Use the given values in problem to answer the following: Find the moment of inertia and radius of gyration of the section of this bar about an axis parallel to x-axis going through the center of gravity of the bar.The bar is symmetrical about the axis parallel to y-axis and going through the center of gravity of the bar and about the axis parallel to z-axis and going through the center of gravity of the bar. The dimensions of the section are:l=51 mm, h=29 mmThe triangle: hT=15 mm, lT=18 mmand the 2 circles: diameter=7.4 mm, hC=8 mm, dC=7 mm.A is the origin of the referential axis.Provide an organized table and explain all your steps to find the moment of inertia and radius of gyration about an axis parallel to x-axis and going through the center of gravity of the bar.Does the radius of gyration make sense?In the box below enter the y position of the center of gravity of the bar in mm with one decimal. Geometric Properties of Line and Area ElementsCentroid LocationCentroid Location[cbyL-rr sin0Circular are segment-L-20rQuarter and semicircle aresaA₂h (a + b)aCTrapezoidal area-b-A-3abSemiparabolic areaAT (++)a-C}L-atExparabolic area-A-ab10 bA=4abParabolic areaThLhyCircular sector areaA-0²sin1-17²VQuarter circle areaySemicircular areaXCircular areazA-²²XA-bhTriangular area-XRectangular area-A-bhhArea Moment of Inertia1-¹ (0-sin 20)1,---r¹(0+ -sin 20)1164-164₂2747774411₂--bh³12124,--bh³36FormulasMoments of Inertia¹x = [ y²dAly¹y = SxTheorem of Parallel Axis1x = 1 +d² Aaxis going through the centroidx' axis parallel to x going through the point of interestd minimal distance (perpendicular) between and x'lyr=15+d² Aaxis going through the centroidy' axis parallel to y going through the point of interestd minimal distance (perpendicular) between y and y'Composite Bodies1-ΣAll the moments of inertia shouldbe about the same axis.x²dARadius of Gyrationk = hyAĽx.XвтеL

Given:I=51 mmh=29 mmhT=15 mmIT=18 mmϕc=7.4 mmhc=8 mmdc=7 mmFind:The moment of inertia and the radius…

Answered: Find the moment of inertia and radius…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Related questions

Concept explainers

Question

Use the given values in problem to answer the following:

Find the moment of inertia and radius of gyration of the section of this bar about an axis parallel to x-axis going through the center of gravity of the bar.

The bar is symmetrical about the axis parallel to y-axis and going through the center of gravity of the bar and about the axis parallel to z-axis and going through the center of gravity of the bar.

The dimensions of the section are:

l=51 mm, h=29 mm
The triangle: h_T=15 mm, l_T=18 mm
and the 2 circles: diameter=7.4 mm, h_C=8 mm, d_C=7 mm.

A is the origin of the referential axis.

Provide an organized table and explain all your steps to find the moment of inertia and radius of gyration about an axis parallel to x-axis and going through the center of gravity of the bar.

Does the radius of gyration make sense?

In the box below enter the y position of the center of gravity of the bar in mm with one decimal.

Geometric Properties of Line and Area Elements
Centroid Location
Centroid Location
[c
b
y
L-r
r sin
0
Circular are segment
-L-20r
Quarter and semicircle ares
aA₂h (a + b)
a
C
Trapezoidal area
-b-A-3ab
Semiparabolic area
AT (++)
a-
C
}
L-at
Exparabolic area
-A-ab
10 b
A=4ab
Parabolic area
T
h
L
h
y
Circular sector area
A-0²
sin
1-17²
V
Quarter circle area
y
Semicircular area
X
Circular area
z
A-²²
X
A-bh
Triangular area
-X
Rectangular area
-A-bh
h
Area Moment of Inertia
1-¹ (0-sin 20)
1,---r¹(0+ -sin 20)
1
16
4-16
4₂
274
7774
4
1
1₂--bh³
12
12
4,--bh³
36
Formulas
Moments of Inertia
¹x = [ y²dA
ly
¹y = Sx
Theorem of Parallel Axis
1x = 1 +d² A
axis going through the centroid
x' axis parallel to x going through the point of interest
d minimal distance (perpendicular) between and x'
lyr=15+d² A
axis going through the centroid
y' axis parallel to y going through the point of interest
d minimal distance (perpendicular) between y and y'
Composite Bodies
1-Σ
All the moments of inertia should
be about the same axis.
x²dA
Radius of Gyration
k =

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 4 steps with 3 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

Similar questions