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  • 1 # 使用者2586955584324

    解:

    (1)若30KN荷載單獨作用:

    ΣMA =0, YBx3.6m -30KNx1.8m =0,

    支反力YB =15KN(向上)

    在AB中點C有最大彎矩Mc =YBx1.8m =15KNx1.8m = 27KNm

    (2)若F力單獨作用:

    ΣMA =0, YB"x3.6m -Fx4.8m =0,

    支反力YB" =Fx4/3(向上)

    在AB中點C的彎矩Mc"= YBx1.8m -Fx3m =Fx(4/3)x1.8m -Fx3m = -(0.6m)xF

    (3)按彎矩疊加原理計算在AB中點C的實際彎矩:

    Mc"= Mc + Mc" = 27KNm -(0.6m)xF

    (4)梁抗彎截面模量:

    Wz = b(h^2)/6 =(0.15m)x(0.3m^2)/6 =0.00225m^3

    (5)最大拉力在梁在AB中點C處:

    若使 σmax =Mc"/Wz ≦ [σ]

    即 [27KNm -(0.6m)xF]/(0.00225m^3)≦ 8500KP

    即 F ≧ 13.125KN

    答:為使梁內應力不超過許用應力,F最小值為13.125KN

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