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Moment of Inertia

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**Type of moment of inertia**

Moment of inertia of Area Moment of inertia of mass Also known as second moment Why need to calculate the moment of Inertia? To measures the effect of the cross sectional shape of a beam on the beam resistance to a bending moment Application Determination of stresses in beams and columns Symbol I – symbol of area of inertia Ix, Iy and Iz

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**Application : Design Steel ( Section properties)**

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**Moment of Inertia of Area**

y y dx dy dy h h C x C x b b Area of shaded element, Area of shaded element, Moment of inertia about x-axis Moment of inertia about y-axis Integration from h/2 to h/2 Integration from b/2 to b/2

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**J = polar moment of inertia**

Table 6.2. Moment of inertia of simple shapes Shape J = polar moment of inertia 1. Triangle 2. Semicircle 3. Quarter circle 4. Rectangle 5. Circle b h y x y x r y x r b h y x x y r

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**Parallel - Axis theorem**

There is relationship between the moment of inertia about two parallel axes which is not passes through the centroid of the area. From Table 6.1; Ix = and Iy = The centroid is, ( , ) = (b/2, h/2) Moment of inertia about x-x axis, Ixx = Ix + Ady2 where dy is distance at centroid y Moment of inertia about y-y axis, Iyy = Iy + Asx2 where sx is distance at centroid x y x b h

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**Determine centroid of composite area**

Example 6.2 y 140 mm 60 mm 160 mm 60mm x y 140 mm 1 3 2 60 mm x 60mm 160 mm 60 mm Determine centroid of composite area PART AREA(mm2) y(mm) x(mm) Ay (103)(mm3) Ax (103) (mm3) 1 60(200) = 12000 200/2 = 100 60/2 = 30 1200 360 2 160(60)=9600 60 + [160/2] = 140 288 1344 3 /2= 250 3000 Σ: Σ: 2688 x 103 Σ: 4704 x 103

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**Second moment inertia PART AREA (A)(mm2) Ix = bh3/12 (106) (mm4)**

dy = |y-y|(mm) Ady2(106)(mm4) 1 60(200) = 12000 60(2003)/12 = 40 |100– 80| = 20 4.8 2 160(60)=9600 160(603)/12 = 2.88 |30 – 80| = 50 24 3 |100 – 80| =20 Σ: [Ix + Ady2]1 + [Ix + Ady2]2 +[Ix + Ady2]3 = [ ] x106 = x 106 mm4 PART AREA(mm2) Iy = b3h/12 (106) (mm4) Sx=|x-x| (mm) Asx2(106)(mm4) 1 60(200) = 12000 603(200)/12 =3.6 |30-140|=110 145.2 2 160(60)=9600 1603(60)/12 =20.48 |140 – 140 |= 0 3 |250 – 140|=110 Σ: [Iy + As2]1 + [Iy + As2]2 +[Iy + As2]3 = [ ] x106 = x 106 mm4

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Worked examples and exercises are in the text STROUD PROGRAMME 20 INTEGRATION APPLICATIONS 3.

Worked examples and exercises are in the text STROUD PROGRAMME 20 INTEGRATION APPLICATIONS 3.

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