STRUKTUR STATIS TAK TENTU METODE CLAPEYRON- CONTINUOUS BEAM-2
Analisis Struktur Statis Tak Tentu
16
TUG AS ANALISIS STRUKTUR STATIS TAK TENTU Dosen : Ir. Soetoyo Suatu portal bergoyang dengan ukuran/dimensi seperti gambar dibawah ini : Pertanyaan : a. Gambar bidang momen p ortal b. Gambar bidang geser c. Gambar bidang normal 1 Tugas Analisis Struktur Tak Tentu Syeni Hastorini
-
Upload
syeni-hastorini -
Category
Documents
-
view
255 -
download
15
description
Portal BergoyangTeknik SipilAnalisis struktur statis tak tentu
Transcript of Analisis Struktur Statis Tak Tentu
TUG AS ANALISIS STRUKTUR STATIS TAK TENTU
Dosen : Ir. Soetoyo
Suatu portal bergoyang dengan ukuran/dimensi seperti gambar dibawah ini :
Pertanyaan : a. Gambar bidang momen portal
b. Gambar
bidang geser
c. Gambar bidang normal
1Tugas Analisis Struktur Tak Tentu
Syeni Hastorini
Inersia
IAB = 14
π r4
= 14
π 0.154
= 81 π640000 m4
IBC = 112
bh3
= 1
12(0.3 ) (0.5 )3
= 1320 m4
ICD = 112
bh3
= 1
12(0.3 ) (0.6 )3
= 27
5000 m4
ICE = 112
b h3
= 1
12(0.3 ) (0.4 )3
= 1625 m4
IDF = 14
π r14−1
4π r2
4
= 14
π 0.24−14
π0.14
= 3 π8000 m4
Faktor Kekakuan
KBA = 3EIL
= 3E3.5
81π640000
= 243 π E2240000
KBC = KCB
= 4EIL
= 4E5
1320
= E400
KCD = KDC
= 4EIL
= 4E6.5
275000
= 27 E8125
KCE = 3EIL
= 3E5
1625
= 3 E3125
KDF = 4EIL
= 4E6
3 π8000
2Tugas Analisis Struktur Tak Tentu
Syeni Hastorini
= π E4000
Faktor Distribusi
µBA =
243 π E2240000
243 π E2240000
+E400
= π
π+5600243
= 0.12
µBC =
E400
243 π E2240000
+E400
= 1243 π5600
+1
= 0.88
µCB =
E400
E400
+27 E8125
+3 E3125
= 16254409
= 0.37
µCD =
27 E8125
E400
+27 E8125
+3 E3125
= 21604409
= 0.49
µCE =
3 E3125
E400
+27 E8125
+3 E3125
= 624
4409
= 0.14
µDC =
27 E8125
27 E8125
+π E4000
= 165 π864
+1
3Tugas Analisis Struktur Tak Tentu
Syeni Hastorini
= 0.81
µDF =
π E4000
27 E8125
+π E4000
= π
π+86465
= 0.19
Momen Primer
MBA =- q L2
8
= -(1.5 )(3.5)2
8
= -2.29 tm
= -2290 kgm
MBC = Pa b2
L2
= (1 ) (2 ) (32 )52
= 0.72 tm
= 720 kgm
MCB = −Pb a2
L2
4Tugas Analisis Struktur Tak Tentu
Syeni Hastorini
= −(1 ) (3 ) (22 )
52
= -0.48 tm
= -480 kgm
MCD = q L2
24
= (2.5 )(6.52)24
= 4.4 tm
= 4400 kgm
MDC = -4400 kgm
MDF = PL8
= (1.5 )(6)8
= 1.125 tm
= 1125 kgm
MFD = -1125 kgm
5Tugas Analisis Struktur Tak Tentu
Syeni Hastorini
Perhitungan gaya dorong balok BCD
Gaya geser kolom AB
6Tugas Analisis Struktur Tak Tentu
Syeni Hastorini
RBA
B
RB = (1500)(3.5) + MBA/L= 5250+555.14= 5805.14 kg
Gaya geser kolom CE
Gaya geser kolom DF
∑D = RB+RC+RD
= 5018.74 kg
Portal diberi simpangan sebesar ∆, sehingga timbul momen primer pada kolom
BA
sebesar : MBA= + 1.000 kgm
MBA = 3E IBA
LBA2 ∆ = 1000 kgm
∆ = LBA
2
3E IBA1000
= 62
3E(81 π640000
)1000
7Tugas Analisis Struktur Tak Tentu
Syeni Hastorini
MCE
RC
E
CRc = MCE/LCE
= 1008/5= 201.6 kg
D
F
RD
RF
RD = -(1/2P+(MDF-MFD)/LDF)= -(750+238)= -988 kg
= 30180492.91
E m
MCE karena ∆ :
MCE = 3E ICE
LCE2 ∆
= 3E(0.0016)25
30180492.91E
= 5794.65 kgm
MDF karena ∆ :
MDF = 6E IDF
LDF2 ∆
= 6E(3 π
8000)
6230180492.91E
= 5925.925 kgm
8Tugas Analisis Struktur Tak Tentu
Syeni Hastorini
Gaya dorong balok BCD akibat pergeseran
Kolom AB
9Tugas Analisis Struktur Tak Tentu
Syeni Hastorini
MBA
Kolom CE
Kolom DF
∑D’ = 274.86-1058.6-1735.83
= -2519.57
∑D + β∑D’ = 0
5018.74 + β(-2519.57) = 0
Β = 1.99
10Tugas Analisis Struktur Tak Tentu
Syeni Hastorini
RBA
B
RB = MBA/LAB
= 962/3.5= 274.86 kg
RC
E
C
MCE
RB = -MCE/LCE= -5293/5= -1058.6 kg
D
F
RD
RF
RD = -(MDF+MFD)/LDF
= -(4971+5444)/6= -1735.83 kg
MDF
MFD
11Tugas Analisis Struktur Tak Tentu
Syeni Hastorini
Diagram Normal
12Tugas Analisis Struktur Tak Tentu
Syeni Hastorini
Diagram Geser
Diagram Momen
13Tugas Analisis Struktur Tak Tentu
Syeni Hastorini
-5250
+274.86
+600 +400
-16250
+16250
+750
+750
-4833.18 -4689.9
14Tugas Analisis Struktur Tak Tentu
Syeni Hastorini
+7813.29
+10182.56
-28.62
+28.62
+9525.07
