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** Introduction**

In this lesson we will study a semi-graphical method refer to as the Moment-Area method developed by Charles E. Greene for finding deflection of beam using moment curvature relation.

\[{d \over {dx}}\left( {{{dy} \over {dx}}} \right)=-{M \over {EI}}\] (5.1)

\[\Rightarrow {\left( {{{dy} \over {dx}}} \right)_A}- {\left( {{{dy} \over {dx}}} \right)_B}=\int\limits_{{x_A}}^{{x_B}} {{{Mdx} \over {EI}}}\] (5.2)

where, \[{\left( {{{dy}/ {dx}}})_A}-{\left( {{{dy} / {dx}}} \right)_B}\] , hereafter referred to as \[{\theta _{AB}}\] is the angle between tangents at A and B as illustrated in Figure 5.1a. Similarly the deflection at B with respect to tangent at A, may be written as,

\[{\delta _{AB}}=\int\limits_{{x_A}}^{xB} {xd\theta }=\int\limits_{{x_A}}^{{x_B}} {{{Mxdx} \over {EI}}}\] (5.3)

It is to be noted that \[\int\limits_{{x_A}}^{{x_B}} {{{Mxdx}/{EI}}}\] represents the statical moment with respect to B of the total bending moment area between A and B, divided by EI. Therefore equation (5.3) may also be written as,

\[{\delta _{AB}}=\left[ {\int\limits_{{x_A}}^{{x_B}} {{{Mdx} \over {EI}}} } \right]\bar x\] , (5.4)

\[\Rightarrow {\delta _{AB}}=\bar x{\theta _{AB}}\] , (5.4)

where \[\bar x\] is the centroidal distance as shown in Figure 5.1a.

**Fig. 5.1.**

Based on equations (5.2) and (5.4) the moment-area theorem may be stated as,

**Theorem 1**

The change in slope between the tangents drawn to the elastic curve at any two points A and B is equal to the area of bending moment diagram between A and B, divided by EI.

**Theorem 2**

The deviation of any point B relative to the tangent drawn to the elastic curve at any other point A, in a direction perpendicular to the original position of the beam, is equal to the moment with respect to B of the area of bending moment diagram between A and B, divided by EI

Applications of the Moment-area theorem will now be demonstrated via several examples.

**2 Example 1**

A simply supported beam AB is subjected to a uniformly distributed load of intensity of *q* as shown in Figure 5.2. Calculate the deflection at the midspan. Flexural rigidity of the beam is EI.

**Fig. 5.2.**

**Solution**

From Example 4.1, bending moment at a distance *x* from A is,

\[{M_x}={{ql} \over 2}x - {{q{x^2}} \over 2}\] (4.4)

Due to symmetry slope of the elastic line at midspan is zero. Therefore

\[{\theta _{AC}} = {\theta _A} = \int\limits_{{x_A}}^{{x_C}} {{{{M_x}dx} \over {EI}}}={1 \over {EI}}\int\limits_0^{{{l/2}}} 2{\left( {{{ql} \over 2}x - {{q{x^2}} \over 2}} \right)dx}\]

\[\Rightarrow {\theta _A}={{q{l^3}} \over {24EI}}\]

Now since δ* *may be considered as the deflection at A with respect to tangent at C, we have,

\[\delta={\theta _A}\bar x = {{5q{l^3}} \over {384EI}}\]

**3**** Example 2**

A cantilever beam AB is subjected to a concentrated load P at its tip as shown in Figure 5.3. Determine deflection and slope at B.

**Fig. 5.3.**

** ****Solution**

\[{\theta _{AB}}=\int\limits_{{x_A}}^{{x_B}} {{{{M_x}dx} \over {EI}}}={{P{l^2}} \over {2EI}}\]

Since slope at A is zero,

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

\[\delta={\theta _B}\bar x={{P{l^2}} \over {2EI}}{{2l} \over 3}={{P{l^3}} \over {3EI}}\]

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