contoh soal mekanika rekayasa IV (soal nomor 1)

 

Soal Nomor 1

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 D “dikunci”

Ø  Menentukan nilai Stifness Faktor

Stifness Faktor

AB

-

-

BA

4(3EI)/6

2.0000

BC

4(4EI)/4

4.0000

CB

4(4EI)/4

4.0000

CD

4(EI)/5

0.8000

DC

-

-

Ø  Menentukan nilai Distribusi Faktor

Distribusi Faktor

Joint A


 

Joint B

0.3333

1.0000

 

0.6667

Joint C

0.8333

1.0000

 

0.1667

Joint D

1.0000

 

 

Ø  Menentukan nilai Fixed and Momen

 


·  

Fixed and Momen

AB

6.1800

BA

-6.1800

BC

7.7133

CB

-7.7133

CD

4.8000

DC

-7.2000

 

Ø  Tabel Metode Cross

Joint

A

B

C

D

Batang

AB

BA

BC

CB

CD

DC

DF

-

0.3333

0.6667

0.8333

0.1667

1.0000

FEM

6.1800

-6.1800

7.7133

-7.7133

4.8000

-7.2000

BALANCE

 

-0.5110

-1.0223

2.4277

0.4856

7.2000

CO

-0.2555

 

1.2138

-0.5111

3.6000

0.2428

BALANCE

 

-0.4046

-0.8093

-2.5740

-0.5149

-0.2428

CO

-0.2023

 

-1.2870

-0.4046

-0.1214

-0.2575

BALANCE

 

0.4290

0.8580

0.4383

0.0877

0.2575

CO

0.2145

 

0.2192

0.4290

0.1287

0.0438

BALANCE

 

-0.0731

-0.1461

-0.4648

-0.0930

-0.0438

CO

-0.0365

 

-0.2324

-0.0731

-0.0219

-0.0465

BALANCE

 

0.0775

0.1549

0.0792

0.0158

0.0465

CO

0.0387

 

0.0396

0.0775

0.0232

0.0079

BALANCE

 

-0.0132

-0.0264

-0.0839

-0.0168

-0.0079

CO

-0.0066

 

-0.0420

-0.0132

-0.0040

-0.0084

BALANCE

 

0.0140

0.0280

0.0143

0.0029

0.0084

CO

0.0070

 

0.0071

0.0140

0.0042

0.0014

BALANCE

 

-0.0024

-0.0048

-0.0152

-0.0030

-0.0014

CO

-0.0012

 

-0.0076

-0.0024

-0.0007

-0.0015

BALANCE

 

0.0025

0.0051

0.0026

0.0005

0.0015

CO

0.0013

 

0.0013

0.0025

0.0008

0.0003

BALANCE

 

-0.0004

-0.0009

-0.0027

-0.0005

-0.0003

CO

-0.0002

 

 

 

 

 

TEM

5.9391

-6.6618

6.6618

-8.3732

8.3732

0.0000

 

Ø  TEM Hasil perhitungan di plotkan ke Struktur dan diselesaikan dengan cara  mekanika rekayasa 1

 

 


Ø  Menghitung Shear Force (SF) dan Bending Momen (BM)

 

·         Bentang A-B

·         Shear Force

∑Fv = 0

RAV - Q - SFx = 0

SFx = RAV - Q

SFx = (6,0596 – 2.06X) t

X (m)

SFx (t)

0

6.0596

0.5

5.0296

1

3.9996

1.5

2.9696

2

1.9396

2.5

0.9096

3

-0.1204

3.5

-1.1504

4

-2.1804

4.5

-3.2104

5

-4.2404

5.5

-5.2704

6

-6.3004

 

·         Bending Momen

∑Mx = 0

-5,9391 + RAV(X) - 2,06X(1/2 X) – Mx = 0

-5,9391 + 6,0596(X) - 2,06.X(1/2 X) – Mx = 0

Mx = (-5,9391 + 6,0596X - 1.03X2) tm

X (m)

Mx (tm)

0

-5.9391

0.5

-3.1668

1

-0.9095

1.5

0.8328

2

2.0601

2.5

2.7724

3

2.9697

3.5

2.652

4

1.8193

4.5

0.4716

5

-1.3911

5.5

-3.7688

6

-6.6615

 

·         Menentukan Momen Maksimum

 

Mx = (- 5,9391 + 6,0596X + 1.03X2) tm

 =

0 = -2,06X + 6,0596

X = 6,0596/2,06

X = 2,9416 m

 

Mmax        = (-5,9391 + 6,0596X - 1.03X2) tm

                  = (-5,9391 + 6,0596(2,9416) - 1.03((2,9416)2) tm

                  = 2,9732 tm

 ·         Bentang B – C (potongan 1)

 

 

·         Shear Force

∑Fv = 0

RBV – Q – SFx = 0

SFx = RBV – Q

SFx = (9,9922 – 4,06X) t

X (m)

SFx (t)

0

9.9922

0.5

7.9622

1

5.9322

1.5

3.9022

2

1.8722

 

·         Bending momen

∑Mx = 0

-6,6618 + RBV(X) – Q(1/2X) – Mx = 0

-6,6618 + 9,9922(X) – 4,06X(1/2 X) – Mx = 0

Mx = (-6,6618 + 9,9922X – 2,03X2) tm

X (m)

Mx (tm)

0

-6.6618

0.5

-2.1732

1

1.3004

1.5

3.759

2

5.2026

 ·         Bentang B – C (potongan 2)

 

 

·         Shear Force

∑Fv = 0

RBV – Q – P – SFx = 0

SFx = RBV – Q – P

SFx = 9,9922 – 4,06X – 4,6

SFx = (5,3922 – 4,06X) t

X (m)

SFx (t)

2

-2.7278

2.5

-4.7578

3

-6.7878

3.5

-8.8178

4

-10.8478

 

·         Bending momen

∑Mx = 0

-6,6618 + RBV(X) – Q(1/2 X) – P(X-2) – Mx = 0

-6,6618 + 9,9922X – 4,06X(1/2 X) – 4,6(X-2) – Mx = 0

-6,6618 + 9,9922X – 2,03X2 – 4,6X + 9,2 – Mx = 0

Mx = (2,5382 + 5,3922X – 2,03X2) tm

X (m)

Mx (tm)

2

5.2026

2.5

3.3312

3

0.4448

3.5

-3.4566

4

-8.373

 

·         Bentang C – D (potongan 1)

 

 ·      Shear Force

∑Fv = 0

RCV – SFx = 0

SFx = RCV

SFx = 5,6746 t

X (m)

SFx (t)

0

5.6746

0.5

5.6746

1

5.6746

1.5

5.6746

2

5.6746

2.5

5.6746

3

5.6746

 

·      Bending momen

∑Mx = 0

-8,3732 + RCV(X) – Mx = 0

Mx = -8,3732 + RCV(X)

Mx = (-8,3732 + 5,6746X) tm

X (m)

Mx (tm)

0

-8.3732

0.5

-5.5359

1

-2.6986

1.5

0.1387

2

2.976

2.5

5.8133

3

8.6506

·      Bentang C – D (potongan 2)

 

 ·         Shear Force

∑Fv = 0

RCV – P – SFx = 0

SFx = RCV – P

SFx = 5.6746 – 10

SFx = - 4,3254 t

X (m)

SFx (t)

3

-4.3254

3.5

-4.3254

4

-4.3254

4.5

-4.3254

5

-4.3254

 

·         Bending momen

∑Mx = 0

-8,3732 + RCV(X) – P(X-3) – Mx =0

Mx = -8,3732 + 5,6746X – 10(X-3)

Mx = -8.3732 + 5,6746X – 10X + 30

Mx = (21,6268 – 4,3254X) tm

X (m)

Mx (tm)

3

8.6506

3.5

6.4879

4

4.3252

4.5

2.1625

5

-0.0002

Ø  Hasil akhir Analisis

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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