Depression of the loaded end of a heavy cantilever:
Let, the beam PQ represents a heavy cantilever of length l, fixed horizontally at the end P and loaded at the free end B with a load W. For this heavy cantilever, the weight of the beam is also effective. Let \( \omega \) be the weight density i.e., weight per unit length of the beam. In addition to the weight W at Q, the weight of the portion \( (l-x) \) of the beam also acts at the mid point or the centre of gravity of this portion at a distance \( \frac{(l-x)}{2} \) from M.
Now the total external bending moment is
\( \displaystyle{W(l-x)+\omega(l-x)\cdot\frac{(l-x)}{2}\\=W(l-x)+\frac{\omega{(l-x)}^2}{2}} \)
and the internal bending moment is \( \displaystyle{\frac{YI}{R}} \), [ Read In Detail] where \( Y \) is Young’s modulus of the material, \( R \) is the radius of curvature of the neutral axis at the point \( M \) and \( I \) is the moment of inertia.
In the equilibrium of the beam, the internal bending moment of the beam is balanced by the external bending moment.
So we can write,
\( \displaystyle{\frac{YI}{R}=W(l-x)+\frac{\omega{(l-x)}^2}{2}}\tag{1} \)
If \( d\theta \) be the angle subtended at the centre of curvature by the portion of the beam of length \( dx \) from the point \( M \), then we can write
\( \frac{1}{R}=\frac{d\theta}{dx} \).
Now putting the value of \( \frac{1}{R} \) in the equation (1) we get,
\( YI\frac{d\theta}{dx}=W(l-x)+\frac{\omega{(l-x)}^2}{2} \)
or, \( \displaystyle{d\theta=\frac{W(l-x)+\frac{\omega{(l-x)}^2}{2}}{YI}dx} \)
So the depression
\( \displaystyle{dy=(l-x)d\theta\\=\frac{W{(l-x)}^2+\frac{\omega{(l-x)}^3}{2}}{YI}dx} \)
Therefore the total depression of the loaded end below the fixed end is
\( \displaystyle{y =\frac{W}{YI}\int_0^l{(l-x)}^2{dx}\\ +\frac{\omega}{2YI}\int_0^l{(l-x)}^3{dx}} \) \( \displaystyle{\int_0^l{(l-x)}^2{dx}\\=\int_0^l{(l^2+x^2-2lx)}dx\\={[l^2x+\frac{x^3}{3}-2l\frac{x^2}{2}]}_0^l\\=(l^3+\frac{l^3}{3}-l^3)\\=\frac{l^3}{3}} \)
\( \displaystyle{\int_0^l{(l-x)}^3{dx}\\=\int_0^l{(l^3-3l^2x+3lx^2-x^3)}dx\\=(l^4-\frac{3}{2}l^4+l^4-\frac{1}{4}l^4)\\=\frac{1}{4}l^4} \)
Now, \( y=\frac{Wl^3}{3YI}+\frac{\omega{l^4}}{8YI} \)
\( W_1=\omega{l} \) is the total weight of the beam.
Therefore, the total depression of the loaded end below the fixed end is given by,
\( \displaystyle{y=\frac{Wl^3}{3YI}+\frac{W_1{l}^3}{8YI}\\=(W+\frac{3W_1}{8})\frac{l^3}{3YI}} \)\( or,\ \displaystyle{y=(W+\frac{3W_1}{8})\frac{l^3}{3YAK^2}} \)
where, \( I=AK^2 \), \( A \) is the cross sectional area of the beam and \( K \) is the raadius of gyration.