SLOPE DEPLECTION PADA PORTAL

CONTOH SOAL ANALISA PORTAL 2 TINGKAT 1t/m1 T 1,5EI

4m1t/m2EI

2EI

2t/m

2T

2EI 2t/m 2EI

5EI

4m

3M 3M

HITUNG DAN GAMBAR BIDANG M, D, N DENGAN METODE SLOPE DEFLECTIONPORTAL MENGALAMI PERGOYANGAN

FORMULA MOMEN AKHIR

A. MOMEN AKHIR

1t/m

1 T

1,5EI

4m

2EI

2EI

2T

2EI

2,5EI

2,5EI

`

4m

3M 3M

B. MOMEN PRIMER

1t/m

1 T

1,5EI

4m

1t/m 2EI

2EI

2t

2t/m

2T

2EI

2,5EI

2,5EI

2t/m

4m

3M 3M

MOMEN PRIMER/MOMEN JEPIT UJUNGMFAC = +1/12.qac.Lac2 = + 1/12.2.42 = +2,667 tmMFCA = - 2,667 tmMFBD = 0 (tidak ada beban)MFDB = 0 (tidak ada beban)MFCE = +1/12.qce.Lce2 = + 1/12.1.42 = +1,333 tm

MFEC = -1,333 tm

MFFD = 0 (tidak ada beban)MFDF = 0 (tidak ada beban)MFCD = +1/8.P.Lcd + (11/192).qcd.Lcd2 = + 1/8.2.6 + (11/192).2.62 = +5,625 tm

MFDC = -1/8.P.Lcd - (5/192).qcd.Lcd2 = - 1/8.2.6 - ( 5/192).2.62 = +3,375 tm

MFFE = - (1/12).qcd.Lcd2 = - 1/12.1.62 = - 3,000 tm

MFEF = +3,000 tm

C. DEFORMASI STRUKTUR1. PUTARAN SUDUT PADA JOINT (SEMUA PUTARAN SUDUT DIASUMSIKAN SEARAH JARUM JAM)

1,5EI

4m

2EI

2EI

2EI

2,5EI

2,5EI

3M 3M

2. TRANSLASI/DEFLEKSI HORISONTAL PADA TIAP TINGKAT (ASUMSI BERGESER KE ARAH KANAN)

1,5EI

4m

2EI

2EI

2EI

2,5EI

2,5EI 4m

3M 3M

PUTARAN SUDUT AKIBAT PERGOYANGAN

Rac=Rca = 1/4 ; Rbd = Rdb = 1/4

Rce=Rec = 2/4 ; Rbf = Rfb = 2/4a = 0 (PERLETAKAN JEPIT) ; b = 0 (PERLETAKAN JEPIT) ; MOMEN AKHIR DALAM FUNGSI DEFORMASI STRUKTUR

MAC = + 2,667 + 2,5/4(-2a - c + 3.Rac) = + 2,667 + 2,5/4 (- c + 3.1/41) = +2,667 0,625c + 0,4691MBD = 0 + 2,5/4(-2b - d + 3. Rbd) = 0 + 2,5/4 (-d + 3.1/41)

= 0,625d + 0,4691MCA = -2,667 + 2,5/4(-2c - a + 3. Rca) = - 2,667 + 2,5/4 (-2c + 3.1/41)

= -2,667 1,25c + 0,4691 MCD = +5,625 + 2/6 (-2c - d)

= +5,625 0,667c 0,333d

MCE = +1,333+ 2/4(-2c - e + 3.Rce)= -1,333+ 2/4(-2c - e + 3.2) =+1,333 - c 0,5e + 0,3752MDB = 0 + 2,5/4(-2d - b + 3.Rbd) = 0 - 1,25d + 2,5/4.3.1/41 = -1,25d + 0,4691MDC = -3,375 + 2/6 (-2d - c)

= -3,375 0,667d 0,333c MDF =0 + 2/4(-2d - f + 3.Rdf) = 0 - d + 2/4.3/42 = -d + 0,3752MEC =-1,333 + 2/4(-2e - c + 3.Rec) = -1,333 - e 0,5c + 2/4. 3.1/42 = -1,333 - e -0,5c + 0,3752MEF = +3 + 1,5/6(- 2e - f ) = +3 0,5e 0,25f MFE =-3 + 1,5/6(-2f - e) = -3 0,5f - 0,25e

MFD =0 + 2/4(-2f - d + 3.Rfd) = - f - 0,5d + 2/4.3.1/42 = - f - 0,5d + 0,3752Untuk menentukan seluruh momen akhir harus diketahui :c, d, e,f dan 1 dan 2(HARUS DISIAPKAN/DISUSUN 6 PERSAMAAN, DENGAN MENGGUNAKAN SYARAT BATAS YAITU : 1) SETIAP JOINT JUMLAH MOMEN =0 KECUALI PERLETAKAN JEPIT, 2) JUMLAH GAYA HORISONTAL TIAP TINGKAT = 0)A. SYARAT BATAS JUMLAH MOMEN SETIAP JOIN = 0MC = 0 ; MCA + MCD + MCE = 0

(-2,667 1,25c + 0,4691) + (+5,625 0,667c 0,333d) + (+1,333 - c 0,5e + 0,3752) = 0

(-2,667 + 5,625 + 1,333) +( 1,25c c - 0,667c) + ( 0,333d) +( 0,5e) +(+ 0,4691+ 0,3752 = 0+4,291 2,917c 0,333d 0,5e + 0,4691+ 0,3752 = 0 -2,917c 0,333d 0,5e + 0,4691+ 0,3752 = - 4,291

+2,917c + 0,333d + 0,5e - 0,4691 - 0,3752 = + 4,291 ................................. (1)

MD = 0 ; MDB + MDC + MDF = 0

(-1,25d + 0,4691)+( -3,375 0,667d 0,333c)+( -d + 0,3752) = 0(-3,375) +( 0,333c) + (-1,25d 0,667d-d) + (0,4691)+ (0,3752) = 0

c 2,917d + 0,4691+ 0,3752 = 0

c 2,917d + 0,4691+ 0,3752 = +3,375 .......................................................(2)

ME = 0 ; MEC + MEF = 0

(-1,333 - e -0,5c + 0,3752)+( +3 0,5e 0,25f) = 0

(-1,333+3) +(-e 0,5e)+( -0,5c) +( 0,25f) + (0,3752) = 0 +1,667 -1,5e 0,5c 0,25f + 0,3752 = 0-1,5e 0,5c 0,25f + 0,3752 = - 1,667+1,5e + 0,5c + 0,25f - 0,3752 = + 1,667 .......................................................................(3)

MF = 0 ; MFE + MFD = 0

(-3 0,5f - 0,25e)+( - f - 0,5d + 0,3752) = 0 0,5f f)+ (- 0,5d)+(- 0,25e) +(0,3752) = 0f 0,5d 0,25e + 0,3752 = 0f 0,5d 0,25e + 0,3752 = + 3

f + 0,5d + 0,25e - 0,3752 = - 3 ...........................................................(4)

SYARAT BATAS JUMLAH GAYA HORISONTAL TIAP TINGKAT = 0H2 = 0 ; (MEC + MCE)/Lec + (MFD + MDF)/Ldf 1 1.4 = 0 (-1,333 - e -0,5c + 0,3752)+( (+1,333 - c 0,5e + 0,3752))/4} +{(- f - 0,5d + 0,3752)+( -d + 0,3752)} -5 = 0[(+1,333-1,333) +(-0,5c-c) + (- 0,5d-d) + ( 0,5e- e) +(- f) + (+ 0,3752+ 0,3752 + 0,3752+ 0,3752)]/4 - 5= 0-5 +(-1,5c 1,5d 1,5e -f + 1,52)/4 = 0(-1,5c 1,5d 1,5e -f + 1,52 )/4= +5

-1,5c 1,5d 1,5e -f + 1,52 = +20 .....................................................(5)H1 = 0 ; (MAC + MCA)/Lac + (MBD + MDB)/Ldb 1 - 2 1.4 - 2.4 = 0

(MAC + MCA)/Lac + (MBD + MDB)/Ldb 15= 0 (MAC + MCA)/Lac + (MBD + MDB)/Ldb = +15

{(+2,667 0,625c + 0,4691)+ (-2,667 1,35c + 0,4691)}/4 +{( 0,625d + 0,4691)+

(-1,25d + 0,4691)}/4 = +15{(+2,667 0,625c + 0,4691)+ (-2,667 1,35c + 0,4691)} +{( 0,625d + 0,4691)+

(-1,25d + 0,4691)} = +60

(2,667-2,667) + ( 0,625c 1,25c) +( 0,625d-1,25d)+( + 0,4691+ 0,4691+ 0,4691+ 0,4691) = + 60

1,875c 1,875d+ 1,8861 = + 60 ................................................(6)

PERSAMAAN MATRIK

[K].[] = [C][K]-1.[K].[] = [K]-1. [C]

[K]-1 dicari dengan bantuan Microsoft Excel

[K]-1 =[K]-1.[K].[] = [K]-1. [C]

Maka :c =14.037

d =13.840

e =8.980

f =5.325

1=60.272

2=53.740

Momen akhir

FREE BODY/BENDA BEBAS/KONSTRUKSI SUDAH MENJADI STATIS TERTENTU

MAB = MFAB + 2EI/L(-2a b + 3Ra)

MBA = MFBA + 2EI/L(-2b a+ 3Rb)

BA

A

MCE

MCD

MDF

MEC

MBD

MDC

MFD

FDA

EDA

MEF

MFE

MFDE

MFFD

MFBD

MFDB

MFDF

MFFE

MFEF

MFCD

MFCE

MFEC

MFAC

MFCA

FDA

DA

EDA

DA

CA

Rca

CA

MDB

MCA

MAC

BA

A

FDA

FDA

d

CA

A

EDA

A

EDA

CA

f

DA

e

c

BA

DA

BA

Rac

Rce

Rbd

Rdf

HARUS BERNILAI 1