Underground Structure Course DesigningA Design of Shield Tunnel Lining College Civil EngineeringMajor Department of Geotechnical Engineering Student No.100xxxName xxxDirector xxxDate6th Sep. 2013Part 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: D 0=9500mmRadius of centroid of segmental lining: R c =4550mm Width of segment: b=1200mm Thickness of segment: t=400mm (2)Ground Conditions Overburden: H=12.3mGroundwater table: G.L.+0.6m =12.3+0.6=12.9m N Value: N=50Unit weight of soil: γ=18kN/m 3Submerged unit weight of soil: γ'=8kN/m 3 Angle of internal friction of soil: φ=30o Cohesion of soil: c =0 kN/m 2Coefficient of reaction: k=50MN/m 3Coefficient of lateral earth pressure: λ=0.4 Surcharge: P 0=39.7kN/m 2 Soil condition: Sandy (3)MaterialsThe grade of concrete: C30Nominal strength: f ck =20.1N/mm 2Allowable compressive strength: f c =14.3N/mm 2 Allowable tensile strength: f t =1.43N/mm 2 The type of steel bars: HRB335Allowable strength: f y = f y ’= 300N/mm 2 Bolt:Yield strength: f By =240N/mm 2 Shear strength: B τ=150N/mm 23 Design Method Requirement(1)How to check member forces: ① Elastic equation method(option)② Force method based on the textbook (must do this)③ Bedded frame model(Beam element with elastic support)(option)0P K =constant of rotation spring for positive moment at joint=18070/kN m rad ⋅ 0N K =constant of rotation spring for negative moment at joint=32100/kN m rad ⋅(2)How to calculate reinforcement for segmental lining:Limit state method①Based on the national code GB50010-2002 for reinforcement concrete design. ②Please choose the grade of concrete and the type of steel rebars.Bolt: yield strength 2240/By f MN m =shear strength 2150/B MN m τ=Part Two: Computation by Force Method1 Load Conditions(1)Judgment of Tunnel Type (by Terzaghi’s formula) c 0⑤Average Self -weight Wherec γ= Unit weight of RC segment 326/kN m =⑥Lateral Resistance Pressure Whereδ= Displacement of lining at tunnel spring η= Reduction factor of model rigidity = 0.8E = Modulus of elasticity of segment = 423.010/N mm ⨯I = Moment of inertia of area of segment =33411.20.4 6.41012m -⨯⨯=⨯ k = Coefficient of reaction = 360/MN mϕ= the angle of measured from the vertical direction around the tunnel2 Computation of Member ForcesFor 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 ,,pj pj pj M N Q are the bending moment, the axial force and the shear force (per unit length) under the j th 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: WhereAnd 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 MA TLAB can be seen in the addendum.)NOTE: 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 4The 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 5Simplified 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 s A and sA ' (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: 7B 14 both in compressive region and tensile region (Actually,21077.6ssA A mm '==). 2 Check of Safety at Section BSupposing that and are unknown, then the value of can be calculated and whether it is smallerthan 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 s A and sA ' Therefore, sectionB 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 segmentalrings, 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 then2212301413.74s s A A mm π'==⨯⨯⨯=.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 A ①Calculation data②Judgment on the type of eccentric compressionThe bolts are be arranged symmetrically.Therefore, this section should be calculated as a large eccentric compression section. ③Safety checkThen suppose that 2160sx a mm '== Therefore, the bolts at section A are safe.(2)Section CAt this section, shear force occurs maximum, equaling to 355.76kN.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)*R^2*sin(theta)^2;Np1 = P(1)*R*sin(theta)^2;Qp1 = P(1)*R*sin(theta)*cos(theta);if theta < pi/2Mp2 = 0;Np2 = 0;Qp2 = 0;elseMp2 = -0.5*(P(2)-P(1))*R^2*(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);endMp3 = -0.5*P(3)*R^2*(1-cos(theta))^2;Np3 = -P(3)*R*(1-cos(theta))*cos(theta);Qp3 = P(3)*R*(1-cos(theta))*sin(theta);Mp4 = -1/12*P(4)*R^2*(1-cos(theta))^3;Np4 = -0.25*P(4)*R*(1-cos(theta))^2*cos(theta);Qp4 = 0.25*P(4)*R*(1-cos(theta))^2*sin(theta);Mp5 = -P(5)*R^2*(cos(theta) + theta*sin(theta)-1);Np5 = P(5)*R*theta*sin(theta);Qp5 = P(5)*R*theta*cos(theta);if theta < pi/4Mp6 = 0;Np6 = 0;Qp6 = 0;else if theta >= pi/4 && theta <= 3*pi/4Mp6 = -1/3*P(6)*R^2*(cos(2*theta) - 2*cos(theta + pi/4));Np6 = 1/3*P(6)*R*(cos(2*theta) - 2*cos(theta + pi/4));Qp6 = -2/3*P(6)*R*(sin(2*theta) - 2*sin(theta + pi/4));elseMp6 = 2*sqrt(2)/3*P(6)*R^2*cos(theta);Np6 = -2*sqrt(2)/3*P(6)*R*cos(theta);Qp6 = 2*sqrt(2)/3*P(6)*R*sin(theta);endendMp = Mp1 + Mp2 + Mp3 + Mp4 + Mp5 + Mp6;Np = Np1 + Np2 + Np3 + Np4 + Np5 + Np6;Qp = Qp1 + Qp2 + Qp3 + Qp4 + Qp5 + Qp6;M1 = 1;N1 = 0;Q1 = 0;M2 = R*(1-cos(theta));N2 = cos(theta);Q2 = -sin(theta);M(k) = M1*x1 + M2*x2 + Mp;N(k) = N1*x1 + N2*x2 + Np;Q(k) = Q1*x1 + Q2*x2 + Qp;end>> MM =Columns 1 through 1038.1000 24.5627 -11.2537 -55.9618 -90.3510 -93.9478 -63.0671 -12.1028 42.3759 85.1985Columns 11 through 19105.8135 100.0987 71.2815 30.6020 -4.3031 -25.9724 -38.8660 -45.7972 -48.0002>> NN =1.0e+003 *Columns 1 through 101.7807 1.7948 1.8343 1.8913 1.95462.0121 2.0562 2.08552.1016 2.1089Columns 11 through 192.1095 2.1054 2.1009 2.0978 2.0952 2.0928 2.0916 2.0912 2.0912>> QQ =Columns 1 through 100 33.0655 54.2683 54.2120 27.8748 403.9668 355.7588 315.5075 284.2149 259.1631Columns 11 through 19234.3567 202.9612 156.9845 88.6816 34.6814 20.9562 12.0903 5.60980.0000。