# analisa struktur metode slope deflection

of 14 /14
KELOMPOK 2 ANALISA STRUKTUR SLOPE DEFLECTION KELOMPOK 2 MUCHAMAD LUQMAN RAHMAWAN (1561122003) WAHYU PURNOMO ADI (1561122004) MADE BAGUS GITA DARMA (1561122005) UNIVERSITAS WARMADEWA FAKULTAS TEKNIK 1

Post on 11-Apr-2017

527 views

Category:

## Engineering

Embed Size (px)

TRANSCRIPT

KELOMPOK 2

ANALISA STRUKTURSLOPE DEFLECTION

KELOMPOK 2

FAKULTAS TEKNIK

JURUSAN TEKNIK SIPIL

1

KELOMPOK 2

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

KELOMPOK 2

2. MENENTUKAN JUMLAH VARIABEL YANG ADA :

- A = rol, θA bukan variabel- B = titik simpul, ada variable θB- C = jepit, θC= 0- D = jepit, θD=0- Jadi variabelnya hanya satu, θB

3. PERHITUNGAN MOMEN PRIMER DAN KEKAKUAN BATANG

Momen primer

MFBA = + 3/16 PL = + 3/16 (3) 6 = + 9 tm

MFBC = - 1/8 PL = - 1/8 (2) 4 = - 1 tm

MFCB = + 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 = M°AB + 1 EI (2θA + θB) = 0 + 2 EIθA + 1 EIθB (1) MBA = M°BA + 1 EI (2θB + θA) = +9 + 2 EIθB + 1 EIθA (2) MBC = M°BC + 1 EI (2θB + θC) = -1 + 2 EIθB + 1 EIθC (3) MCB = M°CB + 1 EI (2θC + θB) = 1 + 2 EIθC + 1 EIθB (4) MBD = M°BD + 0,66 EI (2θB + θD) = 0 + 1,33 EIθB + 0,66 EIθD (5)

MDB = M°DB + 0,66 EI (2θD + θB) = + 0 + 1,33 EIθD + 0,66 EIθB (6)

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

3

KELOMPOK 2

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

5. MENCARI NILAI θ B

Σ Mb = 0 (2a) + (3a) + (5a) = 0(9 + 2 EIθB + 1 EIθA) +(-1 + 2 EIθB ) + (1,33 EIθB ) = 0(+8+ 5,33 EI θB + 1EI θA) = 0 1EI θA + 5,33 EI θB = -8 (7)

Eliminasi – substitusi persamaan 1a dengan persamaan 71EI θA + 5,33 EI θB = -8 x 2 2EI θA + 10,66EI θB = 16 2EIθA + 1 EIθB = 0 x 1 2 EIθA + 1 EIθB = 0 -

9,66 EIθB = 16 EI θ B = 1,553

Substitusi EIθB ke persamaan 1a2 EIθA + 1 EIθB = 02 EIθA + 1( 1,553 )= 02 EIθA + 1,553 = 0 EI θ A = -0,776

6. MOMEN UJUNG

4

KELOMPOK 2

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

ΣMB = 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

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

Batang BC

5

KELOMPOK 2

ΣMB = 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

ΣV= 0

RB1 – P2 + RC = 0

RB1 – 2 + 0,159 = 0

0,159- 2+ RB1= 0

RB1 = 1,841

Batang BD

ΣMB = 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

6

ΣV= 0 RB3 + RD = 0RB3 + 0,514 = 0RB3 =0 ,514

DV =RB + RB1

DV =3,388+ 1,841

DV =5,229

KELOMPOK 2

FreebodySuperposisi

8. KONTROL STRUKTUR

ΣMA = 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

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

ΣH = 0 HC – HD = 0 0,514 – 0,514 = 0 0 = 0

9. MENGHITUNG BIDANG M & D

7

KELOMPOK 2

Bidang M

- Batang AB

- Batang BC

Bidang D

8

Diagram momen (M)

Momen di x = 3

0 + Ra . x

0 + (-0,388) . 3

-1,164 tm

Diagram momen (M)

Momen di x = 2

2,106 + Rb . x

2,106 + 1,841 . 2

5,788 tm

KELOMPOK 2

- 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

10. MenghitungBidang N

9

BIDANG M BIDANG D

KELOMPOK 2

Batang BDN BD = Rb + Rb1

= 3,388 + 1,841= 5,229

A B C

D

5,229(+)

11. Diagram Superposisi BidangM , D , N

10

KELOMPOK 2

- Bidang M

A B C

D

(-)

1,164 11,33

2,106

(-)

5,788

(+)

2,553

1,024

2,065

(-)

- Bidang D

A B C

D

(-)(-)

0,388

6,388

(+)

1,841

(-)

2,159

(-)0,514

- Bidang N

11

KELOMPOK 2

A B C

D

5,229(+)

12