Without actually solving the centroidal moment of inertia about x-axis, how will you tackle this problem? Example answer: Icgx = [(bh^3/12 + Ad^2) of the top rectangle] - [(bh^3/36 + Ad^2) of the triangle] + [pi(r^4)/4 of the circle] Reminder: In your future courses, the location of the centroid/center of gravity can also be referred to as the location of the neutral axis. |Neutral Axis
Without actually solving the centroidal moment of inertia about x-axis, how will you tackle this problem? Example answer: Icgx = [(bh^3/12 + Ad^2) of the top rectangle] - [(bh^3/36 + Ad^2) of the triangle] + [pi(r^4)/4 of the circle] Reminder: In your future courses, the location of the centroid/center of gravity can also be referred to as the location of the neutral axis. |Neutral Axis
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
Is the answer on the right photo correct for the corresponding question?
Transcribed Image Text:[bh³
+ AdA?|+
12
+ Ad
12
+.
12
+ AdA?
(Ctrl) -
![Without actually solving the centroidal moment of inertia about x-axis, how will you tackle this
problem?
Example answer: Icgx = [(bh^3/12 + Ad^2) of the top rectangle] - [(bh^3/36 + Ad^2) of the triangle]
+ [pi(r^4)/4 of the circle]
Reminder: In your future courses, the location of the centroid/center of gravity can also be referred to
as the location of the neutral axis.
|Neutral Axis
TH](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0ccb8bf5-509f-4203-88ab-a015a8a5329e%2Fa72bb440-f16c-4643-a360-c5bfa7148261%2Fhrv5id4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Without actually solving the centroidal moment of inertia about x-axis, how will you tackle this
problem?
Example answer: Icgx = [(bh^3/12 + Ad^2) of the top rectangle] - [(bh^3/36 + Ad^2) of the triangle]
+ [pi(r^4)/4 of the circle]
Reminder: In your future courses, the location of the centroid/center of gravity can also be referred to
as the location of the neutral axis.
|Neutral Axis
TH
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