VECTOR MECHANICS FOR ENGINEERS W/CON >B
VECTOR MECHANICS FOR ENGINEERS W/CON >B
12th Edition
ISBN: 9781260804638
Author: BEER
Publisher: MCG
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Chapter 9.6, Problem 9.181P

(a)

To determine

Find the principal mass moment of inertia at the origin.

(a)

Expert Solution
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Answer to Problem 9.181P

The principal mass moment of inertia at the origin is K1=414×103kgm2,K2=29.8×103kgm2,andK3=32.3×103kgm2_.

Explanation of Solution

Calculation:

Refer to problem 9.145 and 9.149.

Ix=26.4325×103kgm2Iy=31.1726×103kgm2Iz=8.5773×103kgm2

Ixy=2.5002×103kgm2Iyz=4.0627×103kgm2Izx=8.8062×103kgm2

Substitute the values of 26.4325×103kgm2 for Ix, 31.1726×103kgm2 for Iy, 8.5773×103kgm2 for Iz, 2.5002×103kgm2 for Ixy, 4.0627×103kgm2 for Iyz and 8.8062×103kgm2 for Izx into Equation 9.56.

{K3[(26.4325+31.1726+8.5773)103]K2+[26.4325(31.1726)+(31.1726)(8.5773)+(8.5773)26.4325](2.5002)2(4.0627)2(8.8062)2(106)K[(26.4325)(31.1726)(8.5773)(26.4325)(4.0627)231.1726(8.8062)2(8.5773)(2.5002)22(2.5002)(4.0627)(8.8062)109]}=0K3(66.1824×103)K2+(1217.76×106)K(3981.23×109)=0

Solve the above Equation.

K1=4.1443×103kgm2K2=29.7840×103kgm2K3=32.2541×103kgm2

Thus, the principal mass moment of inertia at the origin is K1=414×103kgm2,K2=29.8×103kgm2,andK3=32.3×103kgm2_.

(b)

To determine

Find the principal axis of inertia at the origin.

(b)

Expert Solution
Check Mark

Answer to Problem 9.181P

The principal axis of inertia at the origin for K1 is (θx)1=67.8°,(θy)1=80.1°,and(θz)1=24.6°_.

The principal axis of inertia at the origin for K2 is (θx)2=31.6°,(θy)2=71.4°,and(θz)2=114.5°_.

The principal axis of inertia at the origin for K3 is (θx)3=111.2°,(θy)3=21.2°,and(θz)3=91.5°_.

Explanation of Solution

Calculation:

Use Equation 9.54 and 9.57 to find the direction cosines λx,λy,λz of each principal axis.

For K1:

Use Equation 9.54a and 9.54b.

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

Substitute 4.1443×103kgm2 for K1, 29.7840×103kgm2 for K2, 32.2541×103kgm2 for K3, 26.4325×103kgm2 for Ix, 31.1726×103kgm2 for Iy, 2.5002×103kgm2 for Ixy, 4.0627×103kgm2 for Iyz, and 8.8062×103kgm2 for Izx.

[(26.43254.1443)(103)](λx)1(2.5002×103)(λy)1(8.8062×103)(λz)1=0(2.5002×103)(λx)1+[31.17264.1443](103)[(λy)1(4.0627×103)](λz)1=0

Simplifying,

8.9146(λx)1(λy)13.5222(λz)1=0 (1)

0.0925(λx)1(λy)10.1503(λz)1=0 (2)

Solving the Equations (1) and (2) for (λz)1.

(λz)1=2.4022(λx)1

Substitute into 2.4022(λx)1 for (λz)1 in Equation (1).

(λy)1=[8.91463.5222(2.4022)(λx)1]=0.45357(λx)1

Substitute into Equation 9.57:

(λx)12+[0.45357(λx)1]2+[2.4022(λx)1]2=1(λx)1=0.37861(λz)1=0.90950(λx)1=0.17173

(θx)1=67.8°(θy)1=80.1°(θz)1=24.6°

Thus, the principal axis of inertia at the origin for K1 is (θx)1=67.8°,(θy)1=80.1°,and(θz)1=24.6°_.

For K2:

Use Equation 9.54a and 9.54b.

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

Substitute 4.1443×103kgm2 for K1, 29.7840×103kgm2 for K2, 32.2541×103kgm2 for K3, 26.4325×103kgm2 for Ix, 31.1726×103kgm2 for Iy, 2.5002×103kgm2 for Ixy, 4.0627×103kgm2 for Iyz, and 8.8062×103kgm2 for Izx.

[26.432529.7840×103][(λx)2(2.5002×103)(λy)2(8.8062×103)(λz)2]=0(2.5002×103)(λx)2+[(31.172629.7840)(103)](λy)2(4.0627×103)(λz)2=0

Simplifying

1.3405(λx)2(λy)23.5222(λz)2=0 (3)

1.8005(λx)2+(λy)22.9258(λz)2=0 (4)

Solving Equations (3) and (4) for (λz)2:

(λz)2=0.48713(λx)2

Substitute 0.48713(λx)2 for (λz)2 in Equation (3).

(λy)2=[1.34053.5222(0.48713)](λx)2=0.37527(λx)2

Substitute into Equation 9.57:

(λx)22+[0.37527(λx)2]2+[0.48713(λx)2]2=1(λx)2=0.85184(λz)2=0.41496(λy)2=0.31967

(θx)2=31.6°(θy)2=71.4°(θz)2=114.5°

Thus, the principal axis of inertia at the origin for K2 is (θx)2=31.6°,(θy)2=71.4°,and(θz)2=114.5°_.

For K3:

Use Equation 9.54a and 9.54b.

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

Substitute 4.1443×103kgm2 for K1, 29.7840×103kgm2 for K2, 32.2541×103kgm2 for K3, 26.4325×103kgm2 for Ix, 31.1726×103kgm2 for Iy, 2.5002×103kgm2 for Ixy, 4.0627×103kgm2 for Iyz, and 8.8062×103kgm2 for Izx.

[(26.432532.2541)103](λx)3(2.5002×103)(λy)3(8.8062×103)(λz)3=0[(2.5002×103)(λx)3+[31.172632.2541(103)](λ3)+[31.172632.2541(103)][(λy)3(4.0627×103)(λz)3]=0]

Simplifying

2.3285(λx)3(λy)33.5222(λz)3=0 (5)

2.3118(λx)3(λy)33.7565(λz)3=0 (6)

Solving Equations (5) and (6) for (λz)3.

(λz)3=0.071276(λx)3

Substitute 0.071276(λx)3 for (λz)3 in Equation (5).

(λy)3=[2.32853.5222(0.071276)(λx)3]=2.5795(λx)3

Substitute into Equation 9.57:

(λx)32+[2.5795(λx)3]2+[0.07127(λx)3]2=1

(λx)3=0.36134(λz)3=0.025755(λy)3=0.93208

(θx)3=68.8°(θy)3=158.8°(θz)3=88.5°

Find the direction cosines corresponding to the labelled axis, take the negative root of (λx)3.

(λx)3=0.36134

(θx)3=111.2°(θy)3=21.2°(θz)3=91.5°

Thus, the principal axis of inertia at the origin is (θx)3=111.2°,(θy)3=21.2°,and(θz)3=91.5°_.

Sketch the orientation of the principal axis to the x,y,z axis as shown in Figure 1.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 9.6, Problem 9.181P

Refer to Figure 1.

Principal axis 3 has been labelled so that the principle axes form a right handed set.

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Chapter 9 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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