Solve for the moment of inertia, about the x-axis at the centroid. Using the parallel axis theorem, solve for the moment of inertia, about the x-axis at the top of the shape.

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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Textbook: Strength of Materials

 

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A formula sheet is provided for your reference.

The cross section of a beam is shown below. The formula for the Moment of Inertia
about the x-axis at the centroid of the section is shown below.
Outer Diameter = 50 mm
Inner Diameter = 40 mm
Ixc = (douter - dinner)
dinmer)
64
a) Solve for the moment of inertia, about the x-axis at the centroid.
b) Using the parallel axis theorem, solve for the moment of inertia, about the x-axis at
the top of the shape.
Transcribed Image Text:The cross section of a beam is shown below. The formula for the Moment of Inertia about the x-axis at the centroid of the section is shown below. Outer Diameter = 50 mm Inner Diameter = 40 mm Ixc = (douter - dinner) dinmer) 64 a) Solve for the moment of inertia, about the x-axis at the centroid. b) Using the parallel axis theorem, solve for the moment of inertia, about the x-axis at the top of the shape.
Formula Sheet
Centre of Gravity and Moment of Inertia
ΣΑ
ΣΑ
EA,
ΣΑ
I, = I, + Ad² I¸ =E(!.)+ E(Ad²)
1, = E(!.)+
y
Axial Stress and Internal Forces
ΣΕ, -0 Σ , =0 ΣΜ=0
o = P/ A
T =V | A
S.F. = o alow /o.
allow
actual
Strain
PL
8 =
AE
· E \ateral
V =
E =
E =
L
E axial
Transcribed Image Text:Formula Sheet Centre of Gravity and Moment of Inertia ΣΑ ΣΑ EA, ΣΑ I, = I, + Ad² I¸ =E(!.)+ E(Ad²) 1, = E(!.)+ y Axial Stress and Internal Forces ΣΕ, -0 Σ , =0 ΣΜ=0 o = P/ A T =V | A S.F. = o alow /o. allow actual Strain PL 8 = AE · E \ateral V = E = E = L E axial
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