PARALLEL AXIS THEOREM - PDF Introduction to Mechanics of Materials: Part I

APPENDIX

A.3 PARALLEL AXIS THEOREM

Solution

(a) Polar moment of Inertia. We select, as an element of area, a ring of radius ρ and thickness dρ (Fig. A.12). Using Eq. (A.8), where dA = 2 πρ dρ, we have

/ 2 / 2

2 2 3

0 0

d ^D 2 d 2 ^D d

o A

J =

∫

ð D4 o 32

J = .

Fig. A.12

(b) Moment of Inertia. Because of the symmetry of a circular area I_z = I_y. Recalling Eg. (A.9), we can write

2 32

2^o 2

o z y z z J

J I I I I

D .4 z y 64

I I S

Considering the moment of inertia I_z of an area A with respect to an arbitrary z axis (Fig. A.13). Let us now draw the centroidal z’ axis, i.e., the axis parallel to the z axis which passes though the area’s centroid C. Denoting the distance between the element dA and axis passes though the centroid Cby y’, we write y = y’ + d. Substituting for y in Eq. (A.6), we write

2 d ' d

A A

³

y A

³

y d A^,

2 2

' d 2 ' d d

A A A

I =

∫

y A+ d y A d

∫

A^,

' ' 2

z z z

I =I +Q +Ad (A.14)

where Iz' is the area’s moment of inertia with respect to the centroidal z’ axis and Q_z’ is the first moment of the area with respect to the z’ axis, which is equal to zero since the centroid C of the area is located on that axis. Finally, from Eq. (A.14)we have

' 2

z z

I =I +Ad (A.15)

A similar formula may be derived, which relates the polar moment of inertia J_o of an area to an arbitrary point O and polar moment of inertia J_C of the same area with respect to its centroid C. Denoting the distance between O and Cby d, we write

o C 2

J =J +Ad (A.16)

Example A.05

Determine the moment of inertia I_z of the area shown in Fig. A.14 with respect to the centroidal z axis.

Fig. A.14

Download free eBooks at bookboon.com

157 157

Solution

The first step of the solution is to locate the centroid C of the area. However, this has already been done in Example A.02 for a given area A.

We divide the area A into two rectangular areas A₁ and A₂ (Fig. A.15) and compute the moment of inertia of each area with respect to the z axis. Moment of inertia of the areas are

1 2

z z z

I =I +I ,

where I_z1 is the moment of inertia of A₁ with respect to the z axis. For the solution, we use the parallel-axis theorem (Eq. A.15), and write

2 3 2

1 ' A d1 1 1 b h b h d1 1 1 1 1 z z 12

I =I + = +

3 2

1 1 8t (2t) 8t 2t (7t 4.6t)

z 12

I = × × + × × −

1 97.5 t4

Iz =

Download free eBooks at bookboon.com Click on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read more

360° thinking .

Discover the truth at www.deloitte.ca/careers

360° thinking .

Discover the truth at www.deloitte.ca/careers

360° thinking .

Discover the truth at www.deloitte.ca/careers

360° thinking .

Discover the truth at www.deloitte.ca/careers

Fig. A.15

In a similarly way, we find the moment of inertia I_z2 of A₂ with respect to the z axis and write

2 3 2

2 '' A d2 2 1 b h2 2 b h d2 2 2 z z 12

I =I + = +

3 2

2 1 4t (6t) 4t 6t (4.6t 3t)

z 12

I = × × + × × −

1 133.4 t4

Iz =

The moment of inertia I_z of the area shown in Fig. A.14 with respect to the centroidal z axis is

4 4 4

1 2 97.5t 133.4t 230.9t .

z z z

I =I +I = + =

Example A.06

Fig. A.16

Determine the moment of inertia I_z of the area shown in Fig. A.14 with respect to the centroidal z axis and the moment of inertia I_y of the area with respect to the centroidal y axis.

Download free eBooks at bookboon.com

Fig. A.17

Solution

The first step of the solution is to locate the centroid C of the area. This area has two axis of symmetry, the location of the centroid C is in the intersection of the axes of symmetry.

Fig. A.18

We divide the area A into three rectangular areas A₁, A₂ and A₃. The first way we can divide area A can be seen in Fig. A.17, a second way can be seen in Fig. A.18.

Solution the division of area A by Fig. A.17 (the first way) themoment of inertia I_zis

1 2 3,

z z z z

I =I +I +I

where

2 3 2 4

1 ' A d1 1 1 b h b h d1 1 1 1 1 196t ,

z z 12

I =I + = + ==

2 3 2 4

2 A d2 2 1 b h2 2 b h d2 2 2 36t ,

z z 12

I = +I = + ==

2 3 2 4

3 ''' A d3 3 1 b h b h d3 3 3 3 3 196t .

z z 12

I =I + = + ==

Download free eBooks at bookboon.com

Resulting in

4 4 4 4

1 2 3 196t 36t 196t 428t .

z z z z

I =I +I +I = + + =

For the moment of inertia I_y we have

1 2 3,

y y y y

I =I +I +I

where

3 3 4

1 1 h b1 1 1 2t 6t 36t ,

12 12

y y

I I u u

3 3 4

2 1 h b2 2 1 6t 2t 4t ,

12 12

y y

I I u u

3 3 4

3 1 h b3 3 1 2t 6t 36t .

12 12

y y

I I u u

We will turn your CV into an opportunity of a lifetime

Do you like cars? Would you like to be a part of a successful brand?

We will appreciate and reward both your enthusiasm and talent.

Send us your CV. You will be surprised where it can take you.

Send us your CV on www.employerforlife.com

Resulting in

4 4 4 4

1 2 3 36t 4t 36t 76t .

y y y y

I =I +I +I = + + =

The solution for the division of area A according to Fig. A.18 (by the second way) the moment of inertia I_zis

1 2 3,

z z z z

I =I −I −I

where

3 3 4

1 1 b h1 1 1 6t 10t 500t ,

12 12

z z

I I u u

3 3 4

2 1 b h2 2 1 2t 6t 36t ,

12 12

z z

I I u u

3 3 4

3 1 b h3 3 1 2t 6t 36t .

12 12

z z

I I u u

Resulting in

4 4 4 4

1 2 3 500t 36t 36t 428t .

z z z z

I =I −I −I = − − =

For the moment of inertia I_y we have

1 2 3,

y y y y

I =I −I −I

where

3 3 4

1 1 h b1 1 1 10t 6t 180t ,

12 12

y y

I I u u

3 2

3 2 4

2 1 h b2 2 h b d2 2 2 1 6t 2t 6t 2t 2t 52t ,

12 12

y y

I I u u u u

3 2

3 2 4

3 1 h b h b d3 3 3 3 3 1 6t 2t 6t 2t 2t 52t .

12 12

y y

I I u u u u

Download free eBooks at bookboon.com

Resulting in

4 4 4 4

1 2 3 180t 52t 52t 76t .

y y y y

I =I −I −I = − − =

Example A.07

Fig. A.19

In order to solve the torsion of a rectangular cross-section in Fig. A.19, we defined (See S.P. Thimoshenko and J.N. Goodier, Theory of Elasticity, 3d ed. McGraw-Hill, New York, 1969, sec. 109) the following parameters for b>h:

γ b h,3

J (A.17)

1 b h,2

S D (A.18)

2 b h ,2

S E (A.19)

where parameters α, b and γ are in Tab.A.1.

The shearing stresses at point 1 and 2 are defined as

1 max

ST ,

W W 2

ST , W

(A.20) where T is the applied torque.

Download free eBooks at bookboon.com

Tab.A.1

h/b 1 1.2 1.5 2 3 5 10 >10

D 0.208 0.219 0.231 0.246 0.267 0.291 0.313 1/3 E 0.208 0.196 0.180 0.155 0.118 0.078 0.042 0 γ 0.1404 0.166 0.196 0.229 0.263 0.291 0.313 1/3

A.4 PRODUCT OF INERTIA, PRINCIPAL AXES

Dalam dokumen PDF Introduction to Mechanics of Materials: Part I (Halaman 155-163)