A 卷
计算题(共3小题,第1小题20分,第2、3小题各10分,共40分)
1解:(1)确定f a 进行持力层承载力验算:
因e=0.72,I L =0.42,查表得ηb =0.3,ηd =1.6。b
F k 171.52+γG d =+20⨯1.2=131.2kPa b 1.6
(2)下卧层验算: p k =
先确定软弱下卧层顶面处的p z 和p cz :
E s 17.52.5==3, z /b ==0.9,故压力扩散角θ=23 E s 22.52.6
p z =b (p k -p c ) 1.6⨯(131.2-19⨯1.2) ==70.82kPa b +2z tan θ1.6++2⨯1⨯tan 23)
p cz =19⨯1.2+(19-10) ⨯1=31.8kPa
p z +p cz =102.62kPa
下卧层顶面处承载力:(ηd =1. 0)
f az =f ak +ηd γm (d -0.5) =80+1.0⨯19⨯1.2+(19-10) ⨯1⨯(2.2-0.5) =104.57kPa 2.2
f az >p z +p cz =102.62kPa ,故下卧层满足承载力要求。
(3)求最大弯矩设计值
F 1.35F k 1.35⨯171.52p j ====144.72kPa b b 1.6
11b 1=(b -0.24) =⨯(1.6-0.24)=0.68m 22
11M =p j b 12=⨯144.72⨯0.682=33.46kN ⋅m 22
每延米最大弯矩设计值为33.46kN ⋅m
2. 解:
桩为圆桩,桩径d=0.6 m
R a =q pa A P +u p ∑q sia l i
1 = 800⨯⨯πd 2+πd (42⨯2+38⨯5+50⨯2+52⨯1) 4
=1029kN
3. 解:(1)基本组合荷载效应及桩顶净反力
F=1.35Fk =1.35×2200=2970kN,M=1.35Mk =1.35×200=270kNm,H=1.35Hk =1.35×180=243kN。
F 2970N ===594kN n 5
N max =F (M +Hh ) x max (270+243⨯1) ⨯1+=594+=722.25kN 22n x 4⨯1i
(2)截面弯矩设计值
承台有效高度h 0=1000-70-10=920mm
弯矩:A-A 截面:M y =∑N i x i =2⨯722.25⨯(1-0.4/2)=1155.6kN⋅m B-B 截面:M X =∑N i y i =2⨯594⨯(1-0.4/2)=950.4kN⋅m B 卷
1解:(1)确定f a 并验算持力层承载力:
因粉土粘性颗粒含量
f a =f ak +ηd γm (d -0.5) =232+2.0⨯17.5⨯(1.5-0.5) =267kPa p k =F k 900+γG d =+20⨯1.5=210kPa
基底处总力矩: M k =170+50⨯1=220kN ⋅m 偏心距:e =l 2.5M k 220=0.416m ==0.21m
p k,max =F k +G k 6e 900+1506⨯0.21(1+) =(1+) =315.8kPa
故持力层满足承载力要求。
(2)下卧层验算:
先确定软弱下卧层顶面处的p z 和p cz :
E s181.6==4, z /b ==0.64>0.5,查表插值得压力扩散角θ=24 E s222.5
p k =F k +G k 905+150==211kPa b ⨯l 2⨯2.5
p z =lb (p k -p c ) 2.5⨯2⨯(211-17.5⨯1.5) ==68.73kPa (b +2z tan θ)(l +2z tan θ) (2+2⨯1.6tan 24)(2.5+2⨯1.6tan 24)
p cz =17.5⨯3.1=54.25kPa
p z +p cz =122.98kPa
下卧层顶面处承载力:(ηd =1.0)
f az =f ak +ηd γm (d -0.5) =80+1.0⨯17.5⨯(3.1-0.5) =125.5kPa >p z +p cz =123.52kPa 故下卧层满足承载力要求。
2. 解:
桩为方桩,边长a=0.4 m
R a =q pa A P +u p ∑q sia l i
=800⨯0.42+4⨯0.4⨯(36⨯(4.5-2) +44⨯4+41⨯4.7+54⨯0.8) =931.04kN
3. 解:(1)桩顶净反力设计值
F 2200N ===550kN n 4
N max =F (M +Hh ) x max (200+150⨯1) ⨯1.05+=550+=633.33kN 22n 4⨯1.05x i
(2)最不利截面弯矩设计值
承台有效高度h 0=1000-70-10=920mm
弯矩:A-A 截面:M y =∑N i x i =2⨯633.33⨯(1.05-0.5/2)=1013.3kN⋅m B-B 截面:M X =∑N i y i =2⨯550⨯(1.8/2-0.5/2)=715kN⋅m