2 Type of moment of inertia Moment of inertia of AreaMoment of inertia of massAlso known as second momentWhy 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 momentApplicationDetermination of stresses in beams and columnsSymbolI – symbol of area of inertiaIx, Iy and Iz
4 Moment of Inertia of Area yydxdydyhhCxCxbbArea of shaded element,Area of shaded element,Moment of inertia about x-axisMoment of inertia about y-axisIntegration from h/2 to h/2Integration from b/2 to b/2
5 J = polar moment of inertia Table 6.2. Moment of inertia of simple shapesShapeJ = polar moment of inertia1. Triangle2. Semicircle3. Quarter circle4. Rectangle5. Circlebhyxyxryxrbhyxxyr
6 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 + Ady2where dy is distance at centroid yMoment of inertia about y-y axis, Iyy = Iy + Asx2where sx is distance at centroid xyxbh
7 Determine centroid of composite area Example 6.2y140 mm60 mm160 mm60mmxy140 mm13260 mmx60mm160 mm60 mmDetermine centroid of composite areaPARTAREA(mm2)y(mm)x(mm)Ay (103)(mm3)Ax (103) (mm3)160(200) = 12000200/2 = 10060/2 = 3012003602160(60)=960060 + [160/2] = 14028813443/2= 2503000Σ:Σ: 2688 x 103Σ: 4704 x 103
8 Second moment inertia PART AREA (A)(mm2) Ix = bh3/12 (106) (mm4) dy = |y-y|(mm)Ady2(106)(mm4)160(200) = 1200060(2003)/12 = 40|100– 80| = 204.82160(60)=9600160(603)/12 = 2.88|30 – 80| = 50243|100 – 80| =20Σ:[Ix + Ady2]1 + [Ix + Ady2]2 +[Ix + Ady2]3= [ ] x106= x 106 mm4PARTAREA(mm2)Iy = b3h/12(106) (mm4)Sx=|x-x| (mm)Asx2(106)(mm4)160(200) = 12000603(200)/12=3.6|30-140|=110145.22160(60)=96001603(60)/12=20.48|140 – 140 |= 03|250 – 140|=110Σ:[Iy + As2]1 + [Iy + As2]2 +[Iy + As2]3= [ ] x106= x 106 mm4
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