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 = – 6.7o 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 0mMC = 1.1 x 3 = 3.3 tm

Titik F

MF = RA x (3m + X) – ½qX2

= 3RA + RAX – ½qX2

dMF/dx => RA – qX =01.1   – 2.6X = 0X = (-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).