LEC 6introdcution to managment 101 for engineers PDF

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introdcution to managment 101 for engineers
introdcution to managment 101 for engineersintrodcution to managment 101 for engineers
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2020-09-24

Civil Engineering Mechanics CVG2149

LEC6 Moments of Inertia

Radius of Gyration of an Area • Consider area A with moment of inertia Ix. Imagine that the area is concentrated in a thin strip parallel to the x axis with equivalent Ix. kx = radius of gyration with respect to the x axis ky = radius of gyration with respect to the y axis • Similarly,

Ix 

k x2 A

kx 

Ix A

Radii of gyration: k x, ky, kO:

I y  k y2 A

ky 

J O  k O2 A

kO

Iy

A J  O A

kO2  k x2  k y2

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For a rectangular plate radius of gyration with respect to its base is:

k 2x 

1 bh3 h 2 Ix  3  3 A bh

k x

h 3

Attention: radius of gyration is different then coordinates of Centroid . Not to be confused !

- kx (radius of gyration) depends on second moment - y (centroid ordinate) depends on first moment

I x  k x2 A

kx 

Ix A

Product of inertia For certain problems moments of inertia about rotated axes are required. We determine the product of inertia: I  xy dA xy





Can be positive negative or zero

• When the x axis, the y axis, or both are an axis of symmetry, the product of inertia is zero.

• Parallel axis theorem for products of inertia:

I xy  I xy  x yA

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CVG Dictionary

• • • • • • •

Moment of inertia: Polar moment of inertia: Radii of gyration:

I x   y 2 dA 2

I x  kx A

I y   x2 dA

I y  ky A 2

2

J O  kO A k 2 O  kx2  k2y Centroidal moment of inertia Relation between them (parallel axis th.)

Product of moment of inertia Relation between them (parallel axis th.) I xy   xydA

I  I  Ad 2 J O  J C  Ad 2 2

2

kO  kC  d 2

I xy  I x 'y '  x yA

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Example SOLUTION: • Determine the location of the centroid for the composed section.

Use the parallel axis theorem to determine the moment of inertia of the composed section in regards to the axis passing through the centroid of the composed section. The steel girder W360x44 was reinforced with a plate welded at the upper part, as shown. Determine the moment of the combined section, in regard to an axis parallel with the base of the shape and passing through the centroid C of the entire section.

Section A , mm 2 y , mm yA , mm 3 Plate 4351 185.5 807111 Beam Section 5730 0 0  A  10081  y A  807111

Y  A   yA

Y

 yA  807111 mm  A 10081 mm

3

2

 80.1 mm

Determine Moment inertia in respect with x’, based on parallel axes theorem.

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C1 C2 = origin d1 = 19/2 + 352/2 -Y C

2

d2 = YC

I C,plate  I plate  Ad1

C2

IC , beam section  I beam section  Ad 22

C1

I C , plat e  I plate  Ad 1

2

229 19 3  2   4351185.5  80.1

1 12

d1 = 19/2 + 352/2 -YC

 4.84 10 7 mm 4

d2 = YC C2

I C ,beam sect ion  I beam section  Ad 2

2

122 106  5730 80.1

2

 1.58 10 8 mm4 I C  IC, beam section  IC, plate  1.58 108  4.84 107

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I x  2 I x1  2 I x3  I x2 

C3

2

d3

2

2

 2( Ix1  Ad1 )  2( I x3  Ad 3 )  ( Ix2  Ad2 )

I y  2 I y 1 2 I y 3  I y 2 

C2

2

d2

2

2

2( I y1  Ad1 ) 2( I y3  Ad 3 ) ( Iy2  Ad2 )

d1 C1

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C3 Total area (Atot): y*A=

21.4

C2

YC = Atot/y*A

C1

The coordinates of the centroids are determined in regard with the chosen origin

O O = origin

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