Mekanika Teknik Soal Dan ian Metode Clapeyron
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Transcript of Mekanika Teknik Soal Dan ian Metode Clapeyron
Tugas Mekanika Teknik By: Muh_Desmawan_EWD
Gambarkan moment dan lintangnya,.!!
Penyelesaian:
AKIBAT MUATAN LUAR:
Dukungan C Dukungan D
1 + 2 = 9
384.1 . 2
3
24 +
. 12
16 1 + 2 =
7
384.1 . 2
3
24 +
2 . 33
24
= 9
384.
5. 43
24 +
10. 62
16 =
7
384.
5. 43
24 +
3. 43
16
=7,5
24 +
360
16 =
5,83
24 +
48
16
=22,81
=
2,24
Jepitan B
= 2 . 3
3
24 =
3. 43
24=
2
AKIBAT MOMENT PERALIHAN:
Dukungan C Dukungan D
1+2 = 1
3 ()+
23 ()
+2
6 () 1+2 =
26 ()
+2
3 ()+
33 ()
+3
6 ()
= 6
3 ()+
4
3 ()+
4
6 () =
4
6 ()+
4
3 ()+
4
3 ()+
4
6 ()
= 3,3 3 ()
+0,67 6 ()
= 0,67
+
2,67
+0,67
Jepitan B
= 3
3 ()+
36 ()
= 4
3 ()+
4
6 ()
= 1,3
+
0,67
Tugas Mekanika Teknik By: Muh_Desmawan_EWD
Didapat persamaan belahan sbb:
1 + 2 = 1+2 22,81
=
3,3 3 ()
+0,67 6 ()
1 + 2 = 1+2 2,24
=
0,67
+2,67
+
0,67
= 2
=
1,3
+0,67
Sehingga didapat:
22,81
2,24
2
= 3,3 + 0,67
= 0,67 + 2,67 + 0,67
= 1,3 + 0,67
Dengan cara eliminasi didapat:
= , = , = ,
Bidang Moment
Freebody AC
RAV = RCV1 = 5 kN
MMAX = RAV. l = 5.3m =15kNm (ditengah-tengah batang)
Free body CD
2 = 1 . 2.3
4=
5.2.3
4=
30
4= 7,5
2 = 1 . 2.1
4=
5.2.1
4=
10
4= 2,5
=2
=7,5
5= 1,5 ( )
=2
2
2=
7,52
10= 5,625
Free body DB
1 = = 2 . 4.2
4=
3.4.2
4=
24
4= 6
= . 2 = 6.2
= 12 (
Reaksi gaya lintang pada tiap-tiap sendi:
= 1
=7,23
6= 3,795
= 1 + 2 +1
+2
+2
= 5 + 7,5 +7,23
6+
7,23
4
1,56
4= 15,9025
= 1 + 2 2
+2
+3
3
= 2,5 + 6 7,23
4+
(1,56)
4+
(1,56)
4
2,34
4= 5,3275
= 3
+3
= 6 (1,56)
4+
2,34
4= 6,975
Tugas Mekanika Teknik By: Muh_Desmawan_EWD
Gaya Lintang (D)
= 0
= = 3,795
= = 3,795
= = 3,795 10 = 6,205
= = 6,205
= + = 6,205 + 15,9025 = 9,6975
= 1. 2 = 9,6975 5.2 = 0,3025
= = 0,3025
= = 0,3025
= + = 0,3025 + 5,3275 = 5,025
= 2 . 4 = 5,025 3.4 = 6,975
= + = 6,975 + 6,975 = 0
