筏板基础及侧壁计算书Word文档下载推荐.docx
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ly 5.175
l
①lx=8.400m,ly=5.175m, =
x
8.4
=0.62
E土=rhKa
=8.0×
5.175×
tan245o=41.4KN/m
E水=rh=10.0×
3.475=34.75KN/m
E合=1.27E土+1.27E水=52.6+44.1=96.7KN/m
查静力计算手册,得:
2
Mxmax=0.0072ql
=0.0072×
96.7×
5.1752
=18.6KN·
m
Mymax=0.0209ql
'
=0.0209×
5.175
=54.1KN·
Mxmax
=−0.0354ql
=0.0354×
=−91.7KN·
My=−0.0566ql
=−0.0566×
=−146.6KN·
配筋计算:
取弯矩最大处进行计算。
即取M=146.6kN·
1
混凝土强度等级为C40,fc
α
=1.0;
受力钢筋均采用
HRB400
级,fy=360N/mm2;
ξ
=0.523;
ρ
min
=0.2%;
b
相对受压区高度:
x=h0(1−
1− 2M )
α1fcbh0
=310×
(1−
1− 2×
146600000 )
1.0×
19.1×
1000×
3102
=25.8mm<
xb=ξbh0=0.523×
302=157.9mm
则
As=
α1fcbxfy
=1.0×
25.8
360
=1368.8mm2
实际配筋:
内外均φ16@150
A=1340mm2
s
②lx=7.800m,ly=5.175m, =
7.8
=0.66
Mxmax=0.0081ql
=0.0081×
=21.0KN·
Mymax
M
xmax
=0.0194ql
=−0.0351ql2
=0.01×
96.7×
5.175
=0.0351×
=54.1KN·
=−90.9KN·
My=−0.0542ql
=−0.0542×
=−140.4KN·
即取M=140.4kN·
140400000 )
=24.7mm<
19.1×
24.7
=1310mm2
(2)E~F轴间板:
lx
①lx=4.000m,ly=5.175m,
ly
=4.000
5.175
=0.77
Mxmax=0.0155ql
=0.0155×
=40.1KN·
Mymax=0.0094ql
=0.0094×
=24.3KN·
=−0.0386ql
=0.0386×
=−100.0KN·
My=−0.0394ql
=−0.0394×
=−102.0KN·
即取M=102.0kN·
102000000 )
1000×
=17.7mm<
17.7
=939mm2
三、筏板基础计算:
⑴冲切临界截面周长及极限惯性矩计算:
1、内柱:
c1=hc+h0=800+890=1690mmc2=bc+h0=800+890=1690mm
c =c1=1690=845mm
AB 2 2
um=2c1+2c2=2×
1690+2×
1690=6760mm
3
I=c1h0
c3h chc2
+10+201
s 6 6 6
3 3 2
=1690×
890
+1690
×
890+1690×
890×
1690
=16305.2×
108mm4
6 6 6
2、边柱:
c1=hc
+h0
=800+890=1245mm
c2=bc+h0=800+890=1690mm
x= c1 =
12452
=370.8mm
2c1+c2
2×
1245+1690
cAB=c1−x=1245−370.8=874.2mm
um=2c1+c2=2×
1245+1690=4180mm
c3h c 2
+10+2hc(1−x)2+chx
s 6 6
012 20
=1245×
+1245
890+2×
1245×
(1245−370.8)2+1690×
370.82
6 6 2
=7797.3×
108mm3
3、角柱:
c=b
+h0=800+890=1245mm
2 c 2
2
12452
=415mm
1245+1245
cAB=c1−x=1245−415=830mmum=c1+c2=2×
1245=2490mm
+c1h0+hc(c1−x)2+chx2
s 12 12
012 20
890+890×
(1245−415)2+1245×
4152
12 12 2
=4548.1×
⑵抗冲切及抗剪承载力验算
取柱轴力最大处,即9轴/F轴:
由地下室PKPM竖向导荷,知:
Fl=5206kN
①抗冲切承载力验算:
由上述计算知道
um=6760mm;
c1=1690mm;
c2=1690mm;
cAB=845mm
I=16305.2×
考虑作用在冲切临界截面重心上的不平衡弯矩较小,本计算予以忽略。
即:
Munb=0
h
距柱边0处冲切临界截面的最大剪应力τ :
τ =Fl
+αsMunbcAB=
5206
max
+ =
kNm2
max
umh0