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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:

ΣMB = 0

  • RA x 9.5m – P1 x Sin 45° x 8m – P2 x 5m – q x 1½m – q x ½m = 0
  • 9.5RA – 3.6t x ½√2 x 8m – 2.6t x 5m – 1.6t x 1½m – 1.6t x ½m = 0
  • 9.5RA – 3.6t x ½√2 x 8m – 2.6t x 5m – 1.6t x 1½m – 1.6t x ½m = 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

ΣMA = 0

  • –RB x 9.5m + P1 x Sin 45° x 1½m – P2 x 3½m – Q x 8½ = 0
  • –9.5RB + 3.6t x ½√2 x 1½m + 2.6t x 3½m – 1.6t x 4m x 8½m = 0
  • –9.5RB + 3.6t x ½√2 x 1½m + 2.6t x 3½m + 1.6t x 4m x 8½m = 0
  • –9.5RB + 3.818 + 9.1 + 54.4 = 0
  • –9.5 RB = –67.318
  • RB =  = 7.086 ton

ΣKV = 0

  • RA +  RB – 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

  • DA = RA = 4.059 ton

Titik C

  • DCkiri = RA = 4.059 ton
  • DCkanan = RA – P1 x Sin 45°
    = 4.059 – 3.6 x ½√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 B

    • 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 F

    • 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 1½m
= 4.059 ton x 1½m = 6.088 tm

Titik D

MD = RA 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

ME = RA 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

MG = RA x (6½+ X) – P1 x Cos 45° x (5 + X) – P2 x (3 + X) – ½qX2
= 6½RA + XRA – 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

MB = RA 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).