eBOOK TEORI STRUKTUR

41 Keseimbangan tupang B dan C: MBA + MBC = EI[2.67B + 0.33C + 0.75] – 37.5 = 0 (1) MCB + MCD = EI[0.33B + 1.67C + 0.375] + 37.5 = 0 (2) Keseimbangan ufuk: (MAB + MBA)/hL + (MCD + MDC)/hR = 0 EI[0.75B + 0.375C + 0.563] = 0 (3)  EIB = 20.09 EIC = - 24.11 EI = - 10.71 MAB = 12.05 kNm MBC = - 32.14 kNm MCD = - 28.13 kNm MBA = 32.14 kNm MCB = 28.13 kNm MDC = -16.07 kNm Dari kiraan-kiraan di atas jelas arah anggapan huyung tidak mempengaruhi momen dalaman. 2. Kerangka di bawah ditupang tegar pada A, pin dan rola masing-masing di D dan E. Dapatkan momen dalaman bagi semua titik sambungan. Lukiskan GDR dan GML. Penyelesaian Aganggap kerangka mengalami huyung ke kanan M-: MAB = 4EI 4 [B - 3/4] = EI[B - 0.75] MBA = 4EI 4 [2B - 3/4] = EI[2B - 0.75] MBC = 2EI 6 [2B + C] - 37.5 = EI[0.67B + 0.33C] – 37.5 MCB = 2EI 6 [2C + B] + 37.5 = EI[0.33B + 0.67C] + 37.5 MCD = 3EI 4 [C - /4] = EI[0.75C - 0.188]………. [ k = 3 4 kF = (3 4 ) ( 4EI 4 ) = 3EI 4 ] MCE = 3EI 3 [C] = EI[C] ………. [ k = 3 4 kF = (3 4 ) ( 4EI 3 ) = 3EI 3 ] MDC = 0 MCE = 0 Bentukkan tiga persamaan serentak mengandungi anu B, C dan . Keseimbangan tupang B dan C: MBA + MBC = EI[2.67B + 0.33C – 0.75] – 37.5 = 0 (1) MCB + MCD +MCE = EI[0.33B + 2.42C - 0.188] + 37.5 = 0 (2) 3 m 3 m 3 m 50 kN A D B C E 2I I 4 m I I  C -M -M D  B -M -M 

42 Keseimbangan ufuk: (MAB + MBA)/hL+ (MCD + MDC)/hR = 0 EI[.75B + 0.188C - 0.422] = 0 (3) EIB = 28.13 EIC = - 16.07 EI = 42.86 MAB = - 4.02 kNm MBC = - 24.11 kNm MCD = - 20.09 kNm MBA = 24.11 kNm MCB = 36.16 kNm MDC = 0 MCE = - 16.07 kNm MEC = 0 Tindakbalas, GDR & GML AB 4HA + 4.02 = 24.11 HA = 5.02 kN → HBA = 5.02 kN  BC 6RB + 36.16 = 50(3) + 24.11 RB = 22.99 kN RCB = 27.01 kN CD 4HD = 20.09 HD = 5.02 kN  HCD = 5.02 kN → CE 3RE + 16.07 = 0 RE = - 5.36 kN  Keseimbangan akhir M/h = (24.11 – 4.02 -20.09)/4 = 0 dan S = - HCE = + 5.02 (huyung ke kanan). 3. Kerangka portal di bawah ditupang tegar pada aras yang berbeza di A dan D membawa beban BTS sebesar 10 kN/m. Dapatkan momen dalaman bagi titik-titik sambungan dan lukiskan GDR, GML dan profil lenturan kerangka. Anggapkan huyung B = C. 16.07 5.02 22.99 27.01 5.02 5.36 GDR 44.86 24.11 36.16 GML 20.09 4.02 24.1 1 36.1 6 20.09 16.0 7 5.36 4.02 50 kN 5.02 5.02 5.02 5.02 22.29 27.01

43 Penyelesaian Dengan menganggap kerangka mengalami huyung ke kiri M-: MAB = 2E3I 4 [B + 3/4] = EI[1.5B + 1.125] MBA = 2E3I 4 [2B + 3/4] = EI[3B + 1.125] MBC = 2E2I 6 [2B + C] - 30 = EI[1.33B + 0.67C] -30 MCB = 2E2I 6 [2C + B] + 30 = EI[0.67B + 1.33C] + 30 MCD = 2EI 2 [2C + 3/2] = EI[2C + 1.5] MDC = 2EI 2 [C + 3/2] = EI[C + 1.5] Keseimbangan tupang B dan C: MBA + MBC = EI[4.33B + 0.67C + 1.125] - 30 = 0 (1) MCB + MCD = EI[0.67B + 3.33C + 1.5] + 30 = 0 (2) Keseimbangan ufuk: (MAB + MBA)/hL + (MCD + MDC)/hR = 0 EI[1.125B + 1.5C + 2.062] = 0 (3) EIB = 7.541 EIC = - 12.869 EI = 5.246 MAB = 17.21 kNm MBC = - 28.52 kNm MCD = - 17.87 kNm MBA = 28.52 kNm MCB = 17.87 kNm MDC = - 5.00 kNm Tindakbalas, GDR & GML AB 4HA = 17.81 + 28.52 HA = 11.58 kN → HBA = 11.58 kN  BC 6RBC + 17.87 = 10(62 )/2 + 28.52 RBC = 31.78 kN  RCB = 28.22 kN  D A I 2 m B 10 kN/m C 4 m 3I 2I 3 m 3 m  +M +M  +M +M  28.52 RCB 17.87 HD RBC 17.21 5 10 kN/m HA HBA HCD

44 CD 2HD = 5 + 17.87 HD = 11.44 kN  HCD = 11.44 kN → The final equilibrium H = P = -11.58 + 11.44  0 OK 4. Rajah di bawah menunjukkan kerangka sebuah garaj membawa beban pugak daripada atap sebesar 5 kN/m. Dapatkan momen dalaman bagi semua titik sambungan. Anggapkan huyung B = C. Penyelesaian Beban bersudut tepat kepada paksi rasuk = 5 cos  = 4.93 kN/m Panjang cerun BC = 6.083 m Dengan menganggap kerangka mengalami huyung ke kiri M-: MAB = 2EI 5 [B + 3/5] = EI[0.4B + 0.24]. MBA = 2EI 5 [2B + 3/5] = EI[0.8B + 0.24] MBC = 2EI 6.083 [2B + C] – 15.2 = EI[0.658B + 0.329C] -15.2 MCB = 2EI 6.083 [2C + B] + 15.2 = EI[0.329B + 0.658C] + 15.2 MCD = 2EI 4 [2C + 3/4] = EI[C + 0.375] MDC = 2EI 4 [C + 3/4] = EI[0.5C + 0.375] Keseimbangan tupang B dan C: MBA + MBC = EI[1.458B + 0.33C + 0.24] – 15.2 = 0 (1) MCB + MCD = EI[0.33B + 1.66C + 0.375] + 15.2 = 0 (2) B 5 kN/m 5 m I I  I C 4 m A 6 m D 5 kN/m W90    +M +M +M +M 3.178 m 11.44 11.58 31.78 28.22 SFD 28.52 17.87 17.21 5 21.98 BMD Bent Profile  

45 Keseimbangan ufuk: (MAB + MBA)/hL + (MCD + MDC)/hR = 0 EI[0.24B + 0.375C – 0.284] = 0 (3) EIB = 12.238 EIC = - 13.207 EI = 7.109 Maka MAB = 6.60 kNm MBC = - 11.50 kNm MCD = - 10.54 kNm MBA = 11.50 kNm MCB = 10.54 kNm MDC = - 3.94 kNm Tindakbalas 11.50 11.50 10.54 AB 5HA = 6.60 + 11.50 HA = 3.62 kN → HBA = 3.62 kN  BC 6.083RBC + 10.54 = 4.93(6.0832 )/2 + 11.50 RBC = 15.15 kN  RCB = (4.93 x 6.083) – 15.15 = 14.84 kN  CD 4HD = 3.94 + 10.54 HD = 3.62 kN  HCD = 3.62 kN → Perbezaan daya ufuk di bahagian atas H = HCD – HBC = 3.62 – 3.62 = 0 OK Atau P = - [(MAB+MBA)/5 + (MCD + MDC)/4] = 0 OK LATIHAN HBA HA 6.60 HCD 10.54 3.94 HD 4.93 kN/m RBC 6.083 m RCB

46 RUJUKAN Bedford, A.& W. Fowler. (1985). Statics Engineering, London: Addison-Wesley. Dpencer, W.J. (1988). Fundamental Structural Analysis, Basingtoke: Macmillan Education. Ghali, A. & A.M. Neville. (1982). Structural Analysis. London: Chapman & Hall. Kenneth Leet (2009). Fundamentals Of Structural Analysis (4th Edition): McGraw-Hill. Hibbeler, R.C. (2011). Structural Analysis. (8th Edition). New Jersey: Prentice Hall. Lemas, B.& Gardner, A. (2005). Fundamental Structural Analysis for Design. Sydney: Pearson Logie, K.F. (1991). Structure Basic Theory Including Sample Problems (Malay translation). Kuala Lumpur: Dewan Bahasa dan Pustaka. Marshal, W.T. & H.M. Nelson. (1990). Structures. 3rd Edition.; Terjemahan Prof. Madya Ismail Hassan: Struktur. Teori Asas dengan Contoh Penyelesaian. London; Longman Logie. Russell C. Hibbele (2007) Structural Analysis (7th Edition) [Hardcover], Amazon.com. Smith, M.J. (1972). Theory of Structure. Norwich: Mcdonald And Evans Ltd.

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