TUGAS ANALISA STRUKTUR 2

OLEH

UNIVERSITAS KATOLIK WIDYA MANDIRAKUPANG

###

no reg = 9

1.911

2.5

1 1 4

1.912

2.5

51

1

3 3Hitung dan gambar bidang M, D, dan N

MOMEN PRIMER

1.911

2.5

4

AB

C D

E F

m

t/m

t/m

T

T

m

mm

EI

EI

EI

EI

EIEI

E F

t/m

t/m

T

m

EI

EIEI

P1=P2=

P4=

1.912

2.5

5

3 3

1 Menghitung Momen Primer

= 0 Tidak ada beban

= 0 Tidak ada beban

= 0 Tidak Ada Beban= 0 Tidak Ada Beban= 1 1.9 6 = -5.70 tm

12

= 1 1.9 6 = 5.70 tm12

= 0 tm

MFAC

MFCA

MFBD

MFDB

MFCD

MFDC

MFCE

AB

C D

m

t/mT

mm

EI

EI

EI

**

**2

- 2

= 0 tm

= 1 1.9 6 = -5.70 tm12

= 1 1.9 6 = 5.70 tm12

= 0 Tidak Ada Beban= 0 Tidak Ada Beban

MOMEN AKHIR1.9

112.5

4

1.912

2.5

5

3 3

MFEC

MFEF

MFFE

MFDF

MFFD

2**

2**

-

AB

C D

E F

m

t/m

t/m

T

T

m

mm

EI

EI

EI

EI

EIEI

+

++

+

2 Putaran Sudut akibat PergoyanganRac = Rca = Δ1/5Rdb = Rbd = Δ1/5 Berlaku untuk kolomRce = Rec = Δ2/4 Δ2Rdf = Rfd = Δ2/4 Δ2

= 0 Karena Perletakan Jepit

2.5Rec

4Rdf

2.5

Rac Rdb 5

Rca Rbd

3 3

3 Menghitung Momen Akhir

Δ1Δ1

ɸb dan ɸa

mm

AB

CD

E F

m

m

mm

EI

EI

EI

EI

EIEI

Δ1 Δ2 Δ1 Δ2

Rumus Umum= + 2EI/L= + 2EI/L

= 0.00 1 2 ɸa 3 Δ15 5

= 0.2 ɸc 0.12 Δ1

= 0.00 1 2 ɸc ɸa 3 Δ15 5

= 0.4 ɸc 0.12 Δ1

= 0 1 2 ɸb ɸd 3 Δ15 5

= 0.2 ɸd 0.12 Δ1

= 0 1 2 ɸd ɸb 3 Δ15 5

= 0.4 ɸd 0.12 Δ1

= -5.70 2.5 2 ɸc ɸd6

= -5.70 0.83333 ɸc 0.41667 ɸd

= 5.70 2.5 2 ɸd ɸc6

MAB MFAB ( 2ɸa + ɸb - 3Ra )MBA MFBA ( 2ɸb + ɸa - 3Rb )

MAC ɸc

MCA

MBD

MDB

MCD

MDC

+ + -

-

+ + -

-

+ + -

-

+ + -

-

+ +

+ +

+ +

= 5.70 0.83333 ɸd 0.41667 ɸc

= 0.00 1 2 ɸc ɸe 3 Δ24 4

= 0.5 ɸc 0.25 ɸe 0.1875 Δ2

= 0.00 1 2 ɸe ɸc 3 Δ24 4

= 0.5 ɸe 0.25 ɸc 0.1875 Δ2

= -5.70 2.5 2 ɸe ɸf6

= -5.70 0.83333 ɸe 0.41667 ɸf

= 5.70 2.5 2 ɸf ɸe6

= 5.70 0.83333 ɸf 0.41667 ɸe

= 0 1 2 ɸd ɸf 3 Δ24 4

= 0.5 ɸd 0.25 ɸf 0.1875 Δ2

= 0 1 2 ɸf ɸd 3 Δ24 4

= 0.5 ɸf 0.25 ɸd 0.1875 Δ2

MCE

MEC

MEF

MFE

MDF

MFD

+ + -

-

+ + -

-

+ +

+ +

+ - -

-

+ - -

+ +

+ +

-

+ +

+

+

+ +

+ +

+

+

4 Untuk Menentukan Seluruh momen akhir harus diketahui ɸc ɸd ɸe ɸf Δ1 Δ2

+ += 0.4 ɸc 0.12 Δ1 + -5.70 0.833333 ɸc 0.416666667 ɸd +

= -5.70 1.73333 ɸc 0.41667 ɸd 0.25 ɸe 0.12 Δ1 0.1875 Δ2 == 1.73333 ɸc 0.41667 ɸd 0.25 ɸe 0.12 Δ1 0.1875 Δ2 =

+ += 5.70 0.83333 ɸd 0.41667 ɸc + 0.5 ɸd 0.25 ɸf 0.1875 Δ2 + 0.40

= 5.70 0.41667 ɸc 1.73 ɸd 0.25 ɸf 0.1875 Δ2 0.12 Δ1 = 0= 0.41667 ɸc 1.73333 ɸd 0.25 ɸf 0.1875 Δ2 0.12 Δ1 = -5.70 ......

+= 0.5 ɸe 0.25 ɸc 0.1875 Δ2 + -5.70 0.833333 ɸe 0.416666667 ɸf == -5.70 0.25 ɸc 1.33333 ɸe 0.41667 ɸf 0.1875 Δ2 = 0= 0.25 ɸc 1.33333 ɸe 0.41667 ɸf 0.1875 Δ2 = 5.70 ...... pers 3

ΣMC

MCA MCD MCE

ΣMD

MDC MDF MDB

ΣME

MEC MEF

ΣMF

-

-

- -- -

----

-

---

+

+ +

+ +++ +

+ + +

+ + +++

+ + ++ + +

++

+

= 0.5 ɸf 0.25 ɸd 0.1875 Δ2 + 5.70 0.8333333 ɸf 0.41667 ɸe = 0= 5.70 0.25 ɸd 0.41667 ɸe 1.33333 ɸf 0.1875 Δ2 = 0= 0.25 ɸd 0.41667 ɸe 1.33333 ɸf 0.1875 Δ2 = -5.70

= 0+ + + -11 - 11 4 = 0

0.5 ɸe 0.25 ɸc 0.1875 Δ2 0.5 ɸc 0.25 ɸe 0.1875 Δ2 0.54

0.75 ɸc 0.75 ɸd 0.75 ɸe 0.75 ɸf 0.75 Δ2 = 554

= 0.75 ɸc 0.75 ɸd 0.75 ɸe 0.75 ɸf 0.75 Δ2 = 220 ...... pers

= 0+ + + -11 -12 11 4 12 5

0.2 ɸc 0.12 Δ1 0.4 ɸc 0.12 Δ1 0.25

0.6 ɸc 0.6 ɸd 0.48 Δ1 = 1275

0.6 ɸc 0.6 ɸd 0.48 Δ1 = 635 ...... pers 6

5 REKAP SEMUA PERSAMAAN

MFD MFE

ΣH2

MEC MCE MFD MDF

Lec LDF

ΣH1

MAC MCA MBD MDB

LAC LDB

---

*

- + -+

*- *-

+ +

-

+ + ++ + +

++

+ +

+ + + -

+ + + -

- -

+ -

+

1.73333 ɸc 0.41667 ɸd 0.25 ɸe 0.12 Δ1 0.1875 Δ2 = 5.700.41667 ɸc 1.73333 ɸd 0.25 ɸf 0.1875 Δ2 0.12 Δ1 = -5.70

0.25 ɸc 1.33333 ɸe 0.41667 ɸf 0.1875 Δ2 = 5.700.25 ɸd 0.41667 ɸe 1.33333 ɸf 0.1875 Δ2 = -5.700.75 ɸc 0.75 ɸd 0.75 ɸe 0.75 ɸf 0.75 Δ2 = 220

0.6 ɸc 0.6 ɸd 0.48 Δ1 = 635

6 MENGHITUNG ɸc ɸd ɸe ɸf Δ1 Δ2 DENGAN MENGGUNAKAN MATRIKS

ɸc ɸd ɸe ɸf Δ1 Δ2 ɸc ɸd ɸe ɸf Δ1 Δ2

1.73 0.42 0.25 0.00 -0.12 -0.19 0.74 -0.06 -0.08 0.14 -0.19 -0.170.42 1.73 0.00 0.25 -0.12 -0.19 -0.06 0.74 0.14 -0.08 -0.19 -0.170.25 0.00 1.33 0.42 0.00 -0.19 X -0.08 0.14 0.94 -0.21 -0.20 -0.020.00 0.25 0.42 1.33 0.00 -0.19 0.14 -0.08 -0.21 0.94 -0.20 -0.020.75 0.75 0.75 0.75 0.00 -0.75 0.85 0.85 0.08 0.08 -0.47 -2.510.60 0.60 0.00 0.00 -0.48 0.00 0.75 0.75 0.80 0.80 -2.10 -0.37

1 0 0 0 0 0 ɸc0 1 0 0 0 0 ɸd 5.70 0.74 -0.060 0 1 0 0 0 ɸe -5.70 -0.06 0.74

= 0 0 0 1 0 0 X ɸf = 5.70 X -0.08 0.140 0 0 0 1 0 Δ1 -5.70 0.14 -0.080 0 0 0 0 1 Δ2 220.00 0.85 0.85

635.00 0.75 0.75

7 MENGHITUNG MOMEN AKHIR

MAC = 0.200 Фc - 0.120 Δ1

- -- ---+

+

+ +++

+ ++

+-++

-

= -29.227 + 203.587

= 174.360 tm

MBD = 0.2 Фd - 0.120

= -30.555 + 203.587

= 173.032 tm

MCA = 0.400 Фc - 0.120

= -58.454 + 203.58696

= 145.133 tm

MCD = -5.700 0.833 Фc 0.417 Фd

= -5.700 - 121.780 - 63.657046

= -191.137 tm

MCE = 0.500 Фc 0.250 Фe - 0.188

= -73.068 -12.069 + 131.141= 46.005 tm

MDB = 0.4 Фd - 0.120

= -61.111 + 203.587

= 142.476 tm

MDC = 5.700 0.833 Фd 0.417 Фc

= 5.700 -127.314 -60.890

= -182.504 tm

MDF = 0.5 Фd 0.25 Фf - 0.1875

= -76.388 -14.725 + 131.141

= 40.028 tm

MEC = 0.500 Фe 0.250 Фc - 0.188

Δ1

Δ1

Δ2

Δ1

Δ2

Δ2

+ +

+

+

+

+

+

= -24.137 -36.534 + 131.1413

= 70.470 tm

MEF = -5.700 0.833 Фe 0.417 Фf

= -5.700 -40.229 -24.541

= -70.470 tm

MFD = 0.5 Фf 0.25 Фd - 0.1875

= -29.450 -38.194 + 131.141

= 63.497 tm

MFE = 5.700 0.833 Фf 0.417 Фe

= 5.700 -49.083 -20.114

= -63.497 tm

8 KONTROL SETIAP MOMEN = 0

Σ MA = 0

MAC = 0

174.360 = 0 …………………………. Ok !!!

Σ MC = 0

MCA + MCD + MCE = 0

145.133 + -191.137 + 46.005 = 0

0.000 = 0 …………………………. Ok !!!

Σ MD = 0

MDB + MDC + MDF = 0

142.476 + -182.504 + 40.028 = 0

Δ2

+

+ +

++

+

0.000 = 0 …………………………. Ok !!!

Σ ME = 0

MEC + MEF = 0

70.470 + -70.470 = 0

0.000 = 0 …………………………. Ok !!!

Σ MF = 0

MFD + MFE = 0

63.497 + -63.497 = 0

0.000 = 0 …………………………. Ok !!!

JUMLAH GAYA HORIZONTAL TIAP TINGKAT = 0.

= 0

MCE + MEC+

MDF + MFD - ( q . L ) = 0

46.005 + 70.470+

40.028 + 63.497-8.000 -32.000 = 0

4 4

29.119 + 25.881 -40.000 = 0

55.000 -40.000 = 0

15.000 = 0 …………………………. Ok !!!

= 0

Σ H2

-P2Lce Ldf

Σ H1

MAC + MCA+

MBD + MDB -P2 - P1 = 0

174.360 + 145.133+

173.032 + 142.476-11.000 -12.000 -104.000 = 0

5 5

63.898 + 63.102 -127.000 = 0

127.000 -127.000 = 0

0.000 = 0 …………………………. Ok !!!

9 FREEBODY

MEF 70.470 MFE 63.497

1.911

REF RFE

REC RFDMFD

MEC 70.470

MCE 46.005MDF

RCE

- ( qac . Lac ) - ( qce . Lce )Lac Lbd

E F

T

t/m

t/m

1.9 RDF12

MCD 191.14 182.504 =MDCRCD RDC

RCA RDB

MCA 145.133 MDB

MAC 174.360 MBD

RAC RBD

3 3

10 MENGHITUNG REAKSI PERLETAKAN+ -

REF = 1/2 +MEF

+MFE

= 0.5 1.9 6 + 70.470 + 63.4976 6

= 5.7 + 11.74502 + 10.582881

qef LefLef Lfe

A B

C DT

t/m

= 28.0279 t

RFE = 1/2 -MEF

-MFE

= 0.5 1.9 6 - 70.470 - 63.4976 6

= 5.7 - 11.74502 - 10.582881

= 16.6279 t

Q - REF + RFE = 011.4 - 28.0279 + 16.6279 = -33.2558

11.4 - 11.4 = 00 = 0 OK

RCE = + +MCE

+MEC

= + 11 + 46.005 + 70.4704 4

= + 11 + 11.501163 + 17.6175

= 18.1187 t

qef LefLef Lfe

KONTROL : ΣV = 0

P2Lce Lce

REC = -MEC

-MCE

= - 70.470 - 46.0054 4

= - 17.61753 - 11.501163

= -29.1187 t

+ + RCE - REC = 0

+ 11 + 18.119 - 29.119 = 011.000 + -11.000 = 0

0.000 = 0 ………………..OK !!!

RDF =MDF

+MFD

=40.028

+63.497

4 4

= 10.007 + 15.874

= 25.881 t

RFD = -MFD

-MDF

Lec Lce

KONTROL : ΣH = 0

P2

Ldf Ldf

Lfd Ldf

= -63.497

-40.028

4 4

= - 15.874 - 10.007

= -25.881 t= 25.881 t

RCE + REC = 025.881 + -25.881 = 0

0.000 = 0 ………………..OK !!!

RCD = 1/2 +MCD

+MDC

= 0.5 1.9 6 +191.137

+182.504

6 6

= -5.7 + 31.856 + 30.417

= 56.574 t

RDC = 1/2 -MDC

-MCD

= -0.5 1.9 6 -182.504

-191.137

6 6

KONTROL : ΣH = 0

qcd LcdLcd Lcd

qcd LcdLdc Lcd

= -5.7 - 30.417 - 31.856

= -67.974 t= 67.974 t

Q RDC RCD = 011.4 + -67.974 + 56.574 = 0

67.974 - 67.974 = 00.000 = 0 ………………..OK !!!

RAC = +MAC

+MCA

= 12 +174.360

+145.133

5 5

= 12 + 34.872 + 29.027

= 75.898 t

RCA = -MCA

-MAC

= -145.133

-174.360

5 5

= - 29.027 - 34.872

= -63.898 t

KONTROL : ΣV = 0

P1Lca Lca

Lca Lac

+ + RAC + RCA = 0

+ 12 + 75.898 + -63.898 = 012.000 + 12.000 = 0

0.000 = 0 ………………..OK !!!

RBD =MBD

+MDB

=173.032

+142.476

5 5

= 34.606 + 28.495

= 63.102 t

RDB = -MDB

-MBD

= -142.476

-173.032

5 5

= - 28.495 - 34.606

= -63.102 t= 63.102 t

KONTROL : ΣH = 0

P1

Lbd Ldb

Ldb Lbd

KONTROL : ΣH = 0

RCE + REC = 063.102 + -63.102 = 0

0.000 = 0 ………………..OK !!!

BELUM KERJA11 GAMBAR BIDANG M, D, DAN N

GAMBAR BIDANG MSKLA MOMEN

1 M : 5 CM63.497 tm

8.55 tm70.470 tm

1/8 . q . L2 =

+-

8.55 70.470 = -61.92 tm

70.470 tm

70.470 tm63.497 tm

0 tm

46.005 tm 40.028 tm

182.504 tm

1/8 . q . L2 =

-

E

+

+

F

+

-

+

-

CD

0 tm

191.14 tm0 - 191.14 = -191.14 t m

145.133 - 0 = 145.133 tm145.133 tm 142.476

0 tm

174.360 173.032

GAMBAR BIDANG DSKLA GAYA

1 M : 5 CM

28.0279

1/4 . p . L =

1/8 . q . L2 =

+

+

+

-

+

D

B

C

A

-+

16.6279 8.62710429.1187 25.881

18.1187 25.881

-67.974

0.000

-63.898 63.102

-

-+

-+

-+

E F

DC

75.89863.102

GAMBAR BIDANG NSKLA PANJANG

1 M : 5 CM

-18.119 25.881

E F

28.0279 16.6279

28.0279 16.627963.102

12.62

-+B

A

C -51.898 D

0.000

0.000 -67.974

A -67.9740.000 B

540000 2.5313213333 1

m

m

0.5 ɸc 0.25 ɸe 0.1875 Δ2 = 0

05.70 ...... pers 1

ɸd 0.12 Δ1 = 0

pers 2

0

-

-

+

...... pers 4

ɸf 0.25 ɸd 0.1875 Δ2 0.5 ɸd 0.25 ɸf 0.1875 Δ2 -55 = 04

5

ɸd 0.12 Δ1 0.4 ɸd 0.12 Δ1 -23 -44 -60 = 05

+ -

- +

+ - +

-

Matriks Invers

-0.08 0.14 -0.19 -0.17 ɸc -146.140.14 -0.08 -0.19 -0.17 ɸd = -152.780.94 -0.21 -0.20 -0.02 = ɸe -48.274

-0.21 0.94 -0.20 -0.02 ɸf -58.90.08 0.08 -0.47 -2.51 Δ1 -16970.80 0.80 -2.10 -0.37 Δ2 -699.42

…………………………. Ok !!!

63.497

4

40.028

142.476

5

173.032

1.73 0.42 0.25 0 -0.12 -0.190.42 1.73 0 0.25 -0.12 -0.190.25 0 1.33 0.42 0 -0.19

0 0.25 0.42 1.33 0 -0.190.75 0.75 0.75 0.75 0 -0.750.6 0.6 0 0 -0.48 0

1 0 0 0 0 00 1 0 0 0 00 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1

4.8 0.75 -0.06 -0.08 0.14 -0.19 -0.17-4.8 -0.06 0.75 0.14 -0.08 -0.19 -0.174.8 -0.08 0.14 0.95 -0.21 -0.20 -0.02

-4.8 0.14 -0.08 -0.21 0.95 -0.20 -0.02160 0.86 0.86 0.08 0.08 -0.48 -2.51470 0.75 0.75 0.80 0.80 -2.12 -0.38

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