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## Q. 2.6

Determine the rotation at the free end of the cantilever shown in Figure 2.6. ## Verified Solution

The functions M and m are derived from (i) and (ii) as:

M = wx²/2

and m = 1

The rotation at the free end is given by:

\begin{aligned}1 \times \theta &=\int_{0}^{l} Mm dx /E I \\&=w \int_{0}^{l} x^{2} dx / 2 E I \\&=w l^{3} / 6 E I .\end{aligned}

Alternatively, the solid defined by these functions is shown at (iii); its volume is:

$W l^{2} \times l / 6=w l^{3} / 6$

and the rotation at the free end is given by:

$\theta=w l^{3} / 6 E I .$

From Table 2.3, the value of $\int$ Mm dx/EI is given by:

\begin{aligned}\theta &=l(c+4 d+e) / 6 E I \\&=l\left(w l^{2} / 2+w l^{2} / 2+0\right) / 6 E I \\&=w l^{3} / 6 E I\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