Chapter 2
Q. 2.5
Determine the deflection at the free end of the cantilever shown in Figure 2.5.
Step-by-Step
Verified Solution
The origin of coordinates is taken at the free end and the functions M and m derived from (i) and (ii) as:
M = Wx
and m = x
The deflection at the free end is given by:
\begin{aligned}1 \times \delta &=\int_{0}^{l} M m \mathrm{~d} x / E I \\&=W \int_{0}^{l} x^{2} \mathrm{~d} x / E I \\&=W l^{3} / 3 E I\end{aligned}Alternatively, the solid defined by the functions M and m is shown at (iii); its volume is:
W l^{2} / 2 \times 2 l / 3=W l^{3} / 3and the deflection at the free end is given by:
\delta=W l^{3} / 3 E I .Alternatively, from Table 2.3, the value of \int Mm dx/EI is given by:
\begin{aligned}\delta &=l a c / 3 E I \\&=W l^{3} / 3\end{aligned}| Table 2.3 Volume integrals | |||||
| M
m |
 |
 |
 |
 |
 |
 |
lac | lac/2 | la(c + d)/2 | lac/2 | la(c + 4d + e)/6 |
 |
lac/2 | lac/3 | la(2c + d)/6 | lac (1 + β)/6 | la(c + 2d)/6 |
 |
lac/2 | lac/6 | la(c + 2d)/6 | lac (1 + α)/6 | la (2d + e)/6 |
 |
lc (a + b)/2 | lc(2a + b)/6 | la(2c + d)/6 + lb (c + 2d)/6 |
lac (1 + β)/6 + lbc (1 + α)/6 |
la(c + 2d)/6 + lb(2d + e)/6 |