Mekanika Teknikuntuk gambar dibawah ini kemudian hitung dan gambar Bidang D (lintang), N (normal), dan M (moment).

Soal No. 1 UAS Mektek I PENYELESAIAN Mencari Reaksi Secara Analitis: MB = 0

RA x 9.5m P1 x Sin 45 x 8m P2 x 5m q x 1m q x m = 0 9.5RA 3.6t x 2 x 8m 2.6t x 5m 1.6t x 1m 1.6t x m = 0 9.5RA 3.6t x 2 x 8m 2.6t x 5m 1.6t x 1m 1.6t x m = 0 9.5 RA 20.36 15 2.4 0.8 = 0 9.5 RA 38.56 = 0 9.5 RA = 38.56 RA = 38.56/9.5 => RA = 4.059 ton

MA = 0

RB x 9.5m + P1 x Sin 45 x 1m P2 x 3m Q x 8 = 0 9.5RB + 3.6t x 2 x 1m + 2.6t x 3m 1.6t x 4m x 8m = 0 9.5RB + 3.6t x 2 x 1m + 2.6t x 3m + 1.6t x 4m x 8m = 0 9.5RB + 3.818 + 9.1 + 54.4 = 0 9.5 RB = 67.318 RB = = 7.086 ton

KV = 0

RA + RB P1 x Cos 45 P2 q x 4m = 0 4.059 ton + 7.086 ton 3.6 ton x 2 2.6 ton 1.6ton x 4m = 0 11.145 ton 2.545 2.6 ton 6.4 ton = 0 11.145 ton 11.545 = -0.4 0

Bidang D Titik A

DA = RA = 4.059 ton

Titik C

DCkiri = RA = 4.059 ton DCkanan = RA P1 x Sin 45 = 4.059 3.6 x 2 = 4.059 2.545 = 1.514 ton

Titik D

DDkiri = RA P1 x Sin 45 = 1.514 ton DDkanan = RA P1 x Sin 45 P2 = 1.514 2.6 = -1.086 ton

Titik E

DEkiri = RA P1 x Sin 45 P2 = -1.086 ton DEkanan = RA P1 x Sin 45 P2 (q x 0 m) = -1.086 ton

Titik Bo o

DBkiri = RA P1 x Sin 45 P2 (q x 0 m) = -1.086 ton DBkanan = RA P1 x Sin 45 P2 (q x 3 m) + RB = -1.086 ton (1.6 x 3m) + 7.086

= -1.086 ton 4.8 ton + 7.086 ton = 1.2 ton Titik Fo

DFkanan = [RA P1 x Sin 45 P2 (q x 3 m) + RB] (q x 1 m) = 1.2 ton 1.6 ton

= -0.4 0 Bidang D Titik A

MA = 0 Titik C MC = RA x 1m = 4.059 ton x 1m = 6.088 tm Titik D MD = RA x 3m P1 x Cos 45 x 2m = 4.059 ton x 3m 3.6 x 2 x 2m = 14.206 5.090 = 9.116 tm Titik E ME = RA x 6m P1 x Cos 45 x 5m P2 x 3m = 4.059 ton x 6m 3.6 x 2 x 5m 2.6 x 3m = 26.383 12.726 7.8 = 5.857 tm Titik G MG = RA x (6+ X) P1 x Cos 45 x (5 + X) P2 x (3 + X) qX2 = 6RA + XRA 3.6 x 2 x (5 + X) 2.6 x (3 + X) x 1.6 x X2 = 6 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. = 5.857 + (9.204 x (-0.604)) 0.8 (-0.6042) = 5.857 5.559 0.292 0 Titik B MB = RA x 9m P1 x Cos 45 x 8m P2 x 6m (q x 3m x 1) = 4.059 x 9m 3.6 x 2 x 8m 2.6 x 6m (1.6 x 3m x 1) = 38.560 20.361 15.6 7.2 = -4.601 Gambar kira-kira sebagaimana dibawah ini:

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

Mekanika TeknikCarilah 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: MB = 0

RAH x 4m + RAV x 6m Q x 1m P2 x 2m = 0 2.6 x 4m + 6RAV 2.6 x 3 x 1m 2.6 x 2m = 0 10.4 + 6RAV 11.7 5.2 = 0 6RAV 6.5 = 0 RAV = (6.5/6) = 1.1 t

MA = 0

RB x 6m P x 2m Q x 4m = 0 6RB 2.6 X 2m 2.6 x 3 x 4m = 0 6RB 5.2 35.1 = 0 6RB 40.3 = 0 RB = (40.3/6) = 6.7 t

KV = 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 Co o

DCkiri = RA q x 0 = 1.1 0 = 1.1 DCkanan = RA Q = 1.1 (2.6 x 3)

= 1.1 7.8 = 6.7 Titik Do o

DDkiri = RA Q = 6.7 DCkanan = RA (q x 1) + RB = 1.1 (2.6 x 1) + 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 = R A 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) qX2 = 3RA + RAX qX2 dMF/dx => RA qX =0 1.1 2.6X = 0 X = (-1.1/-2.6) = 0.42 MF = 3RA + RAX qX2 = 3 x 1.1 + 1.1 x 0.42 (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 1m = 1.1 x 6m 2.6 x 3 x 1m = 6.6 11.7 = 5.1 tm

Titik G MG = RA x (6m y) RAH x 4m + P x 2m qy2 = 6 x 1.1 1.1y 2.6 x 4m + 2.6 x 2m (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).