# Analisa struktur metode slope deflection

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KELOMPOK 2

KELOMPOK 2

ANALISA STRUKTURSLOPE DEFLECTION

METODE SLOPE DEFLECTION

ContohSoal Portal BidangTidakBergoyang (Non-Sway Plane Frame)

Diketahuistruktur portal denganpembebanansepertigambarberikut :

Penyelesaian :

1. Derajat kebebasan dalam pergoyangan struktur statis tak tentu :n = 2j (m + 2f + 2h + r)dengan :n = jumlah derajat kebebasan (degree of fredom)j = jumlah titik simpul, termasuk perletakan (joint)m = jumlah batang yang dibatasi oleh dua joint (member)f = jumlah perletakan jepit (fixed)h = jumlah perletakan sendi (hinged)r= jumlah perletakan rol (roll)

Cek:n = 2x4 (3 + 2x2 + 2x0 + 1)= 8 8 = 0 0 tidak ada pergoyangan

2. MENENTUKAN JUMLAH VARIABEL YANG ADA :

- A = rol, A bukan variabel- B = titik simpul, ada variable B- C = jepit, C= 0- D = jepit, D=0- Jadi variabelnya hanya satu, B3. PERHITUNGAN MOMEN PRIMER DAN KEKAKUAN BATANG

Momen primer

MFBA = + 3/16 PL = + 3/16 (3) 6 = + 9 tmMFBC = - 1/8 PL = - 1/8 (2) 4 = - 1 tmMFCB = + 1/8 PL = + 1/8 (2) 4 = + 2 tm

Kekakuan batang

KBA = 3EI/L = 3(2EI)/6 = 1 EIKBC = KCB = 4EI/L = 4(EI)/4 = 1 EIKBD = KDB = 4EI/L = 4(EI)/6 = 0,66 EI

4. PERSAMAAN SLOPE DEFLECTION MAB = MAB + 1 EI (2A + B) = 0 + 2 EIA + 1 EIB (1) MBA = MBA + 1 EI (2B + A) = +9 + 2 EIB + 1 EIA (2) MBC = MBC + 1 EI (2B + C) = -1 + 2 EIB + 1 EIC (3) MCB = MCB + 1 EI (2C + B) = 1 + 2 EIC + 1 EIB (4) MBD = MBD + 0,66 EI (2B + D) = 0 + 1,33 EIB + 0,66 EID (5) MDB = MDB + 0,66 EI (2D + B) = + 0 + 1,33 EID + 0,66 EIB (6)

Syaratbatas (boundary condition) :- C = jepit, C=0- D = jepit, D=0- A = rol, MAB =0

0 =2 EIA + 1 EIB (1a) MBA =9 + 2 EIB + 1 EIA (2a) MBC = -1 + 2 EIB (3a) MCB = 1 + 1 EIB (4a) MBD = 1,33 EIB (5a) MDB = 0,66 EIB (6a)

5. MENCARI NILAI B

Mb = 0 (2a) + (3a) + (5a) = 0(9 + 2 EIB + 1 EIA) +(-1 + 2 EIB ) + (1,33 EIB ) = 0(+8+ 5,33 EI B + 1EI A) = 0 1EI A + 5,33 EI B = -8 (7)

Eliminasi substitusi persamaan 1a dengan persamaan 71EI A + 5,33 EI B = -8 x 2 2EI A + 10,66EI B = 16 2EIA + 1 EIB = 0 x 1 2 EIA + 1 EIB = 0 - 9,66 EIB = 16 EIB =1,553

Substitusi EIB ke persamaan 1a2 EIA + 1 EIB = 02 EIA + 1( 1,553 )= 02 EIA + 1,553 = 0 EIA = -0,776

6. MOMEN UJUNG MAB = 2(-0,776) + 1( 1,553 )= -1,553 + 1,553 = 0 tm (1b) MBA = +9 + 2 (1,553) + 1 (-0,776) = 9+ 3,106 -0,776 = 11 ,33 tm (2b) MBC = -1 + 2 (1,553) = 2,106 tm (3b) MCB = 1 + 1 (1,553) = 2,553 tm (4b) MBD = 1,33 (1,553) =2,065 tm (5b) MDB = 0,66 (1,553) =1,024 tm (6b)

7. URAIAN FREEBODY

Batang AB

MB = 0 RA.6 P1.3 + MBA = 0RA.6 3.3 + 11 ,33 = 0 6RA 9 + 11 ,33 = 0 6RA + 2,33 = 0 RA = -0,388

V= 0 RA - P1 + RB = 0-0,388 - 3 + RB = 0 -3,388 + RB= 0 RB = 3,388

Batang BC

MB = 0 RC.4 + P2.2 MCB - MBC = 0RC.4 + 2 .2 2,553 2,106 = 04 RC + 4 4,639= 04 RC - 0,639 = 0 RC = 0,159

V= 0 RB1 P2 + RC = 0RB1 2 + 0,159 = 0 0,159- 2+ RB1= 0 RB1 = 1,841

Batang BD

V= 0 RB3 + RD = 0RB3 + 0,514 = 0RB3 =0,514 DV =RB + RB1 DV =3,388+ 1,841 DV =5,229 MB = 0 RD.6 MDB - MBD = 0RD.6 1,024 - 2,065 = 06RD 1,024 - 2,065 = 06RD 3,089= 0 RD= 3,089/6 RD=0,514

FreebodySuperposisi

8. KONTROL STRUKTUR

MA = 0 (P1 x 2) (VD x 4) + (HD x 4) + (P2 x 5) (VC x 6) + MCB + MDB = 0(3 X 2 ) (5,229 X 4 ) + (0,514 x 4) +(2 X 5 ) (0,159 X 6 ) + 2,553 + 1,024 = 06 20,916 + 2,056 + 10 -0,954 + 2,553 + 1,024 = 00 = 0

V = 0VA P1 + VD P2 + VC = 0-0,388 3 + 5,229 2 + 0,159 = 00 = 0

H = 0 HC HD = 0 0,514 0,514 = 0 0 = 0

9. MENGHITUNG BIDANG M & D

Bidang M

Batang AB

Diagram momen (M)Momen di x = 30 + Ra . x0 + (-0,388) . 3-1,164 tm

Batang BC

Diagram momen (M)Momen di x = 22,106 + Rb . x2,106 + 1,841 . 25,788 tm

Bidang D Batang AB Gaya lintangdari x = 0 sd x= 6Dx = 0 , Ra= -0,388 tonDx = 3 , Ra P1 = -0,388 - 6= -6,388 tonDx = 6 , Ra P1+Rb =-0,388 6 +6,388 = 0

Batang BCGaya lintangdari x = 0 sd x= 4Dx = 0 ,Rb= 1,841 tonDx = 2 ,Rb P2 = 1,841 4 = -2,159 tonDx = 4 ,Rb P2+Rc = 1,841 4 + 2,159 = 0

Batang BD

BIDANG DBIDANG M

10. MenghitungBidang NBatang BDN BD= Rb + Rb1= 3,388 + 1,841= 5,229

11. Diagram Superposisi BidangM , D , N

Bidang M

Bidang D

Bidang N

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