Question
Determine the moments of inertia and the product of inertia of the$\mathrm{L} 152 \times 102 \times 12.7-\mathrm{mm}$ angle cross section of Prob. 9.78 with respect to new centroidal axes obtained by rotating the $x$ and $y$ axes $30^{\circ}$ clockwise.
Step 1
The formulas are given by: $I_{x'} = I_x \cos^2(\theta) + I_y \sin^2(\theta) - 2I_{xy} \sin(\theta) \cos(\theta)$ $I_{y'} = I_x \sin^2(\theta) + I_y \cos^2(\theta) + 2I_{xy} \sin(\theta) \cos(\theta)$ $I_{x'y'} = (I_y - I_x) \sin(\theta) \cos(\theta) + I_{xy} Show more…
Show all steps
Your feedback will help us improve your experience
Eric Mockensturm and 79 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Using Mohr's circle, determine the moments of inertia and the product of inertia of the L152 $\times 102 \times 12.7-\mathrm{mm}$ angle cross section of Prob. 9.78 with respect to new centroidal axes obtained by rotating the $x$ and $y$ axes $30^{\circ}$ clockwise.
Distributed Forces: Moments of Inertia
Mohr’s Circle for Moments of Inertia
Determine the moments of inertia and the product of inertia of the area of Prob. 9.73 with respect to new centroidal axes obtained by rotating the $x$ and $y$ axes $60^{\circ}$ counterclockwise.
Transformation of Moments of Inertia
Determine the moments of inertia and the product of inertia of the $\mathrm{L} 3 \times 2 \times \frac{1}{4}$ -in angle cross section of Prob. 9.74 with respect to new centroidal axes obtained by rotating the $x$ and $y$ axes $30^{\circ}$ clockwise.
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD