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Moment of Inertia
In classical mechanics, moment of inertia, also called mass moment of inertia, rotational inertia, polar moment of inertia of mass, or the angular mass, (SI units kg m2) is a measure of an object's resistance to changes to its rotation. It is the inertia of a rotating body with respect to its rotation. The moment of inertia plays much the same role in rotational dynamics as mass does in linear dynamics, describing the relationship between angular momentum and angular velocity, torque and angular acceleration, and several other quantities. The symbol I and sometimes J are usually used to refer to the moment of inertia or polar moment of inertia. The moment of inertia of an object about a given axis describes how difficult it is to change its angular motion about that axis. Therefore, it encompasses not just how much mass the object has overall, but how far each bit of mass is from the axis. The farther out the object's mass is, the more rotational inertia the object has, and the more force is required to change its rotation rate. We will limit our application of the Moment of Inertia as it applies to Structural Components identified here. |
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Centroids & Moment of Inertia The centroid of a two dimensional surface (such as the cross-section of a structural shape) is a point that corresponds to the center of gravity of a very thin homogeneous plate of the same area and shape. The planar surface (or figure) may represent an actual area (like a tributary floor area or the cross-section of a beam) or a figurative diagram (like a load or a bending moment diagram). It is often useful for the centroid of the area to be determined in either case. Symmetry can be very useful to help determine the location of the centroid of an area. If the area (or section or body) has one line of symmetry, the centroid will lie somewhere along the line of symmetry. This means that if it were required to balance the area (or body or section) in a horizontal position by placing a pencil or edge underneath it, the pencil would be best laid directly under the line of symmetry. If a body (or area or section) has two (or more) lines of symmetry, the centroid must lie somewhere along each of the lines. Thus, the centroid is at the point where the lines intersect. This means that if it were required to balance the area (or body or section) in a horizontal position by placing a nail underneath it, the point of the nail would best be placed directly below the point where the lines of symmetry meet. This might seem obvious, but the concept of the centroid is very important to understand both graphically and numerically. The position of the center of gravity for some simple shapes is easily determined by inspection. One knows that the centroid of a circle is at its center and that of a square is at the intersection of two lines drawn connecting the midpoints of the parallel sides. The circle has an infinite number of lines of symmetry and the square has four. The centroid of a section is not always within the area or material of the section. Hollow pipes, L shaped and some irregular shaped sections all have thir centroid located outside of the material of the section. This is not a problem since the centroid is really only used as a reference point from which one measures distances. The exact location of the centroid can be determined as described above, with graphic statics, or numerically. The centroid of any area can be found by taking moments of identifiable areas (such as rectangles or triangles) about any axis. This is done in the same way that the center of gravity can be found by taking moments of weights. The moment of an large area about any axis is equal to the algebraic sum of the moments of its component areas. This is expressed by the following equation: Sum MAtotal = MA1 + MA2 + MA3+ ... The moment of any area is defined as the product of the area and the perpendicular distance from the centroid of the area to the moment axis. By means of this principle, we may locate the centroid of any simple or composite area. Center of Gravity: Given: the plate shown in the diagram has a weight of 1 #/in2 (1 pounds per square inch) of horizontal surface. Determine: the center of gravity of the plate knowing that it is symmetrical about the X-X axis. Solution: The principle of moments states that the total weight about an axis is equal to the sum of the moments of the component weights about that same axis. Thus, the first thing to do is to divide the plate into several simple parts. Then, determine the area and the center of gravity (or centroid) for each of the component parts. After this is completed, take the moments of each of the parts around a convenient axis (in this case select the Z-Z axis about which to take these moments). The Z-Z axis is here identified as the Ref Axis. 
Sum MAtotal = MA1 + MA2 + MA3 This simple equation can be rewritten as follows in which each of the component parts are described: (Atotal)(distance from reference axis to centroidal axis) = (A1)(distance from centroid of A1 to reference axis) + (A2)(distance from centroid of A2 to reference axis) + (A3)(distance from centroid of A3 to reference axis) and then solving for y ... the centroidal axis is 7.3 inches from the reference axis. The actual center of gravity occurs midway through the depth of the plate at the point calculated above. As the plate thickness is reduced the line of action of the center of gravity will remain while the center of gravity moves proportionally along this line of action always effective at the midpoint of the depth of the plate. If the plate thickness is reduced to zero it has no weight and the former center of gravity position is now referred to as the centroid of the area. The Moment of Inertia (I) is a term used to describe the capacity of a cross-section to resist bending. It is always considered with respect to a reference axis such as X-X or Y-Y. It is a mathematical property of a section concerned with a surface area and how that area is distributed about the reference axis. The reference axis is usually a centroidal axis. The moment of inertia is also known as the Second Moment of the Area and is expressed mathematically as: Ixx = Sum (A)(y2) In which: Ixx = the moment of inertia around the x axis A = the area of the plane of the object y = the distance between the centroid of the object and the x axis The Moment of Inertia is an important value which is used to determine the state of stress in a section, to calculate the resistance to buckling, and to determine the amount of deflection in a beam. For example, if a designer is given a certain set of constraints on a structural problem (i.e. loads, spans and end conditions) a "required" value of the moment of inertia can be determined. Then, any structural element which has at least that specific moment of inertia will be able to be utilized in the design. Another example could be in the inverse were true: a specific element is given in a design. Then the load bearing capacity of the element could be determined. Let us look at two boards to intuitively determine which will deflect more and why. If two boards with actual dimensions of 2 inches by 10 inches were laid side by side - one on the two inch side and the other on the eight inch side, the board which is supported on its 2" edge is considerably stiffer than that supported along its 10" edge. Both boards have the same cross-sectional area, but the area is distributed differently about the horizontal centroidal axis. 
Ixx = (1/12 ) (b)(h3) = (1/12) x (b) x (h x h x h) In which the value b is always taken to be the side parallel to the reference axis and h the height of the section. This is very important to note! If the wrong value is assumed for the value of b, the calculations will be totally wrong. Moment of Inertia Given: the cross-section. Determine: The Moments of Inertia, Ixx and Iyy of this section. Solution: 
The moment of inertia of a rectangular shape such as this one is easily calculated by using the equation I = 1/12 bh3. However, it is crucial that b and h are assigned correct values. You may simply rotate the member by 90 degrees and recalculate, always remembering the original position of the Member. Ixx= 1/12(4")(10")3 = 333.2 in4 Iyy= 1/12(10")(4")3 = 53.312 in4 In this case, observation will confirm the choices for b and h. It is logical that Ixx is greater than Iyy because a larger amount of the rectangular area lies further from the x-x axis than the y-y axis. This causes the shape to have a greater resistance to rotation around the x-x axis and therefore a larger moment of inertia around that axis. The importance of the distribution of the area around its centroidal axis becomes clear when comparing the values of the moment of inertia of a number of typical beam configurations. All of the members shown below are 2" x 10"; in cross section, equal in length and equally loaded. 
BUILT-UP SECTIONS It is often advantageous to combine a number of smaller members in order to create a beam or column of greater strength. The moment of inertia of such a built-up section is found by adding the moments of inertia of the component parts. This can be done, if and only if the moments of inertia of each component area are taken about a common axis, and if, and only, if the resulting section acts as one unit. Built-Up Sections Given: the following cross sections Determine: Ix of each section considering its component parts. Solution: In this example the Box is broken into 4 separate members, and the procedure of calculating the Ixx is shown. 
Manual Calculation with Computer Generated Calculation Below. 
Example of Results of a Computer Program Available from our Free Software Section TRANSFER FORMULA There are many built-up sections in which the component parts are not symmetrically distributed about the centroidal axis. The easiest way to determine the moment of inertia of such a section is to find the moment of inertia of the component parts about their own centroidal axis and then apply the transfer formula. The transfer formula transfers the moment of inertia of a section or area from its own centroidal axis to another parallel axis. It is known from calculus to be: Ix = Ic + Ad2 Where: Ix = moment of inertia about axis x-x (in4) Ic = moment of inertia about the centroidal axis c-c parallel to x-x (in4) A = area of the section (in2) d = perpendicular distance between the parallel axes x-x and c-c (in) Transfer Formula Given: the glued asymmetric built-up cross-section below. Determine: the moment of inertia of the composite area about the x axis. 
Centroidal Axis & the Transfer Formula 
Example of Results of a Computer Program Available from our Free Software Section 
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