Mekanika Teknik Soal Dan ian Metode Clapeyron

Tugas Mekanika Teknik By: Muh_Desmawan_EWD

Gambarkan moment dan lintangnya,.!! Penyelesaian: AKIBAT MUATAN LUAR: Dukungan C 𝜑𝑐1 + 𝜑𝑐2

Dukungan D =

9 𝑞1 . 𝑙2 3 𝑃. 𝑙1 2 . + 384 24 𝐸𝐼 16 𝐸𝐼

𝜑𝐷1 + 𝜑𝐷2

9 5. 43 10. 62 . + 384 24 𝐸𝐼 16 𝐸𝐼 7,5 360 = + 24 𝐸𝐼 16 𝐸𝐼 22,81 = 𝐸𝐼

=

7 𝑞1 . 𝑙2 3 𝑞2 . 𝑙3 3 . + 384 24 𝐸𝐼 24 𝐸𝐼

7 5. 43 3. 43 . + 384 24 𝐸𝐼 16 𝐸𝐼 5,83 48 = + 24 𝐸𝐼 16 𝐸𝐼 2,24 = 𝐸𝐼

=

=

Jepitan B 𝜑𝐵

𝑞2 . 𝑙3 3 3. 43 2 = = = 24 𝐸𝐼 24𝐸𝐼 𝐸𝐼

AKIBAT MOMENT PERALIHAN: Dukungan C ∝𝐶1 +∝𝐶2

Dukungan D =

𝑀𝑐 𝑙1 𝑀𝑐 𝑙2 𝑀𝐷 𝑙2 + + 3 (𝐸𝐼) 3 (𝐸𝐼) 6 (𝐸𝐼)

=

𝑀𝑐 𝑙2 𝑀𝐷 𝑙2 𝑀𝐷 𝑙3 𝑀𝐵 𝑙3 + + + 6 (𝐸𝐼) 3 (𝐸𝐼) 3 (𝐸𝐼) 6 (𝐸𝐼)

=

𝑀𝑐 6 𝑀𝑐 4 𝑀𝐷 4 + + 3 (𝐸𝐼) 3 (𝐸𝐼) 6 (𝐸𝐼)

=

𝑀𝑐 4 𝑀𝐷 4 𝑀𝐷 4 𝑀𝐵 4 + + + 6 (𝐸𝐼) 3 (𝐸𝐼) 3 (𝐸𝐼) 6 (𝐸𝐼)

=

3,3 𝑀𝑐 0,67 𝑀𝐷 + 3 (𝐸𝐼) 6 (𝐸𝐼)

=

0,67 𝑀𝑐 2,67 𝑀𝐷 0,67 𝑀𝐵 + + 𝐸𝐼 𝐸𝐼 𝐸𝐼

=

𝑀𝐵 𝑙3 𝑀𝐷 𝑙3 + 3 (𝐸𝐼) 6 (𝐸𝐼)

=

𝑀𝐵 4 𝑀𝐷 4 + 3 (𝐸𝐼) 6 (𝐸𝐼)

=

1,3 𝑀𝐵 0,67𝑀𝐷 + 𝐸𝐼 𝐸𝐼

Jepitan B ∝𝐵

∝𝐷1 +∝𝐷2

Tugas Mekanika Teknik By: Muh_Desmawan_EWD Didapat persamaan belahan sbb: 𝜑𝑐1 + 𝜑𝑐2

= ∝𝐶1 +∝𝐶2

↔

𝜑𝐷1 + 𝜑𝐷2

= ∝𝐷1 +∝𝐷2

↔

= ∝𝐵

↔

𝜑𝐵

3,3 𝑀𝑐 0,67 𝑀𝐷 22,81 = + 3 (𝐸𝐼) 6 (𝐸𝐼) 𝐸𝐼 2,24 0,67 𝑀𝑐 2,67 𝑀𝐷 0,67 𝑀𝐵 = + + 𝐸𝐼 𝐸𝐼 𝐸𝐼 𝐸𝐼 2 1,3 𝑀𝐵 0,67𝑀𝐷 = + 𝐸𝐼 𝐸𝐼 𝐸𝐼

Sehingga didapat: 22,81

= 3,3𝑀𝑐 + 0,67𝑀𝐷

2,24

= 0,67𝑀𝑐 + 2,67𝑀𝐷 + 0,67𝑀𝐵

2

= 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 𝑞1 . 2.3 5.2.3 30 𝑅𝐶𝑉2 = = = = 7,5 𝑘𝑁 4 4 4 𝑞1 . 2.1 5.2.1 10 𝑅𝐷𝑉2 = = = = 2,5 𝑘𝑁 4 4 4 𝑅𝐶𝑉2 7,5 𝑥= = = 1,5 (𝑑𝑎𝑟𝑖 𝑡𝑖𝑡𝑖𝑘 𝐶) 𝑞 5 𝑅𝐶𝑉2 2 7,52 𝑀𝑀𝐴𝑋 = = = 5,625 𝑘𝑁𝑚 2𝑞 10 Free body DB 𝑞2 . 4.2 3.4.2 24 𝑅𝐷𝑉1 = 𝑅𝐵𝑉 = = = = 6 𝑘𝑁 4 4 4 𝑀𝑀𝐴𝑋 = 𝑅𝐵𝑉 . 2 = 6.2 = 12 𝑘𝑁𝑚 (𝑑𝑖𝑡𝑒𝑛𝑔𝑎 ℎ−𝑡𝑒𝑛𝑔𝑎 ℎ 𝑏𝑎𝑡𝑎𝑛𝑔 Reaksi gaya lintang pada tiap-tiap sendi: 𝑅𝐷𝐴 𝑅𝐷𝐶

𝑅𝐷𝐷

𝑅𝐷𝐵

𝑀𝐶 7,23 = = 3,795 𝑘𝑁 𝑙1 6 𝑀𝐶 𝑀𝐶 𝑀𝐷 = 𝑅𝐶𝑉1 + 𝑅𝐶𝑉2 + + + 𝑙1 𝑙2 𝑙2 7,23 7,23 −1,56 = 5 + 7,5 + + − = 15,9025 𝑘𝑁 6 4 4 𝑀𝑐 𝑀𝐷 𝑀𝐷 𝑀𝐵 = 𝑅𝐷𝑉1 + 𝑅𝐷𝑉2 − + + − 𝑙2 𝑙2 𝑙3 𝑙3 7,23 (−1,56) (−1,56) 2,34 = 2,5 + 6 − + + − = 5,3275 4 4 4 4 𝑀𝐷 𝑀𝐵 (−1,56) 2,34 = 𝑅𝐵 − + =6− + = 6,975 𝑙3 𝑙3 4 4 = 𝑅𝐴𝑉 −

Tugas Mekanika Teknik By: Muh_Desmawan_EWD Gaya Lintang (D) 𝐷𝐴 𝑘𝑖𝑟𝑖 𝐷𝐴 𝑘𝑎𝑛𝑎𝑛 𝐷𝐴𝑐 𝑘𝑖𝑟𝑖 𝐷𝐴𝐶 𝑘𝑎𝑛𝑎𝑛 𝐷𝐶 𝑘𝑖𝑟𝑖 𝐷𝐶 𝑘𝑎𝑛𝑎𝑛 𝐷𝐶𝐷 𝑘𝑖𝑟𝑖 𝐷𝐶𝐷 𝑘𝑎𝑛𝑎𝑛 𝐷𝐷 𝑘𝑖𝑟𝑖 𝐷𝐷 𝑘𝑎𝑛𝑎𝑛 𝐷𝐵 𝑘𝑖𝑟𝑖 𝐷𝐵 𝑘𝑎𝑛𝑎𝑛

=0 = 𝑅𝐷𝐴 = 3,795 = 𝐷𝐴 𝑘𝑎𝑛𝑎𝑛 = 3,795 = 𝐷𝐴𝐶 𝑘𝑖𝑟𝑖 − 𝑃 = 3,795 − 10 = −6,205 = 𝐷𝐴𝐶 𝑘𝑎𝑛𝑎𝑛 = −6,205 = 𝐷𝐶 𝑘𝑖𝑟𝑖 + 𝑅𝐷𝐶 = −6,205 + 15,9025 = 9,6975 𝑘𝑁

= 𝐷𝐶 𝑘𝑎𝑛𝑎𝑛 − 𝑞1 . 2 = 9,6975 − 5.2 = −0,3025 𝑘𝑁 = 𝐷𝐶𝐷 𝑘𝑖𝑟𝑖 = −0,3025 = 𝐷𝐶𝐷 𝑘𝑎𝑛𝑎𝑛 = −0,3025 = 𝐷𝐷 𝑘𝑖𝑟𝑖 + 𝑅𝐷𝐷 = −0,3025 + 5,3275 = 5,025 𝑘𝑁 = 𝐷𝐷 𝑘𝑎𝑛𝑎𝑛 − 𝑞2 . 4 = 5,025 − 3.4 = −6,975 = 𝐷𝐵 𝑘𝑖𝑟𝑖 + 𝑅𝐷𝐵 = −6,975 + 6,975 = 0