3 - kuliah statika - 3 - grafis balok
TRANSCRIPT
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MENCARI REAKSI MENCARI REAKSI PERLETAKANPERLETAKANCARA GRAFISCARA GRAFISHAL YANG PERLU DIINGAT :HAL YANG PERLU DIINGAT :
KESEIMBANGAN KONSTRUKSI :KESEIMBANGAN KONSTRUKSI :
AKSI = REAKSIAKSI = REAKSI
AKSI = GAYA LUAR / GAYA YANG BEKERJAAKSI = GAYA LUAR / GAYA YANG BEKERJA
PADA KONSTRUKSI AKIBAT BEBANPADA KONSTRUKSI AKIBAT BEBAN
REAKSI = REAKSI OLEH KONSTRUKSIREAKSI = REAKSI OLEH KONSTRUKSI
DI TUMPUANDI TUMPUAN
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SEIMBANG ?SEIMBANG ?JUMLAH GAYA YANG BEKERJA PADA JUMLAH GAYA YANG BEKERJA PADA
KONSTRUKSI = 0KONSTRUKSI = 0
CONTOH :CONTOH :
P1 = 4.9 T
P2 = 3 T
P3 = 3 T
P4 = 4 T
P5 = 4 T
P6 = 12,32 T
P7 = 7,24 T
P1 = 4.9 T
P2 = 3 T
P3 = 3 T
P4 = 4 T
P5 = 4 T
GAYA TIDAK SEIMBANG GAYA SEIMBANG
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Q1 = 3 T
P1 = 5 T
SA
P1 = 5 T
SA
Kasus 1 Kasus 2Kasus 1 Kasus 2
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PENYELESAIAN KASUS 1PENYELESAIAN KASUS 1
Q1 = 3 T
SA
Q1 =
3 TP
1 = 5 T
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Q1 = 3 T
P1 = 5 T
SA
Q1 =
3 TP
1 = 5 T
1
O
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Q1 = 3 T
P1 = 5 T
SA
Q1 =
3 TP
1 =
5 T
1
O
1
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Q1 = 3 T
P1 = 5 T
SA
Q1 =
3 TP
1 = 5 T
1
O
1
2
2
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Q1 = 3 T
P1 = 5 T
SA
Q1 =
3 TP
1 = 5 T
1
O
1
2
2
3
3
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Q1 = 3 T
P1 = 5 T
SA
Q1 =
3 TP
1 = 5 T
1
O
1
2
2
3
3
4 5
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Q1 = 3 T
P1 = 5 T
SA
Q1 =
3 TP
1 = 5 T
1
O
1
2
2
3
3
4 5
6
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Q1 = 3 T
P1 = 5 T
SA
Q1 =
3 TP
1 = 5 T
1
O
1
2
2
3
3
4 5
6
6
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Q1 = 3 T
P1 = 5 T
SAQ
1 = 3 T
P1 =
5 T
1
O
1
2
2
3
3
4 5
6
6
RS
= 4
.9 T
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Q1 = 3 T
P1 = 5 T
SA
Q1 =
3 TP
1 = 5 T
1
O
1
2
2
3
3
4 5
6
6
RS
= 4
.9 T
RA
= 3
.1 T
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q1 = 1 T / m'
3 m 1 m 1 m
P1 = 5
A S
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q1 = 1 T / m'
3 m 1 m 1 m
P1 = 5 T
A S
3 m 1 m 1 m
P1 = 5 T
A S
Q = 3 T
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Q1 = 3 T
P1 = 5 T
SA
3 m 1 m 1 m
Q1 =
3 TP
1 = 5
T
1
2
3
1
O
2 3
4 5
7
8
9
10
910
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P1 = 5 T
SA
PENYELESAIAN PENYELESAIAN KASUS 2KASUS 2
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P1 = 5 T
SA
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P1 = 5 T
SA
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P1 = 5 T
SA
P1 = 5 T
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RA = 1,54 T
Rs = 3,93 T
P1 = 5 T
SA
P1 = 5 T
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P1 = 5 T
SA
P1 = 5 T
RA = 1,54 T
Rs = 3,93 T
RB = 3,93 T
RHS = 3,16 T
RVS = 2,33 T
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LATIHAN :LATIHAN :
P2 = 4 T
P3 = 4 T4 m
C
Q2 = 3 T Q3 = 3 T
BS
RS = 4.9 T
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P2 = 4 T
4 m
B
C
1 m 2 m 3 m 1 m 1 m1 m
Q2 = 3 TQ3 = 3 T
P3 = 4 T
RS = 4.9 T
P2 = 4 T
Q2 =
3 TQ
3 = 3 T
P3 = 4 T
RS
= 4.9 T
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R L
Gar
is
Ker
ja R
L
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