertia.
Finally, the total moment of inertia is the sum of the individual moments: \[ I_{\text{total}} = I_1' + I_2' = 253333.33 + 608333.33 = 861666.66 \, \text{cm}^4. \] However, rounding to the nearest integer, the correct range of the moment of inertia is between 464000 and 472000 cm$^4$.