Mekanika Teknik I (Soal #2)
Posted on 17 Januari 2011
Carilah Reaksi Perletakan dengan Cara Grafis dan Analitis untuk gambar dibawah ini kemudian hitung dan gambar Bidang D (lintang), N (normal), dan M (moment).

Soal No. 2 UAS Mektek I
PENYELESAIAN
Mencari Reaksi Secara Analitis:
ΣMB = 0
• RAH x 4m + RAV x 6m – Q x 1½m – P2 x 2m = 0
• 2.6 x 4m + 6RAV – 2.6 x 3 x 1½m – 2.6 x 2m = 0
• 10.4 + 6RAV – 11.7 – 5.2 = 0
• 6RAV – 6.5 = 0
• RAV = (6.5/6) = 1.1 t
ΣMA = 0
• RB x 6m – P x 2m – Q x 4½m = 0
• 6RB – 2.6 X 2m – 2.6 x 3 x 4½m = 0
• 6RB – 5.2 – 35.1 = 0
• 6RB – 40.3 = 0
• RB = (40.3/6) = 6.7 t
ΣKV = 0
• RB + RAV – Q = 0
• 6.7 + 1.1 – (2.6 x 3m) = 0
• 7.8 t – 7.8 t = 0
Bidang D (untuk Portal)
Titik A
• DA kiri = RA = 1.1
• DA kanan = RA = 1.1
Titik C
o DCkiri = RA – q x 0
= 1.1 – 0 = 1.1
o DCkanan = RA – Q
= 1.1 – (2.6 x 3)
= 1.1 – 7.8 = – 6.7
Titik D
o DDkiri = RA – Q = – 6.7
o DCkanan = RA – (q x 1½) + RB
= 1.1 – (2.6 x 1½) + 6.7
= 1.1 – 3.9 + 6.7
= 3.9
Bidang D (untuk Tegak)
Titik D
• DDatas = RAH = 2.6 ton
Titik E
• DEatas = RAH = 2.6 ton
• DEbawah = RAH – P
= 2.6 – 2.6 = 0
Bidang M
Titik A
MA = RA x 0 = 0
Titik C
MC = RA x 3m – Q x 0m
MC = 1.1 x 3 = 3.3 tm
Titik F
MF = RA x (3m + X) – ½qX2
= 3RA + RAX – ½qX2
dMF/dx => RA – qX =0
1.1 – 2.6X = 0
X = (-1.1/-2.6) = 0.42
MF = 3RA + RAX – ½qX2
= 3 x 1.1 + 1.1 x 0.42 – ½(2.6)0.422
= 3.3 + 0.462 – 0.229
= 3.762 – 0.229 = 3.533 tm
Titik D
MD = RA x 6m – Q x 1½m
= 1.1 x 6m – 2.6 x 3 x 1½m
= 6.6 – 11.7 = – 5.1 tm

Titik G
MG = RA x (6m – y) – RAH x 4m + P x 2m – ½qy2
= 6 x 1.1 – 1.1y – 2.6 x 4m + 2.6 x 2m – ½(2.6)y2
= 6.6 – 1.1y – 10.4 + 5.2 – 1.3y2
= 1.4 – 1.1y – 1.3y2

y1 = -1.54m (tidak dipakai karena minus) ; y2 = 0.70m
MG = 1.4 – 1.1y – 1.3y2
= 1.4 – 1.1(0.7) – 1.3(0.7)2
= 1.4 + 0.77 – 0.63
= 0 tm
Gambar “kira-kira” sebagaimana dibawah ini:

Bidang D (lintang), N (normal), dan M (moment).