连梁跨高比$l/h=2500/900=2.78\ge2.5$，$V_{\mathrm{wb}}\le\frac{1}{\gamma_{\mathrm{RE}}}(0.2\beta_c f_c b h_0)=\frac{1}{0.85}(0.2\times1\times19.1\times400\times830)=1492\mathrm{kN}$。$M_{\mathrm{bua}}=\frac{1}{\gamma_{\mathrm{RE}}}f_{\mathrm{yk}}A_s(h_0-a_s')$，$V_{\mathrm{wb}}=1.1\frac{M_{\mathrm{bua}}^l+M_{\mathrm{bua}}^r}{l_n}+V_{\mathrm{Gb}}=1.1\times\frac{\frac{1}{0.75}\times400\times A_s\times(830-70)\times2}{2500}=1492\times1000$，解得$A_s=4182\mathrm{mm^2}$，选B。