Question

工程力学截面法求此题过程。。

Answer 1

(1)计算支座反力：
ΣFx =0, -FAx +(100KN)(3/5) =0,
FAx =60KN(←)
ΣMB =0, -FAy.4m +(100KN)(4/5)(2m) =0
FAy = 40KN(↑)
ΣFy =0, FBy +40KN -(100KN)(4/5) =0
FBy =40KN(↑)
.
(2)计算C稍左截面内力，取当截面之左梁段为隔离体：
轴力FNLc =|FAx| =60KN(拉力)
剪力FQLc =|FAy| =40KN
弯矩MLc =|FAy.2m| =40KN.2m =80KN.m(逆时针方向）
.
(3)计算C稍右截面内力，取当截面之右梁段为隔离体：
轴力FNRc =|FBx| =0
剪力FQRc = -｜FBy｜= -40KN
弯矩MRc =｜FBy.2m｜=40KN.2m =80KN.m(逆时针方向）
.