Question
Download Solution PDFFor a symmetrical T-section, the moment of inertia through centroidal axes in its plane parallel to the flange Ixx = 2 × 107 mm4, and perpendicular to the flange is Iyy = 1.5 × 107mm4. The moment of inertia about the centroidal axis normal to the planar area works out to (in mm4):
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
Moment of Inertia Analysis for Symmetrical T-Section:
Definition: The moment of inertia is a property of a shape that quantifies its resistance to rotational motion about an axis. For symmetrical sections, the moment of inertia can be calculated about centroidal axes in its plane (Ixx), perpendicular to its plane (Iyy), and normal to the planar area.
Given:
- Moment of inertia about the centroidal axis parallel to the flange (Ixx) = 2 × 107 mm4
- Moment of inertia about the centroidal axis perpendicular to the flange (Iyy) = 1.5 × 107 mm4
Polar Moment of Inertia (J):
- The polar moment of inertia (J) about the centroidal axis normal to the planar area is the sum of the moments of inertia about the two perpendicular centroidal axes in the plane:
J = Ixx + Iyy
Substitute the given values:
- Ixx = 2 × 107 mm4
- Iyy = 1.5 × 107 mm4
J = (2 × 107) + (1.5 × 107)
J = 3.5 × 107 mm4
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