Prof.Kodali Srinivas

The Area Moment Of Inertia or Second Moment of Area is a geometrical property of a beam and depends on a reference axis.  The Moment of Inertia of a beam's cross-sectional area measures the beams ability to resist bending. The larger the Moment of Inertia the less the beam will bend.  The smallest Moment of Inertia about any axis passes through the centroid. The following are the mathematical equations to calculate the Moment of Inertia: I x equ. (1) I y equ. (2) Where , y is the distance from the x axis to an infinitesimal  area dA. x is the distance from the y axis to an infinitesimal  area dA. Perpendicular  axis theorem  The moment of inertia of a plane area about an axis normal to the plane is equal to the sum of the moments of inertia about any two mutually perpendicular axes lying in the plane and passing through the given axis. I zz = I xx + I yy Polar Moment of Inertia The Moment Of Inertia of an area about an axis perpendicular to...

The Center of gravity, of any object is the point within that object from which the force of gravity appears to act. An object will remain at rest if it is balanced on any point along a vertical line passing through its center of gravity. In terms of moments, the center of gravity of any object is the point around which the moments of the gravitational forces completely cancel one another. So Center of gravity is the point at which a object can be suspended Or withstand in perfect equilibrium.  Centroid  of a Section The Centroid of  a two dimensional surface (such as the cross-sectional area 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 area (or figure) may represent an actual area (like a the cross-section of a beam) or a figurative diagram (like a loading diagram). The centroid of the area is to be required often in the analysis structures.  Diffe...

  I If a beam is subjected to Bending Moment 'M' as shown in Fig. Consider an element dx, the Strain Energy in the element is dU is, Problem no 1 A simply supported beam of length l carries a concentrated load W at distances of 'a' and 'b' from the two ends. Find expressions for the total strain energy of the beam and the deflection under load. Solution: The integration for strain energy can only be applied over a length of beam for which a continuous expression for M can be obtained. This usually implies a separate integration for each section between two concentrated loads or reactions. For the section AB, The bending moment M at distance 'x' from Point A is,      The Strain Energy in the section AB is, Similarly by taking a variable X measured from C,The Strain Energy stored In the Section BC is Total Strain Energy is , But if   is the deflection under the load, the strai...

Problem no: 1 Find the Slope at the supports and deflection at the center of the beam shown in fig.  The conjugate beam is also simply supported beam with M/EI diagram as a Loading diagram. There fore the Reactions at supports,  Ra and Rb = 1/EI xTotal load on conjugate beam/2  Ra = Rb = (2x4 + 2x2/2)/ 2EI = 5/EI  The Shear Force at Supports = Ra and Rb  There fore the Slope at the supports = 5 /EI   The B.M. at the Mid point in the conjugate beam = Deflection at mid point.  EI x Mc = 5x2.5 - 1/2x2x1x(2+1/3) - 2x2x1 = 6.167 yc = 6.167/EI  Problem no: 2 Determine the Slope and Deflection at free end of the cantilever beam as shown fig. S olution: The conjugate beam of the actual beam is shown in Figure 4.8(b).  A linearly varying distributed upward  elastic load  with intensity equal to zero at  A, and equal to  PL/EI  at  B.  The free-body diagram for the conj...