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Published byCrystal Black Modified about 1 year ago

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M OMENT OF I NERTIA

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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 I x, I y and I z

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Application : Design Steel ( Section properties)

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

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

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P ARALLEL - A XIS 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; I x = and I y = The centroid is, (, ) = (b/2, h/2) Moment of inertia about x-x axis, I xx = I x + Ad y 2 where d y is distance at centroid y Moment of inertia about y-y axis, I yy = I y + As x 2 where s x is distance at centroid x y x b h

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Example 6.2 PARTAREA(mm 2 )y(mm)x(mm)Ay (10 3 )(mm 3 )Ax (10 3 ) (mm 3 ) 160(200) = /2 = /2 = (60)=960060/2 = [160/2] = (200) = /2 = /2= Σ: Σ: 2688 x 10 3 Σ: 4704 x mm 60 mm 160 mm60mm x y Determine centroid of composite area 140 mm 60 mm 160 mm60mm x y 1 2 3

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PARTAREA (A)(mm 2 ) I x = bh 3 /12 (10 6 ) (mm 4 )d y = |y-y|(mm)Ad y 2 (10 6 )(mm 4 ) 160(200) = (200 3 )/12 = 40 |100– 80| = (60)= (60 3 )/12 = 2.88 |30 – 80| = (200) = (200 3 )/12 = 40 |100 – 80| =204.8 Σ: Second moment inertia [I x + Ad y 2 ] 1 + [I x + Ad y 2 ] 2 +[I x + Ad y 2 ] 3 = [ ] x 10 6 = x 10 6 mm 4 PARTAREA(mm 2 ) I y = b 3 h/12 (10 6 ) (mm 4 )S x =|x-x| (mm)As x 2 (10 6 )(mm 4 ) 160(200) = (200)/12 =3.6 |30-140|= (60)= (60)/12 =20.48 |140 – 140 |= (200) = (200)/12 =3.6 |250 – 140|= Σ: [Iy + As 2 ] 1 + [Iy + As 2 ] 2 +[Iy + As 2 ] 3 = [ ] x 10 6 = x 10 6 mm 4

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