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Transcript of Presensi Penilaian - Just Education Site fileCara Menghitung reaksi gaya Luar pada Balok Menerus...
Statika Strukturselasa, 18:20 s/d 20:30, S01
olehhadi saputra
March 22, 2011 email : [email protected] 1
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.
March 22, 2011 email : [email protected] 2
Isi
• Reaksi Gaya Luar Balok Tak Langsung• Pembebanan terpusat
• Pembebanan Terdistribusi
March 22, 2011 email : [email protected] 3
Reaksi Gaya Luar Balok TakLangsung
(Beban Terpusat, Merata, dan Segitiga)
March 22, 2011 email : [email protected] 4
Syarat Kesetimbangan
Cara Menghitung reaksi gaya Luar pada BalokMenerus
Balok menerus dengan beban terpusat
March 22, 2011 email : [email protected] 5
Caranya Dengan Melepas sambungan sendi
Contoh Soal
• Buat Model sederhananya?
• Buat Free Body Diagramnya?
• Hitung Reaksi tumpuannya?
March 22, 2011 email : [email protected] 6
Contoh Soal
March 22, 2011 email : [email protected] 7
Model
Free Body Diagram
A B C
RVARVB RVB RVC
RHBRHBRHA
MA
8 kN
Contoh Soal
March 22, 2011 email : [email protected] 8
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?
March 22, 2011 email : [email protected] 9
20 m 20 m 20 m 20 m50 m
5 kN/ m
3 kN/ m
Contoh Soal
March 22, 2011 email : [email protected] 10
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
March 22, 2011 email : [email protected] 11
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?
March 22, 2011 email : [email protected] 12
PR No. 2
March 22, 2011 email : [email protected] 13
• 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
March 22, 2011 email : [email protected] 15