Soal 1 annisa

PERANCANGAN BANGUNAN BAJA IFAKULTAS TEKNIK JURUSAN TEKNIK SIPILUNIVERSITAS KRISTEN MARANATHAJl. Prof. Drg. Soeria Soemantri No. 65 Bandung 40164

TUGAS I

SOAL 1

Dik :

- Mutu baja BJ – 37- Profil baja menggunakan siku ganda- Beban :

Beban P1 (kN) P2 (kN) P3 (kN) P4 (kN) P5 (kN)Beban 8 4 5 7 5

Jawab :

I. Reaksi Perletakan

M∑ B = 0

Annisa Fitri Adhalia (0921032)


VA × 15,25 – 0,5 P1 × 15,25 +P2 × 1,352 – P1 × 12,5 +P2 × 2,705 – P1×9,75 +P2× 3,75 – P1 × 7,625 – P5 × 7,625 – P4 × 3,75– P5× 5,5 – P4 × 2,705 – P5 × 2,75 – P4 × 1,352 – P3× 12,5 – P3× 9,75 – P3 × 5,5 – P3× 2,75 = 0

VA = 1

15,25 × (-61+ 5,41- 100+ 10,82 - 78 + 15 - 61 - 38,125 -

38,125 – 26,25 – 27,5 - 18,94 – 13,75 + 9,47 – 62,5 – 48,75 – 27,5 – 13,75)

VA = 1

15,25 × (463,35)

VA = 30.45 kN

M∑ A = 0

– VB × 15,25 + P1 × 2,75+ P2 × 1,352+ P1 × 5,5+ P2 × 2,71 + P1 × 7,625+ P2 ×3,75 + P5 × 7,625 –P4× 3,75 + P5× 9,75-P4 × 2,71 + P5 × 12,5 –P4 × 1,352 + 0,5P5× 15,25 + P3 × 12,5 – P3× 9,75 – P3× 5,5 – P3× 2,75 = 0

VB = 1

15,25 × (22+ 5,41 + 44+ 10,82 + 61 + 15+ 38,125 + 26,25

+48,75 + 18,94 + 62,5 – 9,47 + 38,125 + 62,5 + 48,75- 27,5 – 13,75)

VB = 115 × (376,447)

VB = 24,80 kN

H∑ = 0



HA + (P2 × 2,71) – (P4 × 2,71) = 0

HA = - 10.82+ 18.94

HA = 8.115 kN

Cek

V∑ = 0

VA + VB – 0,5 P1 – P1× 3,75 – P5 × 3,75 – 0,5 P5 = 0

30.45 + 24.80 – 4 – 30 – 2,5 –18,75 = 0

0 = 0 … OK

II. Perhitungan Gaya Batang dengan Cara Manual

Titik A



V = 0∑ H = 0∑

VA – 0,5. 4,4 + S7 sin 31 + S1 sin 11 = 0 HA + S7 cos 31 + S1

cos 11 = 0

26,265 – 2,2 + S7 0,515 + S1 0,190 = 0 2,7 + S7 0,857 + S1

0,981 = 0

S7 0,515 + S1 0,190 = -24,065 … (1) S7 0,857 + S1 0,981

= -2,7 … (2)

Dari persamaan (1) dan (2)

S7 0,515 + S1 0,190 = -24,065

S7 0,857 + S1 0,981 = -2,7

maka didapat,

S1 = 56,245 kN

S7 = -67,562 kN

Titik C



H = 0∑ V = 0∑

-S1 cos 11 + S2 cos 11 = 0 -P3 – S1 sin 11 + S2 sin 11 + S13 = 0

-55,212 + S2 cos 11 = 0 -3,45 – 10,732 + 10,732 +S13 = 0

S2 = 56,245 kN S13 = 3,45 kN

Titik H

V = 0∑ H = 0∑

-P2 – S13 – S7 sin 31 – S14 sin 11 + S8 sin 31 = 0 P1 – S7 cos 31 + S14 cos 11 + S8 cos 31 = 0

-4,4 – 3,45 + 34,797 – S14 0,191 + S8 0,515 = 0 2,8 + 57,912 + S14 0,982 + S8 0,857 = 0

– S14 0,191 + S8 0,515 = -26,947 … (1) S14 0,982 + S8 0,857 = -60,712 … (2)


– S14 0,191 + S8 0,515 = -26,947



S14 0,982 + S8 0,857 = -60,712

maka didapat,

S8 = -56,852 kN

S14 = -12,209 kN

Titik D

H = 0∑ V = 0∑

-S2 cos 11– S14 cos 11 + S3 cos 11 = 0 -P3 + S15 – S2 sin 11 + S14

sin 11 + S3 sin 11 = 0

-55,212 + 11,985 + S3 cos 11= 0 -3,45 + S15 – 10,732 – 2,330 + 8,403 = 0

S3 = 44,036 kN S15 = 8,109 kN

Titik I

V = 0∑ H = 0∑



-P2 - S15 – S8 sin 31- S16 sin 31 + S9 sin 31 = 0 P1 – S8 cos 31+S16 cos 31 + S9 cos 31= 0

-4,4 – 8,109 + 29,281 - S16 sin 31 + S9 sin 31 = 0 2,8 + 48,732 + S16 cos 31 + S9 cos 31= 0

- S16 sin 31 + S9 sin 31 = -16,772 … (1) S16 cos 31 + S9 cos 31= -51,532 … (2)


- S16 sin 31 + S9 sin 31 = -16,772

S16 cos 31 + S9 cos 31= -51,532

maka didapat,

S9 = -46,342 kN

S16 = -13,777 kN

Titik J

H = 0∑ V = 0∑

P1 – P4 – S9 cos 31 + S10 cos 31 = 0 -P2 – P5 – S17 – S9 sin 31– S10 sin 31 = 0

2,8 – 3,7 + 39,723 + S10 0,857 = 0 -4,4 – 6,1 - S17 + 23,868+ 23,327 = 0



S10 = -45,292 kN S17 = 36,695 kN

Titik E

V = 0∑

-P3 + S17 – S3 sin 11 + S16 sin 31 – S4 sin 11 + S18 sin 31 = 0

-3,45 + 36,695 – 8,403 – 7,096 – S4 sin 11 + S18 sin 31 = 0

– S4 sin 11 + S18 sin 31 = -17,746 … (1)

H = 0∑

-S3 cos 11 – S16 cos 31 + S4 cos 11 + S18 cos 31= 0

-43,227 + 11,809 + S4 cos 11 + S18 cos 31= 0

S4 cos 11 + S18 cos 31= 31,418 … (2)


– S4 sin 11 + S18 sin 31 = -17,746

S4 cos 11 + S18 cos 31= 31,418

maka didapat,

S4 = 48,702 kN



S18 = -16,413 kN

Titik K

H = 0∑ V = 0∑

-P4 – S18 cos 31–S10 cos 31+S11 cos 31 = 0 -P5 – S18 sin 31– S19 – S11 sin 31 + S10 sin 31 = 0

-3,7 + 14,069 + 38,823 + S11 0,857= 0 -6,1 + 8,453 – S19 + 29,558 – 23,327 = 0

S11 = -57,389 kN S19 = 8,584 kN

Titik F



V = 0∑ H = 0∑

-P3 + S4 sin 11 + S19 – S5 sin 11 + S20 sin 11 = 0 -S4 cos 11+S5 cos 11+ S20 cos 11 = 0

-3,45 + 9,293 + 8,584 – S5 0,190 + S20 0,190 = 0 S5 0,981+ S20 0,981 = 47,807 … (2)

– S5 0,190 + S20 0,190 = -14,427 … (1)


– S5 0,190 + S20 0,190 = -14,427

S5 0,981+ S20 0,981 = 47,807

maka didapat,

S5 = 62,332 kN

S20 = -13,599 kN

Titik L

H = 0∑ V = 0∑



-P4 – S20 cos 11- S11 cos 31+ S12 cos 31 = 0 -P5 – S2 – S20

sin11 +S11sin31–S12 sin 31 = 0

-3,7 + 13,349 + 49,192 + S12 0,857 = 0 -6,1– S21 + 2,595 – 29,558 + 35,355 = 0

S12 = -68,646 kN S21 = 2,3 kN

Titik G

H = 0∑

-S5 cos 11+ S6 cos 11 = 0

-61,187 + S6 0,982 =0

S6 = 62,332 kN



III. Perhitungan Gaya Batang dengan menggunakan SAP

1. Reaksi Perletakan

2. Gaya Batang





IV. Perbandingan Perhitungan Manual dan Perhitungan SAP

Batang

Gaya Batang HasilManual (kN)

Gaya Batang HasilSAP (KN)

1 56,245 57.2242 56,245 57.2243 44,036 45.0754 48,702 47.9315 62,332 62.9356 62,332 62.9357 -67,562 -68.5878 -56,852 -57.969 -46,342 -47.33310 -45,292 -46.28311 -57,389 -59.12612 -68,646 -71.96913 3,45 3.4514 -12,209 -12.14815 8,109 8.21516 -13,777 -13.89217 36,695 37.66518 -16,413 -17.15819 8,584 9.33520 -13,599 -15.00421 2,3 3.45



V. Pendimensian Batang 15

S15 = 46,342 kN … (Batang Tekan)

L = 2,9155 m

Dicoba Profil ⊥ 65.65.6

Ag = 7,527 cm2 = 752,7 mm2

rx = ry = ix = iy = 1,98 cm = 19,8 mm

x = e = 1,81 cm = 18,1 mm

rmin = iη = 1,27 cm = 12,7 mm

Ix = Iy = 29,4 cm4 = 294000 mm4

BJ37 → fy = 240 MPa

fu = 370 MPa

1. Menentukan dimensi profil berdasarkan syarat kelangsingan

Lrmin ≤ 240

2915,512,7

≤240

229,567 ≤ 240 … OK

2. Pemeriksaan kelangsingan elemen penampang



= bt=656

=¿10,833

r = 200√fy

=200√240 = 12,910

= 10,833 < r = 12,910 … OK

3. Penentuan jarak pelat kopel

Syarat 1

1 50 (SNI… 9.3-7)

L1rmin 50 (SNI… 9.3-4)

L1 50 rmin

L1 50 . 12,7

L1 = 635 mm

Maka batang tekan terbagi menjadi :

2915,5635 = 4,591 5 bagian

Jika pelat kopel membagi panjang batang menjadi 4 bagian, maka jarak antara pelat kopel :

L1 = Lk5 = 2915,55 = 583,1 mm



1 = L1iη

=583,112,7 = 45,913

Syarat 2

Tinjauan terhadap sumbu bahan (sumbu x)

x 1,2 1 (SNI… 9.3-7)

x = Lkxix

≥1,2 L1iη

2915,519,8

≥ 1,2 . 45,913

147,247 55,096 … OK

Syarat 3

Tinjauan terhadap sumbu bebas bahan (sumbu y)

Asumsi jarak antara profil (a) = 10 mm

Iy = 2 [ Iy + Ag (x + ½a )2 ]

Iy = 2 [ 294000 + 752,7 . (18,1 + ½ 10)2 ]

Iy = 1391296,494 mm4



iy = √ Iy2Ag

= √1391296,4942.752,7

iy = 30,401 mm4

y = Lkyiy =2915,530,401 = 95,901

iy = y = 95,901 1,2 . 1

95,901 1,2 . (45,913)

95,901 55,096 … OK

→ Jarak antara pelat kopel adalah 583,1 mm

4. Kuat rencana batang tekan

Profil siku ganda harus direncanakan memikul beban terfaktor x =147,247 > y = 95,901

Tekuk terjadi sumbu bahan ( sumbu x - x )

c = Kc.Lπ.rx √fyE

= 1.2915,5π.19,8 √ 2402.105

= 1,624

c > 1,2 maka:

= 1,25 c2 = 1,25 . 1,624 = 2,03

fcr = fy

ω = 240

2,03 = 118,227 MPa



Ø Nn = Ø . Ag . fcr

= 0,75 . 752,7 . 118,227

= 66742,097 N

= 66,742 kN

Gaya maksimum yang bekerja pada batang S9 dengan nial Nu = 46,342 kN

Nu∅Nn

=46,34266,742

=0,694 < 1 … OK


Date post:	20-Feb-2023
Category:	Documents
Upload:	nathansaputra
View:	0 times
Download:	0 times

Soal 1 annisa

Documents