Question 15.10: Determine the deflection of the free end of the cantilever b...

The temperature, t, at any section x of the beam is given by

t=\frac{x}{L} t_{0}

Thus, substituting for t in Eq. (15.43),

\Delta_{\mathrm{{Te.B}}}=-\int_{0}^{L}M_{1}\,{\frac{\alpha t}{h}}\,\mathrm{d}x       (15.43)

which applies since the variation of temperature through the depth of the beam is identical to that in Fig. 15.19(b), and noting that M_{1}=-1(L-x) we have

\Delta_{\mathrm{Te}, \mathrm{B}}=-\int_{0}^{L}[-1(L-x)] \frac{\alpha}{h} \frac{x}{L} t_{0} \mathrm{~d} x

which simplifies to

\Delta_{\mathrm{Te}, \mathrm{B}}=\frac{\alpha t_{0}}{h L} \int_{0}^{L}\left(L x-x^{2}\right) \mathrm{d} x

whence

\Delta_{\mathrm{Te}, \mathrm{B}}=\frac{\alpha t_{0} L^{2}}{6 h}