Struktur Rangka Batang

Tugas Struktur Rangka Nama

: Ambrosio Martins Nuno

NIM

: 122842820972 12284282097 2

Jurusan

: Teknik Sipil

Hitunglah gaya-gaya batang dari gambar berikut dengan metode : 1. Kesetimbangan titik buhul cara analitis 2. Kesetimbangan titik buhul cara grafis 3. Cremona 16

G

6

7

17

H

9

8

10

2

1 P1

Diketahui :

5 m

=

P2

=

P3

=

P4

=

K

13

14

6

15 5

E

F P3

5 m

1250 N 2500 N

19

4

P2

P1

12

11

D

5 m

J

3

C A

18

I

P4

B

5 m

5 m

E

=

19

2

J

=

E

+

3

J

=

11

2

11

=

19

+

3

22

=

22

2500 N 1250 N

Struktur statis tertentu

1. Menghitung metode Analitis ∑MA =

0

- RB

25 +

- RB

25 +

-

25 R B RB

∑MB =

20 + 1250

+

93750 -93750

=

-25

P3

15 +

20 + =

=

2500

P2

10 +

15 +

P1

5 =

25 00

10 +

P1

5 =

25 00

10 -

0 12 5 0

5 =

0 3750 N

0

+ RA

25 -

+ RA

25 -

+

P4

25 R A RA

=

P4

20 1250

-

93750 -93750 -25

=

P3

15 -

20 =

2 50 0

P2 15 -

10 -

0 12 5 0

5 =

0 3 75 0 N

Kontrol :

Tugas UTS

∑V =

0

RA

+

3750

+

RB

3750

0

P1

-

-

12 5 0

=

0

P2

-

P3

-

-

P4

2500

=

-

0 25 0 0

-

1250

+

-

=

0

ok!!!!

Titik buhul A Sisi miring

S6

=

S6y

2.5

= S1

2

+

6

6.3 +

=

2

36

6.5 m

S6x A

∑V =

0

-

+ =

VA

+

S6 y

3750

+

S6

S6

=

S6x

=

Cos α

S6y

=

sin α

2.5 6.5 6 6.5

=

0.38

=

0.92

=

∑H

0

0.92 0.

=

-3750

=

0.92

S6 0

X

-4062.5 -1562.5

-4062.5 N (tekan)

0 +

S1

0.38 + + S1

S1 =

S1

=

=

0 1562.5 N

0 (tarik)

Titik Buhul G S16

G S6y

S6

S7x S7

S6x

∑V =

S7y

0

S6y

-

=

=

=

0

X

+ 0.38

1562.5

+

0

4062.5

0.92

-3750

=

-0.92 +

S1 6 =

-

S7

0.92

=

0

4062.5 N (tarik)

-

S7

4062.5

+

S7y =

+

S7

∑H

S6

-

+

X

+

S 16 =

4062.5

1562.5

+

0.38

0 +

S 16 = -3125 N (tekan)

S 16

=

0

0

Tugas UTS

Titik buhul C S8 S7y S7

S8y

∑V =

0

S7y

+

S8y -

+

S7x S2

P1

4062.5

2500

S8x S1

+

S8

=

∑H

=

+

=

0

0.92

+

0.92

S8

-2500

=

0.92

S8

0.92

-

1250

=

-2708.33 N (tekan)

C P1

- S7 -4062.5

X

0

+

+

S8

0.38 +

-1562.5

X

-

S1 S1

-2708.33 0.38 +

-1041.67

-4166.67 + S2

=

+

-1562.5

-1562.5

S2

=

+

S2

=

=

0

+

0

S2

0

S2

= 0 4166.67 N (tarik)

Titik buhul H H

S16

∑V =

S17

0

S8y

-

+ S8y S8x

-

S9y =

∑H

S8

S9

=

=

0

X

0

2708.33 0.92

+

=

2708.33 N

X

+

S 16 +

2708.33 0.38 +

0.92

0

(tarik)

-

S9

2708.33 0.38 +

S9

=

-0.92

+

1041.667

-

-2500

S9 S9y

+

1041.667

5208.33 +

+

+

S17 = 3 12 5

3125

+

0

+

S 17 =

S17

=

0

= 0 -5208.33 N (tekan)

S 17 =

S 17

Titik buhul D S10 S9y S9

∑V =

0

S9y

S10y

+

+

S 10 y

+

- P2

=

0

2708.33 0.92

+

S1 S10

+

S8

0 S9x

S10x

S 10

=

∑H

=

0

P2

- S9

X

0 +

+

+ =

2500

=

0

-

S2

0

0.38

0

-

-5208.33 + S3

-

-

S 10 X

-2708.33 0.38 + -1041.67

=

0.92

S3

S2

0.92

0.92

-

+

S3

=

4166.67 +

4166.667

+

S3

0 S3

=

=

0

S3

= 0 5208.33 N (tarik)

Tugas UTS

Titik buhul B

S15 S15y S5

S15x B

∑V =

0

-

+

∑H

VB

+

S15y =

0

3750

+

S15y 0.92

S 15 y

=

=

-3750

0

=

0.92

=

S 15 X 4062.5 1562.5

-4062.5 N (tekan)

0 -

+

-

S5

0.38 S5

S5 =

S5

=

=

0 1562.5 N

0 (tarik)

Titik Buhul K S19

K

S14x

S15

S15y

S14 S14y

∑V =

S15x

0

S15y

-

S14y

+ S1 4

=

=

0

∑H

-

S15 X

0.38

1562.5

-

-3125

=

4062.5

0 0.92

-3750

+ X

-

0.92

=

0

4062.5 N (tarik)

S 19 =

4062.5

1562.5

S1 9 =

S 14

-

-

-

=

-0.92

S14

4062.5

+

-

0

0.38

-

S19

S 19

=

0

S 19 = -3125 N

=

0

0 (tekan)

Tugas UTS

Titik buhul F S14 S13y S13

S14y

∑V =

0

S 14 y

+

S13y -

+

S13x S5

P4

4062.5

2500

S14x S4

+

S 13

=

∑H

=

+

=

0.92

0 +

S1 S13

0.92

-

12 5 0

=

0.92 S13

-2500

=

0.92

-2708.33 N (tekan)

F P4

- S 13 X

0 +

+

-

S 14 X

2708.33 0.38 +

-

4062.5

1041.667

+

0.38

-

-1562.5

1562.5

1041.67 S4

S4

S5

=

-

S4

-1562.5

-

0 =

S4

0 =

0

S4

= 0 1041.67 N (tekan)

=

Titik buhul J J

S18

S19

∑V =

0

S13y

-

+ S13y S13x

S12

-

+

S 12 y =

0

2708.33 0.92

S12

=

=

0

-

-2500

S12

=

-0.92

0.92

=

2708.33 N

0

(tarik)

S12y ∑H

- S 12 X

-

-

S 13 X

-2708.33 0.38 -1041.67

+

-

-

S18 -

2708.33

0.38

1041.667

-

-5208.33 S 18 =

S 19 =

-

3 12 5 3125

-

0

-

S 18 =

S 18

=

0

S 18

= 0 -5208.33 N (tekan)

Titik buhul D S12 S11y S11

=

∑H

- S 11 X

S12y

S1 1 S12x S4

S3

+

+

-

S12 X

0.38 +

-1041.67 S11x

0

+

S3

2708.33

0.38

5208.33

-

3125.00 S 18 =

-

-

-

-

S4

=

0

5208.33 -

104

1041.67

-

S 18

=

S 18

= 0 3125.00 N (tarik)

P3

Tugas UTS

TABEL No

Titik Buhul

Nilai

Satuan

Arah Gaya

Skala

1

S1

1562.5

newton

tarik

1.563

2

S2

4166.67

newton

tarik

4.167

3

S3

5208.33

newton

tarik

5.208

4

S4

1041.67

newton

tekan

1.042

5

S5

1562.5

newton

tarik

1.563

6

S6

4062.5

newton

tekan

4.063

7

S7

4062.5

newton

tarik

4.063

8

S8

2708.33

newton

Tekan

2.708

9

S9

2708.33

newton

tarik

2.708

10

S10

0

-

-

0.000

11

S11

3125.00

newton

tarik

3.125

12

S12

2708.33

newton

tarik

2.708

13

S13

5208.33

newton

tekan

5.208

14

S14

4062.50

newton

tarik

4.063

15

S15

3125

newton

tekan

3.125

16

S16

3125

newton

tekan

3.125

17

S17

5208.33

newton

tekan

5.208

18

S18

5208.33

newton

tekan

5.208

19

S1 9

3125

newton

tekan

3.125

Tugas UTS

m

0

0

Tugas UTS

Tugas UTS

0

0

0

0

Tugas UTS

Tugas UTS

0

0

1.67 =

0

0

Tugas UTS