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Sabtu, 18 Mei 2013
MaNajemen
RENCANA PELAKSANAAN PEMBELAJARAN PKKBS
|
No
|
Kunci jawaban
|
Skor
|
|
|
1
|
a.
Perhitungan
balok S1 – S2
∑MS1 = 0
-RS2 . 4 + P3
. 2 = 0
-4RS2 + 1,5
. 2 = 0
-4RS2 + 3 =
0
RS2 = 0,75 t
( )
∑MS2 = 0
RS1 . 4 – P3
. 2 = 0
4RS1 – 1,5
.2 = 0
RS1 = 0, 75
t ( )
Control ; ∑V = 0
(RS1 + RS2)
– P3 = 0
(0,75 + 0,75) – 1,5 = 0
(OK)
b.
Perhitungan
balok A – B
∑MA = 0
RS.5,5–RB.4,5+P2.3+P1.1,5=
0
0,75x5,5-4,5RB+2.3+2.1,5=
0
4,125-4,5RB+6+3= 0
RB = 2,92 t ( )
∑MB = 0
RA.4,5-P1.3-P2.1,5+RS1.1=
0
4,5RA-2.3-2.1,5+0,75.1=
0
RA = 1,83 t ( )
Control ; ∑V = 0
(RA + RB)
–(P1 + P2 + RS1) = 0
(1,83 + 2,92) –(2 + 2
+0,75) = 0 (OK)
|
c.
Perhitungan
balok C – D
∑MC = 0
-RS2.1+P4.2+P5.4-RD.6=
0
-0,75+2.2+2.4-6RD= 0
-0,75+4+8-6RD= 0
11,25-6RD= 0
∑MD = 0
-RS2..7+RC.6-P4.4-P5.2=
0
-0,75+6RC-2.4-2.2= 0
-0,75+6RC-8-4= 0
RC= 2,88 t ( )
Control ; ∑V = 0
(RA + RB +
RC + RD) –∑P = 0
(1,83 + 2,92 + 2,88
+1,87) –9,5 = 0 (OK)
d.
Perhitungan
gaya lintang
DAkr = 0
DAkn = RA = 1,83 t
DEkr = 1,83 t
DEkn = 1,83 – 2 = -0,17
t
DFkr = - 0,17 t
DFkn = -0,17 – 2 =
-2,17 t
DBkr = -2,17 t
DBkn = -2,17 + 2,92 =
0,75 t
DGkr = 0,75 t
DGkN = 0,75 – 1,5 =
-0,75 t
DCkr = -0,75 t
DCkn = -0,75 +2,88 =
2,13 t
DHkr = 2,13 t
DHkn = 2,13 – 2 = 0,13
t
DIkr = 0,13 t
DIkn = 0,13 -2 = -1,87
t
DDkr = -1,87 t
DDkn = -1,87 + 1,87 = 0
|
50
|
|
e.
Perhitungan
momen
MA = RA .0 = 0
ME = RA . 1,5 = 1,83 .
1,5 = 2,75 tm
MF = RA . 3 – P1
. 1,5 = 1,83 . 3 – 2 . 1,5 = 5,49 – 3 = 2,49 tm
MB = RA . 4,5 – P1
. 3 – P2 . 1,5 = 1,83 . 4,5 – 2 .3 – 2 . 1,5 = -0,76 tm
MG = RA.7,5 – P1.6
– P2 4,5 + RB.3 = 1,83.7,5
– 2.6 – 2.4,5 + 2,92
= 1,49 tm
MH = RA.12,5 – P1.11
– P2.9,5 + RB.8 – P3.5 + RC.2
= 1,83.12,5 – 2.11 – 2.9,5 + 2,92.8 –
1,5.5 + 2,88 . 2
= 3,50 tm
MI = RA.14,5 – P1.13 – P2.11,5
+ RB.10 – P3.7 + RC.4 – P4.2
= 1,83.14,5 – 2.13 – 2.11,5 + 2,92.10
– 1,5.7 + 2,88.4 – 2.2
= 3,76 tm
MD = RA.16,5 – P1.15
– P2.13,5 + RB.12 – P3.9 + RC.6 – P4.4 – P5.2
= 1,83.16,5 – 2.15 – 2.13,5 + 2,92.12
– 1,5.9 + 2,88.6 – 2.4 – 2.2
= 0
|
|||
|
2
|
a.
Perhitungan
balok S – C
Q = q . L
= 1,5 . 5
= 7,5 tm
RS = RC = ½ 7,5 = 3,75
b.
Perhitungan
balok A – B
Q = q . L
= 1,5 . 7
= 10,5 t
∑MA = 0
Q.3,5 – RB.6 + RS.7 = 0
10,5.3,5 – 6RB + 3,75.7
= 0
36,75 – 6RB + 26,25 = 0
RB = 10,5 t ( )
∑MB = 0
RA.6 – Q.2,5 + RS.1 = 0
6RA – 10,5.2,5 + 3,75.1
= 0
6RA – 26,25 + 3,75 =
RA = 3,75 t ( )
Kontrol :
∑V = 0
(RA + RB + RC) – (q.∑L)
= 0
(3,75 + 10,5 + 3,75) –
18 = 0
18 – 18 = 0 (OK)
c.
Perhitungan
gaya lintang
X = 0
DX = RA– (q.X)= 3,75– 0
= 3,75t
X = 6
DX = 3,75 – 1,5.6 =
-5,25 t
DX = 3,75 – 1,5.6 + RB
=
-5,25 + 10,5
= 5,25 t
X = 12
DX = 3,75 – 1,5.12 + RB
= - 14,25 + 10,5
= -3,75 t
DX = 3,75 – 1,5.12 +RB
+ RC
= -3,75 + 3,75
= 0
|
d.
Perhitungan
gaya momen
X = 0; MX = 0
X = 6; MX = 3,75.6 –
½.1,5.62
= -4,5 tm
X = 12
MX = 3,75.12 – ½.1,5.122
+
10,5.6
= 0
Menentukan momen
maksimal
RA – q.X = 0
3,75 – 1,5X = 0
3,75 = 1,5 X
X = 2,5 m dari titik A
RC – q.X = 0
3,75 – 1,5.X = 0
3,75 = 1,5 X
X = 2,5 m dari titik C
Momen max 1
= RA.X – ½ q X2
= 3,75.2,5 – ½. 1,5.2,52
= 4,69 tm
Momen max 2
= RA.X – ½ q X2
= 3,75.2,5 – ½. 1,5.2,52
= 4,69 tm
|
50
|
|
|
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