Problems in Conjugate Beam

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.

Solution: 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 conjugate beam is shown in Figure. The reactions at A of the conjugate beam are given by**


	The slope at A , and the deflection at the free end A of the actual beam in Fig.(d) are respectively, equal to the “shearing force” and the “bending moment” at the fixed end A of the conjugate beam in Figure (d)


	Note that points downward because causes tension in bottom fiber of the beam at A (i.e. sagging moment) . Problem no :3 Find the Deflection at the free end of the over hanging beam as shown in fig.3 Solution: The corresponding conjugate beam and loading are shown in Figure. The loading is upward linearly distributed load with maximum value of at B = Taking moment about point B, the vertical reaction at A in the conjugate beam is given by The bending moment at C (by taking moment about C ) is given by (sagging B.M.) Hence, the deflection of point C will be equal to in the downward direction.

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