Tugas Besar Baja ARJ

Tugas Besar Struktur Baja

Hadi Jatmoko 201210340311160

Arif Rachman Julianto 201210340311186

ANALISA DAN PERHITUNGAN

Direncanakan Bangunan Gedung Industri berbahan baja serta gambar kerjanya dengan data-

data sebagai berikut :

1. Bentang Kuda-kuda (L) : 12,5 m

2. Profil Kuda-kuda : WF (wide flange)

3. Jumlah Kuda-kuda (n) : 6

4. Jarak antar Kuda-kuda : 7 m

5. H1 : 5,5 m

6. Sudut α : 20º

7. Beban Angin : 35 kg/cm2

8. Jenis Atap : Zincalum

9. Dinding Samping : Tertutup (Zincalum)

10. Ikatan Angin Dinding : Portal baja

11. Mutu Baja : A36

12. Jenis Sambungan : Baut (A325)

13. Struktur Balok Crane dengan beban bergerak : 15 ton


Hadi Jatmoko 201210340311160


Tinggi Kuda-kuda

tan 20º = tinggi6,25

tinggi = tan 20º x 6,25 = 2,27 m

Dengan Kemiringan

√6,252+2,272 = 6,65 m

Jarak Antar Gording

6,655

= 1,33 m ≈ 1,40 m

Perhitungan Beban Pada Atap

1. Beban Mati

Jarak antar gording = 1,40 m

Berat penutup atap (Zincalum) = 4,10 kg/m

Berat gording diperkirakan = 6 kg/m

Berat atap (1,40 x 4,10) = 5,74 kg/m

Berat gording = 6 kg/m +

qD = 11,74 kg/m

qD = 11,74 kg/m => RD =

12

. qD . L =

12

.(11 , 74 ).(7 ) = 41,09 kg

RDx = (41,09).cos 20° = 38,61 kg

RDy = (41,09).sin 20° = 14,05 kg


Hadi Jatmoko 201210340311160


qD = 11,74 kg/m =>MD =

18

. qD . L2

=

18

.(11 , 74 ). (7 )2

= 71,91 kgm

MDx = (71,91).cos 20° = 67,57 kgm

MDy = (71,91).sin 20° = 24,59 kgm

2. Beban Hidup

Berat pekerja = 100 kg

PL = 100 kg => RL =

12

.PL=

12

.(100) = 50 kg

RLx = (50).cos 20° = 46,98 kg

RLy = (50).sin 20° = 17,10 kg

PL = 100 kg => ML =

14

. PL . L=

14

.(100) .(7 )= 175 kgm

MLx = (175).cos 20° = 164,45 kgm

MLy = (175).sin 20° = 59,85 kgm

3. Beban Angin

Angin tekan, q = {(0,02).α – (0,4)}. (Wangin) . (Jarak antar gording)

= {(0,02).(20) – (0,4)}.(35).(1,4)

= 0 kg/m

q = 0 kg/m => Rw =

12

.q .L=

12

.(0 ).(7 )= 0 kg

Mw =

18

. q . L2

=

18

.(0 ).(7 )2

= 0 kgm

Angin hisap, q = (- 0,4). (Wangin) . (Jarak antar gording)

= (- 0,4).(35).(1,4)

= - 19,6 kg/m

q = - 19,6 kg/m => Rw =

12

.q .L=

12

.(−19 ,6 ) .(7)= - 68,6 kg

Mw =

18

. q . L2

=

18

.(−19 , 6) .(7 )2

= - 120,05 kgm


Hadi Jatmoko 201210340311160


Kombinasi Pembebanan

A. Pembebanan Sementara 1

1. Arah Tegak Lurus Bidang Atap

RUx1 = (1,2).(RDx) + (0,5).(RLx) + (1,3).(Rw)

= (1,2).( 38,61) + (0,5).( 46,98) + (1,3).(0)

= 69,822 kg

RUx2 = (1,2).(RDx) + (0,5).(RLx) + (1,3).(Rw)

= (1,2).( 38,61) + (0,5).(46,98) + (1,3).( - 68,6)

= -19,358 kg

2. Arah Sejajar Bidang Atap

RUy = (1,2).(RDy) + (0,5).(RLy)

= (1,2).( 14,05) + (0,5).( 17,10)

= 25,410 kg

B. Pembebanan Tetap


RUx = (1,2).(RDx) + (0,5).(RLx)

= (1,2).( 38,61) + (0,5).(46,98)

= 69,822 kg


RUy = (1,2).(RDy) + (0,5).(RLy)

= (1,2).(14,05) + (0,5).(17,10)

= 25,410 kg

C. Pembebanan Sementara 2


RUx 1 = (1,2).(RDx) + (1,6).(RLx) + (0,8).(Rw)

= (1,2).( 38,61) + (1,6).(46,98) + (0,8).(0)

= 121,500 kg

RUx 2 = (1,2).(RDx) + (1,6).(RLx) + (0,8).(Rw)

= (1,2).( 38,61) + (1,6).(46,98) + (0,8).( - 68,6)

= 66,620 kg



Hadi Jatmoko 201210340311160


RUy = (1,2).(RDy) + (1,6).(RLy)

= (1,2).(14,05) + (1,6).(17,10)

= 44,220 kg

D. Pembebanan Sementara 3


RUx = (0,9).(RDx) + (1,3).(Rw)

= (0,9).( 38,61) + (1,3).(0)

= 34,749 kg

RUx = (0,9).(RDx) + (1,3).(Rw)

= (0,9).( 38,61) + (1,3).( - 68,6)

= - 54,431 kg


RUy = (0,9).(RDy)

= (0,9).(14,05)

= 12,645 kg

Kombinasi Momen

A. Momen Akibat Beban Sementara 1


MUx1 = (1,2).(MDx) + (0,5).(MLx) + (1,3).(Mw)

= (1,2).( 67,57) + (0,5).( 164,45) + (1,3).(0)

= 163,309 kgm

MUx2 = (1,2).(MDx) + (0,5).(MLx) + (1,3).(Mw)

= (1,2).( 67,57) + (0,5).( 164,45) + (1,3).(- 120,05)

= 7,244 kgm


MUy = (1,2).(MDy) + (0,5).(MLy)

= (1,2).( 24,59) + (0,5).( 59,85)

= 59,433 kgm

B. Momen Akibat Beban Tetap



Hadi Jatmoko 201210340311160


MUx = (1,2).(MDx) + (0,5).(MLx)

= (1,2).( 67,57) + (0,5).( 164,45)

= 163,309 kgm


MUy = (1,2).(MDy) + (0,5).(MLy)

= (1,2).( 24,59) + (0,5).( 59,85)

= 59,433 kgm

C. Momen Akibat Beban Sementara 2


MUx1 = (1,2).(MDx) + (1,6).(MLx) + (0,8).(Mw)

= (1,2).( 67,57) + (1,6).( 164,45) + (0,8).(0)

= 344,204 kgm

MUx2 = (1,2).(MDx) + (1,6).(MLx) + (0,8).(Mw)

= (1,2).( 67,57) + (1,6).( 164,45) + (0,8).(- 120,05)

= 248,164 kgm


MUy = (1,2).(MDy) + (1,6).(MLy)

= (1,2).( 24,59) + (1,6).( 59,85)

= 125,268 kgm

D. Momen Akibat Beban Sementara 3


MUx1 = (0,9).(MDx) + (1,3).(Mw)

= (0,9).( 67,57) + (1,3).( 0)

= 60,813 kgm

MUx2 = (0,9).(MDx) + (1,3).(Mw)

= (0,9).( 67,57) + (1,3).( -120,05)

= - 95,252 kgm


MUy = (0,9).(MDy)

= (0,9).( 24,59)

= 22,131kgm


Hadi Jatmoko 201210340311160



Hadi Jatmoko 201210340311160


Tabel Hasil Perhitungan Kombinasi Momen dan Reaksi

Keterangan Beban TetapBeban Sementara 1 Beban Sementara 2 Beban Sementara 3

Angin Tekan Angin Hisap Angin Tekan Angin Hisap Angin Tekan Angin HisapMomen

Mx 163,309 kgm 163,309 kgm 7,244 kgm 344,204 kgm 248,164 kgm 60,813 kgm (-) 95,252 kgmMy 59,433 kgm 59,433 kgm 125,268 kgm 22,131 kgm

Reaksi Rx 69,822 kg 69,822 kg (-) 19,358 kg 121,50 kg 66,620 kg 34,749 kgm (-) 54,431 kgRy 25,410 kg 25,410 kg 44,220 kg 12,645 kg


Hadi Jatmoko 201210340311160


Perencanaan Gording Atap

Diambil profil C-Channel 150x65x20 dengan tebal 2,50 mm

q = 5,96 kg/m

t = 2,50 mm Cy = 2,12 cm

A = 7,59 cm2 Xo = 5,15 cm

Ix = 267 cm4 J = 1581 cm4

Iy = 44 cm4 Cw = 2148 cm6

Zx = 35,6 cm3

Zy = 10,0 cm3

rx = 5,93 cm

ry = 2,41 cm

Mutu Baja = A36 = 36 x 1ksi = 36 x 6,875 = 247,5 MPa = 2475 kg/cm2

Momen Nominal Penampang C-Channel

𝑍𝑥 = 14

ℎ𝑡 𝑡2 + 𝑎 𝑡 (ℎ𝑡 – 𝑎) + 𝑡(𝑏 − 2𝑡) (ℎ𝑡 − 𝑡)

= 14

. 150 . 2,52 + 20 x 2,5 (150 - 20) + 2,5 (65 – 2 . 2,5) (150 – 2,5)

= 28,9 cm3

𝑍𝑦 = ℎ𝑡 𝑡 (𝑐 – 12

𝑡) + 2𝑎𝑡 (𝑏 − 𝑐 – 12

𝑡) + 𝑡(𝑐 − 𝑡)2 + 𝑡(𝑏 − 𝑡 − 𝑐)2

= 150 . 2,5 (21,2 - 12

2,5) + 2 . 20 . 2,5 (65 – 21,2 - 12

2,5) + 2,5(21,2 – 2,5)2 + 2,5(65

–

2,5 – 21,2) 2

= 16,9 cm3

Mnx = Zx fy = 28,9 x 2475 kg/cm2 = 71527,5 kgcm

Mny = Zyfy = 16,9 x 2475 kg/cm2 = 41827,5 kgcm

Persyaratan Momen Biaxial


Hadi Jatmoko 201210340311160


MuxØ x Mnx

+ MuyØ x Mny < 1,0

344 , 204 x 1000,9 x71527,5

+125 , 268 x1000,9x 41827 , 5 = 0,87 < 1,0

Lendutan Pada Profil C - Channel

∆max =

L240

=

700240

= 2,92 cm

qD = 11,74 kg/m => qDx = (11,74).cos 20° = 11,032 kg/m

qDy = (11,74).sin 20° = 4,015 kg/m

PD = 100 kg => PDx = (100).cos 20° = 93,97 kg

PDy = (100).sin 20° = 34,20 kg

Lendutan terhadap sumbu x :

qx = (1,2).qDx = (1,2).(11,032) = 13,2384 kg/m

Px = (1,2).PDx = (1,2).(93,97) = 112,764 kg

∆x =

5384

.qx .L4

E . I x +

148

.Px . L3

E . I x

=

5384

.(0 , 132384 ).(7004 )(2000000 ).(267 )

+ 148

.(112 ,764 ) .(7003 )(2000000) .(267 )

= 0,775 + 1,509

= 2,284 cm = 22,84 mm

Lendutan terhadap sumbu y :

qy = (1,2).qDy = (1,2).(4,015) = 4,818 kg/m

Py = (1,2).PDy = (1,2).(34,20) = 41,04 kg

∆y =

5384

.q y . L4

E . I y +

148

.Py . L3

E . I y

=

5384

.(0 , 04818 ).(3504 )(2000000) .(44 )

+ 148

.(41 , 04 ).(3503 )(2000000 ) .(44 )

= 0,107 + 0,417


Hadi Jatmoko 201210340311160


= 0,524 cm = 5,24 mm

∆ = √( Δ x)2+( Δ y )

2

= √(2 , 284 )2+(0 ,524 )2

= 2,343 cm < ∆max = 2,92 cm

Jadi gording dengan profil C 150x65x20, dengan tebal 2,5 mm dapat digunakan karena

telah memenuhi persyaratan.

7 msagrod

gording

1.4 m


Hadi Jatmoko 201210340311160


Perencanaan Sagrod

Cek kelangsingan sagrod

λ =

lkimin

< 300

i =

lk300 =

140300 = 0,47

i =

14 ×d

d = 4 × i

= 4 × 0,47

= 1,88 ≈ 1,9 cm = 19 mm

A =

14 × π × d2 I =

164 × π× d4 i min= √ I

A

=

14 ×π ×1,92 =

164 × π ×1,94 i min= √ 0 ,64

2,84

= 2,84 cm2 = 0,64 cm4 i min= 0,48 cm

λ =

lkimin


Hadi Jatmoko 201210340311160


=

1400 ,48 = 291,67 < 300

Gaya yang bekerja pada sagrod :

Pu = 2 × Ruy

= 2 × 44,220

= 88,440 kg

f =

Pu

Ag

=

88 , 440

0,9×14×π×1,92

= 34,68 kg/cm2 < 2475 kg/cm2

88,440

kg

Ø 19 mm

88,440 kg


Hadi Jatmoko 201210340311160


Perhitungan Beban Pada Dinding

1. Beban Mati

Jarak antar gording = 2,75 m → 5,5 m : 2 = 2,75 m

Berat penutup atap (Zincalum) = 4,10 kg/m

Berat gording diperkirakan = 6 kg/m

Berat dinding = 4,10 kg/m

Berat gording = 6 kg/m +

qD = 10,10 kg/m

qD = 10,10 kg/m => RDx = 0

RDy =

12

. qD . L =

12

.(10 ,10 ) .(5,5) = 27,78 kg

qD = 10,10 kg/m => MDx = 0

MDy =

18

. qD .L2

=

18

.(10 ,10) .(5,5 )2= 38,19 kgm

2. Beban Angin

Angin tekan, q = (0,9).p

= 0,9 x 35

= 31,50 kg/m

q = 31,50 kg/m => My = 0 kgm

Mx =

18

. qD . L2

=

18

.(31 ,50x 2 ,75 ). (5,52)= 327,55 kgm

Angin hisap, q = (-0,4).p

= (-0,4) x 35

= - 14 kg/m

q = - 14 kg/m => My = 0 kgm


Hadi Jatmoko 201210340311160


Mx =

18

. qD . L2

=

18

.(−14 x2 , 75 ).(5,52 )= - 145,58 kgm

A. Momen Akibat Beban Tetap

1. Arah Tegak Lurus Bidang Dinding

MUy = (0,8).(My) MUx = (1,2).(MDx)

= (0,8).( 38,19) = (1,2).( 327,55)

= 30,55 kgm = 393,06 kgm

Perencanaan Gording Dinding

Diambil profil C-Channel 150x65x20 dengan tebal 2,50 mm

q = 5,96 kg/m

t = 2,50 mm Cy = 2,12 cm

A = 7,59 cm2 Xo = 5,15 cm

Ix = 267 cm4 J = 1581 cm4

Iy = 44 cm4 Cw = 2148 cm6

Zx = 35,6 cm3

Zy = 10,0 cm3

rx = 5,93 cm

ry = 2,41 cm

Mutu Baja = A36 = 36 x 1ksi = 36 x 6,875 = 247,5 MPa = 2475 kg/cm2

Momen Nominal Penampang C-Channel

𝑍𝑥 = 14

ℎ𝑡 𝑡2 + 𝑎 𝑡 (ℎ𝑡 – 𝑎) + 𝑡(𝑏 − 2𝑡) (ℎ𝑡 − 𝑡)


Hadi Jatmoko 201210340311160


= 14

. 150 . 2,52 + 20 x 2,5 (150 - 20) + 2,5 (65 – 2 . 2,5) (150 – 2,5)

= 28,9 cm3

𝑍𝑦 = ℎ𝑡 𝑡 (𝑐 – 12

𝑡) + 2𝑎𝑡 (𝑏 − 𝑐 – 12

𝑡) + 𝑡(𝑐 − 𝑡)2 + 𝑡(𝑏 − 𝑡 − 𝑐)2

= 150 . 2,5 (21,2 - 12

2,5) + 2 . 20 . 2,5 (65 – 21,2 - 12

2,5) + 2,5(21,2 – 2,5)2 + 2,5(65

–

2,5 – 21,2) 2

= 16,9 cm3

Mnx = Zx fy = 28,9 x 2475 kg/cm2 = 71527,5 kgcm

Mny = Zyfy = 16,9 x 2475 kg/cm2 = 41827,5 kgcm

Persyaratan Momen Biaxial

MuxØ x Mnx

+ MuyØ x Mny < 1,0

393 , 06 x1000,9 x71527,5

+30 ,55 x 1000,9 x 41827 , 5 = 0,69 < 1,0

Cek Kelangsingan Gording

λ =

lkimin

< 300

=

5502 ,41 < 300

= 228,22 < 300

Jadi gording dengan profil C 150x65x20, dengan tebal 2,5 mm dapat digunakan karena

telah memenuhi persyaratan.


Hadi Jatmoko 201210340311160


Perhitungan Ikatan Angin (Bracing)

1. Pada Atap


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Beban angin samping (bagian atap)

Luas bidang A = 1/2 x 12,5 x 2,27

= 14,188 m2

Tekanan angin P = 35 kg/m2

W = 0,9 x 14,188 x 35

= 446,922 kg

Tiap titik simpul, Wa= 1/8 x 446,922

= 55,865 kg

Beban angin samping (bagian atap)

55,865 kg


Hadi Jatmoko 201210340311160


55,865 kg

55,865 kg

55,865 kg

11 kg11 kg


Hadi Jatmoko 201210340311160


Tan α = 1,33

7

α = 18º

R x 7 = 55,865 x 1,33

R = 10,61 kg = 11 kg

ƩH = 0

55,865 – S cos 18o = 0

55,865 – S (0,951057) = 0

-S (0,951057) = - 55,865

S = 58,740 kg

Digunakan Profil L Siku Sama Sisi 120 x 120 x 8

55,865 kg

11 kgS


Hadi Jatmoko 201210340311160


q = 14,70 kg/m

A = 18,76 cm2

Cx = Cy = 3,20 cm

Ix = Iy = 258 cm4

I min = 106 cm4

I max = 410 cm4

Zx = Zy = 29,50 cm3

ix = iy = 3,710 cm

i min = 2,38 cm

i max = 4,67 cm

Cek Kelangsingan Bracing

Panjang bracing atap , lk = √72+1,332 = 7,125 m

λ = lkimin

< 300

712,52,38

< 300

299,4 < 300

Cek Kekuatan Penampang

Nu = (1,3) x (58,740)

= 76,362 kg

Nu = 76,362 kg ≤ ØNn = 0,9 x 18,76 x 2475

Nu = 76,362 kg ≤ ØNn = 41787,9 kg


Hadi Jatmoko 201210340311160


2. Pada dinding

Beban angin samping (bagian dinding)

Dipakai portal baja

Luas bidang A = 12,5 x 5,5

= 68,75 m2

Tekanan angin P = 35 kg/m2

W = 0,9 x 68,75 x 35

= 2165,63 kg

Tiap titik simpul Wa= 1/4 x 2165,63

= 541,41 kg

ΣMA = 0

(541,41).(5,5) = RB.(7)

RB = 425,39 kg

RA = -425,39 kg

Tan θ =

5,57

ΣV = 0

S.sin (38) = 425,39 kg

S = 690,95 kg

PU = (1,3).( 690,95)

= 898,235 kg

θ = 38o

541,41 kg

541,41 kg

RA RB


Hadi Jatmoko 201210340311160


Dicoba profil WF (125x125x6,5x9)

H = 125 mm

B = 125 mm

t1 = 6,5 mm

t2 = 9 mm

r = 10 mm

A = 30,31 cm2

ix = 5,29 cm

iy = 3,11 cm

Cek kelangsingan

Panjang batang, L = √5,52+72 = 8,90 m

λ =

Limin < 300

=

8903 ,11 < 300

= 286,17 < 300

Cek kekuatan penampang

Nu = (1,3) x (690,95)

= 898,235 kg

Nu = 898,235 kg ≤ ØNn = 0,9 x Ag x 2475

Nu = 898,235 kg ≤ ØNn = 0,9 x 30,31 x 2475

Nu = 898,235 kg ≤ ØNn = 67515,525 kg


Hadi Jatmoko 201210340311160


Spesifikasi KCI – Hoist Crane

Model : 15D.T11

Hoist type : 15D

Trolley type : 15DT

Capacity : 15 ton

Dimension :

L : 8000 mm

H : 785 mm

K : 730 mm

R : 1000 mm

F : 700 mm

E : 740 mm

Weight: 2150 kg

Pcrane : 2150 kg

Pcropcrane : 15000 kg

Pfootwalk : 100 kg +

17250 kg

Perhitungan Pembebanan

λ = 1,04 m’ (Lihat Tabel Hoist Crane)

λ < 0,586 L

λ < 0,586 . 11

λ < 6,446 m’

Mmax = P/2L (L – a/2)2

λ = 104 cm = 1,04

m


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M max=(17,25 ) x1 /2

2(11)×(11−1,04

2)

2

= 43,06 tm’ x 1,15 (ambil impact factor = 15 %)

= 49,52 tm’

V = P ×¿)

= 8,625 ×(2−1,0411

)

= 16,43 ton

Pra-Desain penampang

Lh

< 2511h

< 25 ; h = 45 cm

Lb

< 6511b

< 65 ; b = 18 cm

Di ambil ;

h = 90 cm Lh

= 110,9

= 12,22 < 25 OK

b = 80 cm Lb

= 110,8

= 13,75 < 65 OK

Perencanaan Penampang Balok


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Cek Stabilitas

Tekuk lokal pada sayap (flens)

λ = Btf

= 36,51,5

= 24,33

λp = 1,40 √ Efy

= 1,40 √ 200000247,5

= 39,80( λ ≤ λp )

Tekuk lokal pada badan (web)

λ = Ht

= 862,5

= 34,40

λp = 1,49 √ Efy

= 1,49 √ 200000247,5

= 42,36 ( λ ≤ λp )

(Penampang Balok Girder Crane Melintang Kompak) → OK

* Luas Profil FI = 60 x 1,5 = 90 cm2

FII = 2,5 x 86 = 215 cm2

FIII = 2,5 x 86 = 215 cm2

FIV = 80 x 2,5 = 200 cm2

* Luas Penampang

A = A1 + A2 + A3 + A4

= 90 + 215 + 215 + 200

= 720 cm2

* Garis Berat Penampang

Y = F 1∙ y 1+F 2∙ y 2+F 3 ∙ y3+F 4 ∙ y 4

∑ F

= (90×0,75 )+ (215× 44,50 )+(215× 44,50 )+(200×88,75)

720

= 51,32 cm

X = 40 cm

* Momen Inersia

Ix = Ix1 + Ix2 + Ix3 + Ix4

Ix1 = 1

12∙ b∙ h3+F ∙a2


Hadi Jatmoko 201210340311160


= 1

12∙60 ∙ 1,53+90 ∙(51,32−0,75)2

= 1

12∙60 ∙ 3,38+90 ∙(2557,32)

= 230175,70 cm4

Ix2 = Ix3 = 1

12∙2,5 ∙ 863+215 ∙(51,32−( 1

2∙86+1,5))

2

= 1

12∙2,5 ∙ 863+215 ∙(6,82)2

= 142511,83 cm4

Ix4 = 1

12∙80∙ 2,53+200 ∙(38,68−1,25)2

= 1

12∙80∙ 15,63+200 ∙ 1401

= 280304,20 cm4

Ixtotal = Ix1 + Ix2 + Ix3 + Ix4

= 230175,70 + 142511,83 + 142511,83 + 280304,20

= 795503,56 cm4

Momen inersia arah-y (sumbu lemah penampang)

Iy = 1

12∙603∙ 1,5+ 1

12∙803 ∙ 2,5+{ 1

12∙ 86 ∙ 1,53}2+ {(86 ∙1,5 ) (19,32) }2

= 229817,46 cm4

Tegangan tarik pada web sebesar :

ft = P

7 tw

<σ y

= (17250 )1 /2

7×2,5<σ y

=492,86 kg/cm2<σ y=2475 kg /cm2

→ asumsikan tf = 2,5 cm = 25 mm

fb = 0,75P

t f 2

<σ y

= 0,75 ∙(17250)

2,52 < σ y

= 2070 < σ y=2475 kg /cm2


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A = 720 cm2 = 0,0720 m2

Q bs (profil) = A ∙ σbaja=0,0720 m2×7,850t

m3

= 0,5652 t/m’

Q footwalk = 0,1 t/m’+

Qtotal = 0,6652 t/m’

M max=Qtot ∙ l

2

8+¿ 49,52

¿ 0,6652× 112

8+¿ 49,52

= 59,58 tm’

= 59,58 x 105 kgcm

V = 12

×0,6652 ×11+16,43

= 20,09 ton

Wx = I x

y=795503,56

51,32 = 15500,85 cm3

σ x=Mw x

=59,58 ×105 kgcm15500,85 cm3

= 384,37 kg/cm2 < σ s=σ y

FK =

24751,5

= 1.650 kg/cm2

Tegangan pada profil

My = M x 1,10 x 0,1

= 59,58 x 1,10 x 0,1

= 6,5538 tm’

Wy = I y

x=229817,46

40 = 5745,44 cm3

σ y=M y

w y

=6,5538 ×105 kg cm5754,44 cm3

= 113,89 kg/cm2

σ=√σ x2+σ y

2+¿σ x σ y¿


Hadi Jatmoko 201210340311160


= √384,372+113,892+¿384,37 x 113,89¿

= 452,20 kg/cm2 < σ s=σ y

FK =

24751,5

= 1.650kg/cm2 OK

Profil Balok Crane yang dipakai aman untuk digunakan.

Cek lendutan pada profil

Δijin = L

600 =

11000600

= 18,33 mm = 1,833 cm

Lendutan akibat beban merata

Δ = 5 w L4

384 EI = 5 x 6,65 x11004

384 x2000000 x795503,56 = 0,08 cm

Lendutan akibat beban terpusat

Δ = pxa

24 E Ix [3xl2-4a2]=

17250 x10424 x2000000 x795503,56

[3 x 11002 – 4x1042]= 0,17

cm

Δtotal = 0,08 + 0,17

= 0,25 cm

= 0,25 cm < 1100600

= 1,83 cm → Profil Balok Crane aman digunakan.

Direncanakan kaki balok girder crane → panjang 1,5 m dengan tebal 3 cm

Cek Tahanan Geser

Vn = 0,6 . fy . Aw

Vn = 0,6 . 247,5 . (560x25)

Vn = 207,9 ton

Vu ≤ Ø Vn

20,09 ton ≤ 0,9 x 207,9 ton

20,09 ton ≤ 187,11 ton → OK

Perencanaan Dimensi Balok Crane Memanjang

Lh

< 25 7h

< 25 ; h = 35 cm


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Lb

< 65 7b

< 65 ; b = 35 cm

Di ambil ;

h = 40 cm Lh

= 7

0,4 = 17,5 < 25 OK

b = 40 cm Lb

= 7

0,4 = 17,5 < 65 OK

λ < 0,586 L

λ < 0,586 . 7

λ < 4,102 m’

Reaksi Perletakan

R = ½ x q x L2 + V

= ½ x 0,6652 x 72 + 16,43

= 32,73 ton

Dicoba dengan profil WF 400 x 400 x 13 x 21

λ = 1,50 m


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t2

r

t1

t2

B

H

r

t1 : 13 mm

t2 : 21 mm

r : 22 mm

A : 218,7 cm2

q : 172 kg/m

Ix : 66600 cm4

Iy : 22400 cm4

ix : 17,5 cm

iy : 10,1 cm

zx : 3330 cm3

zy : 1120 cm3

Cek Stabilitas

Tekuk lokal pada sayap (flens)

λ = 12

x Bf

2 x tf =

12

x 400

2x 21 = 4,76

λp = 0,38 √ Efy

= 0,38 √ 200000247,5

= 10,80( λ ≤ λp )

Tekuk lokal pada badan (web)

λ = Ht

= 40010

= 40,0

λp = 3,76 √ Efy

= 3,76 √ 200000247,5

= 106,88 ( λ ≤ λp )

(Penampang Balok Crane Memanjang Kompak) → OK


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Mmax = q total x L2

8 +

P x L4

= (0,172+0,6652) x72

8 +

32,73 x 74

= 62,41 tm

= 62,41 x 1,15

= 71,77 tm

Vmax = ½ x q x L + P

= ½ x(0,172+0,6652) x 7 + 32,73

= 35,66 ton

Cek kelangsingan kolom dan tegangan yang bekerja

λ = lk

imin =

70010,1

= 69,31 < 200

λc = λπ √ fy

E =

69,313,14

√ 247,5200000

= 0,78 ≤ 1,20

ω = 1,43

1,6−0,67 λ c =

1,431,6−0,67 (0 , 78) = 1,33

σ = ω x N

Ø x A +

MØ x zx

= 1,33 x 0

0,85 x218,7 +

717700,9 x3330

= 23,95 kg/cm2 < 2475 kg/cm2

Penampang balok WF cukup aman menerima gaya kombinasi dan tidak terjadi tekuk lentur

Kontrol terhadap lendutan

Δijin = L

600 =

7000600

= 11,67 mm


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Δ = 5w L4

384 EI = 5 x8,37 x 7004

384 x2000000 x 66600 = 0,15 cm

Lendutan akibat beban terpusat

Δ = pxa2

x3 xl 2−4 a2

24 E Ix=32730 x150

2 x 3 x7002 – 4 x1502

24 x2000000 x 66600 =0,94 cm

Δtotal = 0,15 + 0,94

= 1,09 cm

= 1,09 cm < 7000600

= 1,167 cm → Profil Balok Crane aman digunakan.

Mn = Mp

Zx = bf x tf x (d-tf) + ¼ x tw x (d-2tf)2

= 40 x 2,1 x (40-2,1) + ¼ x 1,3 x (40 – 2 x 2,1)2

= 3600,133 cm3

Mp = Zx x fy

= 3600,133 x 2475

= 8910329,175 kg/cm = 89,10 tm

Mu ≤ Ø Mn

39,63 tm ≤ 0,9 x 89,10 tm

71,77 tm ≤ 80,19 tm OK

Cek Tahanan Geser

Vn = 0,6 . fy . Aw

Vn = 0,6 . 247,5 . (358x21)

Vn = 111,64 ton

Vu ≤ Ø Vn

35,66 ton ≤ 0,9 x 111,64 ton

35,66 ton ≤ 100,48 ton → OK

Tugas Besar Baja ARJ

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