Cantilever Beam

Generally, the tangential deviation t is not equal to the beam deflection. In cantilever beams, however, the tangent drawn to the elastic curve at the wall is horizontal and coincidence therefore with the neutral axis of the beam. The tangential deviation in this case is equal to the deflection of the beam as shown below.

From the figure above, the deflection at B denoted as δB is equal to the deviation of B from a tangent line through A denoted as tB/A. This is because the tangent line through A lies with the neutral axis of the beam.

Simply Supported Beam

The deflection δ at some point B of a simply supported beam can be obtained by the following steps:

1. Compute $t_{C/A} = \dfrac{1}{EI}(Area_{AC}) \, \bar{X}_C$

2. Compute $t_{B/A} = \dfrac{1}{EI}(Area_{AB}) \, \bar{X}_B$

3. Solve δ by ratio and proportion (see figure above).

$\dfrac{\delta + t_{B/A}}{x} = \dfrac{t_{C/A}}{L}$

Midspan Deflection in Simply Supported Beams

In simply supported beams, the tangent drawn to the elastic curve at the point of maximum deflection is horizontal and parallel to the unloaded beam. It simply means that the deviation from unsettling supports to the horizontal tangent is equal to the maximum deflection. If the simple beam is symmetrically loaded, the maximum deflection will occur at the midspan.

Finding the midspan deflection of a symmetrically loaded simple beam is straightforward because its value is equal to the maximum deflection. In unsymmetrically loaded simple beam however, the midspan deflection is not equal to the maximum deflection. To deal with unsymmetrically loaded simple beam, we will add a symmetrically placed load for each load actually acting on the beam, making the beam symmetrically loaded. The effect of this transformation to symmetry will double the actual midspan deflection, making the actual midspan deflection equal to one-half of the midspan deflection of the transformed symmetrically loaded beam.

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