地下建筑结构课程设计隧道衬砌设计(英文).docx

此文档收集于网络，如有侵权，请联系网站删除Underground Structure Course DesigningA Design of Shield Tunnel LiningCollege Civil Engineering Major Department of Geotechnical Engineering Student No. 100xxx Name xxx Director xxx Date 6th Sep. 2013 此文档仅供学习与交流Part One: Design Data1 Function of TunnelThe planned tunnel is to be used as a subway tunnel.2 Design Conditions(1)Dimensions of SegmentType of segment: RC, Flat typeDiameter of segmental lining: D0=9500mmRadius of centroid of segmental lining: Rc=4550mmWidth of segment: b=1200mmThickness of segment: t=400mm(2)Ground ConditionsOverburden: H=12.3mGroundwater table: G.L.+0.6m =12.3+0.6=12.9mN Value: N=50Unit weight of soil: =18kN/m3Submerged unit weight of soil: =8kN/m3Angle of internal friction of soil: =30oCohesion of soil: c=0 kN/m2Coefficient of reaction: k=50MN/m3Coefficient of lateral earth pressure: =0.4Surcharge: P0=39.7kN/m2Soil condition: Sandy(3)MaterialsThe grade of concrete: C30Nominal strength: fck=20.1N/mm2Allowable compressive strength: fc=14.3N/mm2Allowable tensile strength: ft=1.43N/mm2The type of steel bars: HRB335Allowable strength: fy= fy= 300N/mm2Bolt:Yield strength: fBy=240N/mm2Shear strength: =150N/mm23 Design Method Requirement (1)How to check member forces: Elastic equation method(option) Force method based on the textbook (must do this) Bedded frame model(Beam element with elastic support)(option) constant of rotation spring for positive moment at joint=18070 constant of rotation spring for negative moment at joint=32100(2)How to calculate reinforcement for segmental lining:Limit state method Based on the national code GB50010-2002 for reinforcement concrete design. Please choose the grade of concrete and the type of steel rebars.Bolt: yield strength shear strength Part Two: Computation by Force Method1 Load Conditions(1)Judgment of Tunnel Type (by Terzaghis formula)Figure 1 Judgment of tunnel typeSo the designed tunnel is a shallow tunnel.(2)Load Types and Partial FactorsTable 1 shows the loads should be considered in the design and corresponding partial factors.Table 1 Load Types and Partial FactorsLoad TypesSurchargeDead LoadWater PressureEarth PressureSubgrade ReactionPartial Factors1.41.21.21.21.2(3)Computation of LoadsComputational element is a 1.2 meter (width of segment) part along the longitudinal direction, and Figure 2 shows the load condition to compute member forces of the segmental lining.Figure 2 Load condition of the designed tunnelVertical Pressure at the Tunnel CrownEarth PressureWater PressureVertical Pressure at the Tunnel BottomLateral Pressure at the Tunnel CrownEarth PressureWater Pressure Lateral Pressure at the Tunnel BottomWhere Computational diameterAverage Self-weightWhere Unit weight of RC segment Lateral Resistance Pressure Where= Displacement of lining at tunnel spring= Reduction factor of model rigidity = 0.8= Modulus of elasticity of segment = = Moment of inertia of area of segment = = Coefficient of reaction = = the angle of measured from the vertical direction around the tunnel2 Computation of Member ForcesFigure 3 Simplified model diagram for calculationFigure 4 shows the simplified model of the segmental lining.(1)Calculation Data (if inside part of lining is tensile) (if outside part of lining is tensile)(2)Coefficients CalculationNOTE: if the joint just located at 180 degree of the half-ring lining, then its stiffness contribution to the whole structure should be considered as half of the total value.The value of coefficients by force-method equations can be summarized as shown in Table 2.Table 2 Coefficients in Force-method EquationsCoefficientsValuesThen the bending moment and axial force (per 1.2 m) acting at the crown can be obtained by the following equations:(3)Member ForcesFor loading case 1, For loading case 2, For loading case 3, For loading case 4, For loading case 5, For loading case 6, So the internal forces caused by surrounding pressures can be determined by accumulating the six loading cases, that is:Where are the bending moment, the axial force and the shear force (per unit length) under the jth loading case, respectively.Then the total internal forces (i.e. the total bending moment, M, the total axial force, N, the total shear force, Q) per unit length (1.2m) along the lining can be obtained by the following equations:Where And with MATLAB software, the maximum (positive and passive) moment, axial force, and shear force, which is shown in the Table 3. (The original code of MATLAB can be seen in the addendum.)Table 3 Member force of the segmental lining038.101780.701024.561794.833.0620-11.251834.354.2730-55.961891.354.2140-90.951954.627.8750-93.952012.1403.9760-63.072056.2355.7670-12.102085.5315.518042.382101.6284.219085.202108.9259.16100105.812109.5234.36110100.102105.4202.9612071.282100.9156.9813030.602097.888.68140-4.302095.234.68150-25.972092.820.95160-38.872091.612.09170-45.802091.25.6118048.002091.20NOTE: The data in bold face represents the member forces on joint sections, and the data in tilt face represents the dominate member forces.According to the Table 3, it is obvious that the maximum positive moment occurs at the section located at 100 degrees from the tunnel crown (Section A), while the maximum negative moment occurs at the section located at 50 degrees (Section B), and that the maximum axial force occurs at the section located at 100 degrees (Section A).Figure 4 The position of Section A, B, and CThe safety of the segmental lining should be checked at Section A, Section B, and the joint parts.Part Three: Arrangement of Steel Bars of Segmental Lining1 Section AFigure 5 Simplified sketch of section A(1)Calculation data(2)Judgment on the type if eccentric compressionThe steel bars can be arranged symmetrically.Thus, this section should be calculated as a large eccentric compression section.(3)Calculation of and (4)Check the out-plane capacitySo the out-plane capacity of this section is safe.Finally, the steel bars of all segments can be chosen primarily as: 7B14 both in compressive region and tensile region (Actually,).2 Check of Safety at Section BSupposing that and are unknown, then the value of can be calculated and whether it is smaller than the result calculated at section A shall also be checked.(1)Calculation data(2)Judgment on the type if eccentric compressionThe steel bars can be arranged symmetrically.Thus, this section should be calculated as a large eccentric compression section.(3)Calculation of and Therefore, section B is safe.3 Arrangement of Steel BarsAs shown in design drawing (see Attachment 2).Part Four: Determination of Bolts of Joint Section1 Bolt TypeBolt (M30) and Bolt (M45) is used between the segment pieces and between the segmental rings, respectively.2 Arrangement of Bolts in Joint SectionsFigure 6 Section of jointFigure 6 shows the primary arrangement of joint section whose safety could be checked later. Four Bolts (M30) are used in one joint between segment pieces, and then .3 Check the Safety of BoltsThe safety of joint sections can be checked at section located at 100 degrees (Section A) with maximum moment and at 60 degrees (Section C) with maximum shear force.(1) Section ACalculation dataJudgment on the type of eccentric compressionThe bolts are be arranged symmetrically.Therefore, this section should be calculated as a large eccentric compression section.Safety checkThen suppose that Therefore, the bolts at section A are safe.(2) Section CAt this section, shear force occurs maximum, equaling to .Therefore, the bolts at section C are safe.ConclusionAccording above analysis, computation and checking, the designed segmental lining is safe against the design loads.Attachment 1:The initial code of MATLAB software k = 0;R = 4.55;x1 = 38.1;x2 = 1780.7;P = 420.6228 467.6713 272.9184 172.9728 14.976 140.1768;for theta = 0:pi/18:pi; k = k + 1; Mp1 = -0.5*P(1)*R2*sin(theta)2; Np1 = P(1)*R*sin(theta)2; Qp1 = P(1)*R*sin(theta)*cos(theta); if theta pi/2 Mp2 = 0; Np2 = 0; Qp2 = 0; else Mp2 = -0.5*(P(2)-P(1)*R2*(1-sin(theta)2; Np2 = -(P(2)-P(1)*R*(1-sin(theta)*sin(theta); Qp2 = -(P(2)-P(1)*R*(1-sin(theta)*cos(theta); end Mp3 = -0.5*P(3)*R2*(1-cos(theta)2; Np3 = -P(3)*R*(1-cos(theta)*cos(thet