B.

TITIK BUHUL B

14
80.
80.
2,39
812

485
120 ,9
.14 .12
.80 0.1
80 0,63 1
4
88

140.140.15
91,92

1. Tinjauan Batang A1 (tekan)

Profil yang dipakai = 80 × 80 × 14
Besar gaya (P) = 8840,63 𝑘𝑔
Tebal pelat = 0,9 𝑐𝑚 = 9 𝑚𝑚
Tegangan tarik yang di ijinkan : r = 0,8 ×1600 = 1280 kg/cm2

Tegangan geser yang di ijinkan : = 0,8 × 1600 = 1280 kg/cm2

Tegangan tumpu yang dijinkan : σtp = 1,6 × 1600 = 2560 kg/cm2

Jarak antar baut ( D = 1,6 cm)

2,5 d < u < 7d

(2,5 × 1,6) < u < (7 × 1,6)

4,25 < u < 11,9

u diambil = 6 cm

1,2 d < u < 3d

(1,2 ×1,7) < u <(3 ×1,7)

2,04 < u < 5,1

u1 diambil = 2 cm

baut bekerja pada 2 irisan

Ng = 2 × ¼ 𝜋 × D2 ×

=2 × ¼ × 3,14 × 1,62 × 1280 = 5144,576 kg

 Ntp = Dlubang × t × σtp

= 1,7 × 0,9 × 2560 = 3916,8 kg

Pilih yang terkecil = 3916,98 kg

𝑃 8840,63
𝐽𝑢𝑚𝑙𝑎ℎ 𝑏𝑎𝑢𝑡 (𝑛) = = = 2,25 ≈ 2 𝑏𝑎𝑢𝑡
𝑁𝑡𝑝 3916,98

Cek tegangan geser untuk 1 baut

𝑅𝑦 = 0

𝑃 8840,63
𝑅𝑥 = = = 4420,325 𝑘𝑔
𝑛 2

Profil 80 × 80 × 14→ e = 2,48 cm

Y
Mx = P × (½ L – e)

= 8840,63× (( ½× 8) – 2,48)

= 134337,788 kgcm
20 20
X
∑𝑟 = ∑ 𝑥 2 + ∑ 𝑦 2 = [(32 + 32) + 02 ] = 18 cm 60

𝑀𝑥 × 𝑦 134337,788× 0
Rx’ = Rx – ∑𝑟
= 4420,325 – 18

= 4420,325 kg
𝑀𝑥 × 𝑥 134337,788× 3
Ry’ = Ry + ∑𝑟
=0+ 18

= 2821,78

R = √𝑅𝑥′2 + 𝑅𝑦′2

= √(4420,325)2 + (2821,78)2

= 3244,207 kg
𝑅
𝜏 = 1 ≤
2 × × 𝜋 × 𝑑2
4

3244,07
= 1 ≤ 0,8 × 1867
2 × × 3,14 × (1,6)2
4

= 807,177𝑘𝑔/𝑐𝑚2 < 1280 kg/cm2

2. Tinjauan Batang A2 (tekan)
Profil yang dipakai = 80 × 80 × 14
Besar gaya (P) = 8122,39 𝑘𝑔
Tebal pelat = 0,9 𝑐𝑚 = 9 𝑚𝑚
Tegangan tarik yang di ijinkan : r = 0,8 ×1600 = 1280 kg/cm2

Tegangan geser yang di ijinkan : = 0,8 × 1600 = 1280 kg/cm2

Tegangan tumpu yang dijinkan : σtp = 1,6 × 1600 = 2560 kg/cm2

Jarak antar baut ( D = 1,6 cm)

2,5 d < u < 7d

(2,5 × 1,6) < u < (7 × 1,6)

4,25 < u < 11,9

u diambil = 6 cm

1,2 d < u < 3d

(1,2 ×1,7) < u <(3 ×1,7)

2,04 < u < 5,1

u1 diambil = 2 cm

Paku Keling bekerja pada 2 irisan

Ng = 2 × ¼ 𝜋 × D2 ×

=2 × ¼ × 3,14 × 1,62 × 1280 = 5144,576 kg

Ntp = Dlubang × t × σtp

= 1,7 × 0,9 × 2560 = 3916,98 kg

Pilih yang terkecil = 3919,98 kg

𝑃 8122,339
𝐽𝑢𝑚𝑙𝑎ℎ 𝑏𝑎𝑢𝑡(𝑛) = = = 2,07 ≈ 2 𝑏𝑎𝑢𝑡
𝑁𝑡𝑝 3916,98

Cek tegangan geser untuk 1 baut

𝑅𝑦 = 0

𝑃 8122,339
𝑅𝑥 = = = 4061,169 𝑘𝑔
𝑛 2
Profil 80 × 80 × 14→ e = 2,48 cm

Mx =P×(½L–e)

= 8122,339× (( ½× 8) – 2,48 )

= 12345,95kgcm

∑𝑟

Rx’
= ∑ 𝑥 2 + ∑ 𝑦 2 = [(32 + 32) + 02 ] = 18 cm

= Rx –
𝑀𝑥 × 𝑦
= 4061,169 –
12345,95× 0
Y
∑𝑟 18

= 4061,169 kg

Ry’ = Ry +
𝑀𝑥 × 𝑥
∑𝑟
=0+
12345,95× 3
18 20 20
X
60
= 2057,65 kg

R = √𝑅𝑥′2 + 𝑅𝑦′2

= √(4061,169)2 + (2057,65)2

= 4552,69𝑘𝑔
𝑅
𝜏 = 1 ≤
2 × × 𝜋 × 𝑑2
4

4552,69
= 1 ≤ 1280
2 × × 3,14 × (1,6)2
4

= 1132,735𝑘𝑔/𝑐𝑚2< 1280 kg/cm2

3. Tinjau batang V1 ( tarik )
Profil yang dipakai =140× 140 × 15

Besar gaya (P) = 91,92 kg

Tebal pelat buhul = 9 mm = 0,9 cm

Tegangan tarik yang di ijinkan : r = 0,8 ×1600 = 1280 kg/cm2

Tegangan geser yang di ijinkan : = 0,8 × 1600 = 1280 kg/cm2

Tegangan tumpu yang dijinkan : σtp = 1,6 × 1600 = 2560 kg/cm2

Jarak antar baut( D = 1,6 cm)

2,5 d < u < 7d

(2,5 ×1,6) < u < (7×1,6))

4,25 < u < 11,9 → u diambil = 6 cm

1,2 d < u < 3d

(1,2 × 1,6) < u <(3 × 1,6)

2,04 < u < 5,1

u diambil = 2 cm

baut bekerja pada 2 irisan

Ng = 2 × ¼ 𝜋 × D2 ×

= 2 × ¼ × 3,14 × 1,62 × 1280 = 5144,576 kg

Ntp = Dlubang × t × σtp

= 1,7 × 0,9 × 2560 = 3916,98 kg

Pilih yang terkecil = 3916,98 kg

𝑃 91,92
Jumlah paku keling (𝑛) = = = 0,023 ≈ 2 𝑏𝑎𝑢𝑡
𝑁𝑡𝑝 3916,98

Cek tegangan geser untuk 1 baut

Ry = 0
𝑃 91,92
Rx = 𝑛 = = 45,96 kg
2

Profil 140 × 140 × 15 → e = 4,00 cm

Y
Mx =P×(½L–e)

=91,92× (( ½ × 14) – 4,00 )

20 20
X
60
 = 321,72 kgcm

∑𝑟 = ∑ 𝑥 2 + ∑ 𝑦 2 = [(32 + 32) + 02] = 18 cm

𝑀𝑥 × 𝑦 321,72 ×0
Rx’ = Rx – ∑𝑟
= 45,96 – 18

= 45,96 kg
𝑀𝑥 × 𝑥 99,514 ×3
Ry’ = Ry + ∑𝑟
=0+ = 53,62 kg
18

R` = √𝑅𝑥′2 + 𝑅𝑦′2

= √(45,96)2 + (53,62)2

= 70,621kg
𝑅
𝜏 = 1
2 × ×𝜋 × 𝑑 2
4

70,621
= 1 ≤ 1280
2 × ×3,14× 1,62
4

= 17,57𝑘𝑔/𝑐𝑚2 < 1280 kg/cm2

4. Tinjau batang D1 ( Tekan )

Profil yang dipakai =120 × 120 × 11

Besar gaya (P) = 485,9 kg

Tebal pelat buhul = 9 mm = 0,9 cm

Tegangan tarik yang di ijinkan : r = 0,8 ×1600 = 1280 kg/cm2

Tegangan geser yang di ijinkan : = 0,8 × 1600 = 1280 kg/cm2

Tegangan tumpu yang dijinkan : σtp= 1,6 × 1600 = 2560kg/cm2

Jarak antar baut ( D = 1,6 cm)

2,5 d < u < 7d

(2,5 ×1,6) < u < (7×1,6))

4,25 < u < 11,9 → u diambil = 6 cm

1,2 d < u < 3d

(1,2 × 1,6) < u <(3 × 1,6)

2,04 < u < 5,1

u diambil = 2 cm

baut bekerja pada 2 irisan

 Ng = 2 × ¼ 𝜋 × D2 ×

= 2 × ¼ × 3,14 × 1,62 × 1280 = 5144,576 kg

Ntp = Dlubang × t × σtp

= 1,7 × 0,9 × 2560 = 3916,98 kg

Pilih yang terkecil = 3916,98 kg

𝑃 485,9
Jumlah baut (𝑛) = = = 0,12 ≈ 2 𝑏𝑎𝑢𝑡
𝑁𝑡𝑝 3916,98

Cek tegangan geser untuk 1 baut

Ry = 0
𝑃 485,9
Rx = = = 242,95 kg
𝑛 2

Profil 120× 120 × 11 → e = 3,36 cm

Mx =P×(½L–e)

= 485,9 × (( ½ × 12) – 3,36 ) Y

= 1282,7 kgcm

∑𝑟 = ∑ 𝑥 2 + ∑ 𝑦 2 = [(32 + 32) + 02] = 18 cm

Rx’ = Rx –
𝑀𝑥 × 𝑦
∑𝑟
= 242,95 –
1282,7 ×0
18 20 20
X
60
= 242,95 𝑘𝑔
𝑀𝑥 × 𝑥 1282,7 ×3
Ry’ = Ry + ∑𝑟
=0+ = 213,87 kg
18

R` = √𝑅𝑥′2 + 𝑅𝑦′2

= √(242,95)2 + (213,78)2

= 323,614 kg
𝑅
𝜏 = 1
2 × ×𝜋 × 𝑑 2
4

323,614
= 1 ≤ 1280
2 × ×3,14× 1,62
4

= 80,51𝑘𝑔/𝑐𝑚2 < 1280kg/cm2

Cek terhadap plat buhul B

80 × 80 ×14 120×120 × 11

A = 20,66 cm2 A = 25,4 cm2

Ix = Iy = 115 cm4 Ix = Iy = 341cm4

e =2,48 cm e = 3,36 cm

Anetto :

Profil 80 × 80 × 14 = 2 × 8 × 1,48 = 23,68 cm2

= 2 × 6,97 × 1,4= 19,516 cm2

Plat Buhul = 27,72 × 0,9 = 24,948 cm2

Profil 120 × 120 × 11 = 2 × 6,8 × 0,2 = 2,72 cm2 +

Abrutto = 70,864 cm2

Lubang = 2 × 1,7× 3,1 = 10,54 cm2 -

Anetto = 60,324 cm2

80
14,8
69,78
277,26
269,86

251,89

134,78
17

68,06

31
120
Statis momen terhadap sisi bawah
𝑙𝑢𝑎𝑠 𝑝𝑟𝑜𝑓𝑖𝑙 × 𝑗𝑎𝑟𝑎𝑘 𝑡𝑖𝑡𝑖𝑘 𝑏𝑒𝑟𝑎𝑡 𝑝𝑟𝑜𝑓𝑖𝑙 𝑘𝑒 𝑏𝑎𝑤𝑎ℎ
Yb = 𝐴 𝑛𝑒𝑡𝑡𝑜

(23,68× 27,72 )+ (19,516× 2,51)+ (24,948× 13,4 )+ (2,72x6,8 )

= 60,324

= 17,76 cm

𝑦𝑎 = h – yb

= 27,72 – 17,76

= 9,966 cm

N1 = −8122,39 × cos 27 = −7257,10 𝑘𝑔 (𝑡𝑒𝑘𝑎𝑛)

N2 = −485,9 𝑘𝑔 (𝑡𝑒𝑘𝑎𝑛)
Ntotal = −8122,339 − 485,9 = −7743 𝑘𝑔 (𝑡𝑒𝑘𝑎𝑛)
D1 = −8122,339 × sin 27 = − 3687 𝑘𝑔 (𝑡𝑒𝑘𝑎𝑛)
D2 = 91,92 𝑘𝑔 (𝑡𝑎𝑟𝑖𝑘)
Dtotal = −3687 + 91,92 = −3595,08 𝑘𝑔 (𝑡𝑒𝑘𝑎𝑛)
M = 𝑁𝑡𝑜𝑡𝑎𝑙 × (𝑌𝑎 – 𝑒1 )
= −7743 × (9,966– 2,48)
= −57964 𝑘𝑔𝑐𝑚

80
14,8
69,78
98,74
170,2
277,26

35,11
31,6

134,78
17
99,66

9
31
I bruto :

Profil 80 × 80 × 14 = 2×((1/12 × 8 × 1,483) + (13,68 × 17,022)) = 3967,15 cm4

= 2×((1/12 × 1,4 × 6,973) + (11,284 ×1,342)) = 162,698 cm4

Plat buhul = (1/12 × 0,9 × 27,7263) + (21,636 × 3,162) = 1569,990 cm4

Profil 120 × 120 × 11 = 2 × (1/12 × 1,1 × 13,43) + (8,4 × 9,962) = 1274,412 cm4 +

Ibrutto = 5047,919 cm4

I lubang = (1/12 × 3,1 × 1,73) + (4,14 × 3,162) = 263,435 cm4 -

Inetto = 4784,484cm4
𝐼𝑛𝑒𝑡𝑡𝑜 4784,484
Wa = = = 493,296 cm3
𝑌𝑎 9,699

𝐼𝑛𝑒𝑡𝑡𝑜 4784,484
Wb = 𝑌𝑏
= 17,76
= 269,396 cm3

Tegangan yang terjadi :

𝑁𝑡𝑜𝑡 −7743
σn = = = - 128,35 kg/cm2
𝐴𝑛𝑒𝑡𝑡𝑜 60,324

𝑀 −57964
σma = = = - 117,5 kg/cm2
𝑤𝑎 493,293

𝑀 57964
σmb = = = 215,162 kg/cm2
𝑤𝑏 269,396

σatas = σn + σma

= - 128,35 + (-117,5) = - 245,85 kg/cm2

σbawah = σn + σmb

= - 128,35 + 215,162 = 86,812 kg/cm2

𝐷𝑡𝑜𝑡 −3595,08
τr = 𝐴𝑛𝑒𝑡𝑡𝑜 = 60,324
= - 59,5 kg/cm2

σi = √𝜎𝑏 2 + 1,56𝜏𝑟2

= √(86,812)2 + 1,56 × (−59,5 )2

= 114,27 kg/cm2< 1280 kg/cm2 …..Ok