50 mm150 mm 150 mm300 mmfind the moment of inertia for thisexercise, with the following formulas:PolarJo= Iox ++ ToyIox - Icx + Adg²Toy = Icg + dobe ²Caculation of centroid attached below. use itto solve the question in red box asapSomAv3comfor Portion A₁ -the figure is symmeabout y-aris.2=0.AL120mmAALDArea of A₁ = 300x50= 15000Y₁ = y₁y= 300for Pation A2Area Az= 50x300= 15000By Considering Centroid, from the Bottom of the section.300+ 50 = ₁ + 3/2= 325Centroid (ā, y) == [0₁SE= 50mmA, Ya+ A₂40,A₁+A₂15000x150 +15000x32515000 15000(0.237.5)

A1=300×50A1=15000 mm2y¯1=3002=150 mm from bottom,A2=300×50A2=15000 mm2y¯2=300+502=325 mm,

Answered: 50 mm 150 mm 150 mm 300 mm *

Engineering

Mechanical Engineering

50 mm 150 mm 150 mm 300 mm *

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter9: Moments And Products Of Inertia Of Areas

Section: Chapter Questions

Problem 9.13P: Figure (a) shows the cross section of a column that uses a structural shape known as W867...

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Concept explainers

Design of Shafts and Shaft Components

The shaft is basically defined as the machine element that rotates, is in circular cross-section, it is majorly used to transfer the power from one part of the system to another, or from a power generating machine to a power absorbing machine. The shaft is supported on bearings and is rotated with the help of a set of gears or pulleys that are used for the power transmission. Bending moment, torsion, and an axial force act upon the shaft. Designing of shaft primarily involves determining stresses at a critical point in the shaft that is arising due to loading.

Question

50 mm
150 mm 150 mm
300 mm
find the moment of inertia for this
exercise, with the following formulas:
Polar
Jo= Iox +
+ Toy
Iox - Icx + Adg²
Toy = Icg + dobe ²
Caculation of centroid attached below. use it
to solve the question in red box asap
Som
Av
3com
for Portion A₁ -
the figure is symme
about y-aris.
2=0.
AL
120mm
AALD
Area of A₁ = 300x50
= 15000
Y₁ = y₁
y
= 300
for Pation A2
Area Az
= 50x300
= 15000
By Considering Centroid, from the Bottom of the section.
300+ 50 = ₁ + 3/2
= 325
Centroid (ā, y) =
= [0₁
SE
= 50mm
A, Ya+ A₂4
0,
A₁+A₂
15000x150 +15000x325
15000 15000
(0.237.5)

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