__What is Moment Area Method?__

* Moment area method,* is a very convenient way to find

*slope*and

*deflection*of a beam.

It is based on * Mohr’s theorem* for use of

*bending moment diagram*and enables a quicker solution to find slope and deflection of a beam.

These theorems are stated as follows –

- Mohr’s theorem – 1
- Mohr’s theorem – 2

__Mohr’s Theorem – 1__

Mohr’s theorem – 1 can be stated as –

*Change of slope between any two points on the elastic curve of a beam is equal to the area of bending moment diagram between these points divided by * ( EI ) *.*

__PROOF –__

Consider about a loaded beam AB whose BM diagram is shown in figure. Now, consider a small element PQ of length ( dx ) of the beam which is at a distance ( x ) from end B .

In figure AP'Q'B is elastic curve of deflected beam. P' and Q' are the corresponding points of element PQ on elastic curve.

Let, ( R ) is the radius of curvature of bent beam and suppose arc P'Q' subtends an angle ( d \theta ) at the centre of curvature. Then, length of P'Q' will be ( R d \theta )

Since, bending occurs very slightly, so ( P'Q' ) = ( PQ ) = ( dx ) . Therefore, \quad dx = ( R d \theta )

Or, \quad d \theta = \left ( \frac {dx}{R} \right ) …….. (1)

In previous article, from *equation of bending stress*, we have got the relation –

\left ( \frac {M}{I} \right ) = \left ( \frac {E}{R} \right )

Or, \quad R = \left ( \frac {EI}{M} \right )

Putting this in equation (1) we get –

d \theta = \left ( \frac {dx}{R} \right )

= \left ( \frac {Mdx}{EI} \right )

= \left ( \frac {1}{EI} \right ) M dx ……. (2)

Hence, total change in slope from point A to B , is find by integrating this equation between the limits ( x = l ) to ( x = 0 )

Therefore, expression for the slope will be –

i = \left ( \frac {1}{EI} \right ) \int\limits_{0}^{l} M dx

But, product ( M dx ) represents the area of bending moment diagram shown as shaded area in figure.

Therefore, slope will be –

i = \left ( \frac {\text {Area of BM diagram between A and B}}{EI} \right )

__Mohr’s Theorem – 2__

Mohr’s theorem – 2 can be stated as –

*Intercepts taken on a vertical reference line of tangents at two points on the elastic curve of a beam, is equal to the moment of area of BM diagram about same reference line divided by * ( EI ) .

__PROOF –__

Consider about the BM diagram of the loaded beam as shown in figure. Let P'J and Q'K are two tangents from points P' and Q' respectively, which intersect at points J and K on a vertical reference line drawn from point B .

From geometry of the figure, the tangents at P' and Q' will also meet at an angle ( d \theta ) .

Therefore, length of intercept is –

JK = x \ d \theta = x \left ( \frac {M dx}{E I} \right ) = \left ( \frac {M x dx}{EI} \right )

Total intercept is then find by integrating the above equation between the limits ( x = l ) to ( x = 0 )

Therefore, *deflection* will be –

y = \left ( \frac {1}{EI} \right ) \int\limits_{0}^{l} M x dx

But, term ( M x dx ) represents the *moment* of area of the bending moment diagram.

Therefore, \quad y = \left ( \frac {\text {Moment of area of BM diagram between A and B}}{EI} \right )

We will learn in detail about the procedure of application of Mohr’s theorem, to find slope and deflection in beams in following numerical problems.

See numerical problems based on this article.

__Conjugate Beam Method__

*A conjugate beam is defined as an imaginary beam whose – (1) length is equal to the length of original beam (2) width is equal to \left ( \frac {1}{EI} \right ) times the width of original beam (3) loaded with the bending moment diagram of the original beam.*

Concept of conjugate beam is used to find the *slope* and *deflection* of original beam in a more simpler and convenient way. It is a modified form of “moment area method”, discussed above.

In conjugate beam method, modified Mohr’s theorems are used which may be stated as –

__Mohr’s Theorem – 1 (For Conjugate beam)__

*Shear force at any section of a conjugate beam is equal to the slope of elastic curve at the corresponding section of original beam.*

__Mohr’s Theorem – 2 (For Conjugate beam)__

*Bending moment at any section of a conjugate beam is equal to the deflection of elastic curve at the corresponding section of original beam.*

We will learn in detail about the procedure of application of “Mohr’s theorem for conjugate beam” to find slope and deflection in numerical problems.

__Features of Conjugate beam__

Comparative features of an actual beam and its corresponding conjugate beam is tabulated below.

Sl. No. | Actual beam. | Conjugate beam. | Remarks. |

1 | Fixed end. | Free end. | Slope and deflection at fixed end of actual beam is zero. SF \ \& \ BM at free end of conjugate beam is zero. |

2 | Free end. | Fixed end. | Slope and deflection at free end of actual beam is zero. SF \ \& \ BM at fixed end of conjugate beam is zero. |

3 | Simply supported or roller supported end. | Simply supported end. | Slope at free end of actual beam exists but deflection is zero. SF at simply supported end of conjugate beam exists but BM is zero. |

See numerical problems based on this article.