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 9.5, Problem 9.145P

Determine the mass moment of inertia of the steel fixture shown with respect to (a) the x axis, (b) the y axis, (c) the z axis. (The density of steel is 7850 kg/m3.)

Chapter 9.5, Problem 9.145P, Determine the mass moment of inertia of the steel fixture shown with respect to (a) the x axis, (b)

Fig. P9.145

(a)

Expert Solution
Check Mark
To determine

Find the mass moment of inertia with respect to x axis.

Answer to Problem 9.145P

The mass moment of inertia with respect to x axis is 26.4×103kgm2_.

Explanation of Solution

Given information:

The density (ρST) of steel is 7,850kg/m3.

Calculation:

Sketch the section of steel fixture as shown in Figure 1.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 9.5, Problem 9.145P

Find the mass (m1) of component 1 as shown below:

(m1)=ρV (1)

Here, V is volume of rectangular section 1.

Modify Equation (1).

(m1)=ρ(wlt) (2)

Here, w is the width of the section 1, l is the length of the rectangular section 1, and t is the thickness of the section 1.

Substitute 80mm for w, 50mm for l, and 16mm for t in Equation (2).

(m1)=7,850×(80mm(1m103mm)×50mm(1m103mm)×16mm(1m103mm))=7,850×0.08×0.05×0.16=5.021kg

Find the mass (m2) of component 2 as shown below:

(m2)=ρV (3)

Here, V is volume of rectangular section 2.

Modify Equation (3).

(m2)=ρ(wlt) (4)

Here, w is the width of the section 2, l is the length of the rectangular section 2, and t is the thickness of the section 2.

Substitute 80mm for w, 38mm for l, and 70mm for t in Equation (4).

(m2)=7,850×[80mm(1m103mm)×38mm(1m103mm)×70mm(1m103mm)]=7,850×(0.08×0.038×0.07)=1.67048kg

Find the mass (m3) of component 3 as shown below:

(m3)=ρV=ρ(π2×r2h) (5)

Substitute 24mm for r and 40mm for h in Equation (5).

(m3)=7,850×(π2×24mm(1m103mm)×40mm(1m103mm))=7,850×(π2×0.0242×0.04)=7,850×3.619×105=0.28410kg

Refer to Figure 9.28, “Mass moment of inertia for common geometric shapes” in the text book.

Find the mass moment of inertia with respect to x axis as shown below:

Ix=(Ix)1(Ix)2(Ix)3=[112×m1[b12+h12]+m1[(b2)2+(h2)2]112(m2[b22+h22])+m2[(b12b22)2+((b12+h22)2)]m3[(14169π2)r2+112h32+m3[b14r3π]+(h1h32)2]] (6)

Convert the dimension in mm to m.

b1=50mm(1m103mm)=0.05m

Similarly calculate the remaining values.

Substitute 0.28410kg for m3, 1.67048kg for m2, 5.021kg for m1, 0.05m for b1, 0.16m for h1, 0.038m for b2, 0.07m for h2, 0.024m for r, 0.04m for h3 in Equation (6).

Ix=[112×5.02400×(0.052)+(0.16)2+5.02400[(0.052)2+(0.162)2]112×1.67048[0.0382+0.072]+1.67048[(0.050.0382)2+(0.050.072)2]0.28410[(14169π2)0.0242+112(0.042)]+0.28410[(0.054×0.0243π)2+(0.160.042)2]]=[11.76456+35.29360.8831+13.67450.0493+6.0187]=26.4325×103kgm2

Thus, the mass moment of inertia with respect to x axis is 26.4×103kgm2_.

(b)

Expert Solution
Check Mark
To determine

Find the mass moment of inertia with respect to y axis.

Answer to Problem 9.145P

The mass moment of inertia with respect to y axis is 31.2×103kgm2_.

Explanation of Solution

Calculation:

Find the mass moment of inertia with respect to y axis as shown below:

Iy=(Iy)1(Iy)2(Iy)3=[112×m1[b12+h12]+m1[(b12)2+(h12)2]112(m2[b22+h22])+m2[(b22)2+((b+h22)2)]112m3[3r3+h32]+m3[(b32)2+(h1+h32)2]] (7)

Convert the dimensions in mm to m.

b1=50mm(1m103mm)=0.05m

Substitute 0.28410kg for m3, 1.67048kg for m2, 5.021kg for m1, 0.08m for b1, 0.16m for h1, 0.08m for b2, 0.07m for h2, 0.024m for r, 0.084m for b3, 0.05m for b, and 0.04m for h3 in Equation (7).

Iy=[112×5.02400[0.082+0.162]+5.02400×[(0.082)2+(0.162)2]112×1.67048[0.082+0.072]+1.67048×[(0.082)2+(0.05+0.072)2]112×0.28410[3(0.0242+0.042)+0.28410[(0.082)2+(0.16+0.042)2]]]=[13.3973+40.19201.5730+14.74200.0788+6.0229]=31.1726×103kgm2

(c)

Expert Solution
Check Mark
To determine

Find the mass moment of inertia with respect to z axis.

Answer to Problem 9.145P

The mass moment of inertia with respect to z axis is 8.58×103kgm2_.

Explanation of Solution

Calculation:

Find the mass moment of inertia with respect to z axis as shown below:

Iz=(Iz)1(Iz)2(Iz)3=[112×m1[b12+h12]+m1[(b12)2+(h12)2]112(m2[b22+h22])+m2[(b22)2+((bh22)2)]m3(12169π2)r2+m3[(b32)2+(h14r3π)2]] (8)

Convert the dimensions in mm to m.

b1=80mm(1m103mm)=0.08m

Substitute 0.28410kg for m3, 1.67048kg for m2, 5.021kg for m1, 0.08m for b1, 0.05m for h1, 0.08m for b2, 0.38m for h2, 0.024m for r, 0.08m for b3, and 0.05m for b, in Equation (8).

Iz=[112×5.0240[0.082+0.052]+5.02400[(0.082)2+(0.052)2]112×1.67048×[0.082+0.382]+1.67048[(0.082)2+(0.050.0382)2](0.28410(12169π2)×0.0242+0.28410[(0.082)2+(0.054×0.0243π)2])]=3.7261+11.17841.0919+4.2781(0.0523+0.9049)=8.5773×103=8.58×103kgm2

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Chapter 9 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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