深基坑手算計(jì)算書

1、處，第二道-11.5m處。土層相關(guān)參數(shù)見下表:表1 土層參數(shù)信息表土層編號(hào)土層名稱重度(kN/m3)黏聚力c(kPa)內(nèi)摩擦角:()土層厚度(m)1-1雜填土160.7156.03-1-2新黃土 219.421.7224.03-2-2古土壤20.224.5204.84-1-2老黃土21.120.3246.120kPa,荷載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 及各土層分布情況見圖1.1。q=20kPaT1.，d H 1t2 1-1雜填土T3 3-1-2 新黃土 23-2-2古土壤4-1-2老黃土圖1.1荷載分布及支護(hù)方案解：1計(jì)算各土層側(cè)壓力系數(shù)(1) 郎肯主動(dòng)土壓

2、力系數(shù)計(jì)算Ka1 =tan2(45 - 1/2) =tan2(45 -15 /2) =0.589.Ka二 0.767Ka2 =tan2 (45 - 2/2) = tan2 (45 -22 /2) =0.455Ka2 = 0.675Katan2(45 - 篤/2) =tan2(45 -20 7 2) =0.490、Ka3 二 0.700Ka4 二 tan2 (45 - I /2) = tan2 (45 一 24 /2) = 0.422-j Ka4 = 0.649(2)郎肯被動(dòng)土壓力系數(shù)計(jì)算Kptan2(451/2) =tan2(4515 /2) =1.698KP1 =1.30320仆200Kp2

3、 =ta n (452/2)=ta n (4522 / 2 2.198Kp2 = 1.483Kp3 =tan2(453 /2) =tan2(4520 /2) = 2.040KP3 二 1.428KP4 二 tan2(45；/2)=tan2(4524 /2) =2.371Kp4 =1.5402各工況土壓力及支撐力計(jì)算(1) 工況1:基坑開挖至-6.0m，并在此處設(shè)置第一道錨桿，地面處的主動(dòng)土壓力為:eaqKa 2c Ka1 =20 0.589 -2 0.7 0.767 =10.706kPa6.0m處的主動(dòng)土壓力：第一層土層：$6 =(qiZjKai 2® . Kai= (2016 6)

4、 0.589 -2 0.7 0.767 = 67.250kPa第二層土層：$6 =(q1乙)心2 -2c2 ： Ka2= (2016 6) 0.455 -2 21.7 0.675 =23.485kPa開挖面處的被動(dòng)土壓力為：ep6 =2c2、Kp2 = 2 21.7 1.483 = 64.362kPa開挖面處主動(dòng)土壓力減去被動(dòng)土壓力為：e6 = ea6 -ep6 =23.485-64.362 - -40.877kPa則所有的主動(dòng)土壓力合力為：Ea1 =0.5 (10.706 67.250) 6 =233.868kN / m10.70640.87767.250t1en圖3.2工況1 土壓力分布圖

5、假設(shè)開挖面以下處剪力為0，該處即是最大彎矩所在截面，則t,m處的主動(dòng)土壓力減去被動(dòng)土壓力為：eti 弋2tiKa2 -( 2tiKp? 2c? JKp?)= 23.485 19.4t, 0.455 - (19.4b 2.198 2 21.7 1.483)= 33.814t40.877)kPa由靜力平衡列方程：Ea =0.5 (40.877 40.877 33.814t1)t1233.868 =40.877116.592t：解得 t2.719m 所以即=33.814 2.719 -40.877 =132.817kPa因此最大彎矩為：Mmax1 =10.706 6 (0.5 6 2.719) 0.

6、5 (67.250-10.706) 6 (13 6 2.719)- -40.877 2.719 0.5 2.719-0.5 (132.817-40.877) 2.719 13 2.719= 827.266kN m/m(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：開挖面處主動(dòng)土壓力：第二層土層：為。二 ea62z2Ka2 =23.485 19.4 4 0.455 = 58.793kPa第三層土層：ea10 =(q1Z12Z2)Ka3 2C3 . Kas= (20 16 6 19.4 4) 0.490 2 24.5 0.700 = 60.564kPa開挖面處的主動(dòng)土壓力為

7、：為1.5 =為0 + ；3Z2Ka3 =60.564+ 20.2x1.5 漢 0.490 = 75.411kPa開挖面處的被動(dòng)土壓力為：ep11.5 =2c3jKp3 =2=<24.5x1.428 = 69.972kPa則主動(dòng)土壓力減去被動(dòng)土壓力為：eh.5 =ea115 ep11.5 =75.41169.972 = 5.439kPa假設(shè)土壓力零點(diǎn)位置為開挖面以下 d2處，則有：ea11.5 3d?Ka3 二 epn.5 3d2Kpd2=0.174mei15_5.4393(Kp3 - 隔廠 20.2 (2.040 -0.490)主動(dòng)土壓力合力為:Ea2 =233.8680.5 (23.

8、485 58.793) 40.5 (75.41160.564) 1.5= 500.878kN /m土壓力零點(diǎn)位置處截面的彎矩為：Ma2 =10.706 6(0.5 641.5 0.174)0.5 (67.250 -10.706) 6 (1364 1.5 0.174)23.4854(0.5 4 1.50.174)0.5 (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 1.5 0.174) 0.5 5.439 0.174 23 0.174 = 2507.943kN m/

9、m則第一道(6m)處錨桿的水平分力為:Ma2a22507.9435.5 0.174=442.006kN / m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,t2m處的土壓力為:-5.439t2- 0.174e2 - -31.259t2由靜力平衡可列方程:T1一丁10.70623.48567.25058.793"60.5645.43975.411d202t2圖3.3工況2 土壓力分布圖Ea20.5 et2 t2500.878=442.006 0.5 31.259t2 t2解得 t2 = 1.941m 所以 気=-31.259 1.941 =-60.674kPa因

10、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 1.5 (0.5 1.5 0.1741.941 -442.006 (5.5 0.1741.941)0.5 (75.411 -60.564) 1.5 (13 1.5 0.174 1.941)0.5 5.439 0.17423 0.174 1.94

11、1)-0.5 1.941 60.6741.941) =270.607kN m/m(3) 工況3:開挖至-17.0m，并在此處設(shè)第三道錨桿:ea14.8 =ea10 +丫3%©3 =60.564+ 20.2 漢 4.8 漢 0.490 =108.074kPa第四層土層:ea14.8 =9&2Z23Z3 )Ka4 -2cKa4-(20 16 6 19.4 4 20.2 4.2)0.422 -2 20.3 0.649= 96.267kPa為7 =ea14.5 + ?4Z4Ka4 =96.267+ 20.2x2.2 漢 0.422 = 115.856kPa 開挖面處的被動(dòng)土壓力為：e

12、p仃=2c4 . Kp4 =2 20.3 1.540 =62.524kpa主動(dòng)土壓力減去被動(dòng)土壓力：% 二 ea17 - ep17 = 115.856 - 62.524 = 53.332kPa假設(shè)土壓力零點(diǎn)位置為開挖面以下d3處，則：10.706T123.48567.250T258.79360.564108.07453.33296.267115.856d3t3圖3.4工況3的土壓力分布圖閒7 4dKa4 - 4dKp4 ' 2c4 , Kp4115856 + 21.1d3 ><0.422=21.1 xd3 漢2.371 + 2江20.3匯1.540解得 d = 1.297m

13、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 (60.564108.074)4.8 0.5 (96.267 115.856) 2.2 0.5 53.332 1.297 =1070.076kN/m土壓力零點(diǎn)處截面的彎矩為：Ma? =10.706 6 (0.5 64 4.82.5 1.297)0.5 (67.250 -10.706) 6(13 644.82.21.297)23.485 4 (0.5 4 4.82.2 1.297)0.5(58.793 -23.485) 4 ( 13 44.82.21.297)60.564

14、 4.8 (0.5 4.8 2.21.297)0.5 (108.074 -60.564) 4.8 (13 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 1.297 23 1.297 =7931.991kN m/m所以第二道（11.5m處）錨桿的水平分力為：_ _ Ma3 -久(17-6)2 一 5.5 1.2977931.991-442.006 (11 1.297)-5.5+1.297= 438.303kN/m假設(shè)土壓力零點(diǎn)位置以下 t3m處剪力為0，即該處

15、彎矩最大，則該界面處的土壓力為:由幾何關(guān)系有：- et3t353.332所以 q3 = -41.120t3則由靜力平衡列方程：Ea3 =£ T2 0.5 41.120 t3 t31070.076 =442.006 438.303 0.5 41.120t3 t3解得 t3 二 3.038me3=-41.120 3.038 = -124.922kPa因此最大彎矩為:Mmax3 =10.706 6 (0.5 6 4 4.8 2.2 1.2973.038) 0.5 (67.250 -10.706)6 ( 13 64 4.8 2.2 1.297 3.038)23.485 4 (0.5 4 4.

16、8 2.21.2973.038)0.5 (58.793 -23.485) 4(134 4.8 2.2 1.297 3.038)60.564 4.84.82.2 1.2(0.5 4.82.21.297 3.038)0.5 (108.074 -60.564) 4.8 97 3.038) 96.267 2.2 (0.5 2.2 1.297 3.038) 0.5 (115.856 -96.267) 2.2(13 2.2 1.297 3.038) 0.5 53.332 1.297 (23 1.297 3.038) - 0.5 124.922 3.03813 3.038 -442.006 (11 1.29

17、73.038)-438.303 (11 -5.5 1.2973.038)二-95.109kN m / m(4) 嵌固深度的計(jì)算Td 二 Ea3 T； T2 =1070.076-442.006-438.303 = 189.767kN/me = -41.120h圖3.5嵌固深度計(jì)算圖對(duì)O點(diǎn)取矩，由力矩平衡可知：二 M (O) - 0 :189.767 h 0.5 (-41.120h) h 13 h =0解得 h = 5.262m則嵌固深度 ha -1.2(h t3) =1.2 (5.262 3.038) = 9.96m因此樁總長L =179.926.96m3灌注樁的內(nèi)力設(shè)計(jì)值及配筋設(shè)計(jì)最大彎矩計(jì)算

18、值：Mc =827.266 1.5 = 1240.899kN m最大彎矩設(shè)計(jì)值：M =1.25 0Mc =1.25 1.1 1240.899 = 1706.236kN m灌注樁的配筋取分段開挖的最大彎矩：M噺=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(cè)附表14-4選用2825，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：: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。根據(jù)土體與錨固體極限摩阻