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上接本刊1965午第1期。 y_0A_3+_0/αB_3+D_3/(α~3EI)=-C/(αE)·(Iocn)/I(y_0A_2+_o/aB_2+D_2/(α~3EI))(34)令 C/(αE)·(Iocn)/I=Kh簡化(34)式,得 y_0(A_3+K_hA_2)+o/α(B_3+K_hB_2)+1/(α~3EI)(D_3+K_hD_2)=0 (35) 令(28)式中Q=Q_h=0(其中A_4、B_4和D_4按z=h由前述表查用)則 y_0A_4+o/α+D_4/(α~3EI)=0 (36)联解(35)式和(36)式,并采取y_0=δнн及_0=-δmн,便可求得:δHH=1/(α~3EI)·((B_3D_4-B_4D_3)+Kh(B_2D_4-B_4D_2))/((A_3B_4-A_4B_3)+K_h(A_2B_4-A_4B_2))(37)δMH=1/(α~2EI)·((A_3D_4-A_4D_3)+K_h(A_2D_4-A_4D_2))/((A_3B_4—A_4B_3)+K_h(A_2B_4-A_4B_2))(38) 用类似方法可以求出当在基础地面处作用单位力矩Mo=1时基础地面处的水平位移δnm和轉角位移δmm的式予(图4b):δHm=1/(α~2EI)·((B_3C_4-B_4C_3)+K_h(B_2C_4-B_4C_2))/((A_3B_4-A_4B_3)+Kh(A_2B_4-A_4B_2))(39)δmm=1/(αEI)
On the first issue of the publication 1965 no. y_0A_3+_0/αB_3+D_3/(α~3EI)=-C/(αE)·(Iocn)/I(y_0A_2+_o/aB_2+D_2/(α~3EI)) (34) Order C/(αE) · (Iocn)/I=Kh Simplify (34), get y_0(A_3+K_hA_2)+o/α(B_3+K_hB_2)+1/(α~3EI)(D_3+K_hD_2)=0 (35) 28) where Q = Q_h = 0 (where A_4, B_4, and D_4 are searched by the preceding table by z = h) then y_0A_4+o/α+D_4/(α~3EI) = 0 (36) concatenation (35) Equations (36) and y_0=δнн and _0=-δmн can be obtained by: δHH=1/(α~3EI)·((B_3D_4-B_4D_3)+Kh(B_2D_4-B_4D_2))/( (A_3B_4-A_4B_3)+K_h(A_2B_4-A_4B_2)) (37) δMH=1/(α~2EI)·((A_3D_4-A_4D_3)+K_h(A_2D_4-A_4D_2))/((A_3B_4_A_4B_3)+K_h( A_2B_4-A_4B_2)) (38) Using a similar method, we can find the formula for the horizontal displacement δnm and the displacement δmm at the base ground when the unit moment of action Mo = 1 at the base ground (Fig. 4b): δHm = 1/( α~2EI)·((B_3C_4-B_4C_3)+K_h(B_2C_4-B_4C_2))/((A_3B_4-A_4B_3)+Kh(A_2B_4-A_4B_2))(39)δmm=1/(αEI)