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Tutorial a 1-A Portal Sendi-rol
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Tutorial a 1-A Portal Sendi-rol
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Nm Setyo Nugroho
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Tutorial ASST T1 PR AS SST 1 Kelaas A 14 Deesember 20111 Dosen n : Ali Awaaludin, S.T., M.Eng., Ph.D. P
Portall Sendi-Rol Soal
Penyeelesaian: Reakssi Tumpuan pada strukttur
tutorial asst 1‐ nmsnugroho ‐ 2011
1
dengan 3
1
3√5
√5
6
2
3√5
√5
Keseimbangan momen di titik A
∑ Ma = 0
(RAv × 0 ) + (RAh × 0 ) + (10 × 13 × 6,5) + (20 × 4 ) + (10 × 13) − (10 × 6 ) − ⎛⎜
⎞ ⎛ 2 ⎞ RBv × 13 ⎟ = 0 ⎠ ⎝ 5 ⎠ ⎝ 5 (0 ) + (0 ) + (845) + (80 ) + (130 ) − (60 ) − ⎛⎜ 2 RBv ⎞⎟ − ⎛⎜ 26 RBv ⎞⎟ = 0 ⎝ 5 ⎠ ⎝ 5 ⎠ 1
RBv × 2 ⎟ − ⎜
(995 ) − ⎛⎜ 28 RBv ⎞⎟ = 0 ⎝
⎠
5
28
RBv = 995
5 RBv = 79, 46 kN
Keseimbangan momen di titik B ∑ MB = 0
(RAv × 13) + (RAh × 2) − (10 × 13 × 6,5) − (20 × 9) + (10 × 0) − (10 × 4) + ⎛⎜ ⎝
1
⎞ ⎛ ⎠ ⎝
RBv × 0 ⎟ + ⎜
5
2
⎞ ⎠
RBv × 0 ⎟ = 0
5
(13RAv ) + (2 RAv ) − (845) − (180) + (0 ) − (40) − (0 ) − (0 ) = 0 (13RAv ) + (2 RAh) − (1.065) = 0 13 RAv + 2 RAh = 1.065
Keseimbangan gaya-gaya horisontal ∑ Fh = 0
(
) ( )
⎛ ⎝
− RAh − 10 − ⎜
(
) ( )
⎛ ⎝
− RAh − 10 − ⎜
(
1 5
1 5
⎞ ⎠
RBv ⎟ = 0
⎞ ⎠
× 79, 46 ⎟ = 0
) ( ) (
)
− RAh − 10 − 35,54 = 0 RAh = −45,54 kN
( )
RAh = 45,54 kN →
tutorial asst 1‐ nmsnugroho ‐ 2011
2
Av + 2 RAh = 1.065 13 RA
(
)
13 RAv + 2 × −45,54 = 1.065 13 RA Av − 91,08 = 1.065 13 RAv = 1.065 + 91,08 RAv = 88,93kN
Cek kesseimbangann gaya-gayaa vertikal ∑ Fh = 0
(RAh ) − (10 × 13) − (200 ) − (10 ) + ⎛⎜ ⎝
(88,93) − (130 ) − (20 ) − (10 ) + ⎛⎜ ⎝
2 5
2 5
⎞ ⎠
RBv ⎟ = 0
⎞ ⎠
× 79, 46 ⎟ = 0
(888,93) − (130 ) − (20 ) − (10 ) + (71,07 ) = 0
( )
0 = 0.......... .... OK
T yanng terjadi Reaksi Tumpuan
tutorial asst 1‐ nmsnugroho ‐ 2011
3
FBD
tutorial asst 1‐ nmsnugroho ‐ 2011
4
NFD, SF FD dan BMD D
(-)
(-) N NFD (kN)
(-)
(+)
( (-)
( (-)
(-) SFD D (kN)
tutorial asst 1‐ nmsnugroho ‐ 2011
5
(-) (+)
MD (kNm) BM
tutorial asst 1‐ nmsnugroho ‐ 2011
6
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