Mekanika Teknik

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. 1 UAS Mektek I

PENYELESAIAN
Mencari Reaksi Secara Analitis:
ΣM_B = 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

ΣM_A = 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
• –9.5R_B + 3.6t x ½√2 x 1½m + 2.6t x 3½m + 1.6t x 4m x 8½m = 0
• –9.5R_B + 3.818 + 9.1 + 54.4 = 0
• –9.5 R_B = –67.318
• R_B =  = 7.086 ton

ΣK_V = 0

• R_A +  R_B – P₁ x Cos 45°– P₂ – q x 4m = 0
• 4.059 ton  + 7.086 ton  – 3.6 ton x ½√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

• D_A = R_A = 4.059 ton

Titik C

• D_Ckiri = R_A = 4.059 ton
• D_Ckanan = R_A – P₁ x Sin 45°
  = 4.059 – 3.6 x ½√2
  = 4.059 – 2.545 = 1.514 ton

Titik D

• D_Dkiri = R_A – P₁ x Sin 45°
  = 1.514 ton
• D_Dkanan = R_A – P₁ x Sin 45°– P₂
  = 1.514 – 2.6 = -1.086 ton

Titik E

• D_Ekiri = R_A – P₁ x Sin 45°– P₂
  = -1.086 ton
• D_Ekanan = R_A – P₁ x Sin 45°– P₂ – (q x 0 m)
  = -1.086 ton

Titik B

• D_Bkiri = R_A – P₁ x Sin 45°– P₂ – (q x 0 m)
  = -1.086 ton
• D_Bkanan = 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
Titik F

• D_Fkanan = [R_A – P₁ x Sin 45°– P₂ – (q x 3 m) + R_B]– (q x 1 m)
  = 1.2 ton – 1.6 ton

= -0.4 ≈ 0
Bidang D
Titik A

M_A = 0

Titik C

M_C = R_A x 1½m
= 4.059 ton x 1½m = 6.088 tm

Titik D

M_D = 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

M_E = 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

M_G = 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.

= 5.857 + (9.204 x (-0.604)) – 0.8 (-0.604²)
= 5.857 – 5.559 – 0.292 ≈ 0

Titik B

M_B = R_A x 9½m – P₁ x Cos 45° x 8m – P₂ x 6m – (q x 3m x 1½)
= 4.059 x 9½m – 3.6 x ½√2 x 8m – 2.6 x 6m – (1.6 x 3m x 1½)
= 38.560 – 20.361 – 15.6 – 7.2
= -4.601

Gambar “kira-kira” sebagaimana dibawah ini:

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