Author: haroonkhan

Consider an element of the arc length rdθ and width dr of the circular disc as shown in Figure 11.19. Now, the moment of inertia of the element about diametral axis x−x is given by, Figure 11.19 Moment of Inertia of a Circular Disc about Its Diametral Axis  IXX = y2 dA = (r sin θ)2 r dθ dr = r3 sin2 θ dθ dr   Now, moment of inertia of entire circle about diametral…

A. Moment of Inertia of a Rectangle Consider an elemental strip of width dy at distance of y from a centroidal axis of a rectangle as shown in Figure 11.17. Figure 11.17 Moment of Inertia of a Rectangle about Centroidal Axis Moment of inertia of the strip is given by,   IXX = y2 dA = y2bdy   Now, moment of inertia of entire rectangle can…

Theorem of parallel axis states that if the moment of inertia of a plane area about an axis in the plane of the area through centroid be represented by IG, then the moment of inertia of the plane about a parallel axis AB (IAB) in the plane at a distance h from the centroid of the area is…

Theorem of perpendicular axis states that if IXX and IYY be the moment of inertia of a plane section about two mutually perpendicular axes X−X and Y−Y in the plane of the section (as shown in Figure 11.15), then the moment of inertia of the section IZZ about the axis Z−Z, perpendicular to the plane and passing through the intersection of axes X−X and Y−Y is given by, Figure 11.15 Perpendicular Axis…

Radius of gyration of a body about an axis is a distance such that its square multiplied by the area gives moment of inertia of the area about the given axis.

Second moment of area is also known as area moment of inertia. Consider a small lamina of area A as shown in Figure 11.14. The second moment of area about x-axis and y-axis can be found by integrating the second moment of area of small element of area dA of the lamina, i.e., ∫ x2 dA and ∫ y2 dA, respectively. The product of the area and square of…

Consider a triangle ABC of base b and height h as shown in Figure 11.13. Let us locate the centroid of the triangle from its base. Let b1 be the width of an elemental strip of thickness dy at a distance y from the base. Since ΔAEF and ΔABC are similar triangles, therefore, Figure 11.13 Centroid of a Triangle Thus, the centroid of a triangle is at a distance h/3 from the base…

Considering a parabolic section of height h and base b as shown in Figure 11.12. Now to find the centroid of this section consider a small element of width dx at a distance of x from the origin O. Figure 11.12 Centroid of a Parabolic Section

Consider a sector of a circular disc of angle 2α as shown in Figure 11.11. Due to symmetry, centroid ‘G’ lies on x-axis. To find its distance from the centre O, consider an elemental area as shown in Figure 11.11. Figure 11.11 Centroid of a Sector of a Circular Disc Now,

Considering a semicircle of radius R as shown in Figure 11.10. Due to symmetry centroid must lie on y-axis. Let its distance from the x-axis be . To find , consider an element at a distance r from the centre O of the semicircle, radial width dr, and bound by radii at θ and θ + dθ. Figure 11.10 Centroid of Circular Section of a Disc Area of the element = rdθ dr. Its…