Question 15.8: Calculate the deflection, υB, at the free end of the cantile...

We shall assume that deflections due to shear are negligible so that v_{\mathrm{B}} is entirely due to bending action in the beam. In this case the total complementary energy of the beam is, from Eq. (15.39)

\delta\left(\sum\limits_{j=1}^{n}\int_{0}^{F_{j}}\delta_{j}\,\mathrm{d}F_{j}-\sum\limits_{k=1}^{r}\Delta_{k}\,P_{k}\right)=0       (15.39)

C=\int_{0}^{L} \int_{0}^{M} \mathrm{~d} \theta \mathrm{d} M-W v_{\mathrm{B}}       (i)

in which M is the bending moment acting on an element, \delta x, of the beam; \delta x subtends a small angle, \delta \theta, at the centre of curvature of the beam. The radius of curvature of the beam at the section is R as shown in Fig. 15.16(b) where, for clarity, we represent the beam by its neutral plane. From the principle of the stationary value of the total complementary energy of the beam

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

whence

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

In Eq. (ii)

\delta \theta=\frac{\delta x}{R}

and from Eq. (9.11)

-\frac{1}{R}=\frac{M}{E I}

(here the curvature is negative since the centre of curvature is below the beam) so that

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

Substituting in Eq. (ii) for \delta \theta we have

v_{\mathrm{B}}=-\int_{0}^{L} \frac{M}{E I} \frac{\partial M}{\partial W} \mathrm{~d} x      (iii)

From Fig. 15.16(a) we see that

M=-W(L-x)

Hence

\frac{\partial M}{\partial W}=-(L-x)

Note: Equation (iii) could have been obtained directly from Eq. (9.21)

U=\int_{L}{\frac{M^{2}}{2E I_{z}}}\,\mathrm{d}x        (9.21)

by using Castigliano’s first theorem (Part II).

Equation (iii) then becomes

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

whence

v_{\mathrm{B}}=-\frac{W L^{3}}{3 E I}       (as in Ex 13.1)

(Note that v_{\mathrm{B}} is downwards and therefore negative according to our sign convention.)