Prof.Kodali Srinivas

Moment - Area Method - Mohr's  Theorems  Otto Mohr (1835-1918)  German Civil Engineer In this method, to determining the slopes and deflections in beams involves the Area and Moment of the ' Bending moment diagram' .   ( B.M.D ) First Theorem : The angle between tangents drawn at any two points on deflected curve, is equal to area of the M/EI diagram between the two points. The angle between the tangents from points A and B  O  AB =  1/EI (Area of AB) Second Theorem : The intercept on a vertical line made by two tangents drawn at the two points on the deflected curve is equal to the moment of the M/EI diagram between two points about the vertical line. The intercepts at point A,and B are, t A/B = Z AB  = 1/EI (Area of AB). X A   and t   B/A  =  Z  BA   =   1/EI (Area of AB). X B Sign Convention : For +ve B.M. ,the area of  M/EI diagram is consider...

The deformation of a beam is usually expressed in terms of its deflection from its original unloaded position.  The deflection is measured from the original neutral surface of the beam to the neutral surface of the deformed beam. The configuration assumed by the deformed neutral surface is known as the elastic curve of the beam.  The angle through which the cross-section rotates with respect to the original position is called the angular rotation of the section.  Generalized  Deflection equation Consider a small element dx in deflected beam as shown below. In Cartesian coordinates, the radius of curvature(R) of a curve and deflection y = f(x) is have the relation given below,  The deflection is very small ,the slope of the curve dy/dx is also very small and squaring of this we may get a negligible valve and may neglect in the above equation.  1/R = d2y/dx2 In the derivation of flexure formula, the radius of curvature of a beam is given as  1/R = M /EI...

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...

A conjugate beam is a fictitious beam that corresponds to the real beam and loaded with M/EI diagram of the real beam so that the shear and moment at any point of the conjugate beam are equal, respectively, to the slope and deflection at the corresponding point on the real beam. The conjugate beam method was developed by Christian Otto Mohr (1835 – 1918), a German civil engineer and one of the most celebrated civil engineer in 19th century. The method is based on the analogy between the relationships among load, shear, and bending moment and the relationships among M/EI, slope, and deflection. Slope on real beam = Shear on conjugate beam  Deflection on real beam = Moment on conjugate beam   Properties of Conjugate Beam The length of a conjugate beam is always equal to the length of the actual beam. The load on the conjugate beam is the M/EI diagram of the loads on the actual beam. A simple support for the real beam remains simple support for the conju...

The deflection  δ  at some point B of a simply supported beam can be obtained by the following steps:   1. Compute   = Z CA 2. Compute   = Z BA 3. Slope at support A =  O   A =  t  CA /L = Z CA / L 4. Slope at support C = O C =  t CA/ L = Z CA / L 5. Solve  δ  by ratio and proportion (see figure above). Problem:1 The middle half of the beam shown in Fig. has a moment of inertia 1.5 times that of the rest of the beam. Find the midspan deflection.  (Hint: Convert the M diagram into an M/EI diagram.)     Solution :     Therefore,             →  answer Problem : 2 Determine the value of EI δ  at the right end of the overhanging beam shown in Fig.      Solution :       ...