Distributed Forces: Moments of Inertia

1. Distributed Forces: Moments of Inertia

2. The second moment of area about an axis is defined as the moment of inertia of area and is given by • Note that elemental area of A must be chosen such that it is parallel to the axis about which i is being -----

3. Moments of inertia are important when dealing with distributed forces that are proportional to both the area over which the forces are distributed as well as the distance of that area from a particular point or axis. • A typical example is beam that is subjected to pure bending. For such a beam, bending stresses are linearly distributed across the entire x-section, such that zero stress exists at the neutral axis of the x-section and maximum tension and compression stresses exist at the outermost force of the section.

4. Hence for a rectangular x-section • From (c), we observe that y ==>  = ky; but  = F/A ==> F = A = KyA • On the elemental area, dA, we have that the force dF is given by dF = kydA • Hence the resultant F of all the entire elementary forces dF is obtained by integrated over the entire x-section, i.e.

5. Because the x-y axes pass through the centroid of x-section area, , hence F must be equal to zero. • Hence the system of forces dF reduce to a couple, called the bending moment. Note that: • shows that the magnitude of the resultant force F depends on the first moment of area .

6. The magnitude of the bending moment is obtained by summing up the bending moments of all the elemental areas along the axis of interest. Hence, • Where, is the second moment of area, i.e. the moment of inertia about the x-axis. • ** To calculate the moment of inertia about a particular axis, selected a strip of elemental area parallel to that axis so that all the points forming that strip are at the same distance from the axis.

7. Polar Moment of Inertia • Of great importance in problems concerning the torsion of cylindrical shafts / bars and in problems dealing with rotation of slabs • For torsion of cylindrical shafts, the shear stress is proportional to both the area of the x-section and the distance from longitudinal axis of the shaft / bar • By definition, the polar moment of inertia Ip is given by the integral where r is distance of an elemental area dA from the longitudinal axis of the shaft.

8. Example • For a circular x-sectional area of radius r, determine the moments of inertia Ix and Iy as well as the polar moment of inertia Ip about the center O of the circle • To determine the polar moment inertia for the circular area, use an elemental circular strip as shown,

9. Radius of Gyration • The radius of gyration of an area with respect to an axis is the distance from the axis to a point where the area is assumed to be concentrated, • i.e. the moment of inertia of the concentrated area about the axis is the same as the moment of inertia of the area about the axis

10. Radius of Gyration(2) • Consider an arbitrary area A with moment of inertia Ix about the x-axis. The radius of gyration about the x-axis, Kx must be such that

11. Radius of Gyration(3) • Similarly, and • Recalling the relationship , we see that ==>

12. Parallel Axis Theorem You can transfer the moment of Inertia from one axis to another axis provided that borh are parallel and one passes through the centroid of the area. The Parallel Axis Theorem is used to compute the moment of inertia of an object about any axis using the known value of the moment of inertia about a parallel centroidal axis.

13. yc dA y d2 x1 x c xc y1 d1 y dA x Is first moment of area about the centroidal axis Is moment of inertia w.r.t centroidal axis Similarly where x = 0

14. Moments of Inertia of Composite Areas 1. Divide/Split the composite areas into regular geometric shapes whose moments of inertia are known/provided or can easily be determined 2. Determine the moments of inertia of each constutuent part about its centoidal axis (which must be parallel to the axis of interest) and use the parallel-axis theorem to compute the moment of inertia about the axis of interest. 3. The momeent of intertia of the composite area is the algebraic sum of the constituent parts/areas talking care to treat holes as negative quantities 4. Tabulating your calculationsmay be helpful in keeping track of things: eg. Given an area, calculate moments of inertia Ix and Iy Jo and Ixy. We know that the following expressions are valid:

15. Mass Moment of Inertia In dynamics, we know that accelerations of an object as a result of a force F acting on it is given by a = F/m ; where m is the mass of the body. Hence acceleratoin of a body is dependant on its mass. The angular acceleration or rotational acceleration of an object about an axis is dependant on a quantity called the Mass Moment of Inertia, which by definition is the second moment of mass about an axis, ie Observe that the unit for mass moment of inertia is kg•m2 Where r is the perpendicular distance from the axis to the elemental mass dm and Io is the mass moment of inertia about the z-axis.

16. x x dx L Mass Moment of Inertia of a Slender Bar Given: A straight slender prismatic bar of mass m, length L and cross-sectional area A and density  Req’d: Mass moment of inertia about a perpendicular axis through its center of mass. solution Determine the location of the center or mass. dm CONTINUED

17. dm=dV= Adr o r dr L Where Ix is the moment of inertia of area of unit about axis perpendicular to length Hence the mass moment of inertia of a slender bar/rod about an axis through its center of mass is given by Io=

18. y r x z Mass Moments of Inertia of Thin Plates Consider a homogeneous flat plate of density, mass m, and uniform thickness T which is shown below: y By definition the mas moment of inertia about the z-axis For the elemental area dA, its volume dV=TdA and its mass dm = dV= TdA x

19. Hence, the mass moment of inertia of a thin homogenious plate of uniform thickness can be expressed in terms of the moments of inertia of the crossectional area up the plate; or in terms of the radii of gyration as shown below: Note these Relationships: If you know the moment of inertia of the x-sectional area of a plate, you can obtain its mass moment of inertia, provied that the plate is thin, has uniform thickness and is made from homogenous material.

20. Parallel Axis Theorem for Mass Moment of Inertia This theorem is used to computer the mass moment of inertia of an object bout any asxis which is parallel to an axis that passes through the centre of mass of the object. The theorem is of the same form as that for moment of inertia of area: Io = Ic + d2m Where Ic is the mas moment of inertia about an axis that passes he centre of mass, m is the mass of the body and di is the distance between the centroidal axis and a parallel axis through point O