Matakuliah
Tahun
Versi
: S0114 / Rekayasa Struktur
: 2006
:1
Pertemuan 12
Slope Deflection Method
1
Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa
akan mampu :
• Mahasiswa dapat menghitung struktur dengan
metode slope deflection method
2
Outline Materi
• Materi 1 Analisa Portal
3
Analisa Portal
Portal ABCD, mempunyai EI sama, ℓ = 4 m,
lihat gambar. Gambarkan diagram M, L, N.
1 t/m’
2t
B
C
ℓ
A
ℓ
D
4
FEM BC= -FEM CB= 1/12.1.16= 1,33 tm
K: AB = I / 4 = 1
BC = I / 4 = 1
CD = I / 4 = 1


B
C
A
D
R: AB =  / 4 = R
BC =
0=0
CD =  / 4 = R
5
Persamaan slope deflection:
MAB = 0+KAB(-2A-B+R) = -B+R
MBA = 0+KBA(-2B-A+R) = -2B+R
MBC =1,33+KBC(-2B-C) = 1,33-2B-C
MCB=-1,33+KCB(-2C-B)= -1,33-2C-B
MCD =0+KCD(-2C-D+R) = -2C+R
MDC =0+KDC(-2D-C+R) = -2C+R
A, D jepit: A, D = 0
MBA + MBC = 0
-2B+R+1,33-2B-C = 0 ….(1)
MCB+ MCD = 0
-1,33-2C+-2C+R = 0
HA =
MAB + MBA
4
HD =
….(2)
H = 0 P - HA - HD = 0 ….(3)
MDC + MCD
4
6
Dengan (1) (2) (3) didapat B, B, R (3 pers ., 3 anu)
(1) -4B-C+R = -1,33
(2)
-B-4C+R = 1,33
(-B+R-2B+R)
(3)
(-C+R-2C+R)
-
4
4
8 + 3B - 2R + 3C - 2R = 0
3B - 3C - 4R = -8
(4) B= - 4C + R - 1,33
-12C + 3R - 3,99 + 3 C - 4R = -8
- 9C - R = -8 + 3,99
16B - 4R + 5,32 - C + R = -1,33
- 3R + 15C = -1,33 - 5,32
3R + 27C = 12
42C = 5,35
C = 0,127
R = 2,857
B = 1, 01
= 0
7
MAB= -1,01 + 2,857 = 1,847
MBA= -2,1,01 + 2,857 = 0,837
MBC= 1,33 - 2,02 - 0,127 = - 0,847
MCB= -1,33 - 2.0,127-1,01 = -2,594
MCD= -2. 0,127 + 2,857 = 2,603
MDC= -0,127 + 2,857 = 2,73
=0
=0
8
Gambar Diagram Gaya Dalam
0,8
P=2
0,8
B
1,15
1 t / m’
2,6 2,6
C
1,33
+
2,85
+
D
+
1,15
A
1,8
2
0,85
1,15
D
2,7
2
0,85
2,85
2,6
0,8
2,6
0,8
M
1,6
2,73
9