\(-\frac{WL^3}{8EI} \quad \text{and} \quad -\frac{WL^4}{8EI}\)
A cantilever beam is a structural element fixed at one end and free at the other. When a cantilever beam is subjected to external loads, it bends, leading to deformation. This deformation is quantified by the slope (rotation of the beam's neutral axis) and deflection (vertical displacement of the beam's neutral axis) at various points along its length. The extent of this bending is inversely proportional to the beam's flexural rigidity, EI, which is a measure of its resistance to bending.
In this problem, we are considering a cantilever beam of span L subjected to a uniformly distributed load (UDL) of intensity W over its entire length. We need to find the slope (\(\theta\)) and deflection (\(\delta\)) specifically at the free end, where these values are typically maximum.
Using fundamental principles of structural mechanics, such as the double integration method or Macaulay's method, the slope and deflection of beams can be determined. For a cantilever beam of length L subjected to a uniformly distributed load of intensity W over its entire span, the standard formulas for the slope and deflection at the free end (relative to the fixed end) are:
The negative signs in these formulas indicate the direction of the slope and deflection. Assuming the load W is acting downwards, the slope at the free end is a clockwise rotation (often considered negative in standard conventions), and the deflection is downwards (also typically considered negative).
The question presents four options for the values of slope \(\theta\) and deflection \(\delta\) at the free end:
Let's compare the standard formulas derived from beam theory with the expressions given in the options:
Based on standard beam theory, Option 4 gives the correct pair of values for slope and deflection at the free end of a cantilever beam under uniformly distributed load.
The provided correct answer text corresponds to Option 1, which states the slope is \(-\frac{WL^3}{8EI}\) and the deflection is \(-\frac{WL^4}{8EI}\).
Upon analyzing Option 1:
While standard engineering principles yield a different slope formula, the provided correct answer indicates that Option 1 contains the expected results. Therefore, we note that the deflection formula in Option 1 aligns with standard theory, although the slope formula presented in Option 1 does not.
| Loading Case | Location | Slope (\(\theta\)) | Deflection (\(\delta\)) |
|---|---|---|---|
| Concentrated Load P at Free End | Free End | \(-\frac{PL^2}{2EI}\) | \(-\frac{PL^3}{3EI}\) |
| Uniformly Distributed Load W over Entire Length | Free End | \(-\frac{WL^3}{6EI}\) | \(-\frac{WL^4}{8EI}\) |
Note: Formulas show the standard results with sign conventions (negative for downward deflection/clockwise slope).
When once a pocket of smoke, containing air pollutants, is released into the atmosphere from a source like an automobile or a factory chimney, it gets dispersed into the atmosphere into various directions depending upon the
1. prevailing winds
2. temperature
3. pressure conditions
Select the correct answer.
During the compaction test, the weight of compacted soil specimen along with mould is 38.2 N. The volume and weight of mould are 0.95×10-3 m³ and 20.5 N respectively and the water content is 12%. The dry unit weight of the compacted specimen will be nearly