SECOND MOMENT, MOMENT OF AREAS - PDF Introduction to Mechanics of Materials: Part I

APPENDIX

A.2 SECOND MOMENT, MOMENT OF AREAS

We selected an element area in Fig. A.5 with a horizontal length u and thickness dy. From thesimilarity in triangles, we have

u h y

b h

= −

u bh y h

= −

and d u dy bh ydy A= = h−

using Eq. (A.1) the first moment is

0 0

h y b

d b dy = hy y dy

h h

z A

³

y A

³

2 3

b y y 1 2

h bh

h 2 3 6

Qz ª º

« »

¬ ¼

(b) Ordinate of the centroid

Recalling the first Eq. (A.4) and observing that 1 bh

A=2 , we get

2 2

1 1 1

y bh bh y y = h

6 2 3

Qz = A ⇒ = = ⇒

Example A.02

Locate the centroid C of the area A shown in Fig. A.6

Fig. A.6

Solution

Selecting the coordinate system shown in Fig. A.7, we note that centroid C must be located on the y axis, since this axis is the axis of symmetry than z 0= .

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Fig. A.7

Dividing A into its component parts A₁ and A₂, determine theyordinate of the centroid, using Eq. (A.5)

1 1 1 2 2

2 1 2

i i i i

i i

A y A y A y A y

y A A A A

= = = +

∑ ∑

1 1 2 2

1 2 2

2t×8t 7t+ 4t×6t 3t 184t 4.6t 2t×8t 4t×6t 40t

A y A y

y A A

u u

Similarly, the second moment, or moment of inertia, of area A with respect to the y axis is

2 d

y A

I =

∫

z A^{. (A.7)}

We now define the polar moment of inertia of area A with respect to point O (Fig. A.8) as the integral

2 d

o A

³

U A^,

(A.8)

Fig. A.8

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where ρ is the distance from O to the element dA. If we use SI units, the moments of inertia are expressed in m⁴ or mm⁴.

An important relation may be established between the polar moment of inertia J_o of a given area and the moment of inertia I_z and I_y of the same area. Noting that U² ^y²^z²_, we write

2 d 2 2 d 2 d 2 d

A A A A

³

U A

³

y z A

³

y A

³

z A

o z y

J = +I I (A.9)

The radius of gyration of area A with respect to the z axis is defined as the quantity r_z, that satisfies the relation

2 ^z

z z z I

I r A r

= ⇒ = A (A.10)

In a similar way, we defined the radius of gyration with respect to the y axis and origin O.

We then have

2 ^y

y y y

I r A r I

= ⇒ = A (A.11)

2 ^o

o o o J

J r A r

= ⇒ = A (A.12)

Substituting for J_o, I_y and I_z in terms of its corresponding radi of gyration in Eg. (A.9), we observe that

2 2 2

0 z y

r =r + r

(A.13) Example A.03

For the rectangular area in Fig. A.9, determine (a) the moment of inertia I_z of the area with respect to the centroidal axis, (b) the corresponding radius of gyration r_z.

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Fig. A.9

Solution

(a) Moment of inertia I_z. We select, as an element area, a horizontal strip with length b and thickness dy (see Fig. A.10). For the solution we use Eq. (A.6), where dA = b dy, we have

/ 2 / 2

2 2 2 3 / 2

/ 2 / 2 / 2

d y b dy b y dy b y

h h h

z h

A h h

I y A

¬ ¼ª º

³ ³ ³

3 3

b h h 1 b h3

3 8 8 12

z z

I § · I

¨ ¸

Fig. A.10

(b) Radius of gyration r_z. From Eq. (A.10), we have

3 2

1 bh h h

bh 12 12

z Iz z

r r

= A = = ⇒ =

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Example A.04

For the circular cross-section in Fig. A.11. Determine (a) the polar moment of inertia J_O, (b) the moment of inertia I_z and I_y.

Fig. A.11

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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

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