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
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