Soal Nomor 3

Diketahui : struktur statis tak tentu seperti tergambar.

 

 

 Ditanya           : selesaikan struktur di atas menggunakan Metode Cross

Catatan            : XYZ adalah 3 digit NPM Terakhir Mahasiswa

                         (Misal: G1B013XYZ) à X, Y, dan Z = 0, 0, dan 6

Penyelesaian

 

 Anggapan Bahwa B dan C “Pinned”

Ø  Menentukan nilai stiffness factor

Stifness Faktor

AD

-

 

DA

4(4EI)/3

5.3333

DB

3(2EI)/8

0.7500

DE

4(5EI)/9

2.2222

BD

-

 

ED

4(5EI)/9

2.2222

EC

3(EI)/7.5

0.4000

CE

-

 

 

Ø  Menentukan nilai distribusi Faktor

 

Distribusi Faktor

Joint A

-

 

Joint D

0.6421

1.0000

 

0.0903

 

0.2676

Joint B

 

 

Joint E

0.8475

1.0000

 

0.1525

Joint C

 

 

 

Ø  Menentukan Nilai Fixed and Momen

 

FIX and MOMEN

AD

1.5450

DA

-1.5450

DB

9.0234

DE

13.7391

BD

0.0000

ED

-24.0159

EC

2.8125

CE

0.0000

 

 

                         

Ø  Tabel Metode Cross

 

Joint

A

D

B

E

C

Batang

AD

DA

DB

DE

BD

ED

EC

CE

DF

-

0.6421

0.0903

0.2676

-

0.8475

0.1525

-

FEM

1.5450

-1.5450

9.0234

13.7391


-24.0159

2.8125


BALANCE


-13.6246

-1.9160

-5.6769


17.9690

3.2344


CO

-6.8123



8.9845


-2.8385



BALANCE


-5.7693

-0.8113

-2.4039


2.4055

0.4330


CO

-2.8847



1.2027


-1.2019



BALANCE


-0.7723

-0.1086

-0.3218


1.0186

0.1833


CO

-0.3862



0.5093


-0.1609



BALANCE


-0.3270

-0.0460

-0.1363


0.1364

0.0245


CO

-0.1635



0.0682


-0.0681



BALANCE


-0.0438

-0.0062

-0.0182


0.0577

0.0104


CO

-0.0219



0.0289


-0.0091



BALANCE


-0.0185

-0.0026

-0.0077


0.0077

0.0014


CO

-0.0093



0.0039


-0.0039



BALANCE


-0.0025

-0.0003

-0.0010


0.0033

0.0006


CO

-0.0012



0.0016


-0.0005



BALANCE


-0.0011

-0.0001

-0.0004


0.0004

0.0001


CO

-0.0005



0.0002


-0.0002



BALANCE


-0.0001

0.0000

-0.0001


0.0002

0.0000


CO

-0.0001








TEM

-8.7346

-22.1043

6.1323

15.9720

0.0000

-6.7003

6.7003

0.0000

 

Ø  TEM hasil perhitungan di plotkan ke struktur dan selesaikan dengan cara mekanika rekayasa 1

 

 

 

Ø  Menghitung shear force, Momen, dan Normal Force

 

·         Bentang A – D

 

·         Shear force

FV = 0

-RAV – 2.06X – SFx = 0

-1.3665 – 2.06X – SFx = 0

SFx = (-1.3665 – 2.06X) t

X (m)

SF (t)

0

-1.3665

0.5

-2.3965

1

-3.4265

1.5

-4.4565

2

-5.4865

2.5

-6.5165

3

-7.5465

 

·         Bending Momen

∑Mx = 0

8.7346 – RAV (X) – 2.06X (1/2 X) – Mx = 0

8.7346 – 1.3665X – 1.03X2 – Mx = 0

Mx = (8.7346 – 1.3665X – 1.03X2) tm

X (m)

Mx (tm)

0

8.7346

0.5

7.79385

1

6.3381

1.5

4.36735

2

1.8816

2.5

-1.11915

3

-4.6349

·         Normal Force

∑FH = 0

5.2849 + NFx = 0

NFx = -5.2849 t

X (m)

NF (t)

0

-5.2849

0.5

-5.2849

1

-5.2849

1.5

-5.2849

2

-5.2849

2.5

-5.2849

3

-5.2849

 

·         Bentang D – B (potongan 1)

 

·         Shear Force

∑FV = 0

RDV – SFx = 0

SFx = RDV

SFx = 3.3915 t

X (m)

SF (t)

0

3.3915

0.5

3.3915

1

3.3915

1.5

3.3915

2

3.3915

2.5

3.3915

3

3.3915

3.5

3.3915

4

3.3915

4.5

3.3915

5

3.3915

·         Bending Momen

∑Mx = 0

-6.1323 + RDV (X) – Mx = 0

-6.1323 + 3.3915X – Mx = 0

Mx = (-6.1323 + 3.3915X) tm

X (m)

Mx (tm)

0

-6.1323

0.5

-4.43655

1

-2.7408

1.5

-1.04505

2

0.6507

2.5

2.34645

3

4.0422

3.5

5.73795

4

7.4337

4.5

9.12945

5

10.8252

 

·         Normal Force

∑FH = 0

15.4442 + NFx = 0

NFx = -15.4442 t

X (m)

NF (t)

0

-15.4442

0.5

-15.4442

1

-15.4442

1.5

-15.4442

2

-15.4442

2.5

-15.4442

3

-15.4442

3.5

-15.4442

4

-15.4442

4.5

-15.4442

5

-15.4442

 

·         Bentang D – B (potongan 2)

 


·         Shear Force

∑FV = 0

RDV – P –SFx = 0

3.3915 – 7 – SFx = 0

SFx = -3.6085 t

X (m)

SF (t)

5

-3.6085

5.5

-3.6085

6

-3.6085

6.5

-3.6085

7

-3.6085

7.5

-3.6085

8

-3.6085

 

·         Bending Momen

∑Mx = 0

-6.1323 + RDV (X) – P (X-5) – Mx = 0

-6.1323 + 3.3915X – 7 (X-5) – Mx = 0

Mx = (28.8677 – 3.6085X) tm

X (m)

Mx (tm)

5

10.8252

5.5

9.02095

6

7.2167

6.5

5.41245

7

3.6082

7.5

1.80395

8

-0.0003

 

 

 

·         Normal Force

∑FH = 0

15.4442 + NFx = 0

NFx = -15.4442 t

X (m)

NF (t)

5

-15.4442

5.5

-15.4442

6

-15.4442

6.5

-15.4442

7

-15.4442

7.5

-15.4442

8

-15.4442

 

·         Bentang D – E (Potongan 1)

 

 ·         Shear force

∑FV = 0

RDV – SFx = 0

SFx = RDV

SFx = 7.8977 t

X (m)

SF (t)

0

7.8977

0.5

7.8977

1

7.8977

1.5

7.8977

2

7.8977

2.5

7.8977

3

7.8977

3.5

7.8977

4

7.8977

4.5

7.8977

 

·         Bending Momen

∑Mx = 0

-15.9720 + RDV (X) – Mx = 0

-15.9720 + 7.8977X – Mx = 0

Mx = (-15.9720 + 7.8977X) tm

X (m)

Mx (tm)

0

-15.972

0.5

-12.0232

1

-8.0743

1.5

-4.12545

2

-0.1766

2.5

3.77225

3

7.7211

3.5

11.66995

4

15.6188

4.5

19.56765

 

·         Normal Force

∑FH = 0

15.9720 + NFx = 0

NFx = -15.9720 t

X (m)

NF (t)

0

-1.8934

0.5

-1.8934

1

-1.8934

1.5

-1.8934

2

-1.8934

2.5

-1.8934

3

-1.8934

3.5

-1.8934

4

-1.8934

4.5

-1.8934


·         Bentang D – E (Potongan 2)

 

·         Shear force

∑FV = 0

RDV -  P – (4.06X – 18.27) –SFx = 0

7.8977 – 4.6 -4.06X + 18.27 – SFx = 0

SFx = (21.5677 – 4.06X) t

X (m)

SF (t)

4.5

3.2977

5

1.2677

5.5

-0.7623

6

-2.7923

6.5

-4.8223

7

-6.8523

7.5

-8.8823

8

-10.9123

8.5

-12.9423

9

-14.9723

 

·         Bending Momen

∑Mx = 0

-15.9720 + RDV (X) – P (X-4.5) – (4.06X – 18.27) (1/2(X-4.5)) – Mx = 0

-15.9720 + 7.8977X – 4.6 (X-4.5) – (4.06X – 18.27) (1/2(X-4.5)) – Mx = 0

-15.9720 + 7.8977X – 4.6X + 20.7 – 2.03 X2 + 18.27X – 41.1075 – Mx = 0

Mx = (-36.3795 + 21.5677X – 2.03 X2) tm

 X (m)


Mx (tm)

4.5

19.56765

5

20.709

5.5

20.83535

6

19.9467

6.5

18.04305

7

15.1244

7.5

11.19075

8

6.2421

8.5

0.27845

9

-6.7002

 

·         Menentukan Momen Maksimum

 Mx = (- 36.3795 + 21.5677X - 2.03X2) tm

 =

0 = 21.5677 – 4.06X

X = 21.5677/4.06

X = 5.3122 m

 

Mmax        = (- 36.3795 + 21.5677X - 2.03X2)) tm

                  = (- 36.3795 + 21.5677(5.3122) - 2.03(5.3122)2)tm

                  = 20.9069 tm

 

·         Normal Force

∑FH = 0

15.9720 + NFx = 0

NFx = -15.9720 t

X (m)

NF (t)

4.5

-1.8934

5

-1.8934

5.5

-1.8934

6

-1.8934

6.5

-1.8934

7

-1.8934

7.5

-1.8934

8

-1.8934

8.5

-1.8934

9

-1.8934

·         Bentang E – C (potongan 1)

 

 ·         Shear force

∑FV = 0

REV – SFx = 0

SFx = REV

SFx = 1.8934 t

X (m)

SF (t)

0

1.8934

0.5

1.8934

1

1.8934

1.5

1.8934

2

1.8934

2.5

1.8934

3

1.8934

3.5

1.8934

3.75

1.8934

 

·         Bending Momen

∑Mx = 0

-6.7003 + REV (X) – Mx = 0

-6.7003 + 1.8934X – Mx = 0

Mx = (-6.7003 + 1.8934X) tm

X (m)

Mx (tm)

0

-6.7003

0.5

-5.7536

1

-4.8069

1.5

-3.8602

2

-2.9135

2.5

-1.9668

3

-1.0201

3.5

-0.0734

3.75

0.39995

·         Normal Force

∑FH = 0

14.9723 + NFx = 0

NFx = -14.9723 t

 

X (m)

NF (t)

0

-14.9723

0.5

-14.9723

1

-14.9723

1.5

-14.9723

2

-14.9723

2.5

-14.9723

3

-14.9723

3.5

-14.9723

3.75

-14.9723

 

 

·        Bentang E – C (potongan 2)

 

 ·         Shear force

∑FV = 0

REV – P – SFX = 0

1.8934 – 2 – SFx = 0

SFx = 1.8934 – 2

SFx = -0.1066 t

X (m)

SF (t)

3.75

-0.1066

4

-0.1066

4.5

-0.1066

5

-0.1066

5.5

-0.1066

6

-0.1066

6.5

-0.1066

7

-0.1066

7.5

-0.1066

 

 

·         Bending Momen

∑Mx = 0

-6.7003 + REV (X) – P (X-3.75) – Mx = 0

-6.7003 + 1.8934X – 2 (X-3.75) – Mx = 0

-6.7003 + 1.8934X – 2X + 7.5 – Mx = 0

Mx = (0.7997 – 0.1066X) tm

X (m)

Mx (tm)

3.75

0.39995

4

0.3733

4.5

0.32

5

0.2667

5.5

0.2134

6

0.1601

6.5

0.1068

7

0.0535

7.5

0.0002

 

·         Normal Force

∑FH = 0

14.9723 + NFx = 0

NFx = -14.9723 t

X (m)

NF (t)

3.75

-14.9723

4

-14.9723

4.5

-14.9723

5

-14.9723

5.5

-14.9723

6

-14.9723

6.5

-14.9723

7

-14.9723

7.5

-14.9723

 

Ø  Hasil Analisis