Vector Mechanics For Engineers
Vector Mechanics For Engineers
12th Edition
ISBN: 9781259977237
Author: BEER
Publisher: MCG
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Textbook Question
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Chapter B, Problem B.56P

Determine the mass moment ofinertia of the steel fixture of Probs. B.35 and B.39 with respect to the axis through the origin that forms equal angles with the x, y, and z axes.

Expert Solution & Answer
Check Mark
To determine

The mass moment of the inertia of the steel fixture passes through the origin.

Answer to Problem B.56P

The mass moment of the inertia of the steel fixture passes through the origin is 3.46×106kgm2.

Explanation of Solution

Given information:

Given information:

The density of the steel is 7850kg/m3.

The following figure represents the given system.

Vector Mechanics For Engineers, Chapter B, Problem B.56P

Figure-(1)

Write the expression for the mass of component 1.

m1=ρV1   ....(I)

Here, the mass of component 1 is m1, the density of component ρ and the volume of component 1 is V1.

Write the expression for the volume of the component 1.

V1=abc   ....(II)

Here, the sides of the component is aorbandc.

Write the expression for the mass moment of inertia for component 1.

(Ix)1=(I¯x)+m1d2   ....(III)

Here, the mass moment of inertia for component 1 is (Ix)1, the mass moment of inertia with respect to centroidal axis is (I¯x) and the distance between the centroidal axis and parallel axis is d.

Write the expression for the mass moment of inertia with respect to centroidal axis.

I¯x=m1(a2+c2)12   ....(IV)

Write the expression for the centroidal axis from the reference axis.

d=(a2+c2)   ....(V)

Write the expression for the mass moment of inertia for component 2.

(Ix)2=(I¯x2)+m2d2   ....(VI)

Here, the mass moment of inertia for component 2 is (Ix)2, the mass moment of inertia with respect to centroidal axis is (I¯x2) and the distance between the centroidal axis and parallel axis is d.

Write the expression for the mass moment of inertia with respect to centroidal axis.

I¯x2=m2(e2+d2)12   ....(VII)

Write the expression for the centroidal axis from the reference axis.

d2=(ce2)2+(c+d2)2   ....(VIII)

Write the expression for the mass moment of inertia with respect to centroidal axis.

(I¯x)3=m3(14169π2)f2+g12   ....(IX)

Here, the mass moment of inertia for component 3 is (Ix)3, the mass moment of inertia with respect to centroidal axis is (I¯x) and the distance between the centroidal axis and parallel axis is d.

Write the expression for the mass moment of inertia for component 3.

(Ix)3=(I¯x)3+m3d2   ....(X)

Write the expression for the mass of component 2.

m2=ρV2   ....(XI)

Here, the mass of component 2 is m2, the density of component ρ and the volume of component 2 is V2.

Write the expression for volume of the component 2.

V2=bde …….. (XII)

Here, the sides of the component is dorbande.

Write the expression for the mass of component 3.

m3=ρV3   ....(XIII)

Here, the mass of component 3 is m3, the density of component ρ and the volume of component 3 is V3.

Write the expression for volume of the component 3.

V3=πf2g4 …….. (XIV)

Here, the diameter of the circle is f and the width the component 3 is g.

Write the expression for the distance of the component 3.

d2=(c(4f3π))2+(ag2)2 …….. (XV)

Write the expression for the mass moment of inertia with respect to x axis.

(Ix)=(Ix)1(Ix)2(Ix)3   ....(XVI)

Write the expression for the mass moment of inertia for component 1.

(Iy)1=(I¯y)+m1d2   ....(XVII)

Here, the mass moment of inertia for component 1 is (Iy)1, the mass moment of inertia with respect to centroidal axis is (I¯y) and the distance between the centroidal axis and parallel axis is d.

Write the expression for the mass moment of inertia with respect to centroidal axis.

I¯y=m1(a2+b2)12   ....(XVIII)

Write the expression for the centroidal axis from the reference axis.

d=(a2+c2)   ....(XIX)

Write the expression for the mass moment of inertia for component 2.

(Iy)2=(I¯y2)+m2d2   ....(XX)

Here, the mass moment of inertia for component 2 is (Iy)2, the mass moment of inertia with respect to centroidal axis is (I¯y2) and the distance between the centroidal axis and parallel axis is d.

Write the expression for the mass moment of inertia with respect to centroidal axis.

I¯y2=m2(b2+d2)12   ....(XXI)

Write the expression for the centroidal axis from the reference axis.

d2=(b2+c+d2)2   ....(XXII)

Write the expression for the mass moment of inertia with respect to centroidal axis.

(I¯y)3=m3(3f2+g2)12   ....(XXIII)

Here, the mass moment of inertia for component 3 is (Iy)3, the mass moment of inertia with respect to centroidal axis is (I¯y) and the distance between the centroidal axis and parallel axis is d.

Write the expression for the mass moment of inertia for component 3.

(Iy)3=(I¯y)3+m3d2   ....(XXIV)

Write the expression for the distance of the component 3.

d2=(b2)+(ag2) …….. (XXV)

Write the expression for the mass moment of inertia with respect to y axis.

(Iy)=(Iy)1(Iy)2(Iy)3   ....(XXVI)

Write the expression for the mass moment of inertia for component 1.

(Iz)1=(I¯z)+m1d2   ....(XXVII)

Here, the mass moment of inertia for component 1 is (Iz)1, the mass moment of inertia with respect to centroidal axis is (I¯z) and the distance between the centroidal axis and parallel axis is d.

Write the expression for the mass moment of inertia with respect to centroidal axis.

I¯z=m1(c2+b2)12   ....(XXVIII)

Write the expression for the centroidal axis from the reference axis.

d=(b2+c2)   ....(XXIX)

Write the expression for the mass moment of inertia for component 2.

(Iz)2=(I¯z2)+m2d2   ....(XXX)

Here, the mass moment of inertia for component 2 is (Iz)2, the mass moment of inertia with respect to centroidal axis is (I¯z2) and the distance between the centroidal axis and parallel axis is d.

Write the expression for the mass moment of inertia with respect to centroidal axis.

I¯z2=m2(b2+e2)12   ....(XXXI)

Write the expression for the centroidal axis from the reference axis.

d2=(b2+c+e2)2   ....(XXXII)

Write the expression for the mass moment of inertia with respect to centroidal axis.

(I¯z)3=m3(12169π2)f2   ....(XXXIII)

Here, the mass moment of inertia for component 3 is (Iz)3, the mass moment of inertia with respect to centroidal axis is (I¯z) and the distance between the centroidal axis and parallel axis is d.

Write the expression for the mass moment of inertia for component 3.

(Iz)3=(I¯z)3+m3d2   ....(XXXIV)

Write the expression for the distance of the component 3.

d2=(b2)2+(c4f3π)2 …….. (XXXV)

Write the expression for the mass moment of inertia with respect to y axis.

(Iz)=(Iz)1(Iz)2(Iz)3   ....(XXXVI)

Write the expression for the mass moment of inertia for component 1.

(Ixy)1=(I¯xy)1+m1(x¯1)(y¯1)   ....(XXXVII)

Here, the products of the inertia of the body with respect to centroidal axis for component 1 is (I¯xy)1, the coordinates for the centre of gravity of the component 1 is (x¯1)and(y¯1).

Write the expression for the mass moment of inertia for component 1.

(Iyz)1=(I¯yz)1+m1(z¯1)(y¯1)   ....(XXXVIII)

Here, the products of the inertia of the body with respect to centroidal axis for component 1 is (I¯yz)1, the coordinates for the centre of gravity of the component 1 is (z¯1)and(y¯1).

Write the expression for the mass moment of inertia for component 1.

(Izx)1=(I¯zx)1+m1(z¯1)(x¯1)   ....(XXXIX)

Here, the products of the inertia of the body with respect to centroidal axis for component 1 is (I¯zx)1, the coordinates for the centre of gravity of the component 1 is (z¯1)and(x¯1).

Write the expression for the mass moment of inertia for component 2.

(Ixy)2=(I¯xy)2+m2(x¯2)(y¯2)   ....(XL)

Here, the products of the inertia of the body with respect to centroidal axis for component 2 is (I¯xy)2, the coordinates for the centre of gravity of the component 2 is (x¯2)and(y¯2).

Write the expression for the mass moment of inertia for component 2.

(Iyz)2=(I¯yz)2+m2(z¯2)(y¯2)   ....(XLI)

Here, the products of the inertia of the body with respect to centroidal axis for component 2 is (I¯yz)2, the coordinates for the centre of gravity of the component 2 is (z¯2)and(y¯2).

Write the expression for the mass moment of inertia for component 2.

(Izx)2=(I¯zx)2+m2(z¯2)(x¯2)   ....(XLII)

Here, the products of the inertia of the body with respect to centroidal axis for component 2 is (I¯zx)2, the coordinates for the centre of gravity of the component 2 is (z¯2)and(x¯2).

Write the expression for the mass moment of inertia for component 3.

(Ixy)3=(I¯xy)3+m3(x¯3)(y¯3)   ....(XLIII)

Here, the products of the inertia of the body with respect to centroidal axis for component 3 is (I¯xy)3, the coordinates for the centre of gravity of the component 3 is (x¯3)and(y¯3).

Write the expression for the mass moment of inertia for component 3.

(Iyz)3=(I¯yz)3+m3(z¯3)(y¯3)   ....(XLIV)

Here, the products of the inertia of the body with respect to centroidal axis for component 3 is (I¯yz)3, the coordinates for the centre of gravity of the component 3 is (z¯3)and(y¯3).

Write the expression for the mass moment of inertia for component 3.

(Izx)3=(I¯zx)3+m3(z¯3)(x¯3)   ....(XLV)

Here, the products of the inertia of the body with respect to centroidal axis for component 3 is (I¯zx)3, the coordinates for the centre of gravity of the component 3 is (z¯3)and(x¯3).

Write the expression for the mass product of inertia.

Ixy=(Ixy)1(Ixy)2(Ixy)3   ....(XLVI)

Here, the mass product of inertia is Ixy.

Write the expression for the mass product of inertia.

Iyz=(Iyz)1(Iyz)2(Iyz)3   ....(XLVII)

Here, the mass product of inertia is Iyz.

Write the expression for the mass product of inertia.

Izx=(Izx)1(Izx)2(Izx)3   ....(XLVIII)

Here, the mass product of inertia is Izx.

Write the expression for the position vector of OA.

OA=40i35j

Write the expression for the magnitude of the position vector OA.

dOA=2825

Write the expression for the unit vector.

λOA=OAdOA   ....(XLIX)

Write the expression for the mass moment of the inertia of the steel fixture passes through the origin.

IOA=Ixλx2+Iyλy22Ixyλxλy   ....(L)

Calculation:

Substitute 160mm for a, 80mm for b and 50mm for c in Equation (II).

V1=(160mm)(80mm)(50mm)=640000mm3(1m3109mm3)=6.4×104m3

Substitute 7850kg/m3 for ρ and 6.4×104m3 for V1 in Equation (I).

m1=(6.4×104m3)(7850kg/m3)=5.024kg

Substitute 38mm for e, 80mm for b and 70mm for d in Equation (XII).

V2=(38mm)(80mm)(70mm)=212800mm3(1m3109mm3)=2.28×104m3

Substitute 7850kg/m3 for ρ and 2.28×104m3 for V2 in Equation (XI).

m2=(2.28×104m3)(7850kg/m3)=1.67048kg

Substitute 24mm for f and 40mm for g in Equation (XIV).

V3=π(24mm)2(40mm)2=36191.147mm3(1m3109mm3)=3.6191×105m3

Substitute 7850kg/m3 for ρ and 3.6191×105m3 for V3 in Equation (XIII).

m3=(3.6191×105m3)(7850kg/m3)=0.2841kg

Substitute 5.024kg for m1, 50mm for c and 160mm for c in Equation (IV).

I¯x=5.024kg×((160mm)2+(50mm)2)12=5.024kg×((160mm(1m1000mm))2+(50mm(1m1000mm))2)12=5.024kg×(0.0281m2)12=0.011764kgm2

Substitute 50mm for c and 160mm for a in Equation (V).

d=(160mm2+50mm2)=(80mm+25mm)=105mm(1m1000mm)=0.105m

Substitute 5.024kg for m1, 0.105m for d and 0.011764kgm2 for I¯x in Equation (III).

(Ix)1=0.011764kgm2+(5.024kg)(0.105m)2=0.011764kgm2+0.05538kgm2=0.06715kgm2

Substitute 50mm for c, 38mm for e and 70mm for d in Equation (VIII).

d2=(50mm38mm2)2+(50mm+70mm2)2=(69mm(1m1000mm))2+(85mm(1m1000mm))2=0.008186m2

Substitute 1.67048kg for m2, 38mm for e and 70mm for d in Equation (VII).

I¯x2=1.67048kg×((38mm)2+(70mm)2)12=1.67048kg×((38mm(1m1000mm))2+(70mm(1m1000mm))2)12=0.0008831kgm2

Substitute 0.008186m2 for d2, 0.0008831kgm2 for I¯x2 and 1.67048kg for m2 in Equation (VI).

(Ix)2=0.0008831kgm2+(1.67048kg)(0.008186m2)=0.0008831kgm2+0.0136745kgm2=0.0145576kgm2

Substitute 24mm for g, 40mm for f and 0.2841kg for m3 in Equation (IX).

(I¯x)3=0.2841kg×[(14169π2)(40mm)2+24mm12]=0.2841kg×[(14169π2)(40mm(1m1000mm))2+(24mm(1m1000mm))212]=0.2841kg×[0.0001117m2+0.000048m2]=0.00059035kgm2

Substitute 24mm for g, 40mm for f, 50mm for c and 160mm for a in Equation (XV).

d2=(50mm(4(24mm)3π))2+(160mm40mm2)2=(39.81mm)2+(140mm)2=(39.81mm(1m1000mm))2+(140mm(1m1000mm))2=0.02118m2

Substitute 0.02118m2 for d2, 0.00059035kgm2 for (I¯x)3 and 0.2841kg for m3 in Equation (X).

(Ix)3=0.00059035kgm2+(0.2841kg)(0.02118m2)=0.00059035kgm2+0.006018kgm2=0.006608kgm2

Substitute 0.006608kgm2 for (Ix)3, 0.0145576kgm2 for (Ix)2 and 0.06715kgm2 for (Ix)1 in Equation (XVI).

(Ix)=0.06715kgm20.0145576kgm20.006608kgm2=0.04598kgm2

Substitute 5.024kg for m1, 160mm for a and 50mm for c in Equation (XVII).

I¯y=(5.024kg)((160mm)2+(50mm)2)12=(5.024kg)((160mm(1m1000mm))2+(80mm(1m1000mm))2)12=0.013877kgm2

Substitute 160mm for a and 80mm for b in Equation (XVIII).

d=(160mm2+80mm2)=80mm+40mm=120mm(1m1000mm)=0.120m

Substitute 5.024kg for m1, 0.120m for d and 0.013877kgm2 for I¯y in Equation (XVI).

(Iy)1=0.013877kgm2+(5.024kg)(0.120m)2=0.013877kgm2+(5.024kg)(0.0144m)2=0.013877kgm2+0.0749376kgm2=0.08632kgm2

Substitute 1.6704kg for m2, 80mm for b and 70mm for d in Equation (XX).

I¯y2=1.6704kg((80mm)2+(70mm)2)12=1.6704kg((80mm(1m1000mm))2+(70mm(1m1000mm))2)12=0.00157296kgm2

Substitute 50mm for c, 80mm for b and 70mm for d in Equation (XXI).

d2=(80mm2+50mm+70mm2)2=(40mm+50mm+35mm)2=(125mm(1m1000mm))2=0.015625m2

Substitute 0.015625m2 for d2, 0.00157296kgm2 for I¯y2 and 1.6704kg for m2 in Equation (XIX).

(Iy)2=0.00157296kgm2+(1.6704kg)(0.015625m2)=0.00157296kgm2+0.0261kgm2=0.02767296kgm2

Substitute 24mm for f, 40mm for g and 0.2841kg for m3 in Equation (XXII).

(I¯y)3=0.2841kg×(3(24mm)2+(40mm)2)12=0.2841kg×(3(24mm(1m1000mm))2+(40mm(1m1000mm))2)12=0.00007879kgm2

Substitute 160mm for a, 80mm for b and 40mm for g in Equation (XXIV).

d2=(80mm2)+(160mm40mm2)=40mm+120mm=160mm(1m1000mm)=0.160m

Substitute 0.160m for d2, 0.00007879kgm2 for (I¯y)3 and 0.2841kg for m3 in Equation (XXIII).

(Iy)3=0.00007879kgm2+(0.2841kg)(0.160m)=0.00007879kgm2+0.045456kgm2=0.04553479kgm2

Substitute 0.04553479kgm2 for (Iy)3, 0.02767296kgm2 for (Iy)2 and 0.08632kgm2 for (Iy)1 in Equation (XXVI).

(Iy)=0.08632kgm20.02767296kgm20.04553479kgm2=0.013116kgm2

Substitute 5.024kg for m1, 80mm for b and 50mm for c in Equation (XXVIII).

I¯z=5.024kg×((50mm)2+(80mm)2)12=5.024kg×((50mm(1m1000mm))2+(80mm(1m1000mm))2)12=5.024kg×(0.0089m2)12=0.003726kgm2

Substitute 80mm for b and 50mm for c in Equation (XXIX).

d=(80mm2+50mm2)=40mm+25mm=65mm(1m1000mm)=0.065m

Substitute 5.024kg for m1, 0.065m for d and 0.003726kgm2 for I¯z in Equation (XXVII).

(Iz)1=0.003726kgm2+(5.024kg)(0.065m)2=0.003726kgm2+0.02122kgm2=0.0249524kgm2

Substitute 80mm for b, 1.6704kg for m2 and 38mm for e in Equation (XXXI).

I¯z2=1.6704kg×((80mm)2+(38mm)2)12=1.6704kg×((80mm(1m1000mm))2+(38mm(1m1000mm))2)12=0.0019188kgm2

Substitute 80mm for b, 50mm for c and 38mm for e in Equation (XXXII).

d2=(80mm2+50mm38mm2)2=(40mm+50mm19mm)2=(71mm(1m1000mm))2=0.005041m2

Substitute 0.005041m2 for d2, 1.6704kg for m2 and 0.0019188kgm2 for I¯z2 in Equation (XXX).

(Iz)2=0.0019188kgm2+(1.6704kg)(0.005041m2)=0.0019188kgm2+0.008420kgm2=0.01032kgm2

Substitute 0.2841kg for m3 and 24mm for f in Equation (XXXIII).

(I¯z)3=0.2841kg×(12169π2)(24mm)2=0.2841kg×(0.31987)(24mm(1m1000mm))2=0.00005234kgm2

Substitute 80mm for b, 50mm for c and 24mm for f in Equation (XXXV).

d2=(80mm2)2+(50mm4×24mm3π)=(40mm(1m1000mm))2+(39.8140mm(1m1000mm))2=0.0031848m2

Substitute 0.0031848m2 for d2, 0.00005234kgm2 for (I¯z)3 and 0.2841kg for m3 in Equation (XXXIV).

(Iz)3=0.00005234kgm2+(0.2841kg)(0.0031848m2)=0.00005234kgm2+0.0009048kgm2=0.000957kgm2

Substitute 0.000957kgm2 for (Iz)3, 0.01032kgm2 for (Iz)2 and 0.0249524kgm2 for (Iz)1 in Equation (XXXVI).

(Iz)=0.0249524kgm20.01032kgm20.000957kgm2=0.013675kgm2

Substitute 5.024kg for m1, 40mm for x¯1, 25mm for y¯1 and 0 for (I¯xy)1 in Equation (XXXVII).

(Ixy)1=0+5.024kg(40mm)(25mm)=5024kgmm2(1m2106mm2)=5.024×103kgm2

Substitute 5.024kg for m1, 80mm for z¯1, 25mm for y¯1 and 0 for (I¯yz)1 in Equation (XXXVIII).

(Iyz)1=0+5.024kg(80mm)(25mm)=10048kgmm2(1m2106mm2)=0.010048kgm2

Substitute 5.024kg for m1, 80mm for z¯1, 40mm for x¯1, and 0 for (I¯zx)1 in Equation (XXXIX).

(Izx)1=0+5.024kg(80mm)(40mm)=16076.8kgmm2(1m2106mm2)=0.0160768kgm2

Substitute 1.67048kg for m2, 40mm for x¯2, 31mm for y¯2 and 0 for (I¯xy)2 in Equation (XL).

(Ixy)2=0+1.67048kg(40mm)(31mm)=2071.5648kgmm2(1m2106mm2)=2.071×103kgm2

Substitute 1.67048kg for m2, 85mm for z¯2, 19mm for y¯2 and 0 for (I¯yz)2 in Equation (XLI).

(Iyz)2=0+1.67048kg(85mm)(31mm)=4400.8252kgmm2(1m2106mm2)=4.4×103kgm2

Substitute 1.67048kg for m2, 85mm for z¯2, 40mm for x¯2, and 0 for (I¯zx)1 in Equation (XLII).

(Izx)2=0+1.67048kg(85mm)(40mm)=5679.632kgmm2(1m2106mm2)=5.679×103kgm2

Substitute 0.2841kg for m3, 40mm for x¯3, 39.8mm for y¯3 and 0 for (I¯xy)3 in Equation (XLIII).

(Ixy)3=0+0.2841kgkg(40mm)(39.8mm)=452.288kgmm2(1m2106mm2)=0.452×103kgm2

Substitute 0.2841kg for m3, 140mm for z¯3, 39.8mm for y¯3 and 0 for (I¯yz)3 in Equation (XLIV).

(Iyz)3=0+0.2841kg(140mm)(39.8mm)=1607.02kgmm2(1m2106mm2)=1.607×103kgm2

Substitute 0.2841kg for m3, 140mm for z¯3, 40mm for x¯3, and 0 for (I¯zx)3 in Equation (XLV).

(Izx)3=0+0.2841kg(140mm)(40mm)=1590.96kgmm2(1m2106mm2)=1.591×103kgm2

Substitute 0.452×103kgm2 for (Ixy)3, 2.071×103kgm2 for (Ixy)2 and 5.024×103kgm2 for (Ixy)1 in Equation (XLVI).

Ixy=[5.024×103kgm22.071×103kgm20.452×103kgm2]=5.024×103kgm22.543×103kgm2=0.002504kgm2

Substitute 0.160×103kgm2 for (Iyz)3, 4.4×103kgm2 for (Iyz)2 and 0.010048kgm2 for (Iyz)1 in Equation (XLVII).

Iyz=[0.010048kgm24.4×103kgm21.607×103kgm2]=0.010048kgm20.006kgm2=0.003996kgm2

Substitute 1.591×103kgm2 for (Izx)3, 5.679×103kgm2 for (Izx)2 and 0.0160768kgm2 for (Izx)1 in Equation (XLVIII).

Izx=[0.0160768kgm25.679×103kgm21.591×103kgm2]=0.0160768kgm20.00727kgm2=0.008801kgm2

Substitute 2825 for dOA and 40i35j for OA in Equation (XLIX).

λOA=40i35j2825=40i282535j2825   ....(LI)

From Equation (LI) the unit vector along the different axis.

λx=40mm2825mm=40mm(1m103mm)2825mm(1m103mm)=0.04m2.825m

λy=30mm2825mm=30mm(1m103mm)2825mm(1m103mm)=0.035m2.825m

Substitute 0.04598kgm2 for Ix, 0.04m2.825m for λx, 0.013116kgm2 for Iy, 0.035m2.825m for λy and 0.002504kgm2 for Ixy in Equation (L).

IOA={(0.04598kgm2)(0.04m2.825m)2+(0.013116kgm2)(0.035m2.825m)22(0.002504kgm2)(0.04m2.825m)(0.035m2.825m)}={(0.04598kgm2)(1.6×103m22.825m)+(0.013116kgm2)(1.225×103m22.825m)2(0.002504kgm2)(1.4×103m22.825m)}=0.260×106kgm2+5.68×106kgm22.48×106kgm2=3.46×106kgm2

Conclusion:

The mass moment of the inertia of the steel fixture passes through the origin is 3.46×106kgm2.

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Chapter B Solutions

Vector Mechanics For Engineers

Ch. B - Prob. B.11PCh. B - Prob. B.12PCh. B - Determine by direct integration the mass moment of...Ch. B - Prob. B.14PCh. B - A thin, rectangular plate with a mass m is welded...Ch. B - A thin steel wire is bent into the shape shown....Ch. B - Prob. B.17PCh. B - Prob. B.18PCh. B - Prob. B.19PCh. B - Prob. B.20PCh. B - Prob. B.21PCh. B - Prob. B.22PCh. B - Prob. B.23PCh. B - Prob. B.24PCh. B - Prob. B.25PCh. B - Prob. B.26PCh. B - Prob. B.27PCh. B - Prob. B.28PCh. B - Prob. B.29PCh. B - Prob. B.30PCh. B - Prob. B.31PCh. B - Determine the mass moments of inertia and the...Ch. B - Prob. B.33PCh. B - Prob. B.34PCh. B - Prob. B.35PCh. B - Prob. B.36PCh. B - Prob. B.37PCh. B - Prob. B.38PCh. B - Prob. B.39PCh. B - Prob. B.40PCh. B - Prob. B.41PCh. B - Prob. B.42PCh. B - Prob. B.43PCh. B - Prob. B.44PCh. B - A section of sheet steel 2 mm thick is cut and...Ch. B - Prob. B.46PCh. B - Prob. B.47PCh. B - Prob. B.48PCh. B - Prob. B.49PCh. B - Prob. B.50PCh. B - Prob. B.51PCh. B - Prob. B.52PCh. B - Prob. B.53PCh. B - Prob. B.54PCh. B - Prob. B.55PCh. B - Determine the mass moment ofinertia of the steel...Ch. B - Prob. B.57PCh. B - Prob. B.58PCh. B - Determine the mass moment of inertia of the...Ch. B - Prob. B.60PCh. B - Prob. B.61PCh. B - Prob. B.62PCh. B - Prob. B.63PCh. B - Prob. B.64PCh. B - Prob. B.65PCh. B - Prob. B.66PCh. B - Prob. B.67PCh. B - Prob. B.68PCh. B - Prob. B.69PCh. B - Prob. B.70PCh. B - For the component described in the problem...Ch. B - Prob. B.72PCh. B - For the component described in the problem...Ch. B - Prob. B.74P
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