TSP UPJ

1. IHSAN ISMAIL
2013091009
Anstruk soal nomer 8.28

2. Metode balok konjugate
10 m
0.3 kn/m
A B
4 kn
𝐼 = 0,5 10−3 𝑚4
𝐸 = 13 𝐺𝑝𝑎 = 13. 106
𝐾𝑁
𝑚2
∆ 𝐵=?
𝜃 𝐵 =?
= 𝐸𝐼
𝐼 = 0,5 10−3 𝑚4
𝐸 = 13 𝐺𝑝𝑎
= 13. 106
𝐾𝑁
𝑚2
x 0,5 10−3
𝑚4
= 13. 106KN x 0,5 10−3 𝑚2
𝑚2
= 6,5 . 103
KN𝑚2

3. 10 m
A B
4 KN
4 KN
40 KN m
10 m
AB
10 m
0.3 KN/m
A B
3 KN
3 KN
5 m
15 KN m
10 m
AB

4. 𝜃 𝑏 = 𝑙𝑢𝑎𝑠 1 + 𝑙𝑢𝑎𝑠 2 ∆ 𝑏= (𝑙𝑢𝑎𝑠 1 𝑥 𝐶𝑂𝐺) + (𝑙𝑢𝑎𝑠 2 𝑥 𝐶𝑂𝐺)

5. 10 m
A B
4 KN
4 KN
40 KN m
A
B
𝑙𝑢𝑎𝑠 𝐼
=
1
2
−40
𝐸𝐼
10
= −
200
𝐸𝐼
𝑙𝑢𝑎𝑠 𝐼 𝑥 𝐶𝑂𝐺
= −
200
𝐸𝐼
𝑥
20
3
= −
4000
𝐸𝐼
2
3
𝑥10 =
20
3

6. 10 m
0.3 KN/m
A B
3 KN
3 KN
5 m
15 KN m
10 m
BA 3𝑏
4
=
30
4
𝑙𝑢𝑎𝑠 𝐼𝐼
=
𝑏ℎ
3
=
−10
15
𝐸𝐼
3
= −
150
3𝐸𝐼
= −
50
𝐸𝐼
𝑙𝑢𝑎𝑠 𝐼𝐼 𝑥 𝐶𝑂𝐺
=
−50
𝐸𝐼
𝑥
30
4
= −
1500
4𝐸𝐼
= −
375
𝐸𝐼

7. 𝜃 𝑏 = 𝑙𝑢𝑎𝑠 1 + 𝑙𝑢𝑎𝑠 2
𝑙𝑢𝑎𝑠 𝐼
= −
200
𝐸𝐼
𝑙𝑢𝑎𝑠 𝐼 𝑥 𝐶𝑂𝐺
= −
4000
𝐸𝐼
𝑙𝑢𝑎𝑠 𝐼𝐼
= −
50
𝐸𝐼
𝑙𝑢𝑎𝑠 𝐼𝐼 𝑥 𝐶𝑂𝐺
= −
375
𝐸𝐼
𝜃 𝑏 = (−
200
𝐸𝐼
) + (−
50
𝐸𝐼
)
𝜃 𝑏 = −
250
6,5 . 103
∆ 𝑏= (𝑙𝑢𝑎𝑠 1 𝑥 𝐶𝑂𝐺) + (𝑙𝑢𝑎𝑠 2 𝑥 𝐶𝑂𝐺)
∆ 𝑏= (−
4000
𝐸𝐼
) + (−
375
𝐸𝐼
)
∆ 𝑏= −
5125
𝐸𝐼
= EI
= 6,5 . 103
KN𝑚2
𝜃 𝑏 = −
250
𝐸𝐼
𝜃 𝑏 = −0,0384 𝑟𝑎𝑑
∆ 𝑏= −
5125
3 . 6,5 . 103
∆ 𝑏= −0,2628 𝑚
∆ 𝑏= −262,8 𝑚𝑚

8. 𝜃 𝑏 = −0,0384 𝑟𝑎𝑑 ∆ 𝑏= −262,8 𝑚𝑚
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9. TERIMAKASIH