Applied Statics and Strength of Materials (6th Edition)
Applied Statics and Strength of Materials (6th Edition)
6th Edition
ISBN: 9780133840544
Author: George F. Limbrunner, Craig D'Allaird, Leonard Spiegel
Publisher: PEARSON
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Chapter 8, Problem 8.1P

Calculate the moment of intertia with respect to the X-X centroidal axes for the areas shown.

Chapter 8, Problem 8.1P, Calculate the moment of intertia with respect to the X-X centroidal axes for the areas shown.

a.

Expert Solution
Check Mark
To determine

Moment of inertia with respect to X-X centroidal axes.

Answer to Problem 8.1P

  IX¯=13623 in.4 .

Explanation of Solution

Given:

Applied Statics and Strength of Materials (6th Edition), Chapter 8, Problem 8.1P , additional homework tip  1

Concept Used: Since the shape is symmetrical about X and Y axis, so the centroidal axis lies in the center of the cross-section. The moment of inertia of the ring will be equal to the moment of inertia of the outer rectangle minus the moment of inertia on the inner circle.

Calculation: Moment of inertia of the Square about the centroid is

  I X ¯1=bh312I X ¯1=12× 24312in.4I X ¯1=13824 in.4

Moment of inertia of the Circle about the centroid is

  I X ¯2=πd464in.4I X ¯2=π×8464in.4I X ¯2=201.062 in.4

Total moment of inertia of the shape is

  IX¯=I X ¯1I X ¯2IX¯=13824 in.4201.062 in.4IX¯=13623 in.4

Conclusion: Moment of inertia with respect to X-X centroidal axes IX¯=13623 in.4

b.

Expert Solution
Check Mark
To determine

Moment of inertia with respect to X-X centroidal axes.

Answer to Problem 8.1P

  IX¯= 1017.82 in4 .

Explanation of Solution

Given:

Applied Statics and Strength of Materials (6th Edition), Chapter 8, Problem 8.1P , additional homework tip  2

Concept Used: Since the shape is symmetrical about X and Y axis, so the centroidal axis lies in the center of the cross-section. The moment of inertia of the ring will be equal to the moment of inertia of the outer circle 12 in. minus the moment of inertia on the inner circle of diameter 10 in.

Calculation: Total moment of inertia of the circular ring is

  IX¯=I X ¯1I X ¯2IX¯=πd1464πd2464in4IX¯=π× 12464π×1464in4IX¯= 1017.82 in4

Conclusion:

Moment of inertia of the circular ring is IX¯= 1017.82 in4

c.

Expert Solution
Check Mark
To determine

Moment of inertia with respect to X-X centroidal axes.

Answer to Problem 8.1P

  IX¯=7190000 cm4 .

Explanation of Solution

Given:

Applied Statics and Strength of Materials (6th Edition), Chapter 8, Problem 8.1P , additional homework tip  3

Concept Used: Since the shape is symmetrical about X and Y axis so the centroidal axis lies in the center of the cross-section. The moment of inertia of the ring will be equal to the moment of inertia of the outer square minus the moment of inertia on the inner rectangle.

Calculation: Moment of inertia of the Outer square is

  IX¯1= bh312IX¯1= 100×100312mm4IX¯1=8333333.33 mm4.

Moment of inertia of the Inner rectangle is

  IX¯2= bh312IX¯2= 40×70312mm4IX¯2=1143333.33 mm4.

Total moment of inertia of the cross-section is

  IX¯ =  I X ¯1-I X ¯2IX¯ =8333333.33 mm4-1143333.33 mm4.IX¯ =7190000 mm4.

Conclusion: Moment of inertia with respect to X-X centroidal axes IX¯=7190000 mm4 .

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Chapter 8 Solutions

Applied Statics and Strength of Materials (6th Edition)

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