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1 分类号 TU46 学号学位论文高墩大跨度预应力混凝土连续刚构桥施工控制研究 Construction Control Studies of Long-span Prestressed Concrete Continuous Rigid-frame Bridge with High Pier 陈波指导教师姓名董军教授北京建筑工程学院赵赤云副教授北京建筑工程学院申请学位级别硕士专业名称结构工程论文提交日期 2008 年 2 月 5 日论文答辩日期 2008 年 2 月 3 日学位授予单位和日期北京建筑工程学院 2009 年月日答辩委员会主席评阅人 2008 年 2 月

5 Construction Control Studies of Long-span Prestressed Concrete Continuous Rigid-frame Bridge with High Pier ABSTRACT With the rapid transit development of China highway and railway, more longspan bridges are needed to pan great rivers and bays, PC continuous rigid-frame bridge emerged as the times require and was developed fleetly in recent years. The PC rigid frame bridge can meet with the carrying capacity of long span bridge. This kind of structure is fit for the comfortable steer with no other expansion joints except forthose at the two ends. It characterized the wholeness of structure, good astigmaticability and load-bearing ability. construction rapidness, bright and pithiness bridge style, so rigid frame bridge have taken an increasingly great proportion in bridges constructed and those under construction at city and county. Through the analysis and summary of the development of PC continuous rigid frame bridge and the developmental tendency of the construction control techniquesin our country and abroad, the paper gives the study of the construction control techniques for long-span prestres sing PC continuous rigid frame bridge. According to its special character of the construction control, the paper discusses that every parameters have effect on the inner force and displacement of the structure.the paper suggests a good way of observation in the course of construction control, and establishes a monitoring system. During the construction control of DuBu first Bridge, the system makes the basic parameters and technical parameters be acquired successfully. And through establishing computer model, the paper analyses the difference between theory and actuality, and discerns and changes parameters. And the paper provides a good way of observation in the course of construction control. Key words: long-span PC continuous rigid frame bridge, the construction control, defection observation, stress measurement

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8 ... - [] [3] (Worms) 4. 2m 3 79

9 T T m (Bendorf) 80 T..2 [5] [6] [8] m(Bendorf) T m988 80m

10 270m 298m (Raft Sundet) 30m(Stolma) [6] [8] []

11 - (m) / (m) (m) Orwell /2.8 / (Gateway) /6.0 / Houston /5.7 / Mooney /7.6 / Schot twien Douter Raft Sundet /20.6 / R= Stolma

12 -2 (m) (m) (m) / /8.2 / R7000m /8. / m /8.2 / m /8.8 / m /7.8 / / / /7.8 / m /8.6 / /20.0 / /8.0 / /6.0 / /20.0 / /20.0 / /7.2 /

13 [] [5] [] 300m 330m m 30m m m 270m 260m

14 : 60 [] 000m [2] [3]

15 [2] m 76.8m [] [] [7] 0 79

16 .3 [5] [0].4 - B4. 79

17 - [] [3] [2] 2 79

18 () % 5% 6% 3 79

21 2.2.7 ( ) 6 79

22 [4] [6] [9] 2-2# 7 79

24 [9] [23] [24] : 9 79

25 3 3. 0, [8] [23] [4] [25] [26] 3- x y x i o j x 90 y 20 79

26 3. e e {} δ = δ i = δ e j u e i v e i T e e e e φ i u j v j φ (3-) j e i { F} = = F F e e j T e e e e e e N i Qi M i N j Q j M j (3-2) x y e = δ δ e i {} δ = e i { } = e = F F e e j j u e i v e i e e T e e φ i u j v j φ (3-3) j X i Y i M i X j Y j M (3-4) j e e e e e e F [ ] T (3-)(3-4) 3- { δ } e =[ T ] { δ } e (3-5) { F } e =[ T ] { F } e (3-6) λ 0 T = (3-7) 0 λ [ ] 2 79

27 cosα sinα 0 λ = sinα con α 0 (3-8) 0 0 [ ] α x x x x 3- [T] [ T ] = [ T ] T (3-9) [ K ] e [ ] { F } e = [ K ] e { δ } e (3-0) K 0 0 K K2 K3 0 K2 K3 0 K 3 K4 0 K3 K4 / 2 K e = (3-) K 0 0 K K 2 K3 0 K2 K3 0 K3 K4 / 2 0 K3 K4 K = 3 2 EA/ L K = 2EI l K = 6EI l K = 4 4EI / l 2 / 3 / E I A,I (3-5)(3-6)(3-9)(3-0) { F } e T e =[ T ] [ K ] [ T ] { δ } e =[ K ] { } e δ e (3-2) [ ] e T e K =[ T ] [ K ] [ T ] (3-3) [K] [ K ] e { P } =[ ]{ δ} K (3-4) {} δ { P} 22 79

28 e (3-4){ δ} { δ} (3-5) {} S = [ ] e F = [ K ] e { δ } e (3-)(3-5) i e i e e N j (2-0) { δ } e (3-5) N = = K u i u ) e Q j e Q = = K v v ) + K ( φ + φ ) i e e ( ( 2 i j 3 M = K v v ) + K ( φ + φ / 2) j e ( 3 i j 4 e e M = K v v ) + K ( φ / 2 + φ ) ( 3 i j 4 e e e i i e e j e i e j e j e j e t n n σ = (3-6) i= o σ i t t n+ n ε n+ c = n i= o ( σ / D) ϕ( t, τ ) i i (3-7) + t ε n c n

29 ε i ε 0i ε T n P + = B D ε n c + c = n [ i= 0 dv i n i= o T = B D ( σ / D) ϕ( t, τ ) dv i i T B σ dv ϕ( t, τ )] (3-8) σ = (3-9) i D( ε i ε 0 i) = D ε i D ε 0i i B T dv T σ i = B ( D ε i ) dv T B D ε 0 idv c = F i + Ri = F i (3-20) T F i = B D ε idv = B T DBdv ε i (3-2) T R i = B D ε 0 idv (3-22) ε 0i (2-20)(2-8) c P ε n+ = n i= 0 c F i ϕ ( t τ (3-23) n+, i ) (3-20)(3-23) F i (3-23) F i c c ϕ( t, τ ) c P ε n

30 (JTJ023-85) ϕ( t, τ ) = ( τ ) + 0.4β ( t τ ) + ϕ [ β ( t) β ( τ )] (3-24) β α d β α β d β f ϕ f h τ ( ) f f 4 qi ϕ( t, τ ) = ( ) ( )[ ] ( t τ β ) α τ + βd + C i τ e (3-25) i= f c i qi ( tτ ) = [ β ( τ ) + β ] W ( e ) c P ε n+ 4 F n a n d + n= (3-26) i= i Wn i q = = i t n c i c W n e + Fn C i ( τ n ) W 0 = F n Ci ( τ 0 ) ( i = ~ 4 ) c c τ 0 τ n F 0 F n pt ε ( t) =ε ( ) ( e ) (3-27) s s ε s( ) p τ 0 n+ c ε s + = ε s( )( tn τ 0) ε s( tn τ 0) p = [ ] ( tn τ 0 ) p ( tn+ τ 0 ε ( ) e e ) s pt = ( ) ( tn τ 0 ) p tn ε e ( e + ) (3-28) s s P ε n+ n = B D ε T + s dv H 25 79

31 H = EA ε (3-29) n+ s = [ H, 0,0, H,0, 0] T s R ε n+ (3-30) ε s ε s P n + = R n + (3-3) 3.2.3,,,, T { S '} = [ H, 0, M, H,0, M (3-32) ] [ h y ) t + y t ] h H E α A / (3-33) = ( 2 M = E α I ( t t ) 2 / h (3-34) α AI E t y t 2 h x, y ε α[ h y ) t + y t ]/ h 0 = ( 2 χ = αt t / h 2 H M E ε A (3-35) = 0 = E χ I (3-36) ε 0 χ ε 0 χ ε 0 χ 26 79

32 3.2.4 M = M 0 + M ' (3-37) M 0 M [26] [32] 3-2 e θ θ A A B f e i e B 27 79

33 3-2 4 f e e 4 f e ( x) = + l l M(x) 2 B A x + e 2 A (2-38) M ( x) i N y e( x) = N y 4 f 2 eb ea 4 f x + + ea (2-39) 2 l l (3-42)(3-43) 2 d M ( x) 8 f ( x) = = N y = (2-40) 2 dx l q 2 (3-40)(2-44) 8 f eb ea 4 f θ ( x) = e'( x) = x + 2 l l (2-4) eb ea 4 f θ A = e' = l (2-42) eb ea + 4 f θ B = e'( l) = l (2-43) 8 f θ A θb = (2-44) l N y N y θ q( x) = ( θ A θb ) = = = q (2-45) l l q θ q N [20] [2] y ( θ θ ) A B 3-3 q l 28 79

34 3-3 AC CB e ( x) = e A ea + d xq a eb + d e2( x) = d + ( x a) b AC CB Q ( Q x) = M ' ( x) = N 2 ( 2 x) = M ' ( x) = N y y e A + d a ea + d b = N θ y y = N θ C P P : B A P = θ θ (2-46) N y B A 29 79

35 4 4. [27] [28] 4.. X=- X + = 0 (4-) X + = 0 (4-2) X X X X X [ X, X (2), ΛΛX ( n) ] = (4-3) [ X, X (2), ΛΛX ( n) ] = (4-4) 30 79

36 X X ( ) ( k) + az ( k) = b ( k {,2ΛΛ }) (4-5) [ X ( k) + X ( ) ] Z ( k) = k ( a, b R, R) 2 X ( k) + az ( k) = b dx dt + ax = b (4-6) ( ) X O ( k) ; Z ( k) ( ) X O ( k) A( ) (A) 2 2 : A( ) = a 2 a a 2 22 ( a) ( a2 ) ( ) ( ) a2 a22 (4-7) ( A ) = (4-8) A X ( t + t) X (t) X ( t t) dx ( t + t, t + t) dt 4..2 dx dt : GM GM(,N) N GM(,) GM(,N) GM(,N) X X X 3 79

37 X ( i =,2,...n) X i GM(,) X n X 0 ( X, Λ, X ( n) ) = (4-9) X X X k X ( k) = X ( m) (4-0) m= ( X, X (2), Λ, X ( n) ) = (4-) Z X Z ( k) = 0.5X ( k) + 0.5X ( k ) (4-2) Z = ( Z (2), Z (3), Λ, Z ( N) (4-3) X, ( ) X X X ( ) Y Bα ) N = (4-4) (4-4)Y (4-4): N X (2) X (3) Y N = M X ( n) (4-5) Z (2) Z (3) B = M Z ( n) (4-6) ) a α = b (4-7) B ) α X ( k) + az ( k) = b k = ( 2,3, Λ, n) 32 79

38 GM () : ) ε = Y N Bα (4-8) ε J = ε T ε = min (4-9) α ) ) T T α = ( B B) B YN (4-20) a = b = n k = 2 n k = 2 Z X ( k) n k = 2 ( n ) ( k) n k = 2 X n ( n ) ( k) ( n ) Z ( k) k = 2 k = 2 k = 2 Z n ( k) Z 2 k = 2 n n Z k = 2 k = 2 n ( k) 2 Z n Z ( k) ( k) X ( k) n k = 2 Z Z 2 ( k) ( k) ( k) X 2 ( k) (4-2) (4-22) dx dt + a ( x ) = b (4-23) a,b (4-23) dx dt + ax = b (4-24) (4-2)(4-24)(4-6) : ) X ( k + ) = ak ( X b a) e + b / a (4-25) (4-24) GM(,): ) X ) ) ( k + ) = X ( k + ) X ( k) (4-26) 33 79

39 GM(,) X = ( X (2), X (3), Λ, X ( n) X : Y = ( Y (2), Y (3), Λ, Y ( n) XY : δ = ( δ (2), δ (3), Λ, δ ( n) δ ( k ) = X ( k) Y ( k) + c (4-27) k =,2, Λ,n c X ( k) Y ( k) δ X (4-0) X (4-2) Z, GM(,)( 4-8)(4-2)(4-22) a,b(4-25) X ) (4-26) ) δ ) ) X ) ( X c) ( X c) ) ) δ = (4-28) ) ( δ ( ), δ (2), Λ, δ ( n), Λ, δ ( m) ) ) ) = ( m > n) ) δ ( n +), Λ, ) δ (m), ) δ (m), δ δ (k) : ; δ ( k) = δ ( k) = 0 T T δ ( k) δ ( k), ) δ (k) c c 34 79

40 δ ( k) = δ ( k) δ ( k) > δ ( k) ) δ (k) : T T δ ( k) = δ ( k) T 0 T k c c c U ( k) = X ( k) + δ ( ) (4-29) X (k) k : i + i GM(,) x y x (k) i y (k) i : : x = ( x, x(2), Λ, x( n)) (4-30) y = ( y, y(2), Λ, y( n)) (4-3) δ = ( δ, δ (2), Λ, δ ( n)) (4-32) δ ( i) = x( k) y( k) ( i =,2, Λ, n) (4-33) 35 79

41 X = ( X, X (2), Λ, X ( n)) (4-34) X ( i) = δ ( i) + c (4-35) c x( i) y( i) (4-34) GM () X : X : 0 X = X (4-36) X ( k + ) = X ( k) + X ( k + ) ( i =,2, Λ, n) (4-37) dx dt a,b + ax = b ) a α = b : ) α = ( B T B) B T Y N ( X + X (2)) 2 B = ( X (2) + X (3)) 2 M ( X ( n ) + X ( n)) 2 Y N M X (2) X (3) = M X ( n) (4-38) : ) X ( k + ) = ( X b ) e a ak + b a (4-39) 36 79

42 # 5# cm 5#0#-4# 4.5) 2.9, 2.4, 2.3, 2.3, ( = x 4) 2.5, 2.5, 2, 2, ( = x 0.5) 0.4, 0.3,0., 0.3, ( = x = 5 c = (4.7,4.7,5.,4.6,4.5) x = (4.7,9.4,4.5,9.,23.6) x = = )) ( ) ( ( 2 (3)) (2) ( 2 (2)) ( 2 n X n X X X X X B M M = = ) ( (3) (2) n X M X X Y N ( ) T T T B B B =

43 0.003 = = ) α = ( B T B) B T Y N = = ) a α = = b dx dt + ax = b dx dt X = X ) ( k + ) = ak k ( X b a) e + b / a = 24.64e k = X ) (2) = k = 2 X ) (3) = k = 3 X ) (4) = k = 4 X ) (5) = k = 5 X ) (6) = X ( k + ) = X ( k + ) X ( k) ( k =,2, Λ, n) k = X ) (2) = = k = 2 X ) (3) = = k = 3 X ) (4) = =

44 = 4 k (5) = = X ) = 5 k (6) = = X ) ) ( ) ( ) ( k X k X k q ) = ( ) c=2 ( ) = q GM() ( ) = q = = )) ( ) ( ( 2 (3)) (2) ( 2 (2)) ( 2 n q n q q q q q A M M ( ) T T T T A A A = = = = ' ' b a 39 79

45 ) q ( k + ) = a' k ( q b' a' ) e + b' / a' ) q k k ( a )( q b a ) e a ' ' ' '.77e ( k + ) = = k = q ) (2) = k = 2 q ) (3) = k = 3 q ) (4) = k = 4 q ) (5) = k = 5 q ) (6) = c=2 k = q ) (2) = k = 2 q ) (3) = k = 3 q ) (4) = k = 4 q ) (5) =. 595 k = 5 q ) (6) = k = X ) (2) = = k = 2 X ) (3) = = k = 3 X ) (4) = = k = 4 X ) (5) = = k = 5 X ) (6) = = # = U = = U cm 5 x 5 = x 5 5# 40 79

46 = 900 % 2% a) b) c)

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49 5.2 BridgeSB BridgeSB BridgeSB

50 [5] [6] 5- (Pa) (Pa) (t/m ) αl (m) (m 2 ) (m 2 ) (MPa) (mm) m T T2 T=-30tT2=80t 45 79

51 T T2 T=-27t T2=84t JTJ

52 / T F T2F T3F T4F T5F T6F T7F

53 26 38 T8F T9 T9'F T0T0' TT' T2T2' T3T3' T4T4' T5T5' B-B5 Z-Z H-H Z-Z H3-H Z-Z H3-H

56 cm -7.6cm 5 79

57 6 20MPa 200MPa 5.3 [30] [3] [30] [33] i i- i i

58 a. i i i b. i c. q H i n=h i +H i +H i n (5-) H i n in(ni H i ni) H i i H i i H i n in H i n=h i +H i +H i n+ H i +H (5-2) 53 79

59 H i n i H i i H i i i H i n ( 2) H i n= H i - H i- H i H H i +H i H i +H i +H i n( ) (5-2)

61 cm cm 4.5m φ20 (DS)

62 :005:00 7:009:00 6:008:

63 %80% 30%80% 60%0% i j j i i j 58 79

64 2cm (mm)

65 cm 5.5cm 0 2cm 0 30%70% 60 79

66 Er a 40mm40mm 0 0 () 2 500m (m) () () () G D G D G D G D 6:

67 8: : : : : : (mm) (m) G D G D G D G D 6: : : : : : : (mm) mm b

68 c. d. e a. 5,, 0mm 2mm H = H + H+2mm (5-3) H 63 79

69 22 b ( ) (cm) (cm) (cm) (: 2cm) c cm.7cm 64 79

70 m

71 JMZX-25T JMZX-200X 66 79

72 Z Z Z Z Z + Z 8 + Z 7 3 Z Z 2 + Z 6 2 Z Z 3 + Z 4 + Z

75 5-23 Z Z Z + Z 2 + Z 3 + Z

77 ( ) () [5] [6] # 3# 2#3# E 4d7d4d28d60d E γ 72 79

78 5.5.2 # 7 L/4 ( 0 ) N = ( F F f f + A) K N (KN) F f 73 79

79 AK L/

80 6 6. [4] [8]

81 76 79

82 [].[M] [2].().:996 [3]..:200 [4]..:999 [5]. [D] [6]..:2000 [7] [8]..:200 [9]. [M]. :, 998 [0] []. 270m.200(5) [2] [3].[J].,2003 [4].[J].2005, 25(3):662. [5]. [R] [6]. 2 [R] [7]. 605 [J]. 2004, (9):4449. [8]. [J] [9]. [J] [20]. [M]. 998 [2]. [M]. 200 [22]. [J]. Vol.0, No. [23]. [J]. 998No.2: 0~5 [24]. [M]., 200 [25]. [M].,

83 [26].[M].200. [27] [28]..993(3) [29].[J] [30]..998(3) [3]. [S]., 2004 [32].[C] [33].((JTJ02-89).:989 [34].(JTJ ).: 989 [] [2] [3],,,,. [J]., 2007, (9). [4],,. [J]., 2007,33(34): [5],,,,, [J]., ( )

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