Nama: KristantoNim: 111031129Kelas: ATugas 2 : Mekanika Kekuatan Material

1. Dik: A1= 8.100 mm2

E1= 103 mm2

Z = 1129 KNL= 10 md2 = 70 m

A2= π ( dz )2

G2= 120 GpaA3= 7500 mm2

L3= 1 mE3= 200 Gpa

Dit: a. Lendutan pada titik A?b. Berapa tegangan geser maksimum?

Jawab:

a. δ 1 = Z . LE . A =

1129×10103×8100

¿ 0,0135323 m

∂ = ZG. A =

1129×10−3

120×109 .3848,451×10−6

¿2,4447×10−8

∂ = 2,4447×10−8× 180

= 0,440046

∂ = ( tan 0,440046 × 0,5 )

¿ 3,840 201287×10−3

b. δ 3= Z . LE . A =

1129×1200×7500

= 7,52667×10−4

δ total = δ 3+δ 3+ ∂

= 0,0135323 + 7,52667×10−4 + 3,840201287×10−3

= 0,05946 m

2.Dik: A1= 32.000 mm2

E1= 30 GPaZ = 1129 KNd2 = 25 mE2= 210 GPaA2= ( π ×r2)

¿3,14 × 1,252 ¿4,9

Dit:a. Lendutan horizontal SHb. Lendutan Vertikal SV

Penyelesaian:

FAC = ABAC

× Z

= 2,74,5×1129

= 677,4

δAC = Z AC× L ACE2× A2

= 677,4×4,5210×4,9

= 2,9623

FBC =BCAC

× FAC

= 3,64,5×677,4

= 541,92

δBC = δy= ZBC x LBCE1 x A1

= 541,92x 3,630x 32,000

= 2,0322

δ = BCAC

× δAC

= 2,74,5×2,9623

= 1,77738

3.Dik:A1= 21244 mm2

A2= 1256 mm2

E1= 24 GPaZ = 1129 KNd2 = 20 mmE2= 200 GpaG1= 9,6 GpaG2= 80 Gpa

F = Z

Atotal =

1129(1256+21244)

= 0,05017F max = G total

G = G1 x A1

A1+A2 + G2 x A2

A1+A2

= 9,6 x21244

22500 + 80 x1256

22500

= 13,53Fmax = G × Atotal = 13,53 . 22500 = 30,4425

Karena Fmax lebih besar dari Z maka kontruksi aman.