Tugas Mekanika Teknik By: Muh_Desmawan_EWD

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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