Centroids and Moment of Inertia PDF

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Summary

-Centroid and Center of Gravity
-Center of Gravity
-Centroids of Areas
-Moment of Inertia (I)...

Description

STATICS OF RIGID BODIES Module 5: Centroids and Moment of Inertia Centroid and Center of Gravity Moment of Area M=A Center of Gravity W1

Z

W

Y W2

x1

c.g. x2

𝒚 𝒙 x

Center of gravity of flat plate

Taking moments of weights about the y-axis, we get W  = w1x1 + w2x2 + …= wixi With respect to x-axis, we have W = w1y1 + w2y2 + …= wiyi If the plate is homogenous, the W may be expressed as the product of its density  (weight per unit volume) multiplied by tA.

tA = ta1x1 + ta2x2 + …= tax A = a1x1 + a2x2 + …= ax where:

=density t= thickness A = area

Centroids of Areas A = ax A = ay Centroid = ax/A = ay/A

Shape Rectangle y

Area

h 𝒚

bh

b/2

h/2

1/2bh

_

1/3h

r2/2

0

4r/3

x b 𝒙

Triangle

y

h 𝑦

b

x

Semi-circle y

𝒚 d=2r

x

Quarter circle y r 𝑦

r2/4

4r/3

4r/3

bh/(n+1)

b/(n+2)

(n+1)(h)/(4n+2)

x 𝑥

Area under spandrel

Examples: 1.) The dimension of T-section of a cast iron beam are shown. How far is the centroid of the area above the base? 1cm

8cm

1cm 6cm

Solution:

1cm

A2

8cm

𝒚 A1

1cm

x

6cm

A1 = 6(1) = 6cm2 A2 = 8(1) = 8cm2 AT = 6 + 8 = 14cm2 Moment about x-axis AT  = A1(0.5) + A2(5) 14  = 6(0.5) + 8(5) = 3.07cm //ans. 2.) Determine the coordinates of the centroid of the area shown, with respect to the given axes. y r=3cm

9cm

x 6cm

Solution:

y A2

9cm

r=3cm

A1 𝒚 x 𝒙

6cm

A1 = ½(9)(6) = 27cm2 A2 = (3)2/2 = 14.14cm2 AT = 41.14cm2 Moment @ x-axis AT = A1(6) + A2[9 + 4(3)/3] 41.14  = 27(6) + 14.14[9 + 4/] = 7.47cm Moment @ y-axis 41.14  = 27(2) + 14.14(3) = 2.34cm Coordinate of centroid is (2.34cm, 7.47cm) //ans.

Moment of Inertia (I) – is the sum of the products obtained by multiplying all the infinitely small areas by the square of their distances from an axis. It is sometimes called second moment of area. I = 2dA

(mathematical definition of moment of inertia)

where: dA = area  = distance

y x

dA y x

About x – axis: Ix = y2dA About y- axis: Iy = x2dA Transfer Formula for a moment of inertia: Ix = Ixo + Ad2 Iy = Iyo + Ad2 Where: Ixo, Iyo =moment of inertia at centroidal axis Iy, Ix = moment of inertia at any parallel axis A = area d = distance between axes

Moment of Inertia of Geometric Shapes Shape Rectangle

Moment of Inertia y Ixo = bh 3/12

c

Xo

h

Ix = bh3/3 X b

Triangle Ixo = bh 3/36 c

h

Xo X

b

Ix = bh3/12

Circle r

Ixo = r4/4 = d4/64

Xo

c

Semi-circle c

Ixo = 0.11r4

Xo X

d=2r Quarter-circle

Ix = r4/8 Ixo = Iyo = 0.055r4

Xo

c

r

X

Ix = Iy = r4/16

Examples: 1.) Find the moment of inertia of the T-section with respect to its centroidal axis (Xo). 2cm

A2 8cm

XO 𝒚

2cm

A1 X 8cm

Solution: A1 = 8(2) = 16cm2 A2 = 2(8) = 16cm2 AT = 32cm2 32  = 16(1) + 16(6)  = 3.5cm

Using transfer formula for moment IXO = IXO1 + A1(d)2 + IXO2 + A2(d)2 IXO = 8(2)3/12 + 16(2.5)2 + 2(8)3/12 + 16(2.5)2 IXO = 290.67cm4 //ans.

2.) Find the moment of inertia about the indicated x-axis for the shaded area.

r=4cm 10cm 𝒚

X

Solution: A1 = 10(8) = 80cm2 A2 = (4)2/2 = -25.13cm2 AT = 54.87cm2 54.87  = 80(5) - 25.13(8.3)  = 3.49cm

IX = 8(10)3/12 + 80(5)2 – [0.11(4)4 + 25.13(8.3)2] IX = 907.30cm4 //ans....