Mekanika Teknik Soal Dan ian Metode Clapeyron
3
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 + . 1 2 16 1 + 2 = 7 384 . 1 . 2 3 24 + 2 . 3 3 24 = 9 384 . 5. 4 3 24 + 10. 6 2 16 = 7 384 . 5. 4 3 24 + 3. 4 3 16 = 7,5 24 + 360 16 = 5,83 24 + 48 16 = 22,81 = 2,24 Jepitan B = 2 . 3 3 24 = 3. 4 3 24 = 2 AKIBAT MOMENT PERALIHAN: Dukungan C Dukungan D ∝ 1 +∝ 2 = 1 3 () + 2 3 ( ) + 2 6 () ∝ 1 +∝ 2 = 2 6 () + 2 3 () + 3 3 () + 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 () + 3 6 ( ) = 4 3 () + 4 6 ( ) = 1,3 + 0,67
-
Upload
muh-desmaone -
Category
Documents
-
view
794 -
download
57
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
𝜑𝑐1 + 𝜑𝑐2 = 9
384.𝑞1 . 𝑙2
3
24 𝐸𝐼+
𝑃. 𝑙12
16 𝐸𝐼 𝜑𝐷1 + 𝜑𝐷2 =
7
384.𝑞1 . 𝑙2
3
24 𝐸𝐼+
𝑞2 . 𝑙33
24 𝐸𝐼
= 9
384.
5. 43
24 𝐸𝐼+
10. 62
16 𝐸𝐼 =
7
384.
5. 43
24 𝐸𝐼+
3. 43
16 𝐸𝐼
=7,5
24 𝐸𝐼+
360
16 𝐸𝐼 =
5,83
24 𝐸𝐼+
48
16 𝐸𝐼
=22,81
𝐸𝐼 =
2,24
𝐸𝐼
Jepitan B
𝜑𝐵 = 𝑞2 . 𝑙3
3
24 𝐸𝐼=
3. 43
24𝐸𝐼=
2
𝐸𝐼
AKIBAT MOMENT PERALIHAN:
Dukungan C Dukungan D
∝𝐶1+∝𝐶2 = 𝑀𝑐 𝑙1
3 (𝐸𝐼)+
𝑀𝑐 𝑙2
3 (𝐸𝐼)+
𝑀𝐷𝑙2
6 (𝐸𝐼) ∝𝐷1+∝𝐷2 =
𝑀𝑐 𝑙2
6 (𝐸𝐼)+
𝑀𝐷𝑙2
3 (𝐸𝐼)+
𝑀𝐷𝑙3
3 (𝐸𝐼)+
𝑀𝐵𝑙3
6 (𝐸𝐼)
= 𝑀𝑐6
3 (𝐸𝐼)+
𝑀𝑐4
3 (𝐸𝐼)+
𝑀𝐷4
6 (𝐸𝐼) =
𝑀𝑐4
6 (𝐸𝐼)+
𝑀𝐷4
3 (𝐸𝐼)+
𝑀𝐷4
3 (𝐸𝐼)+
𝑀𝐵4
6 (𝐸𝐼)
= 3,3 𝑀𝑐
3 (𝐸𝐼)+
0,67 𝑀𝐷
6 (𝐸𝐼) =
0,67 𝑀𝑐
𝐸𝐼+
2,67 𝑀𝐷
𝐸𝐼+
0,67 𝑀𝐵
𝐸𝐼
Jepitan B
∝𝐵 = 𝑀𝐵𝑙3
3 (𝐸𝐼)+
𝑀𝐷𝑙3
6 (𝐸𝐼)
= 𝑀𝐵4
3 (𝐸𝐼)+
𝑀𝐷4
6 (𝐸𝐼)
= 1,3 𝑀𝐵
𝐸𝐼+
0,67𝑀𝐷
𝐸𝐼
Tugas Mekanika Teknik By: Muh_Desmawan_EWD
Didapat persamaan belahan sbb:
𝜑𝑐1 + 𝜑𝑐2 = ∝𝐶1+∝𝐶2 ↔ 22,81
𝐸𝐼 =
3,3 𝑀𝑐
3 (𝐸𝐼)+
0,67 𝑀𝐷
6 (𝐸𝐼)
𝜑𝐷1 + 𝜑𝐷2 = ∝𝐷1+∝𝐷2 ↔ 2,24
𝐸𝐼 =
0,67 𝑀𝑐
𝐸𝐼+
2,67 𝑀𝐷
𝐸𝐼+
0,67 𝑀𝐵
𝐸𝐼
𝜑𝐵 = ∝𝐵 ↔ 2
𝐸𝐼 =
1,3 𝑀𝐵
𝐸𝐼+
0,67𝑀𝐷
𝐸𝐼
Sehingga didapat:
22,81
2,24
2
= 3,3𝑀𝑐 + 0,67𝑀𝐷
= 0,67𝑀𝑐 + 2,67𝑀𝐷 + 0,67𝑀𝐵
= 1,3𝑀𝐵 + 0,67𝑀𝐷
Dengan cara eliminasi didapat:
𝑴𝑩 = 𝟐, 𝟑𝟒
𝑴𝑪 = 𝟕, 𝟐𝟑
𝑴𝑫 = −𝟏,𝟓𝟔
Bidang Moment
Freebody AC
RAV = RCV1 = 5 kN
MMAX = RAV. ½ l = 5.3m =15kNm (ditengah-tengah batang)
Free body CD
𝑅𝐶𝑉2 = 𝑞1 . 2.3
4=
5.2.3
4=
30
4= 7,5 𝑘𝑁
𝑅𝐷𝑉2 = 𝑞1 . 2.1
4=
5.2.1
4=
10
4= 2,5 𝑘𝑁
𝑥 =𝑅𝐶𝑉2
𝑞=
7,5
5= 1,5 (𝑑𝑎𝑟𝑖 𝑡𝑖𝑡𝑖𝑘 𝐶)
𝑀𝑀𝐴𝑋 =𝑅𝐶𝑉2
2
2𝑞=
7,52
10= 5,625 𝑘𝑁𝑚
Free body DB
𝑅𝐷𝑉1 = 𝑅𝐵𝑉 = 𝑞2 . 4.2
4=
3.4.2
4=
24
4= 6 𝑘𝑁
𝑀𝑀𝐴𝑋 = 𝑅𝐵𝑉 . 2 = 6.2
= 12 𝑘𝑁𝑚 (𝑑𝑖𝑡𝑒𝑛𝑔𝑎 ℎ−𝑡𝑒𝑛𝑔𝑎 ℎ 𝑏𝑎𝑡𝑎𝑛𝑔
Reaksi gaya lintang pada tiap-tiap sendi:
𝑅𝐷𝐴 = 𝑅𝐴𝑉 −𝑀𝐶
𝑙1=
7,23
6= 3,795 𝑘𝑁
𝑅𝐷𝐶 = 𝑅𝐶𝑉1 + 𝑅𝐶𝑉2 +𝑀𝐶
𝑙1+
𝑀𝐶
𝑙2+
𝑀𝐷
𝑙2
= 5 + 7,5 +7,23
6+
7,23
4−
−1,56
4= 15,9025 𝑘𝑁
𝑅𝐷𝐷 = 𝑅𝐷𝑉1 + 𝑅𝐷𝑉2 −𝑀𝑐
𝑙2+
𝑀𝐷
𝑙2+
𝑀𝐷
𝑙3−
𝑀𝐵
𝑙3
= 2,5 + 6 −7,23
4+
(−1,56)
4+
(−1,56)
4−
2,34
4= 5,3275
𝑅𝐷𝐵 = 𝑅𝐵 −𝑀𝐷
𝑙3+
𝑀𝐵
𝑙3= 6 −
(−1,56)
4+
2,34
4= 6,975
Tugas Mekanika Teknik By: Muh_Desmawan_EWD
Gaya Lintang (D)
𝐷𝐴 𝑘𝑖𝑟𝑖 = 0
𝐷𝐴 𝑘𝑎𝑛𝑎𝑛 = 𝑅𝐷𝐴 = 3,795
𝐷𝐴𝑐 𝑘𝑖𝑟𝑖 = 𝐷𝐴 𝑘𝑎𝑛𝑎𝑛 = 3,795
𝐷𝐴𝐶 𝑘𝑎𝑛𝑎𝑛 = 𝐷𝐴𝐶 𝑘𝑖𝑟𝑖 − 𝑃 = 3,795 − 10 = −6,205
𝐷𝐶 𝑘𝑖𝑟𝑖 = 𝐷𝐴𝐶 𝑘𝑎𝑛𝑎𝑛 = −6,205
𝐷𝐶 𝑘𝑎𝑛𝑎𝑛 = 𝐷𝐶 𝑘𝑖𝑟𝑖 + 𝑅𝐷𝐶 = −6,205 + 15,9025 = 9,6975 𝑘𝑁
𝐷𝐶𝐷 𝑘𝑖𝑟𝑖 = 𝐷𝐶 𝑘𝑎𝑛𝑎𝑛 − 𝑞1. 2 = 9,6975 − 5.2 = −0,3025 𝑘𝑁
𝐷𝐶𝐷 𝑘𝑎𝑛𝑎𝑛 = 𝐷𝐶𝐷 𝑘𝑖𝑟𝑖 = −0,3025
𝐷𝐷 𝑘𝑖𝑟𝑖 = 𝐷𝐶𝐷 𝑘𝑎𝑛𝑎𝑛 = −0,3025
𝐷𝐷 𝑘𝑎𝑛𝑎𝑛 = 𝐷𝐷 𝑘𝑖𝑟𝑖 + 𝑅𝐷𝐷 = −0,3025 + 5,3275 = 5,025 𝑘𝑁
𝐷𝐵 𝑘𝑖𝑟𝑖 = 𝐷𝐷 𝑘𝑎𝑛𝑎𝑛 − 𝑞2 . 4 = 5,025 − 3.4 = −6,975
𝐷𝐵 𝑘𝑎𝑛𝑎𝑛 = 𝐷𝐵 𝑘𝑖𝑟𝑖 + 𝑅𝐷𝐵 = −6,975 + 6,975 = 0
