深基坑手算計(jì)算書

1、For pers onal use only in study and research; not for commercial use蚅題目：基坑深17.0m，支護(hù)方式為排樁加外錨方案，設(shè)兩道錨桿支護(hù)(第一道設(shè)在-6.0m處，第二道-11.5m處。土層相關(guān)參數(shù)見下表：蝿表1 土層參數(shù)信息表螇土層編號(hào)裊土層名稱蒃重度衿(kN /m3)腿黏聚力c薇(kPa)膂內(nèi)摩擦角罿)()薈土層厚度羅(m)羈1-1聿雜填土罿16螇0.7羄15膈6.0肆 3-1-2膅新黃土 2螃 19.4羋 21.7蕆22袇4.0薂 3-2-2薂古土壤袈 20.2蒞 24.5薅20螞4.8艿 4-1-2肇老黃土莄 21.1螂

2、20.3蝕24蒄6.1肂袂此基坑采用分層開挖的方式，在基坑頂部承受擬定的均布荷載，荷載值為20kPa,荷載及各土層分布情況見圖1.1。祎蚃圖1.1荷載分布及支護(hù)方案肁聿解：襖1計(jì)算各土層側(cè)壓力系數(shù)蒂(1)郎肯主動(dòng)土壓力系數(shù)計(jì)算賺 Katan2(45 -/2) =tan2(45 -15 /2) = 0.589, Ka 0.767膆 Ka2 二 tan2(45 - 2/2) =tan2 (45 - 22 / 2) = 0.455Ka 0.67520心200i薆 Katan (45 - 3/2) =tan (45 -20 /2) =0.490. Ka 0.700芁 Ka4 二 tan2(45 - ；

3、/2) =tan2(45 -24 / 2) =0.422. Ka 0.649芁（2）郎肯被動(dòng)土壓力系數(shù)計(jì)算薇 Kpta n2(451/2) =ta n2(45 15 /2) =1.698Kp1 =1.303肄 Kp2 =tan2(452/2)=tan2(4522 72) =2.198Kp 1.483芄 Kp3 二 tan2(45，3/2)=tan2 (4520 7 2) = 2.040Kp 1.428莁 Kp4 二 tan2(45 J/2) =tan2(4524 / 2) =2.371. Kp4 =1.540羈2各工況土壓力及支撐力計(jì)算螆（1）工況1:基坑開挖至-6.0m，并在此處設(shè)置第一道錨

4、桿，地面處的主動(dòng)土壓力為:肅ea qKa2c Ka1 =20 0.589-2 0.7 0.767 = 10.706kPa蒁6.0m處的主動(dòng)土壓力：荿第一層土層：膃ea6 =（q 1乙）Ka1 2& Kaj螁 =（2016 6） 0.589 -2 0.7 0.767 =67.250kPa薁第二層土層:蝿ea6 =(qrZjKa? _2c2 . Ka2裊-(2016 6) 0.455 -2 21.7 0.675 =23.485kPa襖開挖面處的被動(dòng)土壓力為:薁ep6 二 2c2, Kp2 =2 21.7 1.483 二 64.362kPa羆開挖面處主動(dòng)土壓力減去被動(dòng)土壓力為：蚇= ea6

5、- ep6 = 23.485 - 64.362 - -40.877kPa薃則所有的主動(dòng)土壓力合力為：蟻 Ea1 = 0.5 (10.706 67.250) 6 = 233.868kN /m莇賺t1螇et1莂袀圖3.2工況1 土壓力分布圖t1m處的主動(dòng)土壓力芅假設(shè)開挖面以下t1m處剪力為0，該處即是最大彎矩所在截面，則減去被動(dòng)土壓力為：膄 e 二 e6 2心心2-（ 2恥2 2C2 KP2）= 23.485 19.4 0.455 -(19.4鮎 2.198 2 21.7 1.483 =(-33.814鮎-40.877)kPa芆由靜力平衡列方程:羇 =0.5 (40.877 40.877 33.8

6、14右兒羃233.868 二 40.877E 16.592t：肀解得t 2.719m所以內(nèi)=-33.814 2.719-40.877 =-132.817kPa蚇因此最大彎矩為：Mmaxi =10.706 6 (0.5 6 2.719) 0.5 (67.250-10.706) 6 (13 6 2.719)- 蔻一40.877 2.719 0.5 2.7190.5 (132.817 40.877) 2.719 13 2.719-827.266kN m/m螞(2)工況2:開挖至處-11.5m，并在此處設(shè)第二道錨桿，土壓力分布：膀開挖面處主動(dòng)土壓力：肇第二層土層：膆ea10 =ea62z2Ka2 =2

7、3.485 19.4 4 0.455 = 58.793kPa螅第三層土層：膀ea10 =(q1乙2Z2)Ka3 -2© . Ka3蒈 =(20 16 6 19.4 4) 0.490 -2 24.5 0.700 = 60.564kPa薄開挖面處的主動(dòng)土壓力為：蒃ea11.5 =ea10 +Y3z2Ka3 =60.564+20.2沃 1.5沃 0.490 = 75.411kPa芀開挖面處的被動(dòng)土壓力為:衿epii.5 =2C3；.；Kp3 =2 24.5 1.428 = 69.972kPa莆則主動(dòng)土壓力減去被動(dòng)土壓力為：節(jié) ei.5 = $11.5 epii.5 = 75.41169.9

8、72 = 5.439kPa莀假設(shè)土壓力零點(diǎn)位置為開挖面以下d2處，則有:羆為1.53d2Ka3 二 epn.5sdzKp?©1.55.43920.2 (2.040-0.490)=0.174m肁主動(dòng)土壓力合力為:Ea2 =233.8680.5 (23.485 58.793) 40.5 (75.41160.564) 1.5葿= 500.878kN /m莇土壓力零點(diǎn)位置處截面的彎矩為：蒆Ma2 =10.706 6 (0.5 6 4 1.5 0.174) 0.5 (67.250 -10.706) 6 (打 64 1.5 0.174) 23.485 4 (0.5 4 1.5 0.174) 0.

9、5 (58.793 -23.485) 4 (13 4 1.5 0.174)60.564 1.5 (0.5 1.5 0.174)0.5 (75.411 -60.564)1.5 ( 13 1.50.174)0.5 5.439 0.174 23 0.174 = 2507.943kN m/m肄則第一道(6m)處錨桿的水平分力為：Ma 2a22507.9435.5 0.174= 442.006kN / mt2m處的土螈假設(shè)土壓力零點(diǎn)位置以下 t2m處剪力為0,即彎矩最大，則由幾何關(guān)系可知,壓力為：et2 - -31.259t2-et25.439羄t20.174袃由靜力平衡可列方程：蠆裊蒃d2衿t2腿蚆&

10、#169;2腿圖3.3工況2 土壓力分布圖薇膂 Ea2 =T|0.5 e2 t2罿500.878 =442.006 0.5 31.259t2 t2薈解得 t 1.941m 所以 q2 = 一31.259 1.941 =-60.674kPa羅因此最大彎矩為:Mmax2 =10.706 6 (0.5 6 4 4.5 0.174 1.941) 0.5 (67.250 -10.706) 6 (13 6 4 1.5 0.174 1.941) 23.485 4 (0.5 4 1.5 0.174 1.941) 0.5(58.793 -23.485) 4 ( 4 1.5 0.174 1.941)60.564

11、1.5 (0.5 1.5 0.174(23 0.174 1.941)-0.5 1.941 60.6741.941) =270.607kN m/m1.941) 0.5 (75.411 -60.564) 1.5 (打 1.5 0.174 1.941)0.5 5.439 0.17413 1.941 -442.006 (5.5 0.174聿(3)工況3：開挖至-17.0m，并在此處設(shè)第三道錨桿：罿ea14.8 =ea10 +Y3z3Ka3 =60.564+ 20.2 x 4.8 沢 0.490 =108.074kPa螇第四層土層：羄 ea14.8 =(q憶 2Z23Z3 )心4 2c Ka4= (20

12、 16 6 19.4 4 20.2 4.2) 0.422 -2 20.3 0.649 膈= 96.267kPa肆 ea17 =ea14.5 +Y4z4Ka4 =96.267+ 20.2x2.2x0.422 =115.856kPa膅開挖面處的被動(dòng)土壓力為：螃ep仃=2c4、Kp4 =2 20.3 1.540 = 62.524kpa羋主動(dòng)土壓力減去被動(dòng)土壓力：蕆 $7 二 ea仃一 ep 17 = 115.856 - 62.524 = 53.332kPa袇假設(shè)土壓力零點(diǎn)位置為開挖面以下 d3處，則:袁圖3.4工況3的土壓力分布圖羂芇eai74dKa44dKp4 2cKP4蚄115.856 21.1

13、 d3 0.422=21.1 d3 2.371 2 20.3 1.540襖解得 d = 1.297m肂所有主動(dòng)土壓力合力為：Ea3 =0.5 (10.706 67.250) 60.5 (23.485 58.796) 4 0.5 (60.564蚈108.074) 4.8 0.5 (96.267 115.856) 2.2 0.5 53.332 1.297=1070.076kN/m莆土壓力零點(diǎn)處截面的彎矩為:Ma3 =10.706 6 (0.5 64 4.82.5 1.297)0.5 (67.250 -10.706) 6(13 6 4 4.8 2.2 1.297) 23.485 4 (0.5 4 4

14、.8 2.2 1.297) 0.5 (58.793 -23.485) 4 ( 13 4 4.8 2.2 1.297) 60.564 4.8 (0.5 4.8 2.21.297)0.5 (108.074 -60.564) 4.8 (打 4.8 2.2 1.297)96.267 2.2(0.5 2.2 1.297)0.5 (115.856 -96.267) 2.2 ( 132.2 1.297)0.5 53.332 1.2971.297 =7931.991kN m/m所以第二道（11.5m處）錨桿的水平分力為:TMa3 -(17-6)25.5 1.2977931.991 - 442.006 匯(11

15、 +1.297) 肁5.5+1.297-438.303kN/m聿假設(shè)土壓力零點(diǎn)位置以下 t3m處剪力為0,即該處彎矩最大，則該界面處的土壓力為:襖由幾何關(guān)系有：-et353 . 332蒂一t3d 3賺所以 $3 = _41.120t3膆則由靜力平衡列方程：薆 EaT1 T2 0.5 41.120 t3 t3節(jié) 1070.076 =442.006 438.303 0.5 41.1203 t?節(jié)解得 t 3.038m蚈 q3=41.120 3.038 = 124.922kPa肅因此最大彎矩為:Mmax3 =10.706 6 (0.5 6 4 4.8 2.2 1.2973.038) 0.5 (67.

16、250 -10.706)6 ( 13 6 4 4.8 2.2 1.297 3.038) 23.485 4 (0.5 4 4.8 2.2 1.297 3.038)0.5 (58.793 -23.485) 4 (打 4 4.8 2.2 1.297 3.038)60.564 4.8(0.5 4.8 2.2 1.297 3.038) 0.5 (108.074 -60.564) 4.8 ( 13 4.8 2.2 1.2973.038)96.267 2.2 (0.5 2.2 1.297 3.038)0.5 (115.856 -96.267) 2.2(132.2 1.297 3.038)0.5 53.332

17、 1.297(23 1.297 3.038)-0.5 124.9223.03813 3.038 -442.006 (11 1.2973.038)-438.303 (11 -5.5 1.2973.038)-95.109kN m / m莂(4)嵌固深度的計(jì)算聿Td 二 Ea3 -人-T2 =1070.076-442.006-438.303 = 189.767kN/meh - -41.120h螇蒂圖3.5嵌固深度計(jì)算圖極限摩阻力取73kPa，老黃土的極限摩阻力取68kPa,各層加權(quán)內(nèi)摩擦角為20°。芄對(duì)0點(diǎn)取矩，由力矩平衡可知:打h=0袃' M（°）= 0 :薂 189.

18、767 h 0.5 (-41.120h) h袁解得h = 5.262m羆則嵌固深度 0 =1.2(h ta) =1.2 (5.262 3.038) = 9.96m裊因此樁總長L =179.96 =26.96m螞3灌注樁的內(nèi)力設(shè)計(jì)值及配筋設(shè)計(jì)羇最大彎矩計(jì)算值：Mc =827.266 1.5 = 1240.899kN m蚈最大彎矩設(shè)計(jì)值：M =1.25 0Mc =1.25 1.1 1240.899 = 1706.236kN m蚄灌注樁的配筋取分段開挖的最大彎矩：Mmax =1240.899kN m。彎矩設(shè)計(jì)值為：M =1.25 1.1 1240.899 =2047.0kN m查基坑工程手冊附表14

19、-4選用28G25，截面容許彎矩設(shè)計(jì)值M =2068.0kN m 2047.0kN m，樁體選用C30混凝土，箍筋選用>10150螺旋筋，加強(qiáng)筋叮202000。4錨桿計(jì)算錨桿傾角20°，水平間距2m，鉆孔直徑0.15m。錨索采用1 7 5預(yù)應(yīng)力鍍鋅鋼絞線，強(qiáng)度標(biāo)準(zhǔn)值fptk =1470N/mm2。水平力計(jì)算值：=442.006 2 =884.012kNE =438.303 2 =876.606kN水平力設(shè)計(jì)值：0=1.25=1.25 1.1 884.012 =1215.517kN2 =1.25 0T2 =1.25 1.1 876.606 =1205.333kN錨桿設(shè)計(jì)的軸向拉力

20、為：M1=1215.517/cos20 o=1292.55 kNN2=1205.333/cos20 °=1281.91 kNa1kmN"二d 二 <ili=47ma2錨桿的自由端長度l11lf1 =lt *sin(45 -一 k)/sin(45 k 旳2211=11 sin(4520 )/sin(452020 )=6.5m221 1lf2 =lt *sin(45k)/sin(45 - k 旳2211= 5.5 sin(45 20 )/sin(452020 )= 3.25m22所以，第一道錨桿長度 h =47 6.5 0.15 = 54m第二道錨桿長度12 =45 3.25 0.15 = 49m錨桿桿體材料計(jì)算:Ap1 292 5147058°7228 mm錨桿的錨固長度I所需錨索根數(shù)=879.286/139=6.3取7根。僅供個(gè)人用于學(xué)習(xí)、研究；不得用于商業(yè)用途For personal use only in study and research; not for commercia