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Luận án tiến sĩ Kỹ thuật Địa chất: Ứng xử của đất nền dưới đáy móng công trình biển trọng lực chịu tác động của tổ hợp các dạng tải trọng

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Thông tin cơ bản

Tiêu đề Ứng xử của đất nền dưới đáy móng công trình biển trọng lực chịu tác động của tổ hợp các dạng tải trọng
Tác giả Nguyễn Văn A, Bùi Văn B
Người hướng dẫn PGS.TS. Nguyễn Văn C, PGS.TS. Trần Văn D, TS. Lê Văn E
Trường học Đại học Bách khoa Hà Nội
Chuyên ngành Kỹ thuật Địa chất
Thể loại Luận án tiến sĩ
Năm xuất bản 2019
Thành phố Hà Nội
Định dạng
Số trang 136
Dung lượng 2,25 MB

Cấu trúc

  • 2. ĈӕLWѭӧQJYjSKҥPYLQJKLrQFӭX (16)
  • 7. Phѭѫng pháp nghiên FӭXFӫDOXұQiQ (0)
  • 9. éQJKƭDNKRDKӑFYjWKӵFWLӉQFӫDOXұQiQ (20)
  • 10. Cѫ VӣWjLOLӋXFӫDOXұQiQ (0)
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    • 4.2 Bài toán so sánh ± 7tQKWRiQVӭFFKӏXWҧLFӫDQӅQWKHR&ѫ KӑFÿҩW (102)
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      • 4.2.3 Tính lún cho công trình (103)
    • 4.3 Mô hình bài toán trong PLAXIS (110)
      • 4.3.2 Mô hình bài toán (110)

Nội dung

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35 Hình 1.14 0һWFҳWÿӏDFKҩQPLQKKRҥFiFGҥQJEү\FҩXWU~F [4]

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Lӵc song song vӟLKѭӟQJJLyÿѭӧF[iFÿӏnh bӣi công thӭc: ܨ ஽ ൌͳ ʹܥ ௗ ߩܸ ௓ ଶ ܣ (1.1)

48 /ӵFQj\FNJQJÿѭӧFJӑi là lӵFNpRYjQyFy[XKѭӟng xô nghiêng công trình và gây tác dөng chính tҥi thӡLÿLӇPÿҧo chiӅu

Lӵc vuông góc vӟLKѭӟQJJLyÿѭӧF[iFÿӏnh bӣi công thӭc: ܨ ௅ ൌͳ ʹܥ ௅ ߩܸ ௓ ଶ ܣ (1.2)

/ӵFYX{QJJyFYӟLKѭӟQJJLyFy[XKѭӟQJQkQJF{QJWUuQK

$ 'L QWtFKYX{QJJyFYͣLY̵QW͙FJLy

9ұQWӕFJLyNK{QJOLrQWөFEӣLQKӳQJOӵFFҳWYӟLEӅPһWFӫDWUiLÿҩWQyOjVӕNK{QJӣ EӅPһWYjWăQJWKHRFҩSVӕQKkQYӟLPӝWWӕFÿӝ tӕLÿDJLӟi hҥQÿѭӧc gӑi là gió biӃQÿәi Trên mһWQѭӟc tӕFÿӝ gió ӣ ÿӝ FDRYjÿѭӧFÿҥi diӋn bӣLFiFÿӏnh luұWKjPPNJPӝt phҫn bҧy ܸ ௓ ൌ ܸ ௥௘௙ ൬ ܼ ܴ݂݁൰ భ ళ

9UHI 7͙Fÿ͡JLyͧÿ͡FDRIWPÿ͡FDRWK{QJWK˱ͥQJÿ˱ͫFGQJ

= &KL͉XFDRPRQJPX͙QE̹QJÿ˯QY͓feet (ft)

&iFKLӋXӭQJJLyWUrQWҩWFҧFiFEӝSKұQFӫDNӃWFҩXWUrQPһWQѭӟFSKҧLÿѭӧFWtQKWRiQ+DLORҥLWӕFÿӝJLyWKѭӡQJÿѭӧFWtQKWRiQ

49 7ӕFÿӝJLyÿӅXÿѭӧFÿӏQKQJKƭDQKѭOjWӕFÿӝJLyWUXQJEuQKWURQJNKRҧQJWKӡL JLDQPӝWSK~W

7ӕFÿӝJLyJLұWÿѭӧFÿӏQKQJKƭDQKѭOjWӕFÿӝJLyWUXQJEuQKWURQJPӝWNKRҧQJ WKӡLJLDQJLk\+ӋVӕJLyJLұWOjKӋVӕQKkQÿyÿѭӧFQKkQYӟLWӕFÿӝJLyÿӅXÿӇ FyÿѭӧFWӕFÿӝJLyJLұWKRһFQKDQKQKҩW- Yұn tӕc theo dһm HӋ sӕ gió giұt trung bình (F10), ӣ ÿӝ cao 30ft nҵm trong khoҧng 1,35-1,45

Mӝt công thӭFWKѭӡQJÿѭӧc chҩp nhұn cho viӋc tính toán các áp lӵc gió: ܨ ൌ ܭܸ ଶ ܥ ௌ ܣ (1.4)

A 'LӋQWtFKѭӟFOѭӧQJ

+ӋVӕKuQKGҥQJÿLӇQKuQKFKRPӑLJyFÿӝFӫDSKѭѫQJSKiSWLӃSFұQJLyFyWKӇOj:

&ҩXNLӋQGҥQJGҫPÿӭQJCs = 1,5

Hai WKjQKSKҫQFKtQKFӫDOӵFGzQJFKҧ\OjGzQJWULӅXYjJLyOjPÿәLKѭӟQJGzQJFKҧ\'zQJFKҧ\GRJLyӣEӅPһWFӫDQѭӟFWƭQKÿѭӧF[HPOjFӫDWӕFÿӝJLy ÿӅXӣ 30ft (10m) WUrQPӵFQѭӟFWƭQK9ұQWӕFKLӋQWҥLQrQÿѭӧFEәVXQJWKHRNLӇXYHFWRUYӟLYұQWӕFVyQJ WUѭӟFNKLWtQKWRiQOӵFNpR%ӣLYuOӵFNpRSKөWKXӝFYjREuQKSKѭѫQJYұQWӕFQJDQJYjGzQJFKҧ\JLҧPQKҽWKHRÿӝVkXPӝWGzQJFKҧ\

50 WѭѫQJÿӕLQKӓFyWKӇWăQJOӵFFҧQÿiQJNӇ7URQJWKLӃWNӃFKLӅXFDRVyQJWӕLÿD ÿ{LNKLÿѭӧFNӇÿӃQJLiWUӏJLDWăQJWӯãÿӇWtQKFiFWiFÿӝQJFӫDGzQJFKҧ\ YjOӵFGzQJFKҧ\VӁÿѭӧFEӓTXD, [6]

1KѭYұ\OӵFGzQJFKҧ\ÿDQJÿѭӧFWtQKWRiQWӯWUѭӟFÿӃQQD\OjVӵNӇÿӃQJLiWUӏ JLDWăQJFӫDWҧLWUӑQJVyQJPjEӓTXDOӵFWiFÿӝQJ FӫDGzQJFKҧ\

7ҥLFiF NKXYӵFFyWX\ӃWKӋNӃWFҩXQrQÿѭӧFWKLӃWNӃÿӇFKӏXWҩWFҧFiFWҧLWUӑQJ WƭQKYjWҧLWUӑQJWX\ӃWSKӫ7URQJWҩWFҧFiFQѫLFyEăQJWU{LFyWKӇWӗQWҥLQKӳQJ WҧQJEăQJFyÿӝGj\UҩWOӟQYjYұQWӕFPjEăQJFyWKӇYDFKҥPFiFNӃWFҩXWKu FҫQÿѭӧFWtQKWRiQ[iFÿӏQK9ӟLPҧQJKRһFGҧLEăQJWUҥQJWKiLWӟLKҥQFӫDEăQJ FNJQJÿѭӧF[iFÿӏQKӢPӝWVӕQѫLWX\ӃWFyWKӇWtFKONJ\WUrQNӃWFҩXYjFiFNӃW FҩXWU~FSKtDWUrQ3KҧLFyJLӟLKҥQÿӕLYӟLFiFWiFÿӝQJFӫDVӵWtFKONJ\Qj\WKHR WtQKәQÿӏQKYjӭQJVXҩWWURQJNӃWFҩX&iFOӵFEәVXQJÿӇNKiQJOҥLJLyGRWtFK WөEăQJQKѭYұ\FNJQJÿѭӧFEDRJӗPWURQJJLӟLKҥQQj\, [6]

9ӟLÿһFÿLӇPÿӏDOêFӫD%LӇQĈ{QJQyLFKXQJFNJQJQKѭ7UNJQJ1DP&{Q6ѫQQyL ULrQJWKuÿLӅXNLӋQEăQJWX\ӃWWiFÿӝQJOrQF{QJWUuQKELӇQOjNK{QJWӗQWҥL9u Yұ\WҧLWUӑQJEăQJWX\ӃWÿѭӧFORҥLEӓWURQJWKLӃWNӃÿӕLYӟLNKXYӵFÿDQJ[pWFӫD QJKLrQFӭXQj\

7UѭӡQJKӧSF{QJWUuQKOjPӝWNKXYӵFFyOӏFKVӱÿӝQJÿҩWNӃWFҩXFҫQÿѭӧFWKLӃW NӃÿӇFKӕQJOҥLFiFOӵFQJDQJOӵFGӑFYjQKӳQJPRPHQWSKiWVLQKWӯFiFKRҥW ÿӝQJÿӏDFKҩQ&iFWK{QJWLQÿѭӧFVӱGөQJÿӇWKLӃWOұSPӝWWUұQÿӝQJÿҩWWKLӃWNӃ WKѭӡQJÿѭӧFOҩ\WӯGӳOLӋXJKLQKұQWUұQÿӝQJÿҩWWUѭӟFWҥLKRһFJҫQYӏWUtF{QJ WUuQKQӃXGӳOLӋXÿyÿӅXFyVҹQ

%ӣLYuFyUҩWtWWUҥPJKLQKұQVRYӟLVӕOѭӧQJFiFNKXYӵFFyÿӝQJÿҩWNKҧQăQJFKӑQYӏWUtFyPӝWP{WҧOӏFKVӱÿҫ\ÿӫYӅFiFKRҥWÿӝQJÿӏDFKҩQOjUҩWQKӓ7URQJWUѭӡQJKӧSQKѭYұ\ÿLӅXNLӋQYӏWUtF{QJWUuQKQrQÿѭӧFVRViQKYӟLÿLӅXNLӋQFӫDPӝWYӏWUtFyÿӏDFKҩWYjNLӃQWҥRWѭѫQJWӵQKDXPjFyVӕOLӋX, [6].

1.4 7әQJTXDQYӅӭQJ[ӱFӫDÿҩWQӅQGѭӟLÿi\PyQJCTBTL

1.4.1 Tình hình QJKLrQFӭXYӅӭQJ[ӱFӫDÿҩWQӅQGѭӟLÿi\PyQJF{QJWUuQKELӇQ

Công WUuQKELӇQWUӑQJOӵFÿѭӧF[k\GӵQJÿҫXWLrQWUrQWKӃJLӟi WӯQăP3 YjYүQÿDQJ ÿѭӧFQJKLrQFӭX WKLӃWNӃYj[k\GӵQJ FKRÿӃQQJj\QD\7X\QKLrQORҥLKuQKNӃWFҩX Qj\FKѭDÿѭӧFSKiWWULӇQUӝQJUmLFiFF{QJWUuQKÿm[k\GӵQJWKHRQJX\rQOêWUӑQJOӵF EҵQJNӃWFҩXErW{QJFӕWWKpSFKӫ\ӃXOjFiFGjQNKRDQGҫXNKtÿѭӧF[ây GӵQJFKӫ\ӃX ӣNKXYӵF%LӇQ%ҳF Canada, Na Uyô

3+ặ17ậ&+7+é1*6Ӕ6ẽ1*%Ҵ1*/ộ7+8<ӂ7+Ӗ,48< VÀ QU<+2Ҥ&+7+Ӵ&1*+,ӊ0

ĈӅ[XҩWPӟLFiFK[iFÿӏQKWK{QJVӕVyQJELӇQ

2.2.1 &iFK[iFÿӏQKFiFWK{QJVӕVyQJELӇQ

Trong OXұQiQWiFJLҧÿӅ[XҩW VӱGөQJ PӝWSKѭѫQJSKiS PӟLÿӇ[iFÿӏQKFiFWK{QJVӕ VyQJSKөFYөFKRTXiWUuQKWtQKWRiQYjWKLӃWNӃ+jPKӗLTX\ÿѭӧFVӱGөQJÿӇSKkQWtFK Yj[ӱOêFiFVӕOLӋXTXDQWUҳFÿѭӧFWӯFiFWK{QJVӕVyQJQKѭFKXNǤPDKѭӟQJJLy WҫQVXҩWYY

+ӗLTXy WKѭӡQJÿѭӧF[ӃSYjRORҥLEjLWRiQWӕLѭX vì FK~QJWDQӛOӵFÿӇWuPNLӃPPӝW JLҧLSKiSÿӇFKRVDLVӕYjSKҫQGѭ là WӕWQKҩW3KkQWtFKKӗLTX\ÿѭӧFVӱGөQJWKjQK F{QJWURQJQKLӅXOƭQKYӵFYjFK~QJW{LPXӕQÿӅ[XҩWQKѭPӝWFiFKPӟLÿӇJLҧLTX\ӃWbài toán sóng WKӵF

.ӃWTXҧWtQKWRiQFiFWK{QJVӕVyQJEҵQJSKѭѫQJSKiSWKӕQJNrWUrQFiFJLiWUӏTXDQWUҳFVyQJWKӵFWURQJPӝWNKRҧQJWKӡLJLDQKӳXKҥQWҥLYQJELӇQÿһWF{QJWUuQK ÿmFKRJLiWUӏ ÿiQJWLQFұ\KѫQYuFyFѫVӣWRiQKӑFU}UjQJKѫQWK{QJTXDOêWKX\ӃWKӗLTX\YjTX\KRҥFKWKӵFQJKLӋP

2.2.2 3KѭѫQJSKiS[iFÿӏQKWҧLWUӑQJVyQJFӵFÿҥLWiFÿӝQJOrQWUө&7%7/Fy

WLӃWGLӋQWKD\ÿәLWURQJYQJQѭӟFVkX

7ҧLWUӑQJVyQJWiFGөQJOrQ&7%7/ÿѭӧFOҩ\WURQJÿLӅXNLӋQEҩWOӧLQKҩW.KLÿyWUөÿӥ FӫDF{QJWUuQKOjEӝSKұQJiQKFKӏXFiFWiFÿӝQJFӫDVyQJELӇQ&KRÿӃQQD\YLӋFQJKLrQ FӭXYjWtQKWRiQWҧLWUӑQJVyQJWiFÿӝQJOrQWUөÿӥYүQFzQQKLӅX\ӃXWӕSKӭFWҥSFҫQ SKҧLWLӃSWөFÿѭӧF[HP[pW

9LӋF[iFÿӏQKWҧLWUӑQJVyQJWiFÿӝQJOrQWUөÿӥFKӫ\ӃXGӵDYjRSKѭѫQJWUuQK0RULVRQ 7X\Yұ\SKѭѫQJWUuQK0RULVRQFKӍJLӟLKҥQEӣLWҧLWUӑQJVyQJWiFGөQJOrQWUөWKҷQJ ÿӭQJFyWLӃWGLӋQNK{QJÿәLWUzQKRһFWLӃWGLӋQEҩWNuĈLӅXQj\NK{QJxác WKӵFYӟLÿһF ÿLӇPFҩXWҥRFӫDF{QJWUuQKYuWKӃWURQJWtQKWRiQQJѭӡLWDWKѭӡQJÿѭDYjRFiFKӋVӕ ÿLӅXFKӍQKWKLrQYӅDQWRjQ, [20]

/XұQiQ Qj\WiFJLҧÿӅ[XҩWiSGөQJSKѭѫQJWUuQK0RULVRQYjWKӵFKLӋQFiFELӃQÿәL WRiQKӑFÿӇWtQKWRiQWҧLWUӑQJVyQJWiFÿӝQJOrQWUөÿӥNtFKWKѭӟFOӟQFyWLӃWGLӋQWKD\ ÿәLӣYQJQѭӟFVku

*LҧWKLӃWYүQGQJVyQJ$LU\YjF{QJWKӭF0RULVRQYu

&KLӅXVkXQѭӟFQJKLrQFӭXKP ĈӝGjLVyQJO # 120 m WKѭӡQJOӟQKѫQÿѭӡQJNtQKG, [20]

*LҧLWtFKKӗLTX\OjP{QWRiQKӑFYҥQQăQJÿѭӧFVӱGөQJÿӇQJKLrQFӭXFiFPӕLTXDQ KӋFyWtQKWKӕQJNr6ӵ[XҩWKLӋQFӫDJLҧLWtFKKӗLTX\OLrQTXDQÿӃQYLӋFQJKLrQFӭX FiFPӕLSKөWKXӝFWURQJVLQKKӑFNKLSKiWWULӇQQyÿmOLrQTXDQÿӃQFiFEjLWRiQYӅWKt QJKLӋP

*LҧLWtFKKӗLTX\OjFѫVӣOêWKX\ӃWFKRSKpSSKkQWtFKQKkQWӕYjÿһWQӅQPyQJFKREjLWRiQTX\KRҥFKWKӵFQJKLӋP

59 3KѭѫQJSKiSSKkQWtFKKӗLTX\WҥRÿLӅXNLӋQÿӇ[iFÿӏQKFiFP{WҧWRiQKӑFFӫDFiFÿӕL WѭӧQJFyÿһFWUѭQJFKѭDELӃWWUrQFѫVӣTXDQWUҳFFiFJLiWUӏYjRYjra (hình 2.1)

Hình 2ĈӕLWѭӧQJSKLWX\ӃQQKLӅXFKLӅX [21]

XpWPӝWÿӕLWѭӧQJFyÿҫXYjRx 1 , x 2 , x 3 ôx S YjPӝWÿҫXUDyĈӕLWѭӧQJQj\FKӏXVӵQKLӉX ORҥQz QjRÿy(hình 2.1)9LӃWÿӕLWѭӧQJEӣLPӝWÿһFWUѭQJSKLWX\ӃQFKѭDELӃW y= f(x 1 , x 2 , x 3 ô[ S , z) (2.1)

1KkQWӕJk\QKLӉXORҥQz WKD\ÿәLOLrQWөF9ӟLQKӳQJJLiWUӏ[iFÿӏQKFӫDÿҫXYjRx 1 , x 2 , x 3 ô[ S FyWKӇQKұQÿѭӧFFiFJLiWUӏFӫDÿҫXUDy

7DTXDQWkPPӕLSKөWKXӝF2FӫDÿҫXUDy YjÿҫXYjRx 1 , x 2 , x 3 ô[ S, WKѭӡQJQyNK{QJ SKҧLOjPӕLSKөWKXӝFKjPVӕPjOjPӕLSKөWKXӝFEӏJk\QKLӉXORҥQ

.KiLQLӋPYӅPӕLSKөWKXӝFKjPVӕÿѭӧFGQJWURQJQKLӅXOƭQKYӵFNKRDKӑFYjNƭ WKXұWNKiFQKDXWURQJÿyFyQJjQKJLҧLWtFKWRiQKӑF7KӵFFKҩWPӕLSKөWKXӝFKjPVӕ Qj\OjVӵWKD\ÿәLFӫDy khi x WKD\ÿәLFyWtQKÿѫQWUӏ(hình (2.2)

/ҫQOѭӧWPӛLJLiWUӏFyWKӇFӫDx Qy[iFÿӏQKPӝWJLiWUӏy WӯÿyYLӃWÿѭӧF

7URQJSKpSSKkQWtFKWKӕQJNrWKuNKiLQLӋPy=f(x) Qj\OjUҩWKҽSYuWKӃQJѭӡLWDÿѭD YjRNKiLQLӋPWәQJTXiWKѫQÿyOjPӕLSKөWKXӝFEӏJk\QKLӉXORҥQ7KӵFUDNKLTXDQ WUҳFKjPVӕ2OjtWJһSEӣLWDNK{QJORҥLEӓÿѭӧFQKkQWӕҧQKKѭӣQJFyWtQKQJүX QKLrQ9ӅQJX\rQWҳFWURQJWKӵFWӃFiFPӕLSKөWKXӝFTXDQWUҳFÿѭӧFOjPӕLSKөWKXӝF EӏQKLӉXORҥQGRÿyFiFPӕLSKөWKXӝFQj\OjÿDWUӏ(hình 2.3)

7KӵFFKҩWPӕLSKөWKXӝFEӏQKLӉXORҥQOjPӕLSKөWKXӝFFӫDPӝWELӃQQJүXQKLrQy và ELӃQQJүXQKLrQxQyOjPӕLSKөWKXӝFÿѫQWUӏFӫDVӵSKkQSKӕL[iFVXҩWWӯELӃQQJүX nhiên y SKөWKXӝFYjRJLiWUӏx ÿmFKӑQWK{QJTXDELӃQQJүXQKLrQx

Hình 20ӕLSKөWKXӝFQKLӉXORҥQ [21]

&yWKӇQyLELӃQQJүXQKLrQY OjSKөWKXӝFQKLӉXORҥQYjRELӃQQJүXQKLrQX QӃXQKѭ

SKpSSKkQSKӕLF FӫDELӃQQJүXQKLrQY YӟLÿLӅXNLӋQX=x OjPӝWKjPFӫDJLiWUӏy FӫD ELӃQQJүu nhiên Y PjFӫDFҧJLiWUӏx

.KL[iFÿӏQKFiFKP{WҧWRiQKӑFFӫDÿӕLWѭӧQJQJѭӡLWDTXDQWkPQKҩWÿӃQJLiWUӏNu YӑQJFӫDELӃQQJүXQKLrQY YӟLÿLӅXNLӋQX=x QӃXJLiWUӏNuYӑQJFyÿLӅXNLӋQFӫDELӃQ QJүXQKLrQY OjPӝWKjPFӫDx:

O~FQj\WDQyLFyVӵWѭѫQJTXDQJLӳDELӃQQJүXQKLrQY và X

61 KLQKұQӣÿLӇPnELӃWFiFJLiWUӏÿҫXYjRx 1 , x 2 , x 3 ô[ S YjWѭѫQJӭQJYӟLQyOjJLiWUӏ ÿҫXUDy FӫDÿӕLWѭӧQJÿmWUuQKEj\(hình 2.1)O~FÿyWDQKyPVӕOLӋXQj\WKjQKEҧQJ

7URQJÿyx ns OjJLiWUӏÿҫXYjRFӫDx s WҥLÿLӇPn, y N OjJLiWUӏÿҫXUDFӫDy WҥLÿLӇPn 0өFÿtFKOjSKҧL[iFÿӏQKÿһFWUѭQJWKD\WKӃFyGҥQJÿmFKӑQW\ê

Trong (2.6) cú kFiFKӋVӕFKѭDELӃWb k (k ô.1-1) ĈӕLYӟLPӝWVӕÿLӇPÿmFKRn thì hàm (2QKұQPӝWJLiWUӏ

, ệ n f(x n 1 x n 2 x nS b 0 b 1 b K y (2.7) ĈһFWUѭQJ2[ҩS[ӍÿһFWUѭQJ2PjÿһFWUѭQJ2FKѭDELӃWYuQKLӉXORҥQzӢ FiFWKӡLÿLӇPWLӃSWKHRn (Q ô1) WKѭӡQJFNJQJOjGRFKѭDELӃWÿѭӧFGҥQJFyWtQK FKҩWSKLWX\ӃQFӫDÿӕLWѭӧQJ2.1) ĈһFWUѭQJ2[ҩS[ӍÿһFWUѭQJ2WKѭӡQJÿѭӧFJӑLOjP{KuQKWRiQKӑFFӫDÿӕL WѭӧQJYӁWUrQ(hình 2.1) ĈӇ[iFÿӏQKWӕLѭXFiFKӋVӕb 0, b 1, b k FӫDP{KuQKWRiQKӑF2WDFKӑQP{KuQK Qj\GѭӟLGҥQJKjP

S 1 , 2 , ; ệ 1 ,ệ 2 , ệ , (2.8) Qy[iFÿӏQKNKRҧQJFiFKJLӳDWұSKӧSFiFJLiWUӏÿҫXUD^ y 1 ,y 2 , y N `FӫDÿӕLWѭӧQJ2.1) YjÿҫXUD^ yệ 1 ,yệ 2 , yệ N ` FӫDP{KuQK2.7)

62 7URQJWKӵFWӃFKӍWLrXÿѫQJLҧQQKҩWOjNKRҧQJFiFK(XFOLG trong không gian n FKLӅXPj QyGүQÿӃQEjLWRiQ[iFÿӏQKPӝWQKyPWӕLѭXFiFKӋVӕb 0,b 1, b k FӫDP{KuQK2.7), WӯÿyÿѭDÿӃQEjLWRiQOjPFӵFWLӇXELӇXWKӭF

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7UѭӟFWLrQWDQJKLrQFӭXQKѭOjPӝWWUѭӡQJKӧSULrQJ[iFÿӏQKFiFKӋVӕFӫDKjPKӗi TX\WX\ӃQWtQKQKLӅXFKLӅX

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Các NӃWTXҧÿmFKRWKjQKEҧQJTXDQWUҳFFӫDÿҫXYjRÿҫXUD2GѭӟLGҥQJPDWUұQ ĈҫXYjROj; ằ ằ ằ ẳ º ô ô ô ơ ê

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&iFKӋVӕFKѭDELӃWFNJQJÿѭӧFYLӃWGѭӟLGҥQJYpFWѫ ơ b b b S ẳ b 0 , 1 , (2.20)

7ӯÿyFiFJLiWUӏKjPKӗLTX\2.14) VӁFyGҥQJ

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7DQJKLrQFӭXÿӕLWѭӧQJ(hình 2.1) ÿѭӧFYLӃWEҵQJKjP [22]: y = f(x 1 ,x 2 ô[ s ,z) (2.26)

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65 JLiWUӏÿҫXUDWѭѫQJӭQJy 1 , y 2 ôô\ n FӫDÿӕLWѭӧQJ7UrQFѫVӣFiFWKtQJKLӋPQj\VӁ QKұQÿѭӧFPӝWEҧQJVӕOLӋXQKѭVDX : áá áá á ạ ã n nS n n

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&ӵFWUӏWӯQJSKҫQFy2 2-S WKtQJKLӋP ĈӇQkQJFDRÿӝWLQFұ\WD[pWPӝWYtGөÿѫQJLҧQOұSNӃKRҥFKKDLPӭFÿӇPLQKKӑD SKѭѫQJSKiSWӕLѭXKyD7uPFӵFÿҥLFӫDÿҫXUDFӫDÿӕLWѭӧQJS=3 ÿҥLOѭӧQJYjRQy ÿѭӧFYLӃWEҵQJKjPSKLWX\ӃQ y =f(x 1 , x 2 , x 3 ), (2.30)

WURQJYQJQyÿѭӧFJLӟLKҥQOj ° ¿ ° ắ ẵ d d d d d d

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7ҥLOkQFұQÿLӇP2WuPKjPGѭӟLGҥQJ

%ҧQJ2.ӃWTXҧWKtQJKLӋPFӫD%R[D- Wilsona [22]

67 9LӃWFiFNӃWTXҧWKtQJKLӋPGѭӟLGҥQJPDWUұQYjRYjÿҫXUDFӫDÿӕLWѭӧQJy QKѭVDX ằằ ằằ ằằ ằằ ằằ ẳ º ôô ôô ôô ôô ôô ơ ê

X ; ằằ ằằ ằằ ằằ ằằ ẳ º ôô ôô ôô ôô ôô ơ ê

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éQJKƭDFӫDKjPKӗLTX\

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7URQJOXұQiQWiFJLҧÿӅ[XҩWYjVӱGөQJOêWKX\ӃWKjPKӗLTX\YjTX\KRҥFKWKӵFQJKLӋP ÿӇWtQKWRiQFiFWK{QJVӕWҧLWUӑQJFyWtQKFKҩWQJүXQKLrQYjSKӭFWҥS0ӝWQJKLrQFӭXWKtÿLӇPÿѭӧFWKӵFKLӋQÿyOj[iFÿӏQKFiFWK{QJVӕVyQJELӇQWӯEӝVӕOLӋX

68 TXDQWUҳFSKӭFWҥSQKҵPWtQKWRiQiSOӵFVyQJWiFGөQJOrQF{QJWUuQKÿӇSKөFYөF{QJWiFWKLӃWNӃCTBTL 3KѭѫQJSKiSKjPKӗLTX\ ÿmÿѭӧFӭQJ GөQJKLӋXTXҧWURQJQKLӅX OƭQKYӵF NKL\Ӄu WӕFҫQWәQJKӧSFyQKLӅXFѫVӣÿҫXYjRSKӭFWҥS'RYұ\YLӋF[iFÿӏQKWҧLWUӑQJ PjÿһFELӋWOjWҧLWUӑQJVyQJFҫQWKLӃWSKҧLiSGөQJSKѭѫQJSKiSKjPKӗLTX\

ÈSGөQJSKѭѫQJSKiSKjPKӗLTX\ÿӇWtQKFKLӅXFDRVyQJELӇQ

2.7.1 7әQJKӧSVӕOLӋXNKҧRViW

;HPFKLӅXFDRVyQJH OjKjPYӟLELӃQOjWҫQVXҩWKѭӟQJWKiQJÿmTXDQWUҳFÿѭӧF

%ҧQJ2.2 %ҧQJVӕOLӋXVyQJTXDQWUҳFWҥLNKXYӵFPӓĈҥL+QJSKDVH, [23]

WҫQVXҩW Tháng H (T) +KѭӟQJ H (tháng)

&KLӅXFDRH ÿѭӧFOҩ\WUXQJEuQKFӝWFXӕL+ +7+K˱ͣQJ+WKiQJ

%ҧQJ2.3 %ҧQJFKXNuVyQJTXDQWUҳFWҥLNKXYӵFPӓĈҥL+QJSKDVH, [23]

7tQKFKRWUѭӡQJKӧSÿDQJӣQJX\rQGҥQJVӕOLӋXTXDQWUҳFÿѭӧFYӟLWҫQVXҩWQăP FӝW9uFyҭQQrQSKҧLWKrPSKpSTXDQWUҳFWKӭWӭFFҫQWLӃQKjQK2 S = 8 quan

WUҳFKRһFWKtQJKLӋP%ҧQJWUrQFyWKtQJKLӋPYuWKӃFҫQWKrPWKtQJKLӋPWKӭÿѭӧF Oҩ\EҵQJWKtQJKLӋPWKӭWӯEҧQJWUrQ7ӯÿyFyKDLPDWUұQÿҫ\ÿӫVDX ĈҫXYjROj á á á á á á á á á á á ạ ã ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ © §

Y ĈѭDPDWUұQQj\YjRFKѭѫQJWUuQKWuPKjPWX\ӃQWtQK ǔ E 0 + b 1 x 1 + b 2 x 2 + b 3 x 3 (troQJÿyǔ{ H OjFKLӅXFDRVyQJWKHRELӃQ

7uPÿѭӧFWӯFKѭѫQJWUuQKSKҫQSKөOөF b0=2,75; b1=0,018; b2=0,478; b3=0,103

9ұ\KjPKӗLTX\FyGҥQJWѭӡQJminh là ǔ 2,75 + 0,018x 1 + 0,478x 2 + 0,103x 3

&KӑQÿLӇPT=6 (4.1dTd14,5); KѭӟQJ 3,6dhuongd8,9); tháng = 6 (1dthang d8),

70 Hình 2 ӃWTXҧFKҥ\FKѭѫQJWUuQKWtQKWK{QJVӕVyQJ

&KѭѫQJWUuQKPi\WtQKÿѭӧFLQӣSKҫQSKөOөF 1

2.8 Tính toiQWҧLWUӑQJVyQJWiFGөQJOrQWUөF{QJWUuQKELӇQWUӑQJOӵFFy WLӃWGLӋQWKD\ÿәLWURQJYQJQѭӟFVkX

2.8.1 ĈһFÿLӇPFӫDWҧLWUӑQJVyQJWiFGөQJOrQWUө&7%7/

7ҧLWUӑQJVyQJWiFÿӝQJ OrQ&7%7/ÿѭӧFOҩ\WURQJÿLӅXNLӋQEҩWOӧLQKҩW.KLÿyWUөÿӥ FӫD F{QJ WUuQK OjEӝSKұQ JiQK FKӏXFiF WiF ÿӝQJFӫD VyQJELӇQ &KRÿӃQ QD\ YLӋF QJKLrQFӭXYjWtQKWRiQWҧLWUӑQJVyQJWiFÿӝQJ OrQWUөÿӥYүQFzQQKLӅX\ӃXWӕSKӭFWҥS FҫQSKҧLWLӃSWөFÿѭӧF[HP[pW

9LӋF[iFÿӏQKWҧLWUӑQJVyQJWiFÿӝQJ OrQWUөÿӥFKӫ\ӃXGӵDYjRSKѭѫQJWUuQK0RULVRQ7X\Yұ\SKѭѫQJWUuQK0RULVRQFKӍJLӟL KҥQEӣLWҧLWUӑQJVyQJWiFÿӝQJ OrQWUөWKҷQJ ÿӭQJFyWLӃWGLӋQNK{QJÿәLWUzQKRһFWLӃWGLӋQEҩWNuĈLӅXQj\NK{QJViWWKӵFYӟLÿһF ÿLӇPFҩXWҥRFӫDF{QJWUuQKYuWKӃWURQJWtQKWRiQQJѭӡLWDWKѭӡQJÿѭDYjRFiFKӋVӕ ÿLӅXFKӍQKWKLrQYӅDQWRjQ

2.8.2 7әQJTXDQYӅWuQKKuQK QJKLrQFӭXWҧLWUӑQJVyQJWiFGөQJOrQWUө

&iFQJKLrQFӭXFӫD+DYHORFN (1940), Mac Camy và Fuchs (1954) ÿmÿѭDUDJLҧL pháp ÿiQK giá áp OӵF ÿӝQJ trên PӝW hình WUө tròn WKҷQJ ÿӭQg OӟQ Eӏ sóng WKѭӡQJ [X\rQWiFÿӝQJ>24]

Laird (1955), Priest (1962), Chakrabarti và Tam(1975), Endo và Tosaka (1985),

&KDNUDEDUWLFQJFӝQJVӵ ÿmWLӃQKjQKFiFWKtQJKLӋP YӅ áp OӵFÿӝQJ[XQJ TXDQKWUөWKҷQJÿӭQJ Eӏ sóng WKѭӡQJ [X\rQWiFGөQJYjWUình bày NӃW TXҧ trên PӝW ORҥW các PүX kích WKѭӟFNKiFQKDXFӫD WUө [24]

Hellstorm và Rundgren (1954) ÿR iSOӵF ÿӝQJ[XQJTXDQKPӝW mô hình KuQKWUө FӫD6RHGUD2HODQGV Grund và NӃW OXұQ UҵQJiSOӵF OrQPһW bên là khác nhau theo ÿӝVkXFӫDQѭӟF>24]

Nakamura (1976) ÿm WLӃQ KjQK PӝW QJKLrQ FӭX WKӵF ÿӏD ӣ 7KiL %uQK 'ѭѫQJ và ÿR iS OӵF GR sóng QJүX QKLrQ 0DF Camy Yj Oê WKX\ӃW VyQJ QKLӉX [ҥ WX\ӃQWtQK FӫD)XFKV Fuchs FKRUҵQJFiFJLiWUӏ ÿR QKӓKѫQ ӣSKҫQ sâu YjOӟQ KѫQJҫQPһWQѭӟFWUXQJEuQK6:/ VRYӟL gLiWUӏOêWKX\ӃW [24]

Borgman (1965) ÿӅ QJKӏ PӝW SKѭѫQJ SKiS tính toán PұW ÿӝ TXDQJ SKә FӫD sóng Wӯ PұW ÿӝ TXDQJ SKә FӫD ÿӝ FDR VyQJ EҵQJ FiFK Vӱ GөQJ PӝW KjP FKX\ӇQWLӃS /êWKX\ӃWQj\ ÿѭӧFiSGөQJ ÿӕLYӟLFiFVyQJ JҫQQKѭOjPӝW quang SKә GҧLKҽS>24]

Pierson và Holmes (1965) ÿmQJKLrQFӭXiSOӵF sóng NK{QJÿӅXGӵDWUrQNKiL QLӋP FKXӛLWKӡLJLDQ WҥL các ÿӝVkXNKiFQKDX [24]

+XQWLQJWRQ QJKLrQ FӭX Pӣ UӝQJ Oê WKX\ӃW QKLӉX [ҥ WX\ӃQ tính sóng QJҳQ QJүXQKLrQ và FKRWKҩ\ UҵQJWURQJ WUѭӡQJKӧSFӫD mӝWKuQKWUө OӟQ xuyên WUrQ EӅ PһW , TXDQJ SKә WәQJ OӵF trên WUө có liên quan ÿӇ TXDQJ SKә FKLӅX cao sóng WK{QJTXDPӝW hàm FKӭFQăQJ [24]

5DPDQYj6DPEKX9HQNDWD5DRÿmiSGөQJSKѭѫQJSKiSTXDQJSKәÿӇWҥRUDiSOӵFFӫDVyQJOrQWUөWKҷQJÿӭQJWURQJSKzQJWKtQJKLӋP7X\QKLrQFiF

72 WUѭӡQJVyQJÿѭӧFÿӅFұSFXӕLFQJÿѭӧFWҥRUDEӣLJLyWKәLWUrQEӅPһWQѭӟF [24]

7URQJNKLQKӳQJQӛOӵFÿiQJNӇÿmÿѭӧFGjQKFKRQJKLrQFӭXYӟLFiFWҧLWUӑQJ WiFGөQJOrQWUөWKҷQJÿӭQJQJKLrQFӭXYӅVӵSKkQEӕӭQJVXҩWVyQJEDRJӗPiS OӵFÿӝQJ[XQJTXDQKWUөWKҷQJÿӭQJÿһFELӋWOjGRVyQJJk\UDWKuNӃWTXҧÿҥW ÿѭӧFOjFzQUҩWKҥQFKӃ ÈSOӵFVyQJWiFÿӝQJOrQWUөWUzQWKҷQJÿӭQJFyWLӃWGLӋQWKD\ÿәLWURQJQѭӟFVkX tWÿѭӧFQJKLrQFӭXFKRÿӃQQD\

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+ &KLӅXVkXQѭӟFQJKLrQFӭXKP

+ ĈӝGjLVyQJO#120mWKѭӡQJOӟQKѫQÿѭӡQJNtQKG

7ҧLWUӑQJVyQJWiFÿӝQJ OrQWUөWKҷQJÿӭQJNK{QJFKX\ӇQÿӝQJÿѭӧF0RULVRQYjPӝWVӕ WiFJLҧNKiFQJKLrQFӭX.êKLӋXf OjWҧLWUӑQJVyQJSKkQEӕWKHRSKѭѫQJQJDQJx tác ÿӝQJOrQPӝWÿѫQYӏFKLӅXGjLFӝWFyÿѭӡQJNtQKD (hình 2.5), theo Morison là:

7URQJÿyU PұWÿӝQѭӟFCvt, Cqt FiFKӋVӕFҧQYұQWӕFYjTXiQWtQKv x , a x YұQ WӕFYjJLDWӕFQҵPQJDQJFӫDSKҫQWӱQѭӟFÿDQJ[pWGRVyQJJk\UDÿѭӧFWtQKWKHROê WKX\ӃWVyQJÿmFKӑQWUѭӟF

Hình 2.5 7ҧLWUӑQJVyQJWiFÿӝQJOrQWUөWKҷQJÿӭQJ, [18]

6ӕKҥQJÿҫXFӫD2.37ÿѭӧFJӑLOjWKjQKSKҫQFҧQYұQWӕFVӕKҥQJWKӭKDLOjWKjQK SKҫQFҧQTXiQWtQKC vt ÿѭӧFQKұQWURQJSKҥPYLy1,0; C qt QKұQWӯy2,0

'RYұQWӕFYjJLDWӕFFӫDFiFSKҫQWӱFKҩWOӓQJQyLFKXQJOjJLҧPWKHRÿӝVkXQrQVӵ SKkQEӕFӫDWҧLWUӑQJf GӑFWKHRFӝWFyGҥQJQKѭhình 2.5)

+ӧSOӵFFӫDWҧLSKkQEӕf OrQWUөWURQJÿRҥQWӯÿi\ELӇQ\ ÿӃQÿӝFDR\QjRÿy EҵQg:

YjPRPHQFӫDWҧLWUӑQJQj\ÿӕLYӟLÿi\ELӇQy = 0) là:

&iQKWD\ÿzQFӫDKӧSOӵFÿӕLYӟLÿi\ELӇQWuPÿѭӧFVDXNKLÿm[iFÿӏQKM và F theo F{QJWKӭF

9ӟLQJKLrQFӭXQj\YQJQѭӟFVkXWDGQJF{QJWKӭF Ȧ 2 =gk (2.41)

Có k WtQKYұQWӕFWKHRF{QJWKӭF

2 kx t kh sh ky ch v x ZH Z

2 t kh kx sh ky ch a x Z H Z

7URQJÿyWXQJÿӝ\FӫDSKҫQWӱQѭӟFOҩ\WKHRF{QJWKӭFVDX

2.8.5.1 &iFWK{QJV͙FKͯ\͇XFͯDP{KuQK Ĉ97P

&KLӅXVkXQѭӟFWKLӃWNӃ: 110m

&KLӅXFDRSKҫQErQGѭӟLFӫDVjQF{QJWiFP ĈѭӡQJNtQKWUөSKҫQWLӃWGLӋQNK{QJÿәLP ĈѭӡQJNtQKWUөSKҫQWLӃWGLӋQWKD\ÿәL7ӯP± 15m

&KLӅXFDRSKҫQWUөWLӃWGLӋQNK{QJÿәLP

&KLӅXGj\SKҫQWUөWLӃWGLӋQNK{QJÿәL,5m

&KLӅXFDRSKҫQWUөWLӃWGLӋQWKD\ÿәLP

&KLӅXGj\SKҫQWUөWLӃWGLӋQWKD\ÿәL,0m ĈѭӡQJNtQKFӫDWKQJFKuPFDLVon): 24m

&KLӅXGj\FӫDEҧQQҳSWKQJFKuP,0m

+ӋVӕWҧLWUӑQJJLDWăQJFӫDFKҩWOӓQJFKRFiFWUө,0 +ӋVӕWҧLWUӑQJJLDWăQJFӫDFKҩWOӓQJFKRWKQJFKuP,0 +ӋVӕTXiQWtQKFӫDFKҩWOӓQJ,0

7UӑQJOѭӧQJWKӇWtFKFӫDErW{QJ.500 kG/m 3

7UӑQJOѭӧQJWKӇWtFKFӫDQѭӟFELӇQ.025 kG/m 3

TUөEӏQJұSQѭӟFOrQÿӃQPӵFQѭӟFWƭQK

3KҫQQj\VӱGөQJ KDLWK{QJVӕVyQJOjH = 4,9m YӟLT = 6 sec&iFWK{QJVӕVyQJÿѭӧF VӱGөQJWӯNӃWTXҧWtQK[iFVXҩWWKHRSKѭѫQJSKiS+jPKӗLTX\YjTX\KRҥFKWKӵF QJKLӋPӣSKҫQWUrQ

2.8.5.3 ͇WTX̫WtQKWRiQ Ӣÿk\WtQKWҥLx=0 WUөFy WUQJYӟLWUөF FӫDWUөNKLÿyÿӍQKVyQJFҳWWUQJYӟLWUөFy nên ta có:

9ұ\ӣYQJQѭӟFVkXWDWtQKk WKHRF{QJWKӭFȦ 2 =kh ĈѭD k=0,1118 ; H=4,9m ; h= 110 m ; D 1=30 m PһWFҳWGѭӟLD 2 PPһWFҳWWUrQ Ȧ=1,0473 1/sec; T=6 sec ĈѭD x=YjRF{QJWKӭFWtQKYұQWӕFJLDWӕF có

2 ch ky t ch ky t kh sh t H kh kx sh ky ch v x ZH Z Z Z u Z

2 t ky ch kh t sh ky ch a x Z H Z Z u u

D ÿѭӧFWtQKWKHRD 1 0m và D 2 m EҵQJOұSWUuQKWKHRF{QJWKӭF

H k h là tang JyFQJKLrQJÿӝYiWPһWErQFӫDWUө

9ұQWӕFYjJLDWӕFOjKjPELӃQFӫDÿӝVkXy YjWKӡLJLDQt Oҩ\TXDȦt)

77 'Rÿy WtQKFK~QJÿӗQJWKӡLEӣLFKѭѫQJWUuQKWҥLFiF y và Ȧt NKiFQKDXUӗLÿѭDYjRF{QJ WKӭF0RULVRQf: x qt x x vt D a

7ӯÿk\WtQKKӧSOӵFFӫDELӇXÿӗf và momen M FӫDF x ÿӕLYӟLÿi\WUөWKHRF{QJWKӭF

Simpson PӝWFiFKGӉGjQJEҵQJFKѭѫQJWUuQKPi\WtQKĈk\OjFiFKWtQKPӟLYjÿѫQ giҧQKѫQVRYӟLFiFKOҩ\WtFKSKkQWUӵFWLӃSSKӭFWҥSQKѭWK{QJWKѭӡQJ

&iFJLiWUӏKӧSOӵFF xPRPHQWҥLFKkQWUөM x YjFiQKWD\ÿzQb ÿѭӧF[iFÿӏQKJLiWUӏ EӣLQJKLӋPJLҧLWtFK (xem hình 2.7)

Hình 2.7 ӃWTXҧELӇXÿӗOӵFQJDQJfx theo Morison và tính momen M

.KҧRViWPD[FӫDF x theo ȦtNӃWTXҧFKRWD

F x (max) = 4.7̭Q, khi Ȧt = 7,75WѭѫQJӭQJFy

7URQJFKѭѫQJWUuQKFNJQJFKRELӇXÿӗf theo các Ȧt khác nhau

3KҫQFKѭѫQJWUuQKPi\WtQKÿѭӧFLQӣSKҫQSKө OөF 2

Bài toán [iFÿӏQKWҧLWUӑQJVyQJWiFÿӝQJOrQ WUөCTBTL có WLӃWGLӋQ WKD\ÿәLtrong vùng QѭӟFVkX ÿѭӧF[iFÿӏQKEҵQJSKѭѫQJSKiSJLҧLWtFKNӃWKӧSYӟLOұSWUuQK máy tính và FKRWDPӝWNӃWTXҧÿiQJWLQFұ\

+ӧSOӵFFӫDELӇXÿӗf và momen M FӫDF x ÿӕLYӟLFKkQWUөWKHRF{QJWKӭFSimpson ÿѭӧFSKkQWtFKGӉGjQJEҵQJFiFKOұSWUuQKWUrQ Pi\WtQK1yÿѫQJLҧQKѫQFiFKWtQKWK{QJWKѭӡQJOjOҩ\WtFKSKkQWUӵFWLӃSNKiSKӭFWҥS

%,ӂ1'Ҥ1*Ĉӆ;È&Ĉӎ1+ CHUYӆ19ӎ &Ӫ$1ӄ1ĈҨ7 'ѬӞ,Ĉẩ@ĈRjQĈuQK4XDQJ³Tính toán ÿ͡O~QFͯDÿ̭WQ͉QG˱ͣLPyQJdàn kKRDQWU͕QJO͹F

W̩LNKXY͹F%͋1DP&{Q6˯Q´/XұQYăQWӕWQJKLӋSÿҥLKӑF.KRD.ƭWKXұWĈӏD

FKҩWYj'ҫXNKtĈҥLKӑFBách NKRD73+ӗ&Kt0LQKWKiQJQăP

Rem CHUONG TRINH TIM HAM HOI QUY

Static a(4, 4), ad(4, 4), tich(10, 10), p(10), p1(4, 4) As Double

Static km(4, 4), b(10), xo(10), y(10), dao(4, 4), xt(10, 10) As Double

Static nhan1(4, 8), nhan2(10, 10), hesob(10), x(8, 4), xxdd(1000), xxra(1000) As Double Show

Rem BAI TOAN CUA TRUONG, DUA TRUC TIEP SO LIEU VAO y(i) y(1) = 7.07: y(2) = 6.7: y(3) = 4.87: y(4) = 5.7: y(5) = 6.27: y(6) = 6.33: y(7) = 7.2: y(8) = 6.7 Rem dua ve dang da chuan hoa thi nghiem

Rem So lieu vao ma tran X dong = 8: cot = 4 x(1, 2) = 10.7: x(1, 3) = 8.9: x(1, 4) = 1# x(2, 2) = 11.3: x(2, 3) = 6.5: x(2, 4) = 2# x(3, 2) = 12#: x(3, 3) = 3.6: x(3, 4) = 3# x(4, 2) = 12.6: x(4, 3) = 5#: x(4, 4) = 4# x(5, 2) = 13.4: x(5, 3) = 6.1: x(5, 4) = 5# x(6, 2) = 4.1: x(6, 3) = 5.9: x(6, 4) = 6# x(7, 2) = 14.5: x(7, 3) = 6.7: x(7, 4) = 7# x(8, 2) = 11.3: x(8, 3) = 6.5: x(8, 4) = 8#

Print " So lieu vao ma tran X"

Print " So lieu vao vecto y"

Rem chuyen vi ma tran x

Rem cho ma tran don vi p1(1, 1) = 1: p1(1, 2) = 0: p1(1, 3) = 0: p1(1, 4) = 0 p1(2, 1) = 0: p1(2, 2) = 1: p1(2, 3) = 0: p1(2, 4) = 0 p1(3, 1) = 0: p1(3, 2) = 0: p1(3, 3) = 1: p1(3, 4) = 0 p1(4, 1) = 0: p1(4, 2) = 0: p1(4, 3) = 0: p1(4, 4) = 1 For kkk = 1 To cot

For j = 1 To cot km(i, j) = a(i, j) p(i) = p1(kkk, i)

Rem GIAI HE PHUONG TRINH DAI SO TUYEN TINH mhe = cot

For k = 1 To mhe b(k) = p(k): Rem doi don vi Tan, Tan.m

For i = k To mhe da = km(i, n) / de b(i) = b(i) - b(n) * da

For j = s To mhe km(i, j) = km(i, j) - km(n, j) * da

201 For j = 1 To mhe ac = km(n, j)

Next i xo(mhe) = b(mhe) / km(mhe, mhe) k1 = mhe k2 = mhe - 1

For j = k1 To mhe r = r + xo(j) * km(k4, j)

Rem in vec to nghiem xo(i)

701 uu$ = " dinh thuc bang khong"

Print , " He phuong trinh vo nghiem"

For j = 1 To mhe ad(j, kkk) = xo(j)

'Print " ma tran nghich dao (Xt*X)-1"

For d = 1 To cot tich(i, j) = tich(i, j) + a(i, d) * ad(d, j)

'Print " In ra tich ma tran (Xt*X)-1 va matran (Xt*X phai bang don vi "

Rem NHAN ADAO VOI XT DE THU XEM CO BANG MA TRAN DON VI KHONG 'Print "In ra tich ma tran (Xt*X)-1 va ma tran Xt"

For d = 1 To cot nhan1(i, j) = nhan1(i, j) + ad(i, d) * xt(d, j)

'Print " " & Format(nhan1(ik, jk), "0.00000");

For d = 1 To dong hesob(i) = hesob(i) + nhan1(i, d) * y(d)

Rem ham hoi quy la chieu cao song H=y

Rem Chon diem T=6 (sec); huong=3 ; tháng=6 , ta co: Hmax=4,9 m voi T=6 sec Print " In ra he so b cua ham hoi quy "

Print "y = b(1) + b(2) * x(2)+ b(3) * x(3) + b(4) * b(4)" yy = hesob(1) + hesob(2) * 6 + hesob(3) * 3# + hesob(4) * 6

Rem DOAN TINH SO SONG k t = 6: 't la chu ky

Print "omega ="; omega; " (1/m)" buoc = 1 kk0 = 0.1117 ee = 0.0001

'Print "khi kh=1 thi k = " & Format(kk0, " 0.0000")

'Print " da nhan k= " & Format(kk0, " 0.0000")

100 h = 15# kh = kk0 * h t1 = Exp(kh) t2 = 1 / t1 ssh = (t1 - t2) / 2 cch = (t1 + t2) / 2 tth = ssh / cch kk = omega * omega / (9.81 * tth)

'Print "buoc "; buoc; ": kh = " & Format(kh, " 0.000"); " th = " & Format(tth, " 0.0000"); " da tinh ra k = " & Format(kk, " 0.0000")

If Abs(kk - kk0) < ee Then GoTo 199 kk0 = kk buoc = buoc + 1

'Print " da nhan k= " & Format(kk, " 0.0000")

Rem KHI VUNG NUOC CO DO SAU h LON nghia la kh >3.14 thi lay k = omega * omega / 9.81

Print " Voi vung nuoc sau lay k = " & Format(kk, " 0.0000")

Rem NUOC SAU THI VAN TOC GIAM NHANH THEO DO SAU omega = Sqr(9.81 * k)

Rem giai phuong trinh bac 2

Rem CHUONG TRINH VE DAO DONG KHUNG CONG TRINH BIEN

Static a(4, 4), ad(4, 4), tich(10, 10), p(10), p1(4, 4) As Double

Static km(4, 4), b(10), xo(10), y(10), dao(4, 4), xt(10, 10) As Double

Static nhan1(4, 8), nhan2(10, 10), hesob(10), x(8, 4), xxdd(1000), xxra(1000) As Double

Rem DOAN TINH SO SONG k t = 6: 't la chu ky omega = 2 * 3.1416 / t

'Print "omega ="; omega; " (1/m)" buoc = 1 kk0 = 0.1117 ee = 0.0001

'Print "khi kh=1 thi k = " & Format(kk0, " 0.0000")

'List1.AddItem " da nhan k= " & Format(kk0, " 0.0000")

100 h = 15# kh = kk0 * h t1 = Exp(kh) t2 = 1 / t1 ssh = (t1 - t2) / 2 cch = (t1 + t2) / 2 tth = ssh / cch kk = omega * omega / (9.81 * tth)

'Print "buoc "; buoc; ": kh = " & Format(kh, " 0.000"); " th = " & Format(tth, " 0.0000"); " da tinh ra k = " & Format(kk, " 0.0000")

If Abs(kk - kk0) < ee Then GoTo 199 kk0 = kk buoc = buoc + 1

'List1.AddItem " da nhan k= " & Format(kk, " 0.0000")

Rem KHI VUNG NUOC CO DO SAU h LON nghia la kh >3.14 thi lay

'Print " Voi vung nuoc sau lay k = " & Format(kkk, " 0.0000")

List1.AddItem " Voi vung nuoc sau lay "

Rem NUOC SAU THI VAN TOC GIAM NHANH THEO DO SAU k = 0.1118 ro = 1.024: cvt = 1: cqt = 2: d1 = 30#: d2 = 15#: h = 110#: h0 = 4.9

Dim fx(200), mx(200) ro = 1.024: cvt = 1: cqt = 2: d1 = 30#: d2 = 15#: h = 110#: h0 = 4.9 dentad = (d1 - d2) / 2 tank1 = 2 * (h + h0 / 2) / (d1 - d2) tank2 = 1 / tank1

'List2.AddItem " tang1= " & Format(tank1, " 0.0000")

'List2.AddItem " tang2= " & Format(tank2, " 0.0000") omega = Sqr(9.81 * k)

'List2.AddItem " omega= " & Format(omega, " 0.0000")

Dim omet(100), hopfx(100), hopmx(100) hmax = h + (h0 / 2)

Rem VE DONG VE DONG VE DONG VE DONG VE DONG

Cls: Rem SAU CLS CO 2 CHAM THI LUU LAI DO THI TAN SO j = HScrtanso.Value: 'Mo j

Rem VE TINH VE TINH VE TINH VE TINH VE TINH VE TINH VE TINH

'For j = 1 To 15: 'mo for omet(j) = omega * (6 + j * 0.2)

List2.AddItem " " & " omega(" & j & ") = " & Format(omet(j), "0.00")

'eta = (h0 / 2) * Cos(omet(j)): ' o dinh song Cos(omega)=0 zy = h + (h0 / 2) * Cos(omet(j)): ' tinh o dinh song, o day bien y=0 u1 = Exp(k * h) u2 = Exp(-k * h) ss = (u1 - u2) / 2 hs = (omega * h0 / 2) / ss

'List2.AddItem " hsvx= " & Format(hs, " 0.000000000") hsax = (omega * omega * h0 / 2) / ss

'List2.AddItem " hsax= " & Format(hsax, " 0.000000000")

For i = 0 To hmax dtt = (h + (h0 / 2) * Cos(omet(j)) - i) * tank2 d = d2 + 2 * dtt r1 = Exp(k * i) r2 = 1 / r1 cc = (r1 + r2) / 2 vx = hs * cc * Cos(omet(j)) ax = hs * omega * cc * Sin(omet(j))

Rem in vx ra de so sanh voi vung nuoc nong gan bo

'List2.AddItem " vx= " & i & " " & Format(vx, " 0.0000") & " ax= " & Format(ax, " 0.0000") fx(i) = 0.5 * ro * cvt * d * Abs(vx) * vx + ro * cqt * (3.1416 * d) * (3.1416 * d) * ax / 4 mx(i) = fx(i) * (i - 0.5)

'List2.AddItem " " & (i) & " D=" & Format(d, "0.0") & " fx=" & Format(fx(i), " 0.0000") Scale (-320, 240)-(320, -240) xphai = 30

Rem ve vx de so sanh voi vung nuoc nong gan bo

Rem VE CHO TRUONG HOP MAX la j=7

Rem ve tinh tinh tinh

'PSet (fx(i) / 50 - xphai, i), QBColor(15): 'mo lenh ve

Rem ve dong ve dong

PSet (-fx(i) / 50 - xphai, i), QBColor(15): 'mo lenh ve

'Line (-fx(i) / 50 - xphai, i)-(0 - xphai, i), QBColor(11)

For i = 1 To hmax hopfx(j) = hopfx(j) + fx(i) hopmx(j) = hopmx(j) + mx(i)

PSet (omet(j) * 25 - 50 - 7 * xphai, hopfx(j) / 700), QBColor(11) Rem doi kN ve Tan (chia cho 10)

List1.AddItem " " & " omega(" & j & ") = " & Format(omet(j), "0.00") List1.AddItem " Fx =" & Format(hopfx(j) / 10, "0.00") & " Tan, " List1.AddItem " mz =" & Format(hopmx(j) / 10, "0.00") & " Tm" tay = hopmx(j) / hopfx(j)

Rem tinh cho truong hop Fx cuc dai

'eta = (h0 / 2) * Cos(omet(j)): ' o dinh song Cos(omega)=0 zy = h + (h0 / 2) * Cos(omet(j)): ' tinh o dinh song, o day bien y=0 u1 = Exp(k * h) u2 = Exp(-k * h) ss = (u1 - u2) / 2 hs = (omega * h0 / 2) / ss

'List2.AddItem " heso= " & Format(hs, " 0.000000000")

For i = 0 To hmax dtt = (h + (h0 / 2) * Cos(omet(j)) - i) * tank2 d = d2 + 2 * dtt r1 = Exp(k * i) r2 = 1 / r1

128 cc = (r1 + r2) / 2 vx = hs * cc * Cos(omet(j)) ax = hs * omega * cc * Sin(omet(j))

'List2.AddItem " vx= " & Format(vx, " 0.0000") & " ax= " & Format(ax, " 0.0000") fx(i) = 0.5 * ro * cvt * d * Abs(vx) * vx + ro * cqt * (3.1416 * d) * (3.1416 * d) * ax / 4 mx(i) = fx(i) * (i - 0.5)

List2.AddItem " y= " & i & ", D=" & Format(d, "0.0") & ", fx=" & Format(fx(i) / 10, " 0.0000") Scale (-320, 240)-(320, -240) xphai = 30

For i = 1 To hmax hopfx(j) = hopfx(j) + fx(i) hopmx(j) = hopmx(j) + mx(i)

Rem doi kN ve Tan (chia cho 10)

'List1.AddItem " Fx =" & Format(hopfx(j) / 10, "0.00") & " Tan " & " omegat= " & Format(omet(j), "0.00")

'List1.AddItem " mz =" & Format(hopmx(j) / 10, "0.00") & " Tm" tay = hopmx(j) / hopfx(j)

'List1.AddItem " ome =" & Format(ome, "0.00")

Rem CHI VE DEN d` (voi d` thi chuyen vi=0 tren truc dung) DE LAY 1 DAU CUA CHUYEN VI Rem NEU da THi CHUYEN Vi BAT DAU DOi DAU

Dim vte(1000), uer(1000), urq(1000), tx1(1000), tx2(1000), dx1(1000), dy1(1000), dx2(1000), dy2(1000) As Double

Dim r(1000), rr(1000), ttx(1000), tty(1000), x0(1000), y0(1000) As Double

Dim urp(1000), urch(1000), urt(1000), urpp(1000), xima(900)

Open "d:\cv18.txt" For Output As #1

Textdd.Text = HScrdd.Value jd = HScrdd.Value + 0: 'Gan j de viet goc bang do dd = jd: Rem Dieu kien duong kinh d

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