Patil vishal Prajapati Ashvin Prajapati vivek Pathak prakuti Qureshi zaid Guided by:- Keyur Shah

Terminology  Moment of inertia = second moment of area  Instead of multiplying mass by distance to the first power (which gives the first mass moment), we multiply it by distance to the second power.

1 - 4 Radius of Gyration of an Area Consider area A with moment of inertia I x. Imagine that the area is concentrated in a thin strip parallel to the x axis with equivalent I x. k x =radius of gyration with respect to the x axis

Similarly,

1 - 6 Parallel Axis Theorem Consider moment of inertia I of an area A with respect to the axis AA’ The axis BB’ passes through the area centroid and is called a centroidal axis. parallel axis theorem

1 - 7 Parallel Axis Theorem Moment of inertia I T of a circular area with respect to a tangent to the circle, Moment of inertia of a triangle with respect to a centroidal axis,

1 - 8 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 A 1, A 2, A 3,..., with respect to the same axis.

Perpendicular Axis Theorem  For flat objects the rotational moment of inertia of the axes in the plane is related to the moment of inertia perpendicular to the plane. M I x = (1/12) Mb 2 I y = (1/12) Ma 2 a b I z = (1/12) M(a 2 + b 2 )

The second moment of area is also known as the moment of inertia of a shape. The second moment of area is a measure of the 'efficiency' of a cross-sectional shape to resist bending caused by loading. Symbol is I. Units are mm4 Both beams have the same area and even the same shape. Beam 1 is stronger than Beam 2 because it has a higher second moment of area (I). Orientation can change the second moment of area (I). For a rectangle, Where b is breadth (horizontal) and h is height (vertical) if the load is vertical - e.g. gravity load.

Laminations that do not slip (Glulam) The beam is strong in bending because it is deep. Each piece of wood must be thoroughly glued to ensure they do not slip (shear) against each other. Smaller sizes are easier to dry (season) and any localized imperfections (knots, splits etc) can be carried by adjacent sections. A glulam beam is less likely to bend and warp because the individual pieces are laid up in opposite directions, cancelling out their warping tendencies. Glulam beams can also be formed in curves, and very long lengths can be achieved. Laminations that slip. In a leaf spring, the laminations form a beam, but each lamination (leaf) is designed to slip against each other. This means the second moment of area does not equal the total depth of the beam. A leaf spring with 4 leaves is 4 times as stiff as a single leaf. But if each leaf was welded together somehow, then 4 leaves would be 43 = 64 times stiffer than a single leaf. According to equation..

A single leaf. A composite (fibreglass) leaf spring. Fibreglass not as stiff as steel, yet this composite beam has less depth than a multi-leaf steel spring of the same stiffness. This is because the composite beam is one piece, so the full depth of the beam (h) goes into the second moment of area; Weight saving will be significant.

Moment of Inertia of a Rectangular Section 1. 1st Case. Moment of inertia of the rectangular section about X-X axis passing through the C.G. of the section. Fig shows a rectangular section ABCD having width = b and depth = d. Let X-X is the horizontal axis passing through the C.G. of the rectangular section. We want to determine the moment of inertia of the rectangular section about X-X axis. The moment of inertia of the given section about X-X axis is represent by IXX. Consider a rectangular elementary strip of thickness dy at a distance y from the X-X axis as shown in Fig.

2. 2 nd Case. Moment of inertia of the rectangular section about its base.

3. 3rd Case. Moment of inertia of the hollow rectangular section.

Example Q.1 Find second moment of area Ixx about axis N-N,where b=18,h=4.9 & d=6.2mm. Ic = bh3/12 (This is the natural bending about it's own centroid h/2) = 18*4.9^3/12 = mm4 I = Ic + Ad2 = (18*4.9)*(6.2^2) = mm4 (This is the forced bending about the N-N axis) Note that the Ad2 term increases I dramatically. This is a good way to increase Second Moment of Area.