$p=(F+1.35G_k)/A=(1600+1.35\times20\times2.5^2\times1.5)/6.25=(1600+1.35\times187.5)/6.25=(1600+253.1)/6.25=1853.1/6.25=296.5\mathrm{kPa}$。$p_{\max}=2p-p_{\min}=2\times296.5-230=363\mathrm{kPa}$。柱边处弯矩：$a_1=(2.5-0.5)/2=1.0\mathrm{m}$，$l=2.5$，$a'=0.5$，$p=p_{\min}+(b-a_1)/b\times(p_{\max}-p_{\min})=230+(2.5-1)/2.5\times133=230+1.5/2.5\times133=230+79.8=309.8\mathrm{kPa}$。$M=\frac{1}{12}a_1^2[(2l+a')(p_{\max}+p-2G/A)+(p_{\max}-p)l]$，$G/A=1.35\times20\times1.5=40.5$，$p_{\max}+p-2G/A=363+309.8-81=591.8$，代入得$M=\frac{1}{12}\times1^2[(5+0.5)\times591.8+(363-309.8)\times2.5]=\frac{1}{12}[5.5\times591.8+53.2\times2.5]=\frac{1}{12}[3254.9+133]=3387.9/12=282.3\mathrm{kN\cdot m}$，选C。