hitung menara air

L A M P I R A N

Universitas Sumatera Utara

KONTROL PROFIL RANGKA MENARA

C. Berat Sendiri Profil 2L dengan Jumlah Medan (n) = 8

1. Berat Air

Volume air = 25 m³

Berat Air = 25 m³ x 1000 kg/m³ = 25 Ton

2. Berat Rangka Tangki

d. Berat Rangka Tangki

Keterangan:

Rangka tangki memakai profil L 50 50 5 dengan berat profil = 3,77 kg/m’

Panjang bentang diagonal dinding plat tangki

l² = 3,5² + 2,5²

l = 22 5,25,3 +

l = 4,3 m

Panjang batang diagonal plat datar tangki

l² = 3,5² + 3,5²

l = 22 5,35,3 +

l = 4,95 m

3,50 x 3.77 x 4 x 2 = 105,56 kg

2,50 x 3,77 x 4 = 37,70 kg

4,30 x 3,77 x 4 = 64,844 kg

4,95 x 3,77 x 1 = 18,662 kg

Berat Rangka Tangki = 226,766 kg


= 0,2277 Ton

e. Berat plat tangki dengan tebal 10 mm = 0,01 m

Keterangan :

Tutup tangki dianggap senyawa dengan plat atas tangki

Plat dinding tangki = 2,5 x 3,5 x 0,01 x 7,85 x 4 = 2,7475 T

Plat datar tangki = 3,5 x 3,5 x 0,01 x 7,85 x 2 = 1,9233 T

= 4,6698 T

f. Berat ikutan/manusia (P) = 100 kg = 0,1 T

Jadi Berat sendiri pada tangki

= Berat air + berat rangka tangki + berat plat tangki + berat ikutan

= 25 + 0,2277 + 4,6698 + 0,1

= 29,9975 Ton ≈ 30 Ton

Berat Rangka = 0,8615 Ton

Berat sendiri pada Tangki = 30 Ton

Berat sendiri Total = Berat rangka + Berat sendiri pada tangki

= 0,8615 ton + 30 ton

= 30,8615 ton


L M N

21 22

23 24

19 20

A

C

F

I

B

E

H

K

D

G

J

1

2

3

9

13

17

4

5

18

6

7

14

10

8 11

12

15

1625 26

29 30

27 28

O P Q

R S T

U V W

X Y Z

31 32

33 34

35 36

37 38

39 40

41 42

43 44

45 46

47 48

P = 7,7154 T P = 7,7154

P = 7,7154 T

X

S8

S9

Z

S11

S10

P = 7,7154 T

Jadi Berat sendiri tiap titik (P) = 4

ritotalBeratsendi = 4

8615,30 = 7,7154 Ton

Tan α = 75,1

875,1 = 46,975 º

Titik X

Σ kx = 0 Σ ky = 0

S9 = 0 T S8 - P = 0

S8 = P = 7,71545 T (tekan)

Titik Z

Σ kx = 0 Σ ky = 0

S10 = 0 T S11 - P = 0

S11 = P = 7,71545 T (tekan)


Y

S48S47

S9 S10

U

S7

S45S8 S47

W

S12

S46S11S48

V

S44S43

S45 S46

R

S6

S41S7 S43

Titik Y

Σ kx = 0

S5 + S17 cos α – S18 cos α – S4 = 0

S47 = 0

S48 = 0

Titik U

Σ kx = 0 Σ ky = 0

S45 –S47 cos α = 0 S7 – S8 – S47 sin α = 0

S45 = 0 T S7 = S8 = 7,71545 T (tekan)

Titik W

Σ kx = 0 Σ ky = 0

S48 cos α – S46 = 0 S12 – S11 – S48 sin α = 0

S46 = 0 T S12 = S11 = 7,71545 T (tekan)

Titik V

Σ kx = 0

S46 + S43 cos α – S44 cos α – S45 = 0

S43 = 0 T

S44 = 0 T

Titik R

Σ kx = 0 Σ ky = 0

S41 –S43 cos α = 0 S6 – S7 – S47 sin α = 0

S45 = 0 T S6 = S7 = 7,71545 T (tekan)


T

S13

S42S12S44

S

S40S39

S41 S42

O

S56

S37S6 S39

Q

S14

S38S13S40

P

S36S35

S37 S38

Titik T

Σ kx = 0 Σ ky = 0

S44 cos α – S42 = 0 S13 – S12 – S48 sin α = 0

S42 = 0 T S13 = S12 = 7,71545 T (tekan)

Titik S

Σ kx = 0

S42 + S39 cos α – S40 cos α – S41 = 0

S39 = 0 T

S40 = 0 T

Titik O

Σ kx = 0 Σ ky = 0

S37 –S39 cos α = 0 S5 – S6 – S39 sin α = 0

S37 = 0 T S5 = S6 = 7,71545 T (tekan)

Titik Q

Σ kx = 0 Σ ky = 0

S40cos α – S38= 0 S14 – S13 – S40sin α = 0

S40 = 0 T S14 = S13 = 7,71545 T (tekan)

Titik P

Σ kx = 0

S38 + S35 cos α – S36 cos α – S37 = 0

S36 = 0 T

S35 = 0 T


L

S4

S33S5 S35

N

S15

S34S14S36

M

S32S31

S33 S34

I

S3

S29S4 S31

K

S16

S30S15S32

Titik L

Σ kx = 0 Σ ky = 0

S33 –S35 cos α = 0 S4 – S5 – S35 sin α = 0

S33 = 0 T S4 = S5 = 7,71545 T (tekan)

Titik N

Σ kx = 0 Σ ky = 0

S36cos α – S34= 0 S15 – S14 – S36sin α = 0

S34 = 0 T S15 = S14 = 7,71545 T (tekan)

Titik M

Σ kx = 0

S34 + S31 cos α – S32 cos α – S33 = 0

S31 = 0 T

S32 = 0 T

Titik I

Σ kx = 0 Σ ky = 0

S29 –S31 cos α = 0 S3 – S4 – S31 sin α = 0

S29 = 0 T S3 = S4 = 7,71545 T (tekan)

Titik K

Σ kx = 0 Σ ky = 0

S32cos α – S30= 0 S16 – S15 – S32sin α = 0

S30 = 0 T S16 = S15 = 7,71545 T (tekan)


J

S28S27

S29 S30

F

S2

S25S3 S27

H

S17

S26S16S28

G

S24S23

S25 S26

C

S1

S21S2 S23

Titik J

Σ kx = 0

S30 + S27 cos α – S28 cos α – S29 = 0

S27 = 0 T

S28 = 0 T

Titik F

Σ kx = 0 Σ ky = 0

S25 –S27 cos α = 0 S2 – S3 – S31 sin α = 0

S25 = 0 T S2 = S3 = 7,71545 T (tekan)

Titik H

Σ kx = 0 Σ ky = 0

S28cos α – S26= 0 S17 – S16 – S28sin α = 0

S26 = 0 T S17 = S16 = 7,71545 T (tekan)

Titik G

Σ kx = 0

S26 + S23 cos α – S24 cos α – S25 = 0

S27 = 0 T

S28 = 0 T

Titik C

Σ kx = 0 Σ ky = 0

S21 –S23 cos α = 0 S1 – S2 – S21 sin α = 0

S21 = 0 T S1 = S2 = 7,71545 T (tekan)


E

S18

S22S17S24

D

S20S19

S21 S22

Titik E

Σ kx = 0 Σ ky = 0

S24cos α – S22= 0 S18 – S17 – S24sin α = 0

S22 = 0 T S18 = S17 = 7,71545 T (tekan)

Titik D

Σ kx = 0

S22 + S19 cos α – S20 cos α – S21 = 0

S19 = 0 T

S20 = 0 T


Tabel Gaya Batang Akibat Berat Sendiri Untuk Profil 2L pada Jumlah Medan (n) = 8

No

Batang Berat sendiri (ton)

Tekan Tarik 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

7,71545 7, 71545 7, 71545 7, 71545 7, 71545 7, 71545 7, 71545 7. 71545

- -

7, 71545 7, 71545 7, 71545 7, 71545 7, 71545 7, 71545 7, 71545 7, 71545

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -


Tabel Gaya Batang Akibat Berat Sendiri Untuk Profil 2C pada Jumlah Medan (n) = 8

No


Tekan Tarik 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

7,7428 7,7428 7,7428 7,7428 7,7428 7,7428 7,7428 7,7428

- -

7,7428 7,7428 7,7428 7,7428 7,7428 7,7428 7,7428 7,7428

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -


Tabel Gaya Batang Akibat Berat Sendiri Untuk Profil IWF pada Jumlah Medan (n) = 8

No


Tekan Tarik 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

7,8058 7,8058 7,8058 7,8058 7,8058 7,8058 7,8058 7,8058

- -

7,8058 7,8058 7,8058 7,8058 7,8058 7,8058 7,8058 7,8058

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -


P2

P1

P1

P1

P1

P1

P1

P1

P1

P3

P3

P3

P3

P3

P3

P3

P3

P1

P1

P1+P2

P1

P1

P1

P1

P1

P3

P3

P3

P3

P3

P3

P3

P3

D. Akibat Berat Angin

1. Berat Angin Kiri

Luas bidang yang dikenai angin

Panjang batang tegak = 1,875 m

Panjang batang horizontal = 1,75 m

Panjang batang diagonal = 2,5648 m

Panjang profil yang dikenai angin

Batang tegak 2 . 55 . 55 . 10 = 12 cm = 0,12 m

Batang horizontal 2 . 30 . 30 . 3 = 7 cm = 0,07 m

Batang diagonal 2 . 50 . 50 5 = 11 cm = 0,11 m


Batang tegak = 0,12 x 1,875 x 8 x 2 = 3,6 m²

Batang horizontal = 0,07 x 1,75 x 8 x 2 = 1,96 m²

Batang diagonal = 0,11 x 2,5648 x 8 x 2 = 4,514 m²

Luas total bidang yang dikenai angin = 10,074 m²


L M N

21 22

23 24

19 20A

C

F

I

B

E

H

K

D

G

J

1

2

3

9

13

17

4

5

18

6

7

14

10

8 11

12

15

1625 26

29 30

27 28

O P Q

R S T

U V W

X Y Z

31 32

33 34

35 36

37 38

39 40

41 42

43 44

45 46

47 48

P1

P1

P1+P2

P1

P1

P1

P1

P1

P3

P3

P3

P3

P3

P3

P3

P3

P1 = 16

6,1 anginangdikenaitalbidangyginxLuastoxtekananan

= 16

074,10406,1 xx

= 40,296 kg = 0,0403 T

P2 = 2

6,1 dangginxLuasbixtekananan x 15

25,16

= 2

5,25,3406,1 xxx x 1,083

= 303,334 kg = 0,3034 T

P3 = 16


= 16

074,10402,1 xx

= 30,222 kg = 0,03022 T


X

S8

S9P1+P2

Z

S11

S10P3

Y

S48S47

S9 S10

U

S7

S45

S8 S47

P1

Titik X

Σ kx = 0 Σ ky = 0

P – S9 = 0 S8 = 0 T

S9 = P

S9 = 0,3437 T (tekan)

Titik Z

Σ kx = 0 Σ ky = 0

P – S10 = 0 S11 = 0

S10 = P3

S10 = 0,0302 T (tarik)

Titik Y

Σ kx = 0

S10 + S9 – S47 cos α – S48 cos α = 0

S47 cos α + S48 cos α = S9 + S10

S47 cos α + S48 cos α = 0,3437 + 0,0302

S47 + S48 = αcos

3739,0

S47 = 0,2740 T (tarik)

S48 = 0,2740 T (tekan)

Titik U

Σ kx = 0 Σ ky = 0

P + S47 cos α – S45 = 0 S8 + S47 sin α – S7 = 0

S45 = 0,0403 + 0,274 cos α S7 = 0,274 sin α

S45 = 0,2273 T (tekan) S7 = 0,20 T (tarik)


V

S44S43

S45 S46

R

S6

S41

S7 S43

P1

TS42

S44

P3S13

S12

WS46

S48

P3S12

S11

Titik W

Σ kx = 0 Σ ky = 0

P + S48 cos α – S46 = 0 S12 + S48 sin α – S11 = 0

S46 = 0,0302 + 0,274 cos α S12 = 0,274 sin α

S46 = 0,2172 T (tarik) S12 = 0,20 T (tekan) Titik V

Σ kx = 0

S45 + S46 – S43 cos α – S44 cos α = 0

S43 cos α + S44 cos α = S45 + S46

S43 cos α + S44 cos α = 0,2273 + 0,2172

S43 + S44 = αcos

445,0

S43 = 0,3257 T (tarik)

S44 = 0,3257 T (tekan) Titik R

Σ kx = 0 Σ ky = 0

P + S43 cos α – S41 = 0 S7 + S43 sin α – S6 = 0

S41 = 0,0403 + 0,3257 cos α S6 = 0,2 + 0,3257 sin α

S41 = 0,22625 T (tekan) S6 = 0,4381 T (tarik) Titik T

Σ kx = 0 Σ ky = 0

P + S44 cos α – S42 = 0 S13 + S44 sin α – S12 = 0

S42 = 0,0302 + 0,3257 cos α S12 = 0,20 + 0,3257 sin α

S42 = 0,2524 T (tarik) S12 = 0,4381 T (tekan)


S

S40S39

S41 S42

O

S5

S37

S6 S39

P1

QS38

S40

P3S14

S13

P

S36S35

S37 S38

Titik S

Σ kx = 0

S41 + S42 – S39 cos α – S40 cos α = 0

S39 cos α + S40 cos α = 0,2625 + 0,2524

S39 + S40 = αcos

5149,0

S39 = 0,3773 T (tarik)

S40 = 0,3773 T (tekan) Titik O

Σ kx = 0 Σ ky = 0

P + S39 cos α – S37 = 0 S6 + S39 sin α – S5 = 0

S37 = 0,0403 + 0,3773 cos α S5 = 0,4381 + 0,3773 sinα

S37 = 0,2977 T (tekan) S5 = 0,7139 T (tarik) Titik Q

Σ kx = 0 Σ ky = 0

P + S40 cos α – S38 = 0 S13 + S40 sin α – S14 = 0

S38 = 0,0302 + 0,3773 cos α S14= 0,4381 + 0,3773 sinα

S38 = 0,2876 T (tarik) S14 = 0,7139 T (tekan) Titik P

Σ kx = 0

S37 + S38 – S35 cos α – S36 cos α = 0

S35 cos α + S36 cos α = 0,2977 + 0,2876

S35 + S36 = αcos

5853,0

S35 = 0,429 T (tarik)

S36 = 0,429 T (tekan)


L

S4

S33

S5 S35

P1

NS34

S36

P3S15

S14

M

S32S31

S33 S34

I

S3

S29

S4 S31

P1

KS30

S32

P3S16

S15

Titik L

Σ kx = 0 Σ ky = 0

P + S35 cos α – S33 = 0 S5 + S35 sin α – S4 = 0

S33 = 0,0403 + 0,429 cos α S4 = 0,7139 + 0,429 sinα

S33 = 0,333 T (tekan) S4 = 1,0275 T (tarik) Titik N

Σ kx = 0 Σ ky = 0

P + S36 cos α – S34 = 0 S14 + S36 sin α – S15 = 0

S34 = 0,0302 + 0,429 cos α S15 = 0,7139 + 0,429 sinα

S34 = 0,323 T (tarik) S15 = 1,0275 T (tekan) Titik M

Σ kx = 0

S33 + S34 – S31 cos α – S32 cos α = 0

S31 cos α + S32 cos α = 0,333 + 0,323

S31 + S32 = αcos

656,0

S31 = 0,4807 T (tarik)

S32 = 0,4807 T (tekan) Titik I

Σ kx = 0 Σ ky = 0

P + S31 cos α – S29 = 0 S4 + S31 sin α – S3 = 0

S29 = 0,0403 + 0,4807 cos α S3 = 1,0275 + 0,4807 sinα

S29 = 0,3683 T (tekan) S3 = 1,3789 T (tarik) Titik K Σ kx = 0 Σ ky = 0

P + S32 cos α – S30 = 0 S15 + S36 sin α – S16 = 0

S30 = 0,0302 + 0,4807 cos α S16 =1,0275 + 0,4807 sinα

S30 = 0,3582 T (tarik) S16 = 1,3789 T (tekan)


J

S28S27

S29 S30

F

S2

S25

S3 S27

P1

HS26

S28

P3S17

S16

G

S24S23

S25 S26

Titik J

Σ kx = 0

S29 + S30 – S27 cos α – S28 cos α = 0

S27 cos α + S28 cos α = 0,3683 + 0,33582

S27 + S28 = αcos

7265,0

S27 = 0,5324 T (tarik)

S28 = 0,5324 T (tekan) Titik F

Σ kx = 0 Σ ky = 0

P + S27 cos α – S25 = 0 S3 + S27 sin α – S2 = 0

S25 = 0,0403 + 0,5324 cos α S2= 1,3789 + 0,5324 sinα

S25 = 0,4036 T (tekan) S2 = 1,7681 T (tarik) Titik H

Σ kx = 0 Σ ky = 0

P + S28 cos α – S26 = 0 S16 + S28 sin α – S17 = 0

S26 = 0,0302 + 0,5234 cos α S17 =1,3789 + 0,5234 sinα

S26 = 0,3935 T (tarik) S17 = 1,7681 T (tekan) Titik G

Σ kx = 0

S25 + S26 – S23 cos α – S24 cos α = 0

S23 cos α + S24 cos α = 0,4036 + 0,3935

S23 + S24 = αcos

7971,0

S23 = 0,5841 T (tarik)

S24 = 0,5841 T (tekan)


C

S1

S21

S2 S23

P1

ES22

S24

P3S18

S17

D

S20S19

S21 S22

Titik C

Σ kx = 0 Σ ky = 0

P + S23 cos α – S21 = 0 S2 + S27 sin α – S1 = 0

S21 = 0,0403 + 0,5841 cos α S1 = 1,781 + 0,5841 sinα

S21 = 0,4388 T (tekan) S1 = 2,1951 T (tarik) Titik E

Σ kx = 0 Σ ky = 0

P + S24 cos α – S22 = 0 S17 + S28 sin α – S18 = 0

S22 = 0,0302 + 0,5841 cos α S18 =1,781 + 0,5841 sinα

S22 = 0,4287 T (tarik) S18 = 2,1951 T (tekan) Titik D

Σ kx = 0

S21 + S22 – S19 cos α – S20 cos α = 0

S19 cos α + S20 cos α = 0,4388 + 0,4287

S19 + S20 = αcos

8675,0

S19 = 0,6357 T (tarik)

S20 = 0,6357 T (tekan)


Tabel Gaya Batang Akibat Beban Angin Kiri Untuk Profil 2L

pada Jumlah Medan (n) = 8

No Batang

Berat sendiri (ton) Tekan Tarik

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

- - - - - - - -

0,3437 - -

0,200 0,4381 0,7139 1,0275 1,3789 1,7681 2,1951

- 0,6357 0,4388

- -

0,5841 0,4036

- -

0,5324 0,3683

- -

0,4807 0,333

- -

0,429 0,2977

- -

0,3773 0,2625

- -

0,3257 0,2273

- -

0,274

2,1951 1,7681 1,3789 1,0275 0,7139 0,4381 0,20

- - -

0,04725 - - - - - - -

0,6357 - -

0,4287 0,5841

- -

0,3935 0,5324

- -

0,3582 0,4807

- -

0,323 0,429

- -

0,2876 0,3773

- -

0,2524 0,3257

- -

0,2172 0,274

-


Tabel Gaya Batang Akibat Beban Angin Kiri Untuk Profil 2C


No Batang


1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

- - - - - - - -

0,3354 - -

0,1926 0,4151 0,6676 0,9502 1,2627 1,6052 1,9778

- 0,5506 0,3797

- -

0,5096 0,3517

- -

0,4685 0,3237

- -

0,4275 0,2957

- -

0,3865 0,2677

- -

0,3454 0,2397

- -

0,3044 0,2177

- -

0,2634

1,9778 1,6052 1,2627 0,9502 0,6676 0,4151 0,1926

- -

0,024 - - - - - - - -

0,5506 - -

0,3717 0,5096

- -

0,3437 0,4685

- -

0,3157 0,4275

- -

0,2872 0,3865

- -

0,2597 0,3454

- -

0,2317 0,3044

- -

0,2037 0,2634

-


Tabel Gaya Batang Akibat Beban Angin Kiri Profil IWF


No Batang


1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

- - - - - - - -

0,3531 - -

0,2092 0,4649 0,7673 1,1162 1,5118 1,9539 2,4426

- 0,7322 0,5058

- -

0,6685 0,4623

- -

0,6048 0,4189

- -

0,5411 0,3754

- -

0,4773 0,3319

- -

0,4136 0,2884

- -

0,3498 0,2449

- -

0,2861

2,4426 1,9539 1,5118 1,1162 0,7673 0,4649 0,2092

- -

0,0373 - - - - - - - -

0,7322 - -

0,4934 0,6685

- -

0,450 0,6048

- -

0,4065 0,5411

- -

0,3630 0,4773

- -

0,3195 0,4136

- -

0,2760 0,3498

- -

0,2325 0,2861

-


P2

P3

P3

P1

P1

P1

P3

P3

P3

P3

P3

P3

P1

P1

P1

P1

P1

P3

P3

P1

P1

P1

P3

P3

P3

P3

P3

P2+P3

P1

P1

P1

P1

P1

2. Berat Angin Kanan


Panjang batang tegak = 1,875 m

Panjang batang horizontal = 1,75 m

Panjang batang diagonal = 2,5648 m

Panjang profil yang dikenai angin

Batang tegak 2 . 55 . 55 . 10 = 12 cm = 0,12 m

Batang horizontal 2 . 30 . 30 . 3 = 7 cm = 0,07 m

Batang diagonal 2 . 50 . 50 5 = 11 cm = 0,11 m


Batang tegak = 0,12 x 1,875 x 8 x 2 = 3,6 m²

Batang horizontal = 0,07 x 1,75 x 8 x 2 = 1,96 m²

Batang diagonal = 0,11 x 2,5648 x 8 x 2 = 4,514 m²

Luas total bidang yang dikenai angin = 10,074 m²


L M N

21 2223 24

19 20A

C

F

I

B

E

H

K

D

G

J

1

2

3

9

13

17

4

5

18

6

7

14

10

8 11

12

15

1625 26

29 30

27 28

O P Q

R S T

U V W

X Y Z

31 32

33 3435 36

37 3839 40

41 4243 44

45 4647 48

P2+P3

P3

P3

P3

P3

P3

P3

P3

P1

P1

P1

P1

P1

P1

P1

P1

P1 = 16


= 16

074,10402,1 xx

= 30,222 kg = 0,03022 T

P2 = 2

6,1 dangginxLuasbixtekananan x 15

25,16

= 2

5,25,3406,1 xxx x 1,083

= 303,334 kg = 0,3034 T

P3 = 16


= 16

074,10406,1 xx

= 40,296 kg = 0,0403 T


X

S8

S9P1

Z

S11

S10P2+P3

Y

S47 S48

S10S9

U S45

S47

P1S7

S8

Titik X

Σ kx = 0 Σ ky = 0

S9 - P = 0 S8 = 0 T

S9 = P

S9 = 0,0302 T (tarik)

Titik Z

Σ kx = 0 Σ ky = 0

S10 – P = 0 S11 = 0

S10 = P

S10 = 0,3437 T (tekan)

Titik Y

Σ kx = 0

S47 cos α + S48 cos α - S9 - S10 = 0

S47 cos α + S48 cos α = S9 + S10

S47 cos α + S48 cos α = 0,0302 + 0,3437

S47 + S48 = αcos

3739,0

S47 = 0,2740 T (tekan)

S48 = 0,2740 T (tarik)

Titik U

Σ kx = 0 Σ ky = 0

S45 - S47 cos α – P = 0 S7 - S8 - S47 sin α = 0

S45 = 0,0302 + 0,274 cos α S7 = 0,274 sin α

S45 = 0,2172 T (tarik) S7 = 0,20 T (tekan)


W

S12

S46

S11S48

P3

V

S43 S44

S46S45

R S41

S43

P1S6

S7

T

S13

S42

S12S44

P3

Titik W

Σ kx = 0 Σ ky = 0

S46 – P - S48 cos α = 0 S11 + S48 sin α – S12 = 0

S46 = 0,0403 + 0,274 cos α S12 = 0,274 sin α

S46 = 0,2273 T (tekan) S12 = 0,20 T (tarik)

Titik V

Σ kx = 0

S43 cos α + S44 cos α - S45 - S46 = 0

S43 cos α + S44 cos α = S45 + S46

S43 cos α + S44 cos α = 0,2172 + 0,2273

S43 + S44 = αcos

445,0

S43 = 0,3257 T (tekan)

S44 = 0,3257 T (tarik)

Titik R

Σ kx = 0 Σ ky = 0

S41 - P - S43 cos α = 0 S6 - S7 - S43 sin α = 0

S41 = 0,0302 + 0,3257 cos α S6 = 0,2 + 0,3257 sin α

S41 = 0,2524 T (tarik) S6 = 0,4381 T (tekan) Titik T

Σ kx = 0 Σ ky = 0

S42 - P - S44 cos α = 0 S12 - S13 - S44 sin α = 0

S42 = 0,0403 + 0,3257 cos α S12 = 0,20 + 0,3257 sin α

S42 = 0,2625 T (tekan) S13 = 0,4381 T (tarik)


S

S39 S40

S42S41

OS37

S39

P1S5

S6

Q

S14

S38

S13S40

P3

P

S35 S36

S38S37

Titik S

Σ kx = 0

S39 cos α + S40 cos α - S41 - S42 = 0

S39 cos α + S40 cos α = 0,2524 + 0,2625

S39 + S40 = αcos

5149,0

S39 = 0,3773 T (tekan)

S40 = 0,3773 T (tarik)

Titik O

Σ kx = 0 Σ ky = 0

S37 - P - S39 cos α = 0 S5 - S6 - S39 sin α = 0

S37 = 0,0302 + 0,3773 cos α S5 = 0,4381 + 0,3773 sinα

S37 = 0,2876 T (tarik) S5 = 0,7139 T (tekan)

Titik Q

Σ kx = 0 Σ ky = 0

S38 - P - S40 cos α = 0 S14 - S13 - S40 sin α = 0

S38 = 0,0403 + 0,3773 cos α S14= 0,4381 + 0,3773 sinα

S38 = 0,2977 T (tekan) S14 = 0,7139 T (tarik)

Titik P

Σ kx = 0

S35 cos α + S36 cos α - S37 + S38 = 0

S35 cos α + S36 cos α = 0,2876 + 0,2977

S35 + S36 = αcos

5853,0

S35 = 0,429 T (tekan)

S36 = 0,429 T (tarik)


NS34

S36

P3S15

S14

LS33

S35

P1S4

S5

M

S31 S32

S34S33

I S29

S31

P1S3

S4

K

S16

S30

S15S32

P3

Titik L

Σ kx = 0 Σ ky = 0

S33 - P - S35 cos α = 0 S4 - S5 - S35 sin α = 0

S33 = 0,0302 + 0,429 cos α S4 = 0,7139 + 0,429 sinα

S33 = 0,323 T (tarik) S4 = 1,0275 T (tekan) Titik N

Σ kx = 0 Σ ky = 0

S34 - P - S36 cos α = 0 S15 - S14 - S36 sin α = 0

S34 = 0,0403 + 0,429 cos α S15 = 0,7139 + 0,429 sinα

S34 = 0,333 T (tekan) S15 = 1,0275 T (tarik) Titik M

Σ kx = 0

S31 cos α + S32 cos α - S33 - S34 = 0

S31 cos α + S32 cos α = 0,323 + 0,333

S31 + S32 = αcos

656,0

S31 = 0,4807 T (tekan)

S32 = 0,4807 T (tarik) Titik I

Σ kx = 0 Σ ky = 0

S29 - P - S31 cos α = 0 S3 - S4 - S31 sin α = 0

S29 = 0,0302 + 0,4807 cos α S3 = 1,0275 + 0,4807 sinα

S29 = 0,3582 T (tarik) S3 = 1,3789 T (tekan) Titik K Σ kx = 0 Σ ky = 0

S30 - P - S32 cos α = 0 S15 + S36 sin α – S16 = 0

S30 = 0,0403 + 0,4807 cos α S16 =1,0275 + 0,4807 sinα

S30 = 0,3683 T (tekan) S16 = 1,3789 T (tarik)


J

S27 S28

S30S29

F S25

S27

P1S2

S3

H

S17

S26

S16S28

P3

G

S23 S24

S26S25

Titik J

Σ kx = 0

S27 cos α + S28 cos α - S29 - S30 = 0

S27 cos α + S28 cos α = 0,3582 + 0,3683

S27 + S28 = αcos

7265,0

S27 = 0,5324 T (tekan)

S28 = 0,5324 T (tarik)

Titik F

Σ kx = 0 Σ ky = 0

S25 - P - S27 cos α = 0 S2 - S3 - S27 sin α = 0

S25 = 0,0302 + 0,5324 cos α S2 = 1,3789 + 0,5324 sinα

S25 = 0,3935 T (tarik) S2 = 1,7681 T (tekan)

Titik H

Σ kx = 0 Σ ky = 0

S26 - P - S28 cos α = 0 S17 - S16 - S28 sin α = 0

S26 = 0,0403 + 0,5234 cos α S17 =1,3789 + 0,5234 sinα

S26 = 0,4036 T (tekan) S17 = 1,7681 T (tarik)

Titik G

Σ kx = 0

S23 cos α + S24 cos α - S25 - S26 = 0

S23 cos α + S24 cos α = 0,3935 + 0,4036

S23 + S24 = αcos

7971,0

S23 = 0,5841 T (tekan)

S24 = 0,5841 T (tarik)


C S21

S23

P1S1

S2

E

S18

S22

S17S24

P3

D

S19 S20

S22S21

Titik C

Σ kx = 0 Σ ky = 0

S21 - P - S23 cos α = 0 S1 – S2 - S27 sin α = 0

S21 = 0,0302 + 0,5841 cos α S1 = 1,781 + 0,5841 sinα

S21 = 0,4287 T (tarik) S1 = 2,1951 T (tekan)

Titik E

Σ kx = 0 Σ ky = 0

S22 - P - S24 cos α = 0 S18 - S17 - S28 sin α = 0

S22 = 0,0403 + 0,5841 cos α S18 =1,781 + 0,5841 sinα

S22 = 0,4388 T (tekan) S18 = 2,1951 T (tarik)

Titik D

Σ kx = 0

S19 cos α + S20 cos α - S21 - S22 –= 0

S19 cos α + S20 cos α = 0,4287 + 0,4388

S19 + S20 = αcos

8675,0

S19 = 0,6357 T (tekan)

S20 = 0,6357 T (tarik)


Tabel Gaya Batang Akibat Beban Angin Kanan Untuk Profil 2L


No Batang


1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

2,1951 1,7681 1,3789 1,0275 0,7139 0,4381 0,20

- - -

0,3437 - - - - - - -

0,6357 - -

0,4388 0,5841

- -

0,4036 0,5324

- -

0,3683 0,4807

- -

0,333 0,429

- -

0,2977 0,3773

- -

0,2625 0,3257

- -

0,2273 0,274

-

- - - - - - - -

0,0302 - -

0,20 0,4381 0,7139 1,0275 1,3789 1,7681 2,1951

- 0,6357 0,4287

- -

0,5841 0,3935

- -

0,5324 0,3582

- -

0,4807 0,323

- -

0,429 0,2876

- -

0,3773 0,2524

- -

0,3257 0,2172

- -

0,274


Tabel Gaya Batang Akibat Beban Angin Kanan Untuk Profil 2C pada Jumlah Medan (n) = 8

No


Tekan Tarik 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

1,9778 1,6052 1,2627 0,9502 0,6676 0,4151 0,1926

- -

0,3354 - - - - - - - -

0,5506 - -

0,3797 0,5096

- -

0,3517 0,4685

- -

0,3237 0,4275

- -

0,2957 0,3865

- -

0,2677 0,3454

- -

0,2397 0,3044

- -

0,2117 0,2634

-

- - - - - - - -

0,024 - -

0,1926 0,4151 0,6676 0,9502 1,2627 1,6052 1,9778

- 0,5506 0,3717

- -

0,5096 0,3437

- -

0,4685 0,3157

- -

0,4275 0,2872

- -

0,3865 0,2597

- -

0,3454 0,2317

- -

0,3044 0,2037

- -

0,2634


Tabel Gaya Batang Akibat Beban Angin Kanan Untuk Profil IWF pada Jumlah Medan (n) = 8

No


Tekan Tarik 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

2,4426 1,9539 1,5118 1,1162 0,7673 0,4649 0,2092

- -

0,3531 - - - - - - - -

0,7322 - -

0,5058 0,6685

- -

0,4023 0,6048

- -

0,4189 0,5411

- -

0,3754 0,4773

- -

0,3319 0,4136

- -

0,2884 0,3498

- -

0,2449 0,2861

-

- - - - - - - -

0,0373 - -

0,2092 0,4649 0,7673 1,1162 1,5118 1,9539 2,4426

- 0,7322 0,4934

- -

0,6685 0,450

- -

0,6048 0,4065

- -

0,5411 0,3630

- -

0,4773 0,3195

- -

0,4136 0,2760

- -

0,3498 0,2325

- -

0,2861


Tabel Kombinasi Pembebanan pada Jumlah Medan (n) = 8 Untuk Profil 2L

No Batang

Beban Akibat Angin Kiri (ton)

Beban Akibat Angin Kanan (ton)

Berat sendiri (ton)

Beban Tetap (ton)

tekan tarik tekan tarik tekan tarik tekan tarik 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

- - - - - - - -

0,3437 - -

0,200 0,4381 0,7139 1,0275 1,3789 1,7681 2,1951

- 0,6357 0,4388

- -

0,5841 0,4036

- -

0,5324 0,3683

- -

0,4807 0,333

- -

0,429 0,2977

- -

0,3773 0,2625

- -

0,3257 0,2273

- -

0,274

2,1951 1,7681 1,3789 1,0275 0,7139 0,4381 0,20

- -

0,0302 - - - - - - - -

0,6357 - -

0,4287 0,5841

- -

0,3935 0,5324

- -

0,3582 0,4807

- -

0,323 0,429

- -

0,2876 0,3773

- -

0,2524 0,3257

- -

0,2172 0,274

-

2,1951 1,7681 1,3789 1,0275 0,7139 0,4381

0,20 - -

0,3437 - - - - - - - -

0,6357 - -

0,4388 0,5841

- -

0,4036 0,5324

- -

0,3683 0,4807

- -

0,333 0,429

- -

0,2977 0,3773

- -

0,2625 0,3257

- -

0,2273 0,274

-

- - - - - - - -

0,0302 - -

0,20 0,4381 0,7139 1,0275 1,3789 1,7681 2,1951

- 0,6357 0,4287

- -

0,5841 0,3935

- -

0,5324 0,3582

- -

0,4807 0,323

- -

0,429 0,2876

- -

0,3773 0,2524

- -

0,3257 0,2172

- -

0,274

7,7154 7, 7154 7, 7154 7, 7154 7, 7154 7, 7154 7, 7154 7,7154

- -

7, 7154 7, 7154 7, 7154 7, 7154 7, 7154 7, 7154 7, 7154 7, 7154

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

9,9105 9,4835 9,0943 8,7429 8,4293 8,1535 7,9154 7,7154 0,3437 0, 3437 7, 7154 7,9154 8,1535 8,4293 8,7429 9,0943 9,4835 9,9105 0,6357 0,6357 0,4388 0,4388 0,5841 0,5841 0,4036 0,4036 0,5324 0,5324 0,3683 0,3683 0,4807 0,4807 0,333 0,333 0,429 0,429

0,2977 0,2977 0,3773 0,3773 0,2625 0,2625 0,3257 0,3257 0,2273 0,2273 0,274 0,274

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -


Tabel Kombinasi Pembebanan pada Jumlah Medan (n) = 8 Untuk Profil 2C

No Batang


Beban Akibat Angin Kanan (ton)

Berat sendiri (ton)

Beban Tetap (ton)


- - - - - - - -

0,3354 - -

0,1926 0,4151 0,6676 0,9502 1,2627 1,6052 1,9778

- 0,5506 0,3797

- -

0,5096 0,3517

- -

0,4685 0,3237

- -

0,4275 0,2957

- -

0,3865 0,2677

- -

0,3454 0,2397

- -

0,3044 0,2177

- -

0,2634

1,9778 1,6052 1,2627 0,9502 0,6676 0,4151 0,1926

- -

0,024 - - - - - - - -

0,5506 - -

0,3717 0,5096

- -

0,3437 0,4685

- -

0,3157 0,4275

- -

0,2872 0,3865

- -

0,2597 0,3454

- -

0,2317 0,3044

- -

0,2037 0,2634

-

1,9778 1,6052 1,2627 0,9502 0,6676 0,4151 0,1926

- -

0,3354 - - - - - - - -

0,5506 - -

0,3797 0,5096

- -

0,3517 0,4685

- -

0,3237 0,4275

- -

0,2957 0,3865

- -

0,2677 0,3454

- -

0,2397 0,3044

- -

0,2117 0,2634

-

- - - - - - - -

0,024 - -

0,1926 0,4151 0,6676 0,9502 1,2627 1,6052 1,9778

- 0,5506 0,3717

- -

0,5096 0,3437

- -

0,4685 0,3157

- -

0,4275 0,2872

- -

0,3865 0,2597

- -

0,3454 0,2317

- -

0,3044 0,2037

- -

0,2634

7,7428 7,7428 7,7428 7,7428 7,7428 7,7428 7,7428 7,7428

- -

7,7428 7,7428 7,7428 7,7428 7,7428 7,7428 7,7428 7,7428

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

9,7206 9,3480 9,0055 8,6930 8,4104 8,1579 7,9354 7,7428 0,3354 0,3354 7, 7428 7,9354 8,1579 8,4104 8,6930 9,0055 9,3480 9,7206 0,5506 0,5506 0,3797 0,3797 0,5096 0,5096 0,3517 0,3517 0,4685 0,4685 0,3237 0,3237 0,4275 0,4275 0,2957 0,2957 0,3865 0,3865 0,2677 0,2677 0,3454 0,3454 0,2397 0,2397 0,3044 0,3044 0,2177 0,2177 0,2634 0,2634

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -


Tabel Kombinasi Pembebanan pada Jumlah Medan (n) = 8 Untuk Profil IWF No

Batang


Beban Akibat Angin Kanan

Berat sendiri (ton)

Beban Tetap (ton)


- - - - - - - -

0,3531 - -

0,2092 0,4649 0,7673 1,1162 1,5118 1,9539 2,4426

- 0,7322 0,5058

- -

0,6685 0,4623

- -

0,6048 0,4189

- -

0,5411 0,3754

- -

0,4773 0,3319

- -

0,4136 0,2884

- -

0,3498 0,2449

- -

0,2861

2,4426 1,9539 1,5118 1,1162 0,7673 0,4649 0,2092

- -

0,0373 - - - - - - - -

0,7322 - -

0,4934 0,6685

- -

0,450 0,6048

- -

0,4065 0,5411

- -

0,3630 0,4773

- -

0,3195 0,4136

- -

0,2760 0,3498

- -

0,2325 0,2861

-

2,4426 1,9539 1,5118 1,1162 0,7673 0,4649 0,2092

- -

0,3531 - - - - - - - -

0,7322 - -

0,5058 0,6685

- -

0,4023 0,6048

- -

0,4189 0,5411

- -

0,3754 0,4773

- -

0,3319 0,4136

- -

0,2884 0,3498

- -

0,2449 0,2861

-

- - - - - - - -

0,0373 - -

0,2092 0,4649 0,7673 1,1162 1,5118 1,9539 2,4426

- 0,7322 0,4934

- -

0,6685 0,450

- -

0,6048 0,4065

- -

0,5411 0,3630

- -

0,4773 0,3195

- -

0,4136 0,2760

- -

0,3498 0,2325

- -

0,2861

7,8058 7,8058 7,8058 7,8058 7,8058 7,8058 7,8058 7,8058

- -

7,8058 7,8058 7,8058 7,8058 7,8058 7,8058 7,8058 7,8058

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

10,2484 9,7597 9,3176 8,9220 8,5731 8,2707 8,0150 7,8058 0,3531 0,3531 7,8058 8,0150 8,2707 8,5731 8,9220 9,3176 9,7597

10,2484 0,7322 0,7322 0,5058 0,5058 0,6685 0,6685 0,4623 0,4623 0,6048 0,6048 0,4189 0,4189 0,5411 0,5411 0,3754 0,3754 0,4773 0,4773 0,3319 0,3319 0,4136 0,4136 0,2884 0,2884 0,3498 0,3498 0,2449 0,2449 0,2861 0,2861

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -


2 55 55 10

Pendimensian Profil 2L Untuk Jumlah Medan (n) = 8

d. Batang Vertikal

Dari batang vertikal/batang tegak yang ada (batang

1,2,3,4,5,6,7,8,11,12,13,14,15,16,17,18) diambil yang mempunyai gaya

batang paling besar. Diantara batang-batang tersebut batang 1 dan batang

18 yang mempunyai gaya batang paling besar.

Beban Tetap Nt = 9,9105 x 10³ kg (tekan)

Lk = 187,5 cm

Syarat Batas

Taksir ω = 1,5

Fperlu ≥ 1600

5,1 10³ x 9,9105 x

Fperlu ≥ 9,291 cm²

Dicoba profil 2 L 55 55 10

F = 10,1 cm²

IX = IY = 26,3 cm4

Ix = iy = 1,62 cm

In = 11,3 cm4

ex = 1,72 cm

imin = 1,06 cm


Kontrol tegangan tekan

Melentur ke sumbu bahan (X – X)

λx = ixlk =

62,15,187 = 115,741 ωx = 2,5853

σx = FtotalNtx xω

= 2,20

5853,2 10³ x 9,9105 x

= 1268,400 kg/cm² < 1600 kg/cm² ……..OK

Melentur ke sumbu masing-masing

λi = minnxi

lk ≤ 50

n = 06,1505,187

x = 3,53 ( ambil n = 5 )

λi = 06,155,187

x = 35,38

Melentur ke sumbu idiil

Iyt = 2 ( Iyo + F x a² )

a = ex + ½ δ dimana δ = 10 mm

= 1,72 + 0.5

= 2,22 cm

Iyt = 2 ( 26,3 + 10,1 x 2,222)

= 152,153 cm4

iyt = xFIyt

2 =

2,20153,152 = 2,74 cm

λyt = iytlk =

74,25,187 = 68,43


t

2a

a a

c

3

3

pelat kopel

λiy = 22 i22

yt λλ + = 22 38,3522

43,68 +

= 77,035 ωiy = 1,568

σiy = Ftotal

iyNtxω =

48,3568,1 10³9,9105x x

= 769,30 kg/cm² < 1600 kg/cm² ……..OK

Pelat koppel

g) Stabilitas plat

a

Ipl2

≥ 10 nlk

In

lk/n = 55,187 = 37,5

2a = 2 x 2,22 = 4,44

44,4

312

1 th ≥ 10 5,373,11

h3 t ≥ 160,5504

taksir t = 5 mm = 0,5 cm

h ≥ 6,850 cm ambil h = 12 cm

h) Perhitungan Plat membujur

D = 0,02 x Nt

= 0,02 x 9,9105 x 10³

= 198,21 kg


3

3

2

2

h

sx = F x a

= 10,1 x 2,22

= 22,422 cm3

L = Iyt

nlkxDxS X = 1537,152

5,37422,22 198,21 xx

= 1095,343 kg

Keseimbangan momen

L x 2a = N x C

N = C

aLx2 = 6

44,4 095,3431 x

N = 410,55 kg

i) Kontrol Pelat

Fn = 12 (0,5) – 2 (2) 0,5

= 4 cm2

In = 1/12 h3 t – 2 d (1/2 C)2 t

= 1/12 113 x 0,5 – 2 x 2 (1/2 6)2 x 0,5

= 54 cm4

Wn = h

In2

1 =

12542

1 = 9 cm3

σm = WnNxC2

1 =

96 410,552

1 xx = 136,85 kg/cm2

τd = FnL =

4 1095,343 = 273,83 kg/cm2

τi = 22 dm τσ + = 22 83,273 136,85 +

= 306 kg/cm2 < 1600 kg/cm2 . . . OK

C


2 30 30 3

e. Batang horizontal

Dari batang horizontal yang ada (batang

9,10,21,22,25,26,29,30,33,34,37,38,41,42,45,46) diambil yang mempunyai

gaya batang paling besar. Diantara batang-batang tersebut batang 21 dan

batang 22 yang mempunyai gaya batang paling besar.


Lk = 175 cm

Syarat Batas

Taksir ω = 2,0

Fperlu ≥ 1600

0,2104388,0 3 xx



F = 1,74 cm²

IX = IY = 1,41 cm4

Ix = iy = 0,90 cm

In = 0,57 cm4

ex = 0,84 cm

imin = 0,57 cm




λx = ixlk =

90,0175 = 194,44 ωx = 7,297

σx = FtotalNtx xω

= 48,3

297,7104388,0 3 xx

= 920,092 kg/cm² < 1600 kg/cm² ……..OK


λi = minnxi

lk ≤ 50

n = 57,050

175x

= 6,14 ( ambil n = 7 )

λi = 57,07

175x

= 43,86


Iyt = 2 ( Iyo + F x a² )


= 0,84 + 0.5

= 1,34 cm

Iyt = 2 ( 1,41 + 1,74 x 1,342)

= 9,068 cm4

iyt = xFIyt

2 =

74,12068,9

x = 1,614 cm

λyt = iytlk =

614,1175 = 108,426


t

2a

a a

c

3

3

pelat kopel

λiy = 22 i22

yt λλ + = 22 86,4322

426,108 +

= 116,961 ωiy = 2,640

σiy = Ftotal

iyNtxω =

48,364,2104388,0 3 xx

= 332,883 kg/cm² < 1600 kg/cm² ……..OK

Pelat koppel

a) Stabilitas plat

a

Ipl2

≥ 10 nlk

In

lk/n = 7175 = 25

2a = 2 x 1,34 = 2,68

68,2

312

1 th ≥ 10 2557,0

h3 t ≥ 7,332


h ≥ 2,447 cm ambil h = 7 cm

b) Perhitungan Plat membujur

D = 0,02 x Nt

= 0,02 x 0,4388 x 10³

= 8,776 kg


3

3

2

2

h

sx = F x a

= 1,74 x 1,34

= 2,3316 cm3

L = Iyt

nlkxDxS X = 068,9

253316,2 8,776 xx

= 56,413 kg

Keseimbangan momen

L x 2a = N x C

N = C

aLx2 = 2

68,2 56,413 x

N = 75,593 kg

c) Kontrol Pelat

Fn = 7 (0,5) – (2) 0,5

= 2,5 cm2

In = 1/12 h3 t – d (1/2 C)2 t

= 1/12 73 x 0,5 – 2 (1/2 2)2 x 0,5

= 13,29 cm4

Wn = h

In2

1 =

729,13

21

= 3,797 cm3

σm = WnNxC2

1 =

797,32593,752

1 xx = 19,909 kg/cm2

τd = FnL =

5,2413,56 = 22,565 kg/cm2

τi = 22 dm τσ + = 22 565,22909,19 +

= 30,092 kg/cm2 < 1600 kg/cm2 . . . OK

C


2 50 50 5

f. Batang Diagonal

Dari batang diagonal yang ada (batang

19,20,23,24,27,28,31,32,35,36,39,40,43,44,47,48) diambil yang

mempunyai gaya batang paling besar. Diantara batang-batang tersebut

batang 19 dan batang 20 yang mempunyai gaya batang paling besar.


Lk = 256,479 cm

Syarat Batas

Taksir ω = 2,0

Fperlu ≥ 1600

0,2106357,0 3 xx



F = 4,80 cm²

IX = IY = 11 cm4

Ix = iy = 1,51 cm

In = 4,59 cm4

ex = 1,4 cm

imin = 0,98 cm




λx = ixlk =

51,1479,256 = 169,854 ωx = 5,568

σx = FtotalNtx xω =

6,9568,5106357,0 3 xx

= 368,706 kg/cm² < 1600 kg/cm² ……..OK


λi = minnxi

lk ≤ 50

n = 98,050

479,256x

= 5,23 ( ambil n = 7 )

λi = 98,07

479,256x

= 37,39


Iyt = 2 ( Iyo + F x a² )


= 1,40 + 0.5

= 1,90 cm

Iyt = 2 ( 11,0 + 4,80 x 1,92)

= 56,656 cm4

iyt = xFIyt

2 =

8,42656,56

x = 2,429 cm

λyt = iytlk =

429,2479,256 = 105,590


t

2a

a a

c

3

3

pelat kopel

λiy = 22 i22

yt λλ + = 22 39,3722

59,105 +

= 112,015 ωiy = 2,422

σiy = Ftotal

iyNtxω =

6,9422,2106357,0 3 xx

= 160,382 kg/cm² < 1600 kg/cm² ……..OK

Pelat koppel

a) Stabilitas plat

a

Ipl2

≥ 10 nlk

In

lk/n = 7479,256 = 36,64

2a = 2 x 1,9 = 3,8

8,3

312

1 th ≥ 10 64,3659,4

h3 t ≥ 57,12


h ≥ 5,85 cm ambil h = 11 cm

b) Perhitungan Plat membujur

D = 0,02 x Nt

= 0,02 x 0,6357 x 103

= 12, 714 kg


3

3

2

2

h

sx = F x a

= 4,8 x 1,9

= 9,12 cm3

L = Iyt

nlkxDxS X = 656,56

64,3612,9714,12 xx

= 74,987 kg

Keseimbangan momen

L x 2a = N x C

N = C

aLx2 = 5

80,3987,74 x

N = 56,99 kg

c) Kontrol Pelat

Fn = 11 (0,5) – 2 (2) 0,5

= 3,5 cm2

In = 1/12 h3 t – 2 d (1/2 C)2 t

= 1/12 113 x 0,5 – 2 x 2 (1/2 5)2 x 0,5

= 42,958 cm4

Wn = h

In2

1 =

11958,42

21

= 7,81 cm3

σm = WnNxC2

1 =

81,7599,562

1 xx = 18,243 kg/cm2

τd = FnL =

5,3987,74 = 21,425 kg/cm2

τi = 22 dm τσ + = 22 425,21243,18 +

= 28,14 kg/cm2 < 1600 kg/cm2 . . . OK

C


Tabel Kombinasi Pembebanan dan Berat Total Profil 2L dengan Jumlah Medan ( n ) = 8

Batang Lk Berat Beban Angin Beban Bentuk F q Ix ix Iy iy

n λ ω σ Bobot

Sendiri Kiri Kanan Tetap Propil ( cm² )

( kg/cm )

( cm 4

) ( cm

) ( cm 4

) ( cm

) ( kg/cm²

) ( Kg )

1,2,3,4,5

6,7,8,11,12 187.5 -7715 2195.10 -2195.10

-9910.50 20.2 0.15857 26.30 1.62 152.15 2.740 5 115.74 2.585 1268.40 475.710

13,14,15 16,17,18

9,10,21,22 25,26,29,30 175 -438.80 438.80 -438.80 3.48 0.02732 1.41 0.90 9.068 1.614 7 194.440 7.297 920.092 76.490 33,34,37,38 41,42,45,46

19,20,23,24 27,28,31,32 256.48 635.70 -635.70 -635.70 9.6 0.07536 11.00 1.51 56.656 2.429 7 169.85 5.568 368.706 309.252 35,36,39,40 43,44,47,48

Total 861.453

2 30 30 3

2 55 55 10

2 50 50 5


Tabel Kombinasi Pembebanan dan Berat Total Profil 2C dengan Jumlah Medan ( n ) = 8


n λ ω σ Bobot


( kg/cm )

( cm 4

) ( cm

) ( cm 4

) ( cm

) ( kg/cm²

) ( Kg )

1,2,3,4,5

6,7,8,11,12 187.5 -7743 1977.80 -1977.80

-9720.60 14.2 0.11147 26.40 1.92 67.895 2.186 5 97.66 1.975 1352.26 334.410

13,14,15 16,17,18

9,10,21,22 25,26,29,30 175 -379.70 379.70 -379.70 10.8 0.08478 6.39 1.08 46.042 2.065 5 162.037 5.067 178.143 237.384 33,34,37,38 41,42,45,46

19,20,23,24 27,28,31,32 256.48 550.60 -550.60 -550.60 12.4 0.09734 14.10 1.50 54.886 2.104 5 170.986 5.642 250.527 399.451 35,36,39,40 43,44,47,48

Total 971.245

2 30 33 5

2 40 35 5

2 50 38 5


Tabel Kombinasi Pembebanan dan Berat Total Profil IWF dengan Jumlah Medan ( n ) = 8


n λ ω σ Bobot


( kg/cm )

( cm 4

) ( cm

) ( cm 4

) ( cm

) ( kg/cm²

) ( Kg )

1,2,3,4,5

6,7,8,11,12 187.5 -7806 2442.60 -2442.60

-10248.40 17.85 0.14012 666.00 6.11 49.50 1.660 112.95 2.462 1413.53 420.368

13,14,15 16,17,18

9,10,21,22 25,26,29,30 175 -505.80 505.80 -505.80 11.85 0.09302 187.00 3.98 14.800 1.120 156.250 4.712 201.124 260.463 33,34,37,38 41,42,45,46

19,20,23,24 27,28,31,32 256.48 732.20 -732.20 -732.20 16.84 0.13219 413.00 4.95 29.2 1.32 194.3 7.287 316.837 542.480 35,36,39,40 43,44,47,48

Total 1223.310

100 50

150 75

125 60


1615

1211

810

14 7

618

54

17

13

9

3

2

1

J

G

D

K

H

E

B

I

F

C

A

L M N

21 22

23 24

19 20

A

C

F

I

B

E

H

K

D

G

J

1

2

3

9

13

17

4

5

18

6

7

14

10

8

11 12

15 16

QPO

2827

3029

2625

1615

12

11

8

10

14

76

18

5

4

17

13

9

3

2

1

J

G

D

K

H

E

B

I

F

C

A

2019

2423

2221

NML

Gambar Penomoran Rangka Batang Gambar Penomoran Rangka Batang Gambar Penomoran Rangka Batang

Menara Air dengan Jumlah Medan (n) = 3 Menara Air dengan Jumlah Medan (n) = 4 Menara Air dengan Jumlah Medan (n) = 5


L M N

27 2829 30

25 26

A

C

F

I

B

E

H

K

G

J

1

2

3

9

1923

4

5

24

6

7

20

10

8

11

21 22

31 32

35 36

33 34

O P Q

D

R S T

U V W

X Y Z12

13

14

15

16

17

18

39 40

37 38

43 44

41 42

47 48

45 46

A' B' C'

51 52

49 50

53 54

L M N

27 2829 30

25 26

A

C

F

I

B

E

H

K

G

J

1

2

3

9

19

23

4

5

24

6

7

20

10

8

11

21 22

31 32

35 36

33 34

O P Q

D

R S T

U V W

12

13

14

15

161718

39 40

37 38

41 42

L M N

27 28

29 30

25 26

A

C

F

I

B

E

H

K

G

J

1

2

3

9

19

23

4

5

24

6

7

20

10

8 11

21 22

31 32

35 3633 34

O P Q

D

R S T

U V W

X Y Z

12

13

14

15

16

17

18

39 4037 38

43 4441 42

47 4845 46

Gambar Penomoran Rangka Batang Gambar Penomoran Rangka Batang Gambar Penomoran RangkaBatang Menara Air dengan Jumlah Medan (n) = 7 Menara Air dengan Jumlah Medan (n) = 8 Menara Air dengan Jumlah Medan (n) = 9


hitung menara air

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