Vector Mechanics For Engineers
Vector Mechanics For Engineers
12th Edition
ISBN: 9781259977237
Author: BEER
Publisher: MCG
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Chapter B, Problem B.61P
To determine

The mass moment of inertia of the machine component with respect to the axis through the origin characterized by the unit vector (3i6j+2k)/7.

Expert Solution & Answer
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Answer to Problem B.61P

The mass moment of inertia of the machine component with respect to the axis through the origin by the unit vector is 42.75×103lbfts2.

Explanation of Solution

Given information:

The diameter of the formed steel wire is 18in, unit vector is (3i6j+2k)/7 and the specific weight of the steel is 490lb/ft3.

The below figure represents the schematic diagram of the wire.

Vector Mechanics For Engineers, Chapter B, Problem B.61P

Figure-(1)

Concept used:

Expression of mass of the steel wire.

m=ρSTV ........ (I)

Here, density of the steel wire is ρST and the volume of the steel wire is V.

Expression of density of the steel wire.

ρST=γSTg ........ (II)

Here, the specific weight of the steel wire is γST and the acceleration due to gravity is g.

Substitute γSTg for ρST in Equation (I).

m=(γSTg)V ........ (III)

For section (1).

Substitute m1 for m and V1 for V in Equation (III).

m1=(γSTg)V1 ........ (IV)

Expression of volume of the steel wire after formation for section (1).

V1=(π4d12)(πd) ........ (V)

Here, the diameter of the formed wire is d1 and the diameter of the wire is d.

For section (2).

Substitute m2 for m and V2 for V in Equation (III).

m2=(γSTg)V2 ........ (VI)

Expression of volume of the steel wire after formation for section (2).

V2=(π4d12)(πd) ........ (VII)

For section (3).

Substitute m3 for m and V3 for V in Equation (III).

m3=(γSTg)V3 ........ (VIII)

Expression of volume of the steel wire after formation for section (3).

V3=(π4d12)(d) ........ (IX)

For section (4).

Substitute m4 for m and V4 for V in Equation (III).

m4=(γSTg)V4 ........ (X)

Expression of volume of the steel wire after formation for section (4).

V4=(π4d12)(d) ........ (XI)

Expression of Moment of inertia about x axis of the wire.

Ix=(Ix)1+(Ix)2+(Ix)3+(Ix)4 ........ (XII)

Here, the moment of inertia of section (1) about the x axis is (Ix)1, the moment of inertia of section (2) about the x axis is (Ix)2, the moment of inertia of section (3) about the x axis is (Ix)3 and the moment of inertia of section (4) about the x axis is (Ix)4.

Expression of moment of inertia of section (1) about the x axis.

(Ix)1=12m1d2 ........ (XIII)

Expression of moment of inertia of section (2) about the x axis.

(Ix)2=12m2d2+m2d2 ........ (XIV)

Expression of moment of inertia of section (3) about the x axis.

(Ix)3=112m3d2+m3{(d2)2+d2} ........ (XV)

Expression of moment of inertia of section (4) about the x axis.

(Ix)4=112m4d2+m4{(d2)2+d2} ........ (XVI)

Expression of Moment of inertia about y axis of the wire.

Iy=(Iy)1+(Iy)2+(Iy)3+(Iy)4 ........ (XVII)

Here, the moment of inertia of section (1) about the y axis is (Iy)1, the moment of inertia of section (2) about the y axis is (Iy)2, the moment of inertia of section (3) about the y axis is (Iy)3 and the moment of inertia of section (4) about the y axis is (Iy)4.

Expression of moment of inertia of section (1) about the y axis.

(Iy)1=m1d2 ........ (XVIII)

Expression of moment of inertia of section (2) about the y axis.

(Iy)2=m2d2 ........ (XIX)

Expression of moment of inertia of section (3) about the y axis.

(Iy)3=m3d2 ........ (XX)

Expression of moment of inertia of section (4) about the y axis.

(Iy)4=m4d2 ........ (XXI)

Expression of Moment of inertia about z axis of the wire.

Iz=(Iz)1+(Iz)2+(Iz)3+(Iz)4 ........ (XXII)

Here, the moment of inertia of section (1) about the z axis is (Iz)1, the moment of inertia of section (2) about the z axis is (Iz)2, the moment of inertia of section (3) about the z axis is (Iz)3 and the moment of inertia of section (4) about the z axis is (Iz)4.

Expression of moment of inertia of section (1) about the z axis.

(Iz)1=12m1d2 ........ (XXIII)

Expression of moment of inertia of section (2) about the z axis.

(Iz)2=12m2d2+m2d2 ........ (XXIV)

Expression of moment of inertia of section (3) about the z axis.

(Iz)3=13m3d2 ........ (XXV)

Expression of moment of inertia of section (4) about the z axis.

(Iz)4=13m4d2 ........ (XXVI)

Mass products of inertia Ixy is expressed as follows:

Ixy=(Ixy+mxy) …… (XXVII)

Here. Mass products of inertia is Ixy.

Mass products of inertia Iyz is expressed as follows:

Iyz=(Iyz) …… (XXVIII)

Here. Mass products of inertia is Iyz.

Mass products of inertia Izx is expressed as follows:

Izx=(Izx) ……(XXIX)

Here. Mass products of inertia is Izx.

Calculation:

Substitute 18in for d1 and 18in for d in Equation (V).

V1=(π4(18in)2)(π(18in))=(0.01227in2)(π(18in))=(0.694in3)(1ft12in)(1ft12in)(1ft12in)=4.016×104ft3

Substitute 4.016×104ft3 for V2, 490lb/ft3 for γST and 32.2ft/s2 for g in Equation (IV).

m1=(490lb/ft332.2ft/s2)(4.016×104ft3)=(15.217lbs2/ft4)(4.016×104ft3)=6.1112×103lbs2/ft

Substitute 18in for d1 and 18in for d in Equation (VII).

V2=(π4(18in)2)(π(18in))=(0.01227in2)(π(18in))=(0.694in3)(1ft12in)(1ft12in)(1ft12in)=4.016×104ft3

Substitute 4.016×104ft3 for V2, 490lb/ft3 for γST and 32.2ft/s2 for g in Equation (VI).

m2=(490lb/ft332.2ft/s2)(4.016×104ft3)=(15.217lbs2/ft4)(4.016×104ft3)=6.1112×103lbs2/ft

Substitute 18in for d1 and 18in for d in Equation (IX).

V3=(π4(18in)2)(18in)=(0.01227in2)(18in)=(0.22086in3)(1ft12in)(1ft12in)(1ft12in)=1.2783×104ft3

Substitute 1.2783×104ft3 for V3, 490lb/ft3 for γST and 32.2ft/s2 for g in Equation (VIII).

m3=(490lb/ft332.2ft/s2)(1.2783×104ft3)=(15.217lbs2/ft4)(1.2783×104ft3)=1.945×103lbs2/ft

Substitute 18in for d1 and 18in for d in Equation (XI).

V4=(π4(18in)2)(18in)=(0.01227in2)(18in)=(0.22086in3)(1ft12in)(1ft12in)(1ft12in)=1.2783×104ft3

Substitute 1.2783×104ft3 for V4, 490lb/ft3 for γST and 32.2ft/s2 for g in Equation (X).

m4=(490lb/ft332.2ft/s2)(1.2783×104ft3)=(15.217lbs2/ft4)(1.2783×104ft3)=1.945×103lbs2/ft

Substitute 6.1112×103lbs2/ft for m1 and 18in for d in Equation (XIII).

(Ix)1=12(6.1112×103lbs2/ft)(18in)2=12(6.1112×103lbs2/ft){(18in)(1ft12in)}2=12(6.1112×103lbs2/ft)(2.25ft2)=6.8751×103lbfts2

Substitute 6.1112×103lbs2/ft for m2 and 18in for d in Equation (XIV).

(Ix)2=12(6.1112×103lbs2/ft)(18in)2+(6.1112×103lbs2/ft)(18in)2=[12(6.1112×103lbs2/ft){(18in)(1ft12in)}2+(6.1112×103lbs2/ft){(18in)(1ft12in)}2]=(6.8751×103fts2)+(13.7502×103fts2)=20.6252×103lbfts2

Substitute 1.945×103lbs2/ft for m3 and 18in for d in Equation (XV).

(Ix)3=[112(1.945×103lbs2/ft)(18in)2+(1.945×103lbs2/ft){(18in2)2+(18in)2}]=[112(1.945×103lbs2/ft){(18in)(1ft12in)}2+(1.945×103lbs2/ft)[{(18in2)(1ft12in)}2+{(18in)(1ft12in)}2]]=(0.3647×103lbfts2)+(5.4712×103lbfts2)=5.8359×103lbfts2

Substitute 1.945×103lbs2/ft for m4 and 18in for d in Equation (XVI).

(Ix)4=[112(1.945×103lbs2/ft)(18in)2+(1.945×103lbs2/ft){(18in2)2+(18in)2}]=[112(1.945×103lbs2/ft){(18in)(1ft12in)}2+(1.945×103lbs2/ft)[{(18in2)(1ft12in)}2+{(18in)(1ft12in)}2]]=(0.3647×103lbfts2)+(5.4712×103lbfts2)=5.8359×103lbfts2

Substitute 6.8751×103lbfts2 for (Ix)1, 20.6252×103lbfts2 for (Ix)2, 5.8359×103lbfts2 for (Ix)3 and 5.8359×103lbfts2 for (Ix)4 in Equation (XII).

Ix=[(6.8751×103lbfts2)+(20.6252×103lbfts2)+(5.8359×103lbfts2)+(5.8359×103lbfts2)]=(27.5003×103lbfts2)+(11.6718×103lbfts2)=39.1721×103lbfts2

Thus, the mass moment of inertia of the wire with respect to x axis is 39.1721×103lbfts2

Substitute 6.1112×103lbs2/ft for m1 and 18in for d in Equation (XVIII).

(Iy)1=(6.1112×103lbs2/ft)(18in)2=(6.1112×103lbs2/ft){(18in)(1ft12in)}2=(6.1112×103lbs2/ft)(2.25ft2)=13.7502×103lbfts2

Substitute 6.1112×103lbs2/ft for m2 and 18in for d in Equation (XIX).

(Iy)2=(6.1112×103lbs2/ft)(18in)2=(6.1112×103lbs2/ft){(18in)(1ft12in)}2=(6.1112×103lbs2/ft)(2.25ft2)=13.7502×103lbfts2

Substitute 1.945×103lbs2/ft for m3 and 18in for d in Equation (XX).

(Iy)3=(1.945×103lbs2/ft)(18in)2=(1.945×103lbs2/ft){(18in)(1ft12in)}2=4.3769×103lbfts2

Substitute 1.945×103lbs2/ft for m4 and 18in for d in Equation (XXI).

(Iy)4=(1.945×103lbs2/ft)(18in)2=(1.945×103lbs2/ft){(18in)(1ft12in)}2=4.3769×103lbfts2

Substitute 13.7502×103lbfts2 for (Iy)1, 13.7502×103lbfts2 for (Iy)2, 4.3769×103lbfts2 for (Ix)3 and 4.3769×103lbfts2 for (Ix)4 in Equation (XVII).

Iy=[(13.7502×103lbfts2)+(13.7502×103lbfts2)+(4.3769×103lbfts2)+(4.3769×103lbfts2)]=(27.5004×103lbfts2)+(8.7538×103lbfts2)=36.2542×103lbfts2

Thus, the mass moment of inertia of the wire with respect to y axis is 36.2542×103lbfts2.

Substitute 6.1112×103lbs2/ft for m1 and 18in for d in Equation (XXIII).

(Iz)1=12(6.1112×103lbs2/ft)(18in)2=12(6.1112×103lbs2/ft){(18in)(1ft12in)}2=12(6.1112×103lbs2/ft)(2.25ft2)=6.8751×103lbfts2

Substitute 6.1112×103lbs2/ft for m2 and 18in for d in Equation (XXIV).

(Iz)2=12(6.1112×103lbs2/ft)(18in)2+(6.1112×103lbs2/ft)(18in)2=[12(6.1112×103lbs2/ft){(18in)(1ft12in)}2+(6.1112×103lbs2/ft){(18in)(1ft12in)}2]=(6.8751×103fts2)+(13.7502×103fts2)=20.6252×103lbfts2

Substitute 1.945×103lbs2/ft for m3 and 18in for d in Equation (XXV).

(Iz)3=13(1.945×103lbs2/ft)(18in)2=13(1.945×103lbs2/ft){(18in)(1ft12in)}2=1.4590×103lbfts2

Substitute 1.945×103lbs2/ft for m4 and 18in for d in Equation (XXVI).

(Iz)4=13(1.945×103lbs2/ft)(18in)2=13(1.945×103lbs2/ft){(18in)(1ft12in)}2=1.4590×103lbfts2

Substitute 6.8751×103lbfts2 for (Iz)1, 20.6252×103lbfts2 for (Iz)2, 1.4590×103lbfts2 for (Iz)3 and 1.4590×103lbfts2 for (Iz)4 in Equation (XXII).

Iz=[(6.8751×103lbfts2)+(20.6252×103lbfts2)+(1.4590×103lbfts2)+(1.4590×103lbfts2)]=(27.5003×103lbfts2)+(3.098×103lbfts2)=30.5983×103lbfts2

Product of inertia (Ixy)1,(Ixy)2,(Ixy)3,(Ixy)4,(Iyz)1,(Iyz)2,(Izx)3,(Izx)4,y1,y2,x3andx4 are zero due to symmetry.

Substitute 0 for all product of inertia (Ixy) in equation (XXVII).

Ixy=(0+mxy)Ixy=m2x2y2Ixy=(6.1112×103)×(2×18π×12)(1812)Ixy=8.7536×103lbfts2

Substitute (m3y3z3)+(m4y4z4) for product of inertia (Iyz) in equation (XXVIII).

Iyz=(m3y3z3)+(m4y4z4)Iyz=((1.945×103)×y3z3)+((1.945×103)y3(z3))(Y3=Y4z4=z3)Iyz=0

Substitute 0 for product of inertia (Izx) in equation (XXIX).

Izx=0

Mass moment of inertia of the machine component with respect to the axis through the origin by the unit vector is calculated as follows:

I=Ixλx2+Iyλy2+Izλz22Ixyλxλy2Iyzλzλy2IzxλzλxI=[(39.1721×103)(37)2+(36.2542×103)(67)2+(30.5983×103)(27)22×(8.7536×103)(37)(67)00]×103I=(7.1948+26.6357+2.4978+6.4312)×103I=42.75×103lbfts2

Conclusion:

The mass moment of inertia of the machine component with respect to the axis through the origin by the unit vector is 42.75×103lbfts2.

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Chapter B Solutions

Vector Mechanics For Engineers

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