Question 15.9: Determine the deflection, υB, of the free end of a cantileve...

For this example we use the dummy load method to determine v_{\mathrm{B}} since we require the deflection at a point which does not coincide with the position of a concentrated load; thus we apply a dummy load, P_{\mathrm{f}}, at \mathrm{B} as shown. The total complementary energy, C, of the beam includes that produced by the uniformly distributed load; thus

C=\int_{0}^{L} \int_{0}^{M} \mathrm{~d} \theta \mathrm{d} M-P_{\mathrm{f}} v_{\mathrm{B}}-\int_{0}^{L} v w \mathrm{~d} x          (i)

in which v is the displacement of an elemental length, \delta x, of the beam at any distance x from the built-in end. Then

\frac{\partial C}{\partial P_{\mathrm{f}}}=\int_{0}^{L} \mathrm{~d} \theta \frac{\partial M}{\partial P_{\mathrm{f}}}-v_{\mathrm{B}}=0

so that

v_{\mathrm{B}}=\int_{0}^{L} \mathrm{~d} \theta \frac{\partial M}{\partial P_{\mathrm{f}}}       (ii)

Note that in Eq. (i) v is an actual displacement and w an actual load, so that the last term disappears when C is partially differentiated with respect to P_{\mathrm{f}}. As in Ex. 15.8

\delta \theta=-\frac{M}{E I} \delta x

Also

M=-P_{\mathrm{f}}(L-x)-\frac{w}{2}(L-x)^{2}

in which P_{\mathrm{f}} is imaginary and therefore disappears when we substitute for M in Eq. (ii). Then

\frac{\partial M}{\partial P_{\mathrm{f}}}=-(L-x)

so that

v_{\mathrm{B}}=-\int_{0}^{L} \frac{w}{2 E I}(L-x)^{3} \mathrm{~d} x

whence

v_{\mathrm{B}}=-\frac{w L^{4}}{8 E I}       (see Ex. 13.2)

For a linearly elastic system the bending moment, M_{\mathrm{f}}, produced by a dummy load, P_{\mathrm{f}}, may be written as

M_{\mathrm{f}}=\frac{\partial M}{\partial P_{\mathrm{f}}} P_{\mathrm{f}}

If P_{\mathrm{f}}=1, i.e. a unit load

M_{\mathrm{f}}=\frac{\partial M}{\partial P_{\mathrm{f}}} 1

so that \partial M / \partial P_{\mathrm{f}}=M_{1}, the bending moment due to a unit load applied at the point and in the direction of the required deflection. Thus we could write an equation for deflection, such as Eq. (ii), in the form

v=\int_{0}^{L} \frac{M_{\mathrm{A}} M_{1}}{E I} \mathrm{~d} x     (see Eq. (ii) of Ex. 15.4)       (iii)

W_{\mathrm{i}, M}=\int_{0}^{L} \frac{M_{\mathrm{A}} M_{1}}{E I} \mathrm{~d} x     (ii)

in which M_{\mathrm{A}} is the actual bending moment at any section of the beam and M_{1} is the bending moment at any section of the beam due to a unit load applied at the point and in the direction of the required deflection. Thus, in this example

M_{\mathrm{A}}=-\frac{w}{2}(L-x)^{2} \quad M_{1}=-1(L-x)

so that

v_{\mathrm{B}}=\int_{0}^{L} \frac{w}{2 E I}(L-x)^{3} \mathrm{~d} x

as before. Here v_{\mathrm{B}} is positive which indicates that it is in the same direction as the unit load.