隧道結(jié)構(gòu)力學(xué)計(jì)算-課程設(shè)計(jì)

...wd......wd......wd...隧道構(gòu)造力學(xué)計(jì)算課程設(shè)計(jì)一根本資料：圍巖級(jí)別Ⅳ級(jí)，γ=20kN/m3，彈性抗力系數(shù)K=0.4×106kN/m3,二次襯砌類型C20混凝土45cm，γ=23KN/m3，彈性模量Eh=2.7×107kPa,設(shè)計(jì)時(shí)速100km/m,構(gòu)造斷面如圖1所示。圖1襯砌構(gòu)造斷面〔尺寸單位：cm〕二荷載確定：1.豎向圍巖壓力：q=0.45×2s-1γω式中：s——圍巖類別，此處s=4;γ——圍巖容重，此處γ=20kN/m3;ω——跨度影響系數(shù)，ω=1+i(lm-5)ω=1+0.1×(13.044575-5)=1.8044575mq=0.45×24-1×20×1.8044575=129.92094kPa 考慮到初期支護(hù)承當(dāng)大局部荷載，二次襯砌作為安全儲(chǔ)藏，故對(duì)圍巖壓力進(jìn)展折減，對(duì)本隧道按60%進(jìn)展折減，取為77.952564kPa2.水平圍巖壓力：e=0.25×q=0.25×77.952564=19.488141kPa三襯砌幾何要素1.襯砌幾何尺寸內(nèi)輪廓半徑r1=5.7074m,r2=8.2m,內(nèi)徑r1,r2所畫(huà)圓曲線終點(diǎn)截面與豎直軸的夾角φ1=90o，φ2=98.421132o,拱頂截面厚度d0=0.4m,墻底截面厚度dn=0.8m此處墻底截面為自內(nèi)輪廓半徑為r2的圓心向內(nèi)輪廓墻底做連線并延長(zhǎng)至與外輪廓相交，其交點(diǎn)到內(nèi)輪廓墻底間的連線。內(nèi)輪廓線與外輪廓線相應(yīng)圓心的垂直距離為：m=Δ代入數(shù)值計(jì)算得:m=0.35490916m外輪廓線半徑:R1=m+r1+d0=6.46230916mR2=m+r2+d0=8.95490916m拱軸線與內(nèi)輪廓線相應(yīng)的垂直距離為m＇=0.1759934mQUOTE0.5Δd(r拱軸線半徑：r1＇=m＇+r1+0.5d0=6.0833934mr2＇=m＇+r2+0.5d0=8.5759934m拱軸線各段圓弧中心角θ1=90o，θ2=7.259732o2.半拱軸線長(zhǎng)度S及分段周長(zhǎng)ΔS分段軸線長(zhǎng)度:S1=θ1180oπr1'=90/180QUOTE90oS2=θ2180oπr2'=7.259732/180QUOTE11o半拱軸線長(zhǎng)度：S=S1+S2=9.555772+1.08663176=10.64240376m將半拱軸線等分為8段，每段長(zhǎng)為：ΔS=S8=10.64240376/8QUOTE10.11068=1.330300473.各分塊接縫中心幾何要素：〔1〕與豎直軸夾角αiα1=Δθ1=ΔSr1'×180°πQUOTE1.26383α2=Δθ1+α1=12.52929038o+12.52929038o=25.05858076oα3=Δθ1+α2=12.52929038o+25.05858076o=37.58787114oα4=Δθ1+α3=12.52929038o+37.58787114o=50.11716152oα5=Δθ1+α4=12.52929038o+50.11716152oQUOTE56.1353o=62.6464519oα6=Δθ1+α5=12.52929038o+62.6464519o=75.17574228oΔS1=7ΔS-S1=7*1.33030047-9.555772=-0.24366871mα7=θ1+ΔS1r2'×180°πQUOTE94.0804oα8=α7+ΔSr2'×180°πQUOTE94.0804o+1.2638310.473×180°另一方面α8=90o+7.259732o=97.259732o角度閉合差Δ≈0〔2〕接縫中心點(diǎn)坐標(biāo)計(jì)算x1=r1ˊsinα1=6.0833934×sin12.52929038o=1.31972334mx2=r1ˊsinα2=6.0833934×sin25.05858076o=2.57658888mx3=r1ˊsinα3=6.0833934×sin37.58787114o=3.71073268mx4=r1ˊsinα4=6.0833934×sin50.11716152o=4.66813602mx5=r1ˊsinα5=6.0833934×sin62.6464519o=5.4031982m x6=r1ˊsinα6=6.0833934×sin75.17574228o=5.880908576ma2=〔8.5759934-6.0833934〕×sin90o=2.4926x7=r2ˊsinα7–a2=8.5759934×sin88.3720616249o-a2=6.07963197mx8=r2ˊsinα8–a2=8.5759934×sin97.259732-a2=6.01434395my1=r1ˊ(1-cosα1)=6.0833934×〔1-cos12.52929038o)=0.1448746y2=r1ˊ(1-cosα2)=6.0833934×(1-cos25.05858076o)=0.5725754my3=r1ˊ(1-cosα3)=6.0833934×(1-cos37.58787114o)=1.2627982my4=r1ˊ(1-cosα4)=6.0833934×(1-cos50.11716152o)=2.1826my5=r1ˊ(1-cosα5)=6.0833934×(1-cos62.6464519o)=3.288197m y6=r1ˊ(1-cosα6)=6.0833934×(1-cos75.17574228o)=4.526926ma1=(r2ˊ-r1ˊ)cosθ1=〔8.5759934-6.0833934〕×cos90o=0 y7=r1ˊ-r2ˊcosα7=6.0833934-8.5759934×cos88.3720616249o=5.83976m y8=r1ˊ-r2ˊcosα8=6.0833934-8.5759934×cos97.259732o=7.16712m當(dāng)然也可以直接從圖2中量出xi,yi，以后計(jì)算中只取四位有效數(shù)字。圖2襯砌構(gòu)造計(jì)算圖示四位移計(jì)算1.單位位移用辛普生法近似計(jì)算，按計(jì)算列表進(jìn)展計(jì)算。單位位移的計(jì)算見(jiàn)表1。單位位移值計(jì)算如下：δ11=0sM1E?Ids≈Δδ12=δ21=0sM1M2E?Ids≈δ22=0sM22E?Ids≈計(jì)算精度校核為：δ11+2δ12+δ22=〔28.7659+2×40.62215+140.47023〕×10-6=250.48043×10-6δSS=ΔSE?(1+y)2I=閉合差Δ=0表1單位位移計(jì)算表注：①I——截面慣性矩，I=bd312②不考慮軸力的影響。2.載位移——主動(dòng)荷載在根本構(gòu)造中引起的位移〔1〕每一楔塊上的作用力豎向力：Qi=qbi式中：bi——襯砌外緣相鄰兩截面之間的水平投影長(zhǎng)度，由圖2量得b1=1.36399m,b2=1.30423m,b3=1.18658m,b4=1.01493m,b5=0.79525m,b6=0.53573m,b7=0.24678m,b8=0.01477mΣbi=6.46226m≈B2水平壓力：Ei=ehi式中：——襯砌外緣相鄰兩截面之間的垂直投影長(zhǎng)度，由圖2量得h1=0.14556m,h2=0.43097m,h3=0.69904m,h4=0.93852m,h5=1.13863m,h6=1.28952m,h7=1.38272m,h8=1.40043m,Σhi=7.42539m≈H=7.42543m(校核)自重力： Gi=di?1+di式中：di——接縫i的襯砌截面厚度。 注：計(jì)算G8時(shí)，應(yīng)使第8個(gè)楔體的面積乘以γh 作用在各楔體上的力均列入表2中，各集中力均通過(guò)相應(yīng)圖形的形心?！?〕外荷載在根本構(gòu)造中產(chǎn)生的內(nèi)力 楔塊上各集中力對(duì)下一接縫的力臂由圖2中量得，分別記為aq、ag、ae。表2載位移Mpi?1i?1?x?y??x?yM0000000001.31970.144900-76.7074118.69532.83671.25690.4277-149.18815-1.21325-295.3500233.244711.23551.13410.6902-264.5228-7.75474-624.2267339.621924.85850.95740.9198-325.15404-22.8648-1015.2813434.056743.14850.73511.1056-319.07506-47.705-1412.2484513.176665.33830.47771.2387-245.14446-80.9345-1758.9176574.160690.46860.19871.3128-114.08572-118.767-2005.6718614.8982117.4153-0.06531.327440.1528537-155.857-2137.4790內(nèi)力按下式計(jì)算： 彎矩：Mip0=Mi?1,p0-Δxii?1Σ(Q+G)-Δyii?1Σ 軸力：Nip0=sinαiiΣ(Q+G)- 式中：Δxi，Δyi——相鄰兩接縫中心點(diǎn)的坐標(biāo)增值，按下式計(jì)算Δxi=xi-xi-1Δyi=yi-yi-1Mip0、表3載位移Np截面sinαcosαΣ(Q+G)ΣEsinαΣ(Q+G)cosαΣEN0010000010.216940.97619118.69532.836725.74962.769122.980520.423540.90588233.244711.235598.789510.178088.611630.609980.79242339.621924.8585207.161719.6983187.463440.767360.64122434.056743.1485333.076527.6677305.408850.888190.45948513.176665.3383455.797430.0216425.775860.966720.25586574.160690.4686555.049823.1469531.902970.999600.02841614.8982117.4153614.65003.3357611.314480.99198-0.12637639.6637144.7071634.5358-18.2863652.8221根本構(gòu)造中，主動(dòng)荷載產(chǎn)生彎矩的校核為：M8q0=-qB2(x8?B4M8e0=-e2H2=-19.488141M8g0=-∑Gi(x8-xi+a=-G1(x8-x1+ag1)–G2(x8-x2+ag2)–G3(x8-x3+ag3)–G4(x8-x4+ag4)–G5(x8-x5+ag5)–G6(x8-x6+ag6)–G7(x8-x7+ag7)–G8ag8=-12.3688×〔6.0143-1.3197+0.6573〕-12.8813×〔6.0143-2.5766+0.6058〕-13.8803×〔6.0143-3.7107+0.5575〕-15.3183×〔6.0143-4.6681+0.4684〕-17.1282×〔6.0143-5.4032+0.3579〕-19.2225×〔6.0143-5.8809+0.2305〕-21.5004×〔6.0143-6.0796+0.0917〕-23.6141×(-0.0378)=-205.11575M8P0=M8q0+M8e0+ 另一方面，從表2中查M8P0 閉合差Δ=|2144.39475?2137.479|2144.39475=0.32〔3〕主動(dòng)荷載位移 計(jì)算過(guò)程見(jiàn)表4表4主動(dòng)荷載位移計(jì)算表Δ1P=0sM1MP0E?Ids≈Δ2P=0sM2MP0E?Ids≈計(jì)算精度校核Δ1P+Δ2P=-〔22105.2397+69667.4704〕×10-6=-91772.7101×10-6ΔSP=ΔSE?（1+y）MP閉合差Δ=03.載位移——單位彈性抗力及相應(yīng)的摩擦力引起的位移〔1〕各接縫處的抗力強(qiáng)度抗力上零點(diǎn)假定在接縫3，α3=37.58787114o=αb最大抗力值假定在接縫5，α5=62.6464519o=αhσi=cos2α查表1算得：σ3=0，σ4=0.5201σh，σ5=σh最大抗力值以下各截面抗力強(qiáng)度按下式計(jì)算：σi=〔1-yi'2y式中：yiy?'2——由圖2中量得：y6'=1.15408m，y7'=2.5368m 則：σ6=〔1-1.1540823.937242〕σh=0.σ7=〔1-2.536823.937242〕σhσ8=0 按比例將所求抗力繪于圖2中?！?〕各楔體上抗力集中力R按下式近似計(jì)算：Ri'=〔σ式中；ΔSi外——楔體i外緣長(zhǎng)度,可通過(guò)量取夾角，用弧長(zhǎng)公式求得，Ri〔3〕抗力集中力和摩擦力的合力Ri按下式計(jì)算：Ri=R式中：μ——圍巖與襯砌間的摩擦系數(shù)，此處取μ=0.2則：Ri=Ri'1+0.92=1.0198Ri將Ri的方向線延長(zhǎng)，使之交于豎直軸，量取夾角Ψk,將Ri分解為水平和豎直兩個(gè)力：RH=RisinΨk，RV=RicosΨk表5彈性抗力及摩擦力計(jì)算表〔4〕計(jì)算單位抗力及其相應(yīng)的摩擦力在根本構(gòu)造中產(chǎn)生的內(nèi)力 彎矩：M1σ0=-ΣR 軸力：N1σ0=sinαi∑Rv-cosαi式中：rji——力Rj至接縫中心點(diǎn)ki的力臂，有圖中量得計(jì)算見(jiàn)表表6及表7表6Mσ表7Nσ〔5〕單位抗力及相應(yīng)的摩擦力產(chǎn)生的載位移 計(jì)算見(jiàn)表8表8單位抗力及相應(yīng)的摩擦力產(chǎn)生的載位移計(jì)算表?1σ=0sM1Mσ0E?Id?2σ=0sM2Mσ0E?Id計(jì)算精度校核?1σ+?2σ=-〔34.6258+184.6689〕×10-6=-219.2947?sσ=ΔSE?（1+y）Mσ0I閉合差Δ=04.墻底〔彈性地基上的剛性梁〕位移單位彎矩作用下的轉(zhuǎn)角：βa=1KI8=1主動(dòng)荷載作用下的轉(zhuǎn)角：βap0=M8P0βa=-2137.479×58.5938×10-6單位抗力及相應(yīng)摩擦力作用下的轉(zhuǎn)角：βaσ0=M8σ0βa=-13.9667×58.5938×10-五解力法方程襯砌矢高f=y8=7.16712計(jì)算力法方程的系數(shù)為：a11=δ11+βa=(28.7659+58.5938)×10-6=87.3597×10- a12=δ12+fβa=(40.62215+7.16712×58.5938)×10-6=460.5709×10-a22=δ22+f2βa=(140.47023+7.167122×58.5938)×10-6=3150.2936×10-a10=?1p+βap0+(?1σ=-(22105.2397+125242.9102+34.6258σh+818.3613σh)×10-6 =-(147348.1499+852.9871σ?)×10a20=?2p+fβap0+(?2=-(69667.4704+7.16712×125242.9102+184.6689σh+7.16712×818.3613σh)×10-6=-(967298.43695+6049.9625σ?)×10以上將單位抗力及相應(yīng)的摩擦力產(chǎn)生的位移乘以，即為被動(dòng)荷載的載位移。求解方程為：X1=a=3150.2936=296.1223式中：X1P=296.1223，X1σ=-1.57375X2=a=87.3597=263.7571+2.1505式中：X2P=263.7571，X2σ=2.1505六計(jì)算主動(dòng)荷載和被動(dòng)荷載〔σh=1〕分別產(chǎn)生的襯砌內(nèi)力計(jì)算公式為：MNMN計(jì)算過(guò)程列入表9，表10中表9主、被動(dòng)荷載作用下襯砌彎矩計(jì)算表表10主、被動(dòng)荷載作用下襯砌軸力計(jì)算表七最大抗力值的求解首