Question

The area moment of inertia about the y-axis of a linearly tapered section shown in the figure is ............... m⁴. (Answer in integer)

Hint:

When calculating \(I_y\), using vertical strips (\(dA = h(x)dx\)) is often easiest because all points in the strip are at the same distance \(x\) from the y-axis. When calculating \(I_x\), using horizontal strips is generally easier.

Answer 1

Step 1: Understanding the Concept:
The area moment of inertia of a shape about an axis measures its resistance to bending about that axis. For an area \(A\), the moment of inertia about the y-axis (\(I_y\)) is calculated by integrating the square of the distance from the y-axis over the entire area.
Step 2: Key Formula or Approach:
The formula for the area moment of inertia about the y-axis is: \[ I_y = \int_A x^2 \, dA \] For the given shape, we can consider a thin vertical strip of width \(dx\) at a distance \(x\) from the y-axis. The area of this strip is \(dA = h(x) dx\), where \(h(x)\) is the total height of the section at that \(x\). We will integrate from \(x=0\) to \(x=12\).
Step 3: Detailed Calculation:
1. Determine the height function \(h(x)\):
The section is symmetric about the x-axis. Let the top edge be \(y_{top}(x)\) and the bottom edge be \(y_{bottom}(x)\). The total height is \(h(x) = y_{top}(x) - y_{bottom}(x) = 2y_{top}(x)\).
- At \(x=0\), \(y_{top}(0) = 1.5\) m.
- At \(x=12\), \(y_{top}(12) = 3\) m.
Since the taper is linear, \(y_{top}(x)\) is a straight line: \(y_{top}(x) = mx + c\). - The y-intercept is \(c = 1.5\).
- The slope is \(m = \frac{3 - 1.5}{12 - 0} = \frac{1.5}{12} = \frac{1}{8}\).
- So, \(y_{top}(x) = \frac{1}{8}x + 1.5\).
- The total height is \(h(x) = 2 \cdot y_{top}(x) = 2 \left(\frac{1}{8}x + 1.5\right) = \frac{1}{4}x + 3\).
2. Set up the integral for \(I_y\):
The area of a differential strip is \(dA = h(x) dx = (\frac{1}{4}x + 3)dx\).
\[ I_y = \int_0^{12} x^2 \, dA = \int_0^{12} x^2 \left(\frac{1}{4}x + 3\right) dx \] 3. Evaluate the integral: \[ I_y = \int_0^{12} \left(\frac{1}{4}x^3 + 3x^2\right) dx \] \[ I_y = \left[ \frac{1}{4} \frac{x^4}{4} + 3 \frac{x^3}{3} \right]_0^{12} \] \[ I_y = \left[ \frac{x^4}{16} + x^3 \right]_0^{12} \] \[ I_y = \left( \frac{12^4}{16} + 12^3 \right) - (0) \] \[ I_y = \frac{20736}{16} + 1728 \] \[ I_y = 1296 + 1728 \] \[ I_y = 3024 \text{ m}^4 \] Step 4: Final Answer:
The area moment of inertia about the y-axis is 3024 \(m^4\).
Step 5: Why This is Correct:
The solution correctly determines the linear function describing the height of the tapered section. The integral for the area moment of inertia about the y-axis is set up correctly by considering differential vertical strips, and the definite integral is evaluated accurately.