Presensi Penilaian - Just Education Site fileCara Menghitung reaksi gaya Luar pada Balok Menerus Balok menerus dengan beban terpusat March 22, 2011 email : [email protected] 5 Caranya

Statika Strukturselasa, 18:20 s/d 20:30, S01

olehhadi saputra

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Presensi Penilaian: Kehadiran : 10%UTS dan UAS : 40%Tugas I+II+III+IV : 50%

Komposisi Tugas I : PR Kuliah ke 1 sd 3

Komposisi Tugas II : PR Kuliah ke 4 sd 7Komposisi Tugas III : PR Kuliah ke 8 sd 10Komposisi Tugas IV : PR Kuliah ke 11 sd 14

Preface

The contents and styles of these notes will definitely change fromtime to time, therefore hard copies may become obsolete immediatelyafter they are printed. Readers are welcome to contact the author forany suggestions on improving this e-book and report any mistakes inthe presentations on improving this e-books and to report anymistakes in the presentations of the subjects or typographical errors.The ultimate goals of this e-books on the statics structure is to makeit readily available for students, to help them learn subjects in thestatics structure.

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Isi

• Reaksi Gaya Luar Balok Tak Langsung• Pembebanan terpusat

• Pembebanan Terdistribusi

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Reaksi Gaya Luar Balok TakLangsung

(Beban Terpusat, Merata, dan Segitiga)

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Syarat Kesetimbangan

Cara Menghitung reaksi gaya Luar pada BalokMenerus

Balok menerus dengan beban terpusat

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Caranya Dengan Melepas sambungan sendi

Contoh Soal

• Buat Model sederhananya?

• Buat Free Body Diagramnya?

• Hitung Reaksi tumpuannya?

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Contoh Soal

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Model

Free Body Diagram

A B C

RVARVB RVB RVC

RHBRHBRHA

MA

8 kN

Contoh Soal

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Reaksi Tumpuan

Batang BC

+ ∑Fv = 0 RvB + RvC – 4 KN/m. 2m = 0

+ ∑FH = 0 RHB = 0

+ ∑MC = 0 RvB.0 + RvC . 2 m – 8 kN. 1m = 0

RvC = 4 kN maka RvB = 4 kN

Batang AB

+ ∑FH = 0 RHA – 10(3/5) - RHB = 0

+ ∑Fv = 0 RvA – RvB – 10(4/5) kN = 0

+ ∑MA = 0 MA – RvB . 4 m – 10(4/5) kN. 2m = 0

Maka RHA = 6 kN

Maka RVA = 12 kN

Maka MA = 32 kN.m

Contoh Soal

• Buat Free Body Diagramnya?

• Hitung Reaksi tumpuannya?

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20 m 20 m 20 m 20 m50 m

5 kN/ m

3 kN/ m

Contoh Soal

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5 kN/ m

3 kN/ m

Free Body Diagram

A B C D E F

RHA

RVARVC RVD RVF

5 kN/ m

3 kN/ m

A B C D E F

RHA

RVARVC RVD RVF

3 kN/ m

B E

RVB

RHB

RVE

RVB RVE

RHE

Contoh Soal

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Reaksi Tumpuan

Batang AB

+ ∑FH = 0 RHA = 0

+ ∑MB =0 -RVA . 20 + (5 . 20) . 10 = 0 RVA = 50 kN

Batang EF

+ ∑ME =0 RVF . 20 - (3 . 20) . 10 = 0 RVF = 30 kN

+ ∑MD =0 – 50 . 90 + [5(40)](70)– RVC(50) + [3(90)](5) + 30(40) = 0 RVC = 241 kN

+ ∑Fy =0 50 - 5(40) + 241 - 3(90) + RVCD + 30 = 0 RVD = 149 kN

PR No. 1

• Buat Free Body diagramnya?

• Berapa reaksi tumpuannya?

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PR No. 2

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• Buat Free Body diagramnya?

• Berapa reaksi tumpuannya?

PR No. 3

• Buat Free Body diagramnya?

• Berapa reaksi tumpuannya?

March 22, 2011 email : [email protected] 14

References

• Vector mechanics for statics, ferdinand P. Bear;E. Russel Johnston, Jr.• Mekanika teknik statika dalam analisis struktur berbentuk rangka, Binsar Hariandja

• Engineering mechanics statics, R.C. Hibbeler

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