contoh soal statika struktur kesetimbangan, equilibrium, diagram benda bebasDeskripsi lengkap
contoh soal statika struktur kesetimbangan, equilibrium, diagram benda bebas
soal dan penyelesaian statika struktur 3 dimensi, force systemFull description
soal dan penyelesaian statika struktur 3 dimensi, force systemDeskripsi lengkap
statika struktur free body diagram, diagram benda bebas, gaya dan momen
statika struktur free body diagram, diagram benda bebas, gaya dan momen
Deskripsi lengkap
zadaci iz statikeFull description
Full description
Pengerjaan Statika
silahkan buat belajar
pipa noh
Full description
Teknik Sipil, Tugas Besar Statika atau Mekanika TeknikFull description
Programski zadatak, univerzitet u Bihaću; građevinski odsjekFull description
Riješeni zadaci za ispit iz Građevinske statike 1Full description
mehanika 1
Full description
2 zadatakFull description
Mehanika 1 - Statika, skripta, strojarski fakultet slavonski brod. Skripta s predavanja s nekim riješenim zadacima, pogledajte ako vas zanima
CONTOH SOAL BAHASA INDONESIA PAKET 1Deskripsi lengkap
Search
Home
Saved
0
434 views
Upload
Sign In
RELATED TITLES
0
CONTOH SOAL STATIKA 1 Uploaded by Satria Pradana
Top Charts
Books
Audiobooks Magazines
News
Documents
Sheet Music
Statika
Save
Embed
Share
Print
CONTOH SOAL STATIKA
1
Download
Join
of 4
Perencanaan Batang Tekan
PROBLEM SET MEKANIKA
Search document
Carilah Reaksi Perletakan dengan Cara Grafis dan Analitis untuk gambar dibawah ini kemu hitung dan gambar Bidang D (lintang), N (normal), dan M (moment).
Soal No. 1 UAS Mektek I PENYELESAIAN
Mencari Reaksi Secara Analitis: ΣMB = 0 • • • • • • •
R A x 9.5m – P1 x Sin 45° x 8m – P2 x 5m – q x 1½m – q x ½m = 0 9.5R A – 3.6t x ½√2 x 8m – 2.6t x 5m – 1.6t x 1½m – 1.6t x ½m = 0 9.5R A – 3.6t x ½√2 x 8m – 2.6t x 5m – 1.6t x 1½m – 1.6t x ½m = 0 9.5 R A – 20.36 – 15 – 2.4 – 0.8 = 0 9.5 R A – 38.56 = 0 9.5 R A = 38.56 R A = 38.56/9.5 => R A = 4.059 ton
ΣMA = 0 • • • • • •
–R B x 9.5m + P1 x Sin 45° x 1½m – P2 x 3½m – Q x 8½ = 0 –9.5R B + 3.6t x ½√2 x 1½m + 2.6t x 3½m – 1.6t x 4m x 8½m = 0 to vote=on0 this title –9.5R B + 3.6t x ½√2 x 1½m + 2.6t x 3½m + 1.6t xSign 4mup x 8½m Useful Not useful –9.5R B + 3.818 + 9.1 + 54.4 = 0 –9.5 R B = –67.318 R B = = 7.086 ton
Home
Saved
Top Charts
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
434 views
Sign In
Upload
RELATED TITLES
0
CONTOH SOAL STATIKA 1 Uploaded by Satria Pradana
Top Charts
Books
Audiobooks Magazines
News
Documents
Save
Embed
Share
Print
•
Statika CONTOH SOAL STATIKA
1
Download
Join
of 4
Perencanaan Batang Tekan
PROBLEM SET MEKANIKA
Search document
DA = R A = 4.059 ton
Sheet Music
Titik C • •
DCkiri = R A = 4.059 ton DCkanan = R A – P1 x Sin 45° = 4.059 – 3.6 x ½√2 = 4.059 – 2.545 = 1.514 ton
Titik D •
•
DDkiri = R A – P1 x Sin 45° = 1.514 ton DDkanan = R A – P1 x Sin 45°– P2 = 1.514 – 2.6 = -1.086 ton
Titik E •
•
DEkiri = R A – P1 x Sin 45°– P2 = -1.086 ton DEkanan = R A – P1 x Sin 45°– P2 – (q x 0 m) = -1.086 ton
Titik B o
o
DBkiri = R A – P1 x Sin 45°– P2 – (q x 0 m) = -1.086 ton DBkanan = R A – P1 x Sin 45°– P2 – (q x 3 m) + R B = -1.086 ton – (1.6 x 3m) + 7.086
= -1.086 ton – 4.8 ton + 7.086 ton = 1.2 ton
Sign up to vote on this title
Titik F
Useful
Not useful
Home
Saved
Top Charts
Books
Audiobooks
Magazines
News
Documents
Sheet Music
Upload
Sign In
Join
Search
Home
Saved
0
434 views
Sign In
Upload
RELATED TITLES
0
CONTOH SOAL STATIKA 1 Uploaded by Satria Pradana
Top Charts
Books
Audiobooks Magazines
News
Documents
Sheet Music
Statika
Save
Embed
Share
Print
Download
Join
CONTOH SOAL STATIKA
1
of 4
Perencanaan Batang Tekan
PROBLEM SET MEKANIKA
Search document
MC = R A x 1½m = 4.059 ton x 1½m = 6.088 tm Titik D
MD = R A x 3½m – P1 x Cos 45° x 2m = 4.059 ton x 3½m – 3.6 x ½√2 x 2m = 14.206 – 5.090 = 9.116 tm Titik E
ME = R A x 6½m – P1 x Cos 45° x 5m – P2 x 3m = 4.059 ton x 6½m – 3.6 x ½√2 x 5m – 2.6 x 3m = 26.383 – 12.726 – 7.8 = 5.857 tm Titik G
MG = R A x (6½+ X) – P1 x Cos 45° x (5 + X) – P2 x (3 + X) – ½qX2 = 6½R A + XR A – 3.6 x ½√2 x (5 + X) – 2.6 x (3 + X) – ½ x 1.6 x X2 = 6½ x 4.059+ X x 4.059– 12.726 + 2.545X – 7.8 + 2.6X – 0.8X2 = 26.383+ 4.059X– 12.726 + 2.545X – 7.8 + 2.6X – 0.8X2 = 5.857+ 9.204X– 0.8X2 a = -0.8 ; b = 9.204 ; c = 5.857
X1 = -0.604 X2 = 12.109 karena > 4 m maka X2 tidak dipakai. 2
= 5.857 + (9.204 x (-0.604)) – 0.8 (-0.604 ) = 5.857 – 5.559 – 0.292 ≈ 0 Titik B
