Math Problem Statement

Bestimmen Sie das Flächenträgheitsmoment Ix eines Trägers bzgl. Schwerpunktachse mit der Form y = 2x und y = -2x wie in der folgenden Zeichnung: Wie ändert sich Ix, wenn man die Bezugsachse um zwei Einheiten nach unten schiebt P = (0,-2)?

Solution

The task involves determining the moment of inertia IxI_x of a structure with respect to the xx-axis, given its geometric shape described by the equations y=2xy = 2x and y=2xy = -2x, within the range x1|x| \leq 1. The diagram suggests a symmetric, double-triangle shape centered around the origin.

Step 1: Determine the Geometry

The area described consists of two identical triangles with vertices at (0,0), (1,2), and (-1,-2). The area of one of these triangles is: Area=12×base×height=12×2×2=2\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 2 = 2 Hence, the total area of both triangles is 44.

Step 2: Moment of Inertia about the x-axis (IxI_x)

The moment of inertia of a triangle about an axis through its base (parallel to the base) is given by: I=136×base×height3I = \frac{1}{36} \times \text{base} \times \text{height}^3 For one triangle, using base=2\text{base} = 2 and height=2\text{height} = 2: I=136×2×8=49I = \frac{1}{36} \times 2 \times 8 = \frac{4}{9} Since the axis passes through the centroid (central axis for the whole structure), there is no need to use the parallel axis theorem. The total IxI_x for both triangles is: Ix=2×49=89I_x = 2 \times \frac{4}{9} = \frac{8}{9}

Step 3: Adjust for the Shift in Axis

When the reference axis is moved from (0,0)(0,0) to (0,2)(0, -2), we apply the parallel axis theorem: Ix=Ix+Area×d2I'_x = I_x + \text{Area} \times d^2 where d=2d = 2 is the vertical distance the axis has moved: Ix=89+4×4=89+16=1529I'_x = \frac{8}{9} + 4 \times 4 = \frac{8}{9} + 16 = \frac{152}{9}

Therefore, the moment of inertia IxI'_x of the structure about the new axis y=2y = -2 is 1529\frac{152}{9}.

Do you need further details on any of these steps or have other questions? Here are some related questions you might consider:

  1. How would IxI_x change if the shape was reflected over the y-axis?
  2. What would be the moment of inertia around the y-axis for the original position?
  3. How does the moment of inertia affect the structural stability of the object?
  4. What is the parallel axis theorem and how is it applied in various contexts?
  5. What are other geometric transformations that might affect the moment of inertia?

Tip: When calculating moments of inertia, always verify the axis about which the inertia is calculated, as shifting the axis can significantly affect the results.

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Math Problem Analysis

Mathematical Concepts

Moment of Inertia
Geometry
Parallel Axis Theorem

Formulas

Moment of inertia of a triangle I = (1/36) * base * height^3
Parallel Axis Theorem: I' = I + A * d^2

Theorems

Parallel Axis Theorem

Suitable Grade Level

Undergraduate Engineering