struktur kayu eka wijaya i i dimensi batang

8/9/2019 Struktur Kayu EKA WIJAYA I I Dimensi Batang

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Eka wijaya03071001061

Civil Engineering of Sriwijaya University

PERHITUNGAN GAYA BATANG

Akibat Beban BergerakP

= 8640 kg q = 1750 kg/mcontoh perhitungan = P. A1 = P.7 y . A1

q . A1 = q . A pA = . L . M

BatangPanjang ( 7 y ) Beban Bergerak (P)

+ - + -

A1 = A6 - 1,193 - 10307,52

A2 = A5-

1,815-

15681,6A3 = A4 - 2,699 - 23319,36

B1 = B8 1,313 - 11344,32 -

B2 = B7 1,752 - 15137,28 -

B3 = B6 3,677 - 31769,28 -

B4 = B5 2,092 - 18074,88 -

T1 = T7 - 7 - 60480

T2 = T6 - 0,605 - 5227,2

T3 = T5 1,474 - 12735,36 -

T4 6,052 - 52289,28 -

D1 = D8 - 3,967 - 34274,88

D2 = D7 - 0,356 - 3075,84

D3 = D6 - 7,515 - 64929,6

D4 = D5 9,999 - 86391,36 -


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y Akibat beban sendiriP = 1152,49 kg

Batang Luas GP (A) Beban Bergerak (q)

A1 = A6 2,386 4175,5

A2 = A5 3,63 6352,5

A3 = A4 5,398 9446,5

B1 = B8 2,626 4595,5

B2 = B7 3,504 6132

B3 = B6 7,354 12869,5

B4 = B5 4,184 7322

T1 = T7 14 24500

T2 = T6 1,21 2117,5

T3 = T5 2,948 5159

T4 12,104 21182

D1 = D8 7,934 13884,5

D2 = D7 0,712 1246

D3 = D6 15,03 26302,5D4 = D5 19,998 34996,5

BatangPanjang ( 7 y ) qbs .7 y = P.7 y

+ -

+ -

A1 = A6 - 1,193 - 1374,92

A2 = A5 - 1,815 - 2091,77

A3 = A4 - 2,699 - 3110,57

B1 = B8 1,313 - 1513,22 -


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y Akibat beban AnginP = 87,3 kg

B2 = B7 1,752 - 2019,16 -

B3 = B6 3,677 - 4237,71 -

B4 = B5 2,092 - 2411 -

T1 = T7 - 7 - 8067,43

T2 = T6 - 0,605 - 697,26

T3 = T5 1,474 - 1698,77 -

T4 6,052 - 6974,87 -

D1 = D8 - 3,967 - 4571,93

D2 = D7 - 0,356 - 410,286

D3 = D6 - 7,515 - 8660,96

D4 = D5 9,999 - 11523,75 -

BatangPanjang ( 7 y ) qbs .7 y = P.7 y

+ - + -

A1 = A6 - 1,193 - 104,15

A2 = A5 - 1,815 - 158,45

A3 = A4 - 2,699 - 235,62

B1 = B8 1,313 - 114,62 -

B2 = B7 1,752 - 152,95 -

B3 = B6 3,677 - 321,00 -

B4 = B5 2,092 - 182,63 -

T1 = T7 - 7 - 611,1

T2 = T6 - 0,605 - 52,82

T3 = T5 1,474 - 128,68 -

T4 6,052 - 528,34 -

D1 = D8 - 3,967 - 346,32


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REKAPITULASI GAYA BATANG

Batang

Kombinasi (1+2+3) Kombinasi (1+2+4)

+ - + -

A1 = A6 -15857,9

-14587,2

A2 = A5 - 24125,9 - 22192,6

A3 = A4 - 35876,4 - 33001,5

B1 = B8 17453 - 16054,4 -

B2 = B7 23288,4 - 21422,2 -

B3 = B6 48876,5 - 44959,8 -

B4 = B5 27807,9 - 25579,5 -

T1 = T7 - 93047,4 - 85591,1

T2 = T6 - 8041,96 - 7397,52

T3 = T5 19593,1 - 18023 -

T4 80446,2 - 73999,6 -

D1 = D8 - 52731,3 - 48505,7

D2 = D7 - 4732,13 - 4352,92

D3 = D6 - 99893,1 - 91888,2

D4 = D5 132911,6 - 98788 -

D2 = D7 - 0,356 - 31,08

D3 = D6 - 7,515 - 656,06

D4 = D5 9,999 - 872,91 -


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DIMENSI GELAGAR UTAMA

Batang Atas Gaya batang maksimum = 35876,4 kg Ukuran balok ditaksir = 2 x 20/30 Panjang batang = 167,7 cm

Gaya batang maksimum diambil dari kombinasi ( 1+2+3 ) atau ( 1+2+4 )

y Fbr = 2 x 20 x 30 = 1200 cm2y Ix = 2 .12

1. bh3

= 2 .12

1. (20) . (30)3

= 90000 cm4

y ix =F

Ix =)30)(20(2

cm490000 = 8,66 cm

y it = 2 .12

1. b3 h + 2 { b.h (

2

1b+

2

1H) }

= 2 .12

1. (20)3 30 + 2 { 20.30 (

2

120+

2

130) }

= 70000 cm4

y Iq =12

1. (b+b)3. h

=12

1. (20+20)3. 30

= 160000 cm4

y Iy =4

1. (It + 3Iq)

=4

1. (70000 + 3(160000))

= 137500 cm4


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y iy =Fbr

Iy =)30)(20(2

cm4137500 = 10,70 cm

y =mini

L=

70,10

7,167= 15,67 = 16 , w = 1,12

y lltrW = FbwP.

= )30)(20)(2)(8.0(

12,135876,4x

= 41,86 kg/cm2

y Syarat Aman :elltrW lltrW

41,86 kg/cm2 < 50kg/cm2 AMAN!!!

Kontrol terhadap Tegangan tekan sumbu ximin = 0,289 . b

= 0,289 . ( 20 cm )

= 5,78 cm

br

lltkF

wP.!W

mini

Lk!P = 78,5

7,167= 29,01 $ 29

Dari tabel didapat w = 1,24 untuk P = 29 ( PPKI haL.11 )

hb

wPlltr

..2

.

!W

=30.20.2

24,135876,4x= 37,07 kg/cm2

Syarat Aman :

lltrlltr WW e

37,07 kg/cm2 < 50 kg/cm2................................AMAN !!!

Luas penampangbruto


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Batang Tegak Gaya batang maksimum = 93047,4kg Ukuran balok ditaksir = 2 x 30/40 Panjang batang = 400 cm

y Ix =12

1. bh3

=12

1. (30) . (40)3

= 160000 cm4

y ix =F

Ix =)40)(30(

cm4160000 = 11,55 cm

y it =12

1. b3 h + b.h (

2

1b+

2

1h)

=12

1. 303 40 + 30.40 (

2

130+

2

140)

= 132000 cm4

y Iq =12

1. (b+b)3. h

=12

1. (30+30)3. 40

= 720000 cm4

y Iy =4

1. (It + 3Iq)

=4

1. (132000 + 3(720000))

= 573000 cm4

y iy =Fbr

Iy =)40)(30(

cm4573000 = 21,85 cm

y =mini

L=

85,21

400= 18,31 = 19, w = 1,15


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y lltrW = FbwP.

= )40)(30)(2)(8,0(

15,193047,4x

= 45,73 kg/cm2

Syarat Aman :

lltrlltr WW e

45,73 kg/cm2 < 50 kg/cm2................................AMAN !!!

Batang Diagonal Gaya batang maksimum = 132911,6kg Ukuran balok ditaksir = 2 x 40/50 Panjang batang = 394,7 cm Kontrol terhadap Tegangan tekan sumbu x

- Fb = 2 x 0,8 x b x h= 2 x 0,8 x 40 x 50

= 3200 cm2

-b

lltrF

P!W =

3200

132911,6= 41,53 kg/cm2

Syarat Aman :

lltklltk WW e

41,53 kg/cm2 < 50 kg/cm2................................AMAN !!!

Kontrol Tegangan tekan terhadap sumbu yIt = momen lembam teoritis

Iq = momen lembam geser

y It = 2 ( 1/12 x b x h3 ) + ( 2 b h ) ( a + b2 )= 2 ( 1/12 x 40 x (50)3 ) + ( 2 x 40 x 50 ) ( 50 + (40) 2 )

= 833333,33 + 6600000

= 7433333,33 cm4

y Iq = 1/12 x ( 2xb )3 x 50


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= 1/12 x ( 2 x 40 )3 x 50

= 2133333,33 cm4

y Iy = x ( It + 3Iq )= x (7433333,33 + 3 x 2133333,33 )

= 3458333,33 cm4

y iy =2/1

brF

Iy=

2/1

50.40.2

3458333,33

= 29,4

y P =4,29

394,7= 13,43 $ 14

Dari tabel didapat w = 1,1 untukP

= 14 (PP

KI

haL.

10 )

br

lltkF

wP.!W =

50.402

1,1132911,6

x

x= 36,55 kg/cm2

Syarat Aman :

lltklltk WW e

36,55 kg/cm2 < 50 kg/cm2................................AMAN !!!

Batang Bawah Gaya batang maksimum = 48876,5 kg Ukuran balok ditaksir = 2 x 40/50 Panjang batang = 75 cm Kontrol terhadap Tegangan tekan sumbu x

Fb = 0,8 . 2 . b . h

= 0,8 . 2 . 40 . 50 = 3200 cm2

b

lltrF

P!W = 2

3200

48876,5

cm

kg= 15,27 kg/cm2

Syarat Aman :

lltrlltr WW e


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15,27 kg/cm2 < 50 kg/cm2................................AMAN !!!

Kontrol Profil Tunggal P = x 48876,5 kg

= 24438,25 kg

Fb = 0,8 . 40 . 50= 1600 cm2

b

lltrF

P2/1!W = 2

1600

25,24438

c

kg= 15,27 kg/cm2

Syarat Aman :

lltklltk WW e

15,27 kg/cm2 < 50 kg/cm2................................AMAN !!!

struktur kayu eka wijaya i i dimensi batang

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