参考桥墩抗震计算讲解Word格式文档下载.docx
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5.6×
26=145.6kN
合计:
2776.8+404.0+1094.8+156+145.6=4577.2kN
2、下部恒载计算
1)盖梁加防震挡块重力
P=28.8×
25=720kN
G
2)系梁重力
P=8.1×
25=202.5kN
X
3)一个墩柱重力
π
P=×
1.3
2
×
25=185.8kN
40×
25=1767.1kN
d
4
4)单桩自重力
1.5
z
(三)水平地震力计算
1、顺桥向水平地震力计算
1)上部结构对板式橡胶支座顶面处产生的水平地震荷载
Kitp
E=
ihs
β
CCKh1G
sp
n
i
∑
i=1
式中:
C=1.7,C=0.3,K=0.2
h
根据地质资料分析,桥位所在地土层属Ⅲ类场地,所以有
0.45
β=2.25×
()
0.95
T
对于板式橡胶支座的梁桥
2π
ω1
T=
其中:
GK+(K+K)G−{[GK+(K+K)Gsp]
2−4GGKK}1/2
tpsp12
=gtp
tp
2GGtp
K=K
is
计算采用2孔×
20m为一联,故n=1
GAr
t
s
K=
n=2×
16=32,G=1200kN/m
由橡胶支座计算知
A=×
0.2
r
=0.0314m
∑t=0.042m
1200×
0.0314
∴K=32×
=28708.6kN/m
0.042
K=1×
28708.6=28708.6kN/m
K=Kip
3
ip
li3
3IE1
墩柱采用30号混凝土,则
E=3.00×
10
c
MPa
E=0.8×
3.00×
=2.4×
kN/m
72
按墩高H=7m控制设计,
支座垫石+支座厚度=0.1+0.042=0.142m
l=7+0.142=7.142m
柱惯矩:
I=×
=0.1402m
64
3×
0.1402×
2.4×
7
2=55418.0kN/m
K=
7.142
K=1×
55418.0=55418.0kN/m
G=2×
4577.2=9154.4kN
G=G+ηGp
cp
G=720kN
G=2×
185.8=371.6kN
p
η=0.16(Xf
+2Xf1
+
++1)
XXf1Xf
f
顺桥向作用于支座顶面的单位水平力在支座顶面处的水平位移为:
X=X-φl+XQ
00
l=l=7.142m
0i
l03
3E1I1
X=
Q
=3×
107
=0.0000361
0.1402
桩的计算宽度:
b=0.9(d+1)=0.9×
(1.5+1)=2.25m
mb1
EI
桩在土中的变形系数:
α=5
m=10000kN/m
桩采用25号混凝土,则
E=2.80×
EI=0.8×
2.8×
×
1.5=5.567×
6
10000×
2.25
∴α=55.567×
106
桩长h=40m,
=0.3321
∴αh=0.3321×
40=13.284m>2.5m
取αh=4.0,故K=0
BD4BD3
−
l0
BC4−BC3
34
从而有X=
+α
α
EIAB−AB3
AD4AD3+l0
EIAB−AB
3443
AC4AC3
φ=−(α
αEI×
AB−AB3
)
由公路桥涵地基与基础设计规范(JTJ024-85)附表6.11查得
BD−BD3
=2.441
AB−AB3
BC−BC3
AD−AD3
344
=1.625
=
AC−AC3
=1.751
2.4411.625l0
故X=α
2.441
5.567×
1.625×
7.142
26
0.3321×
0.3321
=0.0000309
5
1.6251.751l0
αEI
1.625
1.751×
0.3321×
10)
=−(
=-0.00000941
X=0.0000309+0.00000941×
7.142+0.0000361
=0.000134
X0
Xd
0.0000309
=0.2306
0.000134
5l03
48E1I1
X=X-φl/2+X=X-φl/2+
H/2
Q/2
5×
=0.0000309+0.00000941×
+
48×
10×
=0.0000758
XH/2
0.0000758
=0.5657
X=
f/2
∴η=0.16×
(0.2306+2×
0.5657+0.2306×
0.5657+0.5657+1)
22
=0.3823
G=720+0.3823×
371.6=862.1kN
∴ω1
862.1×
28708.6+(28708.6+55418)×
9154.4−{[862.1×
28708.6
−4×
9154.4×
28708.6×
55418}1/2
=9.8×
+(28708.6+55418)×
9154.4]2
2×
862.1
=20.021
ω=4.474
T==1.404
4.474
β=2.25×
=0.7634
1.404
KKip
K+Kip
28708.6×
55418.0
=18911.7kN/m
28708.6+55418.0
itp
18911.7
1×
则E=
1.7×
0.3×
0.2×
0.7634×
9154.4=712.8kN
2)墩身自重在板式支座顶面的水平地震荷载
E=CCKβG=1.7×
862.1=67.1kN
hp
支座顶面的水平地震力总和为
E+E=712.8+67.1=779.9kN
ihshp
(四)墩柱截面内力及配筋计算(柱底截面)
1、荷载计算
上部恒载反力:
4577.2kN
下部恒载重力:
720+2×
185.8=1091.6kN
作用于墩柱底面的恒载垂直力为
N=4577.2+1091.6=5668.8kN
恒
水平地震力:
H=779.9kN
水平地震力对柱底截面产生的弯矩为
M=779.9×
7.142=5570.0kN•m
2、荷载组合(单柱)
1)垂直力:
N=5668.8/2=2834.4kN
2)水平力:
H=779.9/2=390.0kN
3)弯矩:
M=5570.0/2=2785.0kN•m
3、截面配筋计算
偏心矩:
e=M/N=2785.0/2834.4=0.9826m
构件计算长度:
l=2l=2×
5.6=11.2m
I
π×
/64
/4
i=
=0.325
A
l/i=11.2/0.325=34.46>
17.5
∴应考虑偏心矩增大系数η
(l
)ξ1ξ2
η=1+
1400e/h0
h=r+r=0.65+0.59=1.24m
0s
h=2r=2×
0.65=1.3m
ξ=0.2+2.7e
=0.2+2.7×
0.9826
1.24
=2.34>
1.0
∴取ξ=1.0
ξ=1.15-0.01l0=1.15-0.01×
=1.064>
11.2
h1.3
1400×
0.9826/1.241.3
11.2)
η=1+
(
1.0×
1.0=1.067
ηe=1.067×
0.9826=1.048m
由公路钢筋混凝土及预应力混凝土桥涵设计规范(JTGD62-2004)附
录C有
配筋率ρ=f
⋅
Br−Ae0
cd
fsd'
Ce−Dgr
f=13.8MPa
f’=280MPa
sd
g=r/r=0.59/0.65=0.9077
假定ξ=0.33,
8
A=0.6631,B=0.4568,C=-0.8154,D=1.7903
ρ=13.8
0.4568×
0.65−0.6631×
1.048
280−0.8154×
1.048−1.7903×
0.9077×
0.65
=0.01027
N≤Arf+Cρrf’
dcdsd
Arf+Cρrf’=0.6631×
0.65×
13.8×
10-0.8154×
2223
cdsd
0.01027×
0.65×
280×
10=2875.5kN>
N=2834.4kN
23
∴纵向钢筋面积
A=ρπr=0.01027×
π×
0.65=0.01363m
222
选用28φ25HRB335钢筋,A=0.001374m>
A=0.01363m
(五)桩身截面内力及配筋计算
1、内力计算
作用于地面处桩顶的外力为
N=2834.4kN,H=390.0kN,M=2785.0kN•m
1)桩身弯矩
φ0
M0
H0
M=α
y
EI(xA+B+α2C+α3D)
03
2.441+M01.625
x=H0α3
α2EI
390.0×
2.441
2785.0×
=0.01204m
1.625+M01.751)
φ=−(H0α2
1.751
=-0.00367
A3、B、C、D3由公路桥涵地基与基础设计规范(JTJ024-85)附表6.12
3
查得,计算见下表
9
桩身弯矩M计算
h=αy
(m)
My
A3
B3
C4
D4
0
(kN*m)
0.000000.000001.000000.000002785.0
0.3010.1-0.00017-0.00001
0.6020.2-0.00133-0.00013
0.9030.3-0.00450-0.00067
1.2040.4-0.01067-0.00213
1.5060.5-0.02083-0.00521
1.8070.6-0.03600-0.01080
2.1080.7-0.05716-0.02001
2.4090.8-0.08532-0.03412
2.7100.9-0.12144-0.05466
1.000000.100002901.2
0.999990.200003010.9
0.999940.300003108.4
0.999740.399983189.6
0.999220.499913251.3
0.998060.599743291.1
0.995800.699353307.8
0.991810.798543300.7
0.98524
0.897053270.5
3.0111
-0.16652-0.08329
0.975010.994453217.4
0.959751.090163142.8
0.937831.183423048.4
0.907271.273202936.1
0.865731.358212808.1
0.815041.436802679.4
0.738591.506952514.8
0.646371.566212354.3
0.529971.611622187.8
0.385031.639692017.7
0.206761.646281846.4
3.3121.1-0.22152-0.12192
3.6131.2-0.28737-0.17260
3.9141.3-0.36496-0.23760
4.2161.4-0.45515-0.31933
4.5171.5-0.55870-0.42039
4.8181.6-0.67629-0.54348
5.1191.7-0.80848-0.69144
5.4201.8-0.95564-0.86715
5.7211.9-1.11796-1.07357
6.0222
-1.29535-1.31361
6.6252.2-1.69334-1.90567-0.27087
7.2272.4-2.14117-2.66329-0.94885
7.8292.6-2.62126-3.59987-1.87734
8.4312.8-3.10341-4.71748-3.10791
1.575381508.0
1.352011187.5
0.91679896.3
0.19729643.1
9.0333
10.5393.5-3.91921-9.54367-10.34040-5.85402109.7
12.0454-1.61428-11.73066-17.91860-15.0755053.1
-3.54058-5.99979-4.68788-0.89126433.4
y=2.108m处,弯矩最大,M=3307.8kN•m
垂直力:
N=2834.4+202.5/2+×
1.5×
2.108×
25=3028.8kN
2、截面配筋计算
e=M/N=3307.8/3028.8=1.092m
l=0.7×
=0.7×
=8.431m
=0.375
l/i=8.431/0.375=22.48>
h=r+r=0.75+0.66=1.41m
0.75=1.5m
=1.09>
1.41
1.092
ξ=1.15-0.01l0=1.15-0.01×
=1.094〉1.0
8.431
h1.5
1.092/1.411.5
8.431)
1.0=1.029
ηe=1.029×
1.092=1.124m
配筋率ρ=f
fsd’Ce0−Dgr
fcd=11.5MPa
fsd’=280MPa
g=rs/r=0.66/0.75=0.88
假定ξ=0.32,
A=0.6351,B=0.4433,C=-0.8656,D=1.7721
11
ρ=11.5
0.4433×
0.75−0.6351×
1.124
=0.00731
280−0.8656×
1.124−1.7721×
0.88×
0.75
N≤Ar2fcd+Cρr2fsd’
Ar2fcd+Cρr2fsd’=0.6351×
0.75×
10-0.8656×
0.00731×
0.752×
103=3111.7kN>
Nd=3028.8Kn
A=ρπr=0.00731×
0.752=0.01292m2
选用28φ25HRB335钢筋,A=0.001374m2>
As=0.01292m2
12