10.01.03.005
- 1. Welcome to presentation Based on moment of inertia Rafiqul islam 10.01.03.005 Course No: CE 416 Course Title :pre-stressed concrete
- 2. Introduction • Forces which are proportional to the area or volume over which they act but also vary linearly with distance from a given axis. - the magnitude of the resultant depends on the first moment of the force distribution with respect to the axis. - The point of application of the resultant depends on the second moment of the distribution with respect to the axis. Herein methods for computing the moments and products of inertia for areas and masses will be presented
- 3. The definition of moment of inertia is mathematical: I y 2 dA where y is the distance of an element dA of an area from an axis about which the moment of inertia is desired. Generally, The axis is in the plane of area , in which case it is called a rectangular moment of inertia. The axis is perpendicular to the area, in which case is called a polar moment of inertia.
- 4. Moment of Inertia of an Area F • Consider distributed forces whose magnitudes are proportional to the elemental areas A on which they act and also vary linearly with the distance of from a given axis. • Example: Consider a beam subjected to pure bending. Internal forces vary linearly with distance from the neutral axis which passes through the section centroid. F ky A R k y dA 0 M k y 2 dA y dA Qx first moment y 2 dA second moment
- 5. Moment of Inertia of an Area by Integration • Evaluation of the integrals is simplified by choosing d to be a thin strip parallel to one of the coordinate axes. • For a rectangular area, Ix 2 h y dA y 2bdy 0 1 bh3 3 • The formula for rectangular areas may also be applied to strips parallel to the axes, dI x 1 y 3dx 3 dI y x 2 dA x 2 y dx
- 6. Polar Moment of Inertia • The polar moment of inertia is an important parameter in problems involving torsion of cylindrical shafts and rotations of slabs. J0 r 2 dA • The polar moment of inertia is related to the rectangular moments of inertia, J0 r 2 dA Iy Ix x2 y 2 dA x 2 dA y 2 dA
- 7. 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. Ix 2 I x kx A kx A kx = radius of gyration with respect to the x axis • Similarly, Iy 2 Iy ky A ky A JO 2 J O kO A kO A 2 kO 2 kx 2 ky
- 8. Examples SOLUTION: • • A differential strip parallel to the x axis is chosen for dA. dI x y 2dA dA l dy • For similar triangles, Determine the moment of inertia of a triangle with respect to its base. l b h y l h b h y h dA b h y h dy • Integrating dIx from y = 0 to y = h, Ix h 2 y dA 2 y b h 0 3 b y h h 3 y h dy bh 2 hy h0 y 3 dy 4 h y 4 Ix 0 bh3 12 8
- 9. SOLUTION: • An annular differential area element is chosen, dJ O u 2 dA dA r JO dJ O 2 u du r 2 u 2 u du 2 0 0 JO a) Determine the centroidal polar moment of inertia of a circular area by direct integration. b) Using the result of part a, determine the moment of inertia of a circular area with respect to a diameter. u 3du 2 r4 • From symmetry, Ix = Iy, JO Ix Iy 2I x 2 r4 2I x I diameter Ix 4 r4 9
- 10. Parallel Axis Theorem • Consider moment of inertia I of an area A with respect to the axis AA’ y 2 dA I • The axis BB’ passes through the area centroid and is called a centroidal axis. y 2 dA I y d 2 dA y 2 dA 2d y dA d 2 dA I I Ad 2 parallel axis theorem 10
- 11. • Moment of inertia IT of a circular area with respect to a tangent to the circle, I Ad 2 5 4 IT r4 1 4 r4 r2 r2 • Moment of inertia of a triangle with respect to a centroidal axis, I AA I BB I BB Ad 2 I AA 2 Ad 1 bh3 12 1 bh 1 h 2 2 3 1 bh3 36 11
- 12. Moments of Inertia of Composite Areas • The moment of inertia of a composite area A about a given axis is obtained by adding the moments of inertia of the component areas A1, A2, A3, ... , with respect to the same axis. 08.12.2013 Dr. Engin Aktaş 12