Vector Mechanics For Engineers
Vector Mechanics For Engineers
12th Edition
ISBN: 9781259977305
Author: BEER, Ferdinand P. (ferdinand Pierre), Johnston, E. Russell (elwood Russell), Cornwell, Phillip J., SELF, Brian P.
Publisher: Mcgraw-hill Education,
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Question
Chapter B, Problem B.37P
To determine

The mass moment of inertia of the wire with respect to x ,y and z axis.

Expert Solution & Answer
Check Mark

Answer to Problem B.37P

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

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

The mass moment of inertia of the wire with respect to z axis is 30.5983×103lbfts2

Explanation of Solution

Given information:

The diameter of the formed steel wire is 18in and the specific weight of the steel is 490lb/ft3.

The below figure represent the schematic diagram of the wire.

Vector Mechanics For Engineers, Chapter B, Problem B.37P

Figure-(1)

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)

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

Thus, the mass moment of inertia of the wire with respect to z axis is 30.5983×103lbfts2

Conclusion:

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

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

The mass moment of inertia of the wire with respect to z axis is 30.5983×103lbfts2

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Chapter B Solutions

Vector Mechanics For Engineers

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