MEKANIKA TEKNIK IB Dosen BU OKTA MEILAWATY
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Transcript of MEKANIKA TEKNIK IB Dosen BU OKTA MEILAWATY
Diketahui gelagar balok sederhana sebagai berikut :
• Hitung reaksi perletakan secara grafis dan analitis ?• Hitung gaya – gaya dalam • Gambar bidang momen ( M ), lintang ( L ) dan normal ( N ) ?
P1 = 600 kg P2 = 400 kg
C D
2.00 m 2.00 m 2.00 mA B
• Reaksi perletakan secara grafis dengan lukisan kutub
A B
d
a
cb
P1 = 600 kg P2 = 400 kg
P1
P2
O
a
b
c
dRB
RA
RA = 2,7 cm x 200 kg = 540 kg
RB = 2,3 cm x 200 kg = 460 kg
Skala gaya 1cm : 200 kg
• Reaksi perletakan secara analitis
)(667,4666
2800
02.6004.4006.
02.4.6.
0
12
kgR
R
PPR
M
B
B
B
A
)(333,5336
3200
032006.
02.4004.6006.
02.4.6.
0
21
kgR
R
R
PPR
M
A
A
A
A
B
Gaya – gaya dalam• Bentang AC
20 X
kgmMxx
Mxx
xMx
xRAMx
MxxRA
Mx
667,10662.333,5332
00
.333,533
.
0.
0
RA
x
Lx
Nx
Mx
0
0
tan)(333,533
0333,533
0
0
Nx
Hx
konskgLx
Lx
LxRA
Vx
• Bentang CD 20 X
kgmMxx
kgmMxx
xMx
Mxxx
Mxxx
MxxPxRA
Mx
333,9332
667,10660
667,1066667,66
0600333,533667,1066
0600)2(333,533
0.1)2.(
0
RA
x
Lx
Nx
MxP1
0
0
tan)(667,66
0600333,533
01
0
Nx
Hx
konskgLx
Lx
LxPRA
Vx
Gaya – gaya dalam• Bentang BD
20 X
kgmMxx
Mxx
xMx
xRBMx
MxxRB
Mx
332,9332.667,4662
00
.667,466
.
0.
0
Lx
x
RB
Nx
Mx
0
0
tan)(667,466
0667,466
0
0
Nx
Hx
konskgLx
Lx
LxRB
Vx
• Gambar bidang M, L dan N
A B
P1 = 600 kg P2 = 400 kg
Bidang M
1066,667 933,332
+
+
- Bidang L
533,333
466,667
66,667