Mekanika Tugas MR I [PDF]

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TUGAS MEKANIKA REKAYASA I FAKULTAS TEKNIK, JURUSAN TEKNIK SIPIL UNIVERSITAS BORNEO



q = 2 ton/m



HB A



B Q1



Q2



VA 2 meter



Q3 VB



6 meter



2 meter



Penyelesaian ; menghitung nilai Q Q1 = 1/2 x q x L = 1/2 x 2 x 2 = 2 ton Q2 = q x L = 2 x 6 = 12 ton Q3 = 1/2 x q x L = 1/2 x 2 x 2 = 2 ton menghitung reaksi perletakannya ∑H = 0 HB = 0 ∑MA = 0 Q1 x (2/3 x 2) + Q2 x (1/2 x 6 + 2) + Q3 x (1/3 x 2 + 6 + 2) - VB x 10 = 0 2 x (2/3 x 2) + 12 x (1/2 x 6 + 2) + 2 x (1/3 x 2 + 6 + 2) - VB x 10 = 0 2,66667 + 60 + 17,33333 - 10 x VB = 0 80 - 10 x VB = 0 VB = 8 ton ∑MB = 0 (-Q3 x 2/3 x 2) - Q2 x (1/2 x 6 + 2) - Q1 x (1/3 x 2 + 6 + 2) + VA x 10 = 0 (-2 x 2/3 x 2) - 12 x (1/2 x 6 + 2) - 2 x (1/3 x 2 + 6 + 2) + VA x 10 = 0 (-2,66667) - 60 - 17,33333 + 10 x VA = 0 (-80) + 10 x VA = 0 VA = 8 ton Kontrol Perhitungan ∑V = 0 VA + VB - Q1 - Q2 - Q3 = 0 8 + 8 - 2 - 12 - 2 = 0 0=0 ……………………………………………..OKE BANGET!!!!!!!!!!!!!!!!!



1 perhitungan mekanika rekayasa I



TUGAS MEKANIKA REKAYASA I FAKULTAS TEKNIK, JURUSAN TEKNIK SIPIL UNIVERSITAS BORNEO



q = 2 ton /m P1 = 5 ton P2 = 3 ton



2 meter



HB A



B



VA



VB 5 meter



5 meter



Penyelesaian ; menghitung nilai Q Q = 1/2 x q x L = 1/2 x 2 x 5 = 5 ton menghitung reaksi perletakannya ∑H = 0 (-HB - P1) = 0



(-HB - 5) = 0



HB = - 5 ton



∑MA = 0 Q x (1/3 x 5) - P1 x 2 + P2 x 10 - VB x 10 = 0 5 x (1/3 x 5) - 5 x 2 + 3 x 10 - VB x 10 = 0 8,33333 - 10 + 30 - 10 x VB = 0 28,33333 - 10 x VB = 0



VB = 2,833333 ton



∑MB = 0 (-Q) x (2/3 x 5 + 5) - P1 x 2 + VA x 10 = 0 (-5) x (2/3 x 5 + 5) - 5 x 2 + VA x 10 = 0 (-41,66667) - 10 + 10 x VA = 0 (-51,66667) + 10 x VA = 0



VA = 5,166667 ton



Kontrol Perhitungan ∑V = 0 VA + VB - Q - P2 = 0 5,166667 + 2,833333 - 5 - 3 = 0 0=0 ……………………………………………..OKE BANGET!!!!!!!!!!!!!!!!!



2 perhitungan mekanika rekayasa I



TUGAS MEKANIKA REKAYASA I FAKULTAS TEKNIK, JURUSAN TEKNIK SIPIL UNIVERSITAS BORNEO



q = r = 1 ton/m



HB A



B



VA



VB π x r² = π



Penyelesaian ; menghitung nilai Q Q = 1/2 x q² x π = 1/2 x 1² x π = 1,5708 ton menghitung reaksi perletakannya ∑H = 0 HB = 0 ∑MA = 0 Q x (1/2 x π) - VB x π = 0 1,5708 x (1/2 x π) - VB x π = 0 2,4674 - π x VB = 0



VB = 0,7854 ton



∑MB = 0 (-Q x 1/2 x π) + VA x π = 0 (-1,5708 x 1/2 x π) + VA x π = 0 (-2,4674) + π x VA = 0



VA = 0,7854 ton



Kontrol Perhitungan ∑V = 0 VA + VB - Q = 0 0,7854 + 0,7854 - 1,5708 = 0 0=0 ……………………………………………..OKE BANGET!!!!!!!!!!!!!!!!!



3 perhitungan mekanika rekayasa I



TUGAS NIKA REKAYASA I



, JURUSAN TEKNIK SIPIL UNIVERSITAS BORNEO



1



an mekanika rekayasa I



TUGAS NIKA REKAYASA I



, JURUSAN TEKNIK SIPIL UNIVERSITAS BORNEO



2



an mekanika rekayasa I



TUGAS NIKA REKAYASA I



, JURUSAN TEKNIK SIPIL UNIVERSITAS BORNEO



3



an mekanika rekayasa I