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, Problem 9.192RP
To determine

Find the moment and product of inertia of the area with respect to x and y axes about through 30° by using Mohr circle method.

Expert Solution
Check Mark

Answer to Problem 9.192RP

The moment of inertia of the area with respect to x about through 30° using Mohr circle method is 5.27in4_

The moment of inertia of the area with respect to y about through 30° using Mohr circle method is 6.71in4_

The product of inertia of the area with respect to x and y axes about through 30° by using Mohr circle method is 4.39in4_.

Explanation of Solution

Sketch the cross section as shown in Figure 1.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 9, Problem 9.192RP , additional homework tip  1

Refer to Figure 9.13.

The moment of inertia I¯x about x axis is 9.43in4.

The moment of inertia I¯y about y axis is 2.55in4.

Refer to Problem 9.191.

Sketch the cross section as shown in Figure 2.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 9, Problem 9.192RP , additional homework tip  2

Express the product of inertia as shown below:

I¯xy=(Ixy)1+(I¯xy)2

Here, [(Ixy)1,(Ixy)2] is product of inertia section 1 and 2.

Applying parallel axis theorem,

Ixy=Ixy+x¯y¯A

When the x and y axis is symmetry.

Ixy=0

Refer to Figure 1.

Find the area of semicircle section 1 as shown below:

A1=(b1h1) (1)

Here, b1 is width of the section 1 and h1 is height of the section 1.

Substitute 3in. for b1 and 0.5in. for h1 in Equation (1).

A1=3×0.5=1.5in2

Find the area of rectangle section 2 as shown below:

A2=(b2h2) (2)

Here, b2 is width of the section 2 and h2 is height of the section 2.

Substitute 4.5in. for b2 and 0.5in. for h2 in Equation (2).

A2=4.5×0.5=2.25in2

Find the centroid for section 1 about x axis (x1) as shown below:

x1=12(3)0.746=1.50.746=0.754in.

Find the centroid for section 1 about x axis (x2) as shown below:

x2=(0.746(0.5×0.5))=0.7460.25=0.496

Find the centroid for section 1 about y axis (y1) as shown below:

y1=(1.7412(12))=1.49in.

Find the centroid for section 2 about x axis (y2) as shown below:

y2=12(4.5)(1.7412)=2.251.24=1.01in.

Find the product of inertia of the area with respect to x and y axes by using parallel axis theorem as shown below:

I¯xy=(x¯y¯A)1+(x¯y¯A)2=(x1y1A)+(x¯2y¯2A) (3)

Substitute 0.754in. for x1, 0.496 for x2, 1.49in. for (y1), 1.01in. for y2, 1.5in2 for A1, and 2.25in2 for A3 in Equation (3).

I¯xy=(0.754×(1.49)×1.5)+(0.496×(1.01)×2.25)=1.685191.12716=2.81in4

The Mohr circle is defined by the diameter XY, where X(9.432.81235) and Y(2.55,2.81235).

Find the average moment of inertia (Iave) using the relation as shown below:

Iave=12(I¯x+I¯y) (4)

Here, I¯x is moment of inertia about x axis and I¯y is moment of inertia about y axis.

Substitute 9.43in4 for I¯x and 2.55in4 for I¯y in Equation (4).

Iave=12(9.43+2.55)=12(11.98)=5.99in4

Find the radius (R) using the relation as shown below:

R=(I¯xI¯y2)2+I¯xy2 (5)

Here, R is radius and I¯xy is product of inertia.

Substitute 9.43in4 for I¯x and 2.55in4 for I¯y,2.81in4 for I¯xy, in Equation (5).

R=(9.432.552)2+(2.81235)2=11.8336+7.9093=19.7429=4.4433in4

Sketch the Mohr circle as shown in Figure 3.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 9, Problem 9.192RP , additional homework tip  3

Refer to Figure 2.

tan2θm=2IxyIxIy (6)

Substitute 9.43in4 for I¯x and 2.55in4 for I¯y,2.81in4 for I¯xy, in Equation (6).

tan2θm=2(2.81)9.432.55=5.626.88tan2θm=0.8172θm=39.267°

Find the angle (α) value using the relation:

α=180°(39.267+60°)=80.733°

Find the moment of inertia of the area with respect to x about through 30° using Mohr circle method as shown below:

Ix=IaveRcosα (7)

Here, Iave is average value of moment of inertia.

Substitute 5.99in4 for Iave, 4.4433in4 for R, and 80.733° for α in Equation (7).

Ix=(5.994.4433cos80.733°)=(5.990.7155)=5.2744in4

Thus, the moment of inertia of the area with respect to x about through 30° using Mohr circle method is 5.27in4_

Find the moment of inertia of the area with respect to y about through 30° using Mohr circle method as shown below:

Iy=Iave+Rcosα (8)

Substitute 5.99in4 for Iave, 4.4433in4 for R, and 80.733° for α in Equation (8).

Iy=(5.99+4.4433cos80.733°)=(5.99+0.7155)=6.71mm4

Thus, the moment of inertia of the area with respect to y about through 30° using Mohr circle method is 6.71in4_

Find the product of inertia of the area with respect to y about through 45° using Mohr circle method as shown below:

Ixy=Rsinα (9)

Substitute 4.4433in4 for R and 80.733° for α in Equation (9).

Ixy=4.4433×sin(80.733°)=4.385=4.39in4

Thus, the product of inertia of the area with respect to x and y axes about through 30° by using Mohr circle method is 4.39in4_.

(b)

To determine

Find the orientation of the principal axes through the centroid and corresponding values.

(b)

Expert Solution
Check Mark

Answer to Problem 9.192RP

The orientation of the principal axes at the origin is 19.63°counterclockwise_.

The maximum moment of inertia is 10.43in4_.

The minimum moment of inertia is 1.547in4_

Explanation of Solution

Calculation:

Find the orientation of the principal axes through at origin as shown below.

Refer part a.

tan2θm=39.267°θm=19.63°

Thus, the orientation of the principal axes at the origin is 19.63°counterclockwise_.

Sketch the orientation axis as shown in Figure 4.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 9, Problem 9.192RP , additional homework tip  4

Find the maximum moment Imax of inertia using the relation:

Imax=Iave+R (10)

Substitute 5.99in4 for Iave and 4.4433in4 for R, in Equation (10).

Imax=5.99+4.4433=10.4333in4

Thus, the maximum moment of inertia is 10.4333in4_.

Find the minimum moment Imin of inertia using the relation:

Imin=IaveR (11)

Substitute 5.99in4 for Iave and 4.4433in4 for R, in Equation (11).

Imin=5.994.4433=1.547in4

Thus, the minimum moment of inertia is 1.547in4_.

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Chapter 9 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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