Cantilever Beams

The stress is given by: $S = \displaystyle\frac{W}{2\;Z\;l}\;\;(l - x)^2$

The Stress at the Support: $S_s = \displaystyle\frac{W\;l}{2\;Z}$

If the cross section is Constant then this is the maximum stress.

The Deflection is given by: $y = \displaystyle\frac{W\;x^2}{24\;E\;I\;l}\;\;[2l^2 + (2l - x)^2]$

The Maximum deflection is at the end and is: $\hat{y} = \displaystyle\frac{W\;l^3}{8\;E\;I}$

The Stress is given by: $S = \displaystyle\frac{W}{Z}\;\;(l - x)$

The Stress at the Support: $S = \displaystyle\frac{W\;l}{Z}$

If the Cross-Section is Constant, then this is the Maximum Stress.

The Deflection at any point is given by: $y = \displaystyle\frac{W\;x^2}{6\;E\;I}\;\;(3\,l - x)$

The Maximum Deflection is at the end and is: $\hat{y} = \displaystyle\frac{W\;l^3}{3\;E\;I}$

Between the Fixed End and the Load: $\displaystyle S = \frac{W}{Z}\;\;(l - x)$

Beyond the Load the stress is zero.

At the Fixed End: $\displaystyle S = \frac{W\;l}{Z}$

If the Cross-section is Constant, then this is the Maximum Stress.

The Deflection of any point between the Fixed end and the Load is: $\displaystyle y = \frac{W\;x^2}{6\;E\;I}\;\;(3\;l - x)$

Beyond the load the Deflection is: $\displaystyle y = \frac{W\;l^2}{6\;E\;I}\;\;(3\;x- l)$

The Maximum Deflection at the "Free" end is: $\displaystyle y_{max} = \frac{W\;l^2}{6\;E\;I}\;\;(2\;l + 3\;b)$

Deflection at the Load: $\displaystyle y = \frac{W\;l^3}{3\;E\;I}$

The Stress at any point is : $S = \displaystyle\frac{W\;(l - x)}{2\;Z\;l}\;\;\left(\displaystyle\frac{1}{4}\;l - x \right)$

The Maximum Stress at the Fixed End: $S = \displaystyle\frac{Wl}{8Z}$

The Stress is zero at $\displaystyle x = \frac{1}{4}\;l$. The greatest negative Stress is at $\displaystyle x = \frac{5}{8}\;l$ and is: $S= - \frac{9}{128}\;\;\frac{Wl}{Z}$

The deflection is given by: $y = \displaystyle\frac{W\;x^2(l - x)}{48EI\;l}\;\;(3\;l - 2\;x)$

The maximum Deflection is at x = 0.5785 l and is: $\hat{y} = \displaystyle\frac{Wl^3}{185\;EI}$

The Deflection at the centre is: $y_c = \displaystyle\frac{Wl^3}{192\;EI}$

The Deflection at the point of greatest negative Stress, i.e. at $x = \displaystyle\frac{5}{8}l$, is: $y = \frac{Wl^3}{187\;EI}$

The Stress between the Fixed Point and the Load: $S = \displaystyle\frac{W}{16\;Z}\left(3\;l - 11\;x \right)$

Between the Support and the Load: $S\;= - \displaystyle\frac{5}{16}\;\displaystyle\frac{W\;v}{Z}$

Stress at the Fixed end. This is the maximum value $S = \displaystyle\frac{3}{16}\;\displaystyle\frac{W\;l}{Z}$

The Stress is Zero at $\displaystyle x = \frac{3}{11}\;l$

The Greatest negative Stress is at the centre and is $\displaystyle - \frac{5}{32}\;\;\frac{W\;l}{Z}$

The Deflection of any Point between the Fixed End and the Load:

$y = \frac{W\;x^2}{96\;E\;I}\;\;(9\;l - 11\;x)$

The Deflection of any Point between the Support and the Load $y = \frac{W\;v}{96\;E\;I}\;\;(3\;l^2 - 5\;v^2)$

The Maximum Deflection is at $\displaystyle v = 0.4472\;l$ $\hat{y} = \frac{W\;l^3}{107.33\;E\;I}$

The Deflection of the Load is: $\displaystyle\frac{7}{768}\;\;\displaystyle\frac{W\;l^3}{E\;I}$

The Stress at any Point $S = \displaystyle\frac{W\;l}{Z}\;\;\left\{\displaystyle\frac{1}{3} - \displaystyle\frac{x}{l}\;+\displaystyle\frac{1}{2}\;\left(\displaystyle\frac{x}{l} \right)^2 \right\}$

The Maximum Stress is at the support and is $\displaystyle \frac{W\;l}{3\;Z}$

The Stress is zero when $\displaystyle x = 0.4227\;l$

The Greatest negative stress is at the free end and is $\displaystyle - \frac{W\;l}{6\;Z}$

The Deflection at any Point is given by: $y = \displaystyle\frac{W\;x^2}{24\;E\;I\;l}\;\;(2l - x)^2$

The Maximum Deflection is at the free end and is: $\hat{y} = \displaystyle\frac{W\;l^3}{24\;E\;I}$

Stress at any Point: $S = \displaystyle\frac{W}{Z}\;\;\;\left(\displaystyle\frac{1}{2}l - x \right)$

The Stress at the Support $S_s = \displaystyle\frac{W\;l}{2\;Z}$

The Stress at the Free End: $S_f\;= - \displaystyle\frac{W\;l}{2\;Z}$

These are the Maximum Stresses and are equal and opposite. The Stress is zero at the Centre.

The Deflection at any Point is: $y = \displaystyle\frac{W\;x^2}{12\;E\;I}\;\;(3l - 2x)$

The Maximum Deflection is at the Free End and is: $\hat{y} = \displaystyle\frac{W\;l^3}{12\;E\;I}$