请教下高手们，节点最小极限受剪承载力，这个问题怎么解决。

四. 按抗震规范8.2.8条进行梁柱连接的极限承载力验算

        Mpx = 229377570.00N·mm

    1. 梁柱连接极限受剪承载力验算

        (1). 连接板和柱翼缘的连接焊缝抗剪极限承载力

            Vu1 = 0.7*hf*(H - 2hf)*0.58*fu*2

            = 0.7*5*(266 - 2*5)*0.58*470*2=488.50kN

        (2). 腹板与梁连接（螺栓或焊缝）抗剪极限承载力

            Nvub = 0.58*nv*Ae*fub

            =0.58*1*245*1040 = 147.66kN

            Ncub = d*ΣT*1.5*fu

            =20*6*1.5*470 = 84.60kN

            取较小者            Nub = 84.60kN

            Vu2 = Nub×n=507.60kN

        (3). 连接板净截面抗剪极限承载力

            Vu3 = (H - n*d0)*T*0.58*fu

            =(266 - 3*24)*10*0.58*470 = 528.84kN

        (4). 梁腹板净截面抗剪极限承载力

            Vu4 = (Hw - n*d0)*Tw*0.58*fu

             =(334 - 3*24)*6*0.58*470 = 428.53kN

        节点最小极限受剪承载力：            Vu = min{Vu1, Vu2, Vu3, Vu4} = 428.53kN

            Vp=1.20*(∑Mp/ln) + Vgb(未包含竖向地震)

              =1.20*(229.38/1.15) + 0.45=479.15KN

            Vu =     428.53 KN    <     Vp      (******不满足******)