Vector Mechanics for Engineers: Statics
Vector Mechanics for Engineers: Statics
12th Edition
ISBN: 9781259977268
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek
Publisher: McGraw-Hill Education
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Chapter 9.6, Problem 9.183P

9.180 through 9.184 For the component described in the problem indicated, determine (a) the principal mass moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.

*9.183 Prob. 9.168

(a)

Expert Solution
Check Mark
To determine

Find the principal mass moment of inertia of the cylinder at the origin O.

Answer to Problem 9.183P

The principal moment of inertia are K1=2.26γtga4_, K2=17.27γtga4_, and K3=19.08γtga4_.

Explanation of Solution

Given information:

Refer Problem 9.168.

Show the moment of inertia as follows:

Ix=18.91335γt8a4Iy=7.68953γt8a4Iz=12.00922γt8a4

Ixy=1.33333γt8a4Iyz=7.14159γt8a4Izx=0.66667γt8a4

Calculation:

Show the Equation 9.56 as follows:

{K3(Ix+Iy+Iz)K2+(IxIy+IyIz+IzIxIxy2Iyz2Izx2)K(IxIyIzIxIyz2IyIzx2IzIxy22IxyIyzIzx)}=0

Substitute 18.91335γt8a4 for Ix, 7.68953γt8a4 for Iy, 12.00922γt8a4 for Iz, 1.33333γt8a4 for Ixy, 7.14159γt8a4 for Iyz, and 0.66667γt8a4 for Izx.

[K3(18.91335γt8a4+7.68953γt8a4+12.00922γt8a4)K2+(18.91335γt8a4×7.68953γt8a4+7.68953γt8a4×12.00922γt8a4+12.00922γt8a4×18.91335γt8a4(1.33333γt8a4)2(7.14159γt8a4)2(0.66667γt8a4)2)K(18.91335γt8a4×7.68953γt8a4×12.00922γt8a418.91335γt8a4×(7.14159γt8a4)27.68953γt8a4×(0.66667γt8a4)212.00922γt8a4×(1.33333γt8a4)22(1.33333γt8a4)(7.14159γt8a4)×(0.66667γt8a4))]=0

Consider the value of K=Kγt8a4.

K338.1210K2+411.69009K744.47027=0

Solve the above Equation and get the value of K1=2.25890, K2=17.27274, and K3=19.08046.

The principal moment of inertia are K1=2.26γtga4, K2=17.27γtga4, and K3=19.08γtga4.

Thus, The principal moment of inertia are K1=2.26γtga4_, K2=17.27γtga4_, and K3=19.08γtga4_.

(b)

Expert Solution
Check Mark
To determine

Find the angles made by the principal axis of inertia at O with the coordinate axis.

Sketch the body and show the orientation of the principal axis of inertia relative to x, y, and z axis.

Answer to Problem 9.183P

The angles made by the principal axis of inertia at O with the coordinate axis is,

(θx)1=85°, (θy)1=36.8°, (θz)1=53.7°

(θx)2=81.7°, (θy)2=54.7°, (θz)1=143.4°

(θx)3=9.70°, (θy)3=99°, (θz)3=86.3°

Explanation of Solution

Given information:

Consider the direction cosines of each principal axis are denoted by λx,λy,λz.

Calculation:

Refer Part (a).

Consider K1.

Show the Equation 9.54 as follows:

(IxK1)(λx)1Ixy(λy)1Izx(λz)1=0Ixy(λx)1Iyz(λz)1+(IyK1)(λy)1=0} (1)

Substitute 18.91335γt8a4 for Ix, 7.68953γt8a4 for Iy, 1.33333γt8a4 for Ixy, 7.14159γt8a4 for Iyz, and 0.66667γt8a4 for Izx and 2.26γtga4 for K1.

(18.91335γt8a42.26γtga4)(λx)11.33333γt8a4(λy)10.66667γt8a4(λz)1=01.33333γt8a4(λx)17.14159γt8a4(λz)1+(7.68953γt8a42.26γtga4)(λy)1=0} (2)

Solve Equation (2).

Get the value of (λz)1=6.74653(λx)1,(λy)1=9.11761(λx)1.

Show the Equation 9.57 as follows:

(λx)12+(λy)12+(λz)12=1(λx)12+[9.11761(λx)1]2+[6.74653(λx)1]2=1

Solve above Equation and get the value of (λx)1=0.087825,(λy)1=0.80075,(λz)1=0.59251.

Show the direction cosines (θx)1,(θy)1,(θz)1 using the relation:

cos(θx)1=(λx)1=0.087825(θx)1=cos1(0.087825)=85°

cos(θz)1=(λz)1=0.59251(θz)1=cos1(0.59251)=53.7°

cos(θy)1=(λy)1=0.80075(θy)1=cos1(0.80075)=36.8°

Consider K2.

Show the Equation 9.54 as follows:

(IxK2)(λx)2Ixy(λy)2Izx(λz)2=0Ixy(λx)2Iyz(λz)2+(IyK2)(λy)2=0} (3)

Substitute 18.91335γt8a4 for Ix, 7.68953γt8a4 for Iy, 1.33333γt8a4 for Ixy, 7.14159γt8a4 for Iyz, and 0.66667γt8a4 for Izx and 17.27γtga4 for K2.

(18.91335γt8a417.27γtga4)(λx)21.33333γt8a4(λy)20.66667γt8a4(λz)2=01.33333γt8a4(λx)27.14159γt8a4(λz)2+(7.68953γt8a417.27γtga4)(λy)2=0} (4)

Solve Equation (4).

Get the value of (λz)2=5.58515(λx)2,(λy)2=4.02304(λx)2.

Show the Equation 9.57 as follows:

(λx)22+(λy)22+(λz)22=1(λx)22+[4.02304(λx)2]2+[5.58515(λx)2]2=1

Solve above Equation and get the value of (λx)2=0.14377,(λy)2=0.0.57839,(λz)2=0.80298.

Show the direction cosines (θx)2,(θy)2,(θz)2 using the relation:

cos(θx)2=(λx)2=0.14377(θx)2=cos1(0.14377)=81.7°

cos(θz)2=(λz)2=0.80298(θz)2=cos1(0.80298)=143.4°

cos(θy)2=(λy)2=0.0.57839(θy)2=cos1(0.0.57839)=54.7°

Consider K3.

Show the Equation 9.54 as follows:

(IxK3)(λx)3Ixy(λy)3Izx(λz)3=0Ixy(λx)3Iyz(λz)3+(IyK3)(λy)3=0} (5)

Substitute 18.91335γt8a4 for Ix, 7.68953γt8a4 for Iy, 1.33333γt8a4 for Ixy, 7.14159γt8a4 for Iyz, and 0.66667γt8a4 for Izx and 19.08γtga4 for K3.

(18.91335γt8a419.08γtga4)(λx)21.33333γt8a4(λy)20.66667γt8a4(λz)2=01.33333γt8a4(λx)27.14159γt8a4(λz)2+(7.68953γt8a419.08γtga4)(λy)2=0} (6)

Solve Equation (6).

Get the value of (λz)3=0.06522(λx)3,(λy)3=0.15794(λx)3.

Show the Equation 9.57 as follows:

(λx)32+(λy)32+(λz)32=1(λx)32+[0.15794(λx)3]2+[0.06522(λx)3]2=1

Solve above Equation and get the value of (λx)3=0.98571,(λy)3=0.15568,(λz)3=0.06429.

Show the direction cosines (θx)3,(θy)3,(θz)3 using the relation:

cos(θx)3=(λx)3=0.98571(θx)3=cos1(0.98571)=9.7°

cos(θz)3=(λz)3=0.06429(θz)3=cos1(0.06429)=86.3°

cos(θy)3=(λy)3=0.15568(θy)3=cos1(0.15568)=86.3°

The angles made by the principal axis of inertia at O with the coordinate axis is,

(θx)1=85°, (θy)1=36.8°, (θz)1=53.7°

(θx)2=81.7°, (θy)2=54.7°, (θz)1=143.4°

(θx)3=9.70°, (θy)3=99°, (θz)3=86.3°

Sketch the body and show the orientation of the principal axis of inertia relative to x, y, and z axis as shown in Figure 1.

Vector Mechanics for Engineers: Statics, Chapter 9.6, Problem 9.183P

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Chapter 9 Solutions

Vector Mechanics for Engineers: Statics

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