Question

钢筋混凝土结构计算题求助

已知钢筋混凝土矩形梁，处于二类环境，承受弯矩设计值M=160kN×m，采用C40混凝土和HRB335级钢筋，试按正截面承载力要求确定截面尺寸及纵向钢筋截面面积。

Answer 1

混凝土C40
梁宽300mm
梁高500mm
纵筋合力作用点至梁距离35
弯矩160KN* M
受拉区钢筋mm 1232.467

Answer 2

肯定是电大的题,还用C15的混凝土,HPB235的钢筋!真是笑死人了!
1．主筋
1．1
相对界限受压区高度
b
cu
=
0.0033
-
(fcu,k
-
50)×10-5
=
0.0033
-
(15
-
50)×10-5
=
0.0037
>
0.0033
取
cu
=
0.0033
按规范公式(7.1.4-1)
b
=
11
+
fyEscu
=
0.801
+
2102.1×105×0.0033
=
0.61
1．2
受压区高度x
按规范公式(7.2.1-1),
A's
=
0,
A'p
=
0
M
=
1fcbxh0
-
x2
x
=
h0
-
h02
-
2M1fcb
=
565
-
5652
-
2×135.00×1061.00×7.20×250
=
154mm
<
bh0
=
0.61×565
=
346.88mm,
按计算不需要配置受压钢筋
1．3
受拉钢筋截面积As
按规范公式(7.2.1-2)
1fcbx
=
fyAs
得
As
=
1fcbxfy
=
1.00×7.20×250×154210
=
1317mm2
1．4
验算配筋率
验算最小配筋率时取
=
Asbh
=
1317250×600×100%
=
0.88%
min
=
0.200%min
=
0.45ft/fy
=
0.45×0.91/210
=
0.195%min
=
0.200%
<

满足最小配筋率要求
实配受拉钢筋(梁底):
3A25,
As
=
1473mm2
受压钢筋按照规范第10.2.15条对架立钢筋的要求,
配置2A10
A's
=
2××52
=
157mm2
实配受压钢筋(梁顶):
2A12,
A's
=
226mm2

Answer 3

fy=M/0.9a1*FC*b*h0

很好算 自己算算 你可以有多种答案