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Similar to Robby mekban (recovered) (20)
Robby mekban (recovered)
- 2. 1.
A 600 300 C
60000kg/𝑐𝑚2 B 60000kg/𝑐𝑚2
P : 5 T
Jawab :
Lakukan Perbandingan 60 : 30 → 2 : 1
Sehingga P = 5 Ton = 5000 Kg , menjadi:
AB = 60000 Kg dan CB = 60000 Kg
30º → 𝜎 =
𝑃
𝐴
60000 =
5000
cos 60⁄
𝐴
60000 =
2500
1
4⁄ 𝜋 𝐷2
=
2500
1
4⁄ .3,14 𝐷2
60000 =
2500
1
4⁄ .3,14 𝐷2
60000 =
2500
0,785
=
2500
0,785 . 60000
= 4,71 cm
D2
= √4,71
𝐷 = 2,17…………………………………..(BC)
- 3. 60º → 𝜎 =
𝑃
𝐴
60000 =
5000
cos 30⁄
𝐴
60000 =
2500.√3
1
4⁄ 𝜋 𝐷2
=
2500
1
4⁄ .3,14 𝐷2
60000 =
4,330
0,785𝐷2
D2
=
4,330
0,785 . 60000
= 9,19 cm
D = √9,19
𝐷 = 3,03…………………………………..(AB)
- 4. 2.
Diket : P : 10000 kg
Jawab :
#Asumsi tegangan = Z
140
140
𝜎 =
𝑅 𝐴
𝐴
= 𝑅 𝐴𝑉 = Z. (140.140)
𝑅 𝐴 = 19600 Z 𝑐𝑚2
𝑅 𝐴 = 0,196Z 𝑚2
160
160
- 5. 𝜎 =
𝑅 𝐵
𝐴
= 𝑅 𝐵 = Z. (160.160)
𝑅 𝐵 = 25600 Z 𝑐𝑚2
𝑅 𝐵 = 0,256 Z 𝑚2
𝜎 =
𝑃
𝐴
𝜀𝑀𝐴 = 0
-𝑅 𝐵. 160+ P. X
-160𝑅 𝐵 + 10000X= 0
-160(0,256Z)+ 10000X = 0
= -31,36 Z + 10000X = 0 ….. (1)
𝜀𝑀 𝐵 = 0 Maka : Jarak X adalah
𝑅 𝐴.140 - P. (15 – X) = 0 27,44𝑍 + 10000𝑋 = 150000
140 𝑅 𝐴 – 1000(15 – x ) = 0 = 6999,944+ 10000 = 150000
140𝑅 𝐴 – 150000+ 10000x= 0 =143000,056: 10000 = 143.000
140𝑅 𝐴 + 150000x= 150000
140(0,196Z)+ 10000x= 150000
27,44Z + 10000x= 150000 ….. (2)
Subtitusikan Pers (1) dan (2) :
-31,36Z + 10000X = 0
27,44 𝑍 + 10000𝑋 = 150000
−58,8𝑍 = −150000
Z = 2,551 (Tegangan)
- 6. 3.
Diketahui :
N = 12 ton = 12000kg
A =
1
4
. 𝜋. 𝐷2
= 0,758 𝐶𝑚2
E = 2.106
L = 20 m = 2000 cm
Jawab :
𝑑 =
𝑁 . 𝐿
𝐴 . 𝐸
𝑑 =
12000 .20000
(
1
4
. 𝜋. 𝑑2) . 2.106
𝑑 =
12000 .20000
(0.785) . 2.106
𝑑 =
240000000
1570000
𝑑 = 152.866 m
- 7. 4.
Jawab :
A = Daerah Elastis
BC = Titik Leleh- Landing
D = Ultimate Strees
E = Daerah Linier
P = Strain Hardening
EP = Daerah Plastis
- 9. 6)
𝐷𝑖𝑘: 𝜎𝑥 = 500 𝑀𝑝𝑎
𝜎 𝑦 = 350 𝑀𝑝𝑎
𝑡 = 100 𝑀𝑝𝑎
𝑠𝑘𝑎𝑙𝑎 1𝑐𝑚 ∶ 20𝑀𝑝𝑎
Dit:
A. Tegangan Maxdan Min
B. Tegangan Utama
C. Hitung secara analisis dan grafik MOHR
Jawab :
𝑇 𝑚𝑎𝑥/𝑚𝑖𝑛 = ±
1
2
√(𝜎𝑥 − 𝜎𝑦 )² + 4𝑡2
= ±
1
2
√(500 − 350)² + 4(100)2
= ±
1
2
√(22500)+ 4(10000)
= ±
1
2
√62500
= ±
1
2
. 250
𝑇𝑚𝑎𝑥 = +125
𝑇𝑚𝑖𝑛 = −55,9
𝜎 𝑚𝑎𝑥/𝑚𝑖𝑛 =
𝜎𝑥 + 𝜎𝑦
2
±
1
2
√(𝜎𝑥 − 𝜎𝑦)² + 4𝑡2
=
500 + 350
2
±
1
2
√(500− 350)² + 4(100)2
=
850
2
±
1
2
√62500
= 425 ±
1
2
.250
= 425 ± 125
𝜎 𝑚𝑎𝑥 = 425 + 125 = 550
𝜎 𝑚𝑖𝑛 = 425 − 125 = 300
