附加压力增量$\Delta p_0=\frac{2136-1080}{4.8\times4}=55\mathrm{kPa}$。$l/b=2.4/2=1.2$，$z_1/b=4/2=2$，$z_2/b=(4+3.6)/2=3.8$，查附录K.0.1-2得$\bar{\alpha}_1=0.1822$，$\bar{\alpha}_2=0.1234$。$s=\psi_s\frac{p_0}{E_s}(z_2\bar{\alpha}_2-z_1\bar{\alpha}_1)=0.69\times55\times4\times(\frac{4\times0.1822}{8}+\frac{7.6\times0.1234-4\times0.1822}{10})=17\mathrm{mm}$，选B。