How is slope obtained in the conjugate beam method compared with the moment-area method?

Conjugate: shear in the conjugate beam; moment-area: change in slope = area of M/EI

Conjugate: moment of the M/EI area; moment-area: shear

Both use the moment of the M/EI area

Conjugate: area of the shear diagram; moment-area: tangent deviation

Which statement about supports is correct?

Conjugate changes supports (fixed becomes free, interior support becomes hinge); moment-area keeps the real beam

Moment-area changes supports; conjugate keeps the real supports

Neither method changes the supports

Both methods convert fixed ends to free ends

For a cantilever with end load P, why can the moment-area method give the tip deflection directly?

The tangent at the fixed end is horizontal, so the deviation equals the deflection: PL^3/(3EI)

The M/EI diagram is rectangular

The shear is constant along the beam

For a simply supported beam with a central load P, what loads the conjugate beam and what gives the mid-span deflection?

A triangular load of peak PL/(4EI); deflection = conjugate moment at mid-span = PL^3/(48EI)

A uniform load PL/4EI; deflection = shear at mid-span

A point load P/EI; deflection = reaction

A triangular load of peak PL/(4EI); deflection = conjugate shear at mid-span

2069 Bhadra · 4 marks

Moment area vs conjugate beam method

Short answer

Moment area method applies two theorems directly on the real beam using tangents, best for cantilevers. Conjugate beam method replaces the real beam with a fictitious beam having modified supports, then uses statics to find shear (slope) and moment (deflection), convenient for simply supported and multiple-point problems. Both yield identical results, e.g., cantilever end deflection PL^3/(3EI) or simply supported central load deflection PL^3/(48EI).

Explain the difference between moment area method and conjugate beam method with suitable examples.

Both methods use the $M/EI$ diagram of the beam to find slope and deflection, but they apply it in different ways.

Point	Moment-area method	Conjugate beam method
Basis	Two theorems (area and moment of $M/EI$ area)	Analogy between $M/EI$ loading and shear and moment
Slope	Change in slope = area of $M/EI$	Slope = shear in conjugate beam
Deflection	Tangent deviation = moment of $M/EI$ area	Deflection = moment in conjugate beam
Beam used	Real beam with tangents	Fictitious (conjugate) beam
Supports	No change	Changed (fixed becomes free, interior support becomes hinge, etc.)
Procedure	Needs reference tangent (horizontal at fixed end or found from deviations)	Routine statics: reactions, shear, moment
Best for	Cantilevers and quick single-point answers	Simply supported and overhanging beams, many points
Variable EI	Possible, but areas get complex	Convenient: loads become $M/EI$ with different EI per segment

Example 1: moment-area (cantilever)

Cantilever with end load $P$ : $\Delta_B = \dfrac{PL^2}{2EI}\times\dfrac{2L}{3} = \dfrac{PL^3}{3EI}$ (tangent horizontal at the fixed end).

Example 2: conjugate beam (simply supported beam, central load)

The conjugate beam is simply supported with a triangular load of peak $\dfrac{PL}{4EI}$ . The conjugate reaction (the slope at the support) is $\dfrac{PL^2}{16EI}$ and the conjugate moment at mid-span is $\dfrac{PL^3}{48EI}$ (the deflection).

Both methods give the same results.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Moment area vs conjugate beam method

Example 1: moment-area (cantilever)

Example 2: conjugate beam (simply supported beam, central load)

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