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** Introduction**

A conjugate beam is a fictitious beam that corresponds to the real beam and loaded with M/EI diagram of the real beam. For instance consider a simply supported beam subjected to a uniformly distributed load as shown in Figure 5.1a. Figure 5.1b shows the bending moment diagram of the beam. Then the corresponding conjugate beam (Figure 5.1c) is a simply supported beam subjected to a distributed load equal to the M/EI diagram of the real beam.

Module 1 Lesson 6 Fig.6.1

**Fig. 6.1.**

Supports of the conjugate beam may not necessarily be same as the real beam. Some examples of supports in real beam and their conjugate counterpart are given in Table 6.1.

**Table 6.1: Real beam and it conjugate counterpart**

Once the conjugate beam is formed, slope and deflection of the real beam may be obtained from the following relationship,

Slope on the real beam = Shear on the conjugate beam

Deflection on the real beam = Moment on the conjugate beam

**2**** Example 1**

A cantilever beam AB is subjected to a concentrated load P at its tip as shown in Figure 6.2. Determine deflection and slope at B.

**Fig. 6.2.**

**Solution**

**Fig. 6.2.**

Real beam and corresponding conjugate beam are shown in Figure 6.2. Now, from the free body diagram of the entire structure (Figure 6.3), we have

\[{B_y}=-{1 \over 2}{{Pl} \over {EI}}l=-{{P{l^2}} \over {2EI}}\]

\[{M_B}={1 \over 2}{{Pl} \over {EI}}l{{2l} \over 3}={{P{l^3}} \over {3EI}}\]

**Fig. 6.3.**

\[{\theta _B}={B_y}=-{{P{l^2}} \over {2EI}}\]

\[{\delta _B}={M_B}=-{{P{l^3}} \over {3EI}}\]

**3 Example 2**

A simply supported beam AB is subjected to a uniformly distributed load of intensity of *q* as shown in Figure 6.4. Calculate Î¸_{A}, Î¸_{B }and the deflection at the midspan. Flexural rigidity of the beam is EI.

**Fig. 6.4.**

**Solution**

Bending moment and conjugate beam are shown in Figure 6.5.

**Fig. 6.5.**

\[{A_y}={B_y}={1 \over 2} \times {\rm{Area of the parabolic load distribution }}\]

\[\Rightarrow {A_y} = {B_y}={1 \over 2} \times {2 \over 3}l{{q{l^2}} \over 8}={{q{l^3}} \over {24}}\]

Shear force of the conjugate beam at A and B are respectively as *A*_{y} and *B*_{y}.

Therefore,

\[{\theta _A}={A_y}={{q{l^3}} \over {24}}\] and \[{\theta _B}={B_y}={{q{l^3}} \over {24}}\]

Now in order to determine bending moment of the conjugate beam at the midspan the following free body diagram is considered.

**Fig. 6.6.**

Applying equilibrium condition we have,

\[M={{q{l^3}} \over {24EI}}{l \over 2}-{{q{l^3}} \over {24EI}}{{3l} \over {16}}={{5q{l^4}} \over {384EI}}\]

Since deflection at the midspan of the real beam is equal to the bending moment at the midspan of the conjugate beam, we have,

\[\delta= M = {{5q{l^4}} \over {384EI}}\]

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