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tabel Putaran Sudut (1).pdf
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Sudut Pandang
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1. Perletakan Sendi ‐ Sendi No
Pembebanan
Rotasi Sudut θA
=
qL3 24
θB
=
qL3 24
θA
=
9 qL3 384
θB
=
7qL3
1
2
384
θA
=
qa2 (3L‐ 2a) 12
3 θB
=
qa2 (3L‐ 2a) 12
θA
=
7 qL3 324
4 θB
=
7 qL3 324
θA
=
qc [3(L2 – a2) – c2] 24
θB
=
qc [3(L2 – a2) – c2]
5
24
6
7
θA
=
qc2 [2L2 – c2] 24L
θB
=
qc2 [2L– c]2 24L
θA
=
θB
=
qb (3L2‐ b2) 48 qb (3L2‐ b2) 48
θA
=
13 qL3 648
θB
=
13 qL3 648
θA
=
qcb [L2 – b2 – c2] 6L 4
θB
=
qca [L2 – a2 – c2] 6L 4
θA
=
θB
=
8qL3 360 7qL3 360
θA
=
qc2 (20L2‐ 15Lc + 3c2) 360L
θB
=
qa2 (10L2‐ 3c2) 360L
θA
=
qc2 (40L2‐ 45c + 12c2) 360 L
θB
=
qc2 (5L‐ 3c2) 90 L
8
9
10
11
12
θA
=
5qL3 192
θB
=
5qL3 192
θA
=
θB
=
3qL3 192 3qL3 192
θA
=
qa2 (2L‐ a) 24
θB
=
qa2 (2L‐ a) 24
θA
=
L3 [ 53q1 + 37q2 ] 5760
θB
=
L3 [ 37q1 + 53q2 ] 5760
θA
=
qL [ L2 – 2a + a3 ] 24 L
θB
=
qL [ L2 – 2a + a3 ] 24 L
13
14
15
16
17
θA
=
L3 ( 7q2 + 8q1 ) 360
θB
=
L3 ( 8q2 + 7q1 ) 360
θA
=
qL3 (1 + b) (7‐ 3b2) 360 L L2
θB
=
qL3 (1 + a) (7‐ 3a2) 360 L L2
18
19
θA
=
qL3 30
θB
=
qL3 30
20
θA
=
PL2 16
θB
=
PL2 16
θA
=
Pab (b + L) 6L
θB
=
Pab (a + L) 6L
θA
=
Pa (L – (L – a) 2
θB
=
Pa (L – (L – a) 2
θA
=
PL2 9
θB
=
PL2 9
θA
=
15PL2 96
θB
=
15 PL2 96
θA
=
19 PL2 144
θB
=
19 PL2 144
21
22
23
24
25
26
27
θA
=
PL2n (1 – (1 – 1 ) 24 n2
θB
=
PL2n (1 – (1 – 1 ) 24 n2 n = L/a (n=ganjil)
28
θA
=
PL2n (1 + 1 ) 24 2n2
θB
=
PL2n (1 + 1 ) 24 2n2 n = L/a (n=ganjil)
θA
=
P (L2 – c2) 8
θB
=
P (L2 – c2) 8
θA
=
Pb [L2 – b2 – 3 c2] 3L 4
θB
=
Pa [L2 – a2 – 3 c2] 3L 4
θA
=
1 ML 24
θB
=
29
30
31 1 ML 24
θA
=
M (L2 – 3b2) 6L
θB
=
M (3a2 ‐ L2) 6L
θA
=
ML 6
θB
=
ML 3
32
33
θA
=
L [ 2M1 + M2 ] 6
θB
=
L [ 2M2 + M1 ] 6
θA
=
EI [ y2 – y1 ] L
θB
=
34
35
EI [ y2 – y1 ] L
*(Penurunan Pada Tumpuan)
2. Perletakan Jepit – Jepit – Sendi dan Jepit – Jepit – Bebas No
1
Pembebanan
Rotasi Sudut
θB
=
PL2 2EI
θB
=
Pa2 2EI
θC
=
Pa2 2EI
θB
=
Pa2 (1‐ a) 4EI L
θC
=
Pa2 (1‐ a) (2‐a)2 4EI L L
θC
=
Pa2 (1‐ a)2 (1‐2a) 2EI L L
θB
=
1 qL3 6 EI
θB
=
q (L3‐ a3) 6EI
θB
=
qa3 6EI
2
3
4
5
6
7
8
9
10
11
12
13
14
15
θB
=
1 qL3 48 EI
θB
=
1 qL3 8 EI
θB
=
1 qL3 80 EI
θB
=
1 qL3 24 EI
θB
=
1 qL3 120 EI
θB
=
5 qL3 384 EI
θB
=
q (L3 ‐ 2a2L + a3) 48 EI
θB
=
ML EI
16
17
18
θB
=
θC
=
θB
= Ma (3a‐2) 4EI L
θC
= Ma (‐3a3 +12a2 –12a+4) 4EI L3 L2 L
θC
= Ma (1 ‐ 4a + 6a2 ‐ 3a3) EI L L2 L3
θB
=
PL2 (1‐ a) (3 – (3 – a) 6EI L L
θC
=
PLa (1‐ a) 3EI L
θB
=
θC
=
PL2 (1‐ a) (2 – (2 – a) 4EI L L PLa (1‐ a) 4EI L
19
20
Ma EI Ma EI
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