ANALISA STRUKTUR I kuliah 1.ppt
- 5. Reaksi perletakan
MA=0→ VB = (4/3)P
MB = 0→ VA = (5/3)P
H =0 → HA =0
Keseimbangan titik A
S4
S1
VA
V = 0 →S4 sin -VA = 0
H = 0 →S4 cos -S1 =0
Didapat:
S4 = (25/9)P
S1 = (20/9)P
Cos =4/5
Sin = 3/5
- 6. Keseimbangan dititik E
S1 S2
S5
V = 0 → S5 = 0
H = 0 → S2 = S1 =(20/9)P
Keseimbangan dititk C
S4
S8
S6
V = 0 → S4sin +S6sin =2P
H = 0 → S4cos=S6cos +S8
Didapat:
S6=(5/9)P
S8=(16/9)P
2P
- 7. Keseimbangan titik D
P
S8
S9
S7
V = 0 → S7+S9sin = P
H = 0 → S8 = S9 cos
Didapat
S9 = (20/9)P
S7 = -(1/3)P →tarik
Keseimbangan titik F
V = 0 → S6sin =S7→OK
H = 0 → S2 –S6cos =S3
S2
S6
S7
S3
Didapat
S3=(16/9)P