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1、Exercise TwoPart One: Design Data(1)Function of TunnelThe planned tunnel is to be used as a subway tunnel.(2)Design ConditionsDimensions of SegmentType of segment: RC, Flat typeDiameter of segmental lining: D0=9500mmRadius of centroid of segmental lining: Rc=4550mmWidth of segment: b=1200mmThickness
2、 of segment: t=400mmGround 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 late
3、ral earth pressure: =0.4Surcharge: P0=39.7kN/m2Soil condition: SandyMaterialsThe 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 stren
4、gth: fBy=240N/mm2Shear strength: =150N/mm2Parameters for joint spring:( constant of rotation spring for positive moment at joint)( constant of rotation spring for negative moment at joint)Design MethodThe shield tunnel shall be designed in accordance with the National Code.How to compute member forc
5、es? Force method (Part two), Elastic equation method (Part three) and Bedded frame model method (Part four) are demonstrated in this section, respectively.How to check the safety of lining? Limit state method.(3) Geometric Design of Shield LiningThe shield lining is fitted with 9 segmental pieces (o
6、ne Key-type segment, two B-type segments and six other segments). Central angle of each segment piece is 40 degrees.Part Two: Computation by Force Method(1) Load ConditionsFigure 1Judgment of Tunnel Type ( by Terzaghis formula )B1=R0cot8+4=4.750cot1808+304=8.23mh0=B11-cB1tan1-exp-HB1tan+P0exp-HB1tan
7、 =8.231-0tan301-exp-12.38.23tan30+39.7exp-12.38.23tan30=24.99mH=12.3mSo the designed tunnel is a shallow tunnel.Load Types and Partial FactorsLaod typesPartial factorsLoad typesPartial factorsSurcharge1.4Earth pressure1.2Dead load1.2Subgrade reaction1.2Water pressure1.2Computation of Loads:Computati
8、onal 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 tunnel Vertical pressure at tunnel crownEarth pressure:pe1=b1.4P0+1.2H=1.21.439.7+1.281
9、2.3=208.4kN/mWater pressure:pw1=1.2bwHw=1.21.21012.9=185.8kN/mq1=pe1+pw1=208.4+185.8=394.2kN/mq2=1.2b0.215R=1.21.20.2159.5218=26.5kN/mp1=q1+q2=394.2+26.5=420.7kN/mVertical pressure at tunnel botto,p2=p1+1.2bct=420.7+1.21.23.14260.4=420.7+47.0=467.7kN/mWhere c=Unit wight of RC segment=26kN/m3Lateral
10、pressure at tunnel crownEarth pressure:qe1=pe1+1.2bt2=0.4208.4+1.21.280.42=84.3kN/mWater pressure:qw1=1.2bwHw+t2=1.21.21012.9+0.42=188.6kN/mp3=qe1+qw1=84.3+188.6=272.9kN/mLateral pressure at tunnel bottomp4=1.2bDc+wDc=1.21.20.489.1+109.1=173.0kN/mp3+4=p3+p4=272.9+173.0=445.9kN/mWhereDc=Computational
11、 diameter=D0-t=9.5-0.4=9.1mAverage self-weightp5=1.2bct=1.21.2260.4=15.0kN/mLateral resistance pressure =2p1-p3-p3+4+p5Rc424EI+0.0454KRc4=2420.7-272.9-445.9+153.14154.55410-3240.83.01046.410-3+0.0454604.554 =2.29410-3mph=kb=501032.29310-31.2=137.6kN/mp6=ph1-2cos2=-68.8kN/m 43 4Where=Displacement of
12、lining at tunnel spring=Reduction factor of model rigidity=0.8E=Modulus of elasticity of segment=3.0104N/mm2I=Moment of inertia of area of segment=1/121.20.4=6.410-3m4k=Coefficient of reaction=50MN/m3（recommend values can be refer to table9.2,after Liu and Hou,1997）K=kb=501.2=60MN/m2=the angle measu
13、red from the vertical direction around the tunnerl(2) Computation of Member ForcesFigure 3 shows the simplified model of the segmental lining.Figure 3 Simplified model diagram for calculationCalculation dataKP=18070kNm/rad (if inside part of lining is tensile)KN=32100kNm/rad(if outside part of linin
14、g is tensile)EI=3.01041036.410-3=192000kNm2Coefficients Calculationn=n1+n2+n3+n4=1+1+1+2=5Note: 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. 11=REI+i=1n1K(i)=REI+2.5KP+2KN=4.55
15、192000+2.518070+232100=2.75110-4/kNm12=R2EI+Ri=1n1Ki1-cosi=R2EI+2.826RKP+1.674RKN =4.552192000+2.8264.5518070+1.6744.5532100=1.28810-3/kN22=3R32EI+R2i=1n1Ki1-cosi2=3R32EI+5.122R2KP+1.627R2KN=34.5532192000+5.1224.55218070+1.6274.55232100=0.92310-2m/kN1p1=-p1R34EI-12p1R2i=1n1Kisin2i=-p1R2R4EI+0.265KP+0.860KN=-420.74.5524.554192000+0.
